A remark on ground state of boundary Izergin-Korepin model

A REMARK ON GR OUND ST A TE OF BOUND AR Y IZER GIN-K OREPIN MODEL June 1, 2018 T ak eo K OJIMA Dep artment of M athematics and Physics, Gr aduate Scho ol of Scienc e and Engine ering, Y amagata U ni v ersity, Jonan 4-3-16, Y onezawa 992-8510, Jap an k o jima@yz.ya magata- u.ac.jp Abstract W e study the ground state of the b oundary Izergin-Ko repin mo del. The b oundary Izerg in- Korepin mo del is defined b y so- called R - matrix and K -matrix for U q ( A (2) 2 ) which sa tisfy Y ang-Ba xter equatio n and b oundary Y ang -Baxter equatio n. The gr ound s tate asso ciated with ident it y K -matrix ¯ K ( z ) = i d w as constructed b y W.-L.Y a ng and Y.-Z.Zhang in earlier study . W e construct the free field realization of the ground s ta te asso ciated with non trivial diagonal K -matrix. 1 In tro duction There ha v e b een many dev elopmen ts in the field of exactly solv able mo d els. V arious metho d s w ere in ven ted to solv e mo d els. T he free field approac h [1] pro vides a p o w erfu l method to study exactly solv able mo dels. This pap er is devo ted to the free field approac h to b oun dary problem of exactly solv able statistical mec h anics [2]. Exactly solv able b oun dary mo del [2, 3] is d efined b y the solutions of the Y ang-Baxte r equation and the b oundary Y ang-Baxte r equation K 2 ( z 2 ) R 2 , 1 ( z 1 z 2 ) K 1 ( z 1 ) R 1 , 2 ( z 1 /z 2 ) = R 2 , 1 ( z 1 /z 2 ) K 1 ( z 1 ) R 1 , 2 ( z 1 z 2 ) K 2 ( z 2 ) . 1 In this pap er w e are going to study the b oundary Izergin-Korepin mo del defined by t he solutions of the Y ang-Baxter equation [4] and the b oundary Y ang-Baxter equ ation [5, 6] for the quan tum group U q ( A (2) 2 ). W e are going to d iagonalize the infinite transfer m atrix T ǫ ( z ) of the b oundary Izergin-Korepin mo del, b y means of the free field approac h . F or b etter under s tanding of the mo del that w e are going to study , w e giv e comments on the solutions of the b oundary Y ang- Baxter equation. The R -matrix asso ciated with non-exceptional affine symmetry exc ept for D (2) n comm u te with eac h other [ b R ( z 1 ) , b R ( z 2 )] = 0 , where b R ( z ) = P R ( z ) and P ( a ⊗ b ) = b ⊗ a . Hence w e kno w that the ident it y K -matrix ¯ K ( z ) = id is a particular solution of the b ound ary Y ang-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 b ound ary Y ang-Baxte r equation. The diagonal solutions of the b ound ary Y ang-Baxter equation for A (1) n , B (1) n , C (1) n , D (1) n and A (2) n are classified in [5]. F or A (1) n there exists the diagonal K -matrix that has on e con tinuous free parameter. Ho wev er for B (1) n , C (1) n , D (1) n and A (2) n there exist only discrete solutions. F or example, for A (2) 2 there exist thr ee isolate d solutions ¯ K ( z ). ¯ K ( z ) =     1 0 0 0 1 0 0 0 1     ,       z 2 0 0 0 ± √ − 1 q 3 2 + z ± √ − 1 q 3 2 + z − 1 0 0 0 1       . In earlier stud y [7] W.-L. Y ang and Y.