Construct Bethe vectors algebraically

How does the commuting transfer family produce actual eigenvectors? Its upper-right monodromy block creates spin reversals, but a generic product of creation blocks is not an eigenstate. The Bethe equations cancel the extra vectors produced when the transfer matrix acts. You will derive that cancellation for one and two magnons, construct a nonzero six-site state, and check its energy and translation phase. The lesson concerns regular finite-chain states; singular roots and completeness need additional arguments.

Required background. Build monodromy and transfer matrices supplies the auxiliary blocks and RTT relation. Derive commuting charges supplies Hamiltonian extraction. You need matrix multiplication, complex numbers and rational functions. Helpful background. One-magnon waves and two-magnon scattering give the coordinate interpretation.

A creation block on the periodic chain

Use the spin-$1/2$ XXX ring with $N\ge3$, $J\gt0$, $\hbar=1$, unit lattice spacing and

$$H=\frac J2\sum_{n=1}^{N}(I-P_{n,n+1}),\qquad N+1\equiv1.$$

The all-up vector $|0\rangle$ has energy zero. With auxiliary basis $(\uparrow,\downarrow)$, keep the established order

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

Here $u$ is a freely varied probe spectral parameter. The fixed parameters $v,w$ placed in $B(v),B(w)$ are creation parameters, also called $\lambda_B$ in the convention reference. All are dimensionless. The coordinate course’s rapidity has the opposite sign when labeling the same one-magnon wave: $\lambda_{\rm coord}=-\lambda_B$. We will derive the relevant phase rather than infer it from energy.

The local block matrix is

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

Since $S_n^+|\uparrow\rangle_n=0$, each factor acts triangularly on the local up state. Define the scalar vacuum eigenvalues

$$a(u)=(u+i/2)^N,\qquad d(u)=(u-i/2)^N.$$

Multiplying the triangular factors gives

$$C(u)|0\rangle=0,\qquad A(u)|0\rangle=a(u)|0\rangle,\qquad D(u)|0\rangle=d(u)|0\rangle.$$

The upper-right block instead creates one down spin. This is the vacuum construction of Faddeev 1996, § 4, equations (82)–(87), PDF. Our task is to move $A$ and $D$ through one or two $B$ factors until they reach this simple vacuum.

Entry check and repair

Does $[\tau(u),\tau(s)]=0$ imply that $B(v)|0\rangle$ is their common eigenvector for every $v$?

Repair. Commutativity constrains the transfer operators, not every proposed vector. Even the commuting matrices $\operatorname{diag}(1,2)$ and $\operatorname{diag}(3,4)$ do not have $(1,1)^{\mathsf T}$ as an eigenvector. Here we must calculate $\tau(u)B(v)|0\rangle$ and identify the condition that leaves only the original vector. We must also check that this vector is nonzero.

Read the needed relations from RTT

For nonzero $s$, abbreviate

$$f(s)=1-\frac{i}{s},\qquad h(s)=1+\frac{i}{s},\qquad g(s)=\frac{i}{s}.$$

The rational RTT relation gives

$$\begin{aligned} [B(u),B(v)]&=0,\\ A(u)B(v)&=f(u-v)B(v)A(u) +g(u-v)B(u)A(v),\\ D(u)B(v)&=h(u-v)B(v)D(u) -g(u-v)B(u)D(v). \end{aligned}$$

These formulas keep physical operator order. For example, one auxiliary matrix entry of RTT is

$$(u-v+i)B(u)A(v) =iB(v)A(u)+(u-v)A(v)B(u).$$

Interchanging $u,v$ and solving for $A(u)B(v)$ gives the displayed $A,B$ relation, including its positive $g$ term. The corresponding $D,B$ entry gives the opposite sign. This is the component calculation in Faddeev, § 4, equations (66)–(79), PDF.

Use these rational identities initially with $u\ne v$. Their poles come from solving the exchange relation by division: the original monodromy entries themselves are polynomials. We will handle equal probe and root parameters by polynomial continuation after cancellation, rather than substitute into a zero denominator.

