# The Bethe Ansatz

##### Supplement: Completeness of the Bethe Ansatz for the $$M = 2$$ case of the $$XXX$$ antiferromagnetc.h.e.M2

If we specialize to $$M = 2$$, we can solve the problem completely for arbitrary $$N$$. Doing this illustrates peculiar features of the Bethe Ansatz which are swept under the rug by the string hypothesis, gives an idea of what is left behind, and justifies making the string approximation in most circumstances.

Let us begin from the Bethe equations

\begin{equation*} ~\mbox{atan}~ 2\lambda_1 - \frac{1}{N} ~\mbox{atan}~ (\lambda_1 - \lambda_2) = \pi \frac{I_1}{N}, \hspace{1cm} ~\mbox{atan}~ 2\lambda_2 + \frac{1}{N} ~\mbox{atan}~ (\lambda_1 - \lambda_2) = \pi \frac{I_2}{N} \end{equation*}

Here, $$I_{1,2}$$ are integers if $$N$$ is odd, and half-odd integers if $$N$$ is even.

Let us restrict to real $$\lambda_1 < \lambda_2$$. For fixed $$\lambda_2$$, taking $$\lambda_1$$ to $$-\infty$$ yields the lower limit

\begin{equation*} -\frac{\pi}{2} + \frac{\pi}{2N} = \pi \frac{I_1^{min}}{N}, \hspace{1cm} I^{min} = -\frac{N - 1}{2} \end{equation*}

If $$\lambda_1 = -\infty$$, the remaining Bethe equation becomes

\begin{equation*} ~\mbox{atan}~ 2\lambda_2 - \frac{\pi}{2N} = \pi \frac{I_2}{N}, \hspace{1cm} \lambda_2 = \frac{1}{2} \tan \frac{\pi}{2N}(2 I_2 + 1) \end{equation*}

For $$I^{min} < I_2 < I^{max} = \frac{N-1}{2}$$ this has a solution with finite $$\lambda_2$$. Using that $$k_1 = 0$$ and $$\phi = -\pi$$ since $$\lambda_1 = -\infty$$, the wavefunction then reads

\begin{equation} \Psi_2 (j_1, j_2) = e^{i k_2 j_2 + i\pi/2} - e^{i k_2 j_1 -i\pi/2} = i (e^{i k_2 j_2} + e^{i k_2 j_1} ). \tag{h.l1mi}\label{h.l1mi} \end{equation}

If we put $$\lambda_2 \rightarrow \infty$$, we get $$k_2 = 0$$ and $$\phi = \pi$$. The Bethe equation for $$\lambda_1$$ becomes

\begin{equation*} ~\mbox{atan}~ 2\lambda_1 + \frac{\pi}{2N} = \pi \frac{I_1}{N}, \hspace{1cm} \lambda_1 = \frac{1}{2} \tan \frac{\pi}{2N}(2 I_1 - 1) \end{equation*}

with solution for finite real $$\lambda_1 < \lambda_2$$ for $$I^{min} < I_1 < I^{max}$$. The wavefunction then becomes

\begin{equation*} \Psi_2 (j_1, j_2) = e^{i k_1 j_1 - i\pi/2} - e^{i k_1 j_2 +i\pi/2} = -i (e^{i k_1 j_1} + e^{i k_1 j_2} ) \end{equation*}

which coincides up to a constant with h.l1mi. In the particular limit $$\lambda_1 \rightarrow -\infty$$, $$\lambda_2 \rightarrow \infty$$, we get $$k_1 \rightarrow 0^-$$, $$k_2 \rightarrow 0^+$$, $$\phi \rightarrow -\pi$$ so the Bethe wavefunction is a constant, which is not a solution of the (second part of the) eigenvalue equation.

All in all, we are thus able to construct

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

distinct Bethe wavefunctions with real rapidities in the interval $$[-\infty, \infty]$$. The first term is from counting wavefunctions with finite rapidities (for which $$I_1, I_2$$ are chosen within the interval $$I^{min} < I < I^{max}$$), and the second term comes from wavefunctions with one infinite rapidity, obtained by choosing $$I_1 = I^{min}$$ and $$I^{min} < I_2 < I^{max}$$ or $$I^{min} < I_1 < I^{max}$$ and $$I_2 = I^{max}$$ (both choices generating the same set of wavefunctions).

