Nested Bethe Ansatz for Spin Ladder Model with ...

Commun. Theor. Phys. (Beijing, China) 43 (2005) pp. 687–694 c International Academic Publishers

Vol. 43, No. 4, April 15, 2005

Nested Bethe Ansatz for Spin Ladder Model with Open Boundary Conditions∗ WU Jun-Fang,1,2,† ZHANG Chun-Min,1 YUE Rui-Hong,3,‡ and LI Run-Ling4 1

Department of Physics, Xi’an Jiao Tong University, Xi’an 710049, China

2

Department of Physics, Xi’an University of Engineering Science and Technology, Xi’an 710048, China

3

Institute of Modern Physics, Northwest University, Xi’an 710069, China

4

Missile Institute, Air Force Engineering University, Xi’an 710051, China

(Received May 28, 2004; Revised October 18, 2004)

Abstract The nested Bethe ansatz (BA) method is applied to find the eigenvalues and the eigenvectors of the transfer matrix for spin-ladder model with open boundary conditions. Based on the reflection equation, we find the general diagonal solution, which determines the general boundary interaction in the Hamiltonian. We introduce the spin-ladder model with open boundary conditions. By finding the solution K± of the reflection equation which determines the nontrivial boundary terms in the Hamiltonian, we diagonalize the transfer matrix of the spin-ladder model with open boundary conditions in the framework of nested BA. PACS numbers: 05.50.+q, 64.60.Cn, 75.10.Hk, 75.10.Jm

Key words: integrable model, coordinate Bethe ansatz, quantum inverse scattering method, Bethe ansatz equation, nested Bethe ansatz

1 Introduction In this paper, the eigenvalue and the two-particle scattering matrix are obtained in studying the electron spinladder model under open boundary condition with coordinate Bethe ansatz (BA) method in integrable systems. In Sec. 2, the eigenvalue equation is formed according to Schr¨odinger equation. The eigenvalue is worked out when the wave function is given. At the end, the twoparticle scattering matrix is calculated from eigenvalue equation and the continuity condition of the wave function. In the third part we get the eigenvalue and scattering matrix under open boundary condition. In the fourth section, we will apply the nested BA method to find the eigenvalue and the eigenvector of the transfer matrix for spin-ladder model with open boundary conditions. First based on the reflection equation, we find the general diagonal solution, which determine the open boundary interaction in the Hamiltonian. Next we introduce the open spin-ladder model by finding the solution K ± of the reflection equation which determines the nontrivial boundary terms in the Hamiltonian. The last section contributes to the diagonalization of the transfer matrix of the model with open boundary conditions in the framework of nested BA. By using the algebraic Bethe ansatz, we find the exact solution of spin-ladder model with open boundary conditions. The BA equation and the eigenvalue of the transfer matrix are obtained. In the last, a summary of our main results is presented. ∗ The

2 Eigenvalue and Two-Particle Scattering Matrix of Spin-Ladder Model with Open Boundary Condition by Coordinate BA 2.1 Introduction of Spin-Ladder Model The Hamiltonian of spin-ladder model can be written as[1] N 3 N X X X 1 1 βα H= Xjαβ Xj+1 − 2J Xj00 + J− N. (1) 2 2 j=1 j=1 α,β=0

For convenience, in the above denotation, we choose 1 |0i = √ (| ↑, ↓i − | ↓, ↑i) , |1i = | ↑, ↑i , 2 1 |2i = √ (| ↑, ↓i + | ↓, ↑i) , |3i = | ↓, ↓i . (2) 2 The first state denotes a singlet rung and the remaining three indicate the spin-triplet states of a rung. OperaPN tors Nα ≡ j=1 Xjαα , which denote the numbers of the α rungs in the whole system, are conserved quantities. The constant 2J in the last term of Eq. (1) indicates a chemical potential applied on N0 and reduces the global SU(4) symmetry of the Hamiltonian to U(1)×SU(3). Xjαβ ≡ |αj ihβj | are the Hubbard operators and the Dirac states |αj i span the Hilbert space of the j-th rung and are orthogonal (hαj | βj i = δαβ ). Next, we assume the Hamiltonian of spinladder model with open boundary condition: H=

N 3 X X j=1 α,β=0

βα Xjαβ Xj+1 − 2J

N X

Xj00

j=1

project supported by National Natural Science Foundation of China under Grant Nos. 90103001, 40375010, and 60278019 [email protected] ‡ E-mail: [email protected] † E-mail:

688

WU Jun-Fang, ZHANG Chun-Min, YUE Rui-Hong, and LI Run-Ling

3

+

X 1 1 J− N+ (P1α X1αα + PLα XLαα ) . 2 2 α=1

(3)

2.2 One-Particle Eigenvalue If we choose the reference state as |Ωi = |01 i ⊗ |02 i ⊗ · · · ⊗ |0N i ,

(4)

the one-particle eigenstate can be assumed to be X |ψ1 i = ψα1 (j1 )Xjα11 0 |Ωi ,

(5)

We use the following denotation in Eq. (13), λ − i/2 e ik = . (14) λ + i/2 The periodic boundary condition governs the momentum to be 2πj kj = (j = 0, 1, . . . , L − 1) . (15) L Substituting the above ansatz (12) into Eq. (11) we can obtain the following relation:

j1 α1

and the one-particle Schr¨ odinger equation is H|ψ1 i = E|ψ1 i .

