The Bethe Ansatz

The ground state of the planar antiferromagnet

Calculations for the ground state of the infinite XXZ antiferromagnet completely parallel those done for the isotropic case. The differential kernel is here

\begin{equation} a_n (\lambda) = \frac{1}{2\pi} \frac{d}{d\lambda} \phi_n (\lambda) = \frac{1}{\pi} \frac{\sin n\zeta}{\cosh 2\lambda - \cos n\zeta}. \tag{}\label{} \end{equation}

Instead of h.anf, we thus get

\begin{equation} a_n (\omega) = \frac{\sinh (\frac{\omega \pi}{2}(1 - n\frac{\zeta}{\pi}))}{\sinh (\frac{\omega \pi}{2})}. \tag{p.anf}\label{p.anf} \end{equation}

For zero field, we again have \(\lambda_F \rightarrow \infty\); the convolution theorem therefore allows to solve the continuum Bethe equations h.rg for the ground state root density distribution,

\begin{equation} \rho_g (\omega) = \frac{1}{2\cosh (\frac{\omega \zeta}{2})}, \hspace{1cm} \rho_g (\lambda) = \frac{1}{2\cosh (\frac{\pi \lambda}{\zeta})}. \tag{p.rg}\label{p.rg} \end{equation}

The ground state energy and momentum can again be calculated by substituting this distribution back into h.ep, yielding 1966.desCloizeaux.JMP.7.1

\begin{equation} E_{\mbox{gs},0} = -N J \frac{\sin \zeta}{4} \int_{-\infty}^{\infty} d\omega \left[ 1 - \frac{\tanh (\frac{\omega \zeta}{2})}{\tanh (\frac{\omega \pi}{2})} \right], \hspace{1cm} P_{\mbox{gs},0} = \pi \frac{N}{2} ~~\mbox{mod}~ 2\pi. \tag{p.epg}\label{p.epg} \end{equation}

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Author: Jean-Sébastien Caux

Created: 2024-01-18 Thu 14:24