I have been trying to understand the proof of the quantum regression theorem or formula for a Lindblad evolution.

If $V(t,0)$ is the propagator, $\rho_S(t) = V(t,0)\rho_S(0)$, the formula is: $\langle B(t) C(s) \rangle = \text{Tr}_S[BV(t,s)C\rho_S(s)]$ for abitrary operators B and C.

In Breuer & Petruccione there is no actual explanation, which might be because it is a natural consequence of the semigroup property of the propagator. However, when I try to prove it I get this:

\begin{equation} \text{Tr}_s[B(t) C(s) \rho_S(0)] = \text{Tr}_s[BV(t,0)\{V^\dagger (s,0)(C) \rho_S(0)\}] = \text{Tr}_S[BV(t,s)V(s,0)\{V^\dagger (s,0)(C) \rho_S(0)\}]. \end{equation} But then I don't know how to show that $V(s,0)\{V^\dagger (s,0)(C) \rho_S(0)\}=C\rho_S(s)$. Does anyone has any suggestion?

I find the explanation given in the book of Carmichael, Statistical Methods in Quantum Optics 1, kind of unsatisfactory. It relies on the approximation $\rho_\text{tot}(t) \approx \rho_S(t) \otimes \rho_E$ for all times. Even though this approximation is often used in the microscopic derivation of Lindblad master equations (e.g. in Bruer & Petruccione) it is not needed and somehow misleading, as can be shown with the Zwanzig projector technique (Markovian master equations: A critical study).

  • $\begingroup$ The Onsager-Lax regression theorem is shown explicitly in Scully-Zubairy(Quantum Optics) section 10.4 for a specific case. This may be clearer. $\endgroup$ – Wouter Sep 21 '18 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.