On zero-remainder conditions in the Bethe ansatz

We prove that physical solutions to the Heisenberg spin chain Bethe ansatz equations are exactly obtained by imposing two zero-remainder conditions. This bridges the gap between different criteria, and proves the correctness of a recently devised algorithm based on QQ relations.


Introduction
The Bethe ansatz [1] enables one to write an energy level of the periodic Heisenberg (XXX) spin chain on L sites as where the K Bethe roots λ i satisfy the Bethe ansatz equations traditionally written in the following form To make this simply stated result precise however necessitates recalling some subtleties in the resolution of (2).
Firstly, the ordering of the λ i 's is irrelevant, and two solutions differing only by a permutation of the roots should be considered equal; secondly, in general a solution with two coinciding roots λ i = λ j should be discarded [1]; thirdly, there are special solutions with λ 1 = i/2 and λ 2 = −i/2, called exact strings, see e.g. [1,2,3,4,5,6], which must be sometimes discarded and sometimes not. It is well known that the first two points are solved by inspection of the expression of the eigenvalue T (λ) of the transfer matrix at spectral parameter λ. It satisfies the T Q relation [7,8] T (λ)Q(λ) = (λ − i/2) L Q(λ + i) where From general arguments T (λ) has to be a polynomial in λ, and thus (3) gives a zero-remainder condition on the coefficients of Q, that permits then to solve for the λ i 's, and that directly removes the non-physical solutions with coincinding roots.
However, the TQ relation alone does not solve the third point. Indeed, any solution (i/2, −i/2, λ 3 , ..., λ K ) where λ 3 , ..., λ K satisfy the Bethe equations (2) (for ±i/2, they are automatically satisfied if both sides are multiplied by the denominators beforehand) does give a polynomial T (λ), but the normalized Bethe state is then expressed in a singular way whose regularization depends on the way the roots λ i of the strings converge to ±i/2. More precisely, denoting the R-matrix by where the monodromy matrix and the transfer matrix read respectively, using the standard notations of the algebraic Bethe ansatz approach [9], we have where the hat B indicates that the corresponding factor is omitted in the product. Although the residues do vanish in case of strings, we have B(λ 1 )...B(λ K )|0 = 0 [9,10], so that the normalized Bethe state is not necessarily an eigenvector. In fact, the solutions to the Bethe equations with exact strings sometimes do yield eigenvalues and eigenvectors of the Hamiltonian, and sometimes not.
In [11,12] a sufficient condition was found under which a Bethe vector can be built which is an eigenvector, by regularizing in a very particular way the roots ±i/2. Another practical way of distinguishing physical from non-physical solutions is to examine the behaviour of the solutions in terms of an additional parameter, such as a twist [13]. But the most important recent advance on this question was an efficient algorithm [14] found by Marboe and Volin to solve the T Q relation while discarding automatically the non-physical solutions among those with an exact string, with only algebraic manipulations and zero-remainder conditions called QQ relations. However, the reason why the algorithm works is arguably mysterious and in our opinion not sufficiently discussed in [14]. It is seemingly related to the fact that (3) being a second-order difference equation actually has two solutions [15], one of which being Q(λ), and it was indeed shown in [14] that the number of pairs of polynomial solutions to (3) matches the required number of eigenstates [16]. However, there are two particularly unclear points, namely (i) why having two polynomial solutions ensures that T (λ) is an eigenvalue of the transfer matrix, which is a priori very far from the reasoning done in [11,13,12], and (ii) why the algorithm of Marboe and Volin [14] actually imposes that the two solutions are polynomial.
In this paper, we answer these points, explain why the algorithm of [14] works, and show that the T Q relation has to be supplemented with another zero-remainder T Q-like relation to yield only physical solutions. It bridges the gap between different approaches and yields an elegant criterion to distinguish between physical and non-physical solutions.
These results and methods established here for the periodic XXX chain will be used in a subsequent paper [17] to generalize and prove QQ relations for the anisotropic Heisenberg (XXZ) spin chain, and to extend the results to the case of open boundary cases.
Before addressing the main results, for sake of completeness we recall here the following known result Proof. Assume that there are two roots such that In the latter case, the same argument can be then repeated with k = i 3 , so that s = ni with n a negative or zero integer, since there is a finite number of roots. On the other hand, (2) for k = i 2 implies that either λ i 2 = −i/2, in which case s = 0, or there exists another λ i 4 such that λ i 2 − λ i 4 = i. The same argument can be then repeated with k = i 4 , implying that s = ni with n a positive or zero integer. Thus s = 0.
