I am trying to calculate the conductivity in the linear response regime of a disordered electron gas. (or eventually of a mean field Heavy fermion system with known one particle green's functions).

I trying to use method in "Quantum field theory of non-equilibium states" by J.Rammer. Specifically I'm looking at section 6.2. Honestly, I don't understand very much of what he is doing. Have somebody done those calculations and could give guidelines on how to think and perhaps some more steps in the derivations?

My question in short would be: If I know the known free Green's functions, how do I calculate linear response when applying an electrical field?

  • 1
    $\begingroup$ Rammer's book isn't quite what you need. In the linear response regime you relate the response to some applied field to equilibrium properties of the system absent the field. While the full theory of the non-equilibrium Green's functions will also let you calculate what you need, what you ought to be looking for is something called the "Kubo formula". $\endgroup$
    – wsc
    Aug 2, 2012 at 4:25
  • $\begingroup$ I have seen the Kubo formula. But it would be nice if one could derive the Kubo formula from the linear regime of the Dyson equations. The current can, as far as i know, be expressed in terms of the kinetic (Keldysh) Green's function. I don't understand how though. $\endgroup$
    – Garvan
    Aug 2, 2012 at 8:24
  • $\begingroup$ A solid reference and a very clean derivation of Kubo in just this context is provided by Eduardo Fradkin in his Field Theory of Condensed Matter Physics, CUP. The second edition just came out. Much of this material is freely available on his web site at UIUC, but the book is very comprehensive. $\endgroup$
    – user27777
    Aug 2, 2013 at 7:34

1 Answer 1


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?

  1. 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 $$\sigma_{\mu\nu}= \frac{\delta j_\mu}{\delta E_\nu}....(2)$$
  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).
  3. 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 $$\sigma_{\mu\nu}=-i\delta_{k_0}\delta_{k_\nu}\delta_{k_\mu}S....(4)$$
  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.


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.

  • $\begingroup$ Nice deduction. I want to ask is there any reference related to your deduction? I know in the article of xiaoliang qi and Bernevig's book, the conductivity is related to the expression: $\sum_{m,k}Tr\left[\frac{\partial h(k)}{\partial k_{i}} G^T(k-\frac{q}{2},\omega_{m}) \frac{\partial h(k)}{\partial k_{j}}G^T(k+\frac{q}{2},\omega_{m}-\nu_r)\right]$, which is different from yours. I want to know the relationship between this expression and yours. Thank you very much! $\endgroup$
    – fbs147
    Jan 26, 2022 at 3:21
  • $\begingroup$ @fbs147 That expression is the current-current correlation, which differs from the conductivity by another frequency derivative. $\sigma_{\mu\nu}(\omega)=\partial_\omega\langle j_\mu(-\omega) j_\nu(\omega)\rangle$ $\endgroup$ Jan 27, 2022 at 23:20
  • $\begingroup$ Thank you. I know it is the current-current correlation, I want to know how to get your conductivity expression from my expression? To do a frequency derivative like what you have written? Could you write more detailed? Thank you very much indeed! $\endgroup$
    – fbs147
    Jan 28, 2022 at 6:00
  • $\begingroup$ @fbs147 yes. take their expression for the current-current correlation, then take the frequency derivative, then you obtain the conductivity that I wrote. Hints: $\partial_\omega G = G \gamma^0 G$ $\endgroup$ Jan 29, 2022 at 10:40
  • $\begingroup$ Thank you. I want to confirm that 1. in my expression, we should take the frequency $\nu_r$ derivative, not frequency $\omega_m$. 2. $\frac{\partial h(k)}{\partial k_i}$ can be written as $-\frac{\partial G^{-1}}{\partial k_i}$, which is independent of frequency. $\endgroup$
    – fbs147
    Jan 29, 2022 at 12:02

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