There are thus

\begin{equation*} \left(\begin{array}{c} N \\ 2 \end{array} \right) - \left(\begin{array}{c} N-1 \\ 2 \end{array} \right) = N - 1 \end{equation*}

wavefunctions which are not of Bethe form with real rapidities and distinct quantum numbers. This counting was done by Bethe himself in his original article 1931.Bethe.ZP.71.

To search for solutions off the real axis, we begin by noting that since the energy must be real, the two momenta $$k_{1,2} = u_{1,2} + i v_{1,2}$$ must be such that

\begin{align*} 0 = \Im (\cos k_1 + \cos k_2) = \Im (\cos (u_1 + i v_1) + \cos (u_2 + i v_2)) \nonumber \\ = - (\sin u_1 \sinh v_1 + \sin u_2 \sinh v_2) \end{align*}

Since the total momentum must also be real, we have

\begin{equation*} 0 = \Im (k_1 + k_2) = v_1 + v_2 \end{equation*}

so $$\sinh v_1 = -\sinh v_2$$ and therefore $$\sin u_1 = \sin u_2$$, so $$u_2 = u_1$$ or $$u_2 = \pi - u_1$$. For the rapidities, with $$k_1 = u + i v$$,

\begin{align*} 2\lambda_1 = \cot \frac{u + iv}{2} = \frac{\cos \frac{u+iv}{2}}{\sin \frac{u+iv}{2}} = \frac{ \cos \frac{u}{2} \cosh \frac{v}{2} - i \sin \frac{u}{2} \sinh \frac{v}{2}}{\sin \frac{u}{2} \cosh \frac{v}{2} + i \cos \frac{u}{2} \sinh \frac{v}{2}} \nonumber \\ = \frac{(\cos \frac{u}{2} \cosh \frac{v}{2} - i \sin \frac{u}{2} \sinh \frac{v}{2}) (\sin \frac{u}{2} \cosh \frac{v}{2} - i \cos \frac{u}{2} \sinh \frac{v}{2})} {\sin^2 \frac{u}{2} \cosh^2 \frac{v}{2} + \cos^2 \frac{u}{2} \sinh^2 \frac{v}{2}} \nonumber \\ = \frac{1/2}{1/4} \frac{\sin u - i \sinh v}{(1-\cos u)(\cosh v + 1) + (1 + \cos u)(\cosh v - 1)} \end{align*} \begin{equation*} 2\lambda_1 = \frac{\sin u - i \sinh v}{\cosh v - \cos u} \end{equation*}

In the first case, we have $$k_2 = k_1^* = u - iv$$ so

\begin{equation*} 2\lambda_2 = \frac{\sin u + i \sinh v}{\cosh v - \cos u} = 2\lambda_1^* \end{equation*}

In the second case, we have $$k_2 = \pi - u - i v$$ so

\begin{equation*} 2\lambda_2 = \frac{\sin u - i \sinh v}{\cosh v + \cos u} \end{equation*}

\subparagraph{First case: $$k_2 = k_1^* = u - iv$$.} Put

\begin{equation*} \lambda_1 = \mu + i \nu = \lambda_2^* \end{equation*}

The Bethe equations can be written

\begin{align*} \ln \frac{1 -2 \nu + 2 i \mu}{1 + 2\nu - 2i\mu} - \frac{1}{N} \ln \frac{1 - 2\nu}{1+2\nu} = 2\pi i \frac{I_1}{N}, \nonumber \\ \ln \frac{1 +2 \nu + 2 i \mu}{1 - 2\nu - 2i\mu} + \frac{1}{N} \ln \frac{1 - 2\nu}{1+2\nu} = 2\pi i \frac{I_2}{N}, \end{align*}

\begin{equation*} \ln \frac{1 + 2\nu + 2i\mu}{1 + 2\nu - 2i\mu} \frac{1 - 2\nu + 2i\mu}{1 - 2\nu - 2i\mu} = 2\pi i \frac{I_1 +I_2}{N} \end{equation*}

whereas subtracting gives

\begin{equation*} \ln \frac{1 -2 \nu + 2 i \mu}{1 + 2\nu - 2i\mu} \frac{1 - 2\nu - 2i\mu}{1 +2 \nu + 2 i \mu} - \frac{2}{N} \ln \frac{1 - 2\nu}{1+2\nu} = 2\pi i \frac{I_1 - I_2}{N} \end{equation*}

