Ising Model on an Infinite Ladder Lattice

Transcription

1 Coun. Theor. Phys. (Beijing, China 48 (2007 pp c International Acaeic Publishers Vol. 48, No. 3, Septeber 15, 2007 Ising Moel on an Infinite Laer Lattice GAO Xing-Ru 2,3, an YANG Zhan-Ru 1,2 1 CCAST (Worl Laboratory, P.O. Box 8730, Beijing , China 2 Departent of Physics an Institute of Theoretical Physics, Beijing Noral University, Beijing , China 3 Funaental Teaching Departent, Beijing Union University, Beijing , China (Receive Noveber 20, 2006 Abstract In this paper we propose an Ising oel on an infinite laer lattice, which is ae of two infinite Ising spin chains with interactions. It is essentially a quasi-one-iessional Ising oel because the length of the laer lattice is infinite, while its with is finite. We investigate the phase transition an ynaic behavior of Ising oel on this quasi-one-iessional syste. We use the generalize transfer atrix etho to investigate the phase transition of the syste. It is foun that there is no nonzero teperature phase transition in this syste. At the sae tie, we are intereste in Glauber ynaics. Base on that, we obtain the tie evolution of the local spin agnetization by exactly solving a set of aster equations. PACS nubers: Ht Key wors: Ising oel, general transfer atrix, phase transition, Glauber ynaics, critical slow own 1 Introuction As we know, the one-iensional Ising oel with nearest neighbor interactions in a ost siple spin oel, the partition function of which has been exactly calculate by spin variables transforation an transfer atrix ethos. It has been foun that there is no finite teperature phase transition, naely, the critical teperature is zero. Since then various variations of the oel were investigate, such as one-iensional long-range interaction Ising oe, [13 non-perio Ising chain, alternate interactions linear Ising chain, [4,5 an so on. On the irreversible ynaical proble, Glauber [6 an Kawasaki [7 have respectively investigate ynaical behaviour of Ising syste uner un-conservation an conservation conition of orer paraeter. In this paper, we only focus on Glauber ynaics. Glauber assue that only single-spin-flip is allowe each tie in the process of evolution, he then exactly solve the evolution equation aster equation, an successfully obtaine the solution exhibiting the critical slowing own phenoena. In the last several ecaes the Glauber ynaics was extensively extene to one-iensional syste with ifferent kins of interactions, such as nonperioic Ising chains [8 an alternating linear chains. [9 Droz, Kaphorst Leal a Silva, Malaspina (DKM investigate the critical ynaics of an Ising chain with two ifferent near-neighbor interaction strength. [10 Z.R. Yang [11 consiere one-iensional Ising oel with not only nearest-neighbor but also next nearest-neighbor interactions in the absence of external fiel. The higher-iensional Ising syste was extensively investigate. [12 Very recently, the Glauber ynaics was also extene to Gauss spin syste. [13 However, for higher-iensional spin syste, the solution of aster equation cannot be exact, this is because the evolution equation becoes non-linear. Usually it is solve by ecoupling approxiation ethos, [14 renoralizationgroup ethos, [15 an coputer siulations. [1619 In this paper we investigate an Ising oel on an infinite laer lattice. The length of this laer lattice is infinite while the with of the laer is finite. Thus it essentially is a quasi-one-iensional lattice. Our purpose is to stuy the phase transition an ynaical behaviour of Ising oel on such a syste. This paper is organize as follows. In Sec. 2, we escribe the oel an investigate equilibriu phase transition. In Sec. 3, we stuy Glauber ynaics, an give the expression of tie evolution of local agnetization. Finally, in Sec. 4, a short conclusion is presente. 