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Let us consider a particle on a plane with uniform magnetic field $B=B\hat{z}$, and hence with the Hamiltonian $H=\frac{1}{2m}(\vec{p}+e\vec{A})^2$. I am concerned with finding the energy eigenstates, and in order to do that let us specify the gauge potenial as $\vec{A}=Bx\hat{y}$ which is known as the Landau gauge, resulting in a Hamiltonian $H=\frac{1}{2m}(p_x^2+(p_y+eBx)^2)$. Notice that there is a translational invariance of $H$ in $y$.

I want to argue that this implies that the eigenstate of $H$ must be an eigenstate of $p_y$.

Although I can intuitively see any eigenstate of $p_y$ should have invariance under translation in $y$, I want to find a way to show this. In order to do this, how should I start?

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I want to argue that this implies that the eigenstate of $H$ must be an eigenstate of $p_y$.

Tough luck. It doesn't.

  • You are guaranteed one complete set of shared eigenstates between $H$ and $p_y$.
  • However, you are not guaranteed that every eigenstate of $H$ will be an eigenstate of $p_y$.

To get a simple counterexample, start off with the usual shared eigenstates, $$ \psi_{k,n}(x,y) = \varphi_n(x-k/eB)e^{iky} $$ (where $\varphi_n(x)$ is an eigenfunction of the harmonic oscillator with mass $m$ and cyclotron frequency $\omega_c=eB/m$), which are eigenstates of $H$ with eigenvalue $\hbar\omega_c(n+\tfrac12)$ and eigenstates of $p_y$ with eigenvalue $k$.

From those, construct the linear combination \begin{align} \tilde\psi(x,y) & = \psi_{k_1,n}(x,y) + \psi_{k_2,n}(x,y) \\ & = \varphi_n(x-k_1/eB)e^{ik_1y} + \varphi_n(x-k_2/eB)e^{ik_2y} , \end{align} with the same $n$ but with different $k$. These are still eigenstates of $H$ but not eigenstates of $p_y$. QED.

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