# Derivation of polarizability

I am currently reading some papers on Dirac and Weyl physics on condensed matter. Very often, the following result for the polarizability is used: $$\Pi(q,\omega) \propto \sum_{k,s,s'} \frac{f_{sk}-f_{s'k'}}{\omega + \epsilon_{sk} - \epsilon_{s'k'} + i\eta}F_{ss'}(k,k'),$$ where $k' = k + q$ and $s,s'=\pm1$ and $F_{ss'}(k,k') = (1+ss'\cos\theta)/2$, with $\theta$ the angle between $k$ and $k'$ and finally $f_{sk} = (\exp(\beta(\epsilon_{sk} - \mu + 1))^{-1}$ the Fermi distribution with linear dispersion $\epsilon_{sk} = s|{\bf k}|$.

1) Does anyone know of a good reference where it is derived in some detail? I do know it is a electronic bubble diagram resulting from the coupling of electrons to photons. I almost managed to derive it, but I have some small things I do not understand yet, like where the function $F_{ss'}(k,k')$ comes from.
2) In these papers they use the calculation of $\Pi(q,\omega)$ to obtain the collective modes from the zeros of the dielectric function $\epsilon(q,\omega) = 1 + v_c(q)\Pi(q,\omega)$ (with $v_c(q)$ the Fourier transformed Coulomb potential). In principle a bubble diagram has two indices, coming from the external photons, however the expression above does not have that. Where did they go? In particle physics one often uses transversality to write $\Pi^{\mu\nu}(q) = (\eta^{\mu\nu}q^2 - q^{\mu}q^{\nu})\Pi(q^2)$. So did they pull out this transversal part here as well? Or is it just the 00-component, because the dielectric function is corrected by the density-density response function?
The factor you mention goes to zero for $\theta = 0$ [$\pi$] so I'm tempted to say it comes from some selection rules regarding the polarization of incoming photons.