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The Kubo formula for Hall conductance in the Quantum Hall Effect is given as $$\sigma_{xy}=i \hbar \sum_{n \neq 0}\frac{\langle{0}|J_y|n\rangle \langle n|J_x|0\rangle - \langle{0}|J_x|n\rangle \langle n|J_y|0\rangle}{(E_n-E_0)^2}$$ I have read (and understood) the derivation. However, I am still looking for an intuitive explanation for the formula, i.e. I am searching for answers to questions like:

1) Why do the energy eigenfunctions which are further away from $E_0$ contribute less to the sum?

2) What exactly is the meaning of 'crossing' behavior of the nominator?

3) And why does the $n=0$ state not contribute to the sum (apart of course from the fact that we had a divergence in the denominator and the nominator is $=0$)? Normally, only the unfilled bands can carry a current...

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1) Why do the energy eigenfunctions which are further away from $E_0$ contribute less to the sum?

Because the contribution is suppressed by the energy denominator $(E_n-E_0)^2$ if $E_n$ is further away from $E_0$.

2) What exactly is the meaning of 'crossing' behavior of the nominator?

It means that in the second order perturbation process, the ground state must be brought to an excited state by the perturbation and then be brought back.

3) And why does the $n=0$ state not contribute to the sum (apart of course from the fact that we had a divergence in the denominator and the nominator is $=0$)? Normally, only the unfilled bands can carry a current...

This is simply because $\langle 0|\boldsymbol{J}|0\rangle=0$. The ground state has no current.


For more discussions about the physical meaning of the Kubo formula for Hall conductance, please visit the following related questions:

Physical Interpretation of Relationship Between Hall Conductivity and Berry Curvature?

Calculating conductivity from Green's functions

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    $\begingroup$ Is there also a more physical way to explain these results which doesn't rely only on the formula itself? $\endgroup$ – Quasar Apr 29 '17 at 12:20
  • $\begingroup$ @Quasar What kind of physical explanation are you expecting? Please take a look at the previous discussions: physics.stackexchange.com/questions/33275/… and physics.stackexchange.com/questions/86365/… $\endgroup$ – Everett You Apr 30 '17 at 3:57
  • $\begingroup$ The second link was very useful! I will award your answer the bounty worth 50 points in 2 days unless there will (unexpectedly) come in another answer. $\endgroup$ – Quasar Apr 30 '17 at 13:22

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