Reproduce a singular Bethe state

Can a nearly vanishing Bethe-equation residual accompany the wrong quantum state? Yes: a singular pair of rapidities on a four-site XXX ring needs a fourth-order correction to produce an eigenvector. You will reconstruct that vector, test it against an independently built Hamiltonian, and explain why the same pair fails on five sites. This is a reproduction of an established finite-system result, with an analytic target and an executable comparison.

Required background. Construct a small spin-chain Hamiltonian supplies the bond action. Build monodromy and transfer matrices supplies auxiliary blocks and the shift convention. You need complex vectors, polynomial limits, and Python with NumPy for the computation.

Helpful background. Construct Bethe vectors algebraically explains ordinary unwanted-term cancellation and a regular nonzero two-magnon state before the singular case here. Solve two-magnon scattering explains the adjacent-pair contact equation. Derive commuting charges relates the transfer matrix to the Hamiltonian. The research article discusses the evidence and boundaries of this reproduction.

A singular pair on a finite XXX ring

Take spin $1/2$, $\mathbf S_n=\boldsymbol\sigma_n/2$, $\hbar=1$, and the periodic Hamiltonian

$$H=\frac J2\sum_{n=1}^N(I-P_{n,n+1}), \qquad J\gt0,\qquad P_{N,N+1}=P_{N,1}.$$

The main calculation uses $N=4$; the changed-setting check uses $N=5$. The all-up vector $|0\rangle=|\uparrow\cdots\uparrow\rangle$ has energy zero. Write $|xy\rangle=S_x^-S_y^-|0\rangle$, with site labels $1\leq x\lt y\leq N$. These two-down-spin vectors form an orthonormal basis. The rapidities, spectral parameters and regulator below are dimensionless; $J$ sets the energy unit.

For two distinct nonsingular creation parameters, denoted by $\lambda_j$ in this project, the rational Bethe equations are

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

The pair $(i/2,-i/2)$ lies outside this formula’s domain. Define its cleared polynomials by

$$\begin{aligned} F_j={}&(\lambda_j+i/2)^N(\lambda_j-\lambda_k-i)\\ &-(\lambda_j-i/2)^N(\lambda_j-\lambda_k+i). \end{aligned}$$

Both $F_j$ vanish at the singular pair. This observation does not restore the excluded denominators or construct a nonzero eigenvector. The distinction and the corrected four-site example are developed in Nepomechie and Wang 2013, pp. 1–2, equations (4)–(11), PDF.

Your concrete deliverables are a convention conversion, a six-component limiting vector with independent checks, a table distinguishing equation and state residuals, and an exact five-site rejection. None requires solving the entire XXX spectrum.

Entry check and repair

Let $H_*=\operatorname{diag}(0,2J)$ and $\psi=(1,1)^{\mathsf T}/\sqrt2$. Its mean energy is $J$. Is it an eigenvector with eigenvalue $J$? Would testing an unnormalized zero vector repair the problem?

Repair. The full equation gives

$$(H_*-J I)\psi=\frac{J}{\sqrt2}(-1,1)^{\mathsf T}, \qquad \frac{\|(H_*-J I)\psi\|_2}{J}=1.$$

The mean energy tests one scalar; the eigenvector equation tests every component. The zero vector satisfies every homogeneous eigenvector equation but represents no state and cannot be normalized. Always establish a nonzero vector, normalize it, and test the full equation.

