The Bethe Ansatz

For the particular case of zero magnetic field, we have (for \(N\) even) that the magnetization is zero, so \(M = N/2 = M_1\), \(M_1^\infty = 0\). In this case, the ground state is given by the set of quantum numbers

\begin{equation*} \left\{ -\frac{M-1}{2}, -\frac{M-1}{2} + 1, ..., \frac{M-1}{2} \right\} = \left\{ -\frac{N}{4} + \frac{1}{2}, ..., \frac{N}{4} - \frac{1}{2} \right\}. \end{equation*}

Since in this case \(I^{\infty}_{N/2} = \frac{N}{4} + \frac{1}{2}\), we see that the {\it only} eigenstate with real, finite rapidities at zero magnetization is the ground state.