For an observable $A$, define the real-time autocorrelation function $$ C(t) = \langle A A(t) \rangle_{\beta} = \dfrac{1}{Z} \mathrm{Tr}\left[ e^{-\beta H} A e^{i H t} A e^{-i H t}\right], $$ with $Z = \mathrm{Tr}\left[ e^{-\beta H}\right]$, and the spectral function $S(\omega)$ as its Fourier transform, $$ S(\omega) = \int_{-\infty}^{\infty} e^{-i \omega t} C(t) \mathrm{d} t. $$ For real $\omega$, we can write this in terms of the spectral representation, $$ S(\omega) = \sum_{n} p_{n} \sum_{m} |A_{nm}|^{2} \delta(\omega - \omega_{nm}), $$ where $n$ indexes the Hamiltonian eigenstates $|E_{n}\rangle$ with eigenvalue $E_{n}$, $p_{n} = e^{-\beta E_{n}} / Z$ is the thermal occupation probability, $A_{nm} = \langle E_{n} | A | E_{m} \rangle$, and $\omega_{nm} = E_{n} - E_{m}$.
I am interested in continuing $S(\omega)$ to the complex plane. Is this possible to do in general, and if so, can we say anything about the analytic structure of this continuation? For example, should we expect any poles?
From the fluctuation-dissipation theorem, $$ S(\omega) = 2(n_{B}(\omega) + 1) \chi^{''}(\omega), $$ where $n_B$ is the Bose occupation function and $\chi^{''}(\omega) = \mathrm{Im} \chi(\omega)$ is the dissipative part of the response function, it seems like we should maybe expect poles along the imaginary axis at the bosonic Matsubara frequencies. But I'm not sure about the analytic structure of $\chi^{''}(\omega)$ across the whole complex plane. (Perhaps the Kramers-Kronig relations can help to define such a continuation?)
