Kubo transform expression for the time correlation function (Optional)

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 $\dot{B}$ 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,

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

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

$$\Phi_{AB}(t) = {1 \over Q}\int_0^{\beta}d\lambda \sum_n \langle n\vert e^{\lambd... ...over i\hbar}\right)[B(0),H_0]e^{-\lambda H_0}A(t) \vert n\rangle e^{-\beta E_n}$$

$$= {1 \over Q}\int_0^{\beta}d\lambda \sum_{m,n} \langle n\vert e^{\l... ...{-\lambda H_0} \vert m\rangle \langle m\vert A(t) \vert n\rangle e^{-\beta E_n}$$

$$= {1 \over Q}\int_0^{\beta}d\lambda \sum_{m,n} e^{\lambda E_n}e^{-\... ...\vert[B(0),H_0]\vert m\rangle \langle m\vert A(t) \vert n\rangle e^{-\beta E_n}$$

$$= {1 \over Q} \sum_{m,n} e^{-\beta E_n} {e^{\beta(E_n-E_m)}-1 \over... ...\vert[B(0),H_0]\vert m\rangle \langle m\vert A(t) \vert n\rangle e^{-\beta E_n}$$

But

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

Therefore,

$$\Phi_{AB}(t) = -{1 \over i\hbar Q}\sum_{m,n}\left(e^{-\beta E_n}-e^{-\beta E_m}\right) \langle n\vert B(0)\vert m\rangle \langle m\vert A(t)\vert n\rangle$$

$$= -{1 \over i\hbar Q}\left[\sum_{m,n}e^{-\beta E_m}\langle m\vert A... ...E_n}\langle n\vert B(0)\vert m\rangle \langle m\vert A(t)\vert n\rangle \right]$$

$$= {i \over \hbar}\langle [A(t),B(0)]\rangle _0$$

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

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