# How to show that translational invariance in $y$ of implies that it's an eigenstate of $p_y$?

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?

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