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.

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

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.