http://www.physicsbook.gatech.edu/api.php?action=feedcontributions&feedformat=atom&user=EcarderPhysics Book - User contributions [en]2026-05-05T07:13:59ZUser contributionsMediaWiki 1.42.7http://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41675Quantum Harmonic Oscillator2022-12-16T06:38:13Z<p>Ecarder: /* Proving the Ground-State Energy Relation Using Uncertainty Principle */</p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Proving the Ground-State Energy Relation Using Uncertainty Principle==<br />
The Uncertainty Principle mathematically limits the ground-state energy level to <math> E_0=\frac{\hbar\omega}{2}</math>.<br><br />
The classical definition of kinetic energy is: <math> KE=\frac{1}{2}mv^2 </math><br><br />
Rewriting in terms of momentum is: <math>KE= \frac{p^2}{2m} </math> <br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewrite the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
In formulating in terms of uncertainties, the energy for the QHO must be at least:<br><br><br />
<math> E=\frac{(\Delta p)^2}{2m} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
In considering the lower bound of the Uncertainty Principle, namely when the relationship is equal such that:<br><br />
<math> \Delta x \Delta p = \frac{\hbar}{2} </math> <br><br><br />
We can rewrite one uncertainty as a function of the other. In this case, we will rewrite the momentum uncertainty as a function of the position uncertainty:<br><br><br />
<math>\Delta p = \frac{\hbar}{2\Delta x} </math> and substituting into the energy relation <br><br><br />
<math> E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
Our goal is to minimize the energy, hence we now differentiate and set it to zero to find the critical point. <br><br />
<math> \frac{d}{d(\Delta x)}\bigg[ E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 \bigg] = -\frac{\hbar^2}{4m(\Delta x)^3} + m\omega^2 \Delta x =0 </math> <br><br><br />
Solving for <math> \Delta x </math> yields: <br><br />
<math> \Delta x = \sqrt{\frac{\hbar}{2m\omega}} </math> <br><br><br />
All that remains now is to substitute the derived positional uncertainty into the Energy equation: <br><br><br />
<br />
<math>E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
<math>E_0=\frac{\hbar^2}{8m(\sqrt{\frac{\hbar}{2m\omega}})^2} + \frac{1}{2}m\omega^2 (\sqrt{\frac{\hbar}{2m\omega}})^2 </math><br><br><br />
<math>E_0=\frac{2m\omega \hbar}{8m\hbar} + \frac{1}{2}m\omega^2 (\sqrt{\frac{\hbar}{2m\omega}})^2</math><br><br><br />
<math>E_0=\frac{\hbar \omega}{4} + \frac{\hbar \omega}{4} </math><br><br><br />
<math> \therefore E_0 = \frac{\hbar \omega}{2} </math> <br><br />
This is substantial as it proves that, unlike the classical harmonic oscillator, the quantum harmonic oscillator cannot have zero energy. This holds true even in the absence of any temperature. So in the modeling of polyatomic molecules in a gas, they still must have energy even at absolute zero. The ground-vibrational state energy is generally referred to as "zero point vibration"<br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
<br />
Rewriting the potential that we derived above previously as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual Jordan.<br />
<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the phenomena that are byproducts of these vibrations.<br />
<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br> <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925) <br><br><br />
Nave, R. (2014). Energy minimum from uncertainty principle. Quantum Harmonic Oscillator. Retrieved 2022, from http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc4.html#c1</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41674Quantum Harmonic Oscillator2022-12-16T06:36:00Z<p>Ecarder: /* Proving the Ground-State Energy Relation Using Uncertainty Principle */</p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Proving the Ground-State Energy Relation Using Uncertainty Principle==<br />
The Uncertainty Principle mathematically limits the ground-state energy level to <math> E_0=\frac{\hbar\omega}{2}</math>.<br><br />
The classical definition of kinetic energy is: <math> KE=\frac{1}{2}mv^2 </math><br><br />
Rewriting in terms of momentum is: <math>KE= \frac{p^2}{2m} </math> <br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewrite the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
In formulating in terms of uncertainties, the energy for the QHO must be at least:<br><br><br />
<math> E=\frac{(\Delta p)^2}{2m} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
In considering the lower bound of the Uncertainty Principle, namely when the relationship is equal such that:<br><br />
<math> \Delta x \Delta p = \frac{\hbar}{2} </math> <br><br><br />
We can rewrite one uncertainty as a function of the other. In this case, we will rewrite the momentum uncertainty as a function of the position uncertainty:<br><br><br />
<math>\Delta p = \frac{\hbar}{2\Delta x} </math> and substituting into the energy relation <br><br><br />
<math> E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
Our goal is to minimize the energy, hence we now differentiate and set it to zero to find the critical point. <br><br />
<math> \frac{d}{d(\Delta x)}\bigg[ E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 \bigg] = -\frac{\hbar^2}{4m(\Delta x)^3} + m\omega^2 \Delta x =0 </math> <br><br><br />
Solving for <math> \Delta x </math> yields: <br><br />
<math> \Delta x = \sqrt{\frac{\hbar}{2m\omega}} </math> <br><br><br />
All that remains now is to substitute the derived positional uncertainty into the Energy equation: <br><br><br />
<br />
<math>E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
<math>E_0=\frac{\hbar^2}{8m(\sqrt{\frac{\hbar}{2m\omega}})^2} + \frac{1}{2}m\omega^2 (\sqrt{\frac{\hbar}{2m\omega}})^2 </math><br><br><br />
<math>E_0=\frac{2m\omega \hbar}{8m\hbar} + \frac{1}{2}m\omega^2 (\sqrt{\frac{\hbar}{2m\omega}})</math><br><br><br />
<math>E_0=\frac{\hbar \omega}{4} + \frac{\hbar \omega}{4} </math><br><br><br />
<math> \therefore E_0 = \frac{\hbar \omega}{2} </math> <br><br />
This is substantial as it proves that, unlike the classical harmonic oscillator, the quantum harmonic oscillator cannot have zero energy. This holds true even in the absence of any temperature. So in the modeling of polyatomic molecules in a gas, they still must have energy even at absolute zero. The ground-vibrational state energy is generally referred to as "zero point vibration"<br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
<br />
Rewriting the potential that we derived above previously as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual Jordan.<br />
<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the phenomena that are byproducts of these vibrations.<br />
<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br> <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925) <br><br><br />
Nave, R. (2014). Energy minimum from uncertainty principle. Quantum Harmonic Oscillator. Retrieved 2022, from http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc4.html#c1</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41673Quantum Harmonic Oscillator2022-12-16T06:35:29Z<p>Ecarder: /* Proving the Ground-State Energy Relation Using Uncertainty Principle */</p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Proving the Ground-State Energy Relation Using Uncertainty Principle==<br />
The Uncertainty Principle mathematically limits the ground-state energy level to <math> E_0=\frac{\hbar\omega}{2}</math>.<br><br />
The classical definition of kinetic energy is: <math> KE=\frac{1}{2}mv^2 </math><br><br />
Rewriting in terms of momentum is: <math>KE= \frac{p^2}{2m} </math> <br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewrite the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
In formulating in terms of uncertainties, the energy for the QHO must be at least:<br><br><br />
<math> E=\frac{(\Delta p)^2}{2m} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
In considering the lower bound of the Uncertainty Principle, namely when the relationship is equal such that:<br><br />
<math> \Delta x \Delta P = \frac{\hbar}{2} </math> <br><br><br />
We can rewrite one uncertainty as a function of the other. In this case, we will rewrite the momentum uncertainty as a function of the position uncertainty:<br><br><br />
<math>\Delta P = \frac{\hbar}{2\Delta x} </math> and substituting into the energy relation <br><br><br />
<math> E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
Our goal is to minimize the energy, hence we now differentiate and set it to zero to find the critical point. <br><br />
<math> \frac{d}{d(\Delta x)}\bigg[ E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 \bigg] = -\frac{\hbar^2}{4m(\Delta x)^3} + m\omega^2 \Delta x =0 </math> <br><br><br />
Solving for <math> \Delta x </math> yields: <br><br />
<math> \Delta x = \sqrt{\frac{\hbar}{2m\omega}} </math> <br><br><br />
All that remains now is to substitute the derived positional uncertainty into the Energy equation: <br><br><br />
<br />
<math>E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
<math>E_0=\frac{\hbar^2}{8m(\sqrt{\frac{\hbar}{2m\omega}})^2} + \frac{1}{2}m\omega^2 (\sqrt{\frac{\hbar}{2m\omega}})^2 </math><br><br><br />
<math>E_0=\frac{2m\omega \hbar}{8m\hbar} + \frac{1}{2}m\omega^2 (\sqrt{\frac{\hbar}{2m\omega}})</math><br><br><br />
<math>E_0=\frac{\hbar \omega}{4} + \frac{\hbar \omega}{4} </math><br><br><br />
<math> \therefore E_0 = \frac{\hbar \omega}{2} </math> <br><br />
This is substantial as it proves that, unlike the classical harmonic oscillator, the quantum harmonic oscillator cannot have zero energy. This holds true even in the absence of any temperature. So in the modeling of polyatomic molecules in a gas, they still must have energy even at absolute zero. The ground-vibrational state energy is generally referred to as "zero point vibration"<br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
<br />
Rewriting the potential that we derived above previously as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual Jordan.<br />
<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the phenomena that are byproducts of these vibrations.<br />
<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br> <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925) <br><br><br />
Nave, R. (2014). Energy minimum from uncertainty principle. Quantum Harmonic Oscillator. Retrieved 2022, from http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc4.html#c1</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41672Fourier Series and Transform2022-12-15T02:10:04Z<p>Ecarder: </p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br><br><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
==History==<br />
Fourier Analysis was not initially well received. In 1807, Joseph Fourier published ''On the Propagation of heat in solid bodies''. It was in this memoir that he laid the groundwork for such analysis. He read it to the Paris Institute and substantials in academia were not impressed. Namely. Lagrange and Laplace objected to the idea. Fourier, however, did win a prize from the Paris Institute as the result of submitting this paper to a competition. It is important to note that he was one of only two submissions in this competition.<br />
==Connectedness==<br />
Fourier Series and Transforms are applicable on a variety of frontiers. They have proven themselves useful in Modern Physics, but there is a good chance anyone in a mathematics-heavy STEM field will encounter Fourier Series and Transforms as they are heavily prevalent in other areas of science and technology. Even if these aren't useful in a given person's industry, it is still useful to understand them and their importance, as they involve a clever manipulation of information. This type of analysis is largely prevalent in signal processing: the interpretation and modification of signals.<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br><br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br><br />
Chodos, A. (Ed.). (2010, March). This Month in physics history. This Month in Physics History. Retrieved 2022, from https://www.aps.org/publications/apsnews/201003/physicshistory.cfm <br><br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41671Quantum Harmonic Oscillator2022-12-15T01:56:42Z<p>Ecarder: </p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Proving the Ground-State Energy Relation Using Uncertainty Principle==<br />
The Uncertainty Principle mathematically limits the ground-state energy level to <math> E_0=\frac{\hbar\omega}{2}</math>.<br><br />
The classical definition of kinetic energy is: <math> KE=\frac{1}{2}mv^2 </math><br><br />
Rewriting in terms of momentum is: <math>KE= \frac{p^2}{2m} </math> <br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
In formulating in terms of uncertainties, the energy for the QHO must be at least:<br><br><br />
<math> E=\frac{(\Delta p)^2}{2m} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
In considering the lower bound of the Uncertainty Principle, namely when the relationship is equal such that:<br><br />
<math> \Delta x \Delta P = \frac{\hbar}{2} </math> <br><br><br />
We can rewrite one uncertainty as a function of the other. In this case, we will rewrite the momentum uncertainty as a function of the position uncertainty:<br><br><br />
<math>\Delta P = \frac{\hbar}{2\Delta x} </math> and substituting into the energy relation <br><br><br />
<math> E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
Our goal is to minimize the energy, hence we now differentiate and set it to zero to find the critical point. <br><br />
<math> \frac{d}{d(\Delta x)}\bigg[ E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 \bigg] = -\frac{\hbar^2}{4m(\Delta x)^3} + m\omega^2 \Delta x =0 </math> <br><br><br />
Solving for <math> \Delta x </math> yields: <br><br />
<math> \Delta x = \sqrt{\frac{\hbar}{2m\omega}} </math> <br><br><br />
All that remains now is to substitute the derived positional uncertainty into the Energy equation: <br><br><br />
<br />
<math>E=\frac{\hbar^2}{8m(\Delta x)^2} + \frac{1}{2}m\omega^2 (\Delta x)^2 </math><br><br><br />
<math>E_0=\frac{\hbar^2}{8m(\sqrt{\frac{\hbar}{2m\omega}})^2} + \frac{1}{2}m\omega^2 (\sqrt{\frac{\hbar}{2m\omega}})^2 </math><br><br><br />
<math>E_0=\frac{2m\omega \hbar}{8m\hbar} + \frac{1}{2}m\omega^2 (\sqrt{\frac{\hbar}{2m\omega}})</math><br><br><br />
<math>E_0=\frac{\hbar \omega}{4} + \frac{\hbar \omega}{4} </math><br><br><br />
<math> \therefore E_0 = \frac{\hbar \omega}{2} </math> <br><br />
This is substantial as it proves that, unlike the classical harmonic oscillator, the quantum harmonic oscillator cannot have zero energy. This holds true even in the absence of any temperature. So in the modeling of polyatomic molecules in a gas, they still must have energy even at absolute zero. The ground-vibrational state energy is generally referred to as "zero point vibration"<br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
<br />
Rewriting the potential that we derived above previously as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual Jordan.<br />
<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the phenomena that are byproducts of these vibrations.<br />
<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br> <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925) <br><br><br />
Nave, R. (2014). Energy minimum from uncertainty principle. Quantum Harmonic Oscillator. Retrieved 2022, from http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc4.html#c1</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41670Quantum Harmonic Oscillator2022-12-15T00:43:08Z<p>Ecarder: /* Connectedness */</p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual Jordan.<br />
<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the phenomena that are byproducts of these vibrations.<br />
<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br> <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925)</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41669Quantum Harmonic Oscillator2022-12-15T00:42:29Z<p>Ecarder: /* History */</p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual Jordan.<br />
<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the byproducts of these vibrations.<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br> <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925)</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41668Quantum Harmonic Oscillator2022-12-14T16:09:11Z<p>Ecarder: /* Comparing the Classic and Quantum */</p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual, Jordan.<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the byproducts of these vibrations.<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br> <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925)</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41667Quantum Harmonic Oscillator2022-12-14T16:08:13Z<p>Ecarder: /* References */</p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual, Jordan.<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the byproducts of these vibrations.<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br> <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925)</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41666Quantum Harmonic Oscillator2022-12-14T16:07:35Z<p>Ecarder: </p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. The QHO does have applications, most notably as an approximation of molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> can be evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==History==<br />
In the case of the quantum harmonic oscillator, there isn't much history for it. It is more so a specific case of the Schrodinger Equation, as it was one of the first cases solved after the famous equation was formulated. The first solution of the QHO I've been able to find was as early as 1925, in a paper written by Max Born and Pascual, Jordan.<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the byproducts of these vibrations.<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator <br><br />
Born, Max; Jordan, Pascual, Zur Quantenmechanik, Z. f. Physik 34, 858-888 (1925). ZBL51.0728.08. (27 September 1925)</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Main_Page&diff=41665Main Page2022-12-14T00:55:07Z<p>Ecarder: /* Week 15 */</p>
