The formula follows from the Kubo formula of conductivity (based on the linear response theory), which is discussed in this question: Kubo Formula for Quantum Hall Effect and in the references therein. Starting from the Kubo formula (set $e=\hbar=1$)
$$\tag{1}\sigma_{xy}=i\sum_{E_m<0<E_n}\frac{\langle m|v_x|n\rangle\langle n|v_y|m\rangle-\langle m|v_y|n\rangle\langle n|v_x|m\rangle}{(E_m-E_n)^2},$$
where $|m\rangle$ is the single particle eigen state of the eigen energy $E_m$, i.e. $$\tag{2} H|m\rangle = E_m|m\rangle.$$
Let us take the momentum derivative $\partial_k$ on both sides of Eq. (2), we have
$$\tag{3}(\partial_{k}H)|m\rangle + H\partial_{k}|m\rangle = (\partial_{k}E_m)|m\rangle + E_m \partial_{k}|m\rangle.$$
Then overlap with $\langle n|$ from left, Eq. (3) becomes
$$\tag{4}\langle n|(\partial_{k}H)|m\rangle + E_n\langle n|\partial_{k}|m\rangle = (\partial_{k}E_m)\langle n|m\rangle + E_m \langle n|\partial_{k}|m\rangle.$$
Here we have used $\langle n|H = E_n\langle n|$. If $|m\rangle$ and $|n\rangle$ are different eigen states (for $E_m\neq E_n$ in Eq. (1)), their overlap should vanish, i.e. $\langle n|m\rangle=0$. Also note that $\partial_k H$ is nothing but the velocity operator $v=\partial_k H$ by definition. So Eq. (4) can be reduced to
$$\tag{5} \langle n|v|m\rangle = (E_m - E_n) \langle n|\partial_{k}|m\rangle.$$
Substitute Eq. (5) to Eq. (1) (restoring the $x$, $y$ subscript), we have
$$\tag{6} \sigma_{xy}=-i\sum_{E_m<0<E_n}\big(\langle m|\partial_{k_x}|n\rangle\langle n|\partial_{k_y}|m\rangle - \langle m|\partial_{k_y}|n\rangle\langle n|\partial_{k_x}|m\rangle\big).$$
On the other hand, the Berry connection is defined as $A_\mu=i\langle m|\partial_{k_\mu}|m\rangle$, and the Berry curvature is $F_{xy}=\partial_{k_x}A_{y}-\partial_{k_y}A_{x}$. Given that $(\partial_k\langle m|)|n\rangle = - \langle m|\partial_k|n\rangle$ (integrate by part), we can see
$$\tag{7} F_{xy}= -i \sum_n\big( \langle m|\partial_{k_x}|n\rangle\langle n|\partial_{k_y}|m\rangle - \langle m|\partial_{k_y}|n\rangle\langle n|\partial_{k_x}|m\rangle\big) +i \langle m|\partial_{k_x}\partial_{k_y}-\partial_{k_y}\partial_{k_x}|m\rangle.$$
The last term will vanish as the partial derivatives commute with each other. So, by comparing with Eq. (6), we end up with
$$\tag{8} \sigma_{xy}=\sum_{E_m<0}F_{xy}\sim\int_{BZ} d^2k F_{xy}.$$
This means the Hall conductance is simply the sum of the Chern numbers (the total Berry flux through the BZ) for all the occupied bands.
So what is the physical meaning of $F_{xy}$? $F_{xy}$ is an effective magnetic field in the momentum space. We know that for the magnetic field $B$ in the real space, a charged particle moving in it will experience the Lorentz force, such that the equation of motion reads $\dot{k}=\dot{r}\times B$. Now to switch to the momentum space, we just need to exchange the momentum $k$ and the coordinate $r$, and replace $B$ by $F$, which leads to
$$\tag{9} \dot{r}=\dot{k}\times F$$
So what is $\dot{r}$? It is the velocity of the electron, which is proportional to the current $j$. And what is $\dot{k}$? It is the force acting on the electron, which is proportional to the electric field strength $E$, so Eq. (9) implies
$$\tag{10} j \sim E\times F.$$
Therefore the Berry curvature $F_{xy}$ simply gives the Hall response of each single electron state. So the Hall conductivity of the whole electron system should be the sum of the Berry curvature over all occupied states, which is stated in Eq. (8).
This post imported from StackExchange Physics at 2014-03-24 04:13 (UCT), posted by SE-user Everett You
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