Conductivity from path integral and Kubo formlism

Hi,

In calculating the conductivity from the Kubo method

$$j_{\mu}=\int dx' K_{\mu \nu} (x,x') A^{\nu}(x')$$

in literature ( e.g. in Condensed Matter Field Theory by Altland and Simons) you find that

$$K_{\mu \nu}(x,x')= Z^{-1} \frac{\delta^2}{\delta A_{\mu}(x) \delta A_{nu}(x')} Z[A] |_{A=0}$$

Now, I have the following questions:

1-Why do I need to put $A=0$? I guess we take the derivatives to find current-current correlation but current can depend on $A$ itself, so why do we put it to zero?

2- Is this $A$ quantum or the classical (background)?

3-If I have a $Z$ with an effective action of the form:

[tex]
Z=\int D[A] D[\psi] \exp{(-S_E[A,\psi])}
[/itex]

Then what does it mean to put $A=0$? At what stage should I put $A=0$. Do I kill the path integral over $A$?

DrDu
1. Is a consequence of using linear response in contrast to non-linear response. If K would itself depend on A, then j wouldn't be linear in A anymore.
2. You can expand Z in powers of hbar. So in principle you can get not only the quantum mechanical response but also the classical.
However, A is treated by definition as a classical external source. This may only be true for A with sufficiently low k values. This may also answer partially your question 3.

Thanks a lot, DrDU! I am just not sure if I understood exactly what you meant for (2)

2. You can expand Z in powers of hbar. So in principle you can get not only the quantum mechanical response but also the classical.
However, A is treated by definition as a classical external source. This may only be true for A with sufficiently low k values. This may also answer partially your question 3.

My effective action is complex and non local but fortunately remains quadratic in terms of $A$ and $\psi$ or combination of these two. Therefore I can technically integrate over both $A$ and $\psi$ in the path integral. So after differentiating two times in terms of $\delta^2 / \delta A(x)\delta A(x')$ and setting $A=0$ I arrive at

[tex]
K_{\mu \nu}(x,x')= Z^{-1}\int D[A] D[\psi] f[\psi;x,x'] \exp{(-S_E[A,\psi])}
[/itex]

Assuming I can integrate it classically, as if it is just an average over free energy instead of effective action. The $A$ is not background and it is my gauge field. Would that give me a correct result?

I would appreciate it if you could tell me that or explain your last comment with more details.

Thank you!

DrDu