Completing the solution for the OSp(1|2) spin chain

The periodic OSp(1|2) quantum spin chain has both a graded and a non-graded version. Naively, the Bethe ansatz solution for the non-graded version does not account for the complete spectrum of the transfer matrix, and we propose a simple mechanism for achieving completeness. In contrast, for the graded version, this issue does not arise. We also explain (for both versions) the degeneracies and multiplicities of the transfermatrix spectrum, and obtain conditions for selecting the physical singular solutions of the Bethe equations.


Introduction
The periodic OSp(1|2) quantum spin chain was first formulated (as a graded model) and solved in [1]. A non-graded version of this model was formulated and solved in [2,3]. These solutions have figured in various subsequent works, see e.g. [4,5,6,7,8,9,10] and references therein. In this note, we argue that -naively -the solution for the non-graded version does not account for the complete spectrum of the transfer matrix, and we propose a simple way to remedy the problem. In contrast, for the graded version, this issue does not arise. We also explain (for both versions) the degeneracies and multiplicities of the transfer-matrix spectrum, and obtain conditions for selecting the physical singular solutions of the Bethe equations. As a byproduct of our investigation, we clarify the signs in the Bethe equations, over which there has been some controversy, see e.g. [3,9].
We treat the non-graded version of the OSp(1|2) model in Sec. 2, and the graded version in Sec. 3. Our conclusions are in Sec. 4. The OSp(1|2) symmetry of the graded transfer matrix is proved in the appendix.

The non-graded version
In Sec. 2.1, we review the construction of the transfer matrix for the non-graded version of the OSp(1|2) quantum spin chain, and we note that it has only SU(2) symmetry. In Sec. 2.2, we briefly review the solution [2,3]. The degeneracies and multiplicities of the transfermatrix spectrum are explained in Sec. 2.3. We give conditions for selecting the physical singular solutions of the Bethe equations in Sec. 2.4. The difficulty with completeness and its resolution are discussed in Sec. 2.5.

The transfer matrix and its symmetries
The OSp(1|2) spin chain has a 3-dimensional vector space at each site. Following [3], we consider here the bosonic (non-graded) formulation of the model. The R-matrix is given by 1 It has the regularity property R(0) = iP, where P is the (non-graded) permutation matrix on This R-matrix is a solution of the (non-graded) Yang-Baxter equation where R 12 = R ⊗ I , R 13 = P 23 R 12 P 23 , R 23 = P 12 R 13 P 12 . (2.5) The transfer matrix t(u) for a closed spin chain of length N with periodic boundary conditions, which is given as usual by the (non-graded) trace of the monodromy matrix has the commutativity property The corresponding Hamiltonian is given by The transfer matrix has SU(2) symmetry 2

The Bethe ansatz solution
Let |Λ (M ) denote the simultaneous eigenvectors of the transfer matrix t(u) (2.6) and of S z (2.10) that are SU(2) highest-weight states 3 and let Λ (M ) (u) and m denote the corresponding eigenvalues (2.14) It was argued in [3] that these eigenvalues are given by and The corresponding Bethe equations for the Bethe roots {u 1 , . . . , u M } are given by , j = 1, . . . , M .

Degeneracies and multiplicities
The degeneracy (the number of times that a given eigenvalue Λ (M ) (u) appears) is given by N − M + 1, since the eigenstates form SU(2) irreducible representations with spin s = m = (N − M)/2, see (2.15), which have dimension 2s + 1. Since the 3-dimensional vector space at each site decomposes as spin-1/2 plus spin-0, the multiplicities in the spectrum follow from the Clebsch-Gordan decomposition of (2 ⊕ 1) ⊗N . For example, Hence, for N = 2, there are two eigenvalues with degeneracy 1 (M = 2), two eigenvalues with degeneracy 2 (M = 1), and one eigenvalue with degeneracy 3 (M = 0); and similarly for higher N.

Physical singular solutions of the Bethe equations
As in the periodic Heisenberg spin chain [12,13,14], the Bethe equations (2.18) have many "singular" solutions containing ± i 2 ; only a subset of these singular solutions, which we call "physical", correspond to actual eigenvalues and eigenstates of the transfer matrix. Starting from the Bethe equations for the model with twisted boundary conditions (see appendix C in [3]) and following [14], we find that the condition for the singular solution (2.20) Moreover, {u 3 , . . . , u M } must also obey However, we find that this Bethe ansatz solution cannot account in this way for all the eigenvalues of the transfer matrix (2.6). Indeed, for N = 2, there is an eigenvalue with M = 1 that cannot be described by this solution, namely 4 We propose a simple way to account for the missing transfer-matrix eigenvalues: admit precisely one infinite Bethe root. Since the Q-functions appear in the expression for the eigenvalues (2.16) only as ratios, the effect in (2.16) of one infinite Bethe root (u M = ∞) is to replace the expression for Q(u) (2.17) by a product over only the finite roots, i.e. We have explicitly verified that all the missing transfer-matrix eigenvalues for N = 2, 3, 4 can be obtained in this way, with u M = ∞. The Bethe roots corresponding to all of the transfer-matrix eigenvalues for N = 2, 3 are given in Table 1. Note that the degeneracies and multiplicities are in accordance with the group-theory predictions in Sec. 2.3. For N = 4, we report only the Bethe roots corresponding to the "missing" transfer-matrix eigenvalues, and the physical singular solutions, see Table 2. The physical singular solutions in Tables 1  and 2

