Wronskian-Type Formula for Inhomogeneous TQ-Equations

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The celebrated Baxter TQ-equation for the closed periodic XXX spin chain of length N 1

T (u) Q(u)=(u+)N Q−−(u)+(u−)N Q++(u) (1.1) can be regarded, for a given transfer-matrix eigenvalue T (u), as a second-order ﬁnite- diﬀerence equation for Q(u). The eigenvalue T (u) is necessarily a polynomial in u (of degree N), since the model is integrable. Eq. (1.1) is well-known to have two independent polynomial solutions [1]: a polynomial Q(u) of degree M ≤ N/2

Q(u)= (u − uk) , (1.2) kY=1

whose zeros {uk} are solutions of the Bethe equations that follow directly from (1.1)

i N M uj + 2 uj − uk + i i = , j =1,...,M ; (1.3) u − uj − uk − i j 2 k=1;Yk=6 j and a second polynomial P (u) of degree N − M +1 > N/2 corresponding to Bethe roots “on the other side of the equator.” These two solutions are related by a discrete Wronskian (or Casoratian) formula

P +(u) Q−(u) − P −(u) Q+(u) ∝ uN , (1.4) where ∝ denotes equality up to a multiplicative constant. The existence of a second polynomial solution of the TQ-equation is equivalent to the admissibility of the Bethe roots [2, 3]. Using the Wronskian formula, the Q-system for this model [4] (which provides an eﬃcient way of computing the admissible Bethe roots) can be succinctly reformulated in terms of Q and P [5, 6]. A generalization of the Wronskian formula (1.4) for the open XXX spin chain with diagonal boundary ﬁelds was recently obtained [7] g(u) P +(u) Q−(u) − f(u) P −(u) Q+(u) ∝ u2N+1 , (1.5) where Q(u) and P (u) are polynomial solutions of a TQ and a dual-TQ equation, respectively. Moreover, the functions f(u) and g(u) are given by

f(u)=(u − iα)(u + iβ) , g(u)= f(−u)=(u + iα)(u − iβ) , [diagonal case] (1.6) where α and β are boundary parameters. This result was used in [7] to formulate a Q-system for the model. The main result of this note is a further generalization of the Wronskian formula for the case of non-diagonal boundary ﬁelds, namely, g(u) P +(u) Q−(u) − f(u) P −(u) Q+(u)= µ(u) u2N+1 , (1.7)

1 ± i ±± We use throughout the convenient notation f (u)= f(u ± 2 ) and f (u)= f(u ± i).

1 where again Q(u) and P (u) are polynomial solutions of a TQ (2.7) and a dual-TQ (2.12) equation, respectively; f(u) and g(u) are now given by (2.9); and – most importantly – µ(u) is a polynomial that satisﬁes the following remarkably simple relation 2

µ+(u) − µ−(u)= γu (Q(u) − P (u)) . (1.8)

In other words, µ(u) is the discrete integral of γu (Q(u) − P (u)). For the diagonal case, γ = 0; it then follows from (1.8) that µ(u) is constant, and therefore (1.7) reduces to (1.5). The appearance of the nontrivial factor µ(u) in the Wronskian-type formula (1.7)-(1.8) is due to the presence of an inhomogeneous term in the model’s TQ-equation (2.7) [9, 10, 11]. We expect that this Wronskian-type formula will be useful for formulating a Q-system for this model, which however remains a challenge. In Section 2, we ﬁrst brieﬂy review the construction of the model and its TQ-equation, and we then obtain a dual TQ-equation. We derive the Wronskian-type formula (1.7)-(1.8) in Section 3.

2 The model and its TQ-equations

We consider the isotropic (XXX) open Heisenberg quantum spin-1/2 chain of length N with boundary magnetic ﬁelds, whose Hamiltonian is given by

− N 1 ξ 1 1 H = ~σ · ~σ − σx − σz + σz , (2.1) k k+1 β 1 β 1 α N Xk=1 where ~σ = (σx , σy , σz) are the usual Pauli matrices, and α, β and ξ are arbitrary real parameters. For the non-diagonal case ξ =6 0, this model is not U(1)-invariant. In order to construct the corresponding transfer matrix, we use the R-matrix (solution of the Yang-Baxter equation) given by the 4 × 4 matrix

R i I P (u)=(u − 2 ) + i , (2.2) where P is the permutation matrix and I is the identity matrix; and the K-matrices (solutions of boundary Yang-Baxter equations) given by the 2 × 2 matrices [12, 13]

1 KR i(α − 2 )+ u 0 (u)= 1 , 0 i(α + 2 ) − u 1 i KL i(β − 2 ) − u −ξ(u + 2 ) (u)= i 1 , (2.3) −ξ(u + 2 ) i(β + 2 )+ u which depend on the boundary parameters α, β and ξ.

2After this work was completed, we became aware of a similar result for the closed XXX spin chain with a non-diagonal twist, see Theorem 4.10 in [8].

2 The transfer matrix T(u)= T(u; α,β,ξ) is given by [14] T KL M KR M (u)=tr0 0 (u) 0(u) 0 (u) 0(u) , (2.4) where M and M are monodromy matrices given by b M R R R b 0(u)= 01(u) 02(u) ... 0N (u) ,

M0(u)= R0N (u) ··· R02(u) R01(u) . (2.5) The transfer matrix is engineeredb to have the fundamental commutativity property [T(u) , T(v)]=0 , (2.6) T T dT(u) and satisﬁes (−u) = (u). The Hamiltonian (2.1) is proportional to du , up to an u=i/2 additive constant.