(6)

Applying the Hamiltonian (3) on this ansatz (5), we can obtain

Aα1 (k)αα1 (−k) = Aα1 (−k)αα1 (k) ,

(16)

αα1 (k) = 1 + P1α e −ik ,

(17)

Aα1 (k) 1 − P1α e −ik . = Aα1 (−k) 1 − P1α e ik

(18)

where so we get

Eψα1 (j1 ) = ψα1 (j1 − 1) + ψα1 (j1 + 1) + 0 ψα1 (j1 ) , (2 < j1 < L − 1) ,

(7)

Eψα1 (1) = ψα1 (2) + (0 + P1α )ψα1 (1) ,

(8)

Eψα1 (L) = ψα1 (L − 1) + (0 + PLα )ψα1 (L) ,

(9)

where

3 (1 − 2J)N − 2 + 2J . (10) 4 So, we get the following equation from Eqs. (7) and (8)

We get the following equation from Eqs. (7) and (9): ψα1 (L + 1) = PLα ψα1 (L) .

ψα1 (j) = Aα1 (k) e

− Aα1 (−k) e

−ikj

.

Aα1 (k)βα1 (k) = Aα1 (−k)βα1 (−k) ,

(20)

βα1 (k) = (1 + PLα e −ik ) e ik(L+1) ,

(21)

where

(11) so we get

We assume the wave function of one-particle state to be ikj

(19)

Substituting the above ansatz (12) into Eq. (19) we can obtain the following relation:

0 =

ψα1 (0) = P1α ψα1 (1) .

Vol. 43

(12)

Substituting the wave function (12) into Eq. (7), we can obtain the eigenvalue of the one-particle state, 1 E=− 2 + 2J + ε0 . (13) λ + 1/4

Aα1 (k) e −ik − PLα −2ikL = ik e . Aα1 (−k) e − PLα We get the following result: (1 − PLα e −ik )(1 − PLα e −ikL ) i2k(L+1) e = 1. (1 − PLα e ik )(1 − PLα e ikL )

(22)

(23)

2.3 Eigenvalue and Scattering Matrix of N0 Particles For the general N0 -particles, the eigenstates of the Hamiltonian can be assumed as |ψN0 i =

N0 X

α

0

ψα1 α2 ...αN0 (j1 , j2 , . . . , jN0 )Xjα11 0 Xjα22 0 · · · XjNN0 |Ωi .

(24)

0

jm ,αm =1

We assume the wave function of N0 -particle state to be ψα1 α2 ···αN0 (j1 , j2 , . . . , jN0 ) =

X

εP εQ AαQ1 αQ2 ··· αQN (kP1 , kP2 , . . . , kPN0 )θ(jQ1 ≤ jQ2 0

N0  X  ≤ . . . jQN0 ) exp i kPl jQl (25)

P,Q

l=1

with  1,   1 θ(j1 ≤ j2 ≤ · · · jN0 ) = ,  2  0,

j1 < j2 < · · · jN0 , jl 6= jm (l 6= m; l, m = 1, 2, . . . , N0 ) ,

(26)

other cases .

Here αi = 1, 2, . . . , n − 1 (i = 1, 2, . . . , N0 ) stands for the different particle states, and jl is the position of the particle. The summation P and Q are taken over all permutations of N0 momenta kj and N0 coordinates jl respectively. The symbols εP and εQ are the parities of two kinds of permutations. The Schr¨odinger equation of N0 -particle is H|ψN0 i = E|ψN0 i .

(27)

Substituting the above ansatz (24) and the wave function of N0 -particle state (25) into the Schr¨odinger equation (27),

No. 4

Nested Bethe Ansatz for Spin Ladder Model with Open Boundary Conditions

689

we can obtain Eψα1 α2 ···αN0 (j1 , j2 , . . . , jN0 ) =

N0 X X

ψα1 α2 ...αN0 (j1 , j2 , . . . , jl + s, . . . , jN0 )

l=1 S=±1 N0   X + ε0 + (2J − 2) ψα1 α2 ···αN0 (j1 , j2 , . . . , jN0 )

(j1 6= j2 ) ,

(28)

(j1 = j2 ) .

(29)

l=1

Eψα1 α2 ···αN0 (j1 , j2 , . . . , jN0 ) =

N0 X X

ψα1 α2 ···αN0 (j1 , j2 , . . . , jl + s, . . . , jN0 )

l=1 S=±1 N0   X + ε0 + (2J − 1) ψα1 α2 ···αN0 (j1 , j2 , . . . , jN0 ) l=1

Imposing the continuity condition of the wave function lim ψ···αm αl ··· (j1 , . . . , jl , jm , . . . , jN0 ) = lim − ψ···αm αl ··· (j1 , . . . , jl , jm , . . . , jN0 ) .

jm →jl+

(30)

jm →jl

Substituting the wave function of N0 -particle state (25) into Eqs. (28) and (29), we can obtain the eigenvalue N0   X 1 − 2J + ε0 . E=− λ2j + 1/4 j=1

(31)

Using Eqs. (28), (29), and (30), we get A···αi αj ··· (· · · ki, kj · · ·) = Sαi αj (sin ki , sin kj )A···αj αi ··· (· · · kj , ki · · ·) , where Sαi αj (sin ki , sin kj ) =

sin ki − sin kj + ipαi αj sin ki − sin kj + i

(32) (33)

is the scattering matrix of two particles. 2.4 Monodromy Matrix of the Spin-Ladder Model with Open Boundary Condition We have the following relations from Secs. 1 and 2, A(−k1 · · · kN ) = U (k1 )S12 (λ − λ1 ) · · · S1N (λ − λN )V (k1 )A(k2 , . . . , kN , −k1 ) −1 −1 × S¯1N (−λ − λN ) · · · S¯12 (−λ − λ1 )A(−k1 , . . . , kN ) ,

where

Letting λ = λ1 in Eq. (42), we get the following relation:

A(−k1 , . . . , kN ) = U (k1 )A(k1 , . . . , kN ) ,

(35)

t(λ1 ) = tr K + (λ1 )T (λ1 )K − (λ1 )T −1 (−λ1 )

A(k1 , . . . , −kN ) = V (kN )A(k1 , . . . , kN ) ,

(36)

= L12 (λ1 , λ2 ) · · · L1N (λ1 , λN )K − (λ1 )

(37)

× L−1 (−λ1 , λN ) · · · L−1 (−λ1 , λ2 ) ¯ ¯ 1N 12

Sij =

(34)

sin ki − sin kj i I+ Pij , sin ki − sin kj + i sin ki − sin kj + i

− sin ki − sin kj i S¯ıj = I+ Pij .(38) − sin ki − sin kj + i − sin ki − sin kj + i We assume

(39)

so, rewriting Eq. (34) as T˜(λ)A = A ,

We assume ˜ + (λ1 ) = tr P01 L−1 (−λ1 , λ1 )K + (λ1 ) , K 01

(43) (44)

from Eqs. (39) and (43), we get

T˜(λ) = U (k1 )S12 (λ, λ1 ) · · · S1N (λ, λN )V (k1 ) −1 −1 × S¯1N (−λ, λN ) · · · S¯12 (−λ, λ1 ) ,

+ × tr P01 L−1 01 (−λ1 , λ1 )K (λ1 ) .

(40)

we assume T (λ) = L01 (λ, λ1 ) · · · L0N (λ, λN ) , −1 T −1 (−λ) = L−1 0N (−λ, λN ) · · · L01 (−λ, λ1 ) ,

(41)

t(λ) = tr K + (λ)T (λ)K − (λ)T −1 (−λ) .

(42)

L1j (λ1 , λj ) = S1j (λ1 − λj ) ,

(45)

−1 L−1 (−λ1 , λj ) = S¯1j (−λ1 − λj ) , ¯ 1j

(46)

where f (λ1 ) and g(λ1 ) are matrix functions, T˜(λ1 ) = f (λ1 )g(λ1 )t(λ1 ) .

(47)

So, the question of solving the eigenvalue of the Hamiltonian is transferred into solving the eigenvalue of the monodromy matrix. In the next section, we will give the method of solving the eigenvalue of the monodromy matrix.

690

WU Jun-Fang, ZHANG Chun-Min, YUE Rui-Hong, and LI Run-Ling

3 Nested Bethe Ansatz for Open Boundary Conditions 3.1 R-matrix of Spin-Ladder Model The R-matrix of the spin-ladder model is defined as aa Raa (u) = u + i ,

a = 1, 2, 3 ,

ab Rab (u)

= u,

a 6= b = 1, 2, 3 ,

ab Rba (u) = i ,

a 6= b = 1, 2, 3 ,

Given a solution K − (u) of Eq. (54), one can also find a solution K + (u) of Eq. (55). But in a transfer matrix of an integrable lattice, K − (u) and K + (u) need not satisfy Eq. (56). In this paper, we will take Eq. (56) to define K + (u). Solving the reflection Eq. (54), we have the following general diagonal solution,[7] K − (u, ξ) =