Build the creation block

Use the site’s polynomial local operator and ordered monodromy:

$$\begin{aligned} L_{an}(\lambda)&=(\lambda-i/2)I+iP_{an},\\ T_a(\lambda)&=L_{aN}(\lambda)\cdots L_{a1}(\lambda) =\begin{pmatrix}A(\lambda)&B(\lambda)\\C(\lambda)&D(\lambda)\end{pmatrix}_a,\\ \tau(\lambda)&=A(\lambda)+D(\lambda). \end{aligned}$$

The auxiliary basis is $(\uparrow,\downarrow)$, so $B$ is the upper-right auxiliary block, a $2^N\times2^N$ operator on physical spins. Locally,

$$L_{an}(\lambda)= \begin{pmatrix} \lambda+iS_n^z&iS_n^-\\ iS_n^+&\lambda-iS_n^z \end{pmatrix}_a.$$

Here $S_n^-|\uparrow\rangle_n=|\downarrow\rangle_n$ and $S_n^+|\uparrow\rangle_n=0$. This fixes which block creates a down spin; reversing the auxiliary basis without changing the extraction would select the wrong block.

For a reproducible construction, start with $T^{(0)}=I$ and multiply $T^{(n)}=L_{an}T^{(n-1)}$ for $n=1,\ldots,N$. For example, its upper-right block obeys

$$B^{(n)}=(\lambda+iS_n^z)B^{(n-1)} +iS_n^-D^{(n-1)}.$$

The other three blocks follow from the same $2\times2$ multiplication. Keep every physical tensor factor in the order $1,\ldots,N$. A useful first check is

$$B(\lambda)|0\rangle =i\sum_{x=1}^N (\lambda+i/2)^{N-x}(\lambda-i/2)^{x-1}|x\rangle.$$

Before the auxiliary down spin flips at site $x$, each visited up spin contributes $\lambda-i/2$; after the flip, the remaining sites contribute $\lambda+i/2$.

Momentum convention. The parameters in this monodromy calculation are creation parameters $\lambda_B$, written simply as $\lambda$ below. The adjacent one-spin coefficients have ratio $(\lambda_B-i/2)/(\lambda_B+i/2)$. Matching the coordinate course’s $e^{ikx}$ wave therefore uses $\lambda_{\rm coord}=-\lambda_B$. On four sites, $B(1/2)|0\rangle$ has coordinate momentum $-\pi/2$, active translation eigenvalue $i$, and energy $J$. The convention reference derives this direction-sensitive check. The singular pair $\pm i/2$ is unchanged as a set by sign reversal; its limiting vector and its translation will be checked directly here.

Applying a second $B$ creates a vector in the two-down-spin sector. At $N=4$, define

$$\Phi(a,b)=B(a)B(b)|0\rangle.$$

Direct substitution gives $\Phi(i/2,-i/2)=0$. The task is to obtain a meaningful direction as this zero is approached.

Match the source normalization

The 2013 paper uses a scalar-normalized local operator. With its quantities denoted by the subscript NW,

$$\begin{aligned} L_{\mathrm{NW}}(\lambda)&=(\lambda+i/2)^{-1}L(\lambda),\\ B_{\mathrm{NW}}(\lambda)&=(\lambda+i/2)^{-N}B(\lambda),\\ H&=-JH_{\mathrm{NW}}. \end{aligned}$$

These conversions follow from Nepomechie and Wang 2013, equation (1), equation (13), and Appendix equations (34)–(37), PDF. A nonzero scalar multiplying a nonzero vector changes its normalization and possibly its phase, but not its physical ray. A scalar that vanishes or diverges in a limit can nevertheless change the unnormalized limit. Compare normalized states at nonzero regulator before interpreting apparent discrepancies between papers.

Keep the term that survives normalization

For real $\epsilon\gt0$, set

$$\lambda_1=i/2+\epsilon+c\epsilon^4, \qquad \lambda_2=-i/2+\epsilon, \qquad w_\epsilon(c)=\epsilon^{-4}\Phi(\lambda_1,\lambda_2).$$

The complex coefficient $c$ is held fixed as $\epsilon\to0^+$. These finite-$\epsilon$ values are a limiting prescription, not exact solutions of the untwisted Bethe equations. The correction and its coefficient appear in Nepomechie and Wang 2013, pp. 2–4, equations (10) and (20)–(22), PDF.