One magnon: cancel one unwanted vector

Set $|\Phi(v)\rangle=B(v)|0\rangle$. Applying the two reordering relations gives

$$\begin{aligned} \tau(u)|\Phi(v)\rangle ={}&\Lambda_1(u;v)|\Phi(v)\rangle\\ &+\frac{i\,[a(v)-d(v)]}{u-v}B(u)|0\rangle,\\[2pt] \Lambda_1(u;v) ={}&a(u)f(u-v)+d(u)h(u-v). \end{aligned}$$

This is an off-shell action: no Bethe equation has yet been imposed. The first term retains the proposed vector. The second replaces its creation parameter by the probe $u$; this is the unwanted term.

The sufficient cancellation condition is

$$a(v)=d(v),\qquad \left(\frac{v+i/2}{v-i/2}\right)^N=1, \qquad v\ne\pm i/2.$$

A regular root satisfying this equation is called on shell. For a nonzero vector it produces a common eigenstate of every $\tau(u)$. The apparent pole of $\Lambda_1$ at $u=v$ has residue $-i[a(v)-d(v)]$, so the same condition makes the eigenvalue polynomial.

Check the wave and its direction

Let $|x\rangle$ have its down spin at site $x$. Follow an auxiliary down spin through the ordered product. Before it flips at $x$, every visited up spin contributes $v-i/2$; the flip contributes $i$; subsequent sites contribute $v+i/2$. Thus

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

For regular real $v$, adjacent coefficients have ratio

$$z_B=\frac{v-i/2}{v+i/2}=e^{ik_{\rm coord}}.$$

The cancellation condition is exactly $z_B^N=1$. The active right shift $U|x\rangle=|x+1\rangle$ then has eigenvalue $z_B^{-1}$. Matching the coordinate ratio $(\lambda_{\rm coord}+i/2)/(\lambda_{\rm coord}-i/2)$ requires $\lambda_{\rm coord}=-v$.

For $N=4$, $v=1/2$ gives

$$\begin{gathered} B(1/2)|0\rangle =\tfrac14(-1-i,-1+i,1+i,1-i)^{\mathsf T},\\ k_{\rm coord}=-\pi/2,\qquad U\Phi=i\Phi,\qquad E=J. \end{gathered}$$

The basis order is $(|1\rangle,\ldots,|4\rangle)$. Reversing $v$ reverses this momentum and shift phase, while leaving the energy unchanged. Testing only energy would miss the reversed wave.

Two magnons: collect all the extra terms

Take distinct finite roots $v,w$, and set

$$|\Phi(v,w)\rangle=B(v)B(w)|0\rangle.$$

The $B$ blocks commute, so this definition is symmetric in the two parameters. Define

$$\begin{aligned} \Lambda_2(u;v,w) ={}&a(u)f(u-v)f(u-w)\\ &+d(u)h(u-v)h(u-w),\\ C_v={}&a(v)f(v-w)-d(v)h(v-w),\\ C_w={}&a(w)f(w-v)-d(w)h(w-v). \end{aligned}$$

For $u\ne v,w$, reordering yields

$$\begin{aligned} \tau(u)|\Phi(v,w)\rangle ={}&\Lambda_2(u;v,w)|\Phi(v,w)\rangle\\ &+\frac{iC_v}{u-v}B(u)B(w)|0\rangle\\ &+\frac{iC_w}{u-w}B(u)B(v)|0\rangle. \end{aligned}$$

Here is the step that makes this more than a guessed formula. Reordering $A(u)$ first through $B(v)$ and then $B(w)$ gives the wanted coefficient $a(u)f(u-v)f(u-w)$. One unwanted coefficient is directly $a(v)g(u-v)f(v-w)$. The coefficient of $B(u)B(v)|0\rangle$ comes from two paths and is

$$a(w)\bigl[f(u-v)g(u-w)+g(u-v)g(v-w)\bigr].$$

Substitution of the rational functions proves

$$f(u-v)g(u-w)+g(u-v)g(v-w) =g(u-w)f(w-v).$$

Thus that coefficient has the same form with $v,w$ exchanged. The $D$ calculation uses $h$ and $-g$ and gives the corresponding negative contribution. Adding the two produces $C_v,C_w$. This is the two-root specialization of Faddeev, § 4, equations (88)–(96), PDF.

