A remark on ground state of boundary Izergin-Korepin model
aa r X i v : . [ n li n . S I] A p r A REMARK ON GROUND STATE OFBOUNDARY IZERGIN-KOREPIN MODEL
June 1, 2018
Takeo KOJIMA
Department of Mathematics and Physics, Graduate School of Science and Engineering,Yamagata University, Jonan 4-3-16, Yonezawa 992-8510, Japan [email protected]
Abstract
We study the ground state of the boundary Izergin-Korepin model. The boundary Izergin-Korepin model is defined by so-called R -matrix and K -matrix for U q ( A (2)2 ) which satisfyYang-Baxter equation and boundary Yang-Baxter equation. The ground state associatedwith identity K -matrix ¯ K ( z ) = id was constructed by W.-L.Yang and Y.-Z.Zhang in earlierstudy. We construct the free field realization of the ground state associated with nontrivialdiagonal K -matrix. There have been many developments in the field of exactly solvable models. Various methodswere invented to solve models. The free field approach [1] provides a powerful method to studyexactly solvable models. This paper is devoted to the free field approach to boundary problemof exactly solvable statistical mechanics [2]. Exactly solvable boundary model [2, 3] is definedby the solutions of the Yang-Baxter equation and the boundary Yang-Baxter equation K ( z ) R , ( z z ) K ( z ) R , ( z /z ) = R , ( z /z ) K ( z ) R , ( z z ) K ( z ) .
1n this paper we are going to study the boundary Izergin-Korepin model defined by the solutionsof the Yang-Baxter equation [4] and the boundary Yang-Baxter equation [5, 6] for the quantumgroup U q ( A (2)2 ). We are going to diagonalize the infinite transfer matrix T ǫ ( z ) of the boundaryIzergin-Korepin model, by means of the free field approach. For better understanding of themodel that we are going to study, we give comments on the solutions of the boundary Yang-Baxter equation. The R -matrix associated with non-exceptional affine symmetry except for D (2) n commute with each other [ b R ( z ) , b R ( z )] = 0 , where b R ( z ) = P R ( z ) and P ( a ⊗ b ) = b ⊗ a . Hence we know that the identity K -matrix¯ K ( z ) = id is a particular solution of the boundary Yang-Baxter equation [3] for A (1) n , B (1) n , C (1) n , D (1) n and A (2) n . There exist more general solutions of the boundary Yang-Baxter equation. The diagonalsolutions of the boundary Yang-Baxter equation for A (1) n , B (1) n , C (1) n , D (1) n and A (2) n are classifiedin [5]. For A (1) n there exists the diagonal K -matrix that has one continuous free parameter.However for B (1) n , C (1) n , D (1) n and A (2) n there exist only discrete solutions. For example, for A (2)2 there exist three isolated solutions ¯ K ( z ).¯ K ( z ) = , z ±√− q + z ±√− q + z −
