# The Bethe Ansatz

##### String statesc.h.e.s

The Bethe equations for the $$XXX$$ magnet allow solutions which are not restricted to the real line. Complex rapidities can occur. These complex rapidities organize themselves into self-conjugate patterns called {\it strings}, rapidities within an $$n$$-string being reparametrized as

\begin{equation} \lambda_\alpha^{n, a} = \lambda_\alpha^n + \frac{i}{2} (n+1 - 2a) + i \delta_\alpha^{n,a}, \hspace{10mm} a = 1, ..., n. \tag{h.ls}\label{h.ls} \end{equation}

The real parameter $$\lambda_\alpha^j$$ represents the {\it string center}, namely the `center of mass' of the composite object represented by the $$n$$ rapidities. The string deviations $$\delta_\alpha^{j,a}$$ are in most circumstances exponentially small in system size, and can typically be neglected (there are important exceptions to this rule). Since complex rapidities represent bound states of downturned spins (as can be seen from the Bethe Ansatz wavefunction itself), a cluster of the form h.ls represents a single bona fide particle. Out of $$n$$ rapidities within a string, we thus get only one independent parameter, the string center $$\lambda_\alpha^{j}$$.

The string hypothesis assumes that eigenstates of the Heisenberg chain are represented by sets of rapidities which organize themselves into (perfect) strings. An eigenstate with $$M$$ down spins is then understood as an eigenstate with $$M_j$$ strings of length $$j$$, the total number of downturned spins obeying

\begin{equation*} \sum_{j=1}^\infty j M_j = M. \end{equation*}

The total number of strings in an eigenstate is given by

\begin{equation*} \sum_{j=1}^\infty M_j \equiv N_s. \end{equation*}

In a given eigenstate, the number of independent parameters is thus $$N_s$$ (the number of string centers) instead of $$M$$. The Bethe equations for strings can be rewritten by taking the product of h.be over all rapidities within a string. Doing this, one finds (for $$N$$ even) the reduced set of Bethe equations

\begin{equation} \bar{e}_j^N (\lambda_{\alpha}^n) = (-1)^{M_j-1}\prod_{(k,\beta) \neq (j,\alpha)} \bar{E}_{jk} (\lambda_{\alpha}^j - \lambda_{\beta}^k), \tag{h.bgt}\label{h.bgt} \end{equation}

where

\begin{equation} \bar{e}_j(\lambda) = -\frac{\lambda + j \frac{i}{2}}{\lambda - j \frac{i}{2}}, \hspace{1cm} \bar{E}_{jk}(\lambda) = \bar{e}_{|j-k|}^{1-\delta_{j k}} (\lambda) \bar{e}_{|j-k|+2}^2(\lambda) ... \bar{e}_{j+k-2}^2 (\lambda) \bar{e}_{j+k}(\lambda). \tag{h.eE}\label{h.eE} \end{equation}

Taking logs, we obtain the following reduced set of equations (which we shall call the Bethe-Gaudin-Takahashi equations)

\begin{equation} \phi_j (\lambda^j_\alpha) - \frac{1}{N} \sum_{k=1}^{N_s} \sum_{\beta = 1}^{M_k} \Phi_{jk} (\lambda^j_\alpha - \lambda^k_\beta) = \frac{2\pi}{N} I^j_\alpha \tag{h.bgtl}\label{h.bgtl} \end{equation}

in which

\begin{equation*} I_\alpha^j \in \left\{ \begin{array}{cc} \mathbb{Z} + \frac{1}{2}, & M_j ~\mbox{even} \\ \mathbb{Z}, & M_j ~\mbox{odd}. \end{array} \right. \end{equation*}

The kernels $$\phi_j$$ are defined as

\begin{equation*} \phi_j(\lambda) = 2\mbox{atan} \frac{2\lambda}{n} \end{equation*}

and the string-string scattering phase shift is

\begin{equation} \Phi_{jk} (\lambda) = (1 - \delta_{jk}) \phi_{|j-k|} (\lambda) + 2\phi_{|j-k|+2} (\lambda) + ... + 2\phi_{j+k-2} (\lambda) + \phi_{j+k} (\lambda). \tag{h.phijk}\label{h.phijk} \end{equation}