3 X

PaA (u, ξ)Eaa ,

(57)

a=1

aa Raa (u) = 0 , other cases . (48) For convenience, we denote a(u) = u + i , b(u) = u , w(u) = u − i , c(u) = i .(49) The R-matrix defined by Eq. (48) satisfies Yang–Baxter equation, R12 (u−v)R13 (u)R23 (v) = R23 (v)R13 (u)R12 (u−v) , (50) where R12 (u), R13 (u), and R23 (u) act in C 3 ⊗ C 3 ⊗ C 3 with R12 (u) = R(u)⊗1, R23 (u) = 1⊗R(u), etc. The local transfer matrix L(u) satisfies the following equation:[2−4] R12 (u − v)L1 (u)L2 (v) = L2 (v)L1 (u)R12 (u − v) , (51) where L1 (u) = L(u) ⊗ 1, L2 (u) = 1 ⊗ L(u). The standard row-to-row monodromy matrix for an N × N square lattice is defined by T (u) = LN (u) · · · L1 (u) = R0N (u) · · · R01 (u) , (52) where L(u) is assumed to be in the fundamental representation. We can prove that T (u) also satisfies the Yang– Baxter equation, R12 (u − v)T1 (u)T2 (v) = T2 (v)T1 (u)R12 (u − v) , (53) where the operator T is a 3 × 3 matrix of the operators acting in the quantum space V3⊗N . Now, we can use Mezincescue and Nepomechie’s generalized formalism to construct integrable systems with open boundary conditions. In our case, the reflection equations take the following form:[5,6] R12 (u − v)K1− (u)R21 (u + v)K2− (v)

= K2− (v)R12 (u + v)K1− (u)R21 (u − v) , R12 (−u + v)K1+ (u)t1 R21 (−u − v − di)K2+ (v)t2 = K2+ (v)t2 R12 (−u − v − di)K1+ (u)t1 R21 (−u

Vol. 43

(54)

+ v) , (55) where d = −3. Obviously, there is an isomorphism between K + (u) and K − (u),  di t . (56) φ : K − (u) −→ K + (u) = K − −u − 2

where PaA (u, ξ) =



ξ + u,

1 ≤ a ≤ A,

ξ − u,

A < a ≤ 3.

(58)

Correspondingly K + (u, ξ˜ ) =

3 X

PaA (u, ξ˜ )Eaa

(59)

a=1

with  1  ξ˜ − u − di , 1≤a≤B, B 2 ˜ = (60) Pa (u, ξ) 1 ˜ ξ + u + di , B < a ≤ 3. 2 ˜ In Eqs. (58) and (60), both ξ and ξ are free parameters. The integer numbers A and B are also free parameters which take values from 1 to 3. Taking Sklyanin’s formalism, the double-row monodromy matrix is defined as U (u) = T (u)K − (u, ξ)T −1 (−u) ,

(61)

where T −1 (u) is the inverse of T (u) in the auxiliary and quantum spaces. The double-row monodromy matrix U (u) satisfies the reflection equation: R12 (u − v)U1 (u)R21 (u + v)U2 (v) = U2 (v)R12 (u + v)U1 (u)R21 (u − v) .

(62)

Then the transfer matrix is defined as ˜ (v) . t(v) = tr K + (v, ξ)U

(63)

Using the reflections equations (55) and (62), one can prove [t(u), t(v)] = 0 .

(64)

So the transfer matrix constitutes a one-parameter commutative family, which ensures the integrability of the model.

3.2 Commutation Rule Between Operators Rewriting Eq. (62) R12 (u− )ac11ca22 U (u)c1 d1 R21 (u+ )db11dc22 U (v)d2 b2 = U (v)a2 c2 R12 (u+ )ac11dc22 U (u)c1 d1 R21 (u− )db11bd22 ,

(65)

where the repeated indices sum over 1 to 3, u+ = u + v, u− = u − v, and introducing a set of notations for convenience A(v) = U (v)11 ,

Ba (v) = U (v)1a ,

Ca (v) = U (v)a1 ,

Dab (v) = U (v)ab ,

(2 ≤ a, b ≤ 3)

(66)

we get the commutation rule between the operators from Eq. (65). The commutation rule between the operators Ba (v) is[8] ˜ 12 (u1 − u2 )d2 d1 Bd (u2 )Bd (u1 ) , Bb1 (u1 )Bb2 (u2 ) = R (67) 2 1 b2 b1

No. 4

Nested Bethe Ansatz for Spin Ladder Model with Open Boundary Conditions

˜ 12 (u)ab is where R cd

691

R12 (u)ab cd ˜ 12 (u)ab . R cd = R(u)11 11

(68)

We can commute the operator B(vk ) with B(vk−1 ) · · · B(v1 ) respectively. we get ···dL Bd1 (vk )Bd2 (v1 ) · · · Bdk (vk−1 )Bdk+1 (vk+1 ) · · · BdL (vL )|vaci , Bb1 (v1 ) · · · BbL (vL )|vaci = S(vk , {vi })db11···b L

where ···dL S(vk , {vi })db11···b = L

L Y

c

d

c2 d3 k−1 k ˜ 12 (v1 − vk )d1 d2 R ˜ ˜ δbj dj R c2 b1 12 (v2 − vk )c3 b2 · · · R12 (vk−1 − vk )bk bk−1 .