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* OpenStax intro physics textbooks: [https://openstax.org/details/books/university-physics-volume-1 Vol1], [https://openstax.org/details/books/university-physics-volume-2 Vol2], [https://openstax.org/details/books/university-physics-volume-3 Vol3]<br />
* The Open Source Physics project is a collection of online physics resources [http://www.opensourcephysics.org/ OSP]<br />
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* The Feynman lectures on physics are free to read [http://www.feynmanlectures.caltech.edu/ Feynman]<br />
* Final Study Guide for Modern Physics II created by a lab TA [https://docs.google.com/document/d/1_6GktDPq5tiNFFYs_ZjgjxBAWVQYaXp_2Imha4_nSyc/edit?usp=sharing Modern Physics II Final Study Guide]<br />
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== Resources ==<br />
* Commonly used wiki commands [https://en.wikipedia.org/wiki/Help:Cheatsheet Wiki Cheatsheet]<br />
* A guide to representing equations in math mode [https://en.wikipedia.org/wiki/Help:Displaying_a_formula Wiki Math Mode]<br />
* A page to keep track of all the physics [[Constants]]<br />
* A listing of [[Notable Scientist]] with links to their individual pages <br />
<br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 1==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====GlowScript 101====<br />
<div class="mw-collapsible-content"><br />
*[[Python Syntax]]<br />
*[[GlowScript]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====VPython====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[VPython]]<br />
*[[VPython basics]]<br />
*[[VPython Common Errors and Troubleshooting]]<br />
*[[VPython Functions]]<br />
*[[VPython Lists]]<br />
*[[VPython Loops]]<br />
*[[VPython Multithreading]]<br />
*[[VPython Animation]]<br />
*[[VPython Objects]]<br />
*[[VPython 3D Objects]]<br />
*[[VPython Reference]]<br />
*[[VPython MapReduceFilter]]<br />
*[[VPython GUIs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Vectors and Units====<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[SI Units]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Types of Interactions and How to Detect Them]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Velocity and Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's First Law of Motion]]<br />
*[[Mass]]<br />
*[[Velocity]]<br />
*[[Speed]]<br />
*[[Speed vs Velocity]]<br />
*[[Relative Velocity]]<br />
*[[Derivation of Average Velocity]]<br />
*[[2-Dimensional Motion]]<br />
*[[3-Dimensional Position and Motion]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Momentum and the Momentum Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Linear Momentum]]<br />
*[[Newton's Second Law: the Momentum Principle]]<br />
*[[Impulse and Momentum]]<br />
*[[Net Force]]<br />
*[[Inertia]]<br />
*[[Acceleration]]<br />
*[[Relativistic Momentum]]<br />
<!-- Kinematics and Projectile Motion relocated to Week 3 per advice of Dr. Greco --><br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
*[[Iterative Prediction]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Analytic Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
<!-- *[[Analytical Prediction]] Deprecated --><br />
*[[Kinematics]]<br />
*[[Projectile Motion]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Varying Force====<br />
<div class="mw-collapsible-content"><br />
*[[Fundamentals of Iterative Prediction with Varying Force]]<br />
*[[Spring_Force]]<br />
*[[Simple Harmonic Motion]]<br />
<!--*[[Hooke's Law]] folded into simple harmonic motion--><br />
<!--*[[Spring Force]] folded into simple harmonic motion--><br />
*[[Iterative Prediction of Spring-Mass System]]<br />
*[[Terminal Speed]]<br />
*[[Predicting Change in multiple dimensions]]<br />
*[[Two Dimensional Harmonic Motion]]<br />
*[[Determinism]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Fundamental Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Gravitational Force]]<br />
*[[Gravitational Force Near Earth]]<br />
*[[Gravitational Force in Space and Other Applications]]<br />
*[[3 or More Body Interactions]]<br />
<!--[[Fluid Mechanics]]--><br />
*[[Electric Force]]<br />
*[[Introduction to Magnetic Force]]<br />
*[[Strong and Weak Force]]<br />
*[[Reciprocity]]<br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Properties of Matter====<br />
<div class="mw-collapsible-content"><br />
*[[Kinds of Matter]]<br />
*[[Ball and Spring Model of Matter]]<br />
*[[Density]]<br />
*[[Length and Stiffness of an Interatomic Bond]]<br />
*[[Young's Modulus]]<br />
*[[Speed of Sound in Solids]]<br />
*[[Malleability]]<br />
*[[Ductility]]<br />
*[[Weight]]<br />
*[[Hardness]]<br />
*[[Boiling Point]]<br />
*[[Melting Point]]<br />
*[[Change of State]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Identifying Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Free Body Diagram]]<br />
*[[Inclined Plane]]<br />
*[[Compression or Normal Force]]<br />
*[[Tension]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Curving Motion====<br />
<div class="mw-collapsible-content"><br />
*[[Curving Motion]]<br />
*[[Centripetal Force and Curving Motion]]<br />
*[[Perpetual Freefall (Orbit)]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Energy Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Energy of a Single Particle]]<br />
*[[Kinetic Energy]]<br />
*[[Work/Energy]]<br />
*[[The Energy Principle]]<br />
*[[Conservation of Energy]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Work by Non-Constant Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Work Done By A Nonconstant Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
*[[Potential Energy of Macroscopic Springs]]<br />
*[[Spring Potential Energy]]<br />
*[[Ball and Spring Model]]<br />
*[[Gravitational Potential Energy]]<br />
*[[Energy Graphs]]<br />
*[[Escape Velocity]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Multiparticle Systems====<br />
<div class="mw-collapsible-content"><br />
*[[Center of Mass]]<br />
*[[Multi-particle analysis of Momentum]]<br />
*[[Potential Energy of a Multiparticle System]]<br />
*[[Work and Energy for an Extended System]]<br />
*[[Internal Energy]]<br />
**[[Potential Energy of a Pair of Neutral Atoms]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Choice of System====<br />
<div class="mw-collapsible-content"><br />
*[[System & Surroundings]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Thermal Energy, Dissipation, and Transfer of Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Thermal Energy]]<br />
*[[Specific Heat]]<br />
*[[Calorific Value(Heat of combustion)]]<br />
*[[First Law of Thermodynamics]]<br />
*[[Second Law of Thermodynamics and Entropy]]<br />
*[[Temperature]]<br />
*[[Transformation of Energy]]<br />
*[[The Maxwell-Boltzmann Distribution]]<br />
*[[Air Resistance]]<br />
*[[The Third Law of Thermodynamics]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Rotational and Vibrational Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Translational, Rotational and Vibrational Energy]]<br />
*[[Rolling Motion]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Different Models of a System====<br />
<div class="mw-collapsible-content"><br />
*[[Point Particle Systems]]<br />
*[[Real Systems]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Friction====<br />
<div class="mw-collapsible-content"><br />
*[[Friction]]<br />
*[[Static Friction]]<br />
*[[Kinetic Friction]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conservation of Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Collisions====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's Third Law of Motion]]<br />
*[[Collisions]]<br />
*[[Elastic Collisions]]<br />
*[[Inelastic Collisions]]<br />
*[[Maximally Inelastic Collision]]<br />
*[[Head-on Collision of Equal Masses]]<br />
*[[Head-on Collision of Unequal Masses]]<br />
*[[Scattering: Collisions in 2D and 3D]]<br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Coefficient of Restitution]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rotations====<br />
<div class="mw-collapsible-content"><br />
*[[Rotational Kinematics]]<br />
*[[Eulerian Angles]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Angular Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Total Angular Momentum]]<br />
*[[Translational Angular Momentum]]<br />
*[[Rotational Angular Momentum]]<br />
*[[The Angular Momentum Principle]]<br />
*[[Angular Impulse]]<br />
*[[Predicting the Position of a Rotating System]]<br />
*[[The Moments of Inertia]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Analyzing Motion with and without Torque====<br />
<div class="mw-collapsible-content"><br />
*[[Torque]]<br />
*[[Torque 2]]<br />
*[[Systems with Zero Torque]]<br />
*[[Systems with Nonzero Torque]]<br />
*[[Torque vs Work]]<br />
*[[Gyroscopes]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Introduction to Quantum Concepts====<br />
<div class="mw-collapsible-content"><br />
*[[Bohr Model]]<br />
*[[Energy graphs and the Bohr model]]<br />
*[[Quantized energy levels]]<br />
*[[Electron transitions]]<br />
*[[Entropy]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 2==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====3D Vectors====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[Right-Hand Rule]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Field and Electric Potential]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric force====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field of a point particle====<br />
<div class="mw-collapsible-content"><br />
*[[Point Charge]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Superposition====<br />
<div class="mw-collapsible-content"><br />
*[[Superposition Principle]]<br />
*[[Superposition principle]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Dipole]]<br />
*[[Magnetic Dipole]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Interactions of charged objects====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Potential]]<br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Tape experiments====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Polarization====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
*[[Polarization of an Atom]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conductors and Insulators====<br />
<div class="mw-collapsible-content"><br />
*[[Conductivity and Resistivity]]<br />
*[[Insulators]]<br />
*[[Potential Difference in an Insulator]]<br />
*[[Conductors]]<br />
*[[Polarization of a conductor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Charging and Discharging====<br />
<div class="mw-collapsible-content"><br />
*[[Charge Transfer]]<br />
*[[Electrostatic Discharge]]<br />
*[[Charged Conductor and Charged Insulator]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged rod====<br />
<div class="mw-collapsible-content"><br />
*[[Field of a Charged Rod|Charged Rod]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Field of a charged ring/disk/capacitor====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Ring]]<br />
*[[Charged Disk]]<br />
*[[Charged Capacitor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged sphere====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Spherical Shell]]<br />
*[[Field of a Charged Ball]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric potential====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]] <br />
*[[Potential Difference in a Uniform Field]]<br />
*[[Potential Difference of Point Charge in a Non-Uniform Field]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sign of a potential difference====<br />
<div class="mw-collapsible-content"><br />
*[[Sign of a Potential Difference]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential at a single location====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Potential Difference at One Location]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Path independence and round trip potential====<br />
<div class="mw-collapsible-content"><br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in an insulator====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Difference in an Insulator]]<br />
*[[Electric Field in an Insulator]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Moving charges in a magnetic field====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Biot-Savart Law====<br />
<div class="mw-collapsible-content"><br />
*[[Biot-Savart Law]]<br />
*[[Biot-Savart Law for Currents]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Moving charges, electron current, and conventional current====<br />
<div class="mw-collapsible-content"><br />
*[[Moving Point Charge]]<br />
*[[Current]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a wire====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Long Straight Wire]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Magnetic field of a current-carrying loop====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Loop]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a Charged Disk====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Disk]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Dipole Moment]]<br />
*[[Bar Magnet]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Atomic structure of magnets====<br />
<div class="mw-collapsible-content"><br />
*[[Atomic Structure of Magnets]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Steady state current====<br />
<div class="mw-collapsible-content"><br />
*[[Steady State]]<br />
*[[Non Steady State]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Kirchoff's Laws====<br />
<div class="mw-collapsible-content"><br />
*[[Kirchoff's Laws]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric fields and energy in circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential Difference]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Macroscopic analysis of circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Series Circuits]]<br />
*[[Parallel Circuits]]<br />
*[[Parallel Circuits vs. Series Circuits*]]<br />
*[[Loop Rule]]<br />
*[[Node Rule]]<br />
*[[Fundamentals of Resistance]]<br />
*[[Problem Solving]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in circuits with capacitors====<br />
<div class="mw-collapsible-content"><br />
*[[Charging and Discharging a Capacitor]]<br />
*[[RC Circuit]] <br />
*[[R Circuit]]<br />
*[[AC and DC]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic forces on charges and currents====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[Motors and Generators]]<br />
*[[Applying Magnetic Force to Currents]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
*[[Right-Hand Rule]]<br />
*[[Analysis of Railgun vs Coil gun technologies]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric and magnetic forces====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[VPython Modelling of Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Velocity selector====<br />
<div class="mw-collapsible-content"><br />
*[[Lorentz Force]]<br />
*[[Combining Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Hall Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Hall Effect]]<br />
<h1><strong>Alayna Baker Spring 2020</strong></h1><br />
[[File:Hall Effect 1.jpg]]<br />
[[File:Hall Effect 2.jpg]]<br />
<br />
*[[Right-Hand Rule]]<br />
*[[Motional Emf]]<br />
*[[Magnetic Force]]<br />
*[[Magnetic Torque]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
]]]====Motional EMF====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Motional Emf]]<br />
<h1><strong>Adeline Boswell Fall 2019</strong></h1><br />
[[File:Motional EMF Example.jpg]]<br />
<br />
*[[Motional Emf using Faraday's Law]]<br />
</div><br />
</div>http://www.physicsbook.gatech.edu/Special:RecentChangesLinked/Main_Page<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
If you have a bar attached to two rails, and the rails are connected by a resistor, you have effectively created a circuit. As the bar moves, it creates an "electromotive force"<br />
<br />
[[File:MotEMFCR.jpg]]<br />
<br />
====Magnetic force====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic torque====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Torque]]<br />
*[[Right-Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Gauss's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Flux Theorem]]<br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Ampere's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Ampere's Law]]<br />
*[[Ampere-Maxwell Law]]<br />
*[[Magnetic Field of Coaxial Cable Using Ampere's Law]]<br />
*[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]<br />
*[[Magnetic Field of a Toroid Using Ampere's Law]]<br />
*[[Magnetic Field of a Solenoid Using Ampere's Law]]<br />
*[[The Differential Form of Ampere's Law]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Semiconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Semiconductor Devices]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Faraday's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Faraday's Law]]<br />
*[[Motional Emf using Faraday's Law]]<br />
*[[Lenz's Law]]<br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Maxwell's equations====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
*[[Ampere's Law]]<br />
*[[Faraday's Law]]<br />
*[[Maxwell's Electromagnetic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Circuits revisited====<br />
<div class="mw-collapsible-content"><br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Inductors====<br />
<div class="mw-collapsible-content"><br />
*[[Inductors]]<br />
*[[Current in an LC Circuit]]<br />
*[[Current in an RL Circuit]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
==== Electromagnetic Radiation ====<br />
<div class="mw-collapsible-content"><br />
*[[Electromagnetic Radiation]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sparks in the air====<br />
<div class="mw-collapsible-content"><br />
*[[Sparks in Air]]<br />
*[[Spark Plugs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Superconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Superconducters]]<br />
*[[Superconductors]]<br />
*[[Meissner effect]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 3==<br />