-Z. Z hang constructed the f ree fi eld realization of th e ground state asso ciated w ith identi t y K -matrix ¯ K ( z ) = id for A (2) 2 . In this pap er we construct the free field r ealiza tion of the ground s tate associated w ith nontrivial diagonal K -matrix ¯ K ( z ) for A (2) 2 . This realization is the fi r st example asso ciated with the discrete solutions of the b ound ary Y ang-Baxter equation. It is th ou ght that the free fi eld app roac h to the b ound ary problem works for eve ry discrete K -matrix of the affin e symmetry . Con trary to B (1) n , C (1) n , D (1) n and A (2) n case, th ere h a ve b een many pap ers on the free field approac h to b ound ary p r oblem for A (1) n symmetry . F or example, the higher-rank generalization [8, 10] and the elliptic d eformation [9, 10] hav e b een solv ed. The plan of th is pap er is as follo ws . In section 2 we giv e ph ysical picture of our problem. W e introdu ce the b ou n dary Izergin-Korepin mo d el and in trod uce the tr an s fer matrix T ǫ ( z ). In section 3 we translate physica l picture of our prob lem in to mathematical p icture. W e construct the free fi eld realiza tion of the ground state | B i ǫ and give the d iagonaliz ation of the transfer matrix T ǫ ( z ). 2 2 Boundary Izergin-Korepin mo del In this section we form ulate physica l picture of our p roblem. 2.1 R -matrix and K -matrix In this section w e in trod uce R -matrix R ( z ) and K -matrix K ( z ) asso ciated with the q u an tum group U q ( A (2) 2 ). W e c h o ose q and z suc h that − 1 < q 1 2 < 0 an d | q 2 | < | z | < | q − 2 | . Let { v + , v 0 , v − } denotes the natural basis of V = C 3 . W e int ro duce th e R -matrix R ( z ) ∈ End( V ⊗ V ) [4]. R ( z ) = 1 κ ( z )                       1 0 0 0 0 0 0 0 0 0 b ( z ) 0 c ( z ) 0 0 0 0 0 0 0 d ( z ) 0 e ( z ) 0 f ( z ) 0 0 0 z c ( z ) 0 b ( z ) 0 0 0 0 0 0 0 − q 2 z e ( z ) 0 j ( z ) 0 e ( z ) 0 0 0 0 0 0 0 b ( z ) 0 c ( z ) 0 0 0 n ( z ) 0 − q 2 z e ( z ) 0 d ( z ) 0 0 0 0 0 0 0 z c ( z ) 0 b ( z ) 0 0 0 0 0 0 0 0 0 1                       . (2.1) Here we h a ve set the elemen ts b ( z ) = q ( z − 1) q 2 z − 1 , c ( z ) = q 2 − 1 q 2 z − 1 , d ( z ) = q 2 ( z − 1)( q z + 1) ( q 2 z − 1)( q 3 z + 1) , e ( z ) = q 1 2 ( z − 1)( q 2 − 1) ( q 2 z − 1)( q 3 z + 1) , f ( z ) = ( q 2 − 1) { ( q 3 + q ) z − ( q − 1) } z ( q 2 z − 1)( q 3 z + 1) , n ( z ) = ( q 2 − 1) { ( q 3 − q 2 ) z + ( q 2 + 1) } ( q 2 z − 1)( q 3 z + 1) , j ( z ) = q 4 z 2 + ( q 5 − q 4 − q 3 + q 2 + q − 1) z − q ( q 2 z − 1)( q 3 z + 1) , and the normalizing function κ ( z ) = z ( q 6 z ; q 6 ) ∞ ( q 2 /z ; q 6 ) ∞ ( − q 5 z ; q 6 ) ∞ ( − q 3 /z ; q 6 ) ∞ ( q 6 /z ; q 6 ) ∞ ( q 2 z ; q 6 ) ∞ ( − q 5 /z ; q 6 ) ∞ ( − q 3 z ; q 6 ) ∞ . (2.2) Here we h a ve used the abbreviation ( z ; p ) ∞ = ∞ Y m =0 (1 − p m z ) . The matrix elemen ts of R ( z ) are giv en by R ( z ) v j 1 ⊗ v j 2 = P k 1 ,k 2 = ± , 0 v k 1 ⊗ v k 2 R ( z ) j 1 ,j 2 k 1 ,k 2 , w here the ordering of the in dex is giv en b y v + ⊗ v + , v + ⊗ v 0 , v + ⊗ v − , v 0 ⊗ v + , v 0 ⊗ v 0 , v 0 ⊗ v − , 3 v − ⊗ v + , v − ⊗ v 0 , v − ⊗ v − . The R -matrix R ( z ) satisfies the Y ang-Baxter equation R 1 , 2 ( z 1 /z 2 ) R 1 , 3 ( z 1 /z 3 ) R 2 , 3 ( z 2 /z 3 ) = R 2 , 3 ( z 2 /z 3 ) R 1 , 3 ( z 1 /z 3 ) R 1 , 2 ( z 1 /z 2 ) , (2.3) the un itarity R 1 , 2 ( z 1 /z 2 ) R 2 , 1 ( z 2 /z 1 ) = id, (2.4) and the crossing sym metry R ( z ) j 1 ,j 2 k 1 ,k 2 = q j 2 − k 2 2 R ( − q − 3 z − 1 ) − k 2 ,j 1 − j 2 ,k 1 . (2.5) W e ha v e s et the normalizing fu nction κ ( z ) suc h that the minimal eigen v alue of the corner transfer matrix b ecomes 1 [11, 12]. W e in tro d uce K -matrix K ( z ) ∈ End( V ) r epresen ting an in teraction at th e b oundary , whic h satisfies the b ou n dary Y ang-Baxter equation K 2 ( z 2 ) R 2 , 1 ( z 1 z 2 ) K 1 ( z 1 ) R 1 , 2 ( z 1 /z 2 ) = R 2 , 1 ( z 1 /z 2 ) K 1 ( z 1 ) R 1 , 2 ( z 1 z 2 ) K 2 ( z 2 ) . (2.6) W e consider only the diagonal solutions K ( z ) = K ǫ ( z ) , ( ǫ = ± , 0) [5, 6]. K 0 ( z ) = ϕ 0 ( z ) ϕ 0 ( z − 1 )     1 0 0 0 1 0 0 0 1     , K ± ( z ) = ϕ ± ( z ) ϕ ± ( z − 1 )       z 2 0 0 0 ± √ − 1 q 3 2 + z ± √ − 1 q 3 2 + z − 1 0 0 0 1       . (2.7) Here we h a ve set the normalizing fu nction ϕ ǫ ( z ) = ( q 8 z ; q 12 ) ∞ ( − q 9 z 2 ; q 12 ) ∞ ( q 12 z ; q 12 ) ∞ ( − q 5 z 2 ; q 12 ) ∞ ×      1 ( ǫ = 0) ( ± √ − 1 q 9 2 z ; q 6 ) ∞ ( ∓ √ − 1 q 7 2 z ; q 6 ) ∞ ( ± √ − 1 q 1 2 z ; q 6 ) ∞ ( ∓ √ − 1 q 3 2 z ; q 6 ) ∞ ( ǫ = ± ) . (2.8) The matrix elements of K ( z ) are defined b y K ( z ) v j = P k = ± , 0 v k K ( z ) j k , wh ere the ordering of the index is given b y v + , v 0 , v − . The K -matrix K ( z ) ∈ En d( V ) satisfies the b oundary unitarit y K ( z ) K ( z − 1 ) = id, (2.9) and the b oun dary crossing symmetry K ( z ) k 2 k 1 = X j , j 2 = ± , 0 q 1 2 ( k 1 − j 1 ) R ( − q − 3 z − 2 ) − k 1 ,k 2 j 2 , − j 1 K ( − q − 3 z − 1 ) j 2 j 1 . (2.10) T o pu t it another wa y w e ha v e c hosen the normalizing fun ction ϕ ǫ ( z ) suc h that the transfer matrix T ǫ ( z ) (2.16) acts on the groun d state as 1 (2.24). 4 2.2 Ph ysical pict ure In this section, follo wing [3 , 2], w e introdu ce the tran s fer matrix T ǫ ( z ) that is generating function of the Hamiltonian H ǫ of our problem. In ord er to consider physica l problem, it is conv enient to in tro duce graph ical inte rpretation of R -matrix and K -matrix. W e presen t the R -matrix R ( z ) j 1 ,j 2 k 1 ,k 2 in Fig.1. j 2 j 1 k 1 z 1 k 2 z 2 Fig.1: R -matrix R ( z 1 /z 2 ) j 1 ,j 2 k 1 ,k 2 = W e present th e K -matrix K ( z ) j k in Fig.2. j k z − 1 z K ( z ) j k = Fig.2: K -matrix Let u s consider the half-infinite sp in c hain · · · ⊗ V ⊗ V ⊗ V . Let us int ro duce th e subsp ace H of the half-infinite spin chain b y H = S pan {· · · ⊗ v p ( N ) ⊗ · · · ⊗ v p (2) ⊗ v p (1) | p ( s ) = 0 ( s >> 0) } . (2.11) 5 W e introd uce the vertex op erator Φ j ( z ) , ( j = ± , 0) acting on the space H by Fig.3. T o put it another wa y , th e v ertex op erator Φ j ( z ) is infin ite-size matrix wh ose matrix elemen ts are giv en b y pro ducts of the R -matrix R ( z ) j 1 ,j 2 k 1 ,k 2 (Φ j ( z )) ··· p ( N ) ′ ··· p (2) ′ p (1) ′ ··· p ( N ) ··· p (2) p (1) = lim N →∞ X ν (1) ,ν (2) , ··· ,ν ( N )= ± , 0 N Y j = 1 R ( z ) ν ( j ) p ( j ) ′ ν ( j − 1) ,p ( j ) , (2.12) where j = ν (0). In ord er to av oid dive rgence we r estrict our consid eration to the subspace H . W e introdu ce the du al v ertex op erator Φ ∗ j ( z − 1 ) , ( j = ± , 0) acting on the space H by Fig.4 . j z 1 · · · · · · Φ j ( z ) = Fig.3: V er tex op erator j 1 · · · · · · z − 1 Fig.4: Dual vertex op erator Φ ∗ j ( z − 1 ) = 2 3 4 2 3 4 F r om the Y an g-Baxter equation, w e hav e the commutatio n relation of the vertex op erator Φ j ( z ) , ( j = ± , 0) Φ j 2 ( z 2 )Φ j 1 ( z 1 ) = X k 1 ,k 2 = ± , 0 R ( z 1 /z 2 ) k 1 ,k 2 j 1 ,j 2 Φ k 1 ( z 1 )Φ k 2 ( z 2 ) . (2.13) F r om the unitarit y and th e crossing symm etry , we ha ve the inv ersion relations g Φ j 1 ( z )Φ ∗ j 2 ( z ) = δ j 1 ,j 2 id, g X j = ± , 0 Φ ∗ j ( z )Φ j ( z ) = id, (2.14) where we h a ve set g = 1 1 + q ( q 2 ; q 6 ) ∞ ( − q 3 ; q 6 ) ∞ ( q 6 ; q 6 ) ∞ ( − q 5 ; q 6 ) ∞ . 