(69)

(70)

j=1+k

The commutation relation between A(u) and Ba (v) is w(v − u)b(v + u) b(2v)c(v − u) c(v + u) ˜ cb (v) . A(u)Bb (v) = Bb (v)A(u) − Bb (u)A(v) − Bc (u)D w(v + u)b(v − u) w(2v)b(v − u) w(v + u) ˜ a b (u) and Ba (v) is The commutation relation between D

(71)

1 1

R12 (u + v − i)ac11dc22 c(u − v) ˜ c d (u) − ˜ a b (u)Bb (v) = R21 (u − v)db11bd22 Bc2 (v)D D 1 1 1 1 2 b(u − v)b(u + v − i) b(2u − i)b(u − v) c(u + v)b(2v) R12 (2u − i)ba21bd12 Bd2 (u)A(v) . b(2u − i)w(u + v)w(2v) We have the commutation relations from Eqs. (71) and (72) (see Ref. [8]), w(v − u)b(v + u) b(2v)c(v − u) c(v + u) ˜ cb (v) , A(u)Bb (v) = Bb (v)A(u) − Bb (u)A(v) − Bc (u)D w(v + u)b(v − u) w(2v)b(v − u) w(v + u) ˜ d b (v) + × R12 (2u − i)ad21 bd11 Bd1 (u)D 2 2

˜ a b (u)Bb (v) = D 1 1 2

(72)

(73)

R12 (u + v − i)ac11dc22 c(u − v) ˜ c d (u) − R21 (u − v)db11bd22 Bc2 (v)D 1 1 b(u − ν)b(u + ν − i) b(2u − i)b(u − v)

c(u + v)b(2v) R12 (2u − i)ba21bd12 Bd2 (u)A(v) , (74) b(2u − i)w(u + v)w(2v) ˜ is defined by where all indices take values from 2 to 3, and the repeated indices sum over 2 to 3. The new operators D ˜ ab (v) = Dab (v) − δab c(2v) A(v) . D (75) w(2v) Introducing the vacuum state ⊗N Y |vaci = (1, 0, . . . , 0)T , (76) ˜ d b (v) + × R12 (2u − i)ad21 bd11 Bd1 (u)D 2 2

where T denotes the transposition. Acting the operators A(u), Ca (u), and Ba (u) on the vacuum state, we can easily find A(u)|vaci = α(u)|vaci , Ca (u)|vaci = 0 , Ba (u)|vaci = 6 0, (77) where N − −1 N α(u) = [R(u)11 (−u)11 11 ] K11 (u)[R 11 ] .

(78)

˜ ab (u) on the vacuum state. From the definition of Dab (u), we have Now, we consider the action of D − − Dab (u)|vaci = T (u)a1 K11 (u)T −1 (−u)1b |vaci + T (u)ac Kcc (u)T −1 (−u)cb |vaci .

(79)

The contribution of the first term cannot be calculated directly but it can be worked out by using the following method. Taking v = −u in the Yang–Baxter equation, we can get T2−1 (−u)R12 (2u)T1 (u) = T1 (u)R12 (2u)T2−1 (−u) . Taking special indices in this relation and applying both sides of this relation to the vacuum state, we find c(2u) (δab α(u) − T (u)ac T −1 (−u)cb )|vaci . T (u)a1 T −1 (−u)1b |vaci = R(2u)11 11 Substituting this relation into Eq. (79), we have n c(2u)   o c(2u) − − − N ˜bN (−u) |vaci . Dab (u)|vaci = δab K (u)α(u) + K (u) − K (u) b (u) aa 11 11 R(2u)11 R(2u)11 11 11 Then we have   c(2u) − − ˜ ab (u)|vaci = δab Kaa (u) − D K (u) bN (u)˜bN (−u)|vaci . 11 R(2u)11 11

(80)

(81)

(82)

(83)

