`Secular approximation'

We note that $\hat{C}(z)= \hat{C}_{12}(z)+\hat{C}_{21}(z)$. In the secular approximation, one sets

$$\hat{C}_{12}(i\Omega) \equiv \int_0^{\infty} d\omega \rho(\omega) e^{-i\omega t}e^{-i\Omega t}(1+n_B(\omega)) \rightarrow 0$$

$$\hat{C}_{21}(-i\Omega) \equiv \int_0^{\infty} d\omega \rho(\omega) e^{i\omega t}e^{i\Omega t} n_B(\omega) \rightarrow 0.$$ (65)

The real parts of $\hat{C}_{12}(i\Omega)$ and $\hat{C}_{21}(-i\Omega)$ are zero because $\delta(\omega+\Omega)$ yields no contribution from the integral (remember that $\Omega>0$). This approximation therefore neglects the imaginary parts of $\hat{C}_{12}(i\Omega)$ and $\hat{C}_{21}(-i\Omega)$ which, however, do not lead to damping but only to a renormalisation of the system Hamiltonian $H_S$. For consistency, we therefore neglect the imaginary parts of $\hat{C}_{12}(-i\Omega)$ and $\hat{C}_{21}(i\Omega)$ as well. Therefore,

$$c_-+c_+ \approx \frac{1}{2}(\gamma_++\gamma) = \pi\rho(\Omega)[1+2n_B(\omega)]$$

$$c_--c_+ \approx \frac{1}{2}(\gamma_+-\gamma) = \pi\rho(\Omega).$$ (66)