6 F r om the crossing sym metry , w e ha v e Φ ∗ j ( z ) = q − j 2 Φ − j ( − q − 3 z ) . (2.15) W e in tro duce the transfer matrix T ǫ ( z ) acting on the space H by Fig.5. T o put it another w a y we introd uce th e transfer matrix T ǫ ( z ) by pro duct of the v ertex op erators. W e d efine the ”renormalized” transfer m atrix T ǫ ( z ) by T ǫ ( z ) = g T ǫ ( z ) = g X j,k = ± , 0 Φ ∗ j ( z − 1 ) K ǫ ( z ) k j Φ k ( z ) , ( ǫ = ± , 0) . (2.16) 1 z z − 1 Fig.5: Boundary transfer m atrix T ǫ ( z ) = 2 3 4 · · · F r om the commutati on relations of the ve rtex op erators Φ j ( z ) , Φ ∗ j ( z ) and the b oundary Y ang- Baxter equation (2.6) , we ha v e the commutativit y relation of the transfer matrix T ǫ ( z ), ( ǫ = ± , 0). [ T ǫ ( z 1 ) , T ǫ ( z 2 )] = 0 , (for any z 1 , z 2 ) . (2.17) The commutati vit y of the transfer matrix ensur es that, if the transfer matrices T ǫ ( z ) are d iag- onalizable, the transfer matrices T ǫ ( z ) are diagonalized by the basis that is indep endent of the sp ectral parameter z . F rom the unitarit y and the crossing symmetry , we hav e T ǫ ( z ) T ǫ ( z − 1 ) = id, T ǫ ( − q − 3 z − 1 ) = T ǫ ( z ) . (2.18) The Hamiltonian H I K of the b ou n dary I zergin-Korepin mo del is obtained b y − 1 4( q − q − 1 )( q 3 2 + q − 3 2 ) H I K = d dz T ǫ ( z )     z =1 + const. (2.19) 7 The Hamiltonian H I K is written as H I K = H b 1 + ∞ X j = 1 H j + 1 ,j , (2.20) where we h a ve set H b ∈ End( V ) and H ∈ E n d( V ⊗ V ) by H b =    0 , ( ǫ = 0) 4( q − q − 1 ) {− ( q 3 2 + q − 3 2 ) λ 3 − √ 3 3 ( q 3 2 − q − 3 2 ± 2 √ − 1) λ 8 } , ( ǫ = ± ) , (2.21) H = ( q 1 2 + q − 1 2 )( q 2 + q − 2 )( λ 1 ⊗ λ 1 + λ 2 ⊗ λ 2 ) + ( q 1 2 + q − 1 2 )( q 2 − q − 2 ) √ − 1( − λ 1 ⊗ λ 2 + λ 2 ⊗ λ 1 ) + 2( q 1 2 + q − 1 2 ) λ 3 ⊗ λ 3 + ( q 3 2 + q − 3 2 )( q + q − 1 )( λ 4 ⊗ λ 4 + λ 5 ⊗ λ 5 + λ 6 ⊗ λ 6 + λ 7 ⊗ λ 7 ) + ( q 3 2 + q − 3 2 )( q − q − 1 ) √ − 1( λ 4 ⊗ λ 5 − λ 5 ⊗ λ 4 + λ 6 ⊗ λ 7 − λ 7 ⊗ λ 6 ) + ( q − q − 1 ) 2 ( λ 4 ⊗ λ 6 + λ 6 ⊗ λ 4 − λ 5 ⊗ λ 7 − λ 7 ⊗ λ 5 ) + ( q 2 − q − 2 ) √ − 1( − λ 4 ⊗ λ 7 + λ 7 ⊗ λ 4 − λ 5 ⊗ λ 6 + λ 6 ⊗ λ 5 ) + 2 3 ( − ( q 1 2 + q − 1 2 ) + 2( q 3 2 + q − 3 2 ) + 2( q 5 2 + q − 5 2 )) λ 8 ⊗ λ 8 + 1 3 √ 3 ( − ( q 1 2 + q − 1 2 ) + 2( q 3 2 + q − 3 2 ) − ( q 5 2 + q − 5 2 ))( λ 8 ⊗ id + id ⊗ λ 8 ) . (2.22) Here we h a ve used Gell-Mann matrices λ 1 , λ 2 , · · · , λ 8 satisfying T r( λ j λ k ) = 2 δ j,k . λ 1 =     0 0 1 0 0 0 1 0 0     , λ 2 =     0 0 √ − 1 0 0 0 − √ − 1 0 0     , λ 3 =     1 0 0 0 0 0 0 0 − 1     , λ 4 =     0 1 0 1 0 0 0 0 0     , λ 5 =     0 − √ − 1 0 √ − 1 0 0 0 0 0     , λ 6 =     0 0 0 0 0 1 0 1 0     , λ 7 =     0 0 0 0 0 √ − 1 0 − √ − 1 0     , λ 8 = 1 √ 3     1 0 0 0 − 2 0 0 0 1     . The diagonalizatio n of the Hamiltonian H I K is reduced to those of th e tr ansfer matrix T ǫ ( z ). Let us consider th e eigen v ector problem T ǫ ( z ) | v i = t ǫ ( z ) | v i . (2.23) 8 W e hav e c hosen the normalizing function ϕ ǫ ( z ) (2.8) such th at the ground state | B i ǫ satisfies T ǫ ( z ) | B i ǫ = | B i ǫ , ( ǫ = ± , 0) . (2.24) W e w ould lik e to constru ct th e ground state | B i ǫ and wo uld lik e to diagonali ze the Hamiltonian H I K . 3 F ree field realization In this section w e giv e mathematical formulatio n of our problem. W e constru ct the free field realizatio n of the groun d state | B i ǫ , and giv e the diagonalization of the transfer m atrix T ǫ ( z ). 