Equating the imaginary parts, we see that either $$I_1 = I_2$$ and $$1-2\nu > 0$$ or $$|I_1 - I_2| = 1$$ and $$1-2\nu < 0$$. Putting $$\nu = 1/2 - \delta$$ with $$\delta < 1/2$$,

\begin{equation*} \ln \frac{\delta + i \mu}{1 - \delta - i\mu} \frac{\delta - i\mu}{1 - \delta + i \mu} - \frac{2}{N} \ln \frac{\delta}{1 -\delta} = 2\pi i \frac{I_1 - I_2}{N} \end{equation*}

so the equations become

\begin{equation} 2 ~\mbox{atan}~ \frac{\mu}{1-\delta} + 2~\mbox{atan}~ \frac{\mu}{\delta} = 2\pi \frac{I_1 + I_2}{N} \tag{h.be2}\label{h.be2} \end{equation} \begin{equation*} \frac{\delta}{1-\delta} = (-1)^{I_1 - I_2} \left[ \frac{\mu^2 + \delta^2}{\mu^2 + (1-\delta)^2} \right]^{N/2} \end{equation*}

If $$\delta \rightarrow 0$$, we must have $$\mu \rightarrow 0$$. Asymptotically, we have $$\mu = \delta^{1/N}$$ for $$\delta \rightarrow 0^+$$. Then, $$\lambda_{1,2} \rightarrow \delta^{1/N} \pm i(1/2 - \delta)$$ so $$\lambda_1 - \lambda_2 \rightarrow i (1-\delta)$$. This means that $$\phi(\lambda_1 - \lambda_2) \rightarrow 2~\mbox{atan}~(i - i\delta) = \frac{1}{i} \ln \frac{\delta}{2} \rightarrow i\infty$$. Also, $$\delta^{1/N} \pm i(1 - 2\delta) = \cot \frac{k_{1,2}}{2} = i \frac{e^{ik_{1,2}/2}+e^{-ik_{1,2}/2}} {e^{ik_{1,2}/2} - e^{-ik_{1,2}/2}}$$. Taking $$k \rightarrow -i\infty$$ we can write $$\frac{e^{ik/2}+e^{-ik/2}}{e^{ik/2} - e^{-ik/2}} = (1 + 2 e^{-ik} + ...)$$ so we can take $$\Re{k_1} = \pi/2$$. Similarly, we find that $$k_2 \rightarrow -i\infty + \pi/2$$. The wavefunction becomes after a rescaling

\begin{equation*} \Psi_2 (j_1, j_2) = (-1)^{j_1} \delta_{j_1+1, j_2}. \end{equation*}

This is a proper wavefunction for even $$N$$ only. Another way of seeing this is to put $$\delta \rightarrow 0$$ directly into h.be2, which becomes $$\pi = 2\pi \frac{I_1 + I_2}{N}$$. Since $$I_1 + I_2$$ must be integer, $$N$$ must be even.

For $$\delta \neq 0$$, the sign of $$\delta$$ is fixed by $$(-1)^{I_1 - I_2}$$. Let's first look at $$\delta > 0$$, so $$I_1 = I_2 \equiv I$$. We call solutions with $$0 < \delta < 1/2$$ narrow pairs''. This last equation can be solved for $$\mu$$,

\begin{equation*} \left[\frac{\delta^2}{(1-\delta)^2}\right]^{1/N} = \frac{\mu^2 + \delta^2}{\mu^2 + (1-\delta)^2}, \hspace{1cm} \mu = \pm \left[ \frac{(1-\delta)^2 \left[\frac{\delta^2}{(1-\delta)^2}\right]^{1/N} - \delta^2}{1 - \left[\frac{\delta^2}{(1-\delta)^2}\right]^{1/N}} \right]^{1/2} \end{equation*}

Note that the parametric curve $$\mu (\delta)$$ is the same as $$\mu(1-\delta)$$. This is fine, since from the definitions we really only need to consider $$\delta < 1/2$$. Defining $$\xi = \frac{\delta}{1-\delta}$$, $$\delta = \frac{\xi}{1+\xi}$$, $$1-\delta = \frac{1}{1+\xi}$$, we get ($$\mu > 0$$ if $$I > 0$$, with similar equations holding for $$\mu < 0$$, which we must choose for $$I < 0$$)

\begin{equation*} \mu = \frac{1}{1+\xi} \left[ \frac{\xi^{2/N} - \xi^2}{1 - \xi^{2/N}} \right]^{1/2} \end{equation*}

