Advanced Solid State Physics



Magnetic effects and
Fermi surfaces


Linear response


Crystal Physics



Structural phase

Landau theory
of second order
phase transitions




Exam questions




Course notes

TUG students


Solutions to the Schrödinger equation for a charged particle in a magnetic field

The Schrödinger equation for a charged particle in a magnetic field is,

\begin{equation} \label{eq:schroedinger2} i\hbar\frac{\partial \psi}{\partial t} = \frac{1}{2m}\left[\left(-i\hbar\nabla - q\vec{A}(\vec{r},t)\right)^{2} + qV(\vec{r},t)\right]\psi. \end{equation}

Consider the case of free particles $(V(\vec{r},t)=0)$. The time independent Schrödinger equation is then,

\begin{equation} \label{eq:schroedinger3} \frac{1}{2m}\left[\left(-i\hbar\nabla - q\vec{A}(\vec{r},t)\right)^{2}\right]\psi = E\psi. \end{equation}

To solve this equation, it is convenient to use Landau gauge for the vector potential $\vec{A}$,

\begin{equation} \label{eq:landau1} \vec{A} = B_{z}x\hat{y}. \end{equation}

This choice of gauge describes a uniform, magnetic field in the $z$-direction, $\vec{B} = B_{z}\hat{z}$, which can be verified by taking the curl $\vec{B}=\nabla\times\vec{A}$.

Using this gauge, the factor $\left(-i\hbar\nabla - q\vec{A}\right)^{2}$ can be written,

\begin{equation} \label{eq:parentheses} -\hbar^{2}\nabla^{2} + i\hbar qB_{z} \nabla\cdot x\hat{y} +i\hbar qB_{z}x\hat{y}\cdot\nabla + q^{2}B_{z}^{2}x^{2}. \end{equation}

Here $\nabla\cdot x\hat{y}= x\hat{y}\cdot\nabla = x\frac{d}{dy}$ so the two middle terms can be combined and the time independent Schrödinger equation becomes,

\begin{equation} \label{eq:schroedinger4} \frac{1}{2m}\left(-\hbar^{2}\nabla^{2} + i2\hbar qB_{z}x \frac{d}{dy} + q^{2}B_{z}^{2}x^{2}\right)\psi = E\psi. \end{equation}

Because $y$ and $z$ do not appear explicitly in this equation, it can be shown that the wave function has the form,

\begin{equation} \label{eq:eigenfunction} \psi (x,y,z) = e^{ik_{y}y}e^{ik_{z}z}\Phi(x), \end{equation}

where $\Phi(x)$ is a function of $x$ that needs to be determined. We can verify that this form for $\psi$ solves \ref{eq:schroedinger4} by substituting it in,

\begin{equation} \label{eq:schroedinger5} \frac{1}{2m}\left(-\hbar^{2}\nabla^{2} + i2\hbar qB_{z}x \frac{d}{dy} + q^{2}B_{z}^{2}x^{2}\right)e^{ik_{y}y}e^{ik_{z}z}\Phi(x) = Ee^{ik_{y}y}e^{ik_{z}z}\Phi(x) . \end{equation}

Here $\nabla^{2}e^{ik_{y}y}e^{ik_{z}z}\Phi(x) = (-k_y^2-k_z^2)e^{ik_{y}y}e^{ik_{z}z}\Phi(x)+e^{ik_{y}y}e^{ik_{z}z}\frac{d^2\Phi(x)}{dx^2}$ and $\frac{d}{dy}e^{ik_{y}y}e^{ik_{z}z}\Phi(x)=ik_ye^{ik_{y}y}e^{ik_{z}z}\Phi(x)$. The exponential factors cancel out and we are left with an equation for $\Phi$,

\begin{equation} \label{eq:schroedinger6} \frac{1}{2m}\left(-\hbar^{2}\frac{d^2}{dx^2} + \hbar^{2}k_{z}^{2} + (\hbar k_{y} - qB_{z}x)^{2}\right)\Phi(x) = E\Phi(x). \end{equation}

