# The Bethe Ansatz

##### States with real rapiditiesc.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 $$M \rightarrow M-1$$, $$M_1^\infty \rightarrow M_1^\infty + 1$$. Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Created: 2023-06-07 Wed 16:02