00 0 1 . In earlier study [7] W.-L. Yang and Y.-Z. Zhang constructed the free field realization of theground state associated with identity K -matrix ¯ K ( z ) = id for A (2)2 . In this paper we constructthe free field realization of the ground state associated with nontrivial diagonal K -matrix ¯ K ( z )for A (2)2 . This realization is the first example associated with the discrete solutions of theboundary Yang-Baxter equation. It is thought that the free field approach to the boundaryproblem works for every discrete K -matrix of the affine symmetry. Contrary to B (1) n , C (1) n , D (1) n and A (2) n case, there have been many papers on the free field approach to boundary problem for A (1) n symmetry. For example, the higher-rank generalization [8, 10] and the elliptic deformation[9, 10] have been solved.The plan of this paper is as follows. In section 2 we give physical picture of our problem.We introduce the boundary Izergin-Korepin model and introduce the transfer matrix T ǫ ( z ). Insection 3 we translate physical picture of our problem into mathematical picture. We constructthe free field realization of the ground state | B i ǫ and give the diagonalization of the transfermatrix T ǫ ( z ). 2 Boundary Izergin-Korepin model
In this section we formulate physical picture of our problem. R -matrix and K -matrix In this section we introduce R -matrix R ( z ) and K -matrix K ( z ) associated with the quantumgroup U q ( A (2)2 ). We choose q and z such that − < q < | q | < | z | < | q − | . Let { v + , v , v − } denotes the natural basis of V = C . We introduce the R -matrix R ( z ) ∈ End( V ⊗ V ) [4]. R ( z ) = 1 κ ( z ) b ( z ) 0 c ( z ) 0 0 0 0 00 0 d ( z ) 0 e ( z ) 0 f ( z ) 0 00 zc ( z ) 0 b ( z ) 0 0 0 0 00 0 − q ze ( z ) 0 j ( z ) 0 e ( z ) 0 00 0 0 0 0 b ( z ) 0 c ( z ) 00 0 n ( z ) 0 − q ze ( z ) 0 d ( z ) 0 00 0 0 0 0 zc ( z ) 0 b ( z ) 00 0 0 0 0 0 0 0 1 . (2.1)Here we have set the elements b ( z ) = q ( z − q z − , c ( z ) = q − q z − ,d ( z ) = q ( z − qz + 1)( q z − q z + 1) , e ( z ) = q ( z − q − q z − q z + 1) ,f ( z ) = ( q − { ( q + q ) z − ( q − } z ( q z − q z + 1) , n ( z ) = ( q − { ( q − q ) z + ( q + 1) } ( q z − q z + 1) ,j ( z ) = q z + ( q − q − q + q + q − z − q ( q z − q z + 1) , and the normalizing function κ ( z ) = z ( q z ; q ) ∞ ( q /z ; q ) ∞ ( − q z ; q ) ∞ ( − q /z ; q ) ∞ ( q /z ; q ) ∞ ( q z ; q ) ∞ ( − q /z ; q ) ∞ ( − q z ; q ) ∞ . (2.2)Here we have used the abbreviation( z ; p ) ∞ = ∞ Y m =0 (1 − p m z ) . The matrix elements of R ( z ) are given by R ( z ) v j ⊗ v j = P k ,k = ± , v k ⊗ v k R ( z ) j ,j k ,k , wherethe ordering of the index is given by v + ⊗ v + , v + ⊗ v , v + ⊗ v − , v ⊗ v + , v ⊗ v , v ⊗ v − ,3 − ⊗ v + , v − ⊗ v , v − ⊗ v − . The R -matrix R ( z ) satisfies the Yang-Baxter equation R , ( z /z ) R , ( z /z ) R , ( z /z ) = R , ( z /z ) R , ( z /z ) R , ( z /z ) , (2.3)the unitarity R , ( z /z ) R , ( z /z ) = id, (2.4)and the crossing symmetry R ( z ) j ,j k ,k = q j − k R ( − q − z − ) − k ,j − j ,k . (2.5)We have set the normalizing function κ ( z ) such that the minimal eigenvalue of the corner transfermatrix becomes 1 [11, 12].We introduce K -matrix K ( z ) ∈ End( V ) representing an interaction at the boundary, whichsatisfies the boundary Yang-Baxter equation K ( z ) R , ( z z ) K ( z ) R , ( z /z ) = R , ( z /z ) K ( z ) R , ( z z ) K ( z ) . (2.6)We consider only the diagonal solutions K ( z ) = K ǫ ( z ) , ( ǫ = ± ,
