Quantize magnon momenta on a ring

Which locally scattered waves actually fit on a finite ring? Transporting one down spin around the periodic chain changes its coordinate ordering and exchanges the two momentum assignments. You will derive the resulting Bethe equations, solve two explicit six-site examples, and separate a valid eigenstate from a merely formal root set.

Required background. Use the wavefunction and amplitude ratio from Solve two-magnon scattering, together with the translation convention from Solve the one-magnon sector.

Helpful background. The Library treatment gives a connected derivation; the convention reference records the reciprocal scattering phases that occur below.

Periodicity exchanges the ordered coordinates

Use the finite periodic spin-$1/2$ XXX model with $N\geq3$, $J\gt0$, $\hbar=1$, lattice spacing one, and zero all-up energy. For ordered down-spin positions,

$$\psi(x,y)=A_{12}z_1^xz_2^y+A_{21}z_2^xz_1^y,\qquad x\lt y,$$

where $z_j=e^{ik_j}$ and, in the regular case,

$$S_{12}=\frac{A_{21}}{A_{12}} =-\frac{1+z_1z_2-2z_2}{1+z_1z_2-2z_1}.$$

The configurations $(x,y)$ and $(y,x+N)$ describe the same pair of occupied sites, with their displayed order changed. Thus require

$$\psi(x,y)=\psi(y,x+N).$$

Expanding the right side gives

$$\begin{aligned} \psi(y,x+N) ={}&A_{12}z_2^N z_2^xz_1^y\\ &+A_{21}z_1^N z_1^xz_2^y. \end{aligned}$$

For distinct momentum assignments in the regular ansatz, coefficient matching therefore gives

$$A_{12}=A_{21}z_1^N,\qquad A_{21}=A_{12}z_2^N,$$

or the two periodic Bethe equations

$$z_1^N=\frac1{S_{12}},\qquad z_2^N=S_{12}.$$

The inverse in the first equation is essential. It follows from the stated ratio $A_{21}/A_{12}$; it cannot be chosen independently. Compare Karbach and Müller 1997, p. 3, equations (14)–(18), arXiv v1 PDF, whose $e^{i\theta}$ is $1/S_{12}$.

Entry check and repair

Starting from $A_{12}=A_{21}z_1^N$, solve for $z_1^N$ when $A_{21}/A_{12}=S_{12}$. Multiply the two periodic equations. What does their product constrain?

Repair. Dividing by $A_{21}$ gives $z_1^N=A_{12}/A_{21}=1/S_{12}$. Multiplication yields

$$(z_1z_2)^N=1, \qquad K=k_1+k_2=\frac{2\pi\ell}{N}\pmod{2\pi},$$

for an integer $\ell$. Total wave number is quantized in the same lattice as a one-magnon wave number, but the individual interacting momenta generally are not. With $T|x,y\rangle$ translating both occupied sites forward, the coefficient convention $e^{+ikx}$ gives translation eigenvalue $e^{-iK}$.

Write the equations in rapidities

For finite rapidities away from poles, use

$$z_j=\frac{\lambda_j+i/2}{\lambda_j-i/2}, \qquad S_{12}=\frac{\lambda_1-\lambda_2-i} {\lambda_1-\lambda_2+i}.$$

The periodic equations become

$$\left(\frac{\lambda_j+i/2}{\lambda_j-i/2}\right)^N =\frac{\lambda_j-\lambda_\ell+i} {\lambda_j-\lambda_\ell-i}, \qquad \{j,\ell\}=\{1,2\}.$$

The energy is

$$E=J(2-\cos k_1-\cos k_2) =\frac J2\left( \frac1{\lambda_1^2+1/4}+ \frac1{\lambda_2^2+1/4}\right).$$

The rational form is often convenient for solving equations, but its denominators and omitted infinite rapidities matter. Do not divide by a vanishing factor or identify every algebraic solution with a physical state.

Solve two six-site states

Take $N=6$ and opposite real momenta $k_1=k$, $k_2=-k$, excluding the coincident cases $k=0,\pi$. The previous lesson gives $S_{12}=e^{-ik}$. Thus the first periodic equation is

$$e^{6ik}=e^{ik}, \qquad e^{5ik}=1.$$

Two distinct unordered momentum pairs are represented by

$$k=\frac{2\pi}{5} \quad\hbox{or}\quad k=\frac{4\pi}{5}.$$

The negative choices just exchange the two momenta. Their exact energies and rapidities are

Positive $k$	$E/J$	$(\lambda_1,\lambda_2)$
$2\pi/5$	$(5-\sqrt5)/2$	$\bigl(\tfrac12\cot(\pi/5),-\tfrac12\cot(\pi/5)\bigr)$
$4\pi/5$	$(5+\sqrt5)/2$	$\bigl(\tfrac12\cot(2\pi/5),-\tfrac12\cot(2\pi/5)\bigr)$

Neither individual momentum lies on the one-magnon grid $2\pi m/6$. Their sum is zero, so both states have translation eigenvalue one.

With $A_{12}=1$, a useful expression for the actual amplitudes is

$$\begin{aligned} \psi(x,y) &=e^{-ik(y-x)}+e^{-ik}e^{ik(y-x)}\\ &=2e^{-ik/2}\cos\!\left[k\left(y-x-\tfrac12\right)\right]. \end{aligned}$$

This vector is not zero: for adjacent sites its coefficient is $2e^{-ik/2}\cos(k/2)\ne0$ for both choices. It also respects the closing contact. If $d=y-x$, then $e^{i(N-1)k}=1$ implies

$$\psi(N-d)=\psi(d).$$

In particular, the separation $N-1$ across the displayed ends has the same amplitude as separation $1$. This is a direct check of the periodic identification, beyond checking the energy formula.

