Solution for a Single Free Particle

Claimed by Carlos M. Silva (Spring 2022)

The Schrödinger Equation is a linear partial differential equation that governs the wave function of a quantum mechanical system. Similar to Newton's Laws, the Schrödinger Equation is an equation of motion, meaning that it's capable of describing the time-evolution of a position analog of a system.

The free particle is the name given to the system consisting of a single particle subject to a null or constant potential everywhere in space. It's the simplest system to which the Schrödinger Equation has a solution with physical meaning.

Although the free-particle solution does not have ample practical use in the field of Physics, the methods and conclusions that come from the solution of this system are of great use in a plethora of other quantum systems.

The General Schrödinger Equation

The general formulation of the Schrödinger Equation for a time-dependent system of non-relativistic particles in Bra-Ket notation is:

[math]\displaystyle{ i \hbar \frac{d}{d t}\vert\Psi(t)\rangle = \hat H\vert\Psi(t)\rangle }[/math]

Here [math]\displaystyle{ i = \sqrt{-1} }[/math] is the imaginary unit. [math]\displaystyle{ \hbar }[/math] is the reduced Planck's constant. [math]\displaystyle{ \vert\Psi(t)\rangle }[/math] is the state vector of the quantum system at time [math]\displaystyle{ t }[/math], and [math]\displaystyle{ \hat H }[/math] is the Hamiltonian operator. For a single particle, the general Schrödinger Equation reduces to a single linear partial differential equation:

[math]\displaystyle{ i\hbar\Psi(\vec{r},t) = - \frac{\hbar^2}{2m} \nabla ^2 \Psi(\vec{r},t) + V(\vec{r},t)\Psi(\vec{r},t) }[/math]

[math]\displaystyle{ \Psi(\vec{r},t) }[/math] becomes the wave function of the particle at position [math]\displaystyle{ \vec{r} }[/math] and time [math]\displaystyle{ t }[/math]. [math]\displaystyle{ V(\vec{r},t) }[/math] is the scalar potential energy of the particle at position [math]\displaystyle{ \vec{r} }[/math] and time [math]\displaystyle{ t }[/math].

Time-independent Potential and Separation of Variables

The potential energy of a system is often not an explicit function of time, that is [math]\displaystyle{ \frac{\partial V}{\partial t} = 0 }[/math]. Implying that, for such systems, the Schrödinger Equation of a single particle may be written as:

[math]\displaystyle{ i\hbar\frac{d}{d t}\Psi(\vec{r},t) = - \frac{\hbar^2}{2m} \nabla ^2 \Psi(\vec{r},t) + V(\vec{r})\Psi(\vec{r},t) }[/math]

This, along with the spectral theorem allows us to assume that the solution for this equation may be obtained through the separation of variables. That is, expressing the wave function as a product of a time-independent and a position-independent function:

[math]\displaystyle{ \Psi(\vec{r},t) \equiv \psi(\vec{r})\phi(t) \rightarrow i\hbar\frac{d}{d t}\psi(\vec{r})\phi(t) = - \frac{\hbar^2}{2m} \nabla ^2 \psi(\vec{r})\phi(t) + V(\vec{r})\psi(\vec{r})\phi(t) }[/math]

Expanding the derivatives through the chain rule yields:

[math]\displaystyle{ i\hbar\psi(\vec{r})\frac{d}{d t}\phi(t) = - \frac{\hbar^2}{2m} \phi(t)\nabla ^2 \psi(\vec{r}) + V(\vec{r})\psi(\vec{r})\phi(t) }[/math]

Dividing both sides of the equation by [math]\displaystyle{ \psi(\vec{r})\phi(t) }[/math]:

[math]\displaystyle{ \frac{i\hbar\frac{d}{d t}\phi(t)}{\phi(t)} = -\frac{\hbar^2}{2m} \frac{\nabla ^2 \psi(\vec{r})}{\psi(\vec{r})} + V(\vec{r}) }[/math]

The left side of this equation is position-independent, and the right side of the equation is time-independent. Therefore, we have successfully separated variables, allowing us to solve for [math]\displaystyle{ \phi(t) }[/math] and [math]\displaystyle{ \psi(\vec{r}) }[/math] separatedely. Doing so yields:

Time-Independent Schrödinger Equation for 1 Particle

[math]\displaystyle{ - \frac {\hbar ^2}{2m} \nabla^2\psi(\vec{r}) + V(\vec{r})\psi(\vec{r})= E \psi(\vec{r}) }[/math]

Space-Independent Schrödinger Equation for 1 Particle

[math]\displaystyle{ \frac{d}{d t}\phi(t)= -\frac{i}{\hbar}\left(E + V \right)\phi(t) }[/math]

A full derivation of this equations can be found below:

Derivation of the Time-Independent and Space-Independent Schrödinger Equation

Start from the Separated Equation for 1 Particle:

[math]\displaystyle{ \frac{i\hbar\frac{d}{d t}\phi(t)}{\phi(t)} = -\frac{\hbar^2}{2m} \frac{\nabla ^2 \psi(\vec{r})}{\psi(\vec{r})} + V(\vec{r}) }[/math]

Take the derivative of both sides in respect to time:

[math]\displaystyle{ i\hbar\left( \frac{\ddot{\phi}}{\phi} -\frac{\dot{\phi}^2}{\phi^2} \right) = 0 }[/math]

From now on, for the sake of simplicity, Newton's Notation will be used to express derivatives. Simplifying the equation above yields:

[math]\displaystyle{ \ddot{\phi}\phi = \dot{\phi}^2 }[/math]

Treat [math]\displaystyle{ \phi }[/math] as an independent variable and define [math]\displaystyle{ v(\phi) = \dot{\phi} }[/math]

[math]\displaystyle{ \frac{d}{dt}\left(\dot{\phi}\right)\phi = \left( \dot{\phi}\right)^2 \rightarrow \frac{d}{dt}\left(v(\phi)\right)\phi = v^2(\phi) }[/math]

By the chain rule:

[math]\displaystyle{ \frac{d\phi}{dt}\frac{dv}{d\phi}\phi = v^2(\phi) \rightarrow \frac{dv}{d\phi}\phi = v(\phi) }[/math]

Re-writting in differential form:

[math]\displaystyle{ \frac{dv}{v}=\frac{d\phi}{\phi} }[/math]

Integrate both sides yields:

[math]\displaystyle{ \int\frac{dv}{v}=\int\frac{d\phi}{\phi} \rightarrow \ln{v} = \ln{\phi} + C_1 }[/math]

Solve for [math]\displaystyle{ v(\phi) }[/math] and simplify arbitrary constants:

[math]\displaystyle{ v(\phi)=C_1\phi }[/math]

Substitute in the definition of [math]\displaystyle{ v(\phi) }[/math]:

[math]\displaystyle{ \frac{d\phi}{dt} = C_1 \phi }[/math]

Rewrite in differential form and integrate:

[math]\displaystyle{ \int \frac{d\phi}{\phi} = \int C_1 dt \rightarrow \ln{\phi} = C_1 t + C_2 }[/math]

Solving for [math]\displaystyle{ \phi }[/math] and simplifying arbitrary constants yields:

[math]\displaystyle{ \phi(t) = C_2 e^{C_1 t} }[/math]

Substituting this relation back in the Schrödinger Equation yields:

[math]\displaystyle{ C_1 = -i\frac{E+V}{\hbar} }[/math]

Therefore:

[math]\displaystyle{ \frac{\dot{\phi}}{\phi} = -i\frac{E+V}{\hbar} }[/math]

Substituting this relation back in the the Separated Equation for 1 Particle:

[math]\displaystyle{ \frac{i\hbar\frac{d}{d t}\phi(t)}{\phi(t)} = -\frac{\hbar^2}{2m} \frac{\nabla ^2 \psi(\vec{r})}{\psi(\vec{r})} + V(\vec{r}) \rightarrow - \frac {\hbar ^2}{2m} \nabla^2\psi(\vec{r}) + V(\vec{r})\psi(\vec{r})= E \psi(\vec{r}) }[/math]

Which is the Time-Independent Schrödinger Equation.