0) [5, 6]. K ( z ) = ϕ ( z ) ϕ ( z − ) , K ± ( z ) = ϕ ± ( z ) ϕ ± ( z − ) z ±√− q + z ±√− q + z −
00 0 1 . (2.7)Here we have set the normalizing function ϕ ǫ ( z ) = ( q z ; q ) ∞ ( − q z ; q ) ∞ ( q z ; q ) ∞ ( − q z ; q ) ∞ × ǫ = 0)( ±√− q z ; q ) ∞ ( ∓√− q z ; q ) ∞ ( ±√− q z ; q ) ∞ ( ∓√− q z ; q ) ∞ ( ǫ = ± ) . (2.8)The matrix elements of K ( z ) are defined by K ( z ) v j = P k = ± , v k K ( z ) jk , where the ordering ofthe index is given by v + , v , v − . The K -matrix K ( z ) ∈ End( V ) satisfies the boundary unitarity K ( z ) K ( z − ) = id, (2.9)and the boundary crossing symmetry K ( z ) k k = X j , j = ± , q ( k − j ) R ( − q − z − ) − k ,k j , − j K ( − q − z − ) j j . (2.10)To put it another way we have chosen the normalizing function ϕ ǫ ( z ) such that the transfermatrix T ǫ ( z ) (2.16) acts on the ground state as 1 (2.24).4 .2 Physical picture In this section, following [3, 2], we introduce the transfer matrix T ǫ ( z ) that is generating functionof the Hamiltonian H ǫ of our problem. In order to consider physical problem, it is convenientto introduce graphical interpretation of R -matrix and K -matrix. We present the R -matrix R ( z ) j ,j k ,k in Fig.1. j j k z k z Fig.1: R -matrix R ( z /z ) j ,j k ,k =We present the K -matrix K ( z ) jk in Fig.2. jkz − zK ( z ) jk = Fig.2: K -matrixLet us consider the half-infinite spin chain · · · ⊗ V ⊗ V ⊗ V . Let us introduce the subspace H of the half-infinite spin chain by H = Span {· · · ⊗ v p ( N ) ⊗ · · · ⊗ v p (2) ⊗ v p (1) | p ( s ) = 0 ( s >> } . (2.11)5e introduce the vertex operator Φ j ( z ) , ( j = ± ,
0) acting on the space H by Fig.3. To put itanother way, the vertex operator Φ j ( z ) is infinite-size matrix whose matrix elements are givenby products of the R -matrix R ( z ) j ,j k ,k (Φ j ( z )) ··· p ( N ) ′ ··· p (2) ′ p (1) ′ ··· p ( N ) ··· p (2) p (1) = lim N →∞ X ν (1) ,ν (2) , ··· ,ν ( N )= ± , N Y j =1 R ( z ) ν ( j ) p ( j ) ′ ν ( j − ,p ( j ) , (2.12)where j = ν (0). In order to avoid divergence we restrict our consideration to the subspace H .We introduce the dual vertex operator Φ ∗ j ( z − ) , ( j = ± ,
0) acting on the space H by Fig.4. jz · · ·· · · Φ j ( z ) = Fig.3: Vertex operator j · · ·· · · z − Fig.4: Dual vertex operatorΦ ∗ j ( z − ) = 234 234From the Yang-Baxter equation, we have the commutation relation of the vertex operatorΦ j ( z ) , ( j = ± ,
