Form-factors in the Baxter-Bazhanov-Stroganov model II: Ising model on the finite lattice
G. von Gehlen, N. Iorgov, S. Pakuliak, V. Shadura, Yu. Tykhyy
aa r X i v : . [ n li n . S I] A p r Form-factors in the Baxter–Bazhanov–Stroganovmodel II: Ising model on the finite lattice
G von Gehlen † , N Iorgov ‡ , S Pakuliak ♯♭ , V Shadura ‡ andYu Tykhyy ‡§ † Physikalisches Institut der Universit¨at Bonn, Nussallee 12, D-53115 Bonn, Germany ‡ Bogolyubov Institute for Theoretical Physics, Kiev 03680, Ukraine ♯ Bogoliubov Laboratory of Theoretical Physics, Joint Institute for Nuclear Research,Dubna 141980, Moscow region, Russia ♭ Institute of Theoretical and Experimental Physics, Moscow 117259, Russia § Laboratoire de Math´ematiques et Physique Th´eorique CNRS/UMR 6083,Universit´e de Tours, Parc de Grandmont, 37200 Tours, FranceE-mail: [email protected], [email protected],[email protected], [email protected], [email protected]
Dedicated to Professor Anatoly Bugrij on the occasion of his 60-th birthday
Abstract.
We continue our investigation of the Baxter–Bazhanov–Stroganov or τ (2) -model using the method of separation of variables [1, 2]. In this paper we derive for thefirst time the factorized formula for form-factors of the Ising model on a finite latticeconjectured previously by A. Bugrij and O. Lisovyy in [3, 4]. We also find the matrixelements of the spin operator for the finite quantum Ising chain in a transverse field. J. Phys. A: Math. Gen.
PACS numbers: 75.10Hk, 75.10Jm, 05.50+q, 02.30Ik
1. Introduction
The Baxter–Bazhanov–Stroganov (BBS) model [5, 6] (also called the τ (2) -model, see e.g.[7, 8]) is associated to the cyclic L -operators [9, 6, 10] which act in a two-dimensionalauxiliary space L j ( λ ) = λ κ j v j λ u − j ( a j − b j v j ) u j ( c j − d j v j ) λ a j c j + v j b j d j / κ j ! , (1)where λ is the spectral parameter, and at each site j = 1 , . . . , n we have five parameters a j , b j , c j , d j and κ j . At each site there is also an ultra-local Weyl algebra with elements u j and v j , obeying u j u k = u k u j , v j v k = v k v j , u j v k = ω δ j,k v k u j , ω = e πi/N , u Nk = v Nk = 1 , N ∈ Z ≥ . orm-factors in the BBS model II: Ising model on the finite lattice ω is a root of unity, the Weyl operators can be represented naturally by matricesacting in the tensor product (cid:0) C N (cid:1) ⊗ n [1, 2]. The monodromy matrix of the model isdefined by T n ( λ ) = L ( λ ) L ( λ ) · · · L n ( λ ) = A n ( λ ) B n ( λ ) C n ( λ ) D n ( λ ) ! , (2)and the transfer matrix is its trace in the auxiliary space t n ( λ ) = tr T n ( λ ) and givesrise to a set of commuting non-local and non-hermitian Hamiltonians of the model: t n ( λ ) = A n ( λ ) + D n ( λ ) = H + H λ + · · · + H n − λ n − + H n λ n . Commutativity follows from the intertwining of the L k ( λ ) by the asymmetric 6-vertexmodel R -matrix at root of unity.In the present paper our focus will be on the case N = 2. As it has been shown in[11], in this case the BBS-model is related to a generalized Ising model with plaquetteBoltzmann weights W ( σ , σ , σ , σ ) = a (cid:16) P ≤ i ≤ j ≤ a ij σ i σ j + a σ σ σ σ (cid:17) , (3)subject to the free-fermion condition a = a a − a a + a a . Generalizing Sklyanin’s method of Separation of Variables (SoV) [12, 13, 14], in[1, 2] we have worked out a method to find common eigenvectors of the Hamiltonians H m . This proceeds in two steps: ‡• Since [ B ( λ ) , B ( µ )] = 0, the off-diagonal element B ( λ ) of the monodromy matrix(2) gives rise to an auxiliary set of commuting operators h m : B ( λ ) = h λ + h λ + · · · + h n λ n . We construct their common right eigenvectors | Ψ ρ i (the lefteigenvectors h Ψ ρ | are obtained analogously) by an inductive procedure over thechain size n , starting from the one site model. The eigenvalues of B ( λ ) form apolynomial in λ of degree n , and from the intertwining relations we can show that B ( λ ) | Ψ ρ i = λ r ω − ρ n − Y k =1 (cid:0) λ + r k ω − ρ k (cid:1) | Ψ ρ i , ρ = ( ρ o , ρ , . . . , ρ n − ) , (4)where the amplitudes ( r , r , . . . , r n − ) can be expressed in terms of the parametersof the model a j , . . . , κ j , and we can use the phases ρ , ρ k ∈ Z N , of the zeros ν k = − r k ω − ρ k , k = 1 , . . . , n − , (5)of the eigenvalue polynomial for labelling the eigenvectors. • Having solved the auxiliary problem, after a Fourier transformation of ρ to ρ ( ρ labels the Z N -charge sectors), the eigenvalue problem of t ( λ ) = A ( λ ) + D ( λ )is reduced to the solution of Baxter equations. Using that A ( ν k ) and D ( ν k ) areraising and lowering operators on the auxiliary states | Ψ ρ i , we find that the periodiceigenstates | Φ ρ, E i in t ( λ ) | Φ ρ, E i = t ( ρ ) ( λ | E ) | Φ ρ, E i , E = { E , . . . , E n − } ‡ We consider a fixed chain length n , mostly suppressing the subscript n , which had been writtenexplicitly in our previous papers, so r k ≡ r n,k and ρ n ≡ ρ n,k of [1, 2], etc. orm-factors in the BBS model II: Ising model on the finite lattice t ( ρ ) ( λ | E ) = E + E λ + · · · + E n λ n (where E and E n are directly known, see (20)), are obtained via the kernels Q R in | Φ ρ, E i = X ρ , ρ ′ ω − ρ · ρ Q R ( ρ ′ | ρ, E ) | Ψ ρ , ρ ′ i , ρ ′ = ( ρ , . . . , ρ n − ) . (6)The crucial fact is now (SoV) that after splitting off a known function f ( ρ ′ ), the n − Q R ( ρ ′ | ρ, E ) factorizes into single variable functions Q R k ( ρ k )(the ρ k are the components of ρ ′ , we often skip the charge index ρ of the Q R k etc.): Q R ( ρ ′ | ρ, E ) = f ( ρ ′ ) Q n − k =1 Q R k ( ρ k ) , and the Q R k ( ρ k ) are determined by the Baxter equations ( k = 1 , . . . , n − t ( ρ ) ( ν k | E ) Q R k ( ρ k ) = ∆ + k ( ν k ) Q R k ( ρ k + 1) + ∆ − k ( ων k ) Q R k ( ρ k − . (7)The corresponding Baxter equations for the left periodic eigenvector read t ( ρ ) ( ν k | E ) Q L k ( ρ k ) = ω n − ∆ − k ( ν k ) Q L k ( ρ k + 1) + ω − n ∆ + k ( ων k ) Q L k ( ρ k − . (8)The functions ∆ ± k are defined by∆ + k ( λ ) = ( ω ρ /χ k ) ( λ/ω ) − n n − Y m =1 F m ( λ/ω ) , ∆ − k ( λ ) = χ k ( λ/ω ) n − F n ( λ/ω ) , (9) F m ( λ ) = ( b m + ωa m κ m λ ) ( λ c m + d m / κ m ) . (10)We shall not need the expression for χ k , see (43) of [2], since this will cancel in ourfinal formulas. The existence of a non-trivial solution to (7), (8) is provided by a set offunctional equations, which determines the still unknown values E .In [2] we calculated the action of u n , the Weyl operator at site n , on an eigenvector | Ψ ρ i of B ( λ ) to have the form u n | Ψ ρ i = g | Ψ ρ i + P n − k =0 g k | Ψ ρ + k i with ρ + k = ( ρ , . . . , ρ k + 1 , . . . , ρ n − ) , (11)and g and g k are certain functions depending on the parameters a j , b j , c j , d j , κ j andthe components of ρ . Since in [1, 2] we also found a factorized expression for thenorm h Ψ ρ | Ψ ρ ′ i , we have the framework for calculating normalized matrix elements h Ψ ρ | u n | Ψ ρ ′ i / h Ψ | Ψ i , where = (0 , , . . . , h Φ ρ | u n | Φ ρ ′ i , in addition we also need the solutions Q ( ρ ) k ( ρ k ) of the Baxterequations. These are available for N = 2, and our main goal is to obtain such periodicmatrix elements in a factorized form. We achieve this by explicitly performing the sumsover the intermediate Z variables.This paper is organized as follows: In the following Section 2 we recall the N = 2spin matrix element calculated in [2] and transform it into a much more compact form byperforming the summation over an intermediate variable, still keeping the model generaland inhomogeneous. In order to proceed beyond this result, in Section 3 we specialize theparameters of the BBS-model such that we get the homogenous Ising model. We discussthe structure of the eigenvalues and the solutions to the Baxter equations. The vanishingof some transfer matrix eigenvalues at the zeros of B ( λ ) requires us to distinguish fourcases when solving the Baxter equations. Then in Section 4 we continue the evaluation of orm-factors in the BBS model II: Ising model on the finite lattice σ z for the finite quantum Ising chain in a transverse magnetic field. In Section 9 wegive our conclusions.
