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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,

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

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

Thus we are left with,

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

To proceed from here will require some care.


Timothy Jones 2006-10-11