3.1 Mathematical picture In order to diagonalize the transfer matrix T ǫ ( z ), we follo ws the strategy p r op osed in [1 , 2]. Th e corner transfer matrix metho d [11, 12] suggests that we iden tify H with the in tegrable highest- w eigh t repr esen tation V (Λ 1 ) of the quan tum group U q ( A (2) 2 ), b ecause th e c haracters of H and V (Λ 1 ) coincide. W e note that the representati on V (Λ 1 ) is the only one lev el-1 int egrable highest- w eigh t rep r esen tatio n of U q ( A (2) 2 ). W e id en tify the line Φ j ( z ) in Fig.3 with the comp onen ts e Φ j ( z ) of the v ertex op erator e Φ( z ), an d the line Φ ∗ j ( z ) in Fig.4 with the comp onen ts e Φ ∗ j ( z ) of the dual v ertex op erator e Φ ∗ ( z ). e Φ( z ) : V (Λ 1 ) → V (Λ 1 ) ⊗ V z , e Φ( z ) = X j = ± , 0 e Φ j ( z ) ⊗ v j , (3.1) e Φ ∗ ( z ) : V (Λ 1 ) ⊗ V z → V (Λ 1 ) , e Φ ∗ j ( z ) | v i = e Φ ∗ ( z )( | v i ⊗ v j ) . (3.2) Here V z is th e ev aluation representat ion. T he ve rtex op erator e Φ j ( z ) for U q ( A (2) 2 ) satisfies exactly the same fun ctional r elatio ns as those of Φ j ( z ). e Φ j 2 ( z 2 ) e Φ j 1 ( z 1 ) = X k 1 ,k 2 = ± , 0 R ( z 1 /z 2 ) k 1 ,k 2 j 1 ,j 2 e Φ k 1 ( z 1 ) e Φ k 2 ( z 2 ) , (3.3) g e Φ j 1 ( z ) e Φ ∗ j 2 ( z ) = δ j 1 ,j 2 id, e Φ ∗ j ( z ) = q − j 2 e Φ − j ( − q − 3 z ) . (3.4) Let us set the transfer matrix e T ǫ ( z ) , ( ǫ = ± , 0) b y e T ǫ ( z ) = g X j,k = ± , 0 e Φ ∗ j ( z − 1 ) K ǫ ( z ) k j e Φ k ( z ) , ( ǫ = ± , 0) . (3.5) W e hav e the commutativit y of the transf er matrix, the unitarit y and th e crossing symmetry . [ e T ǫ ( z 1 ) , e T ǫ ( z 2 )] = 0 , (for an y z 1 , z 2 ) , (3.6) e T ǫ ( z ) e T ǫ ( z − 1 ) = id, e T ǫ ( − q − 3 z − 1 ) = e T ǫ ( z ) . (3.7) 9 Let us set the ground state | e B i ǫ ∈ V (Λ 1 ) by e T ǫ ( z ) | e B i ǫ = | e B i ǫ , ( ǫ = ± , 0) . (3.8) F ollo wing the strategy prop osed in [1, 2], we consider our problem up on the follo wing identifi- cation. H = V (Λ 1 ) , 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 s tu dy the excitations we in tro d uce t yp e-I I ve rtex op erator e Ψ ∗ µ ( z ) , ( µ = ± , 0). e Ψ ∗ ( z ) : V z ⊗ V (Λ 1 ) → V (Λ 1 ) , e Ψ ∗ µ ( z ) | v i = e Ψ ∗ ( z )( v µ ⊗ | v i ) . (3.10) T yp e-I I v ertex op erator 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 = ± , 0) . (3.11 ) Here we h a ve set τ ( z ) = z − 1 Θ q 6 ( q 5 z )Θ q 6 ( − q 4 z ) Θ q 6 ( q 5 z − 1 )Θ q 6 ( − q 4 z − 1 ) , Θ p ( z ) = ( p ; p ) ∞ ( z ; p ) ∞ ( pz − 1 ; p ) ∞ . (3.12) Let us set the v ectors | ξ 1 , ξ 2 , · · · , ξ N i µ 1 ,µ 2 , ··· ,µ N ,ǫ ∈ V (Λ 1 ), ( µ 1 , µ 2 , · · · , µ N , ǫ = ± , 0) by | ξ 1 , ξ 2 , · · · , ξ N i µ 1 ,µ 2 , ··· ,µ N ,ǫ = e Ψ ∗ µ 1 ( ξ 1 ) e Ψ ∗ µ 2 ( ξ 2 ) · · · e Ψ ∗ µ N ( ξ N ) | B i ǫ . (3.13) F r om the comm utation relation (3.11) we h a ve e T ǫ ( z ) | ξ 1 , ξ 2 , · · · , ξ N i µ 1 ,µ 2 , ··· ,µ N ,ǫ = N Y j = 1 τ ( z /ξ j ) τ ( − 1 /q 3 z ξ j ) | ξ 1 , ξ 2 , · · · , ξ N i µ 1 ,µ 2 , ··· ,µ N ,ǫ . (3.14) Comparing with th e Bethe ansatz calculati on [13] we conclude that the v ecto rs | ξ 1 , ξ 2 , · · · , ξ N i µ 1 ,µ 2 , ··· ,µ N ,ǫ are the basis of the space of state of the Izergin-Korepin mo del. In order to construct the ground state | B i ǫ ∈ V (Λ 1 ), it is conv enien t to in tro duce the free fi eld realizat ion. 3.2 V ert ex op erator In this section we giv e the free fi eld realization of th e v ertex op erator e Φ j ( z ) , ( j = ± , 0) [14, 16, 17]. Let us introd u ce the b osons a m , ( m ∈ Z 6 =0 ) as follo w in g [14, 15, 16, 17]. [ a m , a n ] = δ m + n [ m ] q m ([2 m ] q − ( − 1) m [ m ] q ) . ( 3.15) 10 Here we h a ve used q -inte ger [ n ] q = q n − q − n q − q − 1 . (3.16) Let us set the zero-mode op erators P , Q by [ a m , P ] = [ a m , Q ] = 0 , [ P , Q ] = 1 . (3.1 7) The int egrable highest weig h t repr esen tation V (Λ 1 ) of U q ( A (2) 2 ) is realized b y V (Λ 1 ) = C [ a − 1 , a − 2 , · · · ] ⊕ n ∈ Z e nQ | Λ 1 i , | Λ 1 i = e Q 2 | 0 i . (3.18) The v acuum vec tor | 0 i is characte rized by a m | 0 