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`Secular approximation'

We note that $ \hat{C}(z)= \hat{C}_{12}(z)+\hat{C}_{21}(z)$. In the secular approximation, one sets
$\displaystyle \hat{C}_{12}(i\Omega)$ $\displaystyle \equiv$ $\displaystyle \int_0^{\infty}
d\omega \rho(\omega) e^{-i\omega t}e^{-i\Omega t}(1+n_B(\omega)) \rightarrow 0$  
$\displaystyle \hat{C}_{21}(-i\Omega)$ $\displaystyle \equiv$ $\displaystyle \int_0^{\infty}
d\omega \rho(\omega) e^{i\omega t}e^{i\Omega t} n_B(\omega) \rightarrow 0.$ (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,
$\displaystyle c_-+c_+$ $\displaystyle \approx$ $\displaystyle \frac{1}{2}(\gamma_++\gamma) = \pi\rho(\Omega)[1+2n_B(\omega)]$  
$\displaystyle c_--c_+$ $\displaystyle \approx$ $\displaystyle \frac{1}{2}(\gamma_+-\gamma) = \pi\rho(\Omega).$ (66)


next up previous contents index
Next: - Representation Up: Derivation of Master equation Previous: Derivation of Master equation   Contents   Index
Tobias Brandes 2004-02-18