<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Classical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Classical Physics]]<br />
</div><br />
</div><br />
<br />
[[Category:Which Category did you place this in?]]<br />
<br />
===Weeks 2 and 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Special Relativity and the Lorentz Transformation====<br />
<div class="mw-collapsible-content"><br />
*[[Frame of Reference]]<br />
<br />
*[[Einstein's Theory of Special Relativity]]<br />
*[[Time Dilation]]<br />
*[[Lorentz Transformations]]<br />
*[[Relativistic Doppler Effect]]<br />
*[[Einstein's Theory of General Relativity]]<br />
*[[Albert A. Micheleson & Edward W. Morley]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Photons and the Photoelectric Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Spontaneous Photon Emission]]<br />
*[[Light Scattering]]<br />
*[[Lasers]]<br />
*[[Electronic Energy Levels and Photons]]<br />
*[[Quantum Properties of Light]]<br />
*[[The Photoelectric Effect]]<br />
</div><br />
</div><br />
<br />
===Weeks 5 and 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Matter Waves and Wave-Particle Duality====<br />
<div class="mw-collapsible-content"><br />
*[[Wave-Particle Duality]]<br />
*[[Particle in a 1-Dimensional box]]<br />
*[[Heisenberg Uncertainty Principle]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Wave Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Standing Waves]]<br />
*[[Wavelength]]<br />
*[[Wavelength and Frequency]]<br />
*[[Mechanical Waves]]<br />
*[[Transverse and Longitudinal Waves]]<br />
*[[Fourier Series and Transform]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Schrödinger Equation====<br />
<div class="mw-collapsible-content"><br />
*[[The Born Rule]]<br />
*[[Solution for a Single Free Particle]]<br />
*[[Solution for a Single Particle in an Infinite Quantum Well - Darin]]<br />
*[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]<br />
*[[Quantum Harmonic Oscillator]]<br />
*[[Solution for Simple Harmonic Oscillator]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Quantum Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Tunneling through Potential Barriers]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Hydrogen Atom====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rutherford-Bohr Model====<br />
<div class="mw-collapsible-content"><br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Bohr Model]]<br />
*[[Quantized energy levels]]<br />
*[[Energy graphs and the Bohr model]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Many-Electron Atoms====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
*[[Pauli exclusion principle]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Nucleus====<br />
<div class="mw-collapsible-content"><br />
*[[Nucleus]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Molecules====<br />
<div class="mw-collapsible-content"><br />
*[[Molecules]]<br />
*[[Covalent Bonds]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Application of Statistics in Physics]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Temperature & Entropy]]<br />
</div><br />
</div><br />
<br />
===Additional Topics===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Condensed Matter Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Nuclear Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Nuclear Fission]]<br />
*[[Nuclear Energy from Fission and Fusion]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Particle Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Elementary Particles and Particle Physics Theory]]<br />
*[[String Theory]]<br />
</div><br />
</div><br />
</div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Main_Page&diff=41664Main Page2022-12-14T00:54:51Z<p>Ecarder: /* Week 14 */</p>
<hr />
<div>__NOTOC__<br />
= '''Georgia Tech Student Wiki for Introductory Physics.''' =<br />
<br />
This resource was created so that students can contribute and curate content to help those with limited or no access to a textbook. When reading this website, please correct any errors you may come across. If you read something that isn't clear, please consider revising it for future students!<br />
<br />
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#Pick one of the topics from intro physics listed below<br />
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== Source Material ==<br />
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* A physics resource written by experts for an expert audience [https://en.wikipedia.org/wiki/Portal:Physics Physics Portal]<br />
* A wiki written for students by a physics expert [http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes MSU Physics Wiki]<br />
* A wiki book on modern physics [https://en.wikibooks.org/wiki/Modern_Physics Modern Physics Wiki]<br />
* The MIT open courseware for intro physics [http://ocw.mit.edu/resources/res-8-002-a-wikitextbook-for-introductory-mechanics-fall-2009/index.htm MITOCW Wiki]<br />
* An online concept map of intro physics [http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html HyperPhysics]<br />
* Interactive physics simulations [https://phet.colorado.edu/en/simulations/category/physics PhET]<br />
* OpenStax intro physics textbooks: [https://openstax.org/details/books/university-physics-volume-1 Vol1], [https://openstax.org/details/books/university-physics-volume-2 Vol2], [https://openstax.org/details/books/university-physics-volume-3 Vol3]<br />
* The Open Source Physics project is a collection of online physics resources [http://www.opensourcephysics.org/ OSP]<br />
* A resource guide compiled by the [http://www.aapt.org/ AAPT] for educators [http://www.compadre.org/ ComPADRE]<br />
* The Feynman lectures on physics are free to read [http://www.feynmanlectures.caltech.edu/ Feynman]<br />
* Final Study Guide for Modern Physics II created by a lab TA [https://docs.google.com/document/d/1_6GktDPq5tiNFFYs_ZjgjxBAWVQYaXp_2Imha4_nSyc/edit?usp=sharing Modern Physics II Final Study Guide]<br />
<br />
== Resources ==<br />
* Commonly used wiki commands [https://en.wikipedia.org/wiki/Help:Cheatsheet Wiki Cheatsheet]<br />
* A guide to representing equations in math mode [https://en.wikipedia.org/wiki/Help:Displaying_a_formula Wiki Math Mode]<br />
* A page to keep track of all the physics [[Constants]]<br />
* A listing of [[Notable Scientist]] with links to their individual pages <br />
<br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 1==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====GlowScript 101====<br />
<div class="mw-collapsible-content"><br />
*[[Python Syntax]]<br />
*[[GlowScript]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====VPython====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[VPython]]<br />
*[[VPython basics]]<br />
*[[VPython Common Errors and Troubleshooting]]<br />
*[[VPython Functions]]<br />
*[[VPython Lists]]<br />
*[[VPython Loops]]<br />
*[[VPython Multithreading]]<br />
*[[VPython Animation]]<br />
*[[VPython Objects]]<br />
*[[VPython 3D Objects]]<br />
*[[VPython Reference]]<br />
*[[VPython MapReduceFilter]]<br />
*[[VPython GUIs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Vectors and Units====<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[SI Units]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Types of Interactions and How to Detect Them]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Velocity and Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's First Law of Motion]]<br />
*[[Mass]]<br />
*[[Velocity]]<br />
*[[Speed]]<br />
*[[Speed vs Velocity]]<br />
*[[Relative Velocity]]<br />
*[[Derivation of Average Velocity]]<br />
*[[2-Dimensional Motion]]<br />
*[[3-Dimensional Position and Motion]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Momentum and the Momentum Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Linear Momentum]]<br />
*[[Newton's Second Law: the Momentum Principle]]<br />
*[[Impulse and Momentum]]<br />
*[[Net Force]]<br />
*[[Inertia]]<br />
*[[Acceleration]]<br />
*[[Relativistic Momentum]]<br />
<!-- Kinematics and Projectile Motion relocated to Week 3 per advice of Dr. Greco --><br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
*[[Iterative Prediction]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Analytic Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
<!-- *[[Analytical Prediction]] Deprecated --><br />
*[[Kinematics]]<br />
*[[Projectile Motion]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Varying Force====<br />
<div class="mw-collapsible-content"><br />
*[[Fundamentals of Iterative Prediction with Varying Force]]<br />
*[[Spring_Force]]<br />
*[[Simple Harmonic Motion]]<br />
<!--*[[Hooke's Law]] folded into simple harmonic motion--><br />
<!--*[[Spring Force]] folded into simple harmonic motion--><br />
*[[Iterative Prediction of Spring-Mass System]]<br />
*[[Terminal Speed]]<br />
*[[Predicting Change in multiple dimensions]]<br />
*[[Two Dimensional Harmonic Motion]]<br />
*[[Determinism]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Fundamental Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Gravitational Force]]<br />
*[[Gravitational Force Near Earth]]<br />
*[[Gravitational Force in Space and Other Applications]]<br />
*[[3 or More Body Interactions]]<br />
<!--[[Fluid Mechanics]]--><br />
*[[Electric Force]]<br />
*[[Introduction to Magnetic Force]]<br />
*[[Strong and Weak Force]]<br />
*[[Reciprocity]]<br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Properties of Matter====<br />
<div class="mw-collapsible-content"><br />
*[[Kinds of Matter]]<br />
*[[Ball and Spring Model of Matter]]<br />
*[[Density]]<br />
*[[Length and Stiffness of an Interatomic Bond]]<br />
*[[Young's Modulus]]<br />
*[[Speed of Sound in Solids]]<br />
*[[Malleability]]<br />
*[[Ductility]]<br />
*[[Weight]]<br />
*[[Hardness]]<br />
*[[Boiling Point]]<br />
*[[Melting Point]]<br />
*[[Change of State]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Identifying Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Free Body Diagram]]<br />
*[[Inclined Plane]]<br />
*[[Compression or Normal Force]]<br />
*[[Tension]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Curving Motion====<br />
<div class="mw-collapsible-content"><br />
*[[Curving Motion]]<br />
*[[Centripetal Force and Curving Motion]]<br />
*[[Perpetual Freefall (Orbit)]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Energy Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Energy of a Single Particle]]<br />
*[[Kinetic Energy]]<br />
*[[Work/Energy]]<br />
*[[The Energy Principle]]<br />
*[[Conservation of Energy]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Work by Non-Constant Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Work Done By A Nonconstant Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
*[[Potential Energy of Macroscopic Springs]]<br />
*[[Spring Potential Energy]]<br />
*[[Ball and Spring Model]]<br />
*[[Gravitational Potential Energy]]<br />
*[[Energy Graphs]]<br />
*[[Escape Velocity]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Multiparticle Systems====<br />
<div class="mw-collapsible-content"><br />
*[[Center of Mass]]<br />
*[[Multi-particle analysis of Momentum]]<br />
*[[Potential Energy of a Multiparticle System]]<br />
*[[Work and Energy for an Extended System]]<br />
*[[Internal Energy]]<br />
**[[Potential Energy of a Pair of Neutral Atoms]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Choice of System====<br />
<div class="mw-collapsible-content"><br />
*[[System & Surroundings]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Thermal Energy, Dissipation, and Transfer of Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Thermal Energy]]<br />
*[[Specific Heat]]<br />
*[[Calorific Value(Heat of combustion)]]<br />
*[[First Law of Thermodynamics]]<br />
*[[Second Law of Thermodynamics and Entropy]]<br />
*[[Temperature]]<br />
*[[Transformation of Energy]]<br />
*[[The Maxwell-Boltzmann Distribution]]<br />
*[[Air Resistance]]<br />
*[[The Third Law of Thermodynamics]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Rotational and Vibrational Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Translational, Rotational and Vibrational Energy]]<br />
*[[Rolling Motion]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Different Models of a System====<br />
<div class="mw-collapsible-content"><br />
*[[Point Particle Systems]]<br />
*[[Real Systems]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Friction====<br />
<div class="mw-collapsible-content"><br />
*[[Friction]]<br />
*[[Static Friction]]<br />
*[[Kinetic Friction]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conservation of Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Collisions====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's Third Law of Motion]]<br />
*[[Collisions]]<br />
*[[Elastic Collisions]]<br />
*[[Inelastic Collisions]]<br />
*[[Maximally Inelastic Collision]]<br />
*[[Head-on Collision of Equal Masses]]<br />
*[[Head-on Collision of Unequal Masses]]<br />
*[[Scattering: Collisions in 2D and 3D]]<br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Coefficient of Restitution]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rotations====<br />
<div class="mw-collapsible-content"><br />
*[[Rotational Kinematics]]<br />
*[[Eulerian Angles]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Angular Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Total Angular Momentum]]<br />
*[[Translational Angular Momentum]]<br />
*[[Rotational Angular Momentum]]<br />
*[[The Angular Momentum Principle]]<br />
*[[Angular Impulse]]<br />
*[[Predicting the Position of a Rotating System]]<br />
*[[The Moments of Inertia]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Analyzing Motion with and without Torque====<br />
<div class="mw-collapsible-content"><br />
*[[Torque]]<br />
*[[Torque 2]]<br />
*[[Systems with Zero Torque]]<br />
*[[Systems with Nonzero Torque]]<br />
*[[Torque vs Work]]<br />
*[[Gyroscopes]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Introduction to Quantum Concepts====<br />
<div class="mw-collapsible-content"><br />
*[[Bohr Model]]<br />
*[[Energy graphs and the Bohr model]]<br />
*[[Quantized energy levels]]<br />
*[[Electron transitions]]<br />
*[[Entropy]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 2==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====3D Vectors====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[Right-Hand Rule]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Field and Electric Potential]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric force====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field of a point particle====<br />
<div class="mw-collapsible-content"><br />
*[[Point Charge]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Superposition====<br />
<div class="mw-collapsible-content"><br />
*[[Superposition Principle]]<br />
*[[Superposition principle]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Dipole]]<br />
*[[Magnetic Dipole]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Interactions of charged objects====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Potential]]<br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Tape experiments====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Polarization====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
*[[Polarization of an Atom]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conductors and Insulators====<br />
<div class="mw-collapsible-content"><br />
*[[Conductivity and Resistivity]]<br />
*[[Insulators]]<br />
*[[Potential Difference in an Insulator]]<br />
*[[Conductors]]<br />
*[[Polarization of a conductor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Charging and Discharging====<br />
<div class="mw-collapsible-content"><br />
*[[Charge Transfer]]<br />
*[[Electrostatic Discharge]]<br />
*[[Charged Conductor and Charged Insulator]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged rod====<br />
<div class="mw-collapsible-content"><br />
*[[Field of a Charged Rod|Charged Rod]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Field of a charged ring/disk/capacitor====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Ring]]<br />
*[[Charged Disk]]<br />
*[[Charged Capacitor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged sphere====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Spherical Shell]]<br />
*[[Field of a Charged Ball]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric potential====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]] <br />
*[[Potential Difference in a Uniform Field]]<br />
*[[Potential Difference of Point Charge in a Non-Uniform Field]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sign of a potential difference====<br />
<div class="mw-collapsible-content"><br />
*[[Sign of a Potential Difference]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential at a single location====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Potential Difference at One Location]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Path independence and round trip potential====<br />
<div class="mw-collapsible-content"><br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in an insulator====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Difference in an Insulator]]<br />
*[[Electric Field in an Insulator]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Moving charges in a magnetic field====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Biot-Savart Law====<br />
<div class="mw-collapsible-content"><br />
*[[Biot-Savart Law]]<br />
*[[Biot-Savart Law for Currents]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Moving charges, electron current, and conventional current====<br />
<div class="mw-collapsible-content"><br />
*[[Moving Point Charge]]<br />
*[[Current]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a wire====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Long Straight Wire]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Magnetic field of a current-carrying loop====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Loop]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a Charged Disk====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Disk]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Dipole Moment]]<br />
