General scalar products in the arbitrary six-vertex model
aa r X i v : . [ n li n . S I] S e p General scalar products in the arbitrarysix-vertex model
G.A.P. Ribeiro ∗ Departamento de F´ısica, Universidade Federal de S˜ao Carlos13565-905 S˜ao Carlos-SP, BrazilOctober 2, 2018
Abstract
In this work we use the algebraic Bethe ansatz to derive the generalscalar product in the six-vertex model for generic Boltzmann weights.We performed this calculation using only the unitarity property, theYang-Baxter algebra and the Yang-Baxter equation. We have deriveda recurrence relation for the scalar product. The solution of this re-lation was written in terms of the domain wall partition functions.By its turn, these partition functions were also obtained for genericBoltzmann weights, which provided us with an explicit expression forthe general scalar product.
PACS: 75.10.Pq; 02.30.Ik;Keywords: Algebraic Bethe ansatz, integrable systems, scalar products ∗ [email protected] ontents M = 2 M = 3 Introduction
It is well-known that the quantum inverse scattering method is a powerfultool to solve exactly quantum integrable models as well as their classical coun-terparts [1, 2]. Based on this approach one can construct the Bethe states bythe action of pseudo-particle creation operators on a pseudo-vacuum state.These operators are the off-diagonal matrix elements of the monodromy ma-trix T A ( λ ). By its turn the monodromy matrix elements are the generatorsof the Yang-Baxter algebra given by the following quadratic relation, R ( λ, µ ) T ( λ ) T ( µ ) = T ( µ ) T ( λ ) R ( λ, µ ) . (1)The monodromy matrix elements act on the space of states of the quan-tum physical system and the R -matrix elements play the role of the structureconstants of the above Yang-Baxter algebra (1). The latter algebra is asso-ciative thanks to the Yang-Baxter equation [3] R ( λ, µ ) R ( λ, γ ) R ( µ, γ ) = R ( µ, γ ) R ( λ, γ ) R ( λ, µ ) . (2)The trace of the monodromy matrix over the auxiliary space T ( λ ) =Tr A [ T A ( λ )] is the row-to-row transfer matrix. This matrix constitutes afamily of commuting operators [ T ( λ ) , T ( µ )] = 0, provided that the R -matrixis invertible. This condition is granted by the unitarity relation R ( λ, µ ) R ( µ, λ ) = I, (3)where I is the identity matrix.Taking the action of the transfer matrix over the Bethe states, one is ableto obtain the eigenvalues of the transfer matrix and its related conserved3uantities. The eigenvalue expression is usually given in terms of some pa-rameters which should satisfy the associated Bethe ansatz equations [4].After obtaining the eigenvalues, one of the most challenging problems isto compute the scalar products and correlation functions. This was accom-plished, by means of the algebraic Bethe ansatz, for models descending fromthe R -matrix of the symmetric six-vertex model [5, 6, 7].Recently, it was argued that the algebraic formulation of the Bethe statesof the transfer matrix can be done using only the Yang-Baxter algebra (1),the Yang-Baxter equation (2) and the unitarity property (3) satisfied by the R -matrix [8]. This way the on-shell properties (eigenvalues and the Betheansatz equations) as well as the off-shell Bethe vectors are obtained in termsof arbitrary Boltzmann weights.This idea have been shown to be valid also on the computation of