i = 0 , ( m > 0) , P | 0 i = 0 . (3.19) W e give the free field r ealiza tions of the vertex op erators. Let u s set ǫ ( q ) = ([2] q 1 2 ) 1 2 . The h ighest elemen ts of the v ertex op erators are given b y e Φ − ( z ) = 1 ǫ ( q ) e P ( z ) e Q ( z ) e Q ( − z q 4 ) P + 1 2 , (3.20) e Ψ ∗ + ( z ) = 1 ǫ ( q ) e P ∗ ( z ) e Q ∗ ( z ) e − Q ( − q z ) − P + 1 2 . (3.21) Here we h a ve set the auxiliary op erators P ( z ) , Q ( z ) , P ∗ ( z ) , Q ∗ ( z ) by P ( z ) = X m> 0 a − m [2 m ] q − ( − 1) m [ m ] q q 9 m 2 z m , (3.22) Q ( z ) = − X m> 0 a m [2 m ] q − ( − 1) m [ m ] q q − 7 m 2 z − m , (3.23) P ∗ ( z ) = − X m> 0 a − m [2 m ] q − ( − 1) m [ m ] q q m 2 z m , (3.24) Q ∗ ( z ) = X m> 0 a m [2 m ] q − ( − 1) m [ m ] q q − 3 m 2 z − m . (3.25) The other elements of the v ertex op erators are giv en by the in tertwining r elations (3.1) and (3.10) . e Φ 0 ( z ) = e Φ − ( z ) x − 0 − q x − 0 e Φ − ( z ) , (3.26) e Φ + ( z ) = q 1 2 ( e Φ 0 ( z ) x − 0 − x − 0 e Φ 0 ( z )) , (3.27) e Ψ ∗ 0 ( z ) = x + 0 e Ψ ∗ + ( z ) − q − 1 e Ψ ∗ + ( z ) x + 0 , (3.28) e Ψ ∗ − ( z ) = q − 1 2 ( x + 0 e Ψ ∗ 0 ( z ) − e Ψ ∗ 0 ( z ) x + 0 ) . (3.2 9) 11 The elemen ts x ± m , ( m ∈ Z ) are giv en by integ ral of the curr en ts x ± ( w ) = P m ∈ Z x ± m w − m , x ± m = I dw 2 π √ − 1 w m − 1 x ± ( w ) , ( m ∈ Z ) . (3.30) Here the free fi eld realizatio ns of the cur ren ts x ± ( w ) are giv en by x ± ( w ) = ǫ ( q ) e R ± ( w ) e S ± ( w ) e ± Q w ± P + 1 2 , (3.31) where we h a ve set the auxiliary op erators R ± ( w ) , S ± ( w ) b y R ± ( w ) = ± X m> 0 a − m [ m ] q q ∓ m 2 w m , (3.32) S ± ( w ) = ∓ X m> 0 a m [ m ] q q ∓ m 2 w − m . (3.33) 3.3 Ground state In this section we giv e the free field r ealizat ion of the ground state | B i ǫ whic h satisfies T ǫ ( z ) | B i ǫ = | B i ǫ . Theorem 3.1 The f r e e field r e alizations of the gr ound states | B i ǫ ar e give n by | B i ǫ = e F ( ǫ ) | Λ 1 i , ( ǫ = ± , 0) . (3.34) Her e we have set F ( ǫ ) = − 1 2 X m> 0 mq 8 m [2 m ] q − ( − 1) m [ m ] q a 2 − m (3.35) + X m> 0 ( θ m ( q m 2 − q − m 2 − ( √ − 1) m ) q 4 m [2 m ] q − ( − 1) m [ m ] q ! − ( ǫ √ − 1) m q 3 m [2 m ] q − ( − 1) m [ m ] q ) a − m − Q, wher e θ m ( x ) =    x, m : even 0 , m : o dd . Multiplying the typ e-I I vertex op erators e Ψ ∗ µ ( ξ ) to the ground state | B i ǫ , w e get the diagonal- ization of the trans f er matrix T ǫ ( z ) on the sp ace of state. Pr o of. Let us multiply the ve rtex op erator Φ j ( z ) to the relation T ǫ ( z ) | B i ǫ = | B i ǫ from the left and use the inv ersion relation (2.14). T hen we hav e K ǫ ( z ) j j Φ j ( z ) | B i ǫ = Φ j ( z − 1 ) | B i ǫ , ( j, ǫ = ± , 0) . (3.36) 12 W e w ould lik e to calculate the action of the v er tex op erator Φ j ( z ) on the v ector | B i ǫ . Using the normal orderings Φ − ( z ) x − ( x 4 w ) = : Φ − ( z ) x − ( x 4 w ) : − 1 z (1 − q w /z ) , ( | q w | < | z | ) , (3.37) x − ( x 4 w )Φ − ( z ) = : x − ( x 4 w )Φ − ( z ) : 1 w (1 − q z /w ) , ( | q z | < | w | ) , (3 .38) x − ( w 1 ) x − ( w 2 ) = : x − ( w 1 ) x − ( w 2 ) : w 1 (1 − q 2 w 2 /w 1 )(1 − w 2 /w 1 ) (1 + q w 2 /w 1 ) , ( | w 2 | < | w 1 | ) , (3.39) w e ha v e follo wing realizations of the v ertex op erators Φ j ( z ). Φ − ( z ) = 1 ǫ ( q ) e P ( z ) e Q ( z ) e Q ( − z q 4 ) P + 1 2 , (3.40) Φ 0 ( z ) = I C 1 dw 2 π √ − 1 w ( q 2 − 1) q 4 z (1 − q w /z )(1 − q z /w ) : Φ − ( z ) x − ( q 4 w ) : , (3.41) Φ + ( z ) = I I C 2 dw 1 2 π √ − 1 w 1 dw 2 2 π √ − 1 w 2 q − 5 2 (1 − q 2 ) 2 × (1 − w 1 /w 2 )(1 − w 2 /w 1 ) (1 + q w 1 /w 2 )(1 + q w 2 /w 1 ) z − 2 { ( q + q − 1 ) z − ( w 1 + w 2 ) } 2 Y j = 1 (1 − q w j /z )(1 − q z /w j ) × : Φ − ( z ) x − ( q 4 w 1 ) x − ( q 4 w 2 ) : . (3.42) The integrati on conto ur C 1 encircles w = 0 , q z but not w = q − 1 z . T he integ ration cont our C 2 encircles w 1 = 0 , q z , q w 2 but not w 1 = q − 1 z , q − 1 w 2 , and encircles w 2 = 0 , q z , q w 1 but not w 2 = q − 1 z , q − 1 w 1 . The actions of the basic op erators e S − ( w ) , e Q ( z ) on the vecto r | B i ǫ ha v e the follo wing formulae. e Q ( z ) | B i ǫ = ϕ ǫ ( z − 1 ) e P ( z − 1 ) | B i ǫ , (3.43) e S − ( q 4 w ) | B i ǫ = g ǫ ( w ) e R − ( q 4 /w ) | B i ǫ , (3.44) where ϕ ǫ ( z ) is giv en in (2.8) and g ǫ ( w ) is giv en by g ǫ ( w ) =    (1 − w − 2 ) , ( ǫ = 0) (1 − w − 2 )(1 ∓ √ − 1 q − 1 2 w − 1 ) , ( ǫ = ± ) . (3.45) W e hav e the action of the verte x op erators Φ j ( z ) on th e v ector | B i ǫ as follo w ing. Φ − ( z ) | B i ǫ = 1 ǫ ( q ) ϕ ǫ ( z − 1 ) e P ( z )+ P ( z − 1 ) e Q e F ( ǫ ) | Λ 1 i , (3.46) Φ 0 ( z ) | B i ǫ = (1 − q 2 ) ϕ ǫ ( z − 1 ) I e C 1 dw 2 π √ − 1 w z − 1 wg ǫ ( w ) (1 − q w /z )(1 − q z /w )(1 − q /z w ) 13 × e P ( z )+ P ( z − 1 )+ R − ( q 4 w )+ R − ( q 4 /w ) e F ( ǫ ) | Λ 1 i , (3.47) Φ + ( z ) | B i ǫ = (1 − q 2 ) 2 q − 5 2 ǫ ( q ) ϕ ǫ ( z − 1 ) I I e C 2 dw 1 2 π √ − 1 w 1 dw 2 2 π √ − 1 w 2 × (1 − w 1 /w 2 )(1 − w 2 /w 1 )(1 − 1 /w 1 w 2 )(1 − q 2 /w 1 w 2 ) (1 + q w 1 /w 2 )(1 + q w 2 /w 1 )(1 + q /w 1 w 2 ) 2 Y j = 1 w j g ǫ ( w j ) × z − 2 { ( q + q − 1 ) z − ( w 1 + w 2 ) } 2 Y j = 1 (1 − q w j /z )(1 − q z /w j )(1 − q /z w j ) (3.48) × e P ( z )+ P ( z − 1 )+ R − ( q 4 w 1 )+ R − ( q 4 /w 1 )+ R − ( q 4 w 2 )+ R − ( q 4 /w 2 ) e − Q e F ( ǫ ) | Λ 1 i . The inte gration conto ur e C 1 encircles w = 0 , q z , q z − 1 but not w = q − 1 z . The integ ration con tour e C 2 encircles w 1 = 0 , q z , q z − 1 , q w 2 , q w − 1 2 but not w 1 = q − 1 z , q − 1 w 2 , and encircles w 2 = 0 , q z , q z − 1 , q w 1 , q w − 1 1 but not w 2 = q − 1 z , q − 1 w 1 . F or simp licit y w e summarize the case ǫ = ± . The r elation (3.36) is equiv alent to the follo wing three relations. ϕ ± ( z )Φ − ( z ) | B i ± = ϕ ± ( z − 1 )Φ − ( z − 1 ) | B i ± , (3.49) ϕ ± ( z )(1 ∓ √ − 1 q − 3 2 z )Φ 0 ( z ) | B i ± = ϕ ± ( z − 1 )(1 ∓ √ − 1 q − 3 2 z − 1 )Φ 0 ( z − 1 ) | B i ± , (3.50) ϕ ± ( z ) z Φ + ( z ) | B i ± = ϕ ± ( z − 1 ) z − 1 Φ + ( z − 1 ) | B i ± , (3.51) • The relation (3.49). Usin g formula (3.46), w e ha ve ( LH S ) = 1 ǫ ( q ) ϕ ± ( z ) ϕ ± ( z − 1 ) e P ( z )+ P ( z − 1 ) e Q e F ( ǫ ) | Λ 1 i = ( RH S ) . (3.52) • The relation (3.50). Usin g formula (3.47), w e ha ve ( LH S ) − ( R H S ) = (1 − q 2 ) ϕ ± ( z ) ϕ ± ( z − 1 )( z − 1 − z ) (3.53) × I b C 1 dw 2 π √ − 1 w I 1 ( w ) e P ( z )+ P ( z − 1 )+ R − ( q 4 w )+ R − ( q 4 w − 1 ) (1 − q w /z )(1 − q z /w )(1 − q /wz )(1 − q w z ) e F ( ǫ ) | Λ 1 i , where w e ha v e set I 1 ( w ) = ( w − w − 1 )(1 ∓ √ − 1 q − 1 2 w )(1 ∓ √ − 1 q − 1 2 w − 1 ). Here the integrati on con tour b C 1 encircles w = 0 , q z , q z − 1 but n ot w = q − 1 z , q − 1 z − 1 . The integ ration con tour b C 1 and the inte grand e P ( z )+ P ( z − 1 )+ R − ( q 4 w )+ R − ( q 4 w − 1 ) (1 − q w /z )(1 − q z /w )(1 − q /w z )(1 − q w z ) are in v arian t und er w → w − 1 . Hence the relation I 1 ( w ) + I 1 ( w − 1 ) = 0 en s ures ( LH S ) − ( RH S ) = 0. • The relation (3.51). Usin g formula (3.48), w e ha ve ( LH S ) − ( RH S ) = q − 5 2 (1 − q 2 ) 2 ǫ ( q ) ϕ ± ( z ) ϕ ± ( z − 1 ) I