# The Bethe Ansatz

#### The f-sumruled.sr.f

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

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

(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.