The Bethe Ansatz

Integrated intensity

The simplest sum rules which one can find relate to simple integrals of the dynamical correlators. First, the static correlator is obtained by setting the time difference to zero:

\begin{equation*} S_\alpha (j - j') \equiv S_\alpha (j-j', 0) \end{equation*}

or in Fourier space,

\begin{equation*} S_\alpha (k) \equiv \int_{-\infty}^\infty \frac{d\omega}{2\pi} S_\alpha (k, \omega) = \sum_{\alpha'} | \langle \alpha | O_k | \alpha' \rangle|^2 \delta_{k, K_{\alpha'} - K_\alpha} \end{equation*}

where in the last equality we have made use of the Lehmann representation Lr and explicitly enforced the requirement that the momentum difference between the excited state and the averaging state is equal to \(k\).

Separately, we can also consider the autocorrelator

\begin{equation*} S_\alpha (t) \equiv S_\alpha (0, t) \end{equation*}

which in Fourier space reads

\begin{equation*} S_\alpha (\omega) = \frac{1}{N} \sum_k S_\alpha (k, \omega). \end{equation*}

Combining these two ideas, we get the static autocorrelator or static moment

\begin{equation*} S_\alpha \equiv S_\alpha (0, 0) = \langle \alpha | {\cal O}_j {\cal O}_j | \alpha \rangle \end{equation*}

The idea is that if the equal-time product \({\cal O} {\cal O}\) simplifies to a known or easy to evaluate quantity, then we obtain a constraint on the (integral of the) dynamical correlator.

Spin chain

For example, if we are studying a spin-\(1/2\) chain, since we know that \(S^a_j S^a_j = 1/4\) for any component or site, we obtain the sum rule

\begin{equation*} S^{aa}_\alpha = \frac{1}{4} = \frac{1}{N} \sum_k \int_{-\infty}^\infty \frac{d\omega}{2\pi} S_\alpha^{aa} (k, \omega) \end{equation*}

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Author: Jean-Sébastien Caux

Created: 2023-06-07 Wed 16:02