Some context: I am deriving the Lindblad equation following “The Theory of Open Quantum Systems” by Breuer and Petruccione (somebody transcribed the section I am reading in this link).

My question: I am stuck understanding the so-called “secular approximation”, i.e. neglecting the “rapidly oscillating” terms with $\omega' \neq \omega$ in $$ \frac{d}{dt} \rho_s(t) = \sum_{\alpha, \beta} \sum_{\omega’,\omega} e^{i t (\omega’ -\omega)} \Gamma_{\alpha \beta}(\omega)\left(A_{\beta}(\omega)\rho_s(t) A^{\dagger}_{\alpha} (\omega’)-A^{\dagger}_{\alpha}(\omega’) A_{\beta}(\omega) \rho_s(t)\right) + h.c. $$ so that it becomes $$ \frac{d}{dt} \rho_s(t) = \sum_{\alpha, \beta} \sum_{\omega} \Gamma_{\alpha \beta}(\omega)\left(A_{\beta}(\omega)\rho_s(t) A^{\dagger}_{\alpha} (\omega)-A^{\dagger}_{\alpha}(\omega) A_{\beta}(\omega) \rho_s(t)\right) + h.c. \quad . $$ I am new to the field of open quantum systems so I missed Breuer and Petruccione’s arguments of relaxation times being small and terms being rapidly oscillating.

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    $\begingroup$ You may want to consider a simpler case first, as in this question. $\endgroup$ – Anyon Mar 1 at 18:26
  • $\begingroup$ @Anyon , thanks for the link! Following the gist of that post, the argument would be that $$ \rho_s(t) = \sum_{\alpha, \beta} \sum_{\omega’,\omega} \Gamma_{\alpha \beta}(\omega)\left(A_{\beta}(\omega)\int dt e^{i t (\omega’ -\omega)} \rho_s(t) A^{\dagger}_{\alpha} (\omega’)-A^{\dagger}_{\alpha}(\omega’) A_{\beta}(\omega) \int dt e^{i t (\omega’ -\omega)} \rho_s(t)\right) + h.c.,$$ integrals like $$\int dt e^{i t (\omega’ -\omega)} \rho_s(t)$$ are much smaller when $\omega’ \neq \omega$ than when $\omega’ = \omega$ provided (?) $\rho_s(t)$ is slow changing in comparison to the exponentials. $\endgroup$ – FriendlyLagrangian Mar 2 at 8:56

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