# Kubo identity (electrical conductivity) integration

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.

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

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.