2 Moel an Phase Transition We now consier the Ising oel on an infinite laer lattice, which is constructe with two parallel lattice chains an 2(2N + 1 spins locate on the lattice sites, as shown in Fig. 1. Fig. 1 The infinite laer Ising syste an perioic bounary conition σ N = σ N+1, s N = s N+1. E-ail: Gao xingru@yahoo.co

2 554 GAO Xing-Ru an YANG Zhan-Ru Vol. 48 Ising spin σ i an s i (i = N,..., N take the values ±1 only. Suppose the Hailtonian of the syste is βh = k 1 N i=n + k 3 N i=n σ i σ i+1 + k 2 N i=n s i s i+1 σ i s i, (1 where β = 1/k B T, k B is Boltzann constant, an T teperature, k i = βj i (i = 1, 2, 3 are interaction paraeters, in which J i represents the exchange integral between nearest neighbor particles. For ferroagnetic oel, J i > 0, k 1 an k 2 respectively inicate interactions between spins on the sae chain, an k 3 interactions between spins on the ifferent chain. To investigate phase transition property of the syste, we firstly calculate the partition function of the oel Hailtonian (1. As usual, the partition function is efine as Z N = N exp (k 1 σ i σ i+1 + k 2 s i s i+1 + k 3 σ i s i. (2 {σ i,s i} i=n By introucing the following 4 4 transfer atrix: [ f(σ, s, σ, s = exp k 1 σσ + k 2 ss k 3(σs + σ s f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 = f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 f (1,1(1,1 exp (k 1 +k 2 +k 3 exp (k 1 k 2 exp (k 1 +k 2 exp (k 1 k 2 +k 3 exp (k = 1 k 2 exp (k 1 +k 2 k 3 exp (k 1 k 2 k 3 exp (k 1 +k 2 exp (k 1 +k 2 exp (k 1 k 2 k 3 exp (k 1 +k 2 k 3 exp (k 1 k 2. (3 exp (k 1 k 2 +k 3 exp (k 1 +k 2 exp (k 1 k 2 exp (k 1 +k 2 +k 3 The partition function Z can be rewritten as Z N = σ i,s i f(σ N s N σ N+1 s N+1 f(σ 1 s 1 σ 2 s 2 f(σ 2 s 2 σ 3 s 3 f(σ N s N σ N s N = f 2N+1 (σ N s N σ N s N = Tr f 2N+1 = λ 2N λ 2N λ 2N λ 2N+1 4, (4 σ N,s N where we have consiere the perioic bounary conition: σ N+1 = σ N, s N+1 = s N, an λ i (i = 1, 2, 3, 4 are the four eigenvalues of the transfer atrix f, which can be foun through solving the following secular equation: exp (k 1 +k 2 +k 3 λ exp (k 1 k 2 exp (k 1 +k 2 exp (k 1 k 2 +k 3 exp (k 1 k 2 exp (k 1 +k 2 k 3 λ exp (k 1 k 2 k 3 exp (k 1 +k 2 = 0. (5 exp (k 1 +k 2 exp (k 1 k 2 k 3 exp (k 1 +k 2 k 3 λ exp (k 1 k 2 exp (k 1 k 2 +k 3 exp (k 1 +k 2 exp (k 1 k 2 exp (k 1 +k 2 +k 3 λ The explicit expressions of the eigenvalues are given as follows: λ 1 = 2 cosh(k 1 +k 2 cosh k 3 + cosh(2k 1 +2k 2 (cosh 2k cosh(2k 1 2k 2 + cosh 2k 3 + 1, λ 2 = 2 cosh(k 1 +k 2 cosh k 3 cosh(2k 1 +2k 2 (cosh 2k cosh(2k 1 2k 2 + cosh 2k 3 + 1, λ 3 = 2 sinh(k 1 +k 2 cosh k 3 + cosh(2k 1 +2k 2 (cosh 2k cosh(2k 1 2k 2 cosh 2k 3 1, λ 4 = 2 sinh(k 1 +k 2 cosh k 3 cosh(2k 1 +2k 2 (cosh 2k cosh(2k 1 2k 2 cosh 2k 3 1. (6 To copare the four eigenvalues, we set A = 2 cosh(k 1 +k 2 cosh k 3, A = 2 sinh(k 1 +k 2 cosh k 3, B = cosh(2k 1 +2k 2 (cosh 2k 3 1, C = 2 cosh(2k 1 2k 2, D = cosh 2k (7 Thus λ i (i = 1, 2, 3, 4 can be rewritten as λ 1 = A + B + C + D, λ 2 = A B + C + D, λ 3 = A + B + C D, λ 4 = A B + C D. (8 Because of the function cosh x 1, the values of A, B, C, an D are non-negative. The value of A is always larger than A because of cosh x sinh x. Coparing the above four expressions of eigenvalues, it is obvious that the eigenvalue λ 1 is the largest one aong the four eigenvalues. In the theroynaics liit (N, the free energy per site can be expresse as follows: F = li N [ k B T 2(2N + 1 ln Z N = li N [ k B T 2(2N + 1 ln(λ2n λ 2N λ 2N λ 2N+1 4