A sufficient regular-state criterion

Require $v,w\ne\pm i/2$, $v\ne w$, and $v-w\ne\pm i$. These restrictions keep the rational root equations and both scattering ratios well defined. If

$$\begin{aligned} \left(\frac{v+i/2}{v-i/2}\right)^N &=\frac{v-w+i}{v-w-i},\\ \left(\frac{w+i/2}{w-i/2}\right)^N &=\frac{w-v+i}{w-v-i}, \end{aligned}$$

then $C_v=C_w=0$. If also $\|\Phi(v,w)\|^2\gt0$, the normalized vector is a common eigenstate of the transfer family. No assumption that the unwanted vectors are linearly independent is needed: we make both coefficients vanish directly. Conversely, we have not proved that every eigenstate must be represented by these regular finite roots.

The residues of $\Lambda_2$ at $u=v,w$ are $-iC_v,-iC_w$. They vanish under the same conditions, so both apparent probe poles are removable. The resulting polynomial identity extends the eigenvector equation to those probe values too.

The equations are Faddeev’s equation (97), PDF. Their familiar form alone does not identify the coordinate convention: simultaneous sign reversal of both roots reciprocates both equations. Energies also remain unchanged. The creation-block construction and the actual translation phase fix the interpretation.

Read energy and translation from the eigenvalue

For one or two regular on-shell roots $v_j$, write $\Lambda(u)$ for the corresponding eigenvalue above. The established regular-point identities are

$$\tau(i/2)=i^NU,\qquad H=\frac{JN}{2}I-\frac{iJ}{2}\tau(i/2)^{-1}\tau'(i/2).$$

At $u=i/2$, the $d(u)$ part of $\Lambda$ and its first derivative vanish for $N\ge3$. Consequently

$$\begin{aligned} U\Phi&=\rho\Phi, &\rho&=\prod_j\frac{v_j+i/2}{v_j-i/2},\\ \frac{\Lambda'(i/2)}{\Lambda(i/2)} &=-iN+i\sum_j\frac1{v_j^2+1/4},\\ H\Phi&=E\Phi, &E&=\frac J2\sum_j\frac1{v_j^2+1/4}. \end{aligned}$$

The derivative in the middle line acts on the probe $u$, with roots fixed. It includes the derivative of $a(u)$, which cancels the constant $JN/2$ in the Hamiltonian. These are the momentum and energy results of Faddeev, § 4, equations (106)–(110), PDF, with $H_{\rm site}=-JH_{\rm Faddeev}$ and $\rho=e^{-iK_{\rm coord}}$.

Multiplying the two root equations gives $\rho^N=1$, as a ring translation requires. For real roots, set $e^{ik_j}=(v_j-i/2)/(v_j+i/2)$; then $K_{\rm coord}=k_1+k_2$ modulo $2\pi$. These phase and energy statements follow from the transfer eigenvalue without assuming a general equivalence theorem for arbitrary coordinate and algebraic Bethe vectors.

A regular six-site state with nonzero momentum

Take

$$N=6,\qquad v=\frac{-\sqrt3+\sqrt{11}}8,\qquad w=\frac{-\sqrt3-\sqrt{11}}8.$$

Both roots are real, distinct and regular. To verify their equations without choosing logarithm branches, define the root polynomial

$$Q(u)=(u-v)(u-w)=u^2+\frac{\sqrt3}{4}u-\frac18.$$

The eigenvalue formula becomes

$$\Lambda_2(u) =\frac{a(u)Q(u-i)+d(u)Q(u+i)}{Q(u)}.$$

At $u=v$, its numerator is $-i(v-w)C_v$; at $u=w$ it is $-i(w-v)C_w$. Therefore divisibility by this simple-root polynomial is equivalent to both unwanted coefficients vanishing. Direct polynomial multiplication gives

$$\begin{aligned} a(u)Q(u-i)+d(u)Q(u+i)&=Q(u)\Lambda_2(u),\\ \Lambda_2(u)&=2u^6+\frac52u^4-\sqrt3u^3 +\frac{11}{8}u^2\\ &\quad-\frac{5\sqrt3}{4}u-\frac9{32}. \end{aligned}$$

The quotient is a genuine polynomial, including at $u=v,w$. This proves the cancellations; it does not yet prove that the proposed vector is nonzero.