0) Φ j ( z )Φ j ( z ) = X k ,k = ± , R ( z /z ) k ,k j ,j Φ k ( z )Φ k ( z ) . (2.13)From the unitarity and the crossing symmetry, we have the inversion relations g Φ j ( z )Φ ∗ j ( z ) = δ j ,j id, g X j = ± , Φ ∗ j ( z )Φ j ( z ) = id, (2.14)where we have set g = 11 + q ( q ; q ) ∞ ( − q ; q ) ∞ ( q ; q ) ∞ ( − q ; q ) ∞ . ∗ j ( z ) = q − j Φ − j ( − q − z ) . (2.15)We introduce the transfer matrix T ǫ ( z ) acting on the space H by Fig.5. To put it anotherway we introduce the transfer matrix T ǫ ( z ) by product of the vertex operators. We define the”renormalized” transfer matrix T ǫ ( z ) by T ǫ ( z ) = gT ǫ ( z ) = g X j,k = ± , Φ ∗ j ( z − ) K ǫ ( z ) kj Φ k ( z ) , ( ǫ = ± , . (2.16)1 zz − Fig.5: Boundary transfer matrix T ǫ ( z ) = 234 · · · From the commutation relations of the vertex operators Φ j ( z ) , Φ ∗ j ( z ) and the boundary Yang-Baxter equation (2.6), we have the commutativity relation of the transfer matrix T ǫ ( z ), ( ǫ = ± , T ǫ ( z ) , T ǫ ( z )] = 0 , (for any z , z ) . (2.17)The commutativity of the transfer matrix ensures that, if the transfer matrices T ǫ ( z ) are diag-onalizable, the transfer matrices T ǫ ( z ) are diagonalized by the basis that is independent of thespectral parameter z . From the unitarity and the crossing symmetry, we have T ǫ ( z ) T ǫ ( z − ) = id, T ǫ ( − q − z − ) = T ǫ ( z ) . (2.18)The Hamiltonian H IK of the boundary Izergin-Korepin model is obtained by − q − q − )( q + q − ) H IK = ddz T ǫ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) z =1 + const. (2.19)7he Hamiltonian H IK is written as H IK = H b + ∞ X j =1 H j +1 ,j , (2.20)where we have set H b ∈ End( V ) and H ∈ End( V ⊗ V ) by H b = , ( ǫ = 0)4( q − q − ) {− ( q + q − ) λ − √ ( q − q − ± √− λ } , ( ǫ = ± ) , (2.21) H = ( q + q − )( q + q − )( λ ⊗ λ + λ ⊗ λ )+ ( q + q − )( q − q − ) √− − λ ⊗ λ + λ ⊗ λ )+ 2( q + q − ) λ ⊗ λ + ( q + q − )( q + q − )( λ ⊗ λ + λ ⊗ λ + λ ⊗ λ + λ ⊗ λ )+ ( q + q − )( q − q − ) √− λ ⊗ λ − λ ⊗ λ + λ ⊗ λ − λ ⊗ λ )+ ( q − q − ) ( λ ⊗ λ + λ ⊗ λ − λ ⊗ λ − λ ⊗ λ )+ ( q − q − ) √− − λ ⊗ λ + λ ⊗ λ − λ ⊗ λ + λ ⊗ λ )+ 23 ( − ( q + q − ) + 2( q + q − ) + 2( q + q − )) λ ⊗ λ + 13 √ − ( q + q − ) + 2( q + q − ) − ( q + q − ))( λ ⊗ id + id ⊗ λ ) . (2.22)Here we have used Gell-Mann matrices λ , λ , · · · , λ satisfying Tr( λ j λ k ) = 2 δ j,k . λ = , λ = √−
10 0 0 −√− , λ = − ,λ = , λ = −√− √− , λ = ,λ = √− −√− , λ = 1 √ − . The diagonalization of the Hamiltonian H IK is reduced to those of the transfer matrix T ǫ ( z ).Let us consider the eigenvector problem T ǫ ( z ) | v i = t ǫ ( z ) | v i . (2.23)8e have chosen the normalizing function ϕ ǫ ( z ) (2.8) such that the ground state | B i ǫ satisfies T ǫ ( z ) | B i ǫ = | B i ǫ , ( ǫ = ± , . (2.24)We would like to construct the ground state | B i ǫ and would like to diagonalize the Hamiltonian H IK . In this section we give mathematical formulation of our problem. We construct the free fieldrealization of the ground state | B i ǫ , and give the diagonalization of the transfer matrix T ǫ ( z ). In order to diagonalize the transfer matrix T ǫ ( z ), we follows the strategy proposed in [1, 2]. Thecorner transfer matrix method [11, 12] suggests that we identify H with the integrable highest-weight representation V (Λ ) of the quantum group U q ( A (2)2 ), because the characters of H and