2. Spin operator matrix element for the N = 2 inhomogenous BBS model In [2] we derived a formula for the normalized matrix element of the spin operatorbetween arbitrary states of the periodic inhomogenous N = 2 BBS-model. For N = 2there are two Z -charge sectors ρ = 0 , u n is anticommuting with thecharge operator V = v v . . . v n , its only non-vanishing matrix elements are betweenperiodic states | Φ ρ i with different charge ρ . Let Q L( ρ ) and Q R( ρ ) be solutions to theBaxter equations (7), (8) and r k the zeros of the operator polynomial B ( λ ), see (4).Then our result in (58) of [2] can be written: h Φ | u n | Φ ih ˜Ψ , | ˜Ψ , i = X ρ ′ N ( ρ ′ ) R ( ρ ′ ) (cid:18) a n ˜ r ( − ˜ ρ ′ − κ κ · · · κ n − b n r (cid:19) + n − X k =1 R k ( ρ ′ ) ! , (12) N ( ρ ′ ) = ( − n ˜ ρ ′ n − Y l 3. Homogeneous Ising model In all following Sections we consider only the N = 2 case of the model defined by (1)with (23). For a fixed chain length n , we are left with only the two parameters a, b andthe spectral parameter λ . For N = 2 we have ω = − u k , v k by Pauli matrices acting at the k -th site. So, now our model is defined by L k ( λ ) = λ σ xk λ σ zk ( a − b σ xk ) σ zk ( a − b σ xk ) λa + σ xk b ! . (24)Fixing the spectral parameter to the value λ = b/a , the L -operator (24) degenerates L k ( b/a ) = (1 + σ xk b/a ) a σ zk ! (cid:16) , b σ zk (cid:17) orm-factors in the BBS model II: Ising model on the finite lattice t ( b/a ) = tr L ( b/a ) L ( b/a ) · · · L n ( b/a ) = n Y k =1 (1 + σ xk · b/a ) · n Y k =1 (1 + σ zk − σ zk · a b ) ∼ exp ( P nk =1 K ∗ x σ xk ) exp (cid:0)P nk =1 K x σ zk − σ zk (cid:1) , (25)if we use periodic boundary conditions σ zn + k ≡ σ zk and identify e − K y = tanh K ∗ x = b/a , tanh K x = ab . (26)So at λ = b/a we call the model (24) the Ising model. If we don’t fix the spectralparameter to this special value, we shall talk of the “generalized Ising model”. However,transfer matrix eigenstates are independent of the choice of λ . In [1] the eigenvalues of the transfer matrix t ( λ ) = tr L ( λ ) · · · L n ( λ ) with L k ( λ )given by (1) for N = 2 and homogeneous parameters, have been calculated from thefunctional relations. From Z -invariance t ( λ ) commutes with the Z -charge operator V = σ x σ x . . . σ xn . Since V = 1, the space of eigenstates of t ( λ ) decomposes into twosectors according to the eigenvalues ( − ρ (where ρ = 0 , 1) of V . The sector ρ = 0is called the NS-sector, ρ = 1 the R-sector. The 2 n eigenvalues can be written (wespecialize assuming (23)): t ( ρ ) ( λ ) = ( a n + ( − ρ ) Y q ( λ + ( − σ q s q ) , s q = s − q = s b − b cos q + 1 a − a cos q + 1 , (27)where the quasi-momentum q in each sector takes n values q = πn m with m integer(half-integer) for the R (NS)-sectors. If σ q = 0 the quasi-momentum q is calledunexcited, for σ q = 1 it is called excited. In the NS (R) sector, the eigenstates of t ( λ ) have an even (odd) number of excitations: Q q ( − σ q = ( − ρ .For q = 0 (this occurs only for the R-sector) and for q = π we define s = b − a − , s π = b + 1 a + 1 . (28)The quasi-momentum q = π is in the R sector for n even, it is in the NS sector for n odd. The different presence of factors ( λ ± s ) and ( λ ± s π ) in (27) for n even or oddoften makes it necessary to consider the cases of even n and odd n separately.Sometimes we shall use the notation λ q := ( − σ q s q . In order to obtain the eigenvectors of the transfer matrix t ( λ ), we have to solve Baxter’sequations (7) and (8). As input we use the corresponding eigenvalues t ρ ( λ ) which arespecified by the values σ q for all q in the sector ρ , see (27). Solving Baxter’s equations,we should use the values t ρ ( ± r k ) of these polynomials at the values ± r k of the roots ofthe eigenvalue polynomials of the operator B n ( λ ) (which is the off-diagonal element of orm-factors in the BBS model II: Ising model on the finite lattice N = 2 the r k are simply related with the s q : r k = s q k , q k = πk/n, k = 1 , . . . , n − . (29)This means that for our special choice of parameters (23), the zeros of t ρ ( λ ) may coincidewith the r k , giving rise to the vanishing of the left hand sides of (7) and (8). At theparameters (23) all F m are equal: F m ( λ ) = F ( λ ) and from (10) we obtain F ( λ ) = b − a λ , χ k r n − k = ( − n + k +1 F n − ( r k ) . (30)Let us compare two sets: the set { q k } which parameterizes the roots r k , and the setof all possible quasi-momenta { q } . The latter set is divided into two sub-sets: the NSand the R sectors. The NS-sector contains pairs of quasi-momenta { q k , − q k } for odd k and the R-sector includes the pairs { q k , − q k } for even k . The quasi-momentum q = 0always belongs to the R-sector, and q = π belongs to the R-sector for even n and to theNS-sector for odd n .The solutions of Baxter’s equations for the case of the Ising model were found in[2]. Here we recall the final result. For a fixed sector ρ and the eigenvalue polynomial t ρ ( λ ) we have to solve n − k = 1 , . . . , n − 1. With respect to these data we have to distinguish four cases:( − ρ = ( − k :(i) t ρ ( r k ) = 0 and t ρ ( − r k ) = 0: Q L , R k (0) = 1 , Q L , R k (1) = ( − n − t ρ ( − r k )2 χ k r n − k F ( r k ) . The other three cases occur for( − ρ = ( − k − :(ii) t ρ ( r k ) = 0 , t ρ ( − r k ) = 0: t ρ ( λ ) contains a factor ( λ + r k ) (both q = ± q k notexcited), we may normalize Q L , R k (0) = 1 , Q L , R k (1) = 0 . (iii) t ρ ( r k ) = 0 , t ρ ( − r k ) = 0: t ρ ( λ ) contains a factor ( λ − r k ) (both q = ± q k areexcited), we cannot choose Q L , R k (0) = 1, but we may normalize Q L , R k (0) = 0 , Q L , R k (1) = 1 . (iv) t ρ ( r k ) = t ρ ( − r k ) = 0: t ρ ( λ ) contains ( λ − r k ) (either q = + q k or q = − q k is excited): A L’Hˆopital procedure as described in [2] is required (in order to obtaineigenvectors of the translation operator), leading to Q R k (0) = Q L k (0) = 1 , Q R k (1) = − Q L k (1) = ( − n + σ qk +1 2i sin q k t ρ ˇ q k ( − r k ) n χ k r n − k A ( q k ) , where t ρ ( λ ) = t ρ ˇ q k ( λ ) ( λ + ( − σ qk s q k )( λ − ( − σ qk s − q k ) , A ( q ) = a − a cos q + 1 . (31) orm-factors in the BBS model II: Ising model on the finite lattice Q L , R k (0) = 1 , postponing to Section 6 theconsideration of eigenvectors for which t ρ ( λ ) contains factors ( λ − r k ) , i.e. theeigenvectors involving case (iii) above.As already observed at the beginning of Section 2, the non-vanishing spin matrixelements have left and right eigenstates from different sectors. Let t (0) and t (1) be thecorresponding eigenvalue-polynomials. With respect to these two polynomials we define k ∈ ˘ D ( ρ ) if t ρ has a factor ( λ + r k ) , i.e. we have case (ii), k ∈ b D ( ρ ) if t ρ has a factor ( λ − r k ) , case (iii), and k ∈ D ( ρ ) if t ρ has a factor ( λ − r k ), i.e. we have case (iv).By D = |D| we denote the number of elements in D = D (0) ∪ D (1) , similarly for ˘ D , etc. 4. Calculation of the matrix element of σ zn in the homogeneous Ising model We now start to evaluate (21) with (13) and (22) in our simplified model where ζ = b/a, r = ( a − b )( a n − / ( a − . (32)We have to observe that in the derivation of (12) given in [2], generic BBS-parametersleading to t ρ ( r k ) = 0 were assumed, and the solutions to the Baxter equation werenormalized to Q L , R k (0) = 1. As we have seen in Section 3.2, in the case (iii) thisnormalization is not possible for the special parameters (23). In order not to complicatethe derivation, in the following part of this Section we shall simply exclude state vectorscontaining k ∈ b D , adding the changes necessary for k ∈ b D in Section 6. Also in thissection we shall omit the superscripts L and R in the notations of Q L , R( ρ ) k ( ρ k ) supposingthat the left/right eigenvectors are from NS/R-sectors as they appear in (21).Consider R ( ρ ′ ). Always one of the factors in Q (0) l ( ρ l ) Q (1) l ( ρ l ) is from case (i) above,and excluding l ∈ b D , the other is from either (ii) or (iv). So always Q (0) l (0) Q (1) l (0) = 1.For l ∈ ˘ D we have Q (0) l (1) Q (1) l (1) = 0 since either Q (0) l (1) = 0 or Q (1) l (1) = 0 dependingon the parity of l . So, in (21) the summation reduces to the summation over ρ l for l ∈ D , with ρ l = 0 fixed for l ∈ ˘ D . R ( ρ ′ )Let us show that a common factor can be extracted from the two terms of (22). Wefirst consider the case of odd n where q = 0 appears in the R-sector and q = π in theNS-sector.We start with the first term R (0) ( ρ ′ ) in (22). Now from (27) the NS eigenvaluepolynomial t (0) ( λ ) for odd n isNS , n odd: t (0) ( λ ) = ( a n + 1)( λ + ( − σ π s π ) Y k ∈ ˘ D (0) ( λ + r k ) Y l ∈D (0) ( λ − r l ) , (33)(for even n omit the bracket with s π ) since in the NS-sector only odd k appear, andthese fall into one of the classes (ii) and (iv), class (iii) being momentarily excluded. We orm-factors in the BBS model II: Ising model on the finite lattice t (0) ( − ζ ) from (33) and decompose the denominator product over l in its even- l and odd- l parts. We write the odd part as l ∈ D (0) ∪ ˘ D (0) since for ρ = 0 in cases (ii)and (iv) l must be odd (recall, we still exclude case (iii)): R (0) ( ρ ′ ) = ( a n + 1) ( − ζ + ( − σ π s π ) Q l even ( − ζ + ( − ρ l r l ) Q k ∈ ˘ D (0) ( − ζ + r k ) Q l ∈D (0) ( ζ − r l ) Q k ∈ ˘ D (0) ( − ζ + r k ) Q l ∈D (0) ( − ζ + ( − ρ l r l ) × Q m ∈ ˘ D (1) ( ζ + r m ) Q m ∈D (1) ( ζ + ( − ρ m r m ) Q m even ( ζ + ( − ρ m r m ) . (34)In the last line we put a factor unity, written as quotient of upstairs a product over m ∈ D (1) ∪ ˘ D (1) and downstairs over m even. In the ˘ D terms we omitted the factor( − ρ k since from (ii) this contributes only if ρ k = 0. Now several cancellations takeplace, resulting in R (0) ( ρ ′ ) = ( a n + 1) ( − ζ + ( − σ π s π )( − |D (0) | Q l even ( − ζ + r l ) Y k ∈ ˘ D (( − k ζ + r k ) Y l ∈D ( ζ + ( − ρ l r l ) . (35)Observe that now the ρ ′ -dependence appears only in the last product over D . Thishappens because all l -odd terms cancel and because ˘ D (0) allows only ρ l = 0. In thedenominator we use ζ = b/a , (27), (29) and Q l even A ( q l ) = ( a n − / ( a − 1) to obtain Q l even ( − ζ + r l ) = (( ζ − a b − ( n − / ( a − / ( a n − . (36)The second term in (22) can be evaluated analogously. We insert t (1) ( ζ ) fromR, n odd: t (1) ( λ ) = ( a n − λ + ( − σ s ) Y k ∈ ˘ D (1) ( λ + r k ) Y l ∈D (1) ( λ − r l ) (37)and use Q l odd A ( q l ) = ( a n + 1) / ( a + 1) to get finally for n odd: R ( n odd) ( ρ ′ ) = (cid:0) ( − σ π ( a + 1) ( − ζ + ( − σ π s π ) Q l ∈D (( − ρ l r l + ζ ) − ( − σ ( a − 1) ( ζ + ( − σ s ) Q l ∈D (( − ρ l r l − ζ ) (cid:1) R (38)with (using also (32)) R = r α − ( αβ ) − ( n − / a n − Q k ∈ ˘ D (( − k ζ + r k ) , (39)where α = a − b , β = 1 − a b , and ( − |D (0) | = ( − σ π , ( − |D (1) | = − ( − σ in thecase of odd n .The case of n even is less symmetric between R (0) and R (1) since now both q = 0and q = π are in the R sector, none of them in NS. So the term containing s π appearsin t (1) ( ζ ) instead of in t (0) ( − ζ ). Also, ( − |D (0) | = 1, ( − |D (1) | = − ( − σ + σ π in the caseof even n . In the following products, for l odd there are n/ l even wehave n/ − R ( n even) ( ρ ′ ) = (cid:0)Q l ∈D (( − ρ l r l + ζ ) − ( − σ + σ π Q l ∈D (( − ρ l r l − ζ ) ×× ( a − 1) ( ζ + ( − σ s ) ( ζ + ( − σ π s π ) a / ( αβ ) (cid:1) R , (40) R = r α − ( αβ ) − n/ a n − Y k ∈ ˘ D (( − k ζ + r k ) . (41) orm-factors in the BBS model II: Ising model on the finite lattice N ( ρ ) · R ( ρ ′ )In this subsection we shall show that the product N ( ρ ) · R ( ρ ′ ) can be put into the verysimple form (49). Let us start evaluating R ( ρ ′ ).At the beginning of this Section we already discussed that, if we exclude case (iii), l = b D , then Q (0) l (0) Q (1) l (0) = 1, and from (ii) if l ∈ ˘ D we have Q (0) l (1) Q (1) l (1) = 0.So we have to consider only l ∈ D for which we get from (i) and (iv) (momentarily wesuppose that l is odd, but the result (45) is the same for even l ): Q (0) l (1) Q (1) l (1) = − ( − σ ql + l ( − n q l t (0)ˇ q l ( − r l ) n χ l r n − l A ( q l ) · ( − n − t (1) ( − r l )2 χ l r n − l F ( r l )= ( − σ ql + n +1 i sin q l n A ( q l ) F n − ( r l ) t (0)ˇ q l ( − r l ) t (1) ( − r l ) , (42)where in the last step we used (30). The polynomial t (0)ˇ q l ( λ ) is t (0) ( λ ) given by (33)with the factor λ − r l omitted, see (31). In the first line the factor − ( − σ ql + l takescare whether q l or − q l is excited, the minus sign comes because Q (0) l is a left eigenvectorcomponent. Now since we should not use l ∈ ˘ D (if present, such a term leads to avanishing contribution in (21)), we have t (0)ˇ q l ( − r l ) t (1) ( − r l ) = a n − a − S l Y m ∈D ,m = l ( r l − r m )= a n − a − S l n − Y k =1 , k = l ( r l − r k ) Y k ∈ ˘ D (cid:18) − − r l + r k r l + r k (cid:19) . (43)with S l := ( a − − r l + ( − σ π s π )( − r l + ( − σ s ). Inserting (43) into (42) and using,recall (31), r l − r k = 2 (cos q k − cos q l ) F ( r l ) /A ( q k ) , Q n − k =1 A ( q k ) = a n − a − r a − b . (44)and the trigonometric identity2 n − sin q l Q k = l (cos q l − cos q k ) = n ( − l +1 we obtain Q (0) l (1) Q (1) l (1) = ( − σ ql + l + | ˘ D| +1 S l 2i sin q l F ( r l ) Y k ∈ ˘ D (cid:18) − r l + r k r l + r k (cid:19) , (45)valid both for q l in R and for q l in NS.Let us rewrite the ratio S l / (2i sin q l F ( r l )) in a convenient way. By straightforwarduse of the definitions (28), (30) and (29) we find ∓ S l 2i sin q l F ( r l ) = − r l + ( − σ α ± q l r l + ( − σ α ± q l , α q = b − e i q a − e i q for ( − σ = +( − σ π , ∓ S l 2i sin q l F ( r l ) = − r l + ( − σ β ± q l r l + ( − σ β ± q l , β q = b e i q − a − e i q for ( − σ = − ( − σ π , (46)leading to (written such that it is valid for both ρ l = 0 and ρ l = 1): orm-factors in the BBS model II: Ising model on the finite lattice Q (0) l ( ρ l ) Q (1) l ( ρ l ) = ( − ( n − ρ l ( − ρ l r l + ξ l r l + ξ l · Y k ∈ ˘ D ( − ρ l r l + r k r l + r k , (47)where ξ l = (cid:26) ˜ α l = ( − σ α ˜ q l ˜ β