For that final condition, one coefficient suffices. Write $\alpha_v=v+i/2$, $\beta_v=v-i/2$, and similarly for $w$. Direct creation-block multiplication gives

$$\langle1,2|\Phi(v,w)\rangle =-(\alpha_v\alpha_w)^{N-2} (\alpha_v\alpha_w+\beta_v\beta_w).$$

Only the input one-spin states at sites $1$ and $2$ can contribute to this adjacent final pair: later input positions cannot be removed before the auxiliary spin passes them. The two contributions give the two terms in parentheses. For our roots,

$$\langle1,2|\Phi\rangle =\frac{27(-1+i\sqrt3)}{2048}\ne0.$$

The state therefore exists. From either the root formulas or the polynomial,

$$\begin{gathered} E=\frac52J,\qquad \rho=\frac{1+i\sqrt3}{2}=e^{i\pi/3}, \qquad K_{\rm coord}=-\pi/3,\\ \Lambda_2(i/2)=-\rho,\qquad \frac{\Lambda_2'(i/2)}{\Lambda_2(i/2)}=-i. \end{gathered}$$

The factor $-1=i^6$ checks the transfer normalization. The nonzero total momentum makes the direction test informative. Negating both roots conjugates this translation phase and leaves $E$ unchanged.

Reproduce the finite check

The XXX algebra experiment constructs the actual upper-right monodromy blocks, then compares their product with an independently assembled spin Hamiltonian and bit-shift operator. It checks the off-shell action before imposing roots as well as the regular on-shell examples. The complete experiment (ZIP) includes the code, inputs, saved results and instructions.

For a nonzero vector $\Phi$, useful checks are

$$\frac{\|H\Phi-E\Phi\|_2}{J\|\Phi\|_2},\qquad \frac{\|U\Phi-\rho\Phi\|_2}{\|\Phi\|_2}.$$

The transfer test must likewise compare the full vector equation, including the unwanted terms for off-shell roots. A vanishing mean-energy error would check only one scalar. The exact derivation above establishes the identities; finite residuals check their implementation for the stated cases.

The experiment also perturbs the six-site roots while keeping the formal root-energy sum equal to $5J/2$. For the perturbed pair approximately $(0.20807175,-0.61542895)$, the normalized Hamiltonian residual above is about $0.07135$. The full off-shell action still holds, but its unwanted terms do not vanish. This demonstrates why agreement with the root-energy formula is not a substitute for establishing an eigenstate; the downloadable inputs specify the unrounded construction.

Exercises

Guided practice: find a one-magnon root

On three sites, expand $a(v)-d(v)$ for real $v$. Find its finite roots, and give the energy and active translation phase for the positive root. Does $v=1/2$, which worked on four sites, still cancel the unwanted term?

Hint

The odd powers of $i/2$ survive in $(v+i/2)^3-(v-i/2)^3$. Use $\rho=(v+i/2)/(v-i/2)$ after finding a root.

Solution

The difference is

$$a(v)-d(v)=3iv^2-\frac i4, \qquad v=\pm\frac1{2\sqrt3}.$$

For the positive root, $\rho=e^{2\pi i/3}$, $k_{\rm coord}=-2\pi/3$ and $E=3J/2$. Its creation vector is nonzero by the one-magnon coefficient formula. At $v=1/2$, instead $a(v)-d(v)=i/2$, so the unwanted coefficient is $-1/[2(u-1/2)]$. Changing the chain length changes the root equation even though the local $L$ operator is unchanged.

Independent practice: a rational two-root check

On five sites, take $v=1/2$, $w=-1/2$. Verify $C_v=C_w=0$, exhibit a nonzero component of $B(v)B(w)|0\rangle$, and calculate its energy and translation. Which convention error would this example fail to detect by itself?

