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Freeman
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Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field

Suppose we have an electron, mass $m$, charge $-e$, moving in a plane perpendicular to a uniform magnetic field $\vec{B}=(0,0,B)$. Let $\vec{x}=(x_1,x_2,0)$ be its position and $P_i,X_i$ be the position and momentum operators.

The electron has Hamiltonian

$H=\frac{1}{2m}((P_1-\frac{1}{2}eBX_2)^2+(P_2+\frac{1}{2}eBX_1)^2)$

How can I show that this is analogous to the one dimensional harmonic oscillator and then use this fact to describe its energy levels?

I have attempted to expand out the Hamiltonian and found:

$(\frac{P^2_1}{2m}+ \frac{1}{2} m (\frac{eB}{2m}))^2X^2_1+(\frac{P^2_2}{2m}+\frac{1}{2}m(\frac{eB}{2m})^2)X^2_2+\frac{eB}{2m}(X_1P_2-P_1X_2)$

This looks very similar to the 2D harmonic oscillator, if anyone can help/point out where I am wrong I'd much appreciate it!

Freeman
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