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When deriving a Lindblad equation (for example Breuer chapter 3), one crucial assumption is that $\tau_b$, the reservoir correlation function decay time, is (in short) the smallest relevant time scale. What I cannot understand is why these correlation decay at all.

If the bath is composed, for example, of uncoupled harmonic oscillators, and the system is so much smaller than the environment to the point that we assume its state is constant, wouldn't these correlation functions simply oscillate? If not, what determines the decay time?

My calculations of what I understand are the bath correlations in this case might illustrate my misunderstanding. With the interaction to the system being $B_\omega = (a+a^\dagger)/2 \equiv X$:

$$ B_\omega(t) = X_\omega (t) = \cos(\omega t) X + \sin(\omega t) P $$ $$ \langle B_\omega (t) B_\omega(0)\rangle = \cos(\omega t ) (\langle a^\dagger a \rangle + 1/2) +\kappa = \cos(\omega t ) (n(\omega,T) + 1/2) $$ where $\kappa$ are terms with zero average that I didn't write for clarity, and $n$ is the Bose Einstein distribution.

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A single harmonic oscillator will not act as a bath. You have to consider a continuum of oscillator frequencies, so that your bath correlation becomes $$C(t) = \int d\omega\,\langle B_\omega(t) B_\omega(0)\rangle.$$ Mathematically, this will clearly decay even though each individual $\langle B_\omega(t) B_\omega(0)\rangle$ oscillates forever. The timescale of decay is set by the (inverse of the) reservoir bandwidth, i.e. the range of frequencies over which the integrand is non-zero. Physically, the reason for decay is that the open system does not couple to a single eigenmode of the bath but rather generates a highly non-equilibrium excitation that "decays" by propagating away into the bath. For example, an atom excites the radiation field by generating a relatively localised photon. The photon then very quickly flies away. Obviously, the photon itself never disappears (assuming a perfect harmonic bath = free radiation field, i.e. no other sources), it just keeps flying away. But from the atom's perspective, the photon has disappeared and the bath's memory of the initial excitation is lost.

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  • $\begingroup$ Makes sense. I've worked out an example where I get an analytical expression which exponentially decays, and I think it would be nice for other people to see. Should I make another answer or edit this one (or edit my question with the example)? (not familiar with stack's best practices) $\endgroup$
    – peep
    Nov 16, 2022 at 11:48
  • $\begingroup$ @peep Go ahead and post it as a separate answer! Answering your own questions is actively encouraged. $\endgroup$ Nov 16, 2022 at 12:53

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