When deducing Markovian quantum master equation, supposing the total Hamiltonian is the following form:


where $H_{S}$ is the Hamiltonian for the quantum system, $H_{B}$ is for the reservoir, and $H_{I}$ is the interaction between quantum system and the reservoir.

And $H_{I}$ takes the form $H_{I}=\sum_{\alpha} A_{a}\otimes B_{\alpha}$, with $B_{\alpha}$ acting on the reservoir.

In the deduction, $<B_{\alpha}^{\dagger}(s)B_{\alpha}(0)>$ is assumed to decay sufficiently fast.

Now here comes the problem:

In the book The Theory of Open Quantum Systems written by Breuer, the author says," A rapid decay of the reservoir correlations requires a continuum of frequencies: For an infinitely small frequency spacing Poincare recurrence times becomes infinite and irreversible dynamics can emerge".

So my question is: $<B_{\alpha}^{\dagger}(s)B_{\alpha}(0)>$ decays very fast, meaning the hot bath lost the information from the quantum system and revert back to equilibrium fast. So this seems just to be opposite to the claim, Poincare recurrence times be infinitely long, made by the author. Could anyone explain to me about this?


Here the OP has confused two very different timescales: the recurrence time $t_r$ and the correlation time $t_c$. To get Markovian dynamics, we need $t_r$ to be very large and $t_c$ to be very small in comparison to all other timescales relevant for the open system. There is no contradiction in these two requirements, which have very different physical meaning. The recurrence time is the timescale for the global system to revert back to its original state. Meanwhile, the correlation time is the timescale for the local degrees of freedom of the bath to relax back to their original configuration after a perturbation.

A simple physical example might help to clarify this. Let the open system $S$ be a two-level atom, which interacts with an environment $B$ comprising the zero-temperature electromagnetic field. The atom starts in its excited state and undergoes spontaneous emission into the field. In particular, the atom interacts with a certain field mode $\hat{B} = \sum_k g_k \hat{b}_k$ via the interaction Hamiltonian $$ \hat{H}_I = \hat{\sigma}_+ \hat{B} + \hat{\sigma}_-\hat{B}^\dagger.$$ Roughly speaking, a photon emitted in the state $\hat{B}^\dagger|0\rangle$ evolves into the state ${\rm e}^{-{\rm i}\hat{H}_Bt/\hbar}\hat{B}^\dagger|0\rangle$, i.e. the photon flies away from the atom. Since photons fly very fast, the evolved state very quickly becomes orthogonal to the initial state. In particular, we expect the overlap $$\langle 0| \hat{B}{\rm e}^{-{\rm i}\hat{H}_Bt/\hbar}\hat{B}^\dagger|0\rangle = \langle 0| \hat{B}(t)\hat{B}^\dagger(0)|0\rangle,$$ where $\hat{B}(t) = {\rm e}^{{\rm i}\hat{H}_Bt/\hbar}\hat{B}{\rm e}^{-{\rm i}\hat{H}_Bt/\hbar}$, to decay rapidly to zero over some correlation time $t_c$. The fact that $t_c$ is very small means that the local electric field around the atom relaxes very quickly in response to a perturbation, because the speed of light is large and atoms are very small, so the emitted photons fly outside of the atom's sphere of influence almost immediately.

This is quite different from the phenomenon of recurrence. To see recurrence in our example, let us imagine that the entire system is placed inside a spherical, perfectly reflecting cavity of radius $R$. Because of the cavity, the electromagnetic field modes can only take on discrete frequencies separated by $c/R$. This implies a Poincare recurrence time of order $t_r \sim (c/R)^{-1}$. In fact, the recurrence time is exactly $2R/c$, which is the time taken for a spontaneously emitted photon to reflect from the cavity walls, fly back to the atom and be re-absorbed. This returns the entire system to its original state, not just the atom's local electric field. As $R\to \infty$, the spacing between frequencies tends to zero and the recurrence time $t_r\to\infty$. The correlation time, however, remains small.

  • $\begingroup$ Thanks for your clarification. Now I understand the differences between these two. But another question arises, why do we not want the $\italics{entire}$ system to revert to its original state? Will this will imply the information from the atom will go back to the atom after $\bold{\italics{t_{r}}}$, hence breaking the definition for Markovian? (Meanwhile slow decay during $\bold{\italics{t_{c}}}$ means the system can be affected by near history?) $\endgroup$ – xiang sun Mar 13 '19 at 13:29

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.