The Bethe Ansatz

Detailed balance d.b.db

Let's consider the following correlator

\begin{equation*} S (k, \omega) = \frac{1}{N} \sum_{j, j'} e^{-i k (j-j')} \int_{-\infty}^\infty dt e^{i \omega t} \langle {\cal O}_j (t) {\cal O}_{j'} (0) \rangle_T \end{equation*}

in which the average is a thermal trace in the eigenstates basis (with \(\beta \equiv k_B T\)),

\begin{equation*} \langle \cdots \rangle_T \equiv \frac{1}{Z} \sum_{\alpha} \langle \alpha | \cdots | \alpha \rangle e^{-\beta E_\alpha}, \hspace{10mm} Z \equiv \sum_\alpha e^{-\beta E_\alpha} \end{equation*}

The Lehmann representation is easily obtained by using the single-state version Lr and performing the Gibbs sum:

\begin{align*} S(k, \omega) = \sum_{\alpha} \frac{e^{-\beta E_\alpha}}{Z} S_\alpha (k, \omega) = \frac{2\pi}{N} \sum_{\alpha, \alpha'} \frac{e^{-\beta E_\alpha}}{Z} |\langle \alpha | O_k | \alpha' \rangle|^2 ~\delta(\omega - (E_{\alpha'} -E_\alpha)) \end{align*}

Simple manipulations (interchanging the \(\alpha, \alpha'\) summations and using the delta function) then gives the detailed balance condition

\begin{equation} S(-k, -\omega) = e^{-\beta \omega} S(k, \omega) \tag{db}\label{db} \end{equation}

which is valid for any dynamical correlator at thermal equlibrium.

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Author: Jean-Sébastien Caux

Created: 2024-01-18 Thu 14:24