V (Λ ) coincide. We note that the representation V (Λ ) is the only one level-1 integrable highest-weight representation of U q ( A (2)2 ). We identify the line Φ j ( z ) in Fig.3 with the components e Φ j ( z )of the vertex operator e Φ( z ), and the line Φ ∗ j ( z ) in Fig.4 with the components e Φ ∗ j ( z ) of the dualvertex operator e Φ ∗ ( z ). e Φ( z ) : V (Λ ) → V (Λ ) ⊗ V z , e Φ( z ) = X j = ± , e Φ j ( z ) ⊗ v j , (3.1) e Φ ∗ ( z ) : V (Λ ) ⊗ V z → V (Λ ) , e Φ ∗ j ( z ) | v i = e Φ ∗ ( z )( | v i ⊗ v j ) . (3.2)Here V z is the evaluation representation. The vertex operator e Φ j ( z ) for U q ( A (2)2 ) satisfies exactlythe same functional relations as those of Φ j ( z ). e Φ j ( z ) e Φ j ( z ) = X k ,k = ± , R ( z /z ) k ,k j ,j e Φ k ( z ) e Φ k ( z ) , (3.3) g e Φ j ( z ) e Φ ∗ j ( z ) = δ j ,j id, e Φ ∗ j ( z ) = q − j e Φ − j ( − q − z ) . (3.4)Let us set the transfer matrix e T ǫ ( z ) , ( ǫ = ± ,
0) by e T ǫ ( z ) = g X j,k = ± , e Φ ∗ j ( z − ) K ǫ ( z ) kj e Φ k ( z ) , ( ǫ = ± , . (3.5)We have the commutativity of the transfer matrix, the unitarity and the crossing symmetry.[ e T ǫ ( z ) , e T ǫ ( z )] = 0 , (for any z , z ) , (3.6) e T ǫ ( z ) e T ǫ ( z − ) = id, e T ǫ ( − q − z − ) = e T ǫ ( z ) . (3.7)9et us set the ground state | e B i ǫ ∈ V (Λ ) by e T ǫ ( z ) | e B i ǫ = | e B i ǫ , ( ǫ = ± , . (3.8)Following the strategy proposed in [1, 2], we consider our problem upon the following identifi-cation. H = V (Λ ) , Φ j ( z ) = e Φ j ( z ) , Φ ∗ j ( z ) = e Φ ∗ j ( z ) , T ǫ ( z ) = e T ǫ ( z ) , | B i ǫ = | e B i ǫ . (3.9)In order to study the excitations we introduce type-II vertex operator e Ψ ∗ µ ( z ) , ( µ = ± , e Ψ ∗ ( z ) : V z ⊗ V (Λ ) → V (Λ ) , e Ψ ∗ µ ( z ) | v i = e Ψ ∗ ( z )( v µ ⊗ | v i ) . (3.10)Type-II vertex operator e Ψ ∗ µ ( ξ ) satisfies e Φ j ( z ) e Ψ ∗ µ ( ξ ) = τ ( z/ξ ) e Ψ ∗ µ ( ξ ) e Φ j ( z ) , ( j, µ = ± , . (3.11)Here we have set τ ( z ) = z − Θ q ( q z )Θ q ( − q z )Θ q ( q z − )Θ q ( − q z − ) , Θ p ( z ) = ( p ; p ) ∞ ( z ; p ) ∞ ( pz − ; p ) ∞ . (3.12)Let us set the vectors | ξ , ξ , · · · , ξ N i µ ,µ , ··· ,µ N ,ǫ ∈ V (Λ ), ( µ , µ , · · · , µ N , ǫ = ± ,
0) by | ξ , ξ , · · · , ξ N i µ ,µ , ··· ,µ N ,ǫ = e Ψ ∗ µ ( ξ ) e Ψ ∗ µ ( ξ ) · · · e Ψ ∗ µ N ( ξ N ) | B i ǫ . (3.13)From the commutation relation (3.11) we have e T ǫ ( z ) | ξ , ξ , · · · , ξ N i µ ,µ , ··· ,µ N ,ǫ = N Y j =1 τ ( z/ξ j ) τ ( − /q zξ j ) | ξ , ξ , · · · , ξ N i µ ,µ , ··· ,µ N ,ǫ . (3.14)Comparing with the Bethe ansatz calculation [13] we conclude that the vectors | ξ , ξ , · · · , ξ N i µ ,µ , ··· ,µ N ,ǫ are the basis of the space of state of the Izergin-Korepin model. In order to construct the groundstate | B i ǫ ∈ V (Λ ), it is convenient to introduce the free field realization. In this section we give the free field realization of the vertex operator e Φ j ( z ) , ( j = ± ,
