[Physics] Berry curvature of Landau levels

Question

If we consider an electron on a two dimensional surface with a magnetic field normal to the surface, we know the states the electron can occupy are Landau levels. If we additionally impose periodic boundary conditions, i.e., that these states are eigenstates of the magnetic translation operators, then our surface becomes a torus, and the eigenstates are generalized Bloch states $\psi_{\mathbf{k}}(x,y) = e^{i\mathbf{k}\cdot\mathbf{r}}u_{\mathbf{k}}(x,y)$, with $k_i$ taking values in $\frac{2\pi}{L_i}n, 1 < n < N_{\Phi}$, where $L_i$ are the system dimensions, and $N_{\Phi}$ is the number of flux quanta passing through the area $L_x L_y$. For this boundary condition to be meaningful, we further require that the magnetic translation operators in either direction commute, so that $L_x L_y = 2\pi N_{\Phi} \ell^2$ ($\ell$ is the magnetic length). We can write down wavefunctions $\psi_{\mathbf{k}}(x,y)$ that statisfy these boundary conditions, as done, for example, by Yoshioka (http://prb.aps.org/pdf/PRB/v29/i12/p6833_1) and Haldane and Rezayi (http://prb.aps.org/pdf/PRB/v31/i4/p2529_1). To calculate the Berry curvature, one could simply calculate the Berry connection in some gauge,
$$
\mathbf{A} = \int d^2x \,\, u_{\mathbf{k}}^*(\mathbf{r}) \nabla_k u_{\mathbf{k}}^*(\mathbf{r})
$$
and then the curvature
$$
F_{xy} = \partial_{k_x}A_y – \partial{k_y}Ax
$$
We know the Berry curvature for the Landau levels should be constant in $\mathbf{k}$. My question is, how does one do this in practice? It seems that the calculation of the Berry connection (in the Landau gauge $A = -Bx\hat{k_y}$) reduces to calculating the expectation value of the x position $\left<x\right>$. However, while this calculation would be trivial in an infinite plane, with periodic boundary conditions the wavefunctions become much more complicated.

Answer 1

Here is an outline for what you might want to do.

You can define a magnetic unit cell of dimension $a \times b$ such that each cell contains one flux quantum, i.e., $2\pi\hbar c/e = Bab$. (The whole system is infinite in both $x$ and $y$ directions.) If you adopt the gauge choice such that $\textbf{A} = (0, Bx)$, eigenstates of the form $\phi_{n,k}(x,y) = e^{iky} f_{n}(x-l_{B}^{2}k)$, where $f_n$ is an eigenfunction of SHO, naturally follow.

$\phi_{n,k}$ already has translation symmetry in the $y$ direction. By making a suitable linear combination of ..., $\phi_{n, k-2\pi/a}$, $\phi_{n,k}$, $\phi_{n, k+2\pi/a}$, $\phi_{n, k+4\pi/a}$, ...., one can construct an energy eigenfunction having translation symmetry in the $x$ direction as well, viz. a magnetic Bloch function which is an eigenfunction of the magnetic translation operator which translates the guiding center in the $x$ direction by an amount of $a$.

You can then calculate the Berry curvature using this magnetic Bloch function.