- #1
physengineer
- 21
- 0
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!
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!