0) [14, 16, 17].Let us introduce the bosons a m , ( m ∈ Z =0 ) as following [14, 15, 16, 17].[ a m , a n ] = δ m + n [ m ] q m ([2 m ] q − ( − m [ m ] q ) . (3.15)10ere we have used q -integer [ n ] q = q n − q − n q − q − . (3.16)Let us set the zero-mode operators P, Q by[ a m , P ] = [ a m , Q ] = 0 , [ P, Q ] = 1 . (3.17)The integrable highest weight representation V (Λ ) of U q ( A (2)2 ) is realized by V (Λ ) = C [ a − , a − , · · · ] ⊕ n ∈ Z e nQ | Λ i , | Λ i = e Q | i . (3.18)The vacuum vector | i is characterized by a m | i = 0 , ( m > , P | i = 0 . (3.19)We give the free field realizations of the vertex operators. Let us set ǫ ( q ) = ([2] q ) . The highestelements of the vertex operators are given by e Φ − ( z ) = 1 ǫ ( q ) e P ( z ) e Q ( z ) e Q ( − zq ) P + , (3.20) e Ψ ∗ + ( z ) = 1 ǫ ( q ) e P ∗ ( z ) e Q ∗ ( z ) e − Q ( − qz ) − P + . (3.21)Here we have set the auxiliary operators P ( z ) , Q ( z ) , P ∗ ( z ) , Q ∗ ( z ) by P ( z ) = X m> a − m [2 m ] q − ( − m [ m ] q q m z m , (3.22) Q ( z ) = − X m> a m [2 m ] q − ( − m [ m ] q q − m z − m , (3.23) P ∗ ( z ) = − X m> a − m [2 m ] q − ( − m [ m ] q q m z m , (3.24) Q ∗ ( z ) = X m> a m [2 m ] q − ( − m [ m ] q q − m z − m . (3.25)The other elements of the vertex operators are given by the intertwining relations (3.1) and(3.10). e Φ ( z ) = e Φ − ( z ) x − − qx − e Φ − ( z ) , (3.26) e Φ + ( z ) = q ( e Φ ( z ) x − − x − e Φ ( z )) , (3.27) e Ψ ∗ ( z ) = x +0 e Ψ ∗ + ( z ) − q − e Ψ ∗ + ( z ) x +0 , (3.28) e Ψ ∗− ( z ) = q − ( x +0 e Ψ ∗ ( z ) − e Ψ ∗ ( z ) x +0 ) . (3.29)11he elements x ± m , ( m ∈ Z ) are given by integral of the currents x ± ( w ) = P m ∈ Z x ± m w − m , x ± m = I dw π √− w m − x ± ( w ) , ( m ∈ Z ) . (3.30)Here the free field realizations of the currents x ± ( w ) are given by x ± ( w ) = ǫ ( q ) e R ± ( w ) e S ± ( w ) e ± Q w ± P + , (3.31)where we have set the auxiliary operators R ± ( w ) , S ± ( w ) by R ± ( w ) = ± X m> a − m [ m ] q q ∓ m w m , (3.32) S ± ( w ) = ∓ X m> a m [ m ] q q ∓ m w − m . (3.33) In this section we give the free field realization of the ground state | B i ǫ which satisfies T ǫ ( z ) | B i ǫ = | B i ǫ . Theorem 3.1
The free field realizations of the ground states | B i ǫ are given by | B i ǫ = e F ( ǫ ) | Λ i , ( ǫ = ± , . (3.34) Here we have set F ( ǫ ) = − X m> mq m [2 m ] q − ( − m [ m ] q a − m (3.35)+ X m> ( θ m ( q m − q − m − ( √− m ) q m [2 m ] q − ( − m [ m ] q ! − ( ǫ √− m q m [2 m ] q − ( − m [ m ] q ) a − m − Q, where θ m ( x ) = x, m : even0 , m : odd . Multiplying the type-II vertex operators e Ψ ∗ µ ( ξ ) to the ground state | B i ǫ , we get the diagonal-ization of the transfer matrix T ǫ ( z ) on the space of state. Proof.