692

WU Jun-Fang, ZHANG Chun-Min, YUE Rui-Hong, and LI Run-Ling

Vol. 43

So, the action of the double-row monodromy matrix on the vacuum state is N −1 N − A(u)|vaci = [R(u)11 (−u)11 11 ] [R 11 ] K11 (u)|vaci = α(u)|vaci ,

c(2u) − − ˜ ab (u)|vaci = δab (Kaa K11 (u))bN (u)˜bN (−u)|vaci = δab βa (u)|vaci , D (u) − R(2u)11 11 Ca (u)|vaci = 0 , Ba (u)|vaci = 6 0, (84) −1 ab ˜ where b(u) = R (u)ab , a 6= b. Note that the action of Ba (v) on the vacuum state is not proportional to the vacuum state. 3.3 Nested Bethe Ansatz for Open Boundary Conditions Now we construct the eigenvectors of transfer matrix t(u). It takes the form Ψ(v1 , . . . , vL ) = Bb1 (v1 ) · · · BbL (vL )|vaciF b1 ···bL .

(85)

Recalling the definition of the transfer matrix, we rewrite the transfer matrix as 3 3 3 nX o X X c(2u) + + ˜ t(u) = Ka (u)U (u)aa = Ka (u)Daa (u) + Ka+ (u) + K1+ (u) A(u) w(2u) a=1 a=2 a=2 =

3 X

˜ aa (u) + S1 (u)A(u) , Ka+ (u)D

(86)

a=2

where S1 (u) =

3 X a=2

Ka+ (u)

c(2u) + K1+ (u) . w(2u)

(87)

Using Eq. (86), we then obtain the action of t(u) on Ψ(v1 , . . . , vN ) L Y b(vj + u)w(vj − u) t(u)Ψ = α(u)S1 (u) Bb1 (v1 ) · · · BbL (vL )|vaciF b1 ···bL w(v + u)b(v − u) j j j=1 +

+

L Y w(u − vj )w(vj − u) (1) ···dL b1 ···bL β (u) × τ (2) (˜ u, {˜ vi })bd11···b F Bb1 (v1 ) · · · BbL (vL )|vaci L b(u − v )b(u + v + i) j j j=1 L  X −c(vk − u) k=1

b(vk − u)

L Y

×

j=1,j6=k

S1 (u) +

 b(2v ) c(vk + u) k (d ) T1 1 (u) w(vk + u)b(2u + i) w(2vk )

w(vj − vk )b(vj + vk ) ···dL b1 ···bL α(vk )S(vk , {vi })db11···b F L b(vj − vk )w(vj + vk )

× Bd1 (u)Bd2 (v1 ) · · · Bdk (vk−1 )Bdk+1 (vk+1 ) · · · BdL (vL )|vaci −

L  X c(vk + u) k=1

w(vk + u)

L Y

×

j=1,j6=k

S1 (u) +

 iβ (1) (v ) c(u − vk ) k (d ) T1 1 (u) (d ) 1 b(u − vk )b(2u + i) (vk ) T 1

w(vk − vj )w(vj − vk ) ···cL b1 ···bL ···dL (2) S(vk , {vi })dc11···c τ (˜ vk , {˜ vi })cb11 ···b F L L b(vk − vj )b(vj + vk + i)

× Bd1 (u)Bd2 (v1 ) · · · Bdk (vk−1 )Bdk+1 (vk+1 ) · · · BdL (vL )|vaci ,

(88)

where ···cL τ (2) (˜ u, {˜ vi })cb11 ···b = L

3 X

···cL Ka+ (u){(L(1) (˜ u, v˜1 ) · · · L(1) (˜ u, v˜L )K − (˜ u, ξ (1) )L(1) (−˜ u, v˜L )−1 · · · L(1) (−˜ u, v˜1 )−1 )cb11 ···b }aa L

a=2

=

3 X

···cL Ka+ (u){(T (1) (˜ u, {˜ vi })K − (˜ u, ξ (1) )T (1) (−˜ u, {˜ vi })−1 )bc11 ···b }aa , L

(89)

a=2

2u N ˜N b (u)b (−u) , 2u + i 3 X (d ) 1 T1 1 (u) = Ka+ (u)R(2u + i)ad d1 a , β (1) (u) =

a=2

(90) (91)

No. 4

Nested Bethe Ansatz for Spin Ladder Model with Open Boundary Conditions

693

L(1) (˜ v1 , v˜i )aa12 bb12 = R12 (˜ v1 + v˜i )aa12 bb12 , L Y

···dL S(vk , {vi })db11···b = L

(92)

ck−1 dk c2 d3 ˜ 12 (v1 − vk )d1 d2 R ˜ ˜ δbj dj R c2 b1 12 (v2 − vk )c3 b2 · · · R12 (vk−1 − vk )bk bk−1 ,

