The Bethe Ansatz

Consider a model with lattice sites \(j = 1, ..., N\). Let's look at the following generic correlation function of some operator \({\cal O}_j^a\), where \(a\) is some label (for example, for a lattice Bose gas, we could have \({\cal O}_j = \Psi_j\) or \(\Psi^\dagger_j\) or \(\rho_j \equiv \Psi_j^\dagger \Psi_j\)):

\begin{equation*} {\boldsymbol S}^{a \bar{a}} (k, \omega) \equiv \frac{1}{N} \sum_{j, j'} e^{-i k(j-j')} \int_{-\infty}^\infty dt e^{i\omega t} \langle \frac{1}{2} \left[ {\cal O}^a_j (t), ({\cal O}^a_{j'} (0))^\dagger \right] \rangle \end{equation*}

The expectation value \(\langle ... \rangle\) can be any expectation value: with respect to a specific state, \(\langle ... \rangle = \langle \alpha | ... | \alpha \rangle\), a thermal average \(\langle ... \rangle = \frac{1}{\cal Z} \sum_{\gamma} \langle \gamma | ... | \gamma \rangle e^{-\beta (E_{\gamma} - \mu N_{\gamma})}\), it doesn't matter. In the case of a (grand-canonical) thermal average, we can write

\begin{equation*} {\boldsymbol S}^{a \bar{a}} (k, \omega) \equiv \frac{1}{N} \sum_{j, j'} e^{-i k(j-j')} \int_{-\infty}^\infty dt e^{i\omega t} \frac{1}{\cal Z} \sum_\gamma \langle \gamma | \frac{1}{2} \left[ {\cal O}^a_j (t), ({\cal O}^a_{j'} (0))^\dagger \right] | \gamma \rangle e^{-\beta (E_\gamma - \mu N_\gamma)} \end{equation*}

in which \(\mu\) is the chemical potential, \({\cal Z}\) is the grand-canonical partition function, and \(\sum_{\gamma}\) represents a sum over eigenstates.

Consider integrating this correlation over all frequencies, calculating the first moment in frequency

\begin{equation*} I_k^{(1)} \equiv \int_{-\infty}^\infty \frac{d\omega}{2\pi} \omega ~{\boldsymbol S}^{a \bar{a}} (k, \omega) \end{equation*}

This can be manipulated as follows:

\begin{align*} I_k^{(1)} &= \frac{1}{2N} \int_{-\infty}^\infty \frac{d\omega}{2\pi} \omega \int_{-\infty}^\infty dt e^{i\omega t} \frac{1}{\cal Z} \sum_\gamma \langle \gamma | \left[ {\cal O}^a_k (t), ({\cal O}^a_{k'} (0))^\dagger \right] | \gamma \rangle e^{-\beta (E_\gamma - \mu N_\gamma)} \nonumber \\ &= \frac{1}{2N} \int_{-\infty}^\infty \frac{d\omega}{2\pi} \omega \int_{-\infty}^\infty dt e^{i\omega t} \frac{1}{\cal Z} \times \nonumber \\ &\times \sum_\gamma \sum_{\alpha} \left\{ e^{-i\omega_{\alpha \gamma}t} \langle \gamma | {\cal O}^a_k | \alpha \rangle \langle \alpha | ({\cal O}^a_k)^\dagger | \gamma \rangle - e^{i\omega_{\alpha \gamma}t} \langle \gamma | ({\cal O}^a_k)^\dagger | \alpha \rangle \langle \alpha | {\cal O}^a_k | \gamma \rangle \right\} e^{-\beta (E_\gamma - \mu N_\gamma)} \end{align*}

in which we have used the Fourier transforms

\begin{equation*} {\cal O}^a_j = \frac{1}{N} \sum_k e^{ikj} {\cal O}^a_k, \hspace{5mm} {\cal O}^a_k = \sum_j e^{-ikj} {\cal O}^a_j \end{equation*}

and have inserted a resolution of the identity \({\bf 1} = \sum_{\alpha} | \alpha \rangle \langle \alpha |\) in terms of eigenstates of the Hamiltonian. We have used the notation \(\omega_{\alpha \gamma} \equiv E_\alpha - E_\gamma - \mu(N_\alpha - N_\gamma)\), and the Heisenberg relation \({\cal O} (t) = e^{i (H - \mu N) t} {\cal O} e^{-i (H - \mu N)t}\) to resolve the time dependence.