Let us multiply the vertex operator Φ j ( z ) to the relation T ǫ ( z ) | B i ǫ = | B i ǫ from theleft and use the inversion relation (2.14). Then we have K ǫ ( z ) jj Φ j ( z ) | B i ǫ = Φ j ( z − ) | B i ǫ , ( j, ǫ = ± , . (3.36)12e would like to calculate the action of the vertex operator Φ j ( z ) on the vector | B i ǫ . Using thenormal orderingsΦ − ( z ) x − ( x w ) = : Φ − ( z ) x − ( x w ) : − z (1 − qw/z ) , ( | qw | < | z | ) , (3.37) x − ( x w )Φ − ( z ) = : x − ( x w )Φ − ( z ) : 1 w (1 − qz/w ) , ( | qz | < | w | ) , (3.38) x − ( w ) x − ( w ) = : x − ( w ) x − ( w ) : w (1 − q w /w )(1 − w /w )(1 + qw /w ) , ( | w | < | w | ) , (3.39)we have following realizations of the vertex operators Φ j ( z ).Φ − ( z ) = 1 ǫ ( q ) e P ( z ) e Q ( z ) e Q ( − zq ) P + , (3.40)Φ ( z ) = I C dw π √− w ( q − q z (1 − qw/z )(1 − qz/w ) : Φ − ( z ) x − ( q w ) : , (3.41)Φ + ( z ) = I I C dw π √− w dw π √− w q − (1 − q ) × (1 − w /w )(1 − w /w )(1 + qw /w )(1 + qw /w ) z − { ( q + q − ) z − ( w + w ) } Y j =1 (1 − qw j /z )(1 − qz/w j ) × : Φ − ( z ) x − ( q w ) x − ( q w ) : . (3.42)The integration contour C encircles w = 0 , qz but not w = q − z . The integration contour C encircles w = 0 , qz, qw but not w = q − z, q − w , and encircles w = 0 , qz, qw but not w = q − z, q − w .The actions of the basic operators e S − ( w ) , e Q ( z ) on the vector | B i ǫ have the following formulae. e Q ( z ) | B i ǫ = ϕ ǫ ( z − ) e P ( z − ) | B i ǫ , (3.43) e S − ( q w ) | B i ǫ = g ǫ ( w ) e R − ( q /w ) | B i ǫ , (3.44)where ϕ ǫ ( z ) is given in (2.8) and g ǫ ( w ) is given by g ǫ ( w ) = (1 − w − ) , ( ǫ = 0)(1 − w − )(1 ∓ √− q − w − ) , ( ǫ = ± ) . (3.45)We have the action of the vertex operators Φ j ( z ) on the vector | B i ǫ as following.Φ − ( z ) | B i ǫ = 1 ǫ ( q ) ϕ ǫ ( z − ) e P ( z )+ P ( z − ) e Q e F ( ǫ ) | Λ i , (3.46)Φ ( z ) | B i ǫ = (1 − q ) ϕ ǫ ( z − ) I e C dw π √− w z − wg ǫ ( w )(1 − qw/z )(1 − qz/w )(1 − q/zw )13 e P ( z )+ P ( z − )+ R − ( q w )+ R − ( q /w ) e F ( ǫ ) | Λ i , (3.47)Φ + ( z ) | B i ǫ = (1 − q ) q − ǫ ( q ) ϕ ǫ ( z − ) I I e C dw π √− w dw π √− w × (1 − w /w )(1 − w /w )(1 − /w w )(1 − q /w w )(1 + qw /w )(1 + qw /w )(1 + q/w w ) Y j =1 w j g ǫ ( w j ) × z − { ( q + q − ) z − ( w + w ) } Y j =1 (1 − qw j /z )(1 − qz/w j )(1 − q/zw j ) (3.48) × e P ( z )+ P ( z − )+ R − ( q w )+ R − ( q /w )+ R − ( q w )+ R − ( q /w ) e − Q e F ( ǫ ) | Λ i . The integration contour e C encircles w = 0 , qz, qz − but not w = q − z . The integrationcontour e C encircles w = 0 , qz, qz − , qw , qw − but not w = q − z, q − w , and encircles w = 0 , qz, qz − , qw , qw − but not w = q − z, q − w .For simplicity we summarize the case ǫ = ± . The relation (3.36) is equivalent to the followingthree relations. ϕ ± ( z )Φ − ( z ) | B i ± = ϕ ± ( z − )Φ − ( z − ) | B i ± , (3.49) ϕ ± ( z )(1 ∓ √− q − z )Φ ( z ) | B i ± = ϕ ± ( z − )(1 ∓ √− q − z − )Φ ( z − ) | B i ± , (3.50) ϕ ± ( z ) z Φ + ( z ) | B i ± = ϕ ± ( z − ) z − Φ + ( z − ) | B i ± , (3.51) • The relation (3.49). Using formula (3.46), we have(
LHS ) = 1 ǫ ( q ) ϕ ± ( z ) ϕ ± ( z − ) e P ( z )+ P ( z − ) e Q e F ( ǫ ) | Λ i = ( RHS ) . (3.52) • The relation (3.50). Using formula (3.47), we have(