The local free and contact equations, together with this identification, cover the configurations in the ring Hamiltonian. Thus each nonzero vector constructed here is an eigenstate. There are $\binom62=15$ basis states in the full two-magnon sector; exhibiting these two does not enumerate the other thirteen.

Check roots, vectors, and the boundary separately

For a computed candidate, reconstruct the vector in the ordered spin basis before drawing conclusions. Check the contact relation and both periodic equations, establish $\|\psi\|_2\ne0$, and then measure the normalized eigenvector residual

$$r=\frac{\|H\psi-E\psi\|_2}{J\|\psi\|_2}.$$

The finite-chain project compares this test with independently assembled matrices and diagonalization. A small residual for one vector confirms that finite case; it does not establish completeness.

Real scattering momenta are only part of the story. Bound-state solutions can require complex momenta. A zero-momentum excitation corresponds to an infinite rapidity, while singular rapidity values such as $\lambda=\pm i/2$ make the regular formulas undefined. Such cases need limiting or separate constructions, not rejection merely because an ordinary root solver missed them. See the complex and exceptional two-magnon solutions in Karbach and Müller 1997, p. 4, equations (20)–(25) and the following exceptional case, arXiv v1 PDF.

Exercises

Guided practice: free momenta fail at the seam

For $N=6$, propose $k_1=\pi/3$, $k_2=-\pi/3$. Each momentum obeys the one-magnon periodic condition. Find the contact amplitude and evaluate $\lvert z_1^6-1/S_{12}\rvert$. Does the candidate satisfy the two-magnon periodic equation?

Hint

For opposite momenta, $S_{12}=e^{-ik_1}$. Compare $1$ with $e^{i\pi/3}$.

Solution

Contact requires $S_{12}=e^{-i\pi/3}$, while $z_1^6=1$. Therefore

$$\left|z_1^6-\frac1{S_{12}}\right| =|1-e^{i\pi/3}|=1.$$

The periodic condition fails by an order-one amount. The local dispersion would give $E=J$, but that value does not turn this particular wavefunction into a ring eigenstate. Even if some other state has the same energy, energy agreement alone would not check the proposed vector.

Independent practice: a five-site scattering state

Change to $N=5$. Verify that $k_1=\pi/2$, $k_2=-\pi/2$ gives a regular, nonzero two-magnon eigenstate. Find $S_{12}$, both rapidities, the energy, and the total translation eigenvalue.

Hint

Evaluate $i^5$ and $(-i)^5$. Test one adjacent coefficient to rule out the zero vector.

Solution

The local matching gives $S_{12}=-i$, and the periodic equations read

$$i^5=i=\frac1{-i},\qquad (-i)^5=-i=S_{12}.$$

The rapidities are $(1/2,-1/2)$ and $E=2J$. With $A_{12}=1$, the adjacent coefficient is $\psi(1,2)=1-i\ne0$. The local contact calculation from the preceding lesson and these periodic identities establish the eigenstate. Its total wave number is zero, so the active translation eigenvalue is one.

Transfer: an infinite rapidity with a finite state

Let $q=2\pi m/N$ with $m\ne0\pmod N$, and try $k_1=0$, $k_2=q$. Work in the $z$ variables, where $z_1=1$. Find the exchange amplitude and identify the state obtained by applying $S_{\rm tot}^-$ to a one-magnon state. Why would a search restricted to finite rapidities miss this construction?

Hint

Since $e^{iq}\ne1$, the original amplitude ratio is defined. In $S_{\rm tot}^-|q\rangle$, a final pair $(x,y)$ can arise by flipping site $x$ or site $y$.

Solution

Writing $w=e^{iq}\ne1$ gives

$$S_{12}=-\frac{1+w-2w}{1+w-2}=1.$$

The Bethe equations reduce to $1=1$ and $w^N=1$, and the coefficients are

$$\psi(x,y)=w^x+w^y.$$

Indeed,

$$S_{\rm tot}^-|q\rangle =\frac1{\sqrt N}\sum_{x\lt y}(w^x+w^y)|x,y\rangle.$$

For $N\geq3$ this is nonzero. One way to check is to sum its squared coefficients: since $\sum_xw^x=0$, its squared norm is $N-2$. Each bond swap commutes with the sum of spin-lowering operators, so $[H,S_{\rm tot}^-]=0$. The descendant therefore has the one-magnon energy $J(1-\cos q)$, agreeing with the two-magnon formula at $k_1=0$.

The rapidity $\lambda_1=\tfrac12\cot(k_1/2)$ is infinite. A search over finite rapidities would omit this valid state unless the descendant is restored separately. The doubly zero-momentum case has a further $0/0$ in the displayed scattering ratio and must be constructed separately, for example from $(S_{\rm tot}^-)^2|F\rangle$.

Test the constructed eigenstates

You can now derive finite-ring quantization with a fixed scattering convention, construct explicit nonzero solutions, and recognize several limits of the regular-root description. Continue to Test finite-chain Bethe solutions to check the six-site examples against direct spin matrices and deliberate boundary-condition mistakes.

References

• Karbach, Michael, and Gerhard Müller. “Introduction to the Bethe Ansatz I.” Computers in Physics 11, 36–43 (1997). DOI. Author version arXiv:cond-mat/9809162v1 (1998); Open PDF.