l = ( − σ β ˜ q l for ( − σ = ± ( − σ π ; ˜ q l = ( − σ ql + |D| + l q l . (48)Multiplying by N ( ρ ′ ), it is easy to see that the products over k ∈ ˘ D in (47) cancel(recall that ρ k = 0 for k ∈ ˘ D ) and we get finally N ( ρ ) · R ( ρ ′ ) = Y l ∈D ( − ρ l ( − ρ l r l + ξ l r l + ξ l Y m ∈D ,m>l r l + r m ( − ρ l r l + ( − ρ m r m . (49) ρ ′ in (21) In (38), (40) and (49) we have obtained all factors for the calculation of the normalizedmatrix element in such a form that the dependence on the summation indices ρ ′ isexplicit: h Φ | σ zn | Φ ih ˜Ψ | ˜Ψ i = X ρ ′ ∈ Z n − R ν + Y l ∈D (( − ρ l r l + ζ ) + R ν − Y l ∈D (( − ρ l r l − ζ ) ! ×× Y l ∈D ( − ρ l ( − ρ l r l + ξ l r l + ξ l Y m ∈D ,m>l r l + r m ( − ρ l r l + ( − ρ m r m , (50)where the ρ ′ -independent factors R ν ± can be read off from (38), (39) and (40), (41). Thesuperscript ν stands for n odd and even, respectively.Collecting all factors which depend on ρ l , l ∈ D , the problem of performing themultiple summation over ρ l , l ∈ D , reduces to the calculating following sum (proved inthe Appendix): Y D = X ρ l , l ∈D Q l ∈D ( − ρ l ( ( − ρ l r l + ξ l ) ( ( − ρ l r l + ζ ) Q l 1) ( ζ + ( − σ s ) (( − σ a − b ) (cid:0)Q l ∈D e i˜ q l + ( − σ ab (cid:1) = (54)= 2 ( − σ α (1 − ( − σ ab ) Q l ∈D e i˜ q l , we have finally: h Φ | σ zn | Φ ih ˜Ψ , | ˜Ψ , i = r ˜ c n c α Y k ∈ ˘ D (( − k b + a r k ) ×× Q l ∈D r l e i˜ q l f ( D − / l g ( D − / l Q l 5. Product of matrix elements In this section we sketch the calculation of the conjugate matrix elements h Φ | σ zn | Φ i ,where the vectors h Φ | and | Φ i shall have the same eigenvalues as the vectors | Φ i and h Φ | used in the previous sections. This calculation can be performed in the same wayas we did in Section 4. In analogy to (21), (22) we get for the homogeneous case h Φ | σ zn | Φ ih ˜Ψ , | ˜Ψ , i = a r X ρ ′ N ( ρ ′ ) R ∗ ( ρ ′ ) R ∗ ( ρ ′ ) , with R ∗ ( ρ ′ ) = t (1) ( − ζ ) Q n − l =1 ( − ζ + ( − ρ l r l ) + t (0) ( ζ ) Q n − l =1 ( ζ + ( − ρ l r l ) . Now we have to make the same transformations as we made for R ( ρ ′ ) in Section 4.1.The expression for R ∗ ( ρ ′ ) is obtained from R ( ρ ′ ) given by (22) just by substituting b → − b (in particular, ζ → − ζ ). Let us compare R ∗ ( ρ ′ ) and R ( ρ ′ ). From the solutionof the Baxter equations it follows that Q L(1) l ( ρ l ) Q R(0) l ( ρ l ) = Q L(0) l ( ρ l ) Q R(1) l ( ρ l ) unless l ∈ D and ρ l = 1 .Q L(1) l (1) Q R(0) l (1) = − Q L(0) l (1) Q R(1) l (1) if l ∈ D . So, in the final formula we have to substitute ˜ q l → − ˜ q l . orm-factors in the BBS model II: Ising model on the finite lattice σ = σ π , from (55) we get h Φ | σ zn | Φ ih ˜Ψ , | ˜Ψ , i = r ˜ c n c α Y k ∈ ˘ D ( − ( − k b + a r k ) ×× Q l ∈D r l e − i˜ q l ( f ∗ l ) ( D − / ( g ∗ l ) ( D − / Q l 6. Final formula for the square of the matrix element The proper quantity to consider for the matrix element of the spin operator, which doesnot depend on the normalization of the eigenvectors of the transfer-matrix, is h Φ | σ zn | Φ ih Φ | σ zn | Φ i / ( h Φ | Φ ih Φ | Φ i ) . (61)The factor h Φ | Φ ih Φ | Φ i / h ˜Ψ , | ˜Ψ , i (62)providing this change of normalization has been derived in formulas (74) and (75) of [2]for odd and even n , respectively. To obtain (61) we just divide (60) by (62). The finalresult for the square of the matrix element (60) as well as the formulas for the squaresof norm (62) were obtained for the eigenvectors with eigenvalues not containing factors( λ − s q k ) , i.e. up to now we have excluded the case (iii) in Section 3.2. j ∈ b D producing factors ( λ − s q j ) in t ρ ( λ ) . Now we explain how to modify these formulas for the matrix elements and norms ifthe eigenvalue polynomials of states | Φ i and | Φ i contain factors ( λ − s q j ) for some j , so that t ρ ( r j ) = 0 for ρ satisfying ( − ρ = ( − j − . We denote the set of such j ,corresponding to both states | Φ i and | Φ i , by b D . As already mentioned in Section 3.2,case (iii), we cannot normalize the solution to the Baxter equation for the state | Φ ρ i by Q L , R( ρ ) j (0) = 1. However, we may normalize it by Q L , R( ρ ) j (1) = 1. For the other state,the solution to the Baxter equation is managed by the case (i) and there is no problemwith the normalization Q L , R( ρ ) j (0) = 1. We shall use the normalization Q L , R( ρ ) j (1) = 1for all j ∈ b D irrespective of whether we have case (i) or (iii).In [2] the norm (56) and the matrix element (21) was calculated using thenormalization Q L , R( ρ ) j (0) = 1 for all j = 1 , , . . . , n − 1. These formulas were obtainedfor the case of generic parameters for which any normalization is possible. Let us tracethe changes