First set $c=0$. Multiplying the blocks above and collecting coefficients in the ordered basis $(|12\rangle,|13\rangle,|14\rangle,|23\rangle,|24\rangle,|34\rangle)$ gives the exact polynomial

$$w_\epsilon(0)=-2 \begin{pmatrix} (\epsilon+i)^2\\ \epsilon(\epsilon+i)\\ \epsilon^2\\ \epsilon^2+1\\ \epsilon(\epsilon-i)\\ (\epsilon-i)^2 \end{pmatrix}.$$

This is a compact target for a first reproduction: all lower powers than $\epsilon^4$ in the unscaled vector cancel. Taking its limit yields

$$v(0)=2|12\rangle-2|23\rangle+2|34\rangle.$$

Next differentiate the same finite matrix polynomial before taking the singular limit. The block multiplication gives

$$\left.\partial_a\Phi(a,b)\right|_{(a,b)=(i/2,-i/2)} =i|14\rangle.$$

Taylor expansion in the change $c\epsilon^4$ of the first argument therefore implies

$$\begin{aligned} v(c)&=\lim_{\epsilon\to0^+}w_\epsilon(c)\\ &=2|12\rangle-2|23\rangle+2|34\rangle+ic|14\rangle. \end{aligned}$$

The fourth-order change in a root contributes at leading order after dividing the vanishing vector by $\epsilon^4$. It is not negligible for the state, even though it is much smaller than the common first-order displacement of the roots.

Test the vector with the bond Hamiltonian

Build $H$ from spin exchanges, independently of the monodromy. For each adjacent-pair state on four sites,

$$\frac HJ|xy\rangle =|xy\rangle-\frac12\bigl(|13\rangle+|24\rangle\bigr), \quad xy\in\{12,23,34,14\}.$$

For example, only bonds $(2,3)$ and $(4,1)$ act nontrivially on $|12\rangle$. They move one down spin outward to produce $|13\rangle$ and $|24\rangle$. The closing bond is essential.

Adding the four contributions gives

$$\left(\frac HJ-I\right)v(c) =-\left(1+\frac{ic}{2}\right) \bigl(|13\rangle+|24\rangle\bigr).$$

Cancellation requires $c=2i$. Since $\|v(2i)\|_2=4$, the normalized state is

$$\chi=\frac12\bigl(|12\rangle-|23\rangle+|34\rangle-|14\rangle\bigr), \qquad H\chi=J\chi.$$

Two independent checks identify its quantum numbers. The rightward translation sends sites $x\mapsto x+1$ modulo four and hence $U\chi=-\chi$. Also $S^z_{\mathrm{tot}}\chi=0$ and $S^+_{\mathrm{tot}}\chi=0$: in the latter sum every one-down-spin amplitude cancels. Using $\mathbf S_{\mathrm{tot}}^2=S^-_{\mathrm{tot}}S^+_{\mathrm{tot}}+S^z_{\mathrm{tot}}(S^z_{\mathrm{tot}}+1)$ shows that $\chi$ is a spin singlet.

As a polynomial check on the original monodromy, collect powers of $\lambda$ in $\tau(\lambda)\chi$. They give

$$\tau(\lambda)\chi=\Lambda_4(\lambda)\chi, \qquad \Lambda_4(\lambda)=2\lambda^4+3\lambda^2-\frac38.$$

At $\lambda=i/2$ this is $-1$, agreeing with $\tau(i/2)=i^4U$. Its logarithmic derivative also returns $H\chi=J\chi$ under the extraction formula in Derive commuting charges. These checks concern the same finite vector through distinct constructions.