The transfer matrix and its symmetries
The graded R-matrix is given by 5 It has the regularity property R(0) = iP, where P is now the graded permutation matrix where the gradings are given by p(1) = 0 , p(2) = p(3) = 1. This R-matrix is a solution of the graded Yang-Baxter equation, which is given by (2.4) and (2.5), but with the graded permutation matrix. The transfer matrix t(u) satisfying the commutativity property (2.7) is given by (2.6), but with the factors R 0j (u) constructed as in (2.5) using the graded permutation matrix, and with the graded trace (supertrace) str X = 3 a=1 (−1) p(a) X aa . The transfer matrix now has OSp(1|2) symmetry, in contrast with the non-graded version that has only SU(2) symmetry. Indeed, we show in the appendix that where the generators are given by These generators satisfy the OSp(1|2) algebra and all other (anti-)commutators vanish. Note that the fermionic generators J ± have a non-trivial coproduct (3.5) involving the grading involution P , see e.g. [15,16]. and

Degeneracies and multiplicities
The degeneracy (the number of times that a given eigenvalue Λ (M ) (u) appears) is given by 2N − 2M + 1, since the eigenstates form OSp(1|2) irreducible representations with spin s = m = (N − M)/2, see (3.11), which have dimension 4s + 1. Indeed, these irreps, which we denote by [s], consist of a pair of SU(2) irreps that have highest weights |Λ (M ) (with spin s and dimension 2s + 1) and J − |Λ (M ) (with spin s − 1 2 and dimension 2s). Since the 3-dimensional vector space at each site forms an irrep [ 1 2 ], the multiplicities in the spectrum follow from the OSp(1|2) decomposition of [ 1 2 ] ⊗N , which can be easily computed from the fact (see e.g. [17] and references therein) For example, in terms of the dimensions of the irreps, Hence, for N = 2, there is one eigenvalue with degeneracy 1 (M = 2), one eigenvalue with degeneracy 3 (M = 1), and one eigenvalue with degeneracy 5 (M = 0); and similarly for higher N. Evidently, the patterns (3.17) differ significantly from those in the non-graded version (2.19).

Checking completeness
We have explicitly verified for N = 2, 3, 4 that the Bethe ansatz solution (3.11)-(3.13) accounts for all the eigenvalues of the transfer matrix, where all the Bethe roots are finite and pairwise distinct. In contrast with the non-graded version, an infinite Bethe root is not necessary, which is consistent with the fact that the TQ-equation and Bethe equations do not depend explicitly on M.
The Bethe roots for N = 2, 3, 4 are given in Table 3. Note that the degeneracies and multiplicities are in accordance with the group-theory predictions in Sec. 3.3. The physical singular solutions in Table 3 satisfy (3.14)-(3.15).

Conclusions
We have argued that, by allowing for the possibility of an infinite Bethe root (and all other Bethe roots finite and pairwise distinct), the Bethe ansatz solution for the non-graded version of the periodic OSp(1|2) spin chain (2.16)- (2.18) can account for all the distinct eigenvalues of the transfer matrix (2.6).
We emphasize the difference with respect to, say, the Bethe ansatz solution for the periodic SU(2)-invariant Heisenberg (XXX) spin chain, which (as far as has been checked [13]) accounts for all the transfer-matrix eigenvalues by means of finite Bethe roots. 7 The reason that, in the non-graded version, an infinite Bethe root can help give a "missing" transfer-matrix eigenvalue (i.e., an eigenvalue that cannot be obtained with exclusively 7 As is well known, in the algebraic Bethe ansatz approach for the Heisenberg spin chain, the Bethe states are given by B(u 1 ) · · · B(u M )|0 , where B(u) is a certain creation operator and |0 is a reference state. If all the Bethe roots {u 1 , . . . , u M } are finite, then the Bethe state is an SU (2) highest-weight state; and the lower-weight states (which have the same transfer-matrix eigenvalue as the Bethe state) can be obtained by repeatedly acting with the spin-lowering operator S − on the Bethe state. Since B(∞) ∼ S − , some authors prefer to describe such lower-weight states of the Heisenberg spin chain in terms of (multiple) infinite Bethe roots. In contrast, for the non-graded version of the OSp(1|2) spin chain, we find that one infinite Bethe root is necessary to construct certain Bethe states. (As already noted, the algebraic Bethe ansatz for the OSp(1|2) spin chain is sketched in appendix B of [3]; the renormalized creation operator u −N +1 B 1 (u) is finite in the limit u → ∞.) OSp(1|2) symmetry, which is reflected in the degeneracies and multiplicities of its spectrum. This result is in agreement with the observation in [9], and in disagreement with [3], that the Bethe equations for the model with OSp(1|2) symmetry are (3.13), and not (2.18).

Acknowledgments
I thank Etienne Granet for bringing the OSp(1|2) model to my attention, and for reading the draft.

A Proof of OSp(1|2) symmetry
We show here that the graded transfer matrix has OSp(1|2) symmetry (3.4). We first consider the bosonic generators S z , S ± in Sec. A.1, and then consider the fermionic generators J ± in Sec. A.2. For later reference, we introduce the monodromy matrix in terms of which the transfer matrix is given by

A.1 Bosonic generators
The proof for the bosonic generators is simlar to the classic proof of SU(2) symmetry for the Heisenberg spin chain [18]. The SU(2) symmetry of the R-matrix (3.1) means that