Then,

\begin{equation*} \frac{\mu}{1-\delta} = \left[ \frac{\xi^{2/N} - \xi^2}{1 - \xi^{2/N}} \right]^{1/2}, \hspace{1cm} \frac{\mu}{\delta} = \frac{1}{\xi} \left[ \frac{\xi^{2/N} - \xi^2}{1 - \xi^{2/N}} \right]^{1/2}. \end{equation*}

The Bethe equations are solved if

\begin{equation} ~\mbox{atan}~ \left[ \frac{\xi^{2/N} - \xi^2}{1 - \xi^{2/N}} \right]^{1/2} + ~\mbox{atan}~ \frac{1}{\xi} \left[ \frac{\xi^{2/N} - \xi^2}{1 - \xi^{2/N}} \right]^{1/2} = 2\pi \frac{I}{N} \tag{h.np}\label{h.np} \end{equation}

The left-hand side is a monotonously increasing function of $$\xi$$ for $$0 < \xi < 1$$. For $$\xi = 0$$, the value is $$\pi/2$$. The maximum occurs at $$\xi = 1$$. There, we have $$\mu = \frac{\sqrt{N-1}}{2}$$ and $$\delta = 1/2$$, meaning that we are returning to the real axis. Since

\begin{equation*} \lim_{\xi \rightarrow 1} \frac{\xi^{2/N} - \xi^2}{1 - \xi^{2/N}} = \frac{2/N - 2}{-2/N} = N-1 = \lim_{\xi \rightarrow 1} \frac{1}{\xi} \frac{\xi^{2/N} - \xi^2}{1 - \xi^{2/N}} \end{equation*}

we have that the max is equal to $$2 ~\mbox{atan}~ \sqrt{N-1}$$. The $$I$$ we can choose can be written as $$\frac{N-1}{2} - n$$, $$n = 0, 1, ...$$ so we lose the first (pair of, since this also holds for $$I < 0$$, so $$\mu < 0$$) complex solutions for $$N > N_c$$ where

\begin{equation*} 2 ~\mbox{atan}~ \sqrt{N_c - 1} = \pi (1-3/N_c), \hspace{1cm} N_c = 21.8649. \end{equation*}

(this is the equation for $$n = 1$$; the one for $$n = 0$$ has no solution, so we never lose it). The number $$n_{ex} (N)$$ of excluded narrow pairs for given size $$N$$ is given by $$2\lfloor n\rfloor$$, where $$n$$ is solution to

\begin{equation} 2 ~\mbox{atan}~ \sqrt{N - 1} = \frac{\pi}{N} (N - 1 -2n), \hspace{1cm} N \sin^2 \pi \frac{(n + 1/2)}{N} = 1 \tag{h.nex}\label{h.nex} \end{equation}

Therefore,

\begin{equation*} n_{ex} = \lfloor \frac{N - 1}{2} - \frac{N}{\pi} ~\mbox{atan}~ \sqrt{N-1} \rfloor \end{equation*}

The critial values of $$N$$ are thus $$N_{c2} = 21.8649$$, $$N_{c4} = 61.3488$$, $$N_{c6} = 120.568$$, $$N_{c8} = 199.525$$, $$N_{c10} = 246.406$$, $$N_{c12} = 298.222$$, etc. For large $$N$$, we have that

\begin{align*} n_{ex} = \lfloor \frac{N-1}{2} - \frac{N}{2\pi i} \ln \frac{1 + i \sqrt{N-1}}{1-i\sqrt{N-1}} \rfloor = \lfloor -\frac{1}{2} - \frac{N}{2\pi i} \ln \frac{1 - i/\sqrt{N-1}}{1+i/\sqrt{N-1}} \rfloor \nonumber \\ = \lfloor -\frac{1}{2} - \frac{N}{\pi} (- (N-1)^{-1/2} + \frac{1}{3} (N-1)^{-3/2} + O(N^{-5/2}) \rfloor \nonumber \\ = \lfloor -\frac{1}{2} + \frac{\sqrt{N}}{\pi} ( (1-1/N)^{-1/2} - \frac{1}{3} N^{-1} + O(N^{-2}) \rfloor \nonumber \\ = \lfloor -\frac{1}{2} + \frac{\sqrt{N}}{\pi} (1 + \frac{1}{6} N^{-1} + O(N^{-2}) \rfloor = \lfloor \frac{\sqrt{N}}{\pi} - \frac{1}{2} + \frac{N^{-1/2}}{6\pi} + O (N^{-3/2}) \rfloor \end{align*}

This agrees with the result in reference 1992.Essler.JPA.25, where $$n_{max} - 1 = n_{ex}$$ defined here.