Hint

Use $v-w=1$, $\alpha_v\alpha_w=\beta_v\beta_w=-1/2$, and the adjacent-pair coefficient. Each individual shift factor is $i$ or $-i$.

Solution

For $v=1/2$,

$$a(v)=-\frac{1+i}{8},\qquad d(v)=-\frac{1-i}{8},\qquad a(v)(1-i)=d(v)(1+i)=-\frac14.$$

Thus $C_v=0$; the exchanged calculation gives $C_w=0$. The adjacent component is

$$\langle1,2|\Phi\rangle =-(-1/2)^3(-1)=-\frac18\ne0.$$

Both roots contribute energy $J$, giving $E=2J$, and their shift phases multiply to $i(-i)=1$. Total coordinate momentum is zero. The root set is invariant under simultaneous sign reversal, so this example alone cannot detect the coordinate/creation momentum reversal. The asymmetric six-site state supplies that missing check.

Transfer: change the local normalization

Replace every local operator by $\widetilde L(u)=s(u)L(u)$, where $s$ is analytic and nonzero near $u=i/2$ and at both regular roots. Determine the changes in the two-root vector, transfer eigenvalue and unwanted-term cancellation conditions. Derive the constant correction needed to extract the same Hamiltonian from $\widetilde\tau$.

Hint

Every monodromy contains $N$ scalar factors. For the Hamiltonian, differentiate the logarithm of the transfer eigenvalue with respect to the probe, not the roots.

Solution

The new monodromy blocks are $s(u)^N$ times the old ones. Hence

$$\begin{aligned} \widetilde\Phi(v,w)&=s(v)^Ns(w)^N\Phi(v,w),\\ \widetilde\Lambda(u)&=s(u)^N\Lambda(u). \end{aligned}$$

The root-dependent prefactor is nonzero, so the normalized physical state changes at most by a phase. The two terms in each unwanted coefficient receive the same scalar factor; their cancellation conditions are unchanged. Locally near $u_0=i/2$,

$$\frac{\widetilde\Lambda'(u_0)}{\widetilde\Lambda(u_0)} =N\frac{s'(u_0)}{s(u_0)} +\frac{\Lambda'(u_0)}{\Lambda(u_0)}.$$

Thus the same Hamiltonian is

$$H=\left[\frac{JN}{2} +\frac{iJN}{2}\frac{s'(u_0)}{s(u_0)}\right]I -\frac{iJ}{2}\widetilde\tau(u_0)^{-1}\widetilde\tau'(u_0).$$

Dropping the scalar correction would shift every claimed energy. Zeros or poles of $s$ at a root or the regular point invalidate the divisions used here and need separate treatment. This is the state-construction counterpart of the normalization reference.

Separate regular construction from exceptional states

The completed calculation has three independent steps: derive the off-shell action, cancel its unwanted coefficients with admissible roots, and establish a nonzero vector. A polynomial eigenvalue with canceled poles is a useful check, but cannot replace the last step. Nor does this construction count all vectors in a magnetization sector or establish completeness.

Finite roots at $\pm i/2$, coincident roots, excluded differences, and infinite-rapidity limits lie outside the stated regular calculation. In particular, the singular-state project shows why inserting a singular pair into cleared equations can leave a zero vector or the wrong limiting state. Start that project with the regular cancellation mechanism in hand; it explains exactly which steps a regulator must repair.

To turn a state into an observable, continue with normalized matrix elements and then an exact finite correlation. The same regular five-site state becomes the input to both calculations.

To investigate the separate spanning question, A complete four-site XXX sector combines a regular state, a physical singular state and spin descendants into an explicit orthonormal basis.

References

• Faddeev, L. D. “How Algebraic Bethe Ansatz works for integrable model.” Les Houches lecture notes, 1996. Author version arXiv:hep-th/9605187v1, 26 May 1996; open PDF. Section 4, equations (66)–(79), (82)–(97) and (106)–(110). The equations use the same polynomial local operator and monodromy order; the Hamiltonian and coordinate-momentum conversions are stated above. The explicit finite examples here follow from the displayed algebra and independent matrix checks.