themonodromy matrix elements of the arbitrary six-vertex model in the F -basisand the scalar product for arbitrary Boltzmann weights [9].In this paper we shall obtain the scalar product as well as its associateddomain wall partition functions by means of the algebraic Bethe ansatz forthe completely asymmetric six-vertex model for arbitrary Boltzmann weights.The aim of this paper is to show that this can be accomplished using only theYang-Baxter algebra, the Yang-Baxter equation and the unitarity relation.The outline of the article is as follows. In section 2, we define the asym-metric six-vertex model and list the main relations needed in this work. Insection 3, we derive a recurrence relation for the scalar product and sketchits solution. In section 4, we derive the partition function with domain wallboundary conditions which will help on writing a closed formula for the scalar4roduct. Our conclusions are given on section 6. We define the R -matrix of the asymmetric six-vertex model as [3, 10], R ( λ, µ ) = a + ( λ, µ ) 0 0 00 b + ( λ, µ ) c + ( λ, µ ) 00 c − ( λ, µ ) b − ( λ, µ ) 00 0 0 a − ( λ, µ ) . (4)This R -matrix should satisfy the unitarity property (3), which results on thefollowing set of relations for its matrix elements c − ( λ, µ ) b − ( µ, λ ) + b − ( λ, µ ) c + ( µ, λ ) = 0 , (5) b + ( λ, µ ) c − ( µ, λ ) + c + ( λ, µ ) b + ( µ, λ ) = 0 , (6) a + ( λ, µ ) a + ( µ, λ ) = 1 , (7) a − ( λ, µ ) a − ( µ, λ ) = 1 , (8) c − ( λ, µ ) c − ( µ, λ ) + b − ( λ, µ ) b + ( µ, λ ) = 1 , (9) c + ( λ, µ ) c + ( µ, λ ) + b + ( λ, µ ) b − ( µ, λ ) = 1 . (10)Moreover the above R -matrix satisfies the Yang-Baxter equation (2). Thisprovides us with an additional set of functional relations among the Boltz-mann weights given by, b − c ′− b ′′ + + c − a ′ + c ′′− = a + c ′− a ′′ + , (11) c − a ′ + b ′′− + b − c ′− c ′′ + = c − b ′− a ′′ + , (12) b + a ′ + c ′′− + c + c ′− b ′′ + = a + b ′ + c ′′− , (13)5 + a ′ + b ′′− + b − c ′ + c ′′− = c + b ′− a ′′ + , (14) b + c ′− b ′′− + c − a ′− c ′′− = a − c ′− a ′′− , (15) b − a ′− c ′′− + c + c ′− b ′′− = a − b ′− c ′′− , (16) b + a ′ + c ′′ + + c − c ′ + b ′′ + = a + b ′ + c ′′ + , (17) b − c ′ + b ′′ + + c + a ′ + c ′′ + = a + c ′ + a ′′ + , (18) c − a ′− b ′′ + + b + c ′− c ′′ + = c − b ′ + a ′′− , (19) b − a ′− c ′′ + + c − c ′ + b ′′− = a − b ′− c ′′ + , (20) c + a ′− b ′′ + + b + c ′ + c ′′− = c + b ′ + a ′′− , (21) b + c ′ + b ′′− + c + a ′− c ′′ + = a − c ′ + a ′′− , (22)where a ± = a ± ( λ, µ ), a ′± = a ± ( λ, γ ) and a ′′± = a ± ( µ, γ ) and likewise for theother functions.The monodromy matrix T A ( λ, { ν k } ) = R A L ( λ, ν L ) · · · R A ( λ, ν ) satisfiesthe fundamental relation (1) thanks to the Yang-Baxter equation (2). Forthe six-vertex model, the monodromy matrix can be written as a 2 × T ( λ, { ν k } ) = A ( λ, { ν k } ) B ( λ, { ν k } ) C ( λ, { ν k } ) D ( λ, { ν k } ) . (23)These operators have to obey sixteen commutation rules which follow from(1). We only list the commutation rules used in this work. B ( λ ) B ( µ ) = a − ( λ, µ ) a + ( λ, µ ) B ( µ ) B ( λ ) , (24) A ( λ ) B ( µ ) = a + ( µ, λ ) b − ( µ, λ ) B ( µ ) A ( λ ) − c + ( µ, λ ) b − ( µ, λ ) B ( λ ) A ( µ ) , (25) D ( λ ) B ( µ ) = a − ( λ, µ ) b − ( λ, µ ) B ( µ ) D ( λ ) − c − ( λ, µ ) b − ( λ, µ ) B ( λ ) D ( µ ) , (26)6 ( λ ) C ( µ ) = a − ( µ, λ ) a + ( µ, λ ) C ( µ ) C ( λ ) , (27) C ( λ ) A ( µ ) = a + ( λ, µ ) b − ( λ, µ ) A ( µ ) C ( λ ) − c − ( λ, µ ) b − ( λ, µ ) A ( λ ) C ( µ ) , (28) C ( λ ) D ( µ ) = a − ( µ, λ ) b − ( µ, λ ) D ( µ ) C ( λ ) − c + ( µ, λ ) b − ( µ, λ ) D ( λ ) C ( µ ) , (29) C ( λ ) B ( µ ) = b + ( λ, µ ) b − ( λ, µ ) B ( µ ) C ( λ ) + c − ( λ, µ ) b − ( λ, µ ) ( A ( µ ) D ( λ ) − A ( λ ) D ( µ )) . (30)Additionally, the structure of the R -matrix (4) implies that both ferro-magnetic states |⇑i = N Lj =1 j and |⇓i = N Lj =1 j are innateeigenstates of the transfer matrix T ( λ, { ν k } ) = A ( λ, { ν k } ) + D ( λ, { ν k } ).Therefore, they play the role of pseudo-vacuum states.One can see immediately that the action of the monodromy matrix overthe pseudo-vacuum state, e.g. |⇑i , results in a triangular matrix, T ( λ, { ν k } ) |⇑i = a ( λ, { ν k } ) |⇑i d ( λ, { ν k } ) |⇑i , (31)where a ( λ, { ν k } ) and d ( λ, { ν k } ) correspond to some fixed representation andthe symbol B ( λ, { ν k } ) and C ( λ, { ν k } ) operators are the creation and annihilation op-erators over the state |⇑i [4], which means A ( λ, { ν k } ) |⇑i = a ( λ, { ν k } ) |⇑i B ( λ, { ν k } ) |⇑i = , (32) C ( λ, { ν k } ) |⇑i = 0 D ( λ, { ν k } ) |⇑i = d ( λ, { ν k } ) |⇑i , (33)or alternatively, h⇑| A ( λ, { ν k } ) = h⇑| a ( λ, { ν k } ) h⇑| B ( λ, { ν k } ) = 0 , (34) h⇑| C ( λ, { ν k } ) = h⇑| D ( λ, { ν k } ) = h⇑| d ( λ, { ν k } ) . (35)7he action of the creation operators over the pseudo-vacuum is the alge-braic version of the famous Bethe ansatz and is given by | Ψ M i = B ( λ M , { ν k } ) · · · B ( λ , { ν k } ) |⇑i . (36)In order to obtain the eigenvalues of the transfer matrix one has to com-mute the operators A ( λ, { ν k } ) and D ( λ, { ν k } ) with B ( λ, { ν k } ). In doing sowe need to use the relations (20,24,25) for the operator A ( λ, { ν k } ) and the re-lations (12,24,26) for the operator D ( λ, { ν k } )[8], which results (see appendixA) A ( λ, { ν k } ) M Y i =1 B ( λ i , { ν k } ) = M Y i =1 a + ( λ i , λ ) b − ( λ i , λ ) M Y i =1 B ( λ i , { ν k } ) A ( λ, { ν k } ) − M X j =1 c + ( λ j , λ ) b − ( λ j , λ ) M Y i =1 i = j a ( θ )+ ( λ i , λ j ) b − ( λ i , λ j ) B ( λ, { ν k } ) M Y i =1 i = j B ( λ i , { ν k } ) A ( λ j , { ν k } ) , (37) D ( λ, { ν k } ) M Y i =1 B ( λ i , { ν k } ) = M Y i =1 a − ( λ, λ i ) b − ( λ, λ i ) M Y i =1 B ( λ i , { ν k } ) D ( λ, { ν k } ) − M X j =1 c − ( λ, λ j ) b − ( λ, λ j ) M Y i =1 i = j a ( θ )+ ( λ j , λ i ) b − ( λ j , λ i ) B ( λ, { ν k } ) M Y i =1 i = j B ( λ i , { ν k } ) D ( λ j , { ν k } ) , (38)where a ( θ )+ ( λ i , λ j ) = a + ( λ i , λ j ) θ > ( λ i , λ j ) and θ > ( λ i , λ j ) = a − ( λ i ,λ j ) a + ( λ i ,λ j ) , i > j, , i ≤ j. (39)Therefore, we obtain T ( λ ) | Ψ M i = Λ M ( λ, { ν k } ) | Ψ M i + M X j =1 Γ j ( λ ) B ( λ, { ν k } ) M Y i =1 i = j B ( λ i , { ν k } ) |⇑i , (40)8here the eigenvalues Λ M ( λ, { ν k } ) are given byΛ M ( λ, { ν k } ) = a ( λ, { ν k } ) M Y i =1 a + ( λ i , λ ) b − ( λ i , λ ) + d ( λ, { ν k } ) M Y i =1 a − ( λ, λ i ) b − ( λ, λ i ) , (41)and Γ j ( λ ) = a ( λ j , { ν k } ) c + ( λ j , λ ) b − ( λ j , λ ) M Y i =1 i = j a ( θ )+ ( λ i , λ j ) b − ( λ i , λ j ) (42)+ d ( λ j , { ν k } ) c − ( λ, λ j ) b − ( λ, λ j ) M Y i =1 i = j a ( θ )+ ( λ j , λ i ) b − ( λ j , λ i ) . Finally, we have to impose the coefficient Γ j ( λ ) to vanish for arbitrary λ in order to cancel the unwanted terms. Using the unitarity relation (5), onesees that the parameters λ j have to satisfy the Bethe ansatz equations a ( λ j , { ν k } ) d ( λ j , { ν k } ) = M Y i =1 i = j a ( θ )+ ( λ j , λ i ) a ( θ )+ ( λ i , λ j ) b − ( λ i , λ j ) b − ( λ j , λ i ) . (43)Similarly the dual Bethe state can be written as follow, h Ψ M | = h⇑| C ( λ , { ν k } ) · · · C ( λ M , { ν k } ) . (44)However, in order to obtain the transfer matrix eigenvalues one should lookfor commutation rules between the operators A ( λ, { ν k } ) and D ( λ, { ν k } ) with C ( λ, { ν k } ). Here, one has to use the relations (16,27,28) for the operator9 ( λ, { ν k } ) and the relations (14,27,29) for the operator D ( λ, { ν k } ), such that M Y i =1 C ( λ i , { ν k } ) A ( λ, { ν k } ) = M Y i =1 a + ( λ i , λ ) b − ( λ i , λ ) A ( λ, { ν k } ) M Y i =1 C ( λ i , { ν k } ) − M X j =1 c − ( λ j , λ ) b − ( λ j , λ ) M Y i =1 i = j a ( θ )+ ( λ i , λ j ) b − ( λ i , λ j ) A ( λ j , { ν k } ) M Y i =1 i = j C ( λ i , { ν k } ) C ( λ, { ν k } ) , (45) M Y i =1 C ( λ i , { ν k } ) D ( λ, { ν k } ) = M Y i =1 a − ( λ, λ i ) b − ( λ, λ i ) D ( λ, { ν k } ) M Y i =1 C ( λ i , { ν k } ) − M X j =1 c + ( λ, λ j ) b − ( λ, λ j ) M Y i =1 i = j a ( θ )+ ( λ j , λ i ) b − ( λ j , λ i ) D ( λ j , { ν k } ) M Y i =1 i = j C ( λ i , { ν k } ) C ( λ, { ν k } ) . (46) The general scalar product is defined as S M ( { λ } Mi =1 |{ µ } Mi =1 ) = h⇑| C ( µ ) · · · C ( µ M ) B ( λ M ) · · · B ( λ ) |⇑i (47)where we have dropped the dependence on ν k . In the particular case thatboth set of parameters λ j and µ j satisfy the Bethe ansatz equations (43), theproduct (47) becomes the norm of Bethe eigenstates[5].To compute the scalar product (47), we must calculate the action of theoperator B ( λ ) over the state (44) (or alternatively, the action of C ( µ ) overthe state (36)). To this purpose we have to use repeatedly the relation (30),which results h⇑| C ( µ ) · · · C ( µ M ) B ( λ M ) = M Y i =1 b + ( µ i , λ M ) b − ( µ i , λ M ) h⇑| B ( λ M ) C ( µ ) · · · C ( µ M )+ M X j =1 c − ( µ j , λ M ) b − ( µ j , λ M ) h⇑| j − Y i =1 C ( µ i ) ( A ( λ M ) D ( µ j ) − A ( µ j ) D ( λ M )) M Y i = j +1 C ( µ i ) , (48)10here the first term vanishes thanks to (34) and we set λ i = µ M +1 − i , i =1 , · · · , M for convenience.After that, one has to compute the action of A ( λ ) D ( µ j ) − A ( µ j ) D ( λ )over the C ( µ i ), where one should use the algebraic relations (45-46) alongthe same lines as [2, 5], such that h⇑| C ( µ ) · · · C ( µ M ) B ( µ M +1 ) = M +1 X j,k =1 j = k h⇑| A ( µ k ) D ( µ j ) × (49) c − ( µ j , µ M +1 ) c + ( µ M +1 , µ k ) a ( θ )+ ( µ j , µ k ) a ( θ )+ ( µ j , µ M +1 ) a ( θ )+ ( µ M +1 , µ k ) b − ( µ j , µ k ) M +1 Y i =1 i = j i = k a ( θ )+ ( µ i , µ k ) b − ( µ i , µ k ) a ( θ )+ ( µ j , µ i ) b − ( µ j , µ i ) M +1 Y i =1 i = j i = k C ( µ i ) , where we have also used the functional relation (16) and the unitarity rela-tions (7-9) for simplifications.So we multiply the previous expression by additional B ( µ i ) operators, S M ( { µ } Mi = M +1 |{ µ } Mi =1 ) = h⇑| C ( µ ) · · · C ( µ M ) B ( µ M +1 ) Q Mi = M +2 B ( µ i ) |⇑i . Thisresults in a recurrence relation for the scalar product given by, S M ( { µ } Mi = M +1 |{ µ } Mi =1 ) = M +1 X j,k =1 j = k a ( µ k ) d ( µ j ) c − ( µ j , µ M +1 ) c + ( µ M +1 , µ k ) a ( θ )+ ( µ j , µ k ) a ( θ )+ ( µ j , µ M +1 ) a ( θ )+ ( µ M +1 , µ k ) b − ( µ j , µ k ) ×× M +1 Y i =1 i = j i = k a ( θ )+ ( µ i , µ k ) b − ( µ i , µ k ) a ( θ )+ ( µ j , µ i ) b − ( µ j , µ i ) S M − ( { µ } Mi = M +2 |{ µ } M i =1 i = j i = k ) . (50)One can iterate the above recursion relation (50) and represent the re-sulting expression as follows, S M ( { λ }|{ µ } ) = X { λ } = { λ + }∪{ λ −}{ µ } = { µ + }∪{ µ −}| λ + | + | µ −| = n + n + Y i =1 a ( λ + i ) d ( µ + i ) n − Y i =1 a ( µ − i ) d ( λ − i ) K M (cid:18) { λ + }{ λ − } (cid:12)(cid:12)(cid:12) { µ + }{ µ − } (cid:19) , (51)11here we sum with respect to all partitions of the set { λ } ∪ { µ } into twodisjoint sets { λ + } ∪ { µ − } and { λ − } ∪ { µ + } of M of elements. If the numberof elements in the sets { λ + } and { µ + } is n + = | λ + | = | µ + | , then we have n − = | λ − | = | µ − | = M − n + elements in the sets { λ − } and { µ − } .Notice that any coefficient K M is determined only by terms arising fromalgebraic relations among monodromy matrix elements. Therefore, K M isindependent of the representation, i.e. independent of the choice of the func-tions a ( λ, { ν k } ) and d ( λ, { ν k } ) [2].In view of that one can fix the coefficients K M using any special repre-sentation. Let’s consider the case where L = M and a M ( λ ) = M Y i =1 a + ( λ, ν i ) , d M ( λ ) = M Y i =1 b − ( λ, ν i ) , (52)where we have fixed the inhomogeneities as follows, ν i = λ + i , i = 1 , · · · , n + µ − i − n + , i = n + + 1 , · · · , M. (53)Using the fact that the unitarity relation (5) implies that b − ( λ, λ ) = 0,one sees that d M ( λ ) = 0 for any λ ∈ { λ + } ∪ { µ − } and d M ( λ ) = 0 for any λ ∈ { λ − } ∪ { µ + } . Consequently, there is only one non-vanishing term in thesum (51), which allow us to write n + Y i =1 a M ( λ + i ) d M ( µ + i ) n − Y i =1 a M ( µ − i ) d M ( λ − i ) K M ( { λ + } , { λ − }|{ µ + } , { µ − } )= h⇑ M | C ( µ ) · · · C ( µ M ) B ( λ M ) · · · B ( λ ) |⇑ M i , (54)where |⇑ M i is a state with M spins up.12he product of B -operators overturns all M spins resulting in the state |⇓ M i . Therefore, we have h⇑ M | C ( µ ) · · · C ( µ M ) B ( λ M ) · · · B ( λ ) |⇑ M i == Z ( C ) M ( { µ } ; { λ + } ∪ { µ − } ) Z ( B ) M ( { λ } ; { λ + } ∪ { µ − } ) , (55)where the functions Z ( B,C ) M are defined by Z ( B ) M ( { λ } ; { λ + } ∪ { µ − } ) = h⇓ M | B ( λ M ) · · · B ( λ ) |⇑ M i , (56) Z ( C ) M ( { µ } ; { λ + } ∪ { µ − } ) = h⇑ M | C ( µ ) · · · C ( µ M ) |⇓ M i . (57)Finally, the coefficient K M can be written in terms of the above functions asfollows K M (cid:18) { λ + }{ λ − } (cid:12)(cid:12)(cid:12) { µ + }{ µ − } (cid:19) = Z ( C ) M ( { µ } ; { λ + } ∪ { µ − } ) Z ( B ) M ( { λ } ; { λ + } ∪ { µ − } ) Q n + i =1 a M ( λ + i ) d M ( µ + i ) Q n − i =1 a M ( µ − i ) d M ( λ − i ) . (58)The functions (56-57) are usually called domain wall partition function[5] and play an important role on the calculation of scalar products andcorrelation function. We shall compute these partition functions on the nextsection in order to write a closed formula for the scalar product (51). In this section we derive a recurrence relation for the partition function forthe asymmetric six-vertex model with domain wall boundary