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Via path integral molecular dynamics it is possible to measure the Kubo transformed correlation function between two operators $\hat A$ and $\hat B$ \begin{equation*} K_{\hat A\hat B} = \frac 1 {Z_\beta\beta}\int_0^\beta \mathrm d \lambda\; \mathrm{Tr}\left\{ e^{-(\beta-\lambda)\hat H}\hat A e^{-\lambda\hat H} e^{i\hat H t/\hbar}\hat B e^{-i\hat H t/\hbar}\right\} \end{equation*} A relation between this and the standard correlation function \begin{equation*} C_{\hat A\hat B}(t) = \frac 1 {Z_\beta} \mathrm{Tr}\left\{ e^{-\beta\hat H}\hat A e^{i\hat H t/\hbar}\hat B e^{-i\hat H t/\hbar}\right\} \end{equation*} can be determined by expanding the traces on the energy basis. The result as reported in eqn. (1.15) in (Craig, Ian R. 2006. “Ring Polymer Molecular Dynamics.” PhD thesis, University of Oxford) is \begin{equation*} C_{\hat A\hat B}(\omega) = \frac{\beta \hbar \omega}{1-e^{-\beta \hbar \omega}} \tilde K_{\hat A\hat B}(\omega) \end{equation*} In the work is mentioned that the relation above is obtained by expanding the trace in the energy basis. Hence I tried to derive the relation but I could not complete the derivation; follows my attempt: \begin{align*} \tilde K_{\hat A \hat B}(\omega) &= \frac 1 {Z_\beta\beta}\sum_j \int_{-\infty}^{+\infty} \mathrm d t\; e^{i\omega t} \int_0^\beta \mathrm d \lambda \;\langle j |e^{-(\beta-\lambda)\hat H}\hat A e^{-\lambda\hat H} e^{i\hat H t/\hbar}\hat B e^{-i\hat H t/\hbar}|j\rangle = \\ &= \frac 1 {Z_\beta\beta}\sum_j \int_{-\infty}^{+\infty} \mathrm d t\; e^{i\omega t} \int_0^\beta \mathrm d \lambda\; e^{-(\beta-\lambda)E_j} \langle j| \hat A e^{-\lambda\hat H} e^{i\hat H t/\hbar}\hat B | j\rangle e^{-i E_j t/\hbar} =\\ &= \frac 1 {Z_\beta\beta}\sum_{j,k} \int_{-\infty}^{+\infty} \mathrm d t\; e^{i\omega t} \int_0^\beta \mathrm d \lambda\; e^{-(\beta-\lambda)E_j}\langle j| \hat A| k\rangle e^{-\lambda E_k} e^{i\hat E_k t/\hbar} \langle k |\hat B |j\rangle e^{-i E_j t/\hbar} = \\ &= \frac 1 {Z_\beta\beta}\sum_{j,k} \int_{-\infty}^{+\infty} \mathrm d t\; e^{i\omega t} \left[\frac{ e^{\beta(E_j-E_k)}-1}{E_j-E_k}\right] e^{-\beta E_j} A_{jk} e^{i\hat E_k t/\hbar} B_{kj} e^{-i E_j t/\hbar} = \\ &= \sum_j \frac 1 {Z_\beta\beta} \int_{-\infty}^{+\infty} \mathrm d t\; e^{i\omega t} e^{-\beta E_j }A_{jj} \hat B_{jj} +\\ &+ \frac 1 {Z_\beta\beta}\sum_{j\neq k} \int_{-\infty}^{+\infty} \mathrm d t\; e^{i\omega t} \left[\frac{ e^{\beta(E_j-E_k)}-1}{E_j-E_k}\right] e^{-\beta E_j} A_{jk} e^{i (E_k -E_j)t/\hbar} B_{kj} \end{align*} where $\langle i| A | j\rangle = A_{ij}$. Is it possible to proceed from here to conclude the derivatiion? Or shall I follows another pathway?

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  • $\begingroup$ Added the reference to the related equation within the thesis $\endgroup$ – Graz Apr 8 at 14:00
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This is left as an exercise for the reader, for example in Mark Tuckerman's book Statistical Mechanics: theory and molecular simulation, Chapter 14.

I think the key simplifying feature is that, having introduced the eigenstates of the energy, and inserted a complete set of states, as you have done, you can look at the Fourier integral over time. This is going to give you a Dirac delta function involving $\omega$ and the energy difference $(E_j-E_k)/\hbar$.

So this means you can replace that energy difference in the exponential $\exp[\lambda(E_j-E_k)]$ with $\exp[\lambda\hbar\omega]$, which you integrate with respect to $\lambda$, leading to the desired prefactor. Then you can reverse the transformations involved with the expansion over eigenstates, arriving at the expression for $C(\omega)$.

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  • $\begingroup$ Thanks, I missed the obvious decomposition of Dirac's delta over the Fourier basis. $\endgroup$ – Graz Apr 8 at 13:57

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