*[[Bar Magnet]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Atomic structure of magnets====<br />
<div class="mw-collapsible-content"><br />
*[[Atomic Structure of Magnets]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Steady state current====<br />
<div class="mw-collapsible-content"><br />
*[[Steady State]]<br />
*[[Non Steady State]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Kirchoff's Laws====<br />
<div class="mw-collapsible-content"><br />
*[[Kirchoff's Laws]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric fields and energy in circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential Difference]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Macroscopic analysis of circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Series Circuits]]<br />
*[[Parallel Circuits]]<br />
*[[Parallel Circuits vs. Series Circuits*]]<br />
*[[Loop Rule]]<br />
*[[Node Rule]]<br />
*[[Fundamentals of Resistance]]<br />
*[[Problem Solving]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in circuits with capacitors====<br />
<div class="mw-collapsible-content"><br />
*[[Charging and Discharging a Capacitor]]<br />
*[[RC Circuit]] <br />
*[[R Circuit]]<br />
*[[AC and DC]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic forces on charges and currents====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[Motors and Generators]]<br />
*[[Applying Magnetic Force to Currents]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
*[[Right-Hand Rule]]<br />
*[[Analysis of Railgun vs Coil gun technologies]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric and magnetic forces====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[VPython Modelling of Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Velocity selector====<br />
<div class="mw-collapsible-content"><br />
*[[Lorentz Force]]<br />
*[[Combining Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Hall Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Hall Effect]]<br />
<h1><strong>Alayna Baker Spring 2020</strong></h1><br />
[[File:Hall Effect 1.jpg]]<br />
[[File:Hall Effect 2.jpg]]<br />
<br />
*[[Right-Hand Rule]]<br />
*[[Motional Emf]]<br />
*[[Magnetic Force]]<br />
*[[Magnetic Torque]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
]]]====Motional EMF====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Motional Emf]]<br />
<h1><strong>Adeline Boswell Fall 2019</strong></h1><br />
[[File:Motional EMF Example.jpg]]<br />
<br />
*[[Motional Emf using Faraday's Law]]<br />
</div><br />
</div>http://www.physicsbook.gatech.edu/Special:RecentChangesLinked/Main_Page<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
If you have a bar attached to two rails, and the rails are connected by a resistor, you have effectively created a circuit. As the bar moves, it creates an "electromotive force"<br />
<br />
[[File:MotEMFCR.jpg]]<br />
<br />
====Magnetic force====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic torque====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Torque]]<br />
*[[Right-Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Gauss's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Flux Theorem]]<br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Ampere's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Ampere's Law]]<br />
*[[Ampere-Maxwell Law]]<br />
*[[Magnetic Field of Coaxial Cable Using Ampere's Law]]<br />
*[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]<br />
*[[Magnetic Field of a Toroid Using Ampere's Law]]<br />
*[[Magnetic Field of a Solenoid Using Ampere's Law]]<br />
*[[The Differential Form of Ampere's Law]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Semiconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Semiconductor Devices]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Faraday's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Faraday's Law]]<br />
*[[Motional Emf using Faraday's Law]]<br />
*[[Lenz's Law]]<br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Maxwell's equations====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
*[[Ampere's Law]]<br />
*[[Faraday's Law]]<br />
*[[Maxwell's Electromagnetic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Circuits revisited====<br />
<div class="mw-collapsible-content"><br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Inductors====<br />
<div class="mw-collapsible-content"><br />
*[[Inductors]]<br />
*[[Current in an LC Circuit]]<br />
*[[Current in an RL Circuit]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
==== Electromagnetic Radiation ====<br />
<div class="mw-collapsible-content"><br />
*[[Electromagnetic Radiation]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sparks in the air====<br />
<div class="mw-collapsible-content"><br />
*[[Sparks in Air]]<br />
*[[Spark Plugs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Superconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Superconducters]]<br />
*[[Superconductors]]<br />
*[[Meissner effect]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 3==<br />
<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Classical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Classical Physics]]<br />
</div><br />
</div><br />
<br />
[[Category:Which Category did you place this in?]]<br />
<br />
===Weeks 2 and 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Special Relativity and the Lorentz Transformation====<br />
<div class="mw-collapsible-content"><br />
*[[Frame of Reference]]<br />
<br />
*[[Einstein's Theory of Special Relativity]]<br />
*[[Time Dilation]]<br />
*[[Lorentz Transformations]]<br />
*[[Relativistic Doppler Effect]]<br />
*[[Einstein's Theory of General Relativity]]<br />
*[[Albert A. Micheleson & Edward W. Morley]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Photons and the Photoelectric Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Spontaneous Photon Emission]]<br />
*[[Light Scattering]]<br />
*[[Lasers]]<br />
*[[Electronic Energy Levels and Photons]]<br />
*[[Quantum Properties of Light]]<br />
*[[The Photoelectric Effect]]<br />
</div><br />
</div><br />
<br />
===Weeks 5 and 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Matter Waves and Wave-Particle Duality====<br />
<div class="mw-collapsible-content"><br />
*[[Wave-Particle Duality]]<br />
*[[Particle in a 1-Dimensional box]]<br />
*[[Heisenberg Uncertainty Principle]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Wave Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Standing Waves]]<br />
*[[Wavelength]]<br />
*[[Wavelength and Frequency]]<br />
*[[Mechanical Waves]]<br />
*[[Transverse and Longitudinal Waves]]<br />
*[[Fourier Series and Transform]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Schrödinger Equation====<br />
<div class="mw-collapsible-content"><br />
*[[The Born Rule]]<br />
*[[Solution for a Single Free Particle]]<br />
*[[Solution for a Single Particle in an Infinite Quantum Well - Darin]]<br />
*[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]<br />
*[[Quantum Harmonic Oscillator]]<br />
*[[Solution for Simple Harmonic Oscillator]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Quantum Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Tunneling through Potential Barriers]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Hydrogen Atom====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rutherford-Bohr Model====<br />
<div class="mw-collapsible-content"><br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Bohr Model]]<br />
*[[Quantized energy levels]]<br />
*[[Energy graphs and the Bohr model]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Many-Electron Atoms====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
*[[Pauli exclusion principle]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Nucleus====<br />
<div class="mw-collapsible-content"><br />
*[[Nucleus]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Molecules====<br />
<div class="mw-collapsible-content"><br />
*[[Molecules]]<br />
*[[Covalent Bonds]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Application of Statistics in Physics]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
*[[Temperature & Entropy]]<br />
</div><br />
<br />
===Additional Topics===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Condensed Matter Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Nuclear Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Nuclear Fission]]<br />
*[[Nuclear Energy from Fission and Fusion]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Particle Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Elementary Particles and Particle Physics Theory]]<br />
*[[String Theory]]<br />
</div><br />
</div><br />
</div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Main_Page&diff=41663Main Page2022-12-14T00:54:20Z<p>Ecarder: /* Week 13 */</p>
<hr />
<div>__NOTOC__<br />
= '''Georgia Tech Student Wiki for Introductory Physics.''' =<br />
<br />
This resource was created so that students can contribute and curate content to help those with limited or no access to a textbook. When reading this website, please correct any errors you may come across. If you read something that isn't clear, please consider revising it for future students!<br />
<br />
Looking to make a contribution?<br />
#Pick one of the topics from intro physics listed below<br />
#Add content to that topic or improve the quality of what is already there.<br />
#Need to make a new topic? Edit this page and add it to the list under the appropriate category. Then copy and paste the default [[Template]] into your new page and start editing.<br />
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Please remember that this is not a textbook and you are not limited to expressing your ideas with only text and equations. Whenever possible embed: pictures, videos, diagrams, simulations, computational models (e.g. Glowscript), and whatever content you think makes learning physics easier for other students.<br />
<br />
== Source Material ==<br />
All of the content added to this resource must be in the public domain or similar free resource. If you are unsure about a source, contact the original author for permission. That said, there is a surprisingly large amount of introductory physics content scattered across the web. Here is an incomplete list of intro physics resources (please update as needed).<br />
* A physics resource written by experts for an expert audience [https://en.wikipedia.org/wiki/Portal:Physics Physics Portal]<br />
* A wiki written for students by a physics expert [http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes MSU Physics Wiki]<br />
* A wiki book on modern physics [https://en.wikibooks.org/wiki/Modern_Physics Modern Physics Wiki]<br />
* The MIT open courseware for intro physics [http://ocw.mit.edu/resources/res-8-002-a-wikitextbook-for-introductory-mechanics-fall-2009/index.htm MITOCW Wiki]<br />
* An online concept map of intro physics [http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html HyperPhysics]<br />
* Interactive physics simulations [https://phet.colorado.edu/en/simulations/category/physics PhET]<br />
* OpenStax intro physics textbooks: [https://openstax.org/details/books/university-physics-volume-1 Vol1], [https://openstax.org/details/books/university-physics-volume-2 Vol2], [https://openstax.org/details/books/university-physics-volume-3 Vol3]<br />
* The Open Source Physics project is a collection of online physics resources [http://www.opensourcephysics.org/ OSP]<br />
* A resource guide compiled by the [http://www.aapt.org/ AAPT] for educators [http://www.compadre.org/ ComPADRE]<br />
* The Feynman lectures on physics are free to read [http://www.feynmanlectures.caltech.edu/ Feynman]<br />
* Final Study Guide for Modern Physics II created by a lab TA [https://docs.google.com/document/d/1_6GktDPq5tiNFFYs_ZjgjxBAWVQYaXp_2Imha4_nSyc/edit?usp=sharing Modern Physics II Final Study Guide]<br />
<br />
== Resources ==<br />
* Commonly used wiki commands [https://en.wikipedia.org/wiki/Help:Cheatsheet Wiki Cheatsheet]<br />
* A guide to representing equations in math mode [https://en.wikipedia.org/wiki/Help:Displaying_a_formula Wiki Math Mode]<br />
* A page to keep track of all the physics [[Constants]]<br />
* A listing of [[Notable Scientist]] with links to their individual pages <br />
<br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 1==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====GlowScript 101====<br />
<div class="mw-collapsible-content"><br />
*[[Python Syntax]]<br />
*[[GlowScript]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====VPython====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[VPython]]<br />
*[[VPython basics]]<br />
*[[VPython Common Errors and Troubleshooting]]<br />
*[[VPython Functions]]<br />
*[[VPython Lists]]<br />
*[[VPython Loops]]<br />
*[[VPython Multithreading]]<br />
*[[VPython Animation]]<br />
*[[VPython Objects]]<br />
*[[VPython 3D Objects]]<br />
*[[VPython Reference]]<br />
*[[VPython MapReduceFilter]]<br />
*[[VPython GUIs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Vectors and Units====<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[SI Units]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Types of Interactions and How to Detect Them]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Velocity and Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's First Law of Motion]]<br />
*[[Mass]]<br />
*[[Velocity]]<br />
*[[Speed]]<br />
*[[Speed vs Velocity]]<br />
*[[Relative Velocity]]<br />
*[[Derivation of Average Velocity]]<br />
*[[2-Dimensional Motion]]<br />
*[[3-Dimensional Position and Motion]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Momentum and the Momentum Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Linear Momentum]]<br />
*[[Newton's Second Law: the Momentum Principle]]<br />
*[[Impulse and Momentum]]<br />
*[[Net Force]]<br />
*[[Inertia]]<br />
*[[Acceleration]]<br />
*[[Relativistic Momentum]]<br />
<!-- Kinematics and Projectile Motion relocated to Week 3 per advice of Dr. Greco --><br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
*[[Iterative Prediction]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Analytic Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
<!-- *[[Analytical Prediction]] Deprecated --><br />
*[[Kinematics]]<br />
*[[Projectile Motion]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Varying Force====<br />
<div class="mw-collapsible-content"><br />
*[[Fundamentals of Iterative Prediction with Varying Force]]<br />
*[[Spring_Force]]<br />
*[[Simple Harmonic Motion]]<br />
<!--*[[Hooke's Law]] folded into simple harmonic motion--><br />
<!--*[[Spring Force]] folded into simple harmonic motion--><br />
*[[Iterative Prediction of Spring-Mass System]]<br />
*[[Terminal Speed]]<br />
*[[Predicting Change in multiple dimensions]]<br />
*[[Two Dimensional Harmonic Motion]]<br />
*[[Determinism]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Fundamental Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Gravitational Force]]<br />
*[[Gravitational Force Near Earth]]<br />
*[[Gravitational Force in Space and Other Applications]]<br />
*[[3 or More Body Interactions]]<br />
<!--[[Fluid Mechanics]]--><br />
*[[Electric Force]]<br />
*[[Introduction to Magnetic Force]]<br />
*[[Strong and Weak Force]]<br />
*[[Reciprocity]]<br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Properties of Matter====<br />
<div class="mw-collapsible-content"><br />
*[[Kinds of Matter]]<br />
*[[Ball and Spring Model of Matter]]<br />
*[[Density]]<br />
*[[Length and Stiffness of an Interatomic Bond]]<br />
*[[Young's Modulus]]<br />
*[[Speed of Sound in Solids]]<br />
*[[Malleability]]<br />
*[[Ductility]]<br />
*[[Weight]]<br />
*[[Hardness]]<br />
*[[Boiling Point]]<br />
*[[Melting Point]]<br />
*[[Change of State]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Identifying Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Free Body Diagram]]<br />
*[[Inclined Plane]]<br />
*[[Compression or Normal Force]]<br />
*[[Tension]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Curving Motion====<br />
<div class="mw-collapsible-content"><br />
*[[Curving Motion]]<br />
*[[Centripetal Force and Curving Motion]]<br />
*[[Perpetual Freefall (Orbit)]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Energy Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Energy of a Single Particle]]<br />
*[[Kinetic Energy]]<br />
*[[Work/Energy]]<br />
*[[The Energy Principle]]<br />
*[[Conservation of Energy]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Work by Non-Constant Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Work Done By A Nonconstant Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
*[[Potential Energy of Macroscopic Springs]]<br />
*[[Spring Potential Energy]]<br />
*[[Ball and Spring Model]]<br />
*[[Gravitational Potential Energy]]<br />
*[[Energy Graphs]]<br />
*[[Escape Velocity]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Multiparticle Systems====<br />
<div class="mw-collapsible-content"><br />
*[[Center of Mass]]<br />
*[[Multi-particle analysis of Momentum]]<br />
*[[Potential Energy of a Multiparticle System]]<br />
*[[Work and Energy for an Extended System]]<br />
*[[Internal Energy]]<br />
**[[Potential Energy of a Pair of Neutral Atoms]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Choice of System====<br />
<div class="mw-collapsible-content"><br />
*[[System & Surroundings]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Thermal Energy, Dissipation, and Transfer of Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Thermal Energy]]<br />
*[[Specific Heat]]<br />
*[[Calorific Value(Heat of combustion)]]<br />
*[[First Law of Thermodynamics]]<br />
*[[Second Law of Thermodynamics and Entropy]]<br />
*[[Temperature]]<br />
*[[Transformation of Energy]]<br />
*[[The Maxwell-Boltzmann Distribution]]<br />
*[[Air Resistance]]<br />
*[[The Third Law of Thermodynamics]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Rotational and Vibrational Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Translational, Rotational and Vibrational Energy]]<br />
*[[Rolling Motion]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Different Models of a System====<br />
