2 Dimensional quantum particle A quantum particle moves in 2 dimensions with Hamiltonian H:
$ H = \frac1{2m} ((P_1 + \frac12 eBX_2)^2 + (P_2 - \frac12 eBX_1)^2) $
For constants $e,B,m$ with $e$ and $B$ nonzero.
Show that the energy levels are of the form $ (n + \frac12)\bar h |eB|\frac{1}{m}$
The hint given is to define $\bar P$ and $\bar X$ as proportional to $P_1 + \frac12 eBX_2$ and $P_2 - \frac12 eBX_1$ and show that the original Hamiltonian has the form
$\frac1{2m} P^2 + \frac12m\omega^2X^2$ for some $\omega$, where
$P_j = -i\bar h \frac{\partial}{\partial x_j}$ and
$X_j = x_j$
We are given that this has energy levels $(n+\frac12)\bar h \omega$.

Depending on how you want to approach this problem I've seen the following done:
\begin{eqnarray}
\hat{H} &=& \frac{1}{2m}[\hat{p}_x ^2+\hat{p}_x \hat{y}eB+\frac{1}{4}\hat{y}^2e^2B^2-\hat{p}_y ^2+\hat{p}_y \hat{x}eB+\frac{1}{4}\hat{x}^2e^2B^2]\\
&=&\frac{1}{2m}[\hat{p}_x^2+\hat{p}_y^2]+\hat{L}_z \frac{eB}{2m}+\frac{e^2 B^2}{8m}(\hat{x}^2+\hat{y}^2)
\end{eqnarray}
It's provable that $\hat{L}_z$ commutes with $\hat{p}_x ^2 + \hat{p}_y ^2$ and $\hat{x}^2 + \hat{y}^2$. You can thus form a complete set of commuting operators for $L_z$ and what appears to be a spring Hamiltonian. 
You could probably calculate the levels of $L_z$ and the levels of a spring hamiltonian, so I'll leave the rest to you.
