# I Quantum numbers of Landau levels

1. Nov 13, 2016

### spaghetti3451

I have been reading about Landau levels for a two-dimensional system of charged particles in a perpendicular magnetic field and I have trouble understanding why there is degeneracy in the system. Let me provide some background to my problem.

In the presence of a magnetic field, the momentum of a charge particle changes from $p_{i}$ to $\pi_{i}\equiv p_{i}+eA_{i}$, where $e$ is the charge of the particle.

It is also possible to define another kind of momentum $\tilde{\pi}_{i}=p_{i}-eA_{i}$.

For a two-dimensional system with a magnetic field pointing in the $z$-direction, we can work in the symmetric gauge

$${\bf{A}}=\left(-\frac{yB}{2},\frac{xB}{2},0\right),$$

and define two sets of creation and annihilation operators $a, a^{\dagger}$ and $b, b^{\dagger}$ such that a generic state of the system is given by

$$|n,m\rangle=\frac{a^{\dagger n}b^{\dagger}n}{\sqrt{n!m!}}|0,0\rangle,$$

where $|0,0\rangle$ is the unique ground state annihilated by both $a$ and $b$.

Now, $|0,0\rangle$ is the unique ground state annihilated by both $a$ and $b$ since $\pi_{i}$ and $\tilde{\pi}_{i}$ commute in the complex plane defined by $(z=x-iy,\bar{z}=x+iy)$ and commuting observables have a common basis of eigenfunctions.

Does this not mean that the quantum numbers labelled by $n$ and $m$ give the same wavefunctions?

Why does the energy of the state depend only on $n$, but not on $m$?

2. Nov 14, 2016

### vanhees71

It's much easier to solve the energy-eigen problem in the gauge, where $\vec{A}$ depends only on one spatial component, e.g.,
$$\vec{A}=-By \vec{e}_x,$$
because of the higher symmetry of the corresponding Hamiltonian. In this gauge the Hamiltonian reads
$$\hat{H}=\frac{1}{2m} [(-\mathrm{i} \partial_x+q By)^2-\partial_y^2-\partial_z^2],$$
i.e., you can seek for common eigenvectors of $\hat{p}_x$, $\hat{p}_z$ and $\hat{H}$. This leads to the ansatz
$$u_{E,p_x,p_z}(\vec{x})=C \exp(\mathrm{i} p_x x+\mathrm{i} p_y y) Y(y),$$
leading to an effective one-dimennsional harmonic-oscillator energy-eigenvalue problem for $Y(y)$.

Of course, the energy eigenvalues are gauge invariant, and the wave function transforms with $\psi'(t,\vec{x})=\exp[\pm \mathrm{i} q\chi(\vec{x})] \psi(t,\vec{x})$ while $\vec{A}'=\vec{A} + \vec{\nabla} \chi$. I'm a bit unsure about the sign in the phase factor. Just figure it out by direct calculation.