It is known that one can find the Hall conductivity $\sigma_{xy}$ from a lattice model (in a magnetic field, say) using the TKNN formula (PRL 49 405-408 (1982)), i.e. by summing the Chern numbers for bands below the Fermi energy $E_F$: $$\sigma_{xy}=\frac{e^2}{h}\sum_{E_n<E_F}C_n$$ $C_n$ can be straightforwardly computed from the eigenstates.

If one were to cause the bands in such a model to broaden (by adding an external potential, say), could the longitudinal conductivity $\sigma_{xx}$ be computed in a straightforward way in this model?


1 Answer 1


$\sigma_{xx}$ can be computed via the usual Kubo formula for linear response. You can find that for example in Bruus and Flensberg I suppose.

However, as long as you fill the system up to a spectral gap, even broadening the bands (while keeping $E_F$ in a gap) will lead to $\sigma_{xx}=0$. Indeed, this is part of what characterizes the IQHE. When $\sigma_{xx}\neq0$, the Chern number should not be well defined.

  • 1
    $\begingroup$ Hall conductivity should still be well defined even if $E_F$ lies within a band and can be found by integrating the Berry curvature over occupied states. Indeed, the "Chern number" in such a case will not be an integer. $\endgroup$
    – zeta
    Feb 6, 2020 at 18:13
  • $\begingroup$ I disagree with that. The Fermi projection doesn’t have sufficient off-diagonal decay in the position basis to give a finite result for the trace formula of the Chern number (the double commutator formula) when $E_F$ is in the middle of a band. $\endgroup$
    – PPR
    Feb 6, 2020 at 18:17
  • $\begingroup$ If one takes the formula from the paper referenced in the question and integrates only over states below $E_F$ within a band (in k-space), one gets a finite answer. Are you saying the answer should be infinite? $\endgroup$
    – zeta
    Feb 6, 2020 at 19:07
  • $\begingroup$ What do you mean by "states below $E_F$" if $E_F$ cuts states in their middle (at least some of them) as it passes within a band? $\endgroup$
    – PPR
    Feb 7, 2020 at 5:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.