The Bethe Ansatz

States with real rapidities c.h.e.rr

Allowable quantum numbers

Consider looking for set of real rapidities as solution to h.bel. The maximally allowed quantum number \(I^{1,\infty}_M\) associated to a rapidity \(\lambda_a \rightarrow \infty\) in the presence of \(M - 1\) other finite rapidities \(\lambda_b < \infty\) is given by taking the limit \[ \lim_{\lambda_a \rightarrow \infty} 2 ~\mbox{atan}~ 2\lambda_a - \frac{1}{N} \sum_{b=1}^{M} 2~\mbox{atan} ~(\lambda_a - \lambda_b) = \pi (1 - \frac{M-1}{N}) \equiv 2\pi \frac{I^{1,\infty}_M}{N} \] from which we get

\begin{equation} I^{1,\infty}_M = \frac{N-M + 1}{2}, \hspace{1cm} I^{1,\mbox{max}}_M = I^{1,\infty}_M - 1 = \frac{N-M - 1}{2}. \tag{h.im}\label{h.im} \end{equation}

The Bethe equations will therefore have a solution in terms of a set of distinct real, finite rapidities if and only if all quantum numbers obey the inequality \(|I_j| < I^{\infty}_M\).

For a given \(M\), there are thus

\begin{equation*} \left( \begin{array}{c} N - M \\ M \end{array} \right) \end{equation*}

states containing exclusively real, finite rapidities. Since the \(XXX\) chain has a global \(su(2)\) symmetry, its eigenstates can be classified according to the representation of this algebra. Eigenstates containing real, finite rapidities only are highest-weight states of this representation b-Gaudin. Acting with the global spin lowering operator to obtain the other non-highest-weight states is achieved by adding infinite rapidities to a state, which still provides a {\it bona fide} solution to the Bethe equations for the XXX chain. At each value of \(M\), it is possible to add up to \(N/2 - M\) infinite rapidities without crossing the equator. We will denote the number of one-strings as \(M_1\), the number of finite-rapidity one-strings as \(M_1^<\) and the number of infinite rapidities as \(M_1^\infty\), so \(M_1 = M_1^< + M_1^\infty\). From the states with purely real, finite rapidities at a fixed \(M\), we can thus construct

\begin{equation*} \sum_{M_1^\infty=0}^{N/2-M} \left( \begin{array}{c} N-M+M_1^\infty \\ M - M_1^\infty \end{array} \right) \end{equation*}

states which, for a given \(M_1^\infty\) have global \(su(2)\) quantum numbers \(S = N/2-M+M_1^\infty\), \(S^z = N/2 - M\). Let us provide the simplest examples of these.

\(S = 0\), \(S^z = 0\) sector

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.

\(S = 1\), \(S^z = 1\) sector

We simply have \(M_1 = N/2-1\), \(M_1^\infty = 0\). We have \(I^{1,\infty} = \frac{N}{4} + 1\) and \(I^{\mbox{max}} = \frac{N}{4}\). The number of states is thus \(\left( \begin{array}{c} N/2 + 1 \\ N/2 - 1 \end{array} \right) = \frac{N(N+2)}{8}\). These are the \(S = 1\), \(S^z = 1\) two-spinon states.

\(S = 1\), \(S^z = 0\) sector

Here, we put \(M = N/2 = M_1\), \(M_1^< = N/2-1\) and \(M_1^\infty = 1\). The equations for the limiting quantum numbers then fall back onto the \(M \rightarrow M - 1\) ones, so now \(I^{1,\infty} = \frac{N}{4} + 1\) and \(I^{1,\mbox{max}} = \frac{N}{4}\). We thus have to put \(M_1^<\) quantum numbers in \(N/2 + 1\) slots, yielding \(\left( \begin{array}{c} N/2 + 1 \\ N/2 - 1 \end{array} \right) = \frac{N(N+2)}{8}\) states, which are the \(S = 1, S^z = 0\) two-spinon states.

\(S = 2\), \(S^z = 2\) sector

We simply have \(M = N/2-2 = M_1^<\) finite rapidity one-strings. We have \(I^{1,\infty} = \frac{N}{4} + \frac{3}{2}\) and \(I^{\mbox{max}} = \frac{N}{4} + \frac{1}{2}\). The number of states is thus \(\left( \begin{array}{c} N/2 + 2 \\ N/2 - 2 \end{array} \right) = \frac{(N+4) (N+2) N(N-2)}{384}\). These are the \(S = 2\), \(S^z = 2\) four-spinon states.

\(S = 2\), \(S^z = 1\) sector

We put \(M_1^< = N/2 -2\), \(M_1^\infty = 1\). The equations for the limiting quantum number fall back onto the \(M \rightarrow M-1\) ones. The limiting quantum numbers are \(I^{1,\infty} = \frac{N}{4} + \frac{3}{2}\) so \(I^{\mbox{max}} = \frac{N}{4} + \frac{1}{2}\), so we have to put \(N/2 - 2\) quantum numbers in \(N/2 + 2\) slots, yielding \(\frac{(N+4) (N+2) N (N-2)}{384}\) states, which are the four-spinon states in this sector.

\(S = 2\), \(S^z = 0\) sector

Here, we put \(M_1^< = N/2-2\) and \(M_1^\infty = 2\). The equations for the limiting quantum numbers of the one-string fall back onto the \(M \rightarrow M-2\) ones, so now \(I^{1,\infty} = \frac{N}{4} + \frac{3}{2}\) and \(I^{\mbox{max}} = \frac{N}{4} + \frac{1}{2}\). We thus have to put \(N/2-2\) quantum numbers in \(N/2 + 2\) slots, yielding \(\left( \begin{array}{c} N/2 + 2 \\ N/2 - 2 \end{array} \right) = \frac{(N+4) (N+2) N (N-2)}{384}\) states, which are the four-spinon states in this sector.

Note that for a given \(S\), the number of states at any \(S^z\) is the same due to the invariance of counting under the simultaneous shift \(M \rightarrow M-1\), \(M_1^\infty \rightarrow M_1^\infty + 1\).



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Author: Jean-Sébastien Caux

Created: 2024-01-18 Thu 14:24