The term $\frac{\hbar^{2}k_{z}^{2}}{2m}$ is the kinetic energy in the $z$-direction. Motion in the $z$-direction is unaffected by the magnetic field. This problem decouples into motion in the plane perpendicular to the magnetic field motion parallel to the magnetic field. The total energy is the kinetic energy in the $z$-direction plus the energy of the particle in the $x-y$ plane, $E=\frac{\hbar^{2}k_{z}^{2}}{2m}+E'$, where $E'$ is the energy of the particle in the plane. The Schrödinger equation for the motion in the plane is,

\begin{equation} \label{eq:schroedinger7} \frac{1}{2m}\left(-\hbar^{2}\frac{d^2}{dx^2} + (\hbar k_{y} - qB_{z}x)^{2}\right)\Phi(x) = E'\Phi(x). \end{equation}

This equation is mathematically equivalent to the equation for a one-dimensional harmonic oscillator. The equation for a harmonic oscillator with mass $m$ and spring constant $K$ that oscillates around position $x_j$ is,

\begin{equation} \label{eq:harmonicoscillator1} \left(\frac{-\hbar^{2}}{2m}\frac{d^2}{dx^2} + \frac{K}{2}\left(x - x_{j}\right)^{2}\right)\Phi(x) = E'\Phi(x). \end{equation}

Equations \ref{eq:schroedinger7} and \ref{eq:harmonicoscillator1} are equivalent if we make the associations,

\begin{equation} \label{eq:x0} \frac{K}{2} = \frac{q^{2}B_{z}^{2}}{2m}, \quad \omega_{c} = \frac{qB_{z}}{m}, \quad \text{and} \quad x_{j} = \frac{\hbar k_{y}}{qB_{z}}=\frac{2\pi \hbar j}{qB_{z}L_y}. \end{equation}

Here $\omega_{c}$ is the classical cyclotron frequency of a charged particle in a magnetic field. The last form for $x_j$ arises because $k_y$ can take on the values $\frac{2\pi j}{L_y}$ where $j$ is an integer. Since we know that the energy levels of the harmonic oscillator have the form, $E=\hbar\omega(\nu+1/2)$, where $\omega = \sqrt{K/m}$, it follows that the in-plane energy of a charged particle in a magnetic field is,

\begin{equation} E'= \hbar\omega_c\left(\nu+\frac{1}{2}\right)\qquad \nu=0,1,2,\cdots. \end{equation}

Thus the solutions to equation \ref{eq:schroedinger3} are $\psi (x,y,z) = e^{ik_{y}y}e^{ik_{z}z}\Phi(x)$ where $\Phi(x)$ is a solution to the equation for a harmonic oscillator \ref{eq:schroedinger7}. All of the electron states with the same value of $\nu$ are said to be in the same Landau level.

Degeneracy of the Landau levels
We now consider how many states correspond to a Landau level with a single value of $\nu$. There will be one solution for every allowed value of $x_j$. The allowed values of $x_j$ are restricted by the fact that $x_j$ must be inside the sample, $-\frac{L_x}{2} < x_j < \frac{L_x}{2}$ resulting in a restriction on $k_y$ of $-\frac{qB_zL_x}{2\hbar} < k_y < \frac{qB_zL_x}{2\hbar}$. Furthermore, the periodic boundary conditions restrict $k_y$ to values $k_y = \frac{2\pi j}{L_y}$ where $j$ is an integer. Thus the allowed number of states is $\frac{qB_zL_x}{\hbar}$ divided by $\frac{2\pi}{L_y}$,

$$\text{number of states per }\nu = \frac{qB_zL_xL_y}{h}.$$

The density of states per Landau level per spin is,

$$D_0 = \frac{qB_z}{h} = \frac{m\omega_c}{h}\quad\text{J}^{-1}\text{m}^{-2}.$$
  • Festkörperphyisk, Rudolf Gross and Achim Marx, 2.Auflage
  • Introduction to Solid State Physics, Charles Kittel, 8th edition