# Question about the Landau gauge

1. Mar 22, 2008

### lion8172

I was looking at a derivation of the Landau levels in a crystal, and I had a question about the Landau gauge. The situation under consideration is a two dimensional system of non-interacting particles, exposed to a uniform magnetic field B directed along the z-axis (perpendicular to the plane of the two-dimensional system). In the ''Landau gauge,'' it is claimed that the vector potential can be written as
$$\vec{A} = - By \hat{x}$$.
I can see, however, that the vector potential
$$\vec{A} = - \frac{1}{2}B(x \hat{y} - y \hat{x})$$
would produce the same magnetic field. My questions are as follows. First, what is the Landau gauge? Secondly, does the selection of this gauge in this case correspond to the assumption that current flows in the x-direction?

2. Mar 22, 2008

### jpr0

As far as I understand it, the vector potential

$$\vec{A} = - By \hat{x}$$

is the definition of the Landau gauge. It doesn't correspond to any physical assumption - it's chosen just to simplify the calculation. If you choose this gauge, then the operator $\hat{p}_x$ commutes with the Hamiltonian, which means that you can write your solution as $\psi(x,y)= e^{ip_xx/\hbar}\phi(y)$. You can then replace the operator $\hat{p}_x$ in the Hamiltonian by the eigenvalue $p_x$ and you're left with a relatively simple 1D (the variable $y$) differential equation.

3. Mar 28, 2008

### genneth

Just to supplement what jpr0 said, in the Landau gauge you end up with a harmonic oscillator in y. The other gauge you suggested is called the symmetric gauge, and it is also possible to do the maths in that gauge, and you will get different eigenfunctions. The resolution is that the states are massively degenerate in energy, so the different bases are transformable onto each other.

As a further aside, in the symmetric gauge it's useful to use z=x+iy and then employ some complex analysis. You get the rather (mathematically) pretty result that in the lowest Landau level the wavefunctions are psi = f(z) e^-|z|^2 where f(z) is any analytic function.

4. Apr 7, 2008

### lbrits

The potentials mentioned above of course assume that $$\mathbf{B}$$ points in the $$\hat{z}$$ direction. Of course, this is merely a coordinate choice. As far as I know, the second "symmetric gauge" that you quoted gives more information about the problem, i.e. as a 2D harmonic oscillator, and in this gauge it is much easier to calculate degeneracies. But the Landau gauge is useful if you want a quick answer.