<div class="mw-collapsible-content"><br />
*[[Point Particle Systems]]<br />
*[[Real Systems]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Friction====<br />
<div class="mw-collapsible-content"><br />
*[[Friction]]<br />
*[[Static Friction]]<br />
*[[Kinetic Friction]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conservation of Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Collisions====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's Third Law of Motion]]<br />
*[[Collisions]]<br />
*[[Elastic Collisions]]<br />
*[[Inelastic Collisions]]<br />
*[[Maximally Inelastic Collision]]<br />
*[[Head-on Collision of Equal Masses]]<br />
*[[Head-on Collision of Unequal Masses]]<br />
*[[Scattering: Collisions in 2D and 3D]]<br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Coefficient of Restitution]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rotations====<br />
<div class="mw-collapsible-content"><br />
*[[Rotational Kinematics]]<br />
*[[Eulerian Angles]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Angular Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Total Angular Momentum]]<br />
*[[Translational Angular Momentum]]<br />
*[[Rotational Angular Momentum]]<br />
*[[The Angular Momentum Principle]]<br />
*[[Angular Impulse]]<br />
*[[Predicting the Position of a Rotating System]]<br />
*[[The Moments of Inertia]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Analyzing Motion with and without Torque====<br />
<div class="mw-collapsible-content"><br />
*[[Torque]]<br />
*[[Torque 2]]<br />
*[[Systems with Zero Torque]]<br />
*[[Systems with Nonzero Torque]]<br />
*[[Torque vs Work]]<br />
*[[Gyroscopes]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Introduction to Quantum Concepts====<br />
<div class="mw-collapsible-content"><br />
*[[Bohr Model]]<br />
*[[Energy graphs and the Bohr model]]<br />
*[[Quantized energy levels]]<br />
*[[Electron transitions]]<br />
*[[Entropy]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 2==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====3D Vectors====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[Right-Hand Rule]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Field and Electric Potential]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric force====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field of a point particle====<br />
<div class="mw-collapsible-content"><br />
*[[Point Charge]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Superposition====<br />
<div class="mw-collapsible-content"><br />
*[[Superposition Principle]]<br />
*[[Superposition principle]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Dipole]]<br />
*[[Magnetic Dipole]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Interactions of charged objects====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Potential]]<br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Tape experiments====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Polarization====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
*[[Polarization of an Atom]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conductors and Insulators====<br />
<div class="mw-collapsible-content"><br />
*[[Conductivity and Resistivity]]<br />
*[[Insulators]]<br />
*[[Potential Difference in an Insulator]]<br />
*[[Conductors]]<br />
*[[Polarization of a conductor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Charging and Discharging====<br />
<div class="mw-collapsible-content"><br />
*[[Charge Transfer]]<br />
*[[Electrostatic Discharge]]<br />
*[[Charged Conductor and Charged Insulator]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged rod====<br />
<div class="mw-collapsible-content"><br />
*[[Field of a Charged Rod|Charged Rod]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Field of a charged ring/disk/capacitor====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Ring]]<br />
*[[Charged Disk]]<br />
*[[Charged Capacitor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged sphere====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Spherical Shell]]<br />
*[[Field of a Charged Ball]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric potential====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]] <br />
*[[Potential Difference in a Uniform Field]]<br />
*[[Potential Difference of Point Charge in a Non-Uniform Field]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sign of a potential difference====<br />
<div class="mw-collapsible-content"><br />
*[[Sign of a Potential Difference]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential at a single location====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Potential Difference at One Location]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Path independence and round trip potential====<br />
<div class="mw-collapsible-content"><br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in an insulator====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Difference in an Insulator]]<br />
*[[Electric Field in an Insulator]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Moving charges in a magnetic field====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Biot-Savart Law====<br />
<div class="mw-collapsible-content"><br />
*[[Biot-Savart Law]]<br />
*[[Biot-Savart Law for Currents]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Moving charges, electron current, and conventional current====<br />
<div class="mw-collapsible-content"><br />
*[[Moving Point Charge]]<br />
*[[Current]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a wire====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Long Straight Wire]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Magnetic field of a current-carrying loop====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Loop]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a Charged Disk====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Disk]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Dipole Moment]]<br />
*[[Bar Magnet]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Atomic structure of magnets====<br />
<div class="mw-collapsible-content"><br />
*[[Atomic Structure of Magnets]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Steady state current====<br />
<div class="mw-collapsible-content"><br />
*[[Steady State]]<br />
*[[Non Steady State]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Kirchoff's Laws====<br />
<div class="mw-collapsible-content"><br />
*[[Kirchoff's Laws]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric fields and energy in circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential Difference]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Macroscopic analysis of circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Series Circuits]]<br />
*[[Parallel Circuits]]<br />
*[[Parallel Circuits vs. Series Circuits*]]<br />
*[[Loop Rule]]<br />
*[[Node Rule]]<br />
*[[Fundamentals of Resistance]]<br />
*[[Problem Solving]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in circuits with capacitors====<br />
<div class="mw-collapsible-content"><br />
*[[Charging and Discharging a Capacitor]]<br />
*[[RC Circuit]] <br />
*[[R Circuit]]<br />
*[[AC and DC]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic forces on charges and currents====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[Motors and Generators]]<br />
*[[Applying Magnetic Force to Currents]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
*[[Right-Hand Rule]]<br />
*[[Analysis of Railgun vs Coil gun technologies]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric and magnetic forces====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[VPython Modelling of Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Velocity selector====<br />
<div class="mw-collapsible-content"><br />
*[[Lorentz Force]]<br />
*[[Combining Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Hall Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Hall Effect]]<br />
<h1><strong>Alayna Baker Spring 2020</strong></h1><br />
[[File:Hall Effect 1.jpg]]<br />
[[File:Hall Effect 2.jpg]]<br />
<br />
*[[Right-Hand Rule]]<br />
*[[Motional Emf]]<br />
*[[Magnetic Force]]<br />
*[[Magnetic Torque]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
]]]====Motional EMF====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Motional Emf]]<br />
<h1><strong>Adeline Boswell Fall 2019</strong></h1><br />
[[File:Motional EMF Example.jpg]]<br />
<br />
*[[Motional Emf using Faraday's Law]]<br />
</div><br />
</div>http://www.physicsbook.gatech.edu/Special:RecentChangesLinked/Main_Page<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
If you have a bar attached to two rails, and the rails are connected by a resistor, you have effectively created a circuit. As the bar moves, it creates an "electromotive force"<br />
<br />
[[File:MotEMFCR.jpg]]<br />
<br />
====Magnetic force====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic torque====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Torque]]<br />
*[[Right-Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Gauss's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Flux Theorem]]<br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Ampere's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Ampere's Law]]<br />
*[[Ampere-Maxwell Law]]<br />
*[[Magnetic Field of Coaxial Cable Using Ampere's Law]]<br />
*[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]<br />
*[[Magnetic Field of a Toroid Using Ampere's Law]]<br />
*[[Magnetic Field of a Solenoid Using Ampere's Law]]<br />
*[[The Differential Form of Ampere's Law]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Semiconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Semiconductor Devices]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Faraday's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Faraday's Law]]<br />
*[[Motional Emf using Faraday's Law]]<br />
*[[Lenz's Law]]<br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Maxwell's equations====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
*[[Ampere's Law]]<br />
*[[Faraday's Law]]<br />
*[[Maxwell's Electromagnetic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Circuits revisited====<br />
<div class="mw-collapsible-content"><br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Inductors====<br />
<div class="mw-collapsible-content"><br />
*[[Inductors]]<br />
*[[Current in an LC Circuit]]<br />
*[[Current in an RL Circuit]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
==== Electromagnetic Radiation ====<br />
<div class="mw-collapsible-content"><br />
*[[Electromagnetic Radiation]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sparks in the air====<br />
<div class="mw-collapsible-content"><br />
*[[Sparks in Air]]<br />
*[[Spark Plugs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Superconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Superconducters]]<br />
*[[Superconductors]]<br />
*[[Meissner effect]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 3==<br />
<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Classical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Classical Physics]]<br />
</div><br />
</div><br />
<br />
[[Category:Which Category did you place this in?]]<br />
<br />
===Weeks 2 and 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Special Relativity and the Lorentz Transformation====<br />
<div class="mw-collapsible-content"><br />
*[[Frame of Reference]]<br />
<br />
*[[Einstein's Theory of Special Relativity]]<br />
*[[Time Dilation]]<br />
*[[Lorentz Transformations]]<br />
*[[Relativistic Doppler Effect]]<br />
*[[Einstein's Theory of General Relativity]]<br />
*[[Albert A. Micheleson & Edward W. Morley]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Photons and the Photoelectric Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Spontaneous Photon Emission]]<br />
*[[Light Scattering]]<br />
*[[Lasers]]<br />
*[[Electronic Energy Levels and Photons]]<br />
*[[Quantum Properties of Light]]<br />
*[[The Photoelectric Effect]]<br />
</div><br />
</div><br />
<br />
===Weeks 5 and 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Matter Waves and Wave-Particle Duality====<br />
<div class="mw-collapsible-content"><br />
*[[Wave-Particle Duality]]<br />
*[[Particle in a 1-Dimensional box]]<br />
*[[Heisenberg Uncertainty Principle]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Wave Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Standing Waves]]<br />
*[[Wavelength]]<br />
*[[Wavelength and Frequency]]<br />
*[[Mechanical Waves]]<br />
*[[Transverse and Longitudinal Waves]]<br />
*[[Fourier Series and Transform]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Schrödinger Equation====<br />
<div class="mw-collapsible-content"><br />
*[[The Born Rule]]<br />
*[[Solution for a Single Free Particle]]<br />
*[[Solution for a Single Particle in an Infinite Quantum Well - Darin]]<br />
*[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]<br />
*[[Quantum Harmonic Oscillator]]<br />
*[[Solution for Simple Harmonic Oscillator]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Quantum Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Tunneling through Potential Barriers]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Hydrogen Atom====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rutherford-Bohr Model====<br />
<div class="mw-collapsible-content"><br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Bohr Model]]<br />
*[[Quantized energy levels]]<br />
*[[Energy graphs and the Bohr model]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Many-Electron Atoms====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
*[[Pauli exclusion principle]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Nucleus====<br />
<div class="mw-collapsible-content"><br />
*[[Nucleus]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Molecules====<br />
<div class="mw-collapsible-content"><br />
*[[Molecules]]<br />
*[[Covalent Bonds]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
*[[Application of Statistics in Physics]]<br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
*[[Temperature & Entropy]]<br />
</div><br />
<br />
===Additional Topics===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Condensed Matter Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Nuclear Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Nuclear Fission]]<br />
*[[Nuclear Energy from Fission and Fusion]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Particle Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Elementary Particles and Particle Physics Theory]]<br />
*[[String Theory]]<br />
</div><br />
</div><br />
</div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41662Quantum Harmonic Oscillator2022-12-07T17:59:05Z<p>Ecarder: /* Deriving the Energy and Potential for Analytical, Solutions */</p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Considering the Potential for Analytical, Closed-Form Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}=\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-t\frac{du(t)}{dt}e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2u(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
<br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the byproducts of these vibrations.<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41661Quantum Harmonic Oscillator2022-12-07T17:33:53Z<p>Ecarder: </p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Potential for Analytical, Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{\bigg[(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)\bigg]t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar \omega}-1-2n)a_n=0</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the byproducts of these vibrations.<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41660Fourier Series and Transform2022-12-07T17:26:48Z<p>Ecarder: </p>
<hr />
<div>'''Claimed - Eric Carder Fall 2022''' <br><br />
==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br><br><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==Connectedness==<br />
Fourier Series and Transforms are applicable on a variety of frontiers. They has proven themselves useful in Modern Physics, but there is a good chance anyone in a mathematics heavy STEM field will encounter Fourier Series and Transforms as they are heavily prevalent in other areas of science and technology. Even if these aren't useful in a given person's industry, it is still useful to understand them and their importance, as they involve a clever manipulation of information.<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br><br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41659Quantum Harmonic Oscillator2022-12-07T17:23:50Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Potential for Analytical, Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
==Connectedness==<br />
At first glance, applications of the quantum harmonic oscillator seem limited. It's important to realize that it is simply an analog of its classical counterpart. When harmonically oscillating systems approach the atomic scale, they must obey the laws of quantum mechanics as anything else would. An adept understanding of quantum harmonic oscillators can enhance humans' understanding of larger-scale models. A quick example would be the consideration of how an understanding of molecular vibrations can lead to a greater understanding of the byproducts of these vibrations.<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41658Fourier Series and Transform2022-12-07T17:02:45Z<p>Ecarder: </p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br><br><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==Connectedness==<br />