(93)

j=1+k

and u ˜ = u + i/2, ξ (1) = ξ − i/2, v˜i = vi + i/2. From Eq. (88), one can see that the function Ψ is not the eigenstate of t(u) unless F ’s are the eigenstates of τ (2) and the sum of the third and the fourth terms in the above equation are zero, which will give a restriction on the L spectrum parameters {vi }. So, we arrive at the following results. If F is the eigenstate of τ (2) with the eigenvalue Λ(2) (u, {vi }) satisfying Eq. (95), then Ψ is the eigenstate of t(u) with the eigenvalue Λ(1) , Λ(1) (u) = α(1) (u)S1 (u)

L L Y Y w(u − vj )w(vj − u) (2) w(vj − u)b(vj + u) + β (1) (u) Λ (u, {vi }) , b(v − u)w(v + u) b(u − vj )b(u + vj + i) j j j=1 j=1

(94)

L (d ) α(1) (vk )b(2vk )T1 1 (vk ) Y b(vj + vk ) , −w(vk − vj ) β (1) (vk )w(2vk )i j=1,j6=k

(95)

where τ (2) (u, {vi })F = Λ(2) (u, {vi })F ,

Λ(2) (u, {vi }) =

where α(1) (u) = α(u). Therefore, the diagonalization of t(u) is reduced to finding the eigenvalue of τ (2) . The explicit expression of τ (2) (see Eq. (88)) implies that τ (2) can be considered as the transfer matrix of an L-site quantum chain, in which every spin takes 2 values. The related Yang–Baxter equation is the same as the one of t(u), except R being an 22 × 22 matrix. Hence, we can use the same method to find the eigenvalue of τ (2) . This is the well-known nested BA. The eigenvalue and the constraint on the spectral parameters read as (k−1)

Λ(k) (u, {vi

(k)

(k−1)

}, {vi }) = Sk (u)α(k) (u, {vi

})

(k) (k) Pk Y w(vj − u)b(vj + u + (1 − k)i) (k)

j=1 (k−1)

+ β (k) (u, {vi

})

Pk Y j=1

(k)

w(vj b(u −

b(vj

(k)

− u)w(vj

+ u + (1 − k)i)

(k)

− u)w(u − vj )

(k) vj )b(u

+

(k) vj

− ki)

(k)

(k−1)

Λ(k+1) (u, {vi }, {vi

})

(1 ≤ k ≤ 3)

(96)

and (k)

(k)

(k)

(k+1)

Λ(k+1) (vl , {vi }, {vi

}) =

(k−1)

α(k) (vl , {vi

(k)

})b(2vl

Pk Y

(k)

+ (1 − k)i)Tk (vl )

(k) (k−1) (k) β (k) (vl , {vi })iw(2vl

− ki)

j=1,j6=k

(k)

b(vl

(k)

+ vj

(k) w(vl

+ (1 − k)i) (k)

− vj

+ i)

(1 ≤ k ≤ 2) ,

(97)

where

   d  ξ˜ + i + B i − u ,  ξ˜ + (1 ≤ k ≤ B) , 2 Sk (u) =  Tk (u) =   2u + di  ˜ d  ˜ ξ+ i+u , (B < k ≤ 3) , ξ+ 2 2u − ki Pk−1 Y u + vj(k−1) − ki (k−1) α(k) (u, {vi }) = χk (u) , (1 ≤ k ≤ 3) , (k−1) −u−i j=1 vj

 d i + B i − u (2u + di) , (1 ≤ k ≤ B) , 2  d i + u (2u + di) , (B < k ≤ 3) , (98) 2 (99)

Pk−1 (k−1)

β (k) (u, {vi

}) =

Y (u + vj(k−1) + (1 − k)i)(u − vj(k−1) ) 2u + (1 − k)i , (k−1) (k−1) 2u − ki − u + i)b(u − v + i) j=1 b(v j

 χk (u) =

ξ + u,

1 ≤ k ≤ A,

ξ − Ai − u ,

A < k ≤ 3. (1)

In the above representation, vj

(0)

= vj , vj

(k−1)

= 0, P0 = N , P1 = L, Λ(3) = 1, P3 = 0 are assumed. Notice that (k−1)

(k−1)

, {vi

Pk Y

(k) (vj

×

j=1

(100)

(101)

(k−1)

β (k) (u, {vi }) vanishes at the special points vi due to the factor u − vi in formulae (96), we can get another kind of constraints on Λ(k) , Λ(k) (vl

(1 ≤ k ≤ 2) ,

j

(k−1)

(k)

(k)

(k−1)

}, {vi }) = Sk (vl )α(k) (vl

(k−1)

appearing in β (k) . Taking u = vi

(k−1)

, {vi

(k−1) (k) (k−1) − vl − i)(vj + vl + (1 − k)i) (k) (k−1) (k) (k−1) (vj − vl )w(vj + vl − ki)

,

})

(1 ≤ k ≤ 3) .

