1
$\begingroup$

I am deriving Kubo formula using Kubo identity and I am confused that how does the article perform the following steps. On page 8, we have a integration $$ I\equiv\int_0^\beta d\lambda Tr\bigg\{\rho_0 J_\mu (t-i\hbar \lambda ) J_\nu(0) \bigg\} \tag{0} $$ here $\rho_0$ is density matrix operator and $J_i$ are current density operators. They perform the integration through the following steps:

"By contour integration tricks:" $$ I =\frac{i}{\hbar} \int_t^{t-i\hbar\beta} d\tau Tr\bigg\{\rho_0 J_\mu (\tau) J_\nu(0) \bigg\} \tag{1} $$ $$ =\frac{i}{\hbar} \int_t^{\infty} dt'\; Tr\bigg\{\rho_0 \bigg(J_\mu (t')-J_\mu (t'-i\hbar\beta)\bigg) J_\nu(0) \bigg\} \tag{2} $$ $$ =\frac{i}{\hbar} \int_t^{\infty} dt'\; Tr\bigg\{\rho_0 \bigg(J_\mu (t')J_\nu(0)-J_\nu(0)J_\mu (t')\bigg) \bigg\} \tag{3} $$ $$ =\frac{i}{\hbar} \int_t^{\infty} dt'\; Tr\bigg\{\rho_0 [J_\mu (t'),J_\nu(0)] \bigg\} \tag{4} $$ I know how they get $(1)$ from $(0)$. I need help how they get $(2)$ and then $(3)$ from $(1)$? At the end of this equation set they write "where we assumed that the integrand is analytical."

I have tried everything but I could not find any way to perform step $2$ and $3$. A little starting point will be highly appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ Your integration limits do not coincide with those of the reference, no? $\endgroup$ Dec 15, 2021 at 21:11
  • $\begingroup$ @Jakob sorry, that was a typo. Thank you so much $\endgroup$ Dec 15, 2021 at 21:13

1 Answer 1

5
$\begingroup$

Here's what I'm pretty sure they are doing:

enter image description here

As in the picture they say that integration $\int_t^{t-i\hbar\beta}+\int_{t-i\hbar\beta}^{\infty-i\hbar\beta}+\int_{\infty-i\hbar\beta}^{\infty}+\int_{\infty}^{t}=0$. Then they argue that the correlation function vanishes at infinity so the third contour at real infinity is dropped. Rearranging we get to $(2)$.

To get to $(3)$ notice $$\text{Tr}\left(e^{-\beta H}J(t'-i\hbar \beta)J(0)\right)=\text{Tr}\left(e^{-\beta H}e^{+\beta H}J(t')e^{-\beta H}J(0)\right)=\text{Tr}\left(J(t')e^{-\beta H}J(0)\right)=\text{Tr}\left(e^{-\beta H}J(0)J(t')\right).$$ There may be some subtlety with the difference between $H_0$ and $H$ I overlooked, but I am pretty sure this is the manipulation they are doing.

$\endgroup$
1
  • 1
    $\begingroup$ thank you so much. $\endgroup$ Dec 16, 2021 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.