Expansion of the interaction term in the microscopic derivation of the Lindblad equation

Question

I'm reading The Theory of Open Quantum Systems by Heinz-Peter Breuer and Francesco Petruccione, and in chapter 3, I can't understand the decomposition of the interaction Hamiltonian:

We assume that $H=H_S + H_B + H_I$, where we assume that $$ H_I = \sum\limits_{\alpha}A_{\alpha}\otimes B_{\alpha} ~~~\textrm{(Eq. 3.119)}, $$ where $A_{\alpha}$ acts on the system and $B_{\alpha}$ acts on the bath, only. If we assume that the system Hamiltonian, $H_S$ has a discrete energy spectra, then by the spectral theorem, we can write $$ A_{\alpha} = \sum\limits_{\epsilon,\epsilon'} \langle \epsilon | A_{\alpha} | \epsilon '\rangle \cdot |\epsilon\rangle\langle \epsilon'| = \sum\limits_{\epsilon, \epsilon'}\Pi(\epsilon)A_{\alpha}\Pi(\epsilon'). $$ However, in the book, Eq. 3.120 only considers $\epsilon$ and $\epsilon'$ values, for which $\omega = \epsilon'-\epsilon$ is fixed: $$ A_{\alpha}(\omega) = \sum\limits_{\omega = \epsilon'-\epsilon}\Pi(\epsilon)A_{\alpha}\Pi(\epsilon'). $$ What is the motivation for this expansion? Why are we restricting the values of $\epsilon$ and $\epsilon'$ to have a fixed difference?

Answer 1

All this is saying is that there are three important terms to consider: The Hamiltonian $H^{\,}_S$ for the system alone, the Hamiltonian $H^{\,}_B$ for the bath alone, and the terms $H^{\,}_I$ that couple the system and bath.

The motivation I guess is that we often assume that the bath on its own thermalizes, and that the interaction between the system and bath is weak compared to their individual Hamiltonians.

As far as the fixed energy difference, in the final expression for the Liouvillean / Lindbladian / master equation, you will sum over all energy differences $\omega$. So it's not that you're requiring a particular fixed energy difference. It's just that it's easier to collect terms based on how much energy is absorbed from / given to the bath.

I suggest reading ahead until you find the Lindblad equation with the sum over $\omega$. Things will probably be more clear there. For one thing, grouping terms by the amount of energy exchanged between the system and bath leads to a nicer form of the Lindblad/Liouville equation. You're basically working in the basis of the Hamiltonians for the system/bath individually, looking at the system-bath terms in this basis, regrouping them by the energy exchanged, and eventually, tracing out the bath. The point is that terms that exchange energy $\omega$ with the bath couple to one another in the Lindblad equation, and terms that exchange different amounts of energy are not coupled (to leading order).