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.
Answer to the question $\varphi\mapsto 1/\varphi$:
In the strong magnetic field limit, the eigenstates are Landau wavefunctions, i.e. each electron is dancing around one flux quantum. Now turn on the superlattice, we arrive at the magnetic unit cell, which encloses $p$ flux quanta (one unit cell encloses $p/q$), hence $p$ electrons.
In the strong lattice potential limit, Bloch states are eigenstates, the magnetic field enlarges the unit cell $q$ times, and there are $q$ electrons in the magnetic unit cell.
In either case, perturbations are considerd within a single band or level. If there are $p(q)$ electrons in one magnetic unit cell, there are $p(q)$ subbands.
Best Answer
The TKNN (bulk) and Büttiker (edge) explanations for the quantized Hall conductance correspond to different geometries.
In the TKNN theory, the "sample" consists of a torus closed on itself and therefore has no edges at all. In this case the electric potential is uniform, and the electric field is due to the time derivative of the vector potential (it lasts only as long as one varies the magnetic flux inside the torus). In this case, the Hall current is truly a bulk current.
Büttiker, on the other hand, considers a Hall bar with different electrochemical potentials on each side. If one (as does Büttiker) assumes that the electrostatic potential is uniform within the central region of the bar, and rises on the sides, then one finds that the current flows along the edges (in opposite directions), with more current flowing along one edge than along the other because of the different chemical potentials.
In a more realistic description, the electrostatic potential is not uniform in the bulk of the bar, so that the current flow takes place both along the edges and within the bulk of the bar. In any case, the net current is completely independent of the actual profile of the electrostatic potential accross the hall bar, and depends only upon the chemical potential difference. That is why why Büttuker and TKNN obtain the same answer for the (quantized) total current.
A nice discussion of this question is given by Yoshioka.