Let us start with the following property, that generalizes [15] to the exact strings case. Lemma 2. There exist a polynomial P 0 (λ) and complex numbers α s for s ∈ S such that where ψ(x) is the digamma function. Moreover, α 0 = 0 if and only if the following additional Bethe equation is satisfied Proof. It is directly inspired by [15], where the authors (implictly) treated the case S = ∅. Denote We have Since each s ∈ S appears twice in the denominator in R(λ), we can decompose with π(λ), q ± (λ) polynomials of degree less than or equal to n − 1 (since a term of order n in the numerator could be reabsorbed in the constant term of π(λ)), and c s complex numbers. From this one gets Multiplying by (λ − s + i/2) 2 and sending λ → s − i/2, since there is no double pole in s − i/2 on the left-hand side, one gets c s = 0.
For λ j ∈Λ, multiplying by (λ − λ j ) and taking λ → λ j yields meaning that there exists a polynomial σ such that and thus a polynomial q(λ) such that As any polynomial, π(λ) can be decomposed as with ρ(λ) a polynomial, unique up to an additive constant. Denote now It can be decomposed as with a s , b + s , b − s constants. Using the property of the digamma function ψ(x), one can rewrite it as where (25) Therefore with Note that since s ± i/2 is a root of Q(λ), P is a polynomial if and only if a s + b + s + b − s = 0 for all s ∈ S. Recalling (12), one gets as stated in the theorem. Moreover, P (λ) takes the form (10) by virtue of (27) and (25).
To show the second part of the lemma, we proceed as follows. Replacing the (λ ± i/2) L in the T Q relation (3) by relation (28), one gets Evaluating this relation in s − i/2 yields, using the form (27) and the fact that the digamma function has a pole at each nonpositive integer with residue −1 Using now the T Q relation: and relation (28) for λ = s + i, together with the fact that s = 0 is the only possible string center, one gets from (30) and which concludes the proof.
3 Polynomiality of P (λ) and constructability of the Bethe state We remark that (33) is exactly the condition found in [11,12] for having a physical solution of the Bethe equations. However, [11,12] only proves that (33) implies the physicality of the solution. The purpose of Lemma 4 below is to prove the equivalence between polynomiality of P (λ) and physicality of the solution. Let us briefly explain our reasoning. In presence of exact strings, the residues in (7) still vanish as in the case of non-singular Bethe roots; however, the Bethe state B(λ 1 ) · · · B(λ n )|0 vanishes as well (see [9,10] and Lemma 3), and imposing the TQ relation (3) alone is then non-conclusive. We want to show that one can find a regularization such that the residues in (7) vanish faster than B(λ 1 ) · · · B(λ n )|0 when → 0, if and only if P is a polynomial. To that end, we need to understand how fast the Bethe state actually vanishes when → 0, which is the purpose of Lemma 3 (that is in fact needed in [11] for their reasoning to be conclusive).
Lemma 3. Let λ 1 and λ 2 be such that λ 1,2 = ±i/2 + O( ) when → 0. Then Proof. From (5) we have (see e.g. [9]) where σ − k is the Pauli matrix ( 0 0 1 0 ) acting at site k. An efficient way to obtain this expression is to apply each R matrix on the quantum space vector ( 1 0 ) in (5) before taking the tensor products. Thus when calculating the monodromy matrix we take a (regular) matrix product of  but whose coefficients have to be tensorized at each site. Similarly Hence (38) By computing the power sum over p, after a bit of rearangement one gets which is O( L ).
We will denote with F (λ) its perturbed version, involving T (λ). Lemma 3 implies that the condition for the residue terms in (7) to vanish faster than the Bethe state, in the limit → 0, is Let us first build a P corresponding to the Q . Decomposing with and so which has poles at λ k − ni with n a strictly positive integer, with residue −(a + k ( ) + a − k ( ))Q (λ k − ni).
With relation (48), one has which has a pole at every λ k with residue We now pick a k that corresponds to i/2 or −i/2, for example without loss of generality The left-hand side vanishes if and only if a + k ( ) + a − k ( ) vanishes. Indeed, if the left-hand side vanishes, then Q (λ k −i) (λ k −i/2) L cannot vanish on the right hand-side. If a + k ( ) + a − k ( ) vanishes, then Q (λ k −i) (λ k −i/2) L cannot diverge when → 0, otherwise the right-hand side would diverge faster since P (i/2) = 0, see (32); and so the whole left-hand side must vanish.
If P is not a polynomial, according to Lemma 2 it must have a pole at −3i/2, so that a + k ( ) + a − k ( ) does not vanish when → 0, at least for one k such that λ k = i/2 or λ k = −i/2 (we can assume that it is true for i/2; otherwise we could have chosen −i/2 before). Hence the left-hand side of (53) does not vanish and we cannot have r k ( ) → 0 when → 0.
