Case B

$$H_i(t)\rho_{se}(t')H_i(t')=\hbar^2\left(G(t)a^{\dagger}+G^{\dagger}(t)a\right)\rho_{se}(t')\left(G(t')a^{\dagger}+G^{\dagger}(t')a\right)$$

Based on arguments of Case B, we consider, $\alpha =\hbar^2a^{\dagger}\rho_s(t')a$, $\beta =\hbar^2 a\rho_s(t') a^{\dagger}$,

$$\hbar^2 a^{\dagger}\rho_s(t')aG(t)\rho_{e}G^{\dagger}(t') \ \righ... ...\left(\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right) b_j^{\dagger} \right\}$$

$$\hbar^2 a\rho_s(t')a^{\dagger}G^{\dagger}(t)\rho_{e}G(t') \ \righ... ...gger} \exp\left(\frac{-b_i^{\dagger}b_i\hbar \omega_i}{kT}\right) b_j \right\}$$

To these calculations we simply must note that,

$$\langle n \vert b e^{-b^{\dagger}b}b^{\dagger}\vert n\rangle=(n+1)\langle n+1\vert e^{-b^{dagger}b}\vert n+1\rangle$$

Thus 4.3 becomes,

$$\left(1-e^{-\lambda}\right)e^{-\lambda}\sum_n (n+1)e^{-\lambda n} = e^{-\lambda}(1+\bar{n})$$

We notice that,

$$\frac{1}{e^{\lambda}-1}=\bar{\bar{n}} \ \rightarrow \frac{1+\bar{n}}{\bar{n}}=e^{\lambda}, \ \ \frac{\bar{n}}{1+\bar{n}}=e^{-\lambda}$$ (4.7)

$$\hbar^2 a^{\dagger}\rho_s(t')aG(t)\rho_{e}G^{\dagger}(t') \ \rightarrow \ \alpha \sum_jg_j^2\bar{n}_j\gamma_j$$

$$\hbar^2 a\rho_s(t')a^{\dagger}G^{\dagger}(t)\rho_{e}G(t') \ \rightarrow \ \beta \sum_jg_j^2(1+\bar{n}_j)\gamma_j^{\dagger}$$