Kubo Formula for Quantum Hall Effect

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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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? – Matt Reece Dec 15 '10 at 6:45
I'm not sure, but a related observation is that the conductivity tensor should probably be antisymmetric in the first place. – Greg Graviton Dec 15 '10 at 8:49
That doesn't sound right to me; there should be ordinary (non-Hall) conductivities on the diagonal. – Matt Reece Dec 15 '10 at 15:03
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. – Greg Graviton Dec 15 '10 at 21:08
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. – Greg Graviton Jan 13 '11 at 10:58
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1 Answer

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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