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