For $$I_1 = I_2$$, the number of choices of $$I$$ at our disposal is fixed by observing that equation h.np can only be solved for $$\xi > 0$$ if $$\pi/2 < 2\pi \frac{I}{N} < \pi$$, so $$\frac{N}{4} < \frac{N-1}{2} - n < \frac{N}{2}$$, $$\frac{N-2}{4} > n > 1/2$$, and therefore $$n = 1, ..., \lfloor \frac{N-3}{4} \rfloor$$. We thus have $$2⌊ \frac{N-3}{4} ⌋ • 2n_{ex}(N)$$ narrow pair solutions.

Similarly, for the case $$\delta < 0$$ (which we call wide pairs''), where $$|I_1 - I_2| = 1$$, we get (writing $$\delta = -\tilde{\delta}$$)

\begin{equation*} 2 ~\mbox{atan}~ \frac{\mu}{1+\tilde{\delta}} - 2~\mbox{atan}~ \frac{\mu}{\tilde{\delta}} = 2\pi \frac{I_1 + I_2}{N} \end{equation*} \begin{equation*} \frac{\tilde{\delta}}{1+\tilde{\delta}} = \left[ \frac{\mu^2 + {\tilde{\delta}}^2}{\mu^2 + (1+\tilde{\delta})^2} \right]^{N/2} \end{equation*}

Solving for $$\mu$$,

\begin{equation*} \mu = \pm \left[ \frac{(1+\tilde{\delta})^2 \left[\frac{{\tilde{\delta}}^2}{(1+\tilde{\delta})^2}\right]^{1/N} - {\tilde{\delta}}^2}{1 - \left[\frac{{\tilde{\delta}}^2}{(1+\tilde{\delta})^2}\right]^{1/N}} \right]^{1/2} \end{equation*}

Defining $$\tilde{\xi} = \frac{\tilde{\delta}}{1 + \tilde{\delta}}$$, so $$\tilde{\delta} = \frac{\tilde{\xi}}{1-\tilde{\xi}}$$ and $$1 + \tilde{\delta} = \frac{1}{1-\tilde{\xi}}$$. Since we assume $$0 < \tilde{\delta} < \infty$$, we have $$0 < \tilde{\xi} < 1$$. If $$I_1 + I_2 > 0$$, we must choose $$\mu < 0$$,

\begin{equation*} \mu = -\frac{1}{1-\tilde{\xi}} \left[ \frac{{\tilde{\xi}}^{2/N} - {\tilde{\xi}}^2}{1 - {\tilde{\xi}}^{2/N}} \right]^{1/2} \end{equation*}

Then,

\begin{equation*} \frac{\mu}{1+\tilde{\delta}} = -\left[ \frac{{\tilde{\xi}}^{2/N} - {\tilde{\xi}}^2}{1 - {\tilde{\xi}}^{2/N}} \right]^{1/2}, \hspace{1cm} \frac{\mu}{\tilde{\delta}} = -\frac{1}{\tilde{\xi}} \left[ \frac{{\tilde{\xi}}^{2/N} - {\tilde{\xi}}^2}{1 - {\tilde{\xi}}^{2/N}} \right]^{1/2}. \end{equation*}

The Bethe equations are solved if

\begin{equation*} -~\mbox{atan}~ \left[ \frac{{\tilde{\xi}}^{2/N} - {\tilde{\xi}}^2}{1 - {\tilde{\xi}}^{2/N}} \right]^{1/2} + ~\mbox{atan}~ \frac{1}{\tilde{\xi}} \left[ \frac{{\tilde{\xi}}^{2/N} - {\tilde{\xi}}^2}{1 - {\tilde{\xi}}^{2/N}} \right]^{1/2} = \pi \frac{I_1 + I_2}{N} \end{equation*}

For the allowed values of $$\tilde{\xi}$$, the left-hand side takes on all values in the interval $$]0, \pi/2[$$, so $$0 < I_1 + I_2 < N/2$$. But $$I_1 + I_2$$ can here be written as $$N - 2n$$ with $$n \geq 0$$, so we have $$N/2 > n > N/4$$ so $$\lfloor \frac{N+1}{4}\rfloor \leq n \leq \lfloor \frac{N-1}{2} \rfloor$$. There are thus $$\lfloor \frac{N-1}{2} \rfloor - \lfloor \frac{N+1}{4} \rfloor + 1$$ choices for $$n > 0$$. (Note: this is equal to $$\lfloor \frac{N-1}{4} + 1\rfloor$$). We can thus construct $$2 \lfloor \frac{N-1}{2} \rfloor - 2\lfloor \frac{N+1}{4} \rfloor + 2$$ distinct wide pair Bethe states.

The total number of states that we have constructed is thus

\begin{align*} 1 - (N \% 2) + 2 \lfloor \frac{N-3}{4}\rfloor - n_{ex}(N) + 2 \lfloor \frac{N-1}{2} \rfloor - 2\lfloor \frac{N+1}{4} \rfloor + 2 \nonumber \\ = (1 - (N \% 2) + 2 \lfloor \frac{N-1}{2}) + (2 \lfloor \frac{N-3}{4}\rfloor - 2\lfloor \frac{N+1}{4} \rfloor + 2) - n_{ex}(N) \nonumber \\ = N - 1 + 0 - n_{ex}(N) = N - 1 - n_{ex}(N) \end{align*}

\subparagraph{Second case: $$k_2 = \pi - k_1 = \pi - u - iv$$, $$v \neq 0$$.} We have then

\begin{equation*} 2\lambda_2 = \cot \frac{\pi - k_1}{2} = \tan \frac{k_1}{2} = \frac{1}{2\lambda_1} \end{equation*}

The Bethe equations become

\begin{equation*} ~\mbox{atan}~ 2\lambda_1 - \frac{1}{N} ~\mbox{atan}~ \frac{\lambda_1^2 - 1/4}{\lambda_1} = \pi \frac{I_1}{N}, \hspace{0.3cm} ~\mbox{atan}~ \frac{1}{2\lambda_1} + \frac{1}{N} ~\mbox{atan}~ \frac{\lambda_1^2 - 1/4}{\lambda_1} = \pi \frac{I_2}{N} \end{equation*}

Adding these, we get ($$\lambda \equiv \lambda_1$$)

\begin{align*} ~\mbox{atan}~ 2\lambda + ~\mbox{atan}~ \frac{1}{2\lambda} = \frac{1}{2i} \ln \frac{1 + 2i\lambda}{1-2i\lambda} \frac{1 + i/2\lambda}{1-i/2\lambda} = \pi/2 ~~\mbox{mod} ~~\pi = \pi \frac{I_1 + I_2}{N}, \nonumber \\ I_1 + I_2 = N/2 ~~ \mbox{mod} ~N \end{align*}

Subtracting,

\begin{equation*} ~\mbox{atan}~ 2\lambda - ~\mbox{atan}~ \frac{1}{2\lambda} - \frac{2}{N} ~\mbox{atan}~ \frac{\lambda^2 - 1/4}{\lambda} = \pi\frac{I_1 - I_2}{N} \end{equation*}

The left-hand side simplifies by using

\begin{equation*} ~\mbox{atan}~ 2\lambda - ~\mbox{atan}~ \frac{1}{2\lambda} = \frac{1}{2i} \ln \frac{1 + 2i\lambda}{1-2i\lambda} \frac{1 - i/2\lambda}{1+i/2\lambda} %=\frac{1}{2i} \ln \frac{1 + i \frac{\lambda^2 - 1/4}{\lambda}}{1 - i \frac{\lambda^2 - 1/4}{\lambda}} = ~\mbox{atan}~ \frac{\lambda^2 - 1/4}{\lambda} \end{equation*}

so

\begin{align*} ~\mbox{atan}~ \frac{\lambda^2 - 1/4}{\lambda} = \pi \frac{I_1 - I_2}{N-2}, \hspace{1cm} \frac{\lambda^2 - 1/4}{\lambda} = \tan \pi \frac{I_1 - I_2}{N-2} \equiv f, \nonumber \\ \lambda = \frac{f \pm \sqrt{f^2 + 1}}{2} \end{align*}

Since $$f$$ is real, we always have that $$\lambda$$ is real, so $$v = 0$$, which contradicts our hypothesis (since both rapidities here are real, these wavefunctions have already been counted in the previous section).

