# Charged Particles in an Electromagnetic Field

A problem with some relation to the harmonic oscillator is that of the motion of a charged particle in a constant and uniform magnetic field. In classical mechanics, we know that the Hamiltonian for this system is

$H=\frac{1}{2m}\left (\mathbf{p}-\frac{e}{c}\mathbf{A}\right )^2,$

where $e\!$ is the charge of the particle and $\mathbf{A}$ is the vector potential. In fact, to obtain the Hamiltonian for any system in the presence of a magnetic field, we simply make the replacement, $\mathbf{p}\rightarrow\mathbf{p}-\frac{e}{c}\mathbf{A}.$ In quantum mechanics, we introduce the magnetic field in the same way; this process is referred to as minimal coupling.

## Gauge Invariance in Quantum Mechanics

We know from Maxwell's equations that the classical physics of a charged particle in an electromagnetic field is invariant under a gauge transformation, $\Phi\rightarrow\Phi-\frac{1}{c}\frac{\partial\chi}{\partial t}$ and $\mathbf{A}\rightarrow\mathbf{A}+\nabla\chi,$ where $\Phi\!$ is the scalar potential and $\chi(\mathbf{r},t)\!$ is a single-valued real function. We will now show how this is expressed in quantum mechanics.

In the position basis, the Schrödinger equation for a charged particle in an electromagnetic field is

$-i\hbar\frac{\partial\Psi}{\partial t}-e\Phi\Psi=-\frac{\hbar^2}{2m}\left (\nabla+\frac{ie}{\hbar c}\mathbf{A}\right )^2\Psi.$

If we now perform the above gauge transformation on the electromagnetic field, then this equation becomes

$-i\hbar\frac{\partial\Psi}{\partial t}-e\Phi\Psi+\frac{e}{c}\frac{\partial\chi}{\partial t}\Psi-\frac{\hbar^2}{2m}\left (\nabla+\frac{ie}{\hbar c}\mathbf{A}+\frac{ie}{\hbar c}\nabla\chi\right )^2\Psi.$

If we make the substitution, $\Psi\rightarrow e^{-ie\chi/\hbar c}\Psi,$ then we recover the original equation. Therefore, a gauge transformation of the magnetic field effectively introduces a phase factor to the wave function. This does result in a change in the canonical momentum, but it will have no effect on, for example, the probability density for finding the particle at a given position or, as we will see later, on the expectation value of the position or velocity of the particle.

We see that, in quantum mechanics, gauge invariance is expressed as follows. If one multiplies the wave function by a single-valued phase factor, then it may be "canceled out" by a corresponding change in the electromagnetic potentials that the particle is subject to.

For a constant and uniform magnetic field $\mathbf{B}=B\hat{\mathbf{z}},$ we typically work with one of two gauges. One of these is the Laudau gauge,

$\mathbf{A}(\mathbf{r}) = -yB\hat{\mathbf{x}}$ or $xB\hat{\mathbf{y}}.$

The other is the symmetric gauge,

$\mathbf{A}(\mathbf{r})=-\tfrac{1}{2}yB\hat{\mathbf{x}}+\tfrac{1}{2}xB\hat{\mathbf{y}}.$

## Eigenstates of a Charged Particle in a Static and Uniform Magnetic Field

Let us now find the eigenstates of a charged particle in a static and uniform magnetic field. We will be working in the Landau gauge,

$\mathbf{A}=xB\hat{\mathbf{y}}.$

The Schrödinger equation for this system is

$-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}-\frac{\hbar^2}{2m}\left (\frac{\partial}{\partial y}+\frac{ie}{\hbar c}Bx\right )^2\psi-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial z^2}\psi=E\psi.$

In this gauge, the Hamiltonian is translationally invariant along the $y\!$ and $z\!$ axes. Therefore, our wave function will have the form,

$\psi(x,y,z)=e^{i(k_yy+k_zz)}f(x).\!$

Substituting this form into the equation, we obtain

$-\frac{\hbar^2}{2m}\frac{d^2f}{dx^2}+\frac{\hbar^2}{2m}\left (k_y+\frac{e}{\hbar c}Bx\right )^2f+\frac{\hbar^2k_z^2}{2m}f=Ef,$

or

$-\frac{\hbar^2}{2m}\frac{d^2f}{dx^2}+\frac{e^2B^2}{2mc^2}\left (x+\frac{\hbar c}{eB}k_y\right )^2f=\left (E-\frac{\hbar^2k_z^2}{2m}\right )f.$

If we now introduce the shifted position coordinate $x'=x+\frac{\hbar c}{eB}k_y$ and the shifted energy $E'=E-\frac{\hbar^2k_z^2}{2m},$ this becomes

$-\frac{\hbar^2}{2m}\frac{d^2f}{dx'^2}+\frac{e^2B^2}{2mc^2}x'^2f=E'f.$

This is just the equation for a harmonic oscillator with frequency

$\omega=\frac{eB}{mc}.$

We recognize this as the cyclotron frequency of the particle. We may immediately write down the full eigenfunctions and energy levels of the system. The wave functions are

$\psi(x,y,z)=\frac{1}{\sqrt{L_yL_z}}e^{i(k_yy+k_zz)}\frac{1}{\sqrt{2^nn!l_B\sqrt{\pi}}}\exp\left [-\tfrac{1}{2}\left (\frac{x}{l_B}+l_Bk_y\right )^2\right ]H_n\left (\frac{x}{l_B}+l_Bk_y\right ),$

where $L_y\!$ and $L_z\!$ are the dimensions of the system in the $y\!$ and $z\!$ directions and $l_B=\sqrt{\frac{\hbar c}{eB}}$ is known as the magnetic length. The energies are given by

$E=\left (n+\tfrac{1}{2}\right )\hbar\omega+\frac{\hbar^2k_z^2}{2m}.$

For a fixed value of $k_z,\!$ the energy spectrum that we just obtained is referred to as a Landau level spectrum. Note that the above energies do not depend on $k_y;\!$ it only appears in the wave function, where it determines the "guiding center" $l_B^2k_y\!$ of the wave function. This means that they are very highly degenerate. We may approximate the degeneracy of each of these Landau levels as follows. If the system has a finite size, then $k_y\!$ is quantized to integer multiples of $\frac{2\pi}{L_y}$ if we assume that the wave function satisfies periodic boundary conditions; i.e. $k_y=\frac{2\pi n}{L_y}.$ We now determine the range of values of $n\!$ for which the guiding center is within the range $0 where $L_x\!$ is the dimension of the system in the $x\!$ direction. This value is

$n=\frac{L_xL_y}{2\pi l_B^2},$

or

$n=\frac{BL_xL_y}{\Phi_0},$

where $\Phi_0=\frac{hc}{e}$ is known as the "flux quantum" of the particle. This quantity appears frequently in many contexts, such as in the theory of superconductivity. We may therefore think of the degeneracy of the system as just the number of flux quanta contained within a face in the $xy\!$ plane of the box that the particle is contained inside of.

## Problem

Consider the problem of a particle of charge $e\!$ in a uniform magnetic field along the $z\!$ direction again, but now in the symmetric gauge, $\mathbf{A}=-\tfrac{1}{2}By\hat{\mathbf{x}}+\tfrac{1}{2}Bx\hat{\mathbf{y}}.$ Let $\hat{\Pi}_i=\hat{p}_i-\frac{e}{c}A_i.$

(a) Evaluate $\left [{\hat{\Pi}_{x},\hat{\Pi}_{y}} \right ].$

(b) Using the commutation relation obtained in the previous part, obtain the energy eigenvalues.