Quantum Harmonic Oscillator

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]\displaystyle{ E_n, n=1,2,3... }[/math] and is evaluated by the relation:
[math]\displaystyle{ E_n=(n+\frac{1}{2})\hbar\omega }[/math]
Where [math]\displaystyle{ n }[/math] is the principle quantum number, [math]\displaystyle{ \hbar }[/math] is the reduced planks constant, and [math]\displaystyle{ \omega }[/math] is the angular frequency of the oscillator.

Displayed above is a diagram displaying the quantized energy levels for the quantum harmonic oscillator.

Comparing the Classic and Quantum

Below is a comparison of the positional probabilities of the classical and quantum harmonic oscillators for the principal quantum number [math]\displaystyle{ n=3 }[/math]


As [math]\displaystyle{ 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
[math]\displaystyle{ E_n=(n+\frac{1}{2})\hbar\omega }[/math].
If [math]\displaystyle{ n=0 }[/math], then by substitution we see that [math]\displaystyle{ E_0=\frac{\hbar\omega}{2} }[/math]

This is intuitively logical as [math]\displaystyle{ 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

[math]\displaystyle{ p=0 }[/math] and [math]\displaystyle{ x=C }[/math]. Since this hypothetical particle has invariant momentum and position, then [math]\displaystyle{ \Delta x=0 }[/math] and [math]\displaystyle{ \Delta p=0 }[/math] which directly violates [math]\displaystyle{ \Delta p \Delta x \geq \frac{\hbar}{2} }[/math]

Potential for Analytical, Closed-Form Solutions to the Schrodinger Equation for the Quantum Harmonic Oscillator

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:

Classically, we know the spring potential for the SHO: [math]\displaystyle{ U(x)=\frac{1}{2}kx^2 }[/math]
We also know the angular frequency is: [math]\displaystyle{ \omega=\sqrt{\frac{k}{m}} }[/math]

In solving for [math]\displaystyle{ k }[/math], we get [math]\displaystyle{ k=\omega^2m }[/math] Rewriting the potential as:
[math]\displaystyle{ U(x)=\frac{1}{2}mx^2\omega^2 }[/math]

The one-dimensional, time-dependent Schrodinger Equation is:

[math]\displaystyle{ \frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+U(x)\Psi=E\Psi }[/math]

In substituting our derived potential:

[math]\displaystyle{ \frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}mx^2\omega^2\Psi=E\Psi }[/math]

Dividing both sides by [math]\displaystyle{ \frac{-\hbar^2}{2m} }[/math] and setting the differential equation equal to zero yields: [math]\displaystyle{ \frac{d^2\Psi}{dx^2}+(\frac{2mE}{\hbar^2}-\frac{m^2\omega^2 x^2}{\hbar^2})\Psi=0 }[/math] We will now use a power series to simplify the equation. First, let's assign a dimensionless variable:

[math]\displaystyle{ t=x\sqrt{\frac{m\omega}{\hbar}} }[/math]
Substituting into our main equation now gives:

[math]\displaystyle{ \frac{d^2\Psi}{dt^2}+\Psi(t)(\frac{2E}{\hbar\omega}-t^2)=0 }[/math]

The [math]\displaystyle{ \frac{2E}{\hbar\omega} }[/math] is negligible for large [math]\displaystyle{ t }[/math] values

Hence we use the guessing method to solve this second-order differential equation. The wave function should include [math]\displaystyle{ e^{\frac{-t^2}{2}} }[/math]

So let's guess a solution of the form: [math]\displaystyle{ \Psi(t)=u(t)e^{\frac{-t^2}{2}} }[/math]

Now of course we must differentiate and substitute into the differential equation.

[math]\displaystyle{ \frac{d\Psi(t)}{dt}=\frac{du(t)}{dt}e^{\frac{-t^2}{2}}-tu(t)e^{\frac{-t^2}{2}} }[/math]

[math]\displaystyle{ \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]

Time to substitute the derivatives: [math]\displaystyle{ \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]

Which simplifies quite nicely:

[math]\displaystyle{ \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]

Now we omit [math]\displaystyle{ e^{\frac{-t^2}{2}} }[/math] by dividing both sides by the expression:

[math]\displaystyle{ \frac{d^2\Psi(t)}{dt^2}-2t\frac{du(t)}{dt}-u(t)+(\frac{2E}{\hbar\omega}-1)u(t)=0 }[/math]

We can now construct a power series to find a direct solution:

[math]\displaystyle{ u(t)=\sum_{n=0}^\infty{a_nt^n} }[/math]

Now we repeat the above process of differentiation and substitution:

[math]\displaystyle{ \frac{du(t)}{dt}=\sum_{n=0}^\infty{na_nt^{n-1}} }[/math]

[math]\displaystyle{ \frac{d^2u(t)}{dt^2}=\sum_{n=0}^\infty{(n-1)na_nt^{n-2}} }[/math]

Once again we substitute and unintuitively replace [math]\displaystyle{ n }[/math] with [math]\displaystyle{ n+2 }[/math] (This is because the first two terms are physically meaningless):

[math]\displaystyle{ \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]

In simplifying again: [math]\displaystyle{ \sum_{n=0}^\infty{(n+2)(n+1)a_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)a_n)t^{2n}}=0 }[/math]

Notice that the RHS of the equation above is equal to zero. This means that the coefficient of [math]\displaystyle{ t^{2n} }[/math] equals zero. Hence we can make the following equation and solve for [math]\displaystyle{ a_{n+2} }[/math]

[math]\displaystyle{ a_{n+2}=\frac{2n+1+\frac{2E}{\hbar\omega}}{(n+1)(n+2)} }[/math]

Since [math]\displaystyle{ \frac{2}{n} }[/math] decreases faster than [math]\displaystyle{ e^{\frac{-t^2}{2}} }[/math], we must equate the numerator of [math]\displaystyle{ a_{n+2} }[/math] to zero:

[math]\displaystyle{ 2n+1+\frac{2E}{\hbar\omega}=0 }[/math]

Algebraically solving for [math]\displaystyle{ E }[/math] yields:

[math]\displaystyle{ E=\frac{\hbar\omega}{2}(2n+1)=(n+\frac{1}{2})\hbar\omega }[/math]

Which is the energy formula as seen in the introduction.