Fourier Series and Transforms are applicable on a variety of frontiers. They has proven themselves useful in Modern Physics, but there is a good chance anyone in a mathematics heavy STEM field will encounter Fourier Series and Transforms as they are heavily prevalent in other areas of science and technology. Even if these aren't useful in a given person's industry, it is still useful to understand them and their importance, as they involve a clever manipulation of information.<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br><br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41657Quantum Harmonic Oscillator2022-12-07T16:50:48Z<p>Ecarder: /* References */</p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Potential for Analytical, Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41656Quantum Harmonic Oscillator2022-12-07T16:50:33Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Potential for Analytical, Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the formula for energy in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br><br />
<br />
==References==<br />
Dudik, R. (2004, May 5). The Quantum Harmonic Oscillator. Retrieved 2022, from http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/harmonic/ <br><br />
<br />
Libretexts. (2020, August 11). 5.6: The harmonic-oscillator wavefunctions involve hermite polynomials. Chemistry LibreTexts. Retrieved 2022, from https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials <br />
<br><br />
Libretexts. (2022, September 12). 7.6: The Quantum Harmonic Oscillator. Physics LibreTexts. Retrieved 2022, from https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41655Quantum Harmonic Oscillator2022-12-07T16:04:12Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Deriving the Energy and Potential for Analytical, Solutions==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+1)(n+2)}a_n</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero in order to avoid divergence.:<br><br><br />
<math>2n+1-\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction. <br><br><br />
Deriving a closed-form solution for the wave function is more difficult, and requires the inclusion of Hermite Polynomials. The one dimensional solution is:<br><br>:<br />
<math>\Psi(x)=H_n(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\cdot\frac{1}{\sqrt{2^nn!}}\bigg[\bigg(\frac{m\omega}{\hbar}\bigg)^\frac{1}{2}x\bigg]e^\frac{-m\omega x^2}{2\hbar}</math><br><br> <br />
Where <math> H_n </math> are Hermite Polynomials defined as: <br><br><br />
<math>H_n(x)=(-1)^ne^{x^2}\cdot \frac{d^n }{d x^n}e^{-x^2}</math><br><br></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41654Quantum Harmonic Oscillator2022-12-06T17:12:00Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Potential for Analytical, Closed-Form Solutions to the Schrodinger Equation for the Quantum Harmonic Oscillator==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the solution in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating <math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1+\frac{2E}{\hbar\omega}}{(n+1)(n+2)}</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero:<br><br><br />
<math>2n+1+\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41653Quantum Harmonic Oscillator2022-12-06T16:53:19Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principal quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution, we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Potential for Analytical, Closed-Form Solutions to the Schrodinger Equation for the Quantum Harmonic Oscillator==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the solution in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating </math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1+\frac{2E}{\hbar\omega}}{(n+1)(n+2)}</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero:<br><br><br />
<math>2n+1+\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41652Quantum Harmonic Oscillator2022-12-06T16:51:14Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <br><br />
<math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br><br><br />
If <math>n=0</math>, <br><br><br />
then by substitution we see that <br><br><br />
<math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Potential for Analytical, Closed-Form Solutions to the Schrodinger Equation for the Quantum Harmonic Oscillator==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the solution in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br><br><br />
Rewriting the potential as:<br><br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br><br><br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br><br><br />
We will soon use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br />
Hence we use the guessing method to solve this second-order differential equation.<br><br><br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> <br><br><br />
To show how this works, here is an expansion of <math>u(t)</math>:<br><br><br />
<math>u(t)=a_0+a_1t+a_2t^2+a_3t^3...</math><br><br><br />
By inspection (evaluate if you'd like), differentiating </math>u(t)</math> twice yields zero for the first two terms.<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1+\frac{2E}{\hbar\omega}}{(n+1)(n+2)}</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero:<br><br><br />
<math>2n+1+\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41651Quantum Harmonic Oscillator2022-12-06T16:18:55Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br> If <math>n=0</math>, then by substitution we see that <math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Potential for Analytical, Closed-Form Solutions to the Schrodinger Equation for the Quantum Harmonic Oscillator==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the solution in one spacial dimension using spectral analysis:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br />
Rewriting the potential as:<br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br />
We will now use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br><br><br />
Hence we use the guessing method to solve this second-order differential equation.<br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> (This is because the first two terms are physically meaningless):<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1+\frac{2E}{\hbar\omega}}{(n+1)(n+2)}</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero:<br><br><br />
<math>2n+1+\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41650Quantum Harmonic Oscillator2022-12-06T15:52:41Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br> If <math>n=0</math>, then by substitution we see that <math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Potential for Analytical, Closed-Form Solutions to the Schrodinger Equation for the Quantum Harmonic Oscillator==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the solution in one spacial dimension:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br />
Rewriting the potential as:<br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br />
We will now use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br><br><br />
Hence we use the guessing method to solve this second-order differential equation.<br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> (This is because the first two terms are physically meaningless):<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1+\frac{2E}{\hbar\omega}}{(n+1)(n+2)}</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero:<br><br><br />
<math>2n+1+\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
<math>E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br><br><br />
Which is the energy formula as seen in the introduction.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41649Quantum Harmonic Oscillator2022-12-06T15:51:24Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br> If <math>n=0</math>, then by substitution we see that <math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Potential for Analytical, Closed-Form Solutions to the Schrodinger Equation for the Quantum Harmonic Oscillator==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the solution in one spacial dimension:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br />
Rewriting the potential as:<br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br />
We will now use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br><br><br />
Hence we use the guessing method to solve this second-order differential equation.<br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> (This is because the first two terms are physically meaningless):<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1+\frac{2E}{\hbar\omega}}{(n+1)(n+2)}</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero:<br><br><br />
<math>2n+1+\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br />
Which is the energy formula as seen in the introduction.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41648Quantum Harmonic Oscillator2022-12-06T15:50:16Z<p>Ecarder: /* Potential for Analytical, Closed-Form Solution to the Schrodinger Equation for the Quantum Harmonic Oscillator */</p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br> If <math>n=0</math>, then by substitution we see that <math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Potential for Analytical, Closed-Form Solutions to the Schrodinger Equation for the Quantum Harmonic Oscillator==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. The solution requires many unintuitive substitutions that would be difficult to spontaneously arrive at. We will make an attempt to derive the solution in one spacial dimension:<br><br><br />
Classically, we know the spring potential for the SHO: <math>U(x)=\frac{1}{2}kx^2</math><br><br />
We also know the angular frequency is: <math>\omega=\sqrt{\frac{k}{m}}</math><br><br><br />
In solving for <math>k</math>, we get <math>k=\omega^2m</math><br />
Rewriting the potential as:<br><br />
<math>U(x)=\frac{1}{2}mx^2\omega^2</math><br><br><br />
The one-dimensional, time-dependent Schrodinger Equation is:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi</math><br><br><br />
In substituting our derived potential:<br><br><br />
<math>\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi</math><br><br><br />
Dividing both sides by <math>\frac{-\hbar^2}{2m}</math> and setting the differential equation equal to zero yields:<br />
<math>\frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0</math><br />
We will now use a power series to simplify the equation. First, let's assign a dimensionless variable:<br><br><br />
<math>t=x\sqrt{\frac{m\omega}{\hbar}}</math><br><br />
Substituting into our main equation now gives:<br><br><br />
<math>\frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0</math><br><br><br />
The <math>\frac{2E}{\hbar\omega}</math> is negligible for large <math>t</math> values<br><br><br />
Hence we use the guessing method to solve this second-order differential equation.<br />
The wave function should include <math>e^{\frac{-t^2}{2}}</math> <br><br><br />
So let's guess a solution of the form:<br />
<math>\Psi(t)=u(t)e^{\frac{-t^2}{2}} </math><br><br><br />
Now of course we must differentiate and substitute into the differential equation.<br><br><br />
<math>\frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}</math><br><br><br />
<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}}+u(t)(t^2-1)e^{\frac{-t^2}{2}}.</math><br><br><br />
<br />
Time to substitute the derivatives:<br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}+u(t)(t^2-1)e^{\frac{-t^2}{2}}+(\frac{2E}{\hbar\omega}-t^2)u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Which simplifies quite nicely:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}e^{\frac{-t^2}{2}}-2t\frac{du(t)}{dt}-u(t)e^{\frac{-t^2}{2}}+\frac{2E}{\hbar\omega}u(t)e^{\frac{-t^2}{2}}=0</math><br><br><br />
Now we omit <math>e^{\frac{-t^2}{2}}</math> by dividing both sides by the expression:<br><br><br />
<math>\frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0</math><br><br><br />
<br />
We can now construct a power series to find a direct solution:<br><br><br />
<math>u(t)=\sum_{n=0}^\infty{a_nt^n}</math><br><br><br />
Now we repeat the above process of differentiation and substitution:<br><br><br />
<math>\frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}}</math><br><br><br />
<math>\frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}}</math><br><br><br />
<br />
Once again we substitute and unintuitively replace <math>n</math> with <math>n+2</math> (This is because the first two terms are physically meaningless):<br><br><br />
<math>\sum_{n=0}^\infty{((n+2)-1))a_{n+2}t^{(n+2)-2}+(\frac{2E}{\hbar\omega}-1-2n)\sum_{n=0}^\infty{a_nt^n}=0 </math><br><br><br />
In simplifying again:<br />
<math>\sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}=0</math><br><br><br />
Notice that the RHS of the equation above is equal to zero. This means that the coefficient of <math>t^{2n}</math> equals zero. Hence we can make the following equation and solve for <math>a_{n+2}</math><br><br><br />
<math>a_{n+2}=\frac{2n+1+\frac{2E}{\hbar\omega}}{(n+1)(n+2)</math><br><br><br />
Since <math>\frac{2}{n}</math> decreases faster than <math>e^{\frac{-t^2}{2}}</math>, we must equate the numerator of <math>a_{n+2}</math> to zero:<br><br><br />
<math>2n+1+\frac{2E}{\hbar\omega}=0</math><br><br><br />
Algebraically solving for <math>E</math> yields:<br><br><br />
E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega</math><br />
Which is the energy formula as seen in the introduction.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41647Quantum Harmonic Oscillator2022-12-06T14:04:47Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br />
[[File:qho2.png|300px|]]<br><br />
Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.<br />
<br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that the diagram doesn't have an energy level at zero and that <br> <math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br> If <math>n=0</math>, then by substitution we see that <math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math><br><br />
<br />
==Potential for Analytical, Closed-Form Solution to the Schrodinger Equation for the Quantum Harmonic Oscillator==<br />
The quantum harmonic oscillator is one of the few systems modeled by the Schrodinger Equation that has an analytical solution. We will make an attempt to derive the solution in one spacial dimension:<br></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=File:Qho2.png&diff=41646File:Qho2.png2022-12-06T13:59:37Z<p>Ecarder: </p>
<hr />
<div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=File:Qho.png&diff=41645File:Qho.png2022-12-06T13:57:14Z<p>Ecarder: Ecarder uploaded a new version of File:Qho.png</p>
<hr />
<div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=File:Qho1.jpg&diff=41644File:Qho1.jpg2022-12-06T13:56:05Z<p>Ecarder: </p>
<hr />
<div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41643Quantum Harmonic Oscillator2022-12-06T13:53:48Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHO has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br><br />
==Comparing the Classic and Quantum==<br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br><br><br />
[[File:Qhm2.png|300px|]]<br><br />
As <math>n</math> increases, there is more agreement between models. One thing that is interesting to note is that the classical probability is constrained within the dotted lines, as there is a physical limit to the displacement. There is no energy for the displacement of a classical harmonic oscillator to pass the forbidden zone. In the quantum realm, however, recall that there is some probability that the position exceeds these realms as there is some uncertainty given by the Uncertainty Principle. Another notable difference between the two is their differing ground-state energy levels. Classically the minimum energy is simply zero. For the QHO, recall from the introductory section that <br> <math>E_n=(n+\frac{1}{2})\hbar\omega</math>.<br> If <math>n=0</math>, then by substitution we see that <math>E_0=\frac{\hbar\omega}{2}</math><br><br><br />
This is intuitively logical as <math>E_0=0</math> is contradictory to Heisenberg's Uncertainty Principle. If the ground state energy were to be zero, the given particle would be motionless. If the particle is motionless, then <br><br><math>p=0</math> and <math> x=C </math>. Since this hypothetical particle has invariant momentum and position, then <math>\Delta x=0</math> and <math>\Delta p=0</math> which directly violates <math>\Delta p \Delta x \geq \frac{\hbar}{2}</math></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41642Fourier Series and Transform2022-12-06T13:31:54Z<p>Ecarder: /* Fourier Transform in Quantum Mechanics */</p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br><br><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br><br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41641Fourier Series and Transform2022-12-06T13:30:28Z<p>Ecarder: /* Finding Coefficients */</p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br><br><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br><br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41640Fourier Series and Transform2022-12-06T13:30:09Z<p>Ecarder: /* Finding Coefficients */</p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br><br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=File:Qhm2.png&diff=41639File:Qhm2.png2022-12-06T13:18:34Z<p>Ecarder: </p>
<hr />
<div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=File:Qhm111.jpg&diff=41638File:Qhm111.jpg2022-12-06T13:17:36Z<p>Ecarder: </p>
<hr />
<div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41637Quantum Harmonic Oscillator2022-12-06T13:14:37Z<p>Ecarder: </p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHM has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br><br />
Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number <math>n=3</math><br></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=File:Qho.png&diff=41636File:Qho.png2022-12-06T13:02:07Z<p>Ecarder: </p>
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<div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=File:IMG_3953.jpg&diff=41635File:IMG 3953.jpg2022-12-06T13:00:10Z<p>Ecarder: </p>
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<div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Quantum_Harmonic_Oscillator&diff=41634Quantum Harmonic Oscillator2022-12-06T12:58:28Z<p>Ecarder: Created page with "The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHM has numerous applications such as..."</p>
<hr />
