A particle of charge $-e$ and mass $m$ moves in a magnetic field $\boldsymbol{B}(\boldsymbol{x}, t)$ and in an electric potential $\phi(\boldsymbol{x}, t)$. The time-dependent Schrödinger equation for the particle's wavefunction $\Psi(\boldsymbol{x}, t)$ is

$$i \hbar\left(\frac{\partial}{\partial t}-\frac{i e}{\hbar} \phi\right) \Psi=-\frac{\hbar^{2}}{2 m}\left(\nabla+\frac{i e}{\hbar} \boldsymbol{A}\right)^{2} \Psi$$

where $\boldsymbol{A}$ is the vector potential with $\boldsymbol{B}=\boldsymbol{\nabla} \wedge \boldsymbol{A}$. Show that this equation is invariant under the gauge transformations

$$\begin{array}{ll} \boldsymbol{A}(\boldsymbol{x}, t) & \rightarrow \boldsymbol{A}(\boldsymbol{x}, t)+\boldsymbol{\nabla} f(\boldsymbol{x}, t) \\ \phi(\boldsymbol{x}, t) & \rightarrow \quad \phi(\boldsymbol{x}, t)-\frac{\partial}{\partial t} f(\boldsymbol{x}, t) \end{array}$$

where $f$ is an arbitrary function, together with a suitable transformation for $\Psi$ which should be stated.

Assume now that $\partial \Psi / \partial z=0$, so that the particle motion is only in the $x$ and $y$ directions. Let $\boldsymbol{B}$ be the constant field $\boldsymbol{B}=(0,0, B)$ and let $\phi=0$. In the gauge where $\boldsymbol{A}=(-B y, 0,0)$ show that the stationary states are given by

$$\Psi_{k}(\boldsymbol{x}, t)=\psi_{k}(\boldsymbol{x}) e^{-i E t / \hbar}$$

with

$$\psi_{k}(\boldsymbol{x})=e^{i k x} \chi_{k}(y)$$

Show that $\chi_{k}(y)$ is the wavefunction for a simple one-dimensional harmonic oscillator centred at position $y_{0}=\hbar k / e B$. Deduce that the stationary states lie in infinitely degenerate levels (Landau levels) labelled by the integer $n \geqslant 0$, with energy

$$E_{n}=(2 n+1) \frac{\hbar e B}{2 m}$$

A uniform electric field $\mathcal{E}$ is applied in the $y$-direction so that $\phi=-\mathcal{E} y$. Show that the stationary states are given by $(*)$, where $\chi_{k}(y)$ is a harmonic oscillator wavefunction centred now at

$$y_{0}=\frac{1}{e B}\left(\hbar k-m \frac{\mathcal{E}}{B}\right)$$

Show also that the eigen-energies are given by

$$E_{n, k}=(2 n+1) \frac{\hbar e B}{2 m}+e \mathcal{E} y_{0}+\frac{m \mathcal{E}^{2}}{2 B^{2}} .$$

Why does this mean that the Landau energy levels are no longer degenerate in two dimensions?