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.
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