These have the simple limits

\begin{equation*} \lim_{\lambda \rightarrow \infty} \phi_j (\lambda) = \pi, \hspace{10mm} \lim_{\lambda \rightarrow \infty} \Phi_{jk} (\lambda) = (2\min(j,k) - \delta_{jk}) \pi. \end{equation*}
###### Limiting quantum numbers

As for real rapidities, the quantum numbers of strings are limited. We can calculate the maximal string quantum numbers using the same arguments as before:

\begin{equation*} \lim_{\lambda^j_\alpha \rightarrow \infty} (LHS ~\mbox{h.bgtl}) = \frac{2\pi}{N} \frac{1}{2}\left( N - \sum_{k=1}^{N_s} (2\min(j,k) - \delta_{jk}) (M_k - \delta_{jk}) \right) \equiv \frac{2\pi}{N} I^{j,\infty}. \end{equation*}

More precisely, this can be rewritten as

\begin{equation*} I^{j,\infty} = \frac{1}{2}\left( N + 2j-1 - \sum_{k=1}^{N_s} (2\min(j,k) - \delta_{jk}) M_k \right). \end{equation*}

We will make use of this in the following, in order to classify and count eigenstates.

###### One two-string

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.

• $$S = 1$$, $$S^z = 1$$ 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.

• $$S = 1$$, $$S^z = 0$$ 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.

###### Two two-strings

We take $$M = N/2$$, $$M_1 = M_1^< = M-4 = N/2-4$$, $$M_2 = 2$$. The limiting quantum numbers are here

\begin{equation*} \pi (1 - \frac{M-5}{N} - \frac{4}{N}) = \frac{2\pi}{N} \frac{N-M+1}{2} = \frac{2\pi}{N} (\frac{N}{4} + \frac{1}{2}) \rightarrow I^{1,\infty} = \frac{N}{4} + \frac{1}{2}. \end{equation*} \begin{equation*} \pi (1 - 2 \frac{M-4}{N} - \frac{3}{N}) = \frac{2\pi}{N} \frac{N - 2M + 5}{2} \rightarrow I^{2,\infty} = \frac{5}{2} \end{equation*}

There are thus 2 slots for 2 two-strings (1 possibility) and $$N/2$$ slots for $$N/2 - 4$$ one-strings, giving in total $$\frac{N (N-2) (N-4) (N-6)}{384}$$ states, which are the $$S = 0$$, $$S^z = 0$$ four-spinon states.

###### One three-string

We take $$M = N/2$$, $$M_1 = M_1^< = N/2-3$$, $$M_3 = 1$$. The limiting quantum numbers are

\begin{equation*} I^{1,\infty} = \frac{N}{4} + 1, \hspace{5mm} I^{3,\infty} = 3 \end{equation*}

meaning that there is one available slot for the three-string, and $$N/2 +1$$ slots for the $$N/2-3$$ finite rapidity one-strings. This gives $$(N + 2) N (N-2) (N-4)/384$$ states, which are the four-spinon states in this $$S=0$$, $$S^z=0$$ sector.

###### Counting spinon states
• Two-spinon states

Adding up the $$S=0$$ and $$S=1$$ two-spinon states, we obtain

\begin{equation*} \frac{N(N-2)}{8} + 3 \times \frac{N(N+2)}{8} = \frac{N(N+1)}{2} = \left( \begin{array}{c} N+1 \\ 2 \end{array} \right) \end{equation*}

which is the correct total number of two-spinon states.

• Four-spinon states

Adding the numbers of states in all the $$S=0, 1$$ and $$2$$ sectors gives (note: as expected, there are 2 $$S=0$$ representations, 3 $$S=1$$ ones, and one $$S=2$$)

\begin{align*} \frac{N (N-2) (N-4) (N-6)}{384} + \frac{(N+2)N(N-2)(N-4)}{384} + 3\times \frac{(N+2) N (N-2) (N-4)}{128} + \nonumber \\ + 5\times \frac{(N+4)(N+2) N (N-2)}{384} = \left( \begin{array}{c} N+1 \\ 4 \end{array} \right) \hspace{5cm} \end{align*}

which is the expected total number of four-spinon states. Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Created: 2023-06-07 Wed 16:02