If P is a polynomial, for an arbitrary function → λ j , the different poles a + k ( ) + a − k ( ) do not necessarily vanish individually in the limit → 0, since they can compensate each other (for example, 1/(λ − ) − 1/(λ + ) does not have any poles in the limit → 0, even if the residues at = 0 do not vanish in the limit → 0). Coming back to (48) evaluated at λ = λ k − i/2 for λ k = −i/2 and for λ k = i/2 , one sees that the vanishing of the residues is equivalent to The first condition is always satisfied when P is a polynomial, and the second one is an additional condition that has to be satisfied for the Bethe vector to be an eigenvector in the limit → 0. This shows that if P is a polynomial, then the poles r k ( ) can vanish in the limit → 0 with an appropriate choice of roots λ k .
We remark that the second condition in (54), writing the perturbed roots as i/2 + and −i/2 + η( ), can be translated into which was the regularization found in [11].

An additional T Q relation
We can now prove the is a physical solution to the Bethe ansatz equations if and only if the functions T 0 (λ) and T 1 (λ) in the following two T Q relations are polynomials: where Proof. It is straightforward to show that where P (λ) = P (λ + i/2) − P (λ − i/2) with P (λ) the function introduced in Lemma 2, using equation (9). Then Now, from the general form of P in Lemma 2, one has with A(λ) a rational function with a unique simple pole at 0 with residue proportional to α 0 , using ψ(x + 1) − ψ(x) = 1/x. Since ψ has a pole at −1, T 1 has a priori a pole at 0 with residue iα 0 Q (−i)Q (i). From Lemma 1, ±i are never center of strings and so Q (±i) = ±Q(±3i/2) = 0 if α 0 = 0. It follows that T 1 is a polynomial if and only if P is. Then Lemma 4 concludes the proof.

The algorithm of Marboe and Volin
Let us now come back to the algorithm of Marboe and Volin [14]. It consists in introducing functions Q a,s with s = 0, . . . , L − K, for a = 0, 1, 2 if s ≤ K and a = 0, 1 if s > K, satisfying the following QQ relations with the boundary conditions Q 0,0 (λ) = λ L , Q 2,s = 1 for s ≤ K, Q 1,s = 1 for s > K, and imposing that all the Q a,s are polynomials. The Q(λ) is then given by Q 1,0 (λ). The labels (a, s) can be interpreted as the coordinates of corners of boxes in an associated two-row Young diagram Each QQ relation (61) then imposes a constraint on the four Q-functions associated with the corners of the box whose lower left corner is (a, s). The boundary conditions fix in particular Q a,s = 1 for all corners along the top of the diagram.
In this context, it is readily checked that the two T Q relations (56) are exactly the relations obtained when Q a,s are imposed to be polynomials for a = 0, 1, 2 and s = 0, 1. In other words, they are the zero-remainder conditions associated with the two boxes in the leftmost column of the Young diagram. (Note that in the special case K = 1 there cannot be strings and the second equation of (56) is trivially satisfied.) Thus, according to Theorem 1, all the other polynomials Q a,s for s > 1, as well as the corresponding relations (61) fixing them, are actually superfluous.
We also remark that the fact that only one additional T Q relation is needed to discard the unphysical solutions is linked to the fact that there is only one possible exact string (otherwise this T Q relation would only give one equation relating the α s 's).
However, the corresponding T 1 (λ) reads showing that the polynomiality of the solution to the second T Q relation is satisfied for L = 4, but not for L = 5. Besides, the function P (λ) of Lemma 2 reads and is a polynomial if and only if T 1 (λ) is a polynomial. It turns out that T 0 (λ) is indeed an eigenvalue of the transfer matrix for L = 4, but not for L = 5, in agreement with the theorem.
In Figure 1 we plot the roots of all the polynomials Q(λ) solution to the T Q relation (3) in size L = 6, showing in blue those whose solve the two T Q relations (56) and in red those that only solve the first one. Only the solutions (−i/2, i/2) (−i/2, 0, i/2) among the blue ones involve exact strings in Figure 1 (all the red non-physical solutions must exhibit exact strings). We see in this example that the number of admissible solutions with K roots is L K − L K−1 . We recall that in the Heisenberg spin chain the Bethe states are necessarily highestweight states with respect to the underlying su(2) algebra. Taking into account that the eigenvalue corresponding to a solution with K Bethe roots is (L − 2K + 1)-fold degenerate, one obtains 2 6 eigenstates indeed.