LHS ) − ( RHS ) = (1 − q ) ϕ ± ( z ) ϕ ± ( z − )( z − − z ) (3.53) × I b C dw π √− w I ( w ) e P ( z )+ P ( z − )+ R − ( q w )+ R − ( q w − ) (1 − qw/z )(1 − qz/w )(1 − q/wz )(1 − qwz ) e F ( ǫ ) | Λ i , where we have set I ( w ) = ( w − w − )(1 ∓ √− q − w )(1 ∓ √− q − w − ). Here the integrationcontour b C encircles w = 0 , qz, qz − but not w = q − z, q − z − . The integration contour b C andthe integrand e P ( z )+ P ( z − )+ R − ( q w )+ R − ( q w − ) (1 − qw/z )(1 − qz/w )(1 − q/wz )(1 − qwz )are invariant under w → w − . Hence the relation I ( w )+ I ( w − ) = 0 ensures ( LHS ) − ( RHS ) =0. • The relation (3.51). Using formula (3.48), we have(
LHS ) − ( RHS ) = q − (1 − q ) ǫ ( q ) ϕ ± ( z ) ϕ ± ( z − ) I I b C dw π √− w dw π √− w (1 − w /w )(1 − w /w )(1 − w w )(1 − /w w )(1 + qw /w )(1 + qw /w )(1 + qw w )(1 + q/w w ) × q ( z − z − ) I ( w , w ) Y j =1 (1 − qz/w j )(1 − qw j /z )(1 − qzw j )(1 − q/zw j ) (3.54) × e P ( z )+ P ( z − )+ R − ( q w )+ R − ( q /w )+ R − ( q w )+ R − ( q /w ) e − Q e F ( ǫ ) | Λ i , where we have set I ( w , w ) = {− ( q + q − )( z + z − ) w w + ( w + w )( w w + 1) }× (1 + qw w )(1 − w w ) (1 − q /w w ) Y j =1 ( w j − w − j )(1 ∓ √− q − w − j ) . (3.55)Here the integration contour b C encircles w = 0 , qz, qz − , qw , qw − but not w = q − z, q − z − , q − w , q − w − , and encircles w = 0 , qz, qz − , qw , qw − but not w = q − z, q − z − , q − w , q − w − .The integration contour b C and the integrand(1 − w /w )(1 − w /w )(1 − w w )(1 − /w w )(1 + qw /w )(1 + qw /w )(1 + qw w )(1 + q/w w ) × e P ( z )+ P ( z − )+ R − ( q w )+ R − ( q /w )+ R − ( q w )+ R − ( q /w )2 Y j =1 (1 − qz/w j )(1 − qw j /z )(1 − qzw j )(1 − q/zw j )are invariant under ( w , w ) → ( w − , w ) , ( w , w − ) , ( w − , w − ). Hence the relation I ( w , w ) + I ( w − , w ) + I ( w , w − ) + I ( w − , w − ) = 0ensures ( LHS ) − ( RHS ) = 0.Q.E.D.
In this section we give the free field realizations of the dual ground state ǫ h B | ∈ V (Λ ) ∗ , ( ǫ = ± , ǫ h B | T ǫ ( z ) = ǫ h B | , ( ǫ = ± , . (3.56)The dual integrable highest weight representation V (Λ ) ∗ of U q ( A (2)2 ) is realized by V (Λ ) ∗ = h Λ | ⊕ n ∈ Z e nQ C [ a , a , · · · ] , h Λ | = h | e − Q . (3.57)The vacuum vector h | is characterized by h | a − m = 0 , ( m > , h | P = 0 . (3.58)15 heorem 3.2 The free field realizations of the dual ground states ǫ h B | are given by ǫ h B | = h Λ | e G ( ǫ ) , ( ǫ = ± , . (3.59) Here we have set G ( ǫ ) = − X m> mq − m [2 m ] q − ( − m [ m ] q a m (3.60)+ X m> ( − θ m ( q m − q − m − ( √− m ) q − m [2 m ] q − ( − m [ m ] q ! + ( − ǫ √− m q − m [2 m ] q − ( − m [ m ] q ) a m − Q. The proof of (3.56) is given as the same way as those of (2.24). The following relations are usefulfor proof. ǫ h B | e P ( − q − z − ) = ϕ ǫ ( z − ) ǫ h B | e Q ( − q − z ) , (3.61) ǫ h B | e R − ( qw ) = g ∗ ǫ ( w ) ǫ h B | e S − ( qw − ) , (3.62)where ϕ ǫ ( z ) is given in (2.8) and g ∗ ǫ ( w ) is given by g ∗ ǫ ( w ) = (1 − w ) , ( ǫ = 0)(1 − w )(1 ± √− q − w ) , ( ǫ = ± ) . (3.63) Acknowledgments
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