Deriving the correlation function of a system interacting with a bath of harmonic oscillators

I'm working on the book Quantum Effects in Biology by Mohesni et all. My question is however not biology related, it is about a section on quantum master equations in the weak system-bath coupling limit. To set up the problem, we consider the following:

We have a bath of harmonic oscillators $$H_B = \sum_\xi\hbar\omega_\xi( b_\xi^{\dagger}b_\xi+1/2)$$ and a linear system bath interaction $$H_{SB} = \sum_{l,\xi}\hbar\omega_\xi g_{\xi,l}(b_\xi^{\dagger}+b_\xi)S_l$$ where $S_l$ are some system operators, the exact form of which will not prove to be relevant.

Now, making some additional assumptions about the bosonic bath, one can significantly simplify his/her master equation by putting all of the information about the bath into a correlation function of the form $$C_{l l'}(t-t') = \sum_{\xi}{\omega^2_\xi g_{\xi,l}g_{\xi,l'}\mathrm{Tr}\left[(b_\xi e^{-i\omega_\xi t}+b^\dagger_\xi e^{i\omega_\xi t})(b_\xi e^{-i\omega_\xi t'}+b^\dagger_\xi e^{i\omega_\xi t'})\rho_B(0)\right]}$$ where $$\rho_B(0) = \frac{e^{-\beta H_b}}{\mathrm{Tr}(e^{-\beta H_b})}.$$

I mostly understand how this function is constructed, but the interested reader can find more on page 25 of the book.

My question however comes down to the next part. Without any calculations shown in between, the author equates the above to $$\int_0^\infty{d\omega S_{l l'}(\omega)\left[\coth(\beta \hbar \omega/2)\cos(\omega (t-t')) - i \sin(\omega (t-t'))\right]}$$ with $$S_{l l'}(\omega) = \sum_\xi{g_{\xi,l}g_{\xi,l'}\delta(\omega-\omega_\xi)\omega^2}.$$

What I am trying to do is firstly try to reproduce the derivation. It is very unclear to me as to how one would continue. One can work out the product of the creation/annihilation operators to get four terms, but these then all have to act upon $e^{-\beta H_b}$ which is my first hindrance. How does this work, with the exponential containing creation/annihilation operators itself? And subsequently, one needs to through with the trace, and perform a bunch of summations. In between the spectral density gets introduced too, although I would say that intuitively this part makes sense.

So primarily I am looking for some help in getting started with the first part, working out the annihilation/creation operations onto the exponential term, and then taking the trace.

Finally I should say that I am actually interested in re-doing the derivation for $t' = t$. The reason I am not starting with this is because it is not clear to me that this is possible; plugging this into the result definitely does not give anything interesting I'd argue.

• Hint: first consider the single-mode case. Try to prove that $\mathrm{Tr}[b^\dagger b \mathrm{e}^{-\beta\omega b^\dagger b}]/\mathrm{Tr}[\mathrm{e}^{-\beta\omega b^\dagger b}] = (\mathrm{e}^{\beta\omega} - 1)^{-1}$. Why does this result make sense given what you know about Bose-Einstein statistics? You should then tackle the multi-mode case. – Mark Mitchison Apr 18 '16 at 14:41
• It's shameful, but somehow I cannot find my notes on how the operators act on exponentials with operators in them anymore. I know it was not very involved, but the basic theory I find online doesn't really go into it. – user129412 Apr 19 '16 at 8:02
• Think about the trace. The trace can be performed in any basis, in particular the eigenbasis of $b^\dagger b$. – Mark Mitchison Apr 19 '16 at 8:07
• Right, of course. Just use the Fock basis and a geometric series I suppose. I'll give it a try! – user129412 Apr 19 '16 at 14:37
• Okay, so I see how to get to your aforementioned result. Then one does the same for the other three terms ($b b$, $b b^\dagger$, $b^\dagger b^\dagger$), two of which are zero (I think, right?) and the other one gives the same as the above, except it is an $(n+1)$ term instead of $n$ so it gives an additional factor of 1 (as it gets divided out by the denominator of $\rho_0$). But this is still not it I guess, as the factor you mentioned is not a hyperbolic cotangent for example – user129412 Apr 19 '16 at 15:10