in these formulas if instead we choose the normalization Q L , R( ρ ) j (1) = 1 for j from a subset b D ⊂ { , , . . . , n − } .Compare the formulas for the matrix element (21) corresponding to a differentnormalization of the solutions to the Baxter equation, i.e. we compare the termcorresponding to a set ρ ′ in one formula with the term corresponding to the set ρ ′ + b D inthe other formula (the set ρ ′ + b D is obtained from the set ρ ′ by shifts ρ j → ρ j + 1, j ∈ b D ).For N = 2 this just means interchanging ρ j = 0 and ρ j = 1, while the other componentsremain unchanged. Also, we change all r j → − r j , j ∈ b D in the second formula. orm-factors in the BBS model II: Ising model on the finite lattice Q L , R( ρ ) j ( ρ j + 1) normalized by Q L , R( ρ ) j (1) = 1 coincides with Q L , R( ρ ) j ( ρ j ) (cid:12)(cid:12)(cid:12) r j →− r j with the normalization Q L , R( ρ ) j (0) = 1.The only factor which is changing in (21) is N ( ρ ′ ). The denominator is unchanged underthe simultaneous substitutions r j → − r j and ρ j → ρ j + 1. The change in the numeratoris corrected by the division of the matrix elements corresponding to solutions of Baxterequations with Q L , R( ρ ) k (1) = 1, not by h ˜Ψ , | ˜Ψ , i but by h ˜Ψ , + b D | ˜Ψ , + b D i . From thegeneral expression for the norm, (20) of [2] at N = 2, the change of normalizationmeans multiplying our matrix elements by h ˜Ψ , | ˜Ψ , ih ˜Ψ , + b D | ˜Ψ , + b D i = Q n − l 1, will alsochange the eigenvectors entering (61) to the eigenvectors corresponding to eigenvaluepolynomials with factors ( λ − r j ) , j ∈ b D , instead of ( λ + r j ) . This is exactly what weneed.Summarizing: The final result (61) was obtained for eigenvectors with eigenvaluesnot containing factors ( λ − s q j ) and which could be given in terms of λ , λ π and r k , k = 1 , , . . . , n − 1. Now, if the eigenvalue polynomial contains the factors ( λ − s q j ) instead of ( λ + s q j ) for some j , we just have to replace r j → − r j in the final formulafor all such j . λ , λ π , r k , and ˜ q l . So the final formula for the matrix element becomes h Φ | σ zn | Φ ih Φ | σ zn | Φ ih Φ | Φ ih Φ | Φ i = ( λ π − λ ) ( D − δ ) / ( λ + λ π ) δ Y l 7. Bugrij–Lisovyy formula for matrix element In [3, 4], the matrix elements of σ zk between eigenvectors of symmetric Ising transfer-matrix t IsingSym = exp (cid:0) P nk =1 K x σ zk − σ zk (cid:1) exp ( P nk =1 K ∗ x σ xk ) exp (cid:0) P nk =1 K x σ zk − σ zk (cid:1) . (66)were given. Since (25) and (66) are related by a similarity transformation withexp (cid:0) P nk =1 K x σ zk − σ zk (cid:1) , which commutes with σ zm , it is natural to compare (65) with the square of the matrixelement as given in [4]: | NS h q , q , . . . , q K | σ zm | p , p , . . . , p L i R | == ξ ξ T K Y k =1 Q NS q = q k sinh γ ( q k )+ γ ( q )2 n Q R p sinh γ ( q k )+ γ ( p )2 L Y l =1 Q R p = p l sinh γ ( p l )+ γ ( p )2 n Q NS q sinh γ ( p l )+ γ ( q )2 · (cid:18) t y − t − y t x − t − x (cid:19) ( K − L ) / ×× K Y k 8. Matrix elements for the quantum Ising chain in a transverse field In this section we apply the formulas for the matrix elements obtained in Section 6 tothe derivation of the matrix elements for the quantum Ising chain in a transverse field.Let us start from the L -operator (24) with a = g − / and b = 0: L k ( λ ) = λ σ xk λ g − / σ zk g − / σ zk λg − ! . (75) orm-factors in the BBS model II: Ising model on the finite lattice L -operator wehave: t n ( λ ) = − λg ˆ H + · · · , ˆ H = − n X k =1 ( σ zk σ zk +1 + g σ xk ) , where ˆ H is the Hamiltonian of the periodic quantum Ising chain in a transverse field.From (27) we get the spectrum of this Hamiltonian: E = − X q ± ε ( q ) (76)where the energies of the quasi-particle excitations are ε ( q ) = (1 − g cos q + g ) / = (cid:16) ( g − + 4 g sin q (cid:17) / , q = 0 , π ,ε (0) = g − , ε ( π ) = g + 1 . In (76), the sign + / − in the front of ε ( q ) corresponds to the absence/presence ofthe excitation with the momentum q . The NS-sector includes the states with aneven number of excitations, the R-sector includes the states with an odd number ofexcitations. The momentum q runs over the same set as in (27). Then the formula formatrix elements for σ zn is given by (65) with s q = g/ε ( q ). After some simplification weget the analogue of (67) now for the quantum Ising chain: | NS h q , q , . . . , q K | σ zm | p , p , . . . , p L i R | = g ( K − L )22 ξξ T K Y k =1 e η ( q k ) nε ( q k ) L Y l =1 e − η ( p l ) nε ( p l ) ×× K Y k 1. But for the case 0 ≤ g < ε (0) = 1 − g to be positive. From(76), this change of the sign of ε (0) in the ferromagnetic phase leads to a formal changebetween absence-presence of zero-momentum excitation in the labelling of eigenstates.Therefore the number of the excitations in each sector (NS and R) becomes even. Directcalculation shows that the change of the sign of ε (0) in (77) can be absorbed to obtainformally the same formula (77) but with new ε (0), even L (the number of the excitationsin R-sector) and new ξ = (1 − g ) / . orm-factors in the BBS model II: Ising model on the finite lattice ≤ g < 1) and in the thermodynamic limit n → ∞ (when the energies of | vac i NS and | vac i R coincide giving the degenerationof the ground state), we have ξ T → NS h vac | σ zm | vac i R = ξ / = (1 − g ) / . 