<div>The Quantum Harmonic Oscillator is the quantum parallel to the classical simple harmonic oscillator. The key difference between these two is in the name. In the quantum harmonic oscillator, energy levels are ''quantized'' meaning there are discrete energy levels to this oscillator, (it cannot be any positive value as a classical oscillator can have). At low levels of energy, an oscillator obeys the rules of quantum mechanics. So the QHM has numerous applications such as molecular vibrations. These energy levels, denoted by <math>E_n, n=1,2,3... </math> and is evaluated by the relation: <br><math> E_n=(n+\frac{1}{2})\hbar\omega </math><br><br />
Where <math>n</math> is the principle quantum number, <math>\hbar</math> is the reduced planks constant, and <math>\omega</math> is the angular frequency of the oscillator.<br><br></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41633Fourier Series and Transform2022-12-06T12:26:32Z<p>Ecarder: /* Quantum Example */</p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br><br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Main_Page&diff=41632Main Page2022-12-06T12:17:32Z<p>Ecarder: /* Schrödinger Equation */</p>
<hr />
<div>__NOTOC__<br />
= '''Georgia Tech Student Wiki for Introductory Physics.''' =<br />
<br />
This resource was created so that students can contribute and curate content to help those with limited or no access to a textbook. When reading this website, please correct any errors you may come across. If you read something that isn't clear, please consider revising it for future students!<br />
<br />
Looking to make a contribution?<br />
#Pick one of the topics from intro physics listed below<br />
#Add content to that topic or improve the quality of what is already there.<br />
#Need to make a new topic? Edit this page and add it to the list under the appropriate category. Then copy and paste the default [[Template]] into your new page and start editing.<br />
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Please remember that this is not a textbook and you are not limited to expressing your ideas with only text and equations. Whenever possible embed: pictures, videos, diagrams, simulations, computational models (e.g. Glowscript), and whatever content you think makes learning physics easier for other students.<br />
<br />
== Source Material ==<br />
All of the content added to this resource must be in the public domain or similar free resource. If you are unsure about a source, contact the original author for permission. That said, there is a surprisingly large amount of introductory physics content scattered across the web. Here is an incomplete list of intro physics resources (please update as needed).<br />
* A physics resource written by experts for an expert audience [https://en.wikipedia.org/wiki/Portal:Physics Physics Portal]<br />
* A wiki written for students by a physics expert [http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes MSU Physics Wiki]<br />
* A wiki book on modern physics [https://en.wikibooks.org/wiki/Modern_Physics Modern Physics Wiki]<br />
* The MIT open courseware for intro physics [http://ocw.mit.edu/resources/res-8-002-a-wikitextbook-for-introductory-mechanics-fall-2009/index.htm MITOCW Wiki]<br />
* An online concept map of intro physics [http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html HyperPhysics]<br />
* Interactive physics simulations [https://phet.colorado.edu/en/simulations/category/physics PhET]<br />
* OpenStax intro physics textbooks: [https://openstax.org/details/books/university-physics-volume-1 Vol1], [https://openstax.org/details/books/university-physics-volume-2 Vol2], [https://openstax.org/details/books/university-physics-volume-3 Vol3]<br />
* The Open Source Physics project is a collection of online physics resources [http://www.opensourcephysics.org/ OSP]<br />
* A resource guide compiled by the [http://www.aapt.org/ AAPT] for educators [http://www.compadre.org/ ComPADRE]<br />
* The Feynman lectures on physics are free to read [http://www.feynmanlectures.caltech.edu/ Feynman]<br />
* Final Study Guide for Modern Physics II created by a lab TA [https://docs.google.com/document/d/1_6GktDPq5tiNFFYs_ZjgjxBAWVQYaXp_2Imha4_nSyc/edit?usp=sharing Modern Physics II Final Study Guide]<br />
<br />
== Resources ==<br />
* Commonly used wiki commands [https://en.wikipedia.org/wiki/Help:Cheatsheet Wiki Cheatsheet]<br />
* A guide to representing equations in math mode [https://en.wikipedia.org/wiki/Help:Displaying_a_formula Wiki Math Mode]<br />
* A page to keep track of all the physics [[Constants]]<br />
* A listing of [[Notable Scientist]] with links to their individual pages <br />
<br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 1==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====GlowScript 101====<br />
<div class="mw-collapsible-content"><br />
*[[Python Syntax]]<br />
*[[GlowScript]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====VPython====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[VPython]]<br />
*[[VPython basics]]<br />
*[[VPython Common Errors and Troubleshooting]]<br />
*[[VPython Functions]]<br />
*[[VPython Lists]]<br />
*[[VPython Loops]]<br />
*[[VPython Multithreading]]<br />
*[[VPython Animation]]<br />
*[[VPython Objects]]<br />
*[[VPython 3D Objects]]<br />
*[[VPython Reference]]<br />
*[[VPython MapReduceFilter]]<br />
*[[VPython GUIs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Vectors and Units====<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[SI Units]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Types of Interactions and How to Detect Them]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Velocity and Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's First Law of Motion]]<br />
*[[Mass]]<br />
*[[Velocity]]<br />
*[[Speed]]<br />
*[[Speed vs Velocity]]<br />
*[[Relative Velocity]]<br />
*[[Derivation of Average Velocity]]<br />
*[[2-Dimensional Motion]]<br />
*[[3-Dimensional Position and Motion]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Momentum and the Momentum Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Linear Momentum]]<br />
*[[Newton's Second Law: the Momentum Principle]]<br />
*[[Impulse and Momentum]]<br />
*[[Net Force]]<br />
*[[Inertia]]<br />
*[[Acceleration]]<br />
*[[Relativistic Momentum]]<br />
<!-- Kinematics and Projectile Motion relocated to Week 3 per advice of Dr. Greco --><br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
*[[Iterative Prediction]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Analytic Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
<!-- *[[Analytical Prediction]] Deprecated --><br />
*[[Kinematics]]<br />
*[[Projectile Motion]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Varying Force====<br />
<div class="mw-collapsible-content"><br />
*[[Fundamentals of Iterative Prediction with Varying Force]]<br />
*[[Spring_Force]]<br />
*[[Simple Harmonic Motion]]<br />
<!--*[[Hooke's Law]] folded into simple harmonic motion--><br />
<!--*[[Spring Force]] folded into simple harmonic motion--><br />
*[[Iterative Prediction of Spring-Mass System]]<br />
*[[Terminal Speed]]<br />
*[[Predicting Change in multiple dimensions]]<br />
*[[Two Dimensional Harmonic Motion]]<br />
*[[Determinism]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Fundamental Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Gravitational Force]]<br />
*[[Gravitational Force Near Earth]]<br />
*[[Gravitational Force in Space and Other Applications]]<br />
*[[3 or More Body Interactions]]<br />
<!--[[Fluid Mechanics]]--><br />
*[[Electric Force]]<br />
*[[Introduction to Magnetic Force]]<br />
*[[Strong and Weak Force]]<br />
*[[Reciprocity]]<br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Properties of Matter====<br />
<div class="mw-collapsible-content"><br />
*[[Kinds of Matter]]<br />
*[[Ball and Spring Model of Matter]]<br />
*[[Density]]<br />
*[[Length and Stiffness of an Interatomic Bond]]<br />
*[[Young's Modulus]]<br />
*[[Speed of Sound in Solids]]<br />
*[[Malleability]]<br />
*[[Ductility]]<br />
*[[Weight]]<br />
*[[Hardness]]<br />
*[[Boiling Point]]<br />
*[[Melting Point]]<br />
*[[Change of State]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Identifying Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Free Body Diagram]]<br />
*[[Inclined Plane]]<br />
*[[Compression or Normal Force]]<br />
*[[Tension]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Curving Motion====<br />
<div class="mw-collapsible-content"><br />
*[[Curving Motion]]<br />
*[[Centripetal Force and Curving Motion]]<br />
*[[Perpetual Freefall (Orbit)]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Energy Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Energy of a Single Particle]]<br />
*[[Kinetic Energy]]<br />
*[[Work/Energy]]<br />
*[[The Energy Principle]]<br />
*[[Conservation of Energy]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Work by Non-Constant Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Work Done By A Nonconstant Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
*[[Potential Energy of Macroscopic Springs]]<br />
*[[Spring Potential Energy]]<br />
*[[Ball and Spring Model]]<br />
*[[Gravitational Potential Energy]]<br />
*[[Energy Graphs]]<br />
*[[Escape Velocity]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Multiparticle Systems====<br />
<div class="mw-collapsible-content"><br />
*[[Center of Mass]]<br />
*[[Multi-particle analysis of Momentum]]<br />
*[[Potential Energy of a Multiparticle System]]<br />
*[[Work and Energy for an Extended System]]<br />
*[[Internal Energy]]<br />
**[[Potential Energy of a Pair of Neutral Atoms]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Choice of System====<br />
<div class="mw-collapsible-content"><br />
*[[System & Surroundings]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Thermal Energy, Dissipation, and Transfer of Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Thermal Energy]]<br />
*[[Specific Heat]]<br />
*[[Calorific Value(Heat of combustion)]]<br />
*[[First Law of Thermodynamics]]<br />
*[[Second Law of Thermodynamics and Entropy]]<br />
*[[Temperature]]<br />
*[[Transformation of Energy]]<br />
*[[The Maxwell-Boltzmann Distribution]]<br />
*[[Air Resistance]]<br />
*[[The Third Law of Thermodynamics]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Rotational and Vibrational Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Translational, Rotational and Vibrational Energy]]<br />
*[[Rolling Motion]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Different Models of a System====<br />
<div class="mw-collapsible-content"><br />
*[[Point Particle Systems]]<br />
*[[Real Systems]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Friction====<br />
<div class="mw-collapsible-content"><br />
*[[Friction]]<br />
*[[Static Friction]]<br />
*[[Kinetic Friction]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conservation of Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Collisions====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's Third Law of Motion]]<br />
*[[Collisions]]<br />
*[[Elastic Collisions]]<br />
*[[Inelastic Collisions]]<br />
*[[Maximally Inelastic Collision]]<br />
*[[Head-on Collision of Equal Masses]]<br />
*[[Head-on Collision of Unequal Masses]]<br />
*[[Scattering: Collisions in 2D and 3D]]<br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Coefficient of Restitution]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rotations====<br />
<div class="mw-collapsible-content"><br />
*[[Rotational Kinematics]]<br />
*[[Eulerian Angles]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Angular Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Total Angular Momentum]]<br />
*[[Translational Angular Momentum]]<br />
*[[Rotational Angular Momentum]]<br />
*[[The Angular Momentum Principle]]<br />
*[[Angular Impulse]]<br />
*[[Predicting the Position of a Rotating System]]<br />
*[[The Moments of Inertia]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Analyzing Motion with and without Torque====<br />
<div class="mw-collapsible-content"><br />
*[[Torque]]<br />
*[[Torque 2]]<br />
*[[Systems with Zero Torque]]<br />
*[[Systems with Nonzero Torque]]<br />
*[[Torque vs Work]]<br />
*[[Gyroscopes]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Introduction to Quantum Concepts====<br />
<div class="mw-collapsible-content"><br />
*[[Bohr Model]]<br />
*[[Energy graphs and the Bohr model]]<br />
*[[Quantized energy levels]]<br />
*[[Electron transitions]]<br />
*[[Entropy]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 2==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====3D Vectors====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[Right-Hand Rule]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Field and Electric Potential]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric force====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field of a point particle====<br />
<div class="mw-collapsible-content"><br />
*[[Point Charge]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Superposition====<br />
<div class="mw-collapsible-content"><br />
*[[Superposition Principle]]<br />
*[[Superposition principle]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Dipole]]<br />
*[[Magnetic Dipole]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Interactions of charged objects====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Potential]]<br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Tape experiments====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Polarization====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
*[[Polarization of an Atom]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conductors and Insulators====<br />
<div class="mw-collapsible-content"><br />
*[[Conductivity and Resistivity]]<br />
*[[Insulators]]<br />
*[[Potential Difference in an Insulator]]<br />
*[[Conductors]]<br />
*[[Polarization of a conductor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Charging and Discharging====<br />
<div class="mw-collapsible-content"><br />
*[[Charge Transfer]]<br />
*[[Electrostatic Discharge]]<br />
*[[Charged Conductor and Charged Insulator]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged rod====<br />
<div class="mw-collapsible-content"><br />
*[[Field of a Charged Rod|Charged Rod]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Field of a charged ring/disk/capacitor====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Ring]]<br />
*[[Charged Disk]]<br />
*[[Charged Capacitor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged sphere====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Spherical Shell]]<br />
*[[Field of a Charged Ball]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric potential====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]] <br />
*[[Potential Difference in a Uniform Field]]<br />
*[[Potential Difference of Point Charge in a Non-Uniform Field]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sign of a potential difference====<br />
<div class="mw-collapsible-content"><br />
*[[Sign of a Potential Difference]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential at a single location====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Potential Difference at One Location]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Path independence and round trip potential====<br />
<div class="mw-collapsible-content"><br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in an insulator====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Difference in an Insulator]]<br />
*[[Electric Field in an Insulator]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Moving charges in a magnetic field====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Biot-Savart Law====<br />
<div class="mw-collapsible-content"><br />
*[[Biot-Savart Law]]<br />
*[[Biot-Savart Law for Currents]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Moving charges, electron current, and conventional current====<br />
<div class="mw-collapsible-content"><br />
*[[Moving Point Charge]]<br />
*[[Current]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a wire====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Long Straight Wire]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Magnetic field of a current-carrying loop====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Loop]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a Charged Disk====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Disk]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Dipole Moment]]<br />
*[[Bar Magnet]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Atomic structure of magnets====<br />
<div class="mw-collapsible-content"><br />
*[[Atomic Structure of Magnets]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Steady state current====<br />
<div class="mw-collapsible-content"><br />
*[[Steady State]]<br />
*[[Non Steady State]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Kirchoff's Laws====<br />
<div class="mw-collapsible-content"><br />
*[[Kirchoff's Laws]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric fields and energy in circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential Difference]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Macroscopic analysis of circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Series Circuits]]<br />
*[[Parallel Circuits]]<br />
*[[Parallel Circuits vs. Series Circuits*]]<br />
*[[Loop Rule]]<br />
*[[Node Rule]]<br />
*[[Fundamentals of Resistance]]<br />
*[[Problem Solving]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in circuits with capacitors====<br />
<div class="mw-collapsible-content"><br />
*[[Charging and Discharging a Capacitor]]<br />
*[[RC Circuit]] <br />
*[[R Circuit]]<br />
*[[AC and DC]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic forces on charges and currents====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[Motors and Generators]]<br />
*[[Applying Magnetic Force to Currents]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
*[[Right-Hand Rule]]<br />
*[[Analysis of Railgun vs Coil gun technologies]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric and magnetic forces====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[VPython Modelling of Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Velocity selector====<br />
<div class="mw-collapsible-content"><br />
*[[Lorentz Force]]<br />