conditions forarbitrary Boltzmann weights. This way we will proceed along the samelines of [11] and define a couple of auxiliary one-point boundary correlation13unctions as follows, G ( B ) M,N = 1 Z ( B ) M h⇓ M | B ( λ M ) · · · B ( λ N +1 ) p − B ( λ N ) B ( λ N − ) · · · B ( λ ) |⇑ M i , (59) H ( B ) M,N = 1 Z ( B ) M h⇓ M | B ( λ M ) · · · B ( λ N +1 ) p − B ( λ N ) p +1 B ( λ N − ) · · · B ( λ ) |⇑ M i , (60)where p ± = (1 ± σ z ) are the local spin up and down projectors and G ( B ) M,M =1. The above boundary correlations are related by G ( B ) M,N = N X j =1 H ( B ) M,j , (61) G ( B ) M,N = H ( B ) M,N + G ( B ) M,N − . (62)One can use the above relations in order to write the domain wall partitionfunction Z ( B ) M in terms of the boundary correlations Z ( B ) M ( { λ } ; { ν } ) = M X j =1 h⇓ M | B ( λ M ) · · · B ( λ j +1 ) p − B ( λ j ) p +1 B ( λ j − ) · · · B ( λ ) |⇑ M i . (63)To derive the recurrence relation for the partition function we use thetwo-site model decomposition. More specifically, one has to decompose themonodromy matrix into two parts and introduce two monodromy matrices T A ( λ ) = T (2) A ( λ ) T (1) A ( λ ) = A ( λ ) B ( λ ) C ( λ ) D ( λ ) , (64)such that T (1) A = R A ( λ, ν ) = A ( λ ) B ( λ ) C ( λ ) D ( λ ) , (65) T (2) A = R A M ( λ, ν M ) · · · R A ( λ, ν ) = A ( λ ) B ( λ ) C ( λ ) D ( λ ) , (66)14hich act on the states |⇑ i i , i = 1 , |⇑ i = , |⇑ i = N Mj =2 j . (67)In particular, the required monodromy matrix elements are readily ob-tained B ( λ ) = A ( λ ) B ( λ ) + B ( λ ) D ( λ ) , (68) B ( λ ) = c + ( λ, ν ) σ − , (69) D ( λ ) = b − ( λ, ν ) p +1 + a − ( λ, ν ) p − , (70)and the action of the operator A ( λ ) on the state |⇑ i is given by A ( λ ) |⇑ i = M Y i =2 a + ( λ, ν i ) |⇑ i . (71)Using (68-70), we can rewrite (63) in terms of the operators A ( λ ) and B ( λ ) as follows, Z ( B ) M ( { λ } ; { ν } ) = M X j =1 c + ( λ j , ν ) j − Y i =1 b − ( λ i , ν ) M Y i = j +1 a − ( λ i , ν ) (72) × h⇓ | B ( λ M ) · · · B ( λ j +1 ) A ( λ j ) B ( λ j − ) · · · B ( λ ) |⇑ i . At this point one should use the algebraic relations (24,37) [11] in order toobtain the recurrence relation for the domain wall partition function, Z ( B ) M ( { λ } Mi =1 ; { ν } Mi =1 ) = M X j =1 c + ( λ j , ν ) M Y i =1 i = j b − ( λ i , ν ) M Y i =1 i =1 a + ( λ j , ν i ) × M Y i =1 i = j a ( θ )+ ( λ i , λ j ) b − ( λ i , λ j ) Z ( B ) M − ( { λ } M i =1 i = j ; { ν } M i =1 i =1 ) . (73)15terating the above relation M − P of { λ } Mi =1 Z ( B ) M ( { λ } Mi =1 ; { ν } Mi =1 ) = X P M Y i =1 c + ( λ P i , ν i ) M Y i,j =1 i 16s follows, K M (cid:18) { λ + }{ λ − } (cid:12)(cid:12)(cid:12) { µ + }{ µ − } (cid:19) = n + Y j =1 n − Y k =1 a + ( µ + j , µ − k ) b − ( µ + j , µ − k ) a − ( λ − k , λ + j ) b − ( λ − k , λ + j ) n + Y j,k =1 j Acknowledgments The author thanks F. G¨ohmann for introducing him to this topic and M.J.Martins for discussions. This work has been supported by FAPESP andCNPq. Appendix A: two and three-particle state In this appendix we work out in detail the commutation of the operator A ( λ )with two and three B -operators in order to exemplify the use of Yang-Baxterrelations (11-22). In fact, to pass the A -operator over any number of B ’s oneneed only to use the Yang-Baxter relation (20).18 .1 M = 2 Let’s start with the two B -operators