9. Conclusions In this paper we calculated the normalized spin matrix element between arbitrary statesof the Ising model, the main result being the formulas (64),(65) and (77). We startedwith the result (12) obtained in our previous paper [2] using the Sklyanin method ofSeparation of Variables by which we obtained explicit wave functions in terms of thesolutions of Baxter equations. The result (12) was obtained for the general N = 2BBS-model which is related to a generalized free-fermion Ising-type model (3). Forthis general model we were able to get the much simpler formula (21) which, howeverstill involves a multiple sum over intermediate spins. Since for the general model wecannot perform the summation, for the further calculation we restricted ourselves tothe Ising model with parameters (23). Performing a number of technical manipulations,we succeed in calculating the multiple spin sums explicitly. Although the intermediateformulas are quite involved, a number of surprising cancellations take place which leadto the rather simple formula (60) for the spin matrix element square. This comes stillnormalized to the auxiliary states involved in the method of Separation of Variables, butit is not difficult to convert this result into the properly normalized matrix elements forthe model with periodic boundary condition. The final formula becomes more lengthydue to normalization factors. We show by which transformations we get the recentlyconjectured formula of A. Bugrij and O. Lisovyy. Our derivation provides a first proof ofthese formulas. Another application of the formulas obtained in this article is the result(77) for the spin matrix elements for the finite quantum Ising chain in a transverse field.The presence of degenerations in the spectrum for the special Ising parameter valuesforced us in this case to normalize the Baxter equation solutions differently for differentexcited states. The complexity of the formulas gives little hope that for more generalparameter values the multiple spin summations can be done in the near future, even ifthen the degeneration problems may be avoided. Acknowledgments The authors thank A. Bugrij and O. Lisovyy for helpful discussions. This work has beensupported by the Heisenberg–Landau exchange program HLP-2007. SP was supportedby the RFBR grant 06-02-17383 and the grant for support of scientific schools NSh-8065.2006.2. NI and VS were supported by the INTAS grant 05-1000008-7865, by theProgram of Fundamental Research of the Physics and Astronomy Division of the NAS orm-factors in the BBS model II: Ising model on the finite lattice A. Proof of the summation formula The aim of this Appendix is to find a factorized expression for the sum Y D = X ρ l , l ∈D Q l ∈D ( − ρ l (( − ρ l r l + ξ l ) (( − ρ l r l + ζ ) Q l 2. Then we verify (79) for small D . Finally, by explicitly insertingour conjectured solutions (79) into the recursion relation we prove the expression. Thecases of D odd and even have to be treated separately, since we are dealing with atwo-step relation. A.1. Derivation of the recursion relation This recurrence relation for Y D is obtained by using the identity, compare (17), X i Q k ( x i − y k ) Q j = i ( x i − x j ) = 0 , (81) orm-factors in the BBS model II: Ising model on the finite lattice x i exceeds the number of y k at least by two. Fix any index s ∈ D and consider D + 1 values of x i and two values of y i : { y , y } = {− ξ s , − ζ } , { x , x , . . . , x D } = { r s , − r s , − ( − ρ r , − ( − ρ r , . . . , − ( − ρ s r s | {z } omitted , . . . , − ( − ρ D r D } . Since two parameters y k are chosen, we must have D ≥ 3. Now we separate the twoterms in (81) which correspond to i = 0 , ρ s ∈ { , } for { x , x } = r s ( − ρ s . Then (81) becomes X ρ s ( − ρ s ( r s ( − ρ s + ξ s )( r s ( − ρ s + ζ ) Q k = s ( r s ( − ρ s + r k ( − ρ k )= − X k = s r s ( − r k ( − ρ k + ξ s )( − r k ( − ρ k + ζ )( r k − r s ) Q l = k,s ( − r k ( − ρ k + r l ( − ρ l ) . 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