The eigenvalues T (u) of the transfer matrix T(u) are polynomials in u (as a consequence of (2.6)), and satisfy the TQ-equation [9, 10, 11] − u T (u) Q(u)= g−(u)(u+)2N+1Q−−(u)+ f +(u)(u−)2N+1Q++(u) − γu u−u+ 2N+1 , (2.7) where Q(u) is an even polynomial of degree 2N N

Q(u)= (u − uk)(u + uk) , (2.8) kY=1 the functions f(u) and g(u) are given by f(u)=(u − iα) u 1+ ξ2 + iβ , g(u)= f(−u)=(u + iα) u 1+ ξ2 − iβ , (2.9) p p and we define γ by γ = −2 1 − 1+ ξ2 . (2.10) p Note the presence of an inhomogeneous term (proportional to γ) in the TQ-equation (2.7). For the diagonal case ξ = 0, we see from (2.10) that γ = 0, hence the inhomogeneous term disappears; moreover, the functions f(u) and g(u) (2.9) reduce to (1.6). The transfer matrix transforms under charge conjugation by reflection (negation) of all the boundary parameters C T(u; α,β,ξ) C = T(u; −α, −β, −ξ) , C =(σx)⊗N . (2.11) We therefore obtain a dual TQ-equation from (2.7) as in [7] by making the replacement Q(u) 7→ P (u), and by reflecting the boundary parameters (α 7→ −α , β 7→ −β,ξ 7→ −ξ), which implies that f(u) and g(u) become interchanged −u T (u) P (u)= f −(u)(u+)2N+1P −−(u)+g+(u)(u−)2N+1P ++(u)−γu u−u+ 2N+1 , (2.12) where P (u) is also an even polynomial of degree 2N N

P (u)= (u − u˜k)(u +˜uk) , (2.13) Yk=1 whose zeros can be regarded as dual Bethe roots. We emphasize that the same eigenvalue T (u) appears in both (2.7) and (2.12).

3 3 The Wronskian-type formula

We now look for a relation between Q(u) (a solution of the TQ-equation (2.7) for some transfer-matrix eigenvalue T (u)) and the corresponding P (u) (a solution of the dual TQ- equation (2.12) for the same transfer-matrix eigenvalue T (u)). To this end, we make the ansatz g(u) P +(u) Q−(u) − f(u) P −(u) Q+(u)= µ(u) u2N+1 , (3.1) where f(u) and g(u) are given by (2.9), and the function µ(u) is still to be determined. This ansatz is motivated by the result (1.5) for the diagonal case. We next deﬁne R(u), along the lines of [1], by

u2N+1 1 P + P − R = = g − f , (3.2) Q+ Q− µ Q+ Q− where the second equality follows from the ansatz (3.1). Dividing both sides of the TQ- equation (2.7) by Q Q++ Q−−, we obtain u T − = f +R− + g−R+ − γuQR+R− . (3.3) Q++ Q−− Substituting in (3.3) for R using the second equality in (3.2), and then multiplying both sides by µ+µ−Q++Q−−, we obtain P Q++Q−− −uTµ+µ− = f +g− (µ+ − µ− + γuP ) − f +f −P −−Q++(µ+ + γuP ) Q + g+g−P ++Q−−(µ− − γuP )+ f −g+P −−P ++Q (γu) . (3.4)

Similarly, we deﬁne S(u) by

u2N+1 1 Q− Q+ S = = g − f , (3.5) P + P − µ P − P + and we divide both sides of the dual TQ-equation (2.12) by PP ++ P −−, thereby obtaining u T − = f −S+ + g+S− − γuPS+S− . (3.6) P ++ P −− Substituting in (3.6) for S using the second equality in (3.5), and then multiplying both sides by µ+µ−P ++P −−, we obtain

−uTµ+µ− = f +g−P Q++Q−−(γu) − f +f −P −−Q++(µ− + γ u Q) P −−P ++Q + g+g−P ++Q−−(µ+ − γ u Q) − f −g+ (µ+ − µ− − γ u Q) . (3.7) P Equating the right-hand-sides of (3.4) and (3.7), we arrive at the constraint

1 − µ+ − µ− + γu (P − Q) gP +Q− − fP −Q+ + gP +Q− − fP −Q+ =0 . (3.8) QP 4 This constraint can evidently be satisﬁed by setting

µ+ − µ− = γu (Q − P ) , (3.9)

as claimed in (1.8). For given polynomials Q(u) and P (u), (3.9) can be solved for a polynomial function µ(u), up to an arbitrary additive constant. The result (3.9) can in fact be obtained in a more straightforward way:3 deﬁne µ(u) by 1 µ(u)= g(u) P +(u) Q−(u) − f(u) P −(u) Q+(u) , (3.10) u2N+1 P (u) which is equivalent to (1.7). Multiply the TQ-equation (2.7) by (u−u+)2N+1 , multiply the Q(u) dual TQ-equation (2.12) by (u−u+)2N+1 , and subtract the second equation from the ﬁrst. The result, when expressesed in terms of µ (3.10), is exactly (3.9).

Acknowledgments

The author thanks the organizers of the CQIS-2019 workshop in St. Petersburg for their kind invitation. He was supported in part by a Cooper fellowship. Conﬂict of Interest: The author declares that he has no conﬂict of interest.

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