Using the identity

\begin{equation*} \int_{-\infty}^\infty dt e^{i (\omega - \omega') t} = 2\pi \delta (\omega - \omega'), \end{equation*}

we get

\begin{align*} I_k^{(1)} &= \frac{1}{2N} \int_{-\infty}^\infty d\omega \omega \frac{1}{\cal Z} \sum_\gamma \sum_\alpha \left\{ \delta (\omega - \omega_{\alpha \gamma}) \langle \gamma | {\cal O}^a_k | \alpha \rangle \langle \alpha | ({\cal O}^a_k)^\dagger | \gamma \rangle \right. \nonumber \\ & \left. \hspace{4cm}- \delta (\omega + \omega_{\alpha \gamma}) \langle \gamma | ({\cal O}^a_k)^\dagger | \alpha \rangle \langle \alpha | {\cal O}^a_k | \gamma \rangle \right\} e^{-\beta (E_\gamma - \mu N_\gamma)} \nonumber \\ &= \frac{1}{2N} \frac{1}{\cal Z} \sum_\gamma \sum_\alpha \omega_{\alpha \gamma} \left\{ \langle \gamma | {\cal O}^a_k | \alpha \rangle \langle \alpha | ({\cal O}^a_k)^\dagger | \gamma \rangle + \langle \gamma | ({\cal O}^a_k)^\dagger | \alpha \rangle \langle \alpha | {\cal O}^a_k | \gamma \rangle \right\} e^{-\beta (E_\gamma - \mu N_\gamma)} \nonumber \\ &= \frac{1}{2N} \frac{1}{\cal Z} \sum_\gamma \sum_\alpha \left\{ \langle \gamma | \left[ {\cal O}^a_k, H \right] | \alpha \rangle \langle \alpha | ({\cal O}^a_k)^\dagger | \gamma \rangle + \langle \gamma | ({\cal O}^a_k)^\dagger | \alpha \rangle \langle \alpha | \left[ H, {\cal O}^a_k \right] | \gamma \rangle \right\} e^{-\beta (E_\gamma - \mu N_\gamma)} \nonumber \\ &= \frac{-1}{2N} \frac{1}{\cal Z} \sum_\gamma \langle \gamma | \left[ \left[ H, {\cal O}^a_k \right], ({\cal O}^a_k)^\dagger \right] | \gamma \rangle e^{-\beta (E_\gamma - \mu N_\gamma)} \nonumber \\ &= \frac{-1}{2N} \langle \left[ \left[ H, {\cal O}^a_k \right], ({\cal O}^a_k)^\dagger \right] \rangle_\beta \end{align*}

in which we have used the fact that we're working in a basis of energy eigenstates (so \(H | \alpha \rangle = (E_\alpha - \mu N_\alpha) | \alpha \rangle\)) and the resolution of the identity \(\sum_\alpha | \alpha \rangle \langle \alpha | = {\bf 1}\).

We thus get

\begin{equation} \int_{-\infty}^\infty \frac{d\omega}{2\pi} \omega ~{\boldsymbol S}^{a \bar{a}} (k, \omega) = \frac{-1}{2N} \langle \left[ \left[ H, {\cal O}^a_k \right], ({\cal O}^a_k)^\dagger \right] \rangle \tag{fsr}\label{fsr} \end{equation}

(note: this holds irrespective of how the average \(\langle ... \rangle\) is defined, provided \({\boldsymbol S}(k,\omega)\) and the right-hand side of this equation are averaged in precisely the same way). This relation holds for any correlator of any model.

Now: the magic trick comes from the fact that for specific cases, the right-hand side can be calculated exactly.