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

$H=\frac{1}{2m}((P_1-\frac{1}{2}eB X_2)^2+(P_2+\frac{1}{2}eBX_1)^2)$

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:

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

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

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You cn get displayed equations using double dollar signs, and proper sizes for the parentheses using \left and \right. – joriki Nov 20 '12 at 14:49

Denoting $P_1-\frac12eBX_2$ by $O_1$ and $P_2+\frac12eBX_1$ by $O_2$, we have

$$[O_1,O_2]=\left[P_1-\frac12eBX_2,P_2+\frac12eBX_1\right]=\left[P_1,\frac12eBX_1\right]+\left[P_2,\frac12eBX_2\right]=\mathrm i\hbar eB\;.$$

Thus $O_1$ and $O_2$ have the canonical commutation relation, up to a constant factor $eB$, so you can treat the Hamiltonian $H=(O_1^2+O_2^2)/(2m)$ in analogy to the Hamiltonian $H=(X^2+P^2)/2$.

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I'm rather confused by your answer, would you mind being a bit more explicit? – Freeman Nov 20 '12 at 16:35
Why does the canonical commutation relation imply this? – Freeman Nov 20 '12 at 16:52
@LHS: See the Wikipedia section on the algebraic solution of the quantum harmonic oscillator using ladder operators. Note that the only property of the position and momentum operators that this uses is their commutation relation, so it works for any Hamiltonian that can be written as the sum of the squares of two operators with that commutation relation. The ground state in the present case is a rotationally symmetric Gaussian; the remaining states can be obtained by successively applying the creation operator. – joriki Nov 20 '12 at 20:30