Separate three numerical questions

For a nonzero $w_\epsilon$, normalize $\psi_\epsilon=w_\epsilon/\|w_\epsilon\|_2$. Measure

$$\begin{aligned} r_{\mathrm{root}}&=\max(|F_1|,|F_2|),\\ r_H&=\frac{\|H\psi_\epsilon-J\psi_\epsilon\|_2}{J},\\ d&=\min_{|\zeta|=1}\|\psi_\epsilon-\zeta\chi\|_2. \end{aligned}$$

The first is an unnormalized residual of the displayed cleared polynomials. Multiplying those polynomials by a small scalar would change it; it is neither a relative root error nor a rational-equation residual. The second tests an energy-$J$ eigenvector with the Euclidean norm. The third compares normalized states after removing an irrelevant global phase. Evaluate $d$ by aligning the overlap phase and subtracting vectors; the equivalent formula $\sqrt{2-2|\langle\chi,\psi_\epsilon\rangle|}$ loses precision near agreement.

For the naive prescription, exact substitution gives $F_1=F_2=-2i\epsilon^4$, so $r_{\mathrm{root}}=2\epsilon^4\to0$. Yet

$$\left\langle\frac{v(0)}{\sqrt{12}}, H\frac{v(0)}{\sqrt{12}}\right\rangle=J, \qquad r_H\longrightarrow\frac1{\sqrt6}\ne0.$$

The wrong state passes both a small cleared-equation residual and the limiting mean-energy check. It fails the full vector equation. For $c=2i$, by contrast, the stable computation has $r_H\to0$, $d\to0$, and $r_{\mathrm{root}}=8\epsilon^5+O(\epsilon^7)$.

Reproduce the comparison

Download and extract the complete singular-pair experiment (ZIP), then open its singular-xxx folder. Individual files are also available: Python experiment, inputs, saved results, requirements, and instructions. With the specified dependencies installed, run:

python3 experiment.py --check

Inside the website checkout, use:

python3 public/computations/singular-xxx/experiment.py --check

The experiment forms $B$ from monodromy blocks and $H$ from separate bond actions. It compares three calculations: corrected polynomial evaluation after removing the common $\epsilon^4$, direct binary64 evaluation at the corrected rapidities, and stable evaluation with the correction omitted. In the stable calculation, coefficient arithmetic exposes the common zero before evaluating the remaining polynomial; simply dividing an already inaccurate vector by $\epsilon^4$ cannot recover lost digits.

The recorded Python 3.9.6 / NumPy 2.0.2 run uses $J=1$ and $\epsilon=10^{-1},10^{-2},\ldots,10^{-8}$. Selected Hamiltonian residuals are:

$\epsilon$	Corrected, stable $r_H$	Corrected, direct $r_H$	Naive, stable $r_H$
$10^{-2}$	$1.4140\times10^{-2}$	$1.4140\times10^{-2}$	$0.40819$
$10^{-4}$	$1.4142\times10^{-4}$	$0.037884$	$0.40825$
$10^{-6}$	$1.4142\times10^{-6}$	$0.40841$	$0.40825$
$10^{-8}$	$1.4142\times10^{-8}$	$0.67102$	$0.40825$

The direct column is an illustration of arithmetic breakdown in the recorded environment; its last digits are platform-sensitive. When $2i\epsilon^4$ is added to a rapidity whose imaginary part is near $1/2$, binary64 can no longer resolve the correction at sufficiently small $\epsilon$. Cancellation in the subsequent matrix products adds further error. Decreasing the regulator then improves the analytic approximation while worsening this implementation.

The stable corrected distance behaves as $d\sim2\epsilon$, and its Hamiltonian residual as $r_H\sim\sqrt2\epsilon$. Thus the nonzero residual at a finite regulator is expected; the family was never asserted to solve the untwisted equations exactly. Independent identity checks use tolerance $2\times10^{-12}$. Stable saved-result comparisons use absolute tolerance $2\times10^{-12}$ and relative tolerance $10^{-8}$; small-regulator direct results are recomputed without requiring agreement with their cancellation-sensitive saved values. The instructions detail those checks. The analytic limiting identities and the finite-precision regulator study answer different questions.

