Current in linear combination of landau levels

Question

I’m working through Girvin’s Quantum Hall Effect notes (https://arxiv.org/abs/cond-mat/9907002) And I came across the following problem:

It is interesting to note that the exact eigenstates in the presence of the electric field can be viewed as displaced oscillator states in the original (zero E field) basis. In this basis the displaced states are linear combinations of all the Landau level excited states of the same k. Use first-order perturbation theory to find the amount by which the n = 1 Landau level is mixed into the n = 0 state. Because the displaced state is a linear combination of more than one Landau level, it can carry a finite current.

Can someone please help me understand why the displaced state can carry finite current? Is this just because previously the expectation value of the current was antisymmetric about the center of the Gaussian and so it was zero whereas with the displaced state the expectation value is non-zero? Is there a more intuitive explanation for this (this seems more like a result than an explanation)

Any help greatly appreciated!

Answer 1

The symmetry-based reasoning that you've quoted from Girvin is correct. For a more intuitive way to understand the result, think about the semi-classical picture. Semi-classically, in a single Landau level the electrons are simply moving in circular orbits. The average velocity is zero, and since the current is proportional to the electron velocity the average current will be zero.

Once we apply an electric field, the electrons will acquire a drift velocity in the direction of $\mathbf{E} \times \mathbf{B}$. Now there is an average velocity and so a non-zero average current. Quantum mechanically, the state with non-zero drift velocity is a mixture of Landau levels. For more details, see Chapter 2 of D. Yoshioka, The Quantum Hall Effect. Here is an illustrative picture from that book:

Edit in response to the comment:

The velocity operator is \begin{align} \mathbf{v} = \frac{i}{\hbar}\left[H, \mathbf{r}\right]. \end{align} For the case of the Landau hamiltonian $H = \pi^2 / (2m)$ ($\boldsymbol{\pi} = \mathbf{p} - e\mathbf{A}$), we get $\mathbf{v} = \boldsymbol{\pi}/m$. We diagonalize the Landau hamiltonian by introducing cyclotron ladder operators $a$, $a^{\dagger}$ such that $H = \hbar /(m\ell^2) (a^{\dagger}a + 1/2)$. The ladder operators applied to a Landau level state take the Landau level number up or down by 1. In terms of the ladder operators, the components of the velocity operator are $v_x \sim a + a^{\dagger}$, $v_y \sim a - a^{\dagger}$ (up to constant factors and $SU(1,1)$ Bogoliubov transformations). So the velocity operator applied to a Landau level state will change the Landau level number. Expectation values of $\mathbf{v}$ in single-Landau level states will therefore be zero.

Applying an electric field changes the hamiltonian such that the new eigenstates are mixtures of the zero-field Landau levels. So a quantum mechanical treatment says that the electric field changes the states, which changes the expectation values, which yields a non-zero current — not the other way around.