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I was asked to find conductivity $ \sigma (\omega, T) $ using methods of qft, or more exactly using Matsubara Green's function, for the following system: qed with massless dirac fermions in case $ T \neq 0 $. As I understand, coductivity is a linear response coefficient to applying electric field. $$ j_i (t) = \int_0^{\infty} \sigma_{ij} (\tau, T) E_j (t - \tau) d \tau $$ So I need to use Kubo formula: $$ \sigma (\omega, T) = \frac{i}{\hbar} \int_{0}^{\infty} < [\hat{X}(t), \hat{Y}(0)] > e^{i \omega t} dt $$

I think $ \hat{X} $ should be current operator: $$ \hat{X_{\mu}} = e \bar{\psi} \gamma_{\mu} \psi $$ I can write excitation to electromagnetic field as: $$ \hat{H}_{ext} = i e \bar{\psi} A_{\mu} \gamma_{\mu} \psi $$ But if I use this as $ \hat{Y} $, I will find linear response to the vector potential, not electric field. Maybe, it is still correct and I just need to recalculate conductivity from this linear response coefficient in someway? So that is the first question: what operator I should use as $ \hat{Y} $?

After that, I want to use Wick theorem to express $ <[\hat{X}(t), \hat{Y}(t)]> $ through Matsubara Green's function. Can I use free Green's function $ S(i\omega_n, p) = \frac{1}{i\omega_n \gamma_0 + \bar{p} \bar{\gamma} + i \epsilon} $, or I need to calculate first loop correction?

Anatoliy Lotkov
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Conductivity can be written as polarization operator divided by frequency (what is known as Kubo formula). As soon as you calculate the polarization operator (one-loop diagram where the Matsubara green functions do enter) at finite T you obtain the conductivity.

Please have a look at the following papers where this very problem is solved for 2+1 dimensional fermions https://arxiv.org/abs/1608.03261, https://arxiv.org/abs/1111.3017

However, in the QED in 3+1 dimensions you will have infrared divergences in the calculation which have to handled somehow.

good luck,

Ignat Fialkovskiy
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This is not a simple task, and I am not certain that a complete calculation can be found in the literature.

The Kubo formula is well known (see Derivation of Ohm's Law ), and derivations can be found in many standard text books.

The retarded correlation function contains two fermion bi-linears $\bar\psi\gamma_\mu\psi$, so the leading order diagram is a one-loop graph. This graph corresponds to non-interacting electrons, and it does not contribute to the $\omega\to 0$ conductivity. Indeed, the calculation of the conductivity (even in weak coupling) requires the summation of an infinite number of diagrams, among them the set of all ladder diagrams with the (screened) Coulomb interaction.

In the case of the electron-phonon interaction the calculation can be found, for example, in Mahan's book (chapter 8 of D. Mahan, Many-Particle Physics).

In the case of a QED plasma the answer is known from solutions of the Boltzmann equation (the Boltzmann equation effectively sums an infinite set of diagrams), but I am not aware of an explicit calculation using the Kubo formula.

There are many approximate calculations using the Kubo formula. A common approximation is to compute the one-loop diagram with dressed electron propagators (which include a finite width in the fermion self energy). This will give a non-zero conductivity, controlled by the fermion quasi-particle lifetime.

P.S.: I think the calculation can be found in https://arxiv.org/abs/hep-ph/0209048

Thomas
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