# Kubo Transform Expression for the Time Correlation Function


We shall derive the following expression for the quantum time correlation function

$\Phi_{AB}(t) = \int_0^{\beta}d\lambda\;\langle \dot{B}(-i\hbar\lambda)A(t)\rangle _0$

known as a Kubo transform relation. Since $$\underline {\dot{B} }$$ is given by the Heisenberg equation:

$\dot{B} = {1 \over i\hbar}[B,H_0]$

it follows that

$\dot{B}(t) = -{1 \over i\hbar}e^{iH_0t/\hbar}[H_0,B(0)]e^{-iH_0t/\hbar}$

Evaluating the expression at $$t=-i\hbar\lambda$$ gives

$\dot{B}(-i\hbar\lambda) = e^{\lambda H_0}{1 \over i\hbar}[B(0),H_0]e^{-\lambda H_0}$

Thus,

$\Phi_{AB}(t) = \int_0^{\beta} d\lambda \langle e^{\lambda H_0} \left ({1 \over i\hbar}[B(0),H_0]\right)e^{-\lambda H_0}A(t)\rangle _0$

By performing the trace in the basis of eigenvectors of $$H_0$$, we obtain

\begin{align*} \Phi_{AB}(t) &= {1 \over Q}\int_0^{\beta}d\lambda\sum_n \langle n\vert e^{\lambda H_0} \left ({1 \over i\hbar}\right)[B(0),H_0]e^{-\lambda H_0}A(t)\vert n\rangle e^{-\beta E_n} \\[4pt] &= {1 \over Q}\int_0^{\beta}d\lambda\sum_{m,n} \langle n\vert e^{\lambda H_0}\left ( {1 \over i \hbar } \right ) \left [ B (0), H_0 \right ] e^{-\lambda H_0}\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \\[4pt] &= {1 \over Q}\int_0^{\beta}d\lambda\sum_{m,n} e^{\lambda E_n}e^{-\lambda E_m} {1 \over i \hbar } \langle n \vert [B(0),H_0]\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \\[4pt] &= {1 \over Q}\sum_{m,n} e^{-\beta E_n}{e^{\beta(E_n-E_m)}-1 \over (E_n - E_m) } {1 \over i\hbar } \langle n\vert[B(0),H_0]\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \end{align*}

But

$\langle n\vert[B(0),H_0]\vert m\rangle = \langle n\vert B(0) H_0 - H_0 B (0)\vert m\rangle = (E_m-E_n)\langle n\vert B(0)\vert m\rangle$

Therefore,

\begin{align*} \Phi_{AB}(t) &= -{1 \over i\hbar Q}\sum_{m,n}\left(e^{-\beta E_n}-e^{-\beta E_m}\right)\langle n\vert B(0)\vert m\rangle \langle m\vert A(t)\vert n\rangle \\[4pt] &= -{1 \over i\hbar Q}\left[\sum_{m,n}e^{-\beta E_m}\langle m\vert A (t) \vert n \rangle \langle n \vert B (0) \vert m \rangle - \sum _{m,n} e^{-\beta E_n}\langle n\vert B(0)\vert m\rangle \langle m\vert A(t)\vert n\rangle \right] \\[4pt] &= \underline { {i \over \hbar}\langle [A(t),B(0)]\rangle _0 } \end{align*}

which proves the relation. The classical limit can be deduced easily from the Kubo transform relation:

$\Phi_{AB}(t) \longrightarrow \beta\langle \dot{B}(0)A(t)\rangle _0$

Note further, by using the cylic properties of the trace, that

$\langle \dot{B}(-i\hbar\lambda)B(t)\rangle _0 = -{d \over dt}\langle B(-i\hbar\lambda)B(t)\rangle _0$