The Bethe Ansatz

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{p.an}\label{p.an} \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}