The Trace

To extract $\rho_s$ we now need to trace over the environmental variables, i.e. average out the environmental degrees of freedom. We can most quickly see that,

$$Tr_e[H_i(t),\rho_{se}]=0$$ (4.1)

as $H_i$ is linear with $b$ and $b^{\dagger}_j$, $\rho_{se}=\rho_s rho_e$, and it is self evident that

$$Tr_{e_j}b_j\exp(-\hbar \omega_j b_j^{\dagger}b_j/kT)=0$$

Thus we are left with,

$$\frac{d\rho_s(t)}{dt}=-\frac{1}{\hbar^2}\int^t_0 Tr_e\left[H_i(t),[H_i(t'),\rho_{se}(t')]\right]dt'$$ (4.2)

To proceed from here will require some care.