*[[Combining Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Hall Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Hall Effect]]<br />
<h1><strong>Alayna Baker Spring 2020</strong></h1><br />
[[File:Hall Effect 1.jpg]]<br />
[[File:Hall Effect 2.jpg]]<br />
<br />
*[[Right-Hand Rule]]<br />
*[[Motional Emf]]<br />
*[[Magnetic Force]]<br />
*[[Magnetic Torque]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
]]]====Motional EMF====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Motional Emf]]<br />
<h1><strong>Adeline Boswell Fall 2019</strong></h1><br />
[[File:Motional EMF Example.jpg]]<br />
<br />
*[[Motional Emf using Faraday's Law]]<br />
</div><br />
</div>http://www.physicsbook.gatech.edu/Special:RecentChangesLinked/Main_Page<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
If you have a bar attached to two rails, and the rails are connected by a resistor, you have effectively created a circuit. As the bar moves, it creates an "electromotive force"<br />
<br />
[[File:MotEMFCR.jpg]]<br />
<br />
====Magnetic force====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic torque====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Torque]]<br />
*[[Right-Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Gauss's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Flux Theorem]]<br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Ampere's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Ampere's Law]]<br />
*[[Ampere-Maxwell Law]]<br />
*[[Magnetic Field of Coaxial Cable Using Ampere's Law]]<br />
*[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]<br />
*[[Magnetic Field of a Toroid Using Ampere's Law]]<br />
*[[Magnetic Field of a Solenoid Using Ampere's Law]]<br />
*[[The Differential Form of Ampere's Law]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Semiconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Semiconductor Devices]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Faraday's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Faraday's Law]]<br />
*[[Motional Emf using Faraday's Law]]<br />
*[[Lenz's Law]]<br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Maxwell's equations====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
*[[Ampere's Law]]<br />
*[[Faraday's Law]]<br />
*[[Maxwell's Electromagnetic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Circuits revisited====<br />
<div class="mw-collapsible-content"><br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Inductors====<br />
<div class="mw-collapsible-content"><br />
*[[Inductors]]<br />
*[[Current in an LC Circuit]]<br />
*[[Current in an RL Circuit]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
==== Electromagnetic Radiation ====<br />
<div class="mw-collapsible-content"><br />
*[[Electromagnetic Radiation]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sparks in the air====<br />
<div class="mw-collapsible-content"><br />
*[[Sparks in Air]]<br />
*[[Spark Plugs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Superconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Superconducters]]<br />
*[[Superconductors]]<br />
*[[Meissner effect]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 3==<br />
<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Classical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Classical Physics]]<br />
</div><br />
</div><br />
<br />
[[Category:Which Category did you place this in?]]<br />
<br />
===Weeks 2 and 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Special Relativity and the Lorentz Transformation====<br />
<div class="mw-collapsible-content"><br />
*[[Frame of Reference]]<br />
<br />
*[[Einstein's Theory of Special Relativity]]<br />
*[[Time Dilation]]<br />
*[[Lorentz Transformations]]<br />
*[[Relativistic Doppler Effect]]<br />
*[[Einstein's Theory of General Relativity]]<br />
*[[Albert A. Micheleson & Edward W. Morley]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Photons and the Photoelectric Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Spontaneous Photon Emission]]<br />
*[[Light Scattering]]<br />
*[[Lasers]]<br />
*[[Electronic Energy Levels and Photons]]<br />
*[[Quantum Properties of Light]]<br />
*[[The Photoelectric Effect]]<br />
</div><br />
</div><br />
<br />
===Weeks 5 and 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Matter Waves and Wave-Particle Duality====<br />
<div class="mw-collapsible-content"><br />
*[[Wave-Particle Duality]]<br />
*[[Particle in a 1-Dimensional box]]<br />
*[[Heisenberg Uncertainty Principle]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Wave Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Standing Waves]]<br />
*[[Wavelength]]<br />
*[[Wavelength and Frequency]]<br />
*[[Mechanical Waves]]<br />
*[[Transverse and Longitudinal Waves]]<br />
*[[Fourier Series and Transform]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Schrödinger Equation====<br />
<div class="mw-collapsible-content"><br />
*[[The Born Rule]]<br />
*[[Solution for a Single Free Particle]]<br />
*[[Solution for a Single Particle in an Infinite Quantum Well - Darin]]<br />
*[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]<br />
*[[Quantum Harmonic Oscillator]]<br />
*[[Solution for Simple Harmonic Oscillator]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Quantum Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Tunneling through Potential Barriers]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Hydrogen Atom====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rutherford-Bohr Model====<br />
<div class="mw-collapsible-content"><br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Bohr Model]]<br />
*[[Quantized energy levels]]<br />
*[[Energy graphs and the Bohr model]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Many-Electron Atoms====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
*[[Pauli exclusion principle]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Nucleus====<br />
<div class="mw-collapsible-content"><br />
*[[Nucleus]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Molecules====<br />
<div class="mw-collapsible-content"><br />
</div><br />
*[[Molecules]]<br />
*[[Covalent Bonds]]<br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
*[[Application of Statistics in Physics]]<br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
*[[Temperature & Entropy]]<br />
</div><br />
<br />
===Additional Topics===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Condensed Matter Physics====<br />
<div class="mw-collapsible-content"><br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Nuclear Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Nuclear Fission]]<br />
*[[Nuclear Energy from Fission and Fusion]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Particle Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Elementary Particles and Particle Physics Theory]]<br />
*[[String Theory]]<br />
</div><br />
</div><br />
</div></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41631Fourier Series and Transform2022-12-06T12:12:48Z<p>Ecarder: /* References */</p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br><br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41630Fourier Series and Transform2022-12-06T12:12:21Z<p>Ecarder: /* References */</p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==References==<br />
Branson, J. (2013, April 22). Position Space and Momentum Space. Position space and Momentum Space. Retrieved 2022, from https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html <br><br />
Cheever, E. (2022). Introduction to the Fourier transform. Linear Physical Systems - Erik Cheever. Retrieved 2022, from https://lpsa.swarthmore.edu/Fourier/Xforms/FXformIntro.html <br> <br />
Dawkins, P. (2022). Fourier Series. Differential equations - fourier series. Retrieved 2022, from https://tutorial.math.lamar.edu/classes/de/fourierseries.aspx <br><br />
Greco, E. (2022). GPS Week 7. Modern Physics. Retrieved 2022.</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41629Fourier Series and Transform2022-12-06T11:57:05Z<p>Ecarder: </p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
==Quantum Example==<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==References==</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41628Fourier Series and Transform2022-12-06T11:55:57Z<p>Ecarder: </p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
=Quantum Example=<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math><br />
<br />
==References==</div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41627Fourier Series and Transform2022-12-06T11:52:37Z<p>Ecarder: </p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
=Quantum Example=<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br><br />
<br />
In taking a factor of <math>i</math> from the brackets, and recalling that <math>i^2=-1</math>, we obtain the following result:<br><br><br />
<br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\bigg[\frac{1}{k_0-k}\sin(\frac{L}{2}(k_0-k))+\frac{1}{k_0+k}\sin(\frac{L}{2}(k_0+k))\bigg]</math></div>Ecarderhttp://www.physicsbook.gatech.edu/index.php?title=Fourier_Series_and_Transform&diff=41626Fourier Series and Transform2022-12-06T11:46:35Z<p>Ecarder: </p>
<hr />
<div>==Fourier Series==<br />
A Fourier Series is an expansion of trigonometric functions to model periodic functions. This method proves useful in the study of harmonic systems as the analysis in a more familiar domain may be simpler than in its original domain. It has a variety of applications ranging from signal processing to quantum mechanics. The Fourier Series is defined as: <br><math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br />
==Intuition==<br />
Many physical systems can be modeled by square waves. Consider systems with on-off behavior, similar to an on-and-off switch. A sine wave and square wave looks like this respectively:<br><br />
[[File:sinewave.png|400px|]][[File:Arrow.png|200px|]][[File:squarewave.png|300px|]]<br><br />
As the diagram hints, we can use the Fourier Series to get from the sine wave to the square wave. Consider this progression of solely sine functions. We know what <math>f(x)=\sin(x)</math> looks like. If we keep adding a term in the partial sum for all odd integers of <math>f(x)=\frac{\sin(nx)}{n}</math>, the development of the square wave is noticeable as n increases:<br><br />
[[File:Squarewavee_(3).png|300px|]] <br>The darker function is <math>f(x)=\sin(x)</math> and the lighter function is the partial sum of the series to the fifth term. As you can see, the series function is beginning to look more like a square wave as <math>n \to \infty</math>. In using a summation of sines (and/or cosines), we can eventually reach the square wave.<br />
<br />
==Finding Coefficients==<br />
If we take the formula for a general Fourier Series, we can manipulate it to derive the formulas for <math> a_1, a_n, </math> and <math>b_n</math><br><br />
We have <math>f(x)=\sum_{n=1}^{\infty}{a_n\cos{(\frac{n\pi x}{L}})}+\sum_{n=1}^{\infty}{b_n\sin{(\frac{n\pi x}{L}})}</math><br> for <math>x\in[-L,L]</math><br><br />
<br>Multiply both sides by <math>\cos{(\frac{m\pi x}{L})}</math> and integrate for <math> x\in[-L,L] </math>:<br />
<br><br><math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\cos{(\frac{m\pi x}{L})}}}dx</math><br />
The integral in the first series is always zero if <math> n\neq m</math> and the integral in the second series is always zero due to the mutual orthogonality of sines and cosines.<br />
<br><br>Hence we have the reduced equation:<br />
<math>\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})}}dx=\left\{<br />
\begin{array}{lr}<br />
a_m \cdot (2L), & \text{if } m=n=0\\<br />
a_m \cdot (L), & \text{if } m=n\neq0<br />
\end{array}<br />
\right\}</math><br />
In noting the case that <math>m=n=0</math>, recall that <math>\cos(0)=1 </math> which yields the following when solving for <math>a_0</math>:<br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math> <br><br />
<br>And in the case where <math>m=n\neq0</math>, we cannot eliminate the cosine term. Solving for <math>A_m</math> yields:<br><br><br />
<math> a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br />
<br><br>For <math>a_n</math>, we can repeat the process except for his time we multiply both sides of the Fourier Series equation by <math>\sin{(\frac{n\pi x}{L})}</math>:<br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}}dx=\sum_{n=1}^{\infty}{a_n\int_{-L}^{L}{\cos{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L}})}}dx+\sum_{n=1}^{\infty}{b_n\int_{-L}^{L}{\sin{(\frac{n\pi x}{L}})\sin{(\frac{m\pi x}{L})}}}dx</math><br><br />
<br> For the same reason as before, the first series is always equal to zero and the other series is always equal to zero if <math>m\neq m</math><br><br />
Hence the equality simplifies to:<br><br />
<math>\int_{-L}^{L}{f(x)\sin{(\frac{m\pi x}{L})}dx}=L\cdot b_m</math><br><br><br />
Which yields: <math>b_m=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br />
<br />
==Example==<br />
Below is an example of finding the Fourier series of a given function:<br><br />
Find the Fourier Series for <math>f(x)=x</math>.<br />
This is simply a matter of using the derived equations to evaluate the coefficients:<br><br />
<math> a_0=\frac{1}{2L}\int_{-L}^{L}{f(x)dx} </math><br><br><br />
<math>a_0=\frac{1}{2L}\int_{-L}^{L}{(x)dx}=\frac{1}{2L}(Lx-\frac{x^2}{2})\bigg|^L_{-L}</math><br><br><br />
<math>a_0=0<br />
</math><br><br><br />
For <math>a_n</math>:<br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{f(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=\frac{1}{L}\int_{-L}^{L}{(x)\cos{(\frac{m\pi x}{L})dx}}</math><br><br><br />
<math>a_n=0 </math> (via the opposite bounds of integration over an odd function)<br><br><br />
For <math>b_n</math>:<br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{f(x)\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{1}{L}\int_{-L}^L{x\sin{(\frac{m\pi x}{L})}dx}</math><br><br><br />
<math>b_n=\frac{-Lx}{\pi n}\cos(\frac{n\pi x}{L})\bigg|^L_{-L}+\int_{-L}^{L}{\frac{L}{\pi n}\cos(\frac{n\pi x}{L})dx}</math> (via integration by parts)<br><br><br />
<math>b_n=\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)\bigg|^L_{-L}</math><br><br><br />
Solving for the boundaries gives:<br><br />
<math>\lim_{x\to-L^+}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}L{})\bigg)}=\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<br />
<math>\lim_{x\to L^-}{\bigg(-\frac{Lx}{n\pi}\cos(\frac{n\pi x}{L})+\frac{L^2}{\pi^2n^2}\sin(\frac{n\pi x}{L})\bigg)}=-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-\frac{(-1)^nL^2}{\pi n}-\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
<math>b_n=-2\frac{(-1)^nL^2}{\pi n}</math><br><br><br />
Now we have all the information we need to construct the Fourier Series:<br><br />
<math>F(x)=\frac{1}{2L}\cdot0 + \sum_{n=1}^{\infty}{\frac{1}{L}(-2\frac{(-1)^nL^2}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
<math>F(x)=\sum_{n=1}^{\infty}{(-2\frac{(-1)^nL}{\pi n})\sin(\frac{n\pi x}{L})}</math><br><br><br />
<br />
==Fourier Transform==<br />
The Fourier Transform is a mathematical method that converts a function of a given domain into the frequency domain. This is a particularly useful method, as it allows an analysis of systems through the frequency domain, even when we only have information for a different domain. If we consider how the Fourier Series could bring us from the trigonometric functions to the square wave, the Fourier Transform can directly convert a square wave to the frequency domain and vice versa (via inverse FT). Similar to the Fourier Series, this transformation has a variety of applications in numerous fields including electrical engineering and physics. For the time and frequency domains, the Fourier Transform is generally defined as:<br><br><br />
<math>X(\omega)=\int_{-\infty}^{\infty}{x(t)e^{-jwt}dt}</math><br><br><br />
Or inversely:<br><br><br />
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\omega)e^{jwt}dw}</math><br />
<br />
==Fourier Transform in Quantum Mechanics==<br />
The Fourier Transform is an essential tool in quantum mechanics, particularly when analyzing the wave function. For example, let's say for some arbitrary hypothetical reason,<br />
we have a wave function of position, but we need to express this wave function as a function of the wave constant. We can employ the Fourier Transform to deal with this. For these variables, the Fourier Transform and inverse Fourier Transform are respectfully defined as:<br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math> f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{A(k)e^{ikx}dk}</math><br><br><br />
<br />
<br />
=Quantum Example=<br />
Below is a slightly adjusted problem written by Dr. Ed Greco of Georgia Tech. The solution process will follow:<br><br />
Determine the Fourier transform <math>A(k)</math> of the function <math> f(x)=\cos(k_0x)</math> for <math>x\in\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math> and <math> f(x)=0 </math> for <math> x\notin\bigg[\frac{-L}{2},\frac{L}{2}\bigg]</math><br><br><br />
Use the equation that is a function of <math>k</math><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{1}{\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{\cos(k_0x)e^{-ikx}dx}</math><br><br><br />
Via Euler's formula, recall that <math>\cos(q)=\frac{e^{qi}+e^{-qi}}{2}</math>. Using this in our problem allows us to rewrite our equation:<br />
<br />
<math>A(k)=\frac{1}{2\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}{(e^{ik_0x}+e^{-ik_0x})e^{-ikx}dx}</math><br><br><br />
<math>A(k)=\frac{-i}{2\sqrt{2\pi}}\bigg[\frac{e^{ix(k-k_0)}}{i(k-k_0)}-\frac{e^{ix(k_0+k)}}{i(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
Simplifying yields: <br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{ix(k-k_0)}-(\frac{k_0-k}{k_0+k})e^{-ix(k_0+k)}\bigg]\bigg|^{\frac{L}{2}}_{\frac{-L}{2}}</math><br><br><br />
<br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[e^{i\frac{L}{2}(k-k_0)}-e^{-i\frac{L}{2}(k-k_0)}-\frac{k-k_0}{k+k_0}\big(e^{-i\frac{L}{2}(k_0+k)}-e^{i\frac{L}{2}(k_0+k)}\big)\bigg]</math><br><br><br />
Once again using Euler:<br><br><br />
<math>A(k)=\frac{-i}{(k_0-k)2\sqrt{2\pi}}\bigg[2i\sin(\frac{L}{2}(k-k_0))+\big(\frac{k_0-k}{k_0+k}\big)2i\sin(\frac{L}{2}(k+k_0))\bigg]</math><br><br></div>Ecarder