case and use the commutation rule (25)twice, which results A ( λ ) B ( λ ) B ( λ ) = a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) A ( λ ) − c + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) A ( λ ) (A.1) − h a + ( λ , λ ) b − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) | {z } eq. ( ) − c + ( λ , λ ) b − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) | {z } eq. ( ) B ( λ ) B ( λ ) i A ( λ ) . We can further manipulate the third term of the above expression usingthe commutation rule (24) and the unitarity relation (5) as indicated. Thisprovides us with the following expression A ( λ ) B ( λ ) B ( λ ) = a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) A ( λ ) − c + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) A ( λ ) (A.2) − h eq. ( ) z }| { b − ( λ , λ ) a − ( λ , λ ) c + ( λ , λ ) + c − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) i B ( λ ) B ( λ ) A ( λ ) . Finally, one can see that the resulting expression for the third term coincideswith the left hand side of the Yang-Baxter relation (20). After substituting(20) in (A.2), we obtain A ( λ ) B ( λ ) B ( λ ) = a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) A ( λ ) − c + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) A ( λ ) (A.3) − c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) A ( λ ) . .2 M = 3 Now we turn to the case where we have three B -operators. We use theprevious result (A.4) and the commutation rule (24), such that A ( λ ) B ( λ ) B ( λ ) B ( λ ) = a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) A ( λ ) − c + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) A ( λ ) − c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) A ( λ ) (A.4)+ h − c + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) eq. ( ) z }| { B ( λ ) B ( λ ) B ( λ ) | {z } eq. ( ) + c + ( λ , λ ) b − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) | {z } eq. ( ) + c + ( λ , λ ) b − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) | {z } eq. ( ) a − ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) i A ( λ ) . Again one should use the commutation rule (24) and unitarity relation (5)as indicated. So the last term of (A.4) can be rewritten as follows, − h c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) − c + ( λ , λ ) b − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ )+ c + ( λ , λ ) b − ( λ , λ ) c − ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) i B ( λ ) B ( λ ) B ( λ ) A ( λ ) . (A.5)At this point, we have to simplify the term inside the square bracket ( I ).This can be done by joining the first and the last term on (A.5) and using20he equation (20) a couple of times, such that I = [ b − ( λ , λ ) a − ( λ , λ ) c + ( λ , λ )] a − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ )+ eq. ( ) z }| { [ a − ( λ , λ ) b − ( λ , λ ) c + ( λ , λ )] c − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) (A.6) − c + ( λ , λ ) b − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) , and I = a − ( λ , λ ) b − ( λ , λ ) eq. ( ) z }| { [ b − ( λ , λ ) a − ( λ , λ ) c + ( λ , λ ) + c − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ )] b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ )+ c − ( λ , λ ) b − ( λ , λ ) | {z } eq. ( ) c + ( λ , λ ) b − ( λ , λ ) c − ( λ , λ ) b − ( λ , λ ) − c + ( λ , λ ) b − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) . (A.7)Then we again have to use the equation (20) twice, I = c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) (A.8) − c + ( λ , λ ) b − ( λ , λ ) eq. ( ) z }| { [ c − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) + b − ( λ , λ ) a − ( λ , λ ) c + ( λ , λ )] b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) , = a − ( λ , λ ) b − ( λ , λ ) h c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) − c + ( λ , λ ) b − ( λ , λ ) | {z } eq. ( ) c + ( λ , λ ) b − ( λ , λ ) i , (A.9)and I = a − ( λ , λ ) b − ( λ , λ ) eq. ( ) z }| { [ b − ( λ , λ ) a − ( λ , λ ) c + ( λ , λ ) + c − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ )] b − ( λ , λ ) b − ( λ , λ ) b − ( λ , λ ) , (A.10)= a − ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) c + ( λ , λ ) b − ( λ , λ ) . (A.11)21inally, we can substitute the final result for the term I (A.11) in Eq.(A.4), A ( λ ) B ( λ ) B ( λ ) B ( λ ) = a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) A ( λ ) − c + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) A ( λ ) − c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) a + ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) A ( λ ) (A.12) − c + ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) a − ( λ , λ ) b − ( λ , λ ) B ( λ ) B ( λ ) B ( λ ) A ( λ ) . This expression coincides with the formula (37) for the case M = 3. Notethat we have only used the commutation rules (24,25), the unitarity relation(5) and the Yang-Baxter relation (20). It is remarkable that the same appliesto any M -values, where we again would need only the mentioned relations.Similarly, the expression (38) can be obtained using the commutationrules (24,26), the unitarity relation (5) and the Yang-Baxter relation (12).Alternatively, one could have obtained the formula (37) in a shorter way.Note that the first term in (37) is obtained straightforwardly using the firstterm of the algebraic relation (25) M -times. On the other hand, the secondterm of (37) can be very complicated, as we have seen above. The coefficientof the term not containing the first B -operator, B ( λ M ), is very simple though.This coefficient is obtained using the second term of (25) in order to exchangethe A ( λ ) and B ( λ M ) and in all other steps one has to use only the first term22f (25) [4], which results in the following expression A ( λ, { ν k } ) M Y i =1 B ( λ i , { ν k } ) = M Y i =1 a + ( λ i , λ ) b − ( λ i , λ ) M Y i =1 B ( λ i , { ν k } ) A ( λ, { ν k } ) − c + ( λ M , λ ) b − ( λ M , λ ) M Y i =1 i = M a + ( λ i , λ j ) b − ( λ i , λ j ) B ( λ, { ν k } ) M Y i =1 i = M B ( λ i , { ν k } ) A ( λ M , { ν k } ) (A.13)+ sum of other terms not containing B ( λ j ), j = 1 , · · · , M − . However, one could have done similar analysis for any B ( λ j ). In doingso, one has to use the relation (24) in order to move B ( λ j ) to the leftmostposition, B ( λ M ) · · · B ( λ j +1 ) B ( λ j ) | {z } eq. ( ) · · · B ( λ ) = B ( λ j ) M Y i =1 i = j B ( λ i ) θ > ( λ i , λ j ) (A.14)and then repeat the above described procedure. This would result in themissing terms in (A.13) and finally we should obtain the expression (37). References [1] L.A. Takhtajan and L.D. Faddeev, Russian Math. Surveys, 34 (1979)11 [2] V.E. Korepin, G. Izergin and N.M. Bogoliubov,