Change the chain length before generalizing

The cleared equations also vanish at the singular pair on five sites. To test whether this is a physical state, set $Q(u)=u^2+1/4$ and form the candidate from the polynomial relation

$$\Lambda_N(u)Q(u) =(u+i/2)^NQ(u-i)+(u-i/2)^NQ(u+i).$$

Factoring $Q(u\mp i)$ cancels $Q(u)$ and gives a polynomial:

$$\begin{aligned} \Lambda_N(u)={}&(u+i/2)^{N-1}(u-3i/2)\\ &+(u-i/2)^{N-1}(u+3i/2). \end{aligned}$$

A polynomial candidate is still not an eigenvalue certificate. If a nonzero common eigenvector with this transfer eigenvalue existed, regularity would require

$$\Lambda_N(i/2)=-i^N, \qquad U\psi=-\psi.$$

But $U^N=I$. For odd $N$, applying $U$ $N$ times would give both $\psi$ and $-\psi$, which is impossible for a nonzero vector. This recovers the two-root parity obstruction discussed in Nepomechie and Wang 2014, pp. 2 and 4, equations (6) and (14), PDF. That paper also obtains the four-site state through a twist limit, equations (17)–(18), rather than the present regulator path.

An independent five-site check uses the ten-dimensional two-down-spin Hamiltonian. The candidate would have energy $J$; exact bond-matrix arithmetic instead gives

$$\det(H/J-I)=-\frac1{256}\ne0.$$

Thus energy $J$ is absent from this sector. This is a finite rejection of this candidate, not a claim that all other Bethe states have been classified.

Exercises

Guided practice: reconcile two normalizations

For the four-site regulator above, express $\Phi_{\mathrm{NW}}=B_{\mathrm{NW}}(\lambda_1)B_{\mathrm{NW}}(\lambda_2)|0\rangle$ in terms of $w_\epsilon(c)$. Determine its unnormalized limit. Explain why the source can report a finite vector while the site’s polynomial vector tends to zero. Convert the source energy $E_{\mathrm{NW}}=-1$ to the site’s energy.

Hint

There are four scalar factors in each creation block. Here $\lambda_2+i/2=\epsilon$ and $\lambda_1+i/2=i+\epsilon+c\epsilon^4$.

Solution

Multiplying the two scalar conversions yields

$$\Phi_{\mathrm{NW}} =\frac{\Phi(\lambda_1,\lambda_2)} {\epsilon^4(i+\epsilon+c\epsilon^4)^4} =\frac{w_\epsilon(c)}{(i+\epsilon+c\epsilon^4)^4}.$$

Since $i^4=1$, this tends to $v(c)$. Meanwhile $\Phi=\epsilon^4w_\epsilon\to0$. At each sufficiently small nonzero regulator, the two vectors differ by a nonzero scalar and define the same normalized ray. Neither a zero limit nor a finite limit by itself proves that ray approaches an eigenstate; the Hamiltonian residual supplies that test. Finally $H=-JH_{\mathrm{NW}}$ sends the source energy $-1$ to $J$.

Independent practice: determine the missing coefficient

Starting from $v(c)$ and the adjacent-pair bond action, determine $c$ without matching a published vector. For $c=0$, calculate the norm, mean energy and normalized energy-$J$ residual. Also find the phase-independent distance between the normalized naive vector and $\chi$.

Hint

Every adjacent-pair amplitude contributes to the same two separated-pair amplitudes under $H/J-I$. The naive vector is orthogonal to its defect, but that defect is not zero. For the distance use the overlap of the two unit vectors.

