I'm trying to understand the Kubo Formula for the electrical conductivity in the context of the Quantum Hall Effect.

My problem is that several papers, for instance the famous TKNN (1982) paper, or an elaboration by Kohmoto (1984), write the diagonal entries of the conductivity tensor in the form

$$ \sigma_{xy}(\omega \to 0) = \frac{ie^2}{\hbar} \sum_{E^a < E_F < E^b} \frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle - \langle a|v_y|b \rangle \langle b|v_x|a \rangle}{(E^a - E^b)^2} .$$

This is the static limit $\omega\to 0$ and low temperature $T\to 0$. The sum goes over all eigenstates $|a\rangle$ and $|b\rangle$ of the single-particle Hamiltonian. $E_F$ is the Fermi energy. $v_x$ and $v_y$ are the single-particle velocity operators.

However, these papers don't derive this equation, which is unfortunate because the Kubo formula is usually not presented in this form. I have found (and succeeded in rederiving) the following variation instead

$$ \sigma_{xy}(\omega+i\eta) = \frac{-ie^2}{V(\omega + i\eta)} \sum_{a,b} f(E^a) \left( \frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle}{\hbar\omega + i\eta + E^a - E^b} + \frac{\langle a|v_y|b \rangle \langle b|v_x|a \rangle}{-\hbar\omega - i\eta + E^a - E^b} \right).$$

This is formula (13.37) from Ashcroft, Mermin, though they don't actually prove it. $f(E)$ is the Fermi distribution. A nice derivation is given in Czycholl (german).

Now, my question is, obviously

How to derive the first formula from the second?

I can see that the first equation arises as the linear term when writing the sum as a power series in $\omega$, but why doesn't the constant term diverge?

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    $\begingroup$ I'm not at all sure of this, but: I think the issue may be that the Hall conductivity is defined as an antisymmetrized component of the conductivity tensor, i.e. the quantity that the first formula applies to may actually be $\sigma_{xy} - \sigma_{yx}$. Does this sound plausible? $\endgroup$
    – Matt Reece
    Dec 15, 2010 at 6:45
  • $\begingroup$ I'm not sure, but a related observation is that the conductivity tensor should probably be antisymmetric in the first place. $\endgroup$ Dec 15, 2010 at 8:49
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    $\begingroup$ That doesn't sound right to me; there should be ordinary (non-Hall) conductivities on the diagonal. $\endgroup$
    – Matt Reece
    Dec 15, 2010 at 15:03
  • $\begingroup$ I have no clue. Could you give an example of a material with diagonal conductivities? Or any general pointers on this stuff? After all, diagonal conductivities are strange because the current flows perpendicular to the applied electric field. $\endgroup$ Dec 15, 2010 at 21:08
  • $\begingroup$ Found it! A slight variation of an argument by Czycholl can be used to show that the diverging term actually vanishes. I'll write it up soon. $\endgroup$ Jan 13, 2011 at 10:58

2 Answers 2


The first formula indeed follows from the second formula if we let $\omega\to0$. To see that, expand the fractions as

$$ \frac1{\pm\hbar\omega + E^a - E^b} = \frac1{E^a-E^b}\left(1 \mp \frac{\hbar\omega}{E^a-E^b}\right) + \mathcal O(\omega^2)$$

to obtain $\sigma_{xy} = \sigma^1 + \sigma^2$ as the sum of a potentially divergent term

$$ \sigma^1 = \frac{-ie^2}{V\omega} \sum_{a,b} f(E^a) \frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle + \langle a|v_y|b \rangle \langle b|v_x|a \rangle}{E^a - E^b} $$

and a term that looks like the first formula

$$ \sigma^2 = \frac{-ie^2\hbar}{V} \sum_{a,b} f(E^a) \frac{- \langle a|v_x|b \rangle \langle b|v_y|a \rangle + \langle a|v_y|b \rangle \langle b|v_x|a \rangle}{(E^a - E^b)^2} .$$