\subparagraph{Coinciding quantum numbers} If we put $$I_1 = I_2 \equiv I$$, we get

\begin{equation*} ~\mbox{atan}~ 2\lambda_1 - \frac{1}{N} ~\mbox{atan}~ (\lambda_1 - \lambda_2) = \pi \frac{I}{N} = ~\mbox{atan}~ 2\lambda_2 + \frac{1}{N} ~\mbox{atan}~ (\lambda_1 - \lambda_2) \end{equation*}

so

\begin{equation*} ~\mbox{atan}~ 2\lambda_1 + ~\mbox{atan}~ 2\lambda_2 = 2\pi \frac{I}{N}, \hspace{1cm} ~\mbox{atan}~ 2\lambda_1 - ~\mbox{atan}~ 2\lambda_2 = \frac{2}{N} ~\mbox{atan}~ (\lambda_1 - \lambda_2) \end{equation*}

The second equation can be rewritten (using $$\lambda_1 \equiv \lambda$$ and $$\lambda_2 \equiv \lambda + \delta$$)

\begin{align*} 2\lambda = \tan \left( ~\mbox{atan}~ (2\lambda +2\delta) - \frac{2}{N} ~\mbox{atan}~ \delta \right) \nonumber \\ = \tan \left( \frac{\pi}{2} - ~\mbox{atan}~ \frac{1}{2\lambda +2\delta} - \frac{2}{N} ~\mbox{atan}~ \delta \right) \nonumber \\ = \frac{1}{\tan \left(~\mbox{atan}~ \frac{1}{2\lambda +2\delta} + \frac{2}{N} ~\mbox{atan}~ \delta \right)} = \frac{1 - \frac{1}{2\lambda + 2\delta} \tan(\frac{2}{N}~\mbox{atan}~\delta)}{\tan(\frac{2}{N}~\mbox{atan}~\delta) + \frac{1}{2\lambda + 2\delta}} \end{align*}

so, defining $$t = \tan (\frac{2}{N} ~\mbox{atan}~ \delta)$$,

\begin{equation*} 2\lambda (t (2\lambda + 2\delta) + 1) = 2\lambda + 2\delta -t, \hspace{0.3cm} \lambda^2 + \lambda \delta + \frac{1}{4} - \frac{\delta}{2t} = 0, \hspace{0.3cm} 2\lambda = -\delta + \sqrt{\delta^2 + \frac{2\delta}{t} - 1} \end{equation*}

Therefore, $$2\lambda_1 = g - \delta$$ and $$2\lambda_2 = g + \delta$$, with $$g = \sqrt{\delta^2 + \frac{2\delta}{t} - 1}$$. The first Bethe equation gives

\begin{equation*} ~\mbox{atan}~(g - \delta) + ~\mbox{atan}~(g + \delta) = 2\pi \frac{I}{N} \end{equation*}

The left-hand side takes its minimum value at $$\delta = 0$$. Then, $$\lim_{\delta \rightarrow 0} \frac{t}{\delta} = \frac{2}{N}$$. We thus find, writing $$I = \frac{N-1}{2} - n$$,

\begin{equation*} 2 ~\mbox{atan}~ \sqrt{N-1} = \frac{\pi}{N} (N - 1 - 2n) \end{equation*}

which is the same condition as h.nex. Therefore, each time we lose a narrow pair, we gain an extra real solution obtained from using the same quantum numbers for the two rapidities.

The total number of solutions we have found is thus

\begin{equation*} \left(\begin{array}{c} N \\ 2 \end{array} \right) \end{equation*}

which coincides with the dimension of the Hilbert subspace. There are $$\left(\begin{array}{c} N - 2 \\ 2 \end{array} \right)$$ wavefunctions with real, finite rapidities with distinct quantum numbers, $$N-2$$ wavefunctions with one infinite rapidity, one wavefunction with $$\lambda_{1,2} = \pm i/2$$ for $$N$$ even, $$2\lfloor \frac{N-3}{4} \rfloor - n_{ex}(N)$$ narrow pair eigenstates, $$2\lfloor \frac{N-1}{2} \rfloor - 2\lfloor \frac{N+1}{4} \rfloor + 2$$ wide pair eigenstates and $$n_{ex}(N)$$ extra real pairs with coinciding quantum numbers. The two-particle sector of the isotropic model is thus fully understood.

Exercise: extend this to arbitrary $$M \leq N/2$$ (email me if you manage). Some steps in this direction are taken in 2007.Hagemans.JPA.40. Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Created: 2023-06-07 Wed 16:02