I am having trouble proving that the Hall conductivity is equal to the integral over the Berry curvature in momentum space. In the TKNN (1982) paper, using the Kubo formula

$$ \sigma_{xy} = \frac{ ie^2}{\hbar} \sum_{\epsilon_\alpha < 0 < \epsilon_\beta}{\frac{ \langle \alpha | J_x | \beta \rangle \langle \beta | J_y | \alpha \rangle - \langle \alpha | J_y | \beta \rangle \langle \beta | J_x | \alpha \rangle}{(\epsilon_\alpha - \epsilon_\beta)^2}} \, . $$

We may use the relation $J_\mu = \partial H / \partial k^\mu$, transfer the derivatives to the wavefunction, in the process cancelling out the energy denominators to get $$ \sigma_{xy} = \frac{ e^2}{\hbar} \sum_{\epsilon_\alpha < 0 < \epsilon_\beta}{ \langle \alpha | \frac{ \partial}{\partial k_x} | \beta \rangle \langle \beta | \frac{ \partial}{\partial k_y} | \alpha \rangle - \langle \alpha | \frac{ \partial}{\partial k_y} | \beta \rangle\langle \beta | \frac{ \partial}{\partial k_x} | \alpha\rangle} \, . $$

I suspect TKNN then use the completeness relation $|\alpha \rangle \langle \alpha | + \sum_{\beta \neq \alpha}{|\beta \rangle \langle \beta |} = 1$ to make this equal to the sum of Berry curvatures over filled bands $\alpha$. However, in the sum you only take states $\beta$ in the unfilled bands - in order to use the completeness relation I would have thought you would have to have terms in the sum involving states $\beta \neq \alpha$ which are also in filled bands.

What do we do with the condition $\epsilon_\beta > 0$ when using the completeness relation to express the conductivity in terms of the Berry curvature? Any help would be appreciated!

NOTE: Just to clarify, my question is about the fact that the summation of matrix elements is taken over states $\alpha$ below the Fermi energy and $\beta$ above the Fermi energy. Clearly you cannot directly apply the completeness relation, since doing so you would be including matrix elements between states $\alpha$ below the Fermi energy and other states $\gamma$ also below the Fermi energy. In general these matrix elements are not vanishing, but they do not appear in the conductivity.

The conductivity is a non-equilibrium property - it only involves transitions from filled to unfilled states. The completeness relation does not care about filled or unfilled states. This is the point of my question. In the end TKNN obtain an expression equal to the sum of Chern numbers of bands $\alpha$ below the Fermi energy. However the sum involves transitions from bands below the Fermi energy to bands above the Fermi energy. How did they get rid of the summation over states $\beta$ above the Fermi energy if not by using a completeness relation?

  • 2
    $\begingroup$ Who or what is TKNN? $\endgroup$
    – ACuriousMind
    Jul 4 '15 at 12:44
  • $\begingroup$ It's the original paper demonstrating the Quantum Hall Effect was related to Berry curvature, Thouless, et. al. Phys. Rev. Lett. 49, 405 (1982) $\endgroup$
    – user38184
    Jul 4 '15 at 13:53
  • $\begingroup$ $\varepsilon_\alpha<0<\varepsilon_\beta$ simply means $\alpha$ is filled, $\beta$ is not. $0$ is where the Fermi energy sits. $\endgroup$
    – Meng Cheng
    Jul 4 '15 at 15:12
  • $\begingroup$ Yes, thanks, that is my point. In the second equation in my question the sum is over states $\beta$ above the Fermi energy. You cannot apply the completeness relation since states $\beta \neq\alpha$ below the Fermi energy are not accounted for (clearly there are multiple bands above and below the Fermi energy which are in general connected by operators $J_x, J_y$). How do you then close the sum over $\beta$ states? $\endgroup$
    – user38184
    Jul 4 '15 at 15:16
  • $\begingroup$ I think the completeness relation you need to apply is $\sum_{\varepsilon_{\gamma}<0}|\gamma\rangle\langle \gamma|+\sum_{\varepsilon_{\beta}>0}|\beta\rangle\langle \beta|=1$. $\endgroup$
    – Meng Cheng
    Jul 4 '15 at 15:35

Here is a derivation: Let me focus on the sum in the expression of $\sigma_{xy}$. First notice that $\langle \alpha|(\partial_{k_x}|\beta\rangle)=-(\partial_{k_x}\langle \alpha|)|\beta\rangle$ because $\langle \alpha|\beta\rangle=0$ since they have different energies. We can rewrite the expression as

$\sum_{\varepsilon_{\alpha}<0}\sum_{\varepsilon_{\beta}>0}\Big[-(\partial_{k_x}\langle \alpha|)|\beta\rangle\langle\beta|(\partial_{k_y}|\alpha\rangle)+(\partial_{k_y}\langle \alpha|)|\beta\rangle\langle\beta|(\partial_{k_x}|\alpha\rangle)\Big]$

To make it more clear,

$\sum_{\varepsilon_{\alpha}<0}\Big[-(\partial_{k_x}\langle \alpha|)\Big(\sum_{\varepsilon_{\beta}>0}|\beta\rangle\langle\beta|\Big)(\partial_{k_y}|\alpha\rangle)+(\partial_{k_y}\langle \alpha|)\Big(\sum_{\varepsilon_{\beta}>0}|\beta\rangle\langle\beta|\Big)(\partial_{k_x}|\alpha\rangle)\Big]$

Now we can substitute in the completeness relation $\sum_{\varepsilon_{\beta}>0}|\beta\rangle\langle\beta|=1-\sum_{\varepsilon_{\gamma}<0}|\gamma\rangle\langle\gamma|$, and group the terms:

$\sum_{\varepsilon_{\alpha}<0}\Big[-(\partial_{k_x}\langle \alpha|)(\partial_{k_y}|\alpha\rangle)+(\partial_{k_y}\langle \alpha|)(\partial_{k_x}|\alpha\rangle)\Big]+\sum_{\varepsilon_{\alpha},\varepsilon_{\gamma}<0}\big(\langle \alpha|\partial_{k_x}|\gamma\rangle\langle \gamma|\partial_{k_y}|\alpha\rangle-\langle \alpha|\partial_{k_y}|\gamma\rangle\langle \gamma|\partial_{k_x}|\alpha\rangle\big) $

The second term indeed involves matrix elements between the filled states, however the sum is zero because $\alpha,\gamma$ now both run over the filled states and you can certainly relabel these dummy variables. The first term is the Berry curvature.


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