# The Bethe Ansatz

### ABA for the isotropic $$S=1/2$$ antiferromagnet ($$XXX$$ model)a.h

For convenience, we restate here all important formulas relevant for our further study of the isotropic $$XXX$$ model, in the final parametrization which we have settled upon.

\begin{equation} R (\lambda) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & b (\lambda) & c (\lambda) & 0 \\ 0 & c (\lambda) & b (\lambda) & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), \hspace{1cm} b (\lambda) = \frac{\lambda}{\lambda + i}, \hspace{1cm} c (\lambda) = \frac{i}{\lambda + i}. \tag{h.r}\label{h.r} \end{equation} \begin{equation} a (\lambda) = 1, \hspace{1cm} d(\lambda) = b(\lambda - i/2)^N = \left[ \frac{\lambda - i/2}{\lambda + i/2} \right]^N. \tag{h.ad}\label{h.ad} \end{equation} \begin{equation*} \left[\frac{\lambda_j + i/2}{\lambda_j - i/2} \right]^N = \prod_{l (\neq j) = 1}^M \frac{\lambda_j - \lambda_k + i}{\lambda_j - \lambda_k - i} \end{equation*} \begin{equation} \tau (\lambda | \{ \lambda_j \}) = \prod_{j=1}^M \frac{\lambda_j - \lambda + i}{\lambda_j - \lambda} + \left[\frac{\lambda - i/2}{\lambda + i/2}\right]^N \prod_{j=1}^M \frac{\lambda_j - \lambda - i}{\lambda_j - \lambda}. \tag{h.te}\label{h.te} \end{equation} \begin{equation} P = -i \ln \tau (\lambda)|_{\lambda = i/2} \tag{h.q1}\label{h.q1} \end{equation} \begin{equation} P (\{ \lambda_j \}_M) = -i \sum_{j=1}^M \ln \frac{\lambda_j +i/2}{\lambda_j - i/2} = \pi M - \sum_{j=1}^M 2~\mbox{atan}~ 2\lambda_j. \tag{h.q1e}\label{h.q1e} \end{equation} \begin{equation} H_{XXX} = \sum_{j=1}^N {\bf S}_j \cdot {\bf S}_{j+1} - \frac{1}{4} = \frac{i}{2} \frac{d}{d\lambda} \ln \tau (\lambda) |_{\lambda = i/2} \tag{h.q2}\label{h.q2} \end{equation} \begin{equation} E (\{ \lambda_j \}_M) = \sum_{j=1}^M \frac{-2}{4\lambda_j^2 + 1}. \tag{h.q2e}\label{h.q2e} \end{equation}

A few overall comments are in order here. The form of the $$R$$-matrix we settled upon is not unitary. The construction we ended up with conforms to (some) historical conventions. This choice has as consequence that the monodromy matrix $$A,B,C,D$$ operator entries are not simply related by the usual complex conjugation. This slight annoyance is of no consequence for our further purposes. Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Created: 2023-06-07 Wed 16:02