The Bethe Ansatz
One two-stringc.h.e.s.2
We here take \(M_1 = M - 2\), \(M_2 = 1\). For the one-strings, the limiting quantum numbers become
\begin{align*} \lim_{\lambda^1_{M-2} \rightarrow \infty} \phi_1 (\lambda^1_{M-2}) - \frac{1}{N} \sum_{\alpha = 1}^{M-2} \Phi_{11} (\lambda^1_{M-2} - \lambda^1_\alpha) - \frac{1}{N} \Phi_{12} (\lambda^1_{M-2} - \lambda^2_1) \nonumber \\ = \pi (1 - \frac{M-3}{N} - \frac{2}{N}) \equiv \frac{2\pi}{N} I^{1,\infty}. \end{align*}Similarly, for the two-strings,
\begin{equation*} \lim_{\lambda^2_1 \rightarrow \infty} \phi_2 (\lambda^2_1) - \frac{1}{N} \sum_{\alpha = 1}^{M-2} \Phi_{21} (\lambda^2_1 - \lambda^1_\alpha) = \pi (1 - 2\frac{M-2}{N}) \equiv \frac{2\pi}{N} I^{2,\infty}. \end{equation*}We thus find
\begin{equation*} I^{1,\infty} = \frac{N - M + 1}{2}, \hspace{5mm} I^{2,\infty} = \frac{N - 2M + 4}{2}. \end{equation*}We require strings of length greater than one to have strictly finite rapidities. The maximal quantum number turns out to be given by
\begin{equation*} I^{j,\mbox{max}} = I^{j,\infty} - j. \end{equation*}-
\(S = 0\), \(S^z = 0\) sector
We set \(M = N/2\), \(M_1 = M_1^< = N/2 - 2\) and \(M_2 = 1\). We thus obtain
\begin{equation*} I^{1,\infty} = \frac{N}{4} + \frac{1}{2}, \hspace{5mm} I^{2,\infty} = 2. \end{equation*}There is thus a single allowable quantum number for the two-string, \(I^{2}_1 = 0\). There are \(\frac{N}{2}\) available quantum number for finite rapidity one-strings, of which there are \(\frac{N}{2} - 2\), giving us \(\left( \begin{array}{c} N/2 \\ N/2 - 2 \end{array} \right) = \frac{N(N-2)}{8}\) states which are the \(S = 0, S^z = 0\) two-spinon states.
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\(S = 1\), \(S^z = 1\) sector
Here, we take \(M = N/2-1\), \(M_1 = M-2 = N/2-3 = M_1^<\), \(M_2 = 1\). The limiting quantum numbers are
\begin{equation*} I^{1,\infty} = \frac{N}{4} + 1, \hspace{5mm} I^{2,\infty} = 3 \end{equation*}so there are 3 slots for the two-string, and thus \(3\times \left( \begin{array}{c} N/2 + 1 \\ N/2 - 3 \end{array} \right) = (N + 2) N (N-2) (N-4)/128\) such states. These are the four-spinon states in this sector.
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\(S = 1\), \(S^z = 0\) sector
Here, we take \(M = N/2\), \(M_1^< = N/2-3\), \(M_1^\infty = 1\) and \(M_2 = 1\). The equations for the limiting quantum numbers then fall back onto the \(M \rightarrow M-1\) ones, so now \(I^{1,\infty} = \frac{N}{4} + 1\), \(I^{2\infty} = 3\). There are thus \(3\) available positions for the two-string, and \(\frac{N}{2} + 1\) available slots for the \(\frac{N}{2} - 3\) remaining finite rapidity one-strings. In total, there are thus \(3\times \left( \begin{array}{c} N/2 + 1 \\ N/2 - 3 \end{array} \right) = (N + 2) N (N-2) (N-4)/128\) such states. These are the four-spinon states in this sector.
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Created: 2026-08-26 Wed 11:07