I can try to sketch out the general idea here, but my deduction may not be very precise (please refer to your books to check out the coefficients and sign conventions).
So I guess the question is that given a free Fermion system described by the following action
$$S=-\sum_k\psi_k^\dagger G^{-1}(k)\psi_k, ...(1)$$
where $G(k)=-\langle\psi_k\psi_k^\dagger\rangle$ is the Green's function at the momentum-frequency $k=(i\omega,\vec{k})$, what is the DC electric conductivity?
- Start from the definition of the conductivity $j_\mu=\sigma_{\mu\nu} E_{\nu}$ (where $j_\mu$ is the current and $E_\nu$ is the electric field). Consider the linear (differential) response
$$\sigma_{\mu\nu}= \frac{\delta j_\mu}{\delta E_\nu}....(2)$$
- Introduce the gauge potential $A_\mu$. By definition, current is the source of the gauge potential, i.e. $j_\mu = \delta S/\delta A_\mu$, and the electric field is the conjugate momentum of the gauge potential, which means $E_\nu=\partial_t A_\nu$ according to the equation of motion. Plug into the expression Eq. (2) for $\sigma$
$$\sigma_{\mu\nu} = \frac{\delta}{\delta \partial_t A_\nu}\frac{\delta S}{\delta A_\mu}=-i\delta_{A_0}\delta_{A_\nu}\delta_{A_\mu}S....(3)$$
Because $i\partial_t$ means the frequency $i\omega$. For DC conductivity, we should send the frequency $i\omega\to 0$, which means we are actually varying with respect to $i\omega$. But because $i\omega$ always appears with the chemical potential $A_0$ in the form of $(i\omega+A_0)$ in the action, so it will be equivalent to just varying with respect to $A_0$. This is how we can replace $1/\partial_t$ by $-i\delta_{A_0}$ (there are more rigors deduction for this replacement but let us take this simple argument for now).
- You may be wondering how the gauge potential $A_\mu$ entered the action $S$ (note that the original Fermion action does not even containing the field $A_\mu$). This is done by the minimal coupling procedure, which simply replace every $k$ by $k+A$. So the action in Eq. (1) is actually $S=-\sum_k\psi_k^\dagger G^{-1}(k+A)\psi_k$. The idea is then integrate out the Fermion field $\psi$, and obtain the effective action for the gauge field $S[A]$, then can calculate the conductivity by Eq. (3). But all these can be done in a simpler way by noticing that $k$ always appears with $A$, so $\delta_A=\delta_k$, and hence Eq. (3) becomes
$$\sigma_{\mu\nu}=-i\delta_{k_0}\delta_{k_\nu}\delta_{k_\mu}S....(4)$$
- To calculate the momentum-frequency variation, we should first integrate out the Fermion field $\psi$:
$$S\xrightarrow{\int\mathcal{D}[\psi]}S=-\sum_k\text{Tr}\ln(-G^{-1}(k))....(5)$$
Using the differentiation rule $\delta G = G (\delta G^{-1}) G$ which follows from the definition $G G^{-1}\equiv 1$ (and by varying both sides), it is not hard to show from Eqs. (4) and (5) that
$$\sigma_{\mu\nu}=-i\sum_k \text{Tr} G(k)\gamma_0 G(k)\gamma_\nu G(k)\gamma_\mu, ...(6)$$
where the $\gamma$ matrices are defined as
$$\gamma_\mu=-\partial_{k_\mu}G^{-1}(k)....(7)$$
Ok. Eq. (6) is already the Kubo formula written in terms of the Green's function. You may just plug in your Green's function and complete the momentum-frequency summation to get the electric conductivity.
EXAMPLE:
To demonstrate how this works, please allow me to present a simple example. Consider the Hall conductance of a two band system $$G(k)=(i\omega\sigma_0-k_1\sigma_1-k_2\sigma_2-m\sigma_3)^{-1}=\frac{i\omega\sigma_0+k_1\sigma_1+k_2\sigma_2+m\sigma_3}{(i\omega)^2-E^2},$$
with $E=\sqrt{k_1^2+k_2^2+m^2}$. From Eq.(7), $\gamma_0=-\sigma_0$, $\gamma_1=\sigma_1$, $\gamma_2=\sigma_2$. Then plugging into Eq. (6).
$$\sigma_{\mu\nu}=\sum_{k}\frac{2m}{((i\omega)^2-E^2)^2}=\sum_{\vec{k}}\frac{m}{2E^3}=\frac{m}{2|m|},$$
which is what we expect for a single Dirac cone.