To see that the first term vanishes instead of diverging, we have to use the Heisenberg equation of motion $v_x = \frac{d}{dt}x = [H_0,x]$ which gives

$$ \langle a | v_x | b \rangle = \langle a | H_0 x - x H_0 | b \rangle = (E^a-E^b) \langle a | x | b \rangle $$

and thus

$$ \langle a|v_x|b \rangle \langle b|v_y|a \rangle + \langle a|v_y|b \rangle \langle b|v_x|a \rangle = (E^a-E^b) (\langle a|x|b \rangle \langle b|v_y|a \rangle - \langle a|v_y|b \rangle \langle b|x|a \rangle) .$$

The factors $(E^b-E^b)$ cancel and the remaining sum over $b$ becomes a sum over the identity $\sum_b |b\rangle\langle b| = 1$. Thus, we arrive at

$$ \sigma^1 = \frac{-ie^2}{V\omega} \sum_{a,b} f(E^a) \left(\langle a|xv_y - v_yx |a \rangle \right) = 0 .$$

since the commutator $[x,v_y]$ vanishes.

To see that the second term is correct, we have to get the summation indices right. To do that, we have to rearrange the summation to obtain

$$ \sigma^2 = \frac{ie^2\hbar}{V} \sum_{a,b} (f(E^a)-f(E^b))\frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle}{(E^a - E^b)^2} .$$

In the limit $T\to0$, the difference of Fermi-Dirac distributions $f(E^a)-f(E^b)$ will be equal to

  • $1$ if $E^a < E_F < E^b$
  • $-1$ if $E^b < E_F < E^a$
  • $0$ otherwise

Using this and rearranging the summation again gives the Kubo formula in the first form.

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    $\begingroup$ Hi, I think your proof of the vanishing of the first term is incorrect. For example, if we have a Fermi surface, then the first term is apparently non-zero--it is physically sensible for a conductivity to have 1/omega divergence in a metal. $\endgroup$
    – Xu Yang
    May 7, 2019 at 22:44
  • $\begingroup$ @XuYang Which term do you think is apparently non-zero? Note that this calculation applies only to the transverse conductivity, not the longitudinal conductivity, which may indeed diverge. $\endgroup$ May 9, 2019 at 10:39
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    $\begingroup$ See equation 1.12 in the following paper and the derivation thereafter : journals.aps.org/prb/pdf/10.1103/PhysRevB.48.11705 In this paper it is actually shown that the first term only vanishes for an insulator, not for a metal. $\endgroup$
    – Xu Yang
    May 9, 2019 at 20:28
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    $\begingroup$ @Dwagg No idea. In fact, I don't buy it: A general argument based on gauge invariance would likely also apply to the longitudinal conductivity. But it is known that in a clean metal without disorder, the longitudinal conductivity is actually infinite in the limit $\omega\to 0$. Either the argument you mention contains another ingredient that makes it inapplicable to the longitudinal conductivity, or it's likely not valid. $\endgroup$ May 15, 2020 at 15:39
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    $\begingroup$ @symanzik138 Oh my, that's a good point. In a tight-binding model, the Hamiltonian is a sum of exponentials of the form $\exp(ik_x n + ik_y m)$, and $v_x = ∂H/∂k_x$, so your argument stands. I need to think about this, the real reason is probably that the system is insulating, i.e. the distribution function $f(E^a)$ contains only full bands. $\endgroup$ Feb 24, 2021 at 22:07

A nice derivation of the second formula is given in http://www.damtp.cam.ac.uk/user/tong/kintheory/four.pdf

  • 1
    $\begingroup$ Welcome to the Physics.Stackexchange. Please note that for this site it is better to avoid link-only answers. Please try to include the essential part of the linked text into your answer. The goal is to make SE a sourse of information and not just a list of links. $\endgroup$
    – mpv
    Dec 29, 2015 at 22:48

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