The Free Particle

The free particle is a special case of the Schrödinger Equation where the potential is null (or constant) everywhere in space:

[math]\displaystyle{ V(\vec{r},t) = 0 }[/math]

This potential is time and space independent, therefore we may use the equations found in the section above:

[math]\displaystyle{ \Psi(\vec{r},t) = A\psi(\vec{r})e^{-i\frac{E}{\hbar}t} }[/math]

[math]\displaystyle{ - \frac {\hbar ^2}{2m} \nabla^2\psi(\vec{r}) = E \psi(\vec{r}) }[/math]

The latter is a form of the Helmholtz Equation. There exists a general solution for an unbounded geometry in Spherical Coordinates:

[math]\displaystyle{ \psi (r, \theta, \varphi)= \sum_{\ell=0}^\infty \sum_{n=-\ell}^\ell \left( a_{\ell n} j_\ell \left( \frac{\sqrt{2mE}}{\hbar} r \right) + b_{\ell n} y_\ell \left(\frac{\sqrt{2mE}}{\hbar}r\right) \right) Y^n_\ell(\theta,\varphi) }[/math]

Here [math]\displaystyle{ n }[/math] and [math]\displaystyle{ \ell }[/math] are quantum numbers. [math]\displaystyle{ a_{\ell n} }[/math] and [math]\displaystyle{ b_{\ell n} }[/math] are amplitudes that define the wave packet. [math]\displaystyle{ j_\ell }[/math] and [math]\displaystyle{ y_\ell }[/math] are the spherical Bessel functions, and [math]\displaystyle{ Y^n_\ell(\theta,\varphi) }[/math] are the spherical harmonics.

General Solution in 1 Dimension

In one dimension our equation takes the form:

[math]\displaystyle{ - \frac {\hbar^2}{2m} \frac{\partial^2\psi(x)}{\partial x^2} = E \psi(x) }[/math]

Define [math]\displaystyle{ k^2 \equiv \frac{2mE}{\hbar^2} }[/math] and rewrite:

[math]\displaystyle{ \frac{\partial^2\psi(x)}{\partial x^2} = -k^2 \psi(x) }[/math]

Assume [math]\displaystyle{ \psi }[/math] as an independent variable, then define [math]\displaystyle{ v \equiv \frac{\partial\psi(x)}{\partial x} }[/math]. Then:

[math]\displaystyle{ \frac{\partial v}{\partial x} = -k^2 \psi \rightarrow \frac{\partial v}{\partial \psi} \frac{\partial \psi}{\partial x}= -k^2 \psi \rightarrow \frac{\partial v}{\partial \psi}v = - k^2\psi }[/math]

Rewrite in differential form and integrate:

[math]\displaystyle{ \int v \partial v = -k^2 \int \psi \partial \psi \rightarrow \frac{v^2}{2}= -k^2\left( \frac{\psi^2}{2} + C_1 \right) }[/math]

Solving for [math]\displaystyle{ v }[/math] and simplifying arbitrary constants:

[math]\displaystyle{ v = \pm i k \sqrt{\psi^2 + C_1} }[/math]

Substituting the definition of [math]\displaystyle{ v }[/math]:

[math]\displaystyle{ \frac{\partial\psi(x)}{\partial x} = \pm i k \sqrt{\psi^2 + C_1} }[/math]

Re-writing in differential form and integrating:

[math]\displaystyle{ \int\frac{\partial\psi}{\sqrt{\psi^2 + C_1}} = \pm i k \int \partial x \rightarrow \ln{\left(\psi + \sqrt{\psi^2+C_1}\right)}= \pm i k x + C_2 }[/math]

Solving for [math]\displaystyle{ \psi }[/math] and simplifying arbitrary constants:

[math]\displaystyle{ \psi(x)= C_1 e^{ikx} + C_2 e^{-ikx} }[/math]

Notice, however, that since [math]\displaystyle{ k }[/math] is a function of the energy, this means a more general solution may be obtained by adding the contribution of many different energy levels:

[math]\displaystyle{ \psi(x)= \sum_n A_n e^{i k_n x} + B_n e^{-ik_n x} }[/math]

Furthermore it is possible to show that, if [math]\displaystyle{ \Psi(x,0) }[/math] is known, then:

[math]\displaystyle{ \Psi(x,t) = \sqrt{\frac{m}{2\pi\hbar i}\frac{1}{t}}\int_{-\infty}^{\infty}\psi\left(y\right)e^{-\frac{m}{2\hbar i}\frac{\left(x-y\right)^2}{t}}dy }[/math]