Perturbative solution of the Liouville equation

As in the classical case, we assume a solution of the form

$$\rho(t) = \rho_0(H_0) + \Delta \rho(t)$$

where

$$[H_0,\rho_0]=0\;\;\;\;\;\Rightarrow\;\;\;\;\;{\partial \rho_0 \over \partial t}=0$$

and we will assume

$$\rho_0(H_0) = {e^{-\beta H_0} \over Q(N,V,T)}$$

Substituting into the Liouville equation and working to first order in small quantities, we find

$${\partial \Delta \rho \over \partial t} = {1 \over i\hbar}[H_0,\Delta \rho] - {1 \over i\hbar} [B,\rho_0]F_e(t)$$

which is a first order inhomogeneous equation that can be solved by using an integrating factor:

$$\Delta \rho(t) = -{1 \over i\hbar}\int_{-\infty}^t\;ds\;e^{-iH_0(t-s)/\hbar}[B,\rho_0]e^{iH_0(t-s)/\hbar} F_e(s)$$

(Note that we have chosen the origin in time to be $t=-\infty$, which is an arbitrary choice.)

For an observable $A$, the expectation value is

$$\langle A(t)\rangle = {\rm Tr}(\rho A) = \langle A\rangle _0 + {\rm Tr}(\Delta \rho(t) A)$$

when the solution for $\Delta \rho$ is substituted in, this becomes

$$\langle A(t) \rangle = \langle A \rangle _0 - {1 \over i\hbar} \int_{-\infty}^t\;ds\;{\rm Tr}\left[Ae^{-iH_0(t-s)/hbar}[B,\rho_0]e^{iH_0(t-s)/\hbar}\right]F_e(s)$$

$$= \langle A \rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[e^{iH_0(t-s)/\hbar} Ae^{-iH_0(t-s)/hbar} [B,\rho_0]\right]F_e(s)$$

$$= \langle A\rangle _0 - {1 \over i\hbar} ds\;{\rm Tr}\left[A(t-s)[B,\rho_0]\right]F_e(s)$$

where the cyclic property of the trace has been used and the Heisenberg evolution for $A$ has been substituted in. Expanding the commutator gives

$$\langle A(t) \rangle = \langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\; {\rm Tr}\left[A(t-s)B\rho_0 - A(t-s)\rho_0B\right]F_e(s)$$

$$= \langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\; {\rm Tr}\left[\rho_0\left(A(t-s)B - BA(t-s)\right)\right]F_e(s)$$

$$= \langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\; F_e(s)\langle [A(t-s),B(0)]_0\rangle$$

where the cyclic property of the trace has been used again. Define a function

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

called the after effect function. It is essentially the antisymmetric quantum time correlation function, which involves the commutator between $A(t)$ and $B(0)$. Then the linear response result can be written as

$$\langle A(t)\rangle = \langle A \rangle _0 + \int_{-\infty}^t\;ds F_e(s)\Phi_{AB}(t-s)$$

which is the starting point for the theory of quantum transport coefficients. If we choose to measure the operator $B$, then we find

$$\langle B(t)\rangle = \langle B\rangle _0 + \int_{-\infty}^t\;ds\;F_e(s)\Phi_{BB}(t-s)$$