Case A

$$H_i(t)H_i(t')\rho_{se}(t') =$$

$$\hbar^2\left(G(t)a^{\dagger}+G^{\dagger}(t)a\right)\left(G(t')a^{... ...i}{kT}\right)\right)\exp\left(\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right)$$

The first term we can extract is,

$$\hbar^2\left(G(t)a^{\dagger}G^{\dagger}(t')a\right)\rho_s(t')\pro... ...i}{kT}\right)\right)\exp\left(\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right)$$

We recall that

$$G(t)=\sum_j g_jb_j\exp(i(\omega-\omega_j)t)$$

so we can rewrite the term as (with $\zeta=\hbar^2a^{\dagger}a\rho_s(t')$ and $\chi_i = \left[1-\exp\left(\frac{-\hbar \omega_i}{kT}\right)\right]$),

$$\zeta\left[\sum_jg_jb_j\exp(i(\omega-\omega_j)t)\right] \left[\su... ...ight] \prod_i \chi_i\exp\left(\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right)$$

It is easily seen that when we take the trace, all terms will involve $b_jb^{\dagger}_k$ and will vanish unless $j=k$, so we can write the term as,

$$\zeta\left[\sum_jg_j^2b_jb_j^{\dagger}\exp(i(\omega-\omega_j)(t-t... ...right]\prod_i \chi_i\exp\left(\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right)$$

But again, we can use the same argument to write the term as,

$$\zeta \sum_j\chi_jg_j^2b_jb_j^{\dagger}\exp(i(\omega-\omega_j)(t-t'))\exp\left(\frac{-b_j^{\dagger}b_j\hbar \omega_j}{kT}\right)$$

We can now easily take the trace, noting that $Tr[B\rho] = <B>$, and with $n=b^{\dagger}b$ with $[b,b^{\dagger}]=1$.

$$Tr\left\{\zeta \sum_j\chi_jg_j^2b_jb_j^{\dagger}\exp(i(\omega-\omega_j)(t-t'))\exp\left(\frac{-b_j^{\dagger}b_j\hbar \omega_j}{kT}\right)\right\}$$

$$=\zeta \sum_jg^2_j(1+\bar{n}_j)\exp(i(\omega-\omega_j)(t-t'))$$

For the reader interested in the total details of the previous calculation, [1] is a good reference, where it is also demonstrated that, for example,

$$\frac{\sum_0^{\infty}\langle n \vert b^{\dagger}be^{-\lambda b^{\... ...1-e^{-\lambda})\sum_0^{\infty}ne^{-\lambda n} = \frac{1}{e^{\lambda}-1}=\bar{n}$$ (4.3)

In any case, we now have a general formula we can use, that is,

$$\langle G(t)G(t')\rangle = \langle G^{\dagger}(t)G^{\dagger}(t')\rangle=0$$ (4.4)

$$\langle G(t)G^{\dagger}(t')\rangle = \sum_j(1+\bar{n}_j)\exp(i(\omega-\omega_j)(t-t'))$$ (4.5)

$$\langle G^{\dagger}(t)G(t')\rangle = \sum_j(\bar{n}_j)\exp(-i(\omega-\omega_j)(t-t'))$$ (4.6)

The other non zero trace will be, with $\eta=\hbar^2 a a^{\dagger}\rho_s(t')$ and $\gamma_j=\exp(i(\omega-\omega_j)(t-t'))$

$$Tr\left[\eta \left(G(t)^{\dagger}G(t')\right)\prod_i \chi_i\exp\l... ...\hbar \omega_i}{kT}\right)\right] = \eta \sum_jg_j^2\bar{n}_j\gamma_j^{\dagger}$$