Conductivity from path integral and Kubo formlism

  • #1
Hi,

In calculating the conductivity from the Kubo method

[tex]
j_{\mu}=\int dx' K_{\mu \nu} (x,x') A^{\nu}(x')
[/tex]

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

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

Now, I have the following questions:

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

2- Is this [itex] A [/itex] quantum or the classical (background)?

3-If I have a [itex]Z[/itex] 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 [itex] A=0[/itex]? At what stage should I put [itex]A=0[/itex]. Do I kill the path integral over [itex]A[/itex]?

Thank a lot in advance!
 

Answers and Replies

  • #2
DrDu
Science Advisor
6,210
866
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.
 
  • #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 [itex]A[/itex] and [itex]\psi[/itex] or combination of these two. Therefore I can technically integrate over both [itex]A[/itex] and [itex]\psi[/itex] in the path integral. So after differentiating two times in terms of [itex]\delta^2 / \delta A(x)\delta A(x') [/itex] and setting [itex]A=0[/itex] 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 [itex]A[/itex] 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!
 
  • #4
DrDu
Science Advisor
6,210
866
I am not a specialist in path integrals so I don't know whether I can help you too much.
But the A's in the formula for Conductivity are due to external sources. In the case of longitudinal response you could alternatively assume an external classical charge distribution. So you somehow have to split the field into a part which you want to describe classically and a quantum mechnanical part which you integrate over in your path integral.
You could also try to shift the boundary between the classical and qm modes. This leads to a kind of renormalization procedure.
The following article may be interesting, although it uses the traditional formulation and not path integrals:
http://www.informaworld.com/smpp/content~db=all?content=10.1080/00018736100101281
 
  • #5
I am not...

Thanks, DrDu! I appreciate your help.
 

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