Solution

The adjacent-pair coefficients sum to $2+ic$, so

$$(H/J-I)v(c)=-\frac{2+ic}{2}(|13\rangle+|24\rangle).$$

Its vanishing requires $c=2i$. For $c=0$, the squared norm is $4+4+4=12$. The defect $-(|13\rangle+|24\rangle)$ has squared norm two and is orthogonal to $v(0)$. Consequently the mean energy is $J$, while

$$r_H=\sqrt{\frac2{12}}=\frac1{\sqrt6}.$$

The overlap is $\langle\chi,v(0)/\sqrt{12}\rangle=\sqrt3/2$, positive and real. The squared phase-independent distance is therefore $2-\sqrt3$, not zero. For this exact calculation there is no small-difference numerical cancellation. The mean energy alone has missed a finite error in the state.

Transfer: reject the five-site pair

Evaluate the candidate $\Lambda_N$ and its logarithmic derivative at $u=i/2$ for $N\geq3$. Use $\tau(i/2)=i^NU$ and the Hamiltonian extraction formula to obtain its proposed translation eigenvalue and energy. Reject $N=5$ in two ways: translation consistency and the ten-dimensional bond Hamiltonian. Would the translation test alone establish the state for $N=6$?

Hint

The second term of $\Lambda_N$ and its first derivative vanish at $i/2$. Use

$$E=\frac{JN}{2}-\frac{iJ}{2} \frac{\Lambda_N'(i/2)}{\Lambda_N(i/2)}.$$

To construct the independent Hamiltonian, enumerate the ten unordered pairs on five sites. Each unequal bond contributes $1/2$ to the diagonal of $H/J$ and $-1/2$ to the swapped configuration.

Solution

The first term gives

$$\Lambda_N(i/2)=-i^N,\qquad \frac{\Lambda_N'(i/2)}{\Lambda_N(i/2)}=-i(N-2).$$

Thus the proposed eigenvalues are $U=-1$ and $E=J$. On five sites, $U^5=I$ forbids $-1$ on any nonzero vector.

The independently assembled five-site block has characteristic polynomial

$$\det(xI-H/J)=\frac{x(x-2)}{256} (4x^2-16x+11)^2(4x^2-10x+5)^2.$$

Substituting $x=1$ gives $-1/256$. Because the block dimension is even, this also equals $\det(H/J-I)$; hence $J$ is not an energy in this sector. Both tests reject the candidate without relying on a numerical root search.

For $N=6$, $(-1)^6=1$, so this necessary translation test passes. Passing it alone does not construct a vector, determine its regulator correction, or establish its eigenvector equation. Those checks must still be performed for the changed chain.

Assemble the reproduction

Keep your final result short enough that another reader can check its logic:

1. State the Hamiltonian sign, site order, auxiliary block and source conversion.
2. Derive the common fourth-order zero and the coefficient $c=2i$, then verify energy, translation and spin directly on the limiting vector.
3. Record the three residual definitions, the regulator grid, numerical environment and stable/direct comparison. Include the naive prescription as a deliberate failed control.
4. Give the odd-chain argument and the independent five-site determinant, with the exact source edition and equation locators.

This establishes a reproducible four-site singular state and a five-site exclusion. It does not test spectral completeness, all singular configurations, the thermodynamic limit, or a general regularization theorem. The research article places these finite conclusions beside the cited results and describes bounded extensions.

The separate Library proof A complete four-site XXX sector uses the state reproduced here as one member of an explicit six-state basis. Follow it to see the additional orthogonality and spanning argument required for that finite completeness claim.

References

• Nepomechie, Rafael I., and Chunguang Wang. “Algebraic Bethe ansatz for singular solutions.” Journal of Physics A: Mathematical and Theoretical 46(32) (2013), 325002. DOI. Author version arXiv:1304.7978v3 [hep-th]; Open PDF. The cited printed pages and equation numbers refer to this version.
• Nepomechie, Rafael I., and Chunguang Wang. “Twisting singular solutions of Bethe’s equations.” Journal of Physics A: Mathematical and Theoretical 47(50) (2014), 505004. DOI. Author version arXiv:1409.7382v3 [math-ph]; Open PDF. Equation locators refer to this version.