I b C 2 dw 1 2 π √ − 1 w 1 dw 2 2 π √ − 1 w 2 14 × (1 − w 1 /w 2 )(1 − w 2 /w 1 )(1 − w 1 w 2 )(1 − 1 /w 1 w 2 ) (1 + q w 1 /w 2 )(1 + q w 2 /w 1 )(1 + q w 1 w 2 )(1 + q /w 1 w 2 ) × q 2 ( z − z − 1 ) I 2 ( w 1 , w 2 ) 2 Y j = 1 (1 − q z /w j )(1 − q w j /z )(1 − q z w j )(1 − q /z w j ) (3.54) × e P ( z )+ P ( z − 1 )+ R − ( q 4 w 1 )+ R − ( q 4 /w 1 )+ R − ( q 4 w 2 )+ R − ( q 4 /w 2 ) e − Q e F ( ǫ ) | Λ 1 i , where we h a ve set I 2 ( w 1 , w 2 ) = {− ( q + q − 1 )( z + z − 1 ) w 1 w 2 + ( w 1 + w 2 )( w 1 w 2 + 1) } × (1 + q w 1 w 2 ) (1 − w 1 w 2 ) (1 − q 2 /w 1 w 2 ) 2 Y j = 1 ( w j − w − 1 j )(1 ∓ √ − 1 q − 1 2 w − 1 j ) . (3.55) Here the integ ration cont our b C 2 encircles w 1 = 0 , q z , q z − 1 , q w 2 , q w − 1 2 but not w 1 = q − 1 z , q − 1 z − 1 , q − 1 w 2 , q − 1 w − 1 2 , and encircles w 2 = 0 , q z , q z − 1 , q w 1 , q w − 1 1 but not w 2 = q − 1 z , q − 1 z − 1 , q − 1 w 1 , q − 1 w − 1 1 . The int egration con tour b C 2 and the int egrand (1 − w 1 /w 2 )(1 − w 2 /w 1 )(1 − w 1 w 2 )(1 − 1 /w 1 w 2 ) (1 + q w 1 /w 2 )(1 + q w 2 /w 1 )(1 + q w 1 w 2 )(1 + q /w 1 w 2 ) × e P ( z )+ P ( z − 1 )+ R − ( q 4 w 1 )+ R − ( q 4 /w 1 )+ R − ( q 4 w 2 )+ R − ( q 4 /w 2 ) 2 Y j = 1 (1 − q z /w j )(1 − q w j /z )(1 − q z w j )(1 − q /z w j ) are inv arian t und er ( w 1 , w 2 ) → ( w − 1 1 , w 2 ) , ( w 1 , w − 1 2 ) , ( w − 1 1 , w − 1 2 ). Hence the r elatio n I 2 ( w 1 , w 2 ) + I 2 ( w − 1 1 , w 2 ) + I 2 ( w 1 , w − 1 2 ) + I 2 ( w − 1 1 , w − 1 2 ) = 0 ensures ( LH S ) − ( RH S ) = 0. Q.E.D. 3.4 Dual ground state In this section we give the free fi eld realizations of the dual ground state ǫ h B | ∈ V (Λ 1 ) ∗ , ( ǫ = ± , 0), w hic h satisfies ǫ h B | T ǫ ( z ) = ǫ h B | , ( ǫ = ± , 0) . (3.56) The dual integ rable h ighest w eight representa tion V (Λ 1 ) ∗ of U q ( A (2) 2 ) is realized by V (Λ 1 ) ∗ = h Λ 1 | ⊕ n ∈ Z e nQ C [ a 1 , a 2 , · · · ] , h Λ 1 | = h 0 | e − Q 2 . (3.57) The v acuum vec tor h 0 | is characte rized by h 0 | a − m = 0 , ( m > 0) , h 0 | P = 0 . (3.58) 15 Theorem 3.2 The fr e e field r e alizations of the dual gr ound states ǫ h B | ar e given by ǫ h B | = h Λ 1 | e G ( ǫ ) , ( ǫ = ± , 0) . (3.59) Her e we have set G ( ǫ ) = − 1 2 X m> 0 mq − 2 m [2 m ] q − ( − 1) m [ m ] q a 2 m (3.60) + X m> 0 ( − θ m ( q m 2 − q − m 2 − ( √ − 1) m ) q − m [2 m ] q − ( − 1) m [ m ] q ! + ( − ǫ √ − 1) m q − 2 m [2 m ] q − ( − 1) m [ m ] q ) a m − Q. The pro of of (3.56) is giv en as th e same w a y as those of (2.24). The f ollo wing r elatio ns are useful for pro of. ǫ h B | e P ( − q − 3 z − 1 ) = ϕ ǫ ( z − 1 ) ǫ h B | e Q ( − q − 3 z ) , (3.61) ǫ h B | e R − ( q w ) = g ∗ ǫ ( w ) ǫ h B | e S − ( q w − 1 ) , (3.62) where ϕ ǫ ( z ) is giv en in (2.8) and g ∗ ǫ ( w ) is giv en by g ∗ ǫ ( w ) =    (1 − w 2 ) , ( ǫ = 0) (1 − w 2 )(1 ± √ − 1 q − 1 2 w ) , ( ǫ = ± ) . (3.63) Ac kn o wledgmen ts This work is supp orted by th e Grant -in-Aid for S cien tific Researc h C (21540228) from Japan So ciet y for Pr omotion of Science. References [1] M.Jim b o and T.Miw a, Algebr aic Ana lysis of Solvable L attic e Mo dels , (CBMS Regional Con - ference Series 85 , American Mathematical S o ciet y 1995). [2] M.Jim b o, R.Kedem, T.Ko jima, H.Konno and T.Miw a, Nucl.Phys. B441 437 (1995). [3] E.K.Skly anin, J.Phys. A21 2375 (1988). [4] A.G.Izergin and V.E.Korepin, Commun.Math.Phys. 79 303 (1981). [5] M.T.Batc helor, V.F ridkin, A.Kuniba and Y.K.Zhou, P hys.L e tt. B376 266 (1996). 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