(102)

694

WU Jun-Fang, ZHANG Chun-Min, YUE Rui-Hong, and LI Run-Ling

Vol. 43

Now, changing the index k into k + 1 in the above formulae, we can obtain constraints on Λ(k+1) . Comparing these with Eqs. (96) and (97), one can derive the following BA equations: Pk−1

(k)

(vl

Y

(k)

(vl

j=1

(k−1)

− vj

(k−1)

− vj

(vl

j=1,j6=l

(vl

(k)

)(vl

(k)

Pk Y

×

(k)

− i)(vl

(k)

(k−1)

+ vj

(k−1)

+ vj

(k)

− vj

(k)

− vj

Pk+1 (k) (k+1) (k) (k+1) (vl − vj )(vl + vj − ki) Y − (k + 1)i)

+ (1 − k)i)

(k)

+ i)(vl

(k)

− i)(vl

(k)

+ vj

(k)

+ vj

(k)

j=1

(vl

+ (1 − k)i) − (k + 1)i)

=

(k+1)

− vj

(k)

(k+1)

+ i)(vl

+ vj

− ki)

−(2vlk − (1 + k)i)Sk+1 (vlk )χk+1 (vlk ) , Tk (u)χk (vlk )

(1 ≤ k ≤ 2) . (103)

The above BA equations can be simplified by introducing the following new variables: 1 (k) (k) vl → vl + ki , (1 ≤ k ≤ 3) . 2 Then the BA equations take the following form:

(104)

(k) (k+1) (k) (k+1) k+1 Y (vl(k) − vj(k−1) − i/2)(vl(k) + vj(k−1) − i/2) PY (vl − vj − i/2)(vl + vj − i/2)

Pk−1

(k)

j=1

(vl ×

(k−1)

− vj

Pk Y

(k)

(vl

(k) j=1,j6=l (vl

(k)

+ i/2)(vl (k)

− vj −

(k) vj

(k−1)

+ vj (k)

+ i)(vl −

(k) i)(vl

where θk (vlk ) =

+ i/2)

(k)

+ vj +

(k) vj

+ i) − i)

(k)

j=1

(vl

= θk (vlk ) ,

(k+1)

− vj

(k)

+ i/2)(vl

(k+1)

+ vj

+ i/2)

(1 ≤ k ≤ 2) ,

−(2vlk − i)Sk+1 (vlk + ki/2)χk+1 (vlk + ki/2) , Tk (vlk + ki/2)χk (vlk + ki/2)

(105)

(1 ≤ k ≤ 2) .

(106)

4 Summary In this paper, we solve spin-ladder model with open boundary conditions. Substituting eigenvalue of monodromy matrix into Eq. (39), we get the eigenvalue N0   X 1 − 2J + ε0 , (107) E=− λ2j + 1/4 j=1 and BA equations M

M

j6=l

l=1

(1 − P1α e −ik )(1 − PLα e −ikL ) i2k(L+1) Y1 (vj − vl − i)(vj + vl − i) Y2 (vj − µl + i/2)(vj + µl + i/2) e = , (1 − P1α e ik )(1 − PLα e ikL ) (vj − vl + i)(vj + vl + i) (vj − µl − i/2)(vj + µl − i/2)

(108)

M1 M Y (µα − λj − i/2)(µα + λj − i/2) Y3 (µα − vδ − i/2)(µα + vδ − i/2) (µα − λj + i/2)(µα + λj + i/2) (vδ − µα + i/2)(vδ + µα + i/2) j=1 δ=1

×

M2  Y β6=α

µα − µβ + i  µα + µβ + i  = θ1 (µα ) , µα − µβ − i µα + µβ − i

(109)

M M3  Y vδ − vγ − i  vδ − vγ − i  Y2  vδ − µα + i/2  vδ − µα + i/2  = θ2 (vδ ) . vδ − vγ + i vδ − vγ + i α=1 vδ − µα − i/2 vδ − µα − i/2

(110)

γ6=δ

Using the BA equations, we can study the boundary contributions to the thermodynamic quantities, such as magnetic susceptibility and specific heat, etc.

References [1] [2] [3] [4] [5]

Y.P. Wang, Phys. Rev. B60 (1999) 13. H.J. de Vega and E. Lopes, Phys. Rev. Lett. 67 (1991) 489. J.H. Perk and C.L. Schultz, Phys. Lett. A84 (1981) 3759. E. Lopes, Nucl. Phys. B370 (1992) 636. L. Mezincescue and R.I. Nepomechie, J. Phys. A24 (1991) L19.

[6] L. Mezincescue and R.I. Nepomechie, Int. J. Mod. Phys. A6 (1991) 2497. [7] R.H. Yue and H. Liang, High Energy Physics and Nuclear Physics 20 (1996) 514. [8] R.H. Yue, H. Fan, and B.Y. Hou, Nucl. Phys. B462 (1996) 167.