quantum mechanics - 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!

Answer

Starting from the canonical momenta:

$\Pi_1= P_1 -\frac{1}{2}eBX_2$ and $\Pi_2= P_2 +\frac{1}{2}eBX_1$,

We get

$[\Pi_1, \Pi_2] = ieB\hbar$

Thus the operators:

$a = \frac{1}{\sqrt{2eB\hbar}}(\Pi_1+i\Pi_2)$ and

$a^{\dagger} = \frac{1}{\sqrt{2eB\hbar}}(\Pi_1-i\Pi_2)$

satisfy the canonical commutation relation $[a, a^{\dagger}] = 1$

Making the substitution in the Hamiltonian we obtain:

$H =\hbar \omega (a a^{\dagger}+\frac{1}{2})$, with $\omega=\frac{eB}{m}$.

The difference from the Harmonic oscillator lies in the fact that now the energy levels are infinitely degenerate for example, the ground states equation:

$a\Psi(X_1,X_2) = 0$

has an infinite number of solutions and any function of the form:

$\Psi = f(X_1+iX_2) \mathrm{exp}(-eB(X_1^2+X_2^2)/4\hbar)$

is a ground state. These are the lowest Landau levels.