3 No. 3 Ising Moel on an Infinite Laer Lattice 555 k B T [ ( λ2 2N+1 ( λ3 2N+1 ( λ4 2N+1 ( = li ln λ2n = k BT ln λ 1. (9 N 2(2N + 1 λ 1 λ 1 λ 1 2 Obviously, function F is only relate to the largest eigenvalue λ 1. Fro Eq. (7, it is very evient that the values of A, B, C, an D are continuous an graually increase with the increase of k 1, k 2, an k 3. Therefore accoring to the forula (6, the value λ 1 is also continuously increase with the increase of k 1, k 2, an k 3, an approaches infinite at k 1, k 2, k 3. Fro Eq. (9, it is clear that the free energy is ivergent just only at zero teperature (T C = 0. Therefore we can obtain the conclusion that there is no nonzero teperature phase transition in our oel. If we choose k 1 = k, k 2 = 0, k 3 = 0, fro Eq. (6, the four eigenvalues reuce to λ 1 = 4 cosh k, λ 2 = 0, λ 3 = 4 sinh k, λ 4 = 0, (10 which reprouces precisely the result of one-iensional Ising chain. [20 3 Irreversible Dynaics In this section, we stuy the tie evolution of the syste fro the nonequilibriu state to the equilibriu state. The evolution originates fro interaction between spins an interaction between the syste an heat resource. Since the latter is stochastic, the process of evolution is consiere as a stochastic one, in which the spin can ranoly flip between values ± Master Equation Glauber [6 assue that in the transition process only single spin can change spin value each tie. Therefore the transition probability fro a spin configuration {σ N (t,... σ j (t,..., σ N (t, s N (t,..., s j (t,..., s N (t} to another one {σ N (t,..., σ j (t,..., σ N (t, s N (t,..., s j (t,..., s N (t} can be written as W j (σ j (t, which eans that only spin σ j (t flips, while the others reain unchange. Now let us suppose the probability istribution function of the syste being in configuration {σ N (t,..., σ j (t,..., σ N (t, s N (t,..., s j (t,..., s N (t} at tie t is P (σ N (t,..., σ N (t, s N (t,..., s N (t, then the tie erivative of P (σ N (t,..., σ N (t, s N (t,..., s N (t can be expresse as t P (σ N(t,..., σ N (t, s N (t,..., s N (t [ = W j (σ j (t + W j (s j (t P (σ N (t,..., σ N (t, s N (t,..., s N (t + j j W j (σ j (tp (σ N (t,..., σ j (t,..., σ N (t, s N (t,..., s N (t + j W j (s j (tp (σ N (t,..., σ N (t, s N (t,..., s j (t,..., s N (t, (11 where the first an secon ters on the right-han sie enote the ecrease of probability istribution function P (σ N (t,..., σ N (t, s N (t,..., s N (t per unit tie ue to transition of spin fro σ j (t to σ j (t an fro s j (t to s j (t, respectively, an the thir an fourth ters enote the increase of P (σ 1 (t,..., σ N (t, s 1 (t,..., s N (t per unit tie ue to transition of spin fro σ j (t to σ j (t an fro s j (t to s j (t, respectively. Equation (11 is usually calle aster equation. It is ost iportant that to solve the equation we ust firstly fin the transition probability W j (σ j (t (or W j (s j (t. In principle, it can be foun by solving quantu any-boy proble. However it is so ifficult that it cannot be one. Usually, we eploy soe general principle an arguent to eterinate the transition probability W j (σ j (t. They are, for exaple, W j (σ j shoul be positive an satisfies the etaile balance conition. Base on the above, we can write the following: W j (σ j W j (σ j = P e(σ N,..., σ j,..., σ N, s N,..., s N = exp[k 1σ j (σ j1 + σ j+1 k 3 σ j s j, (12 P e (σ N,..., σ j,..., σ N, s N,..., s N exp[k 1 σ j (σ j1 + σ j+1 + k 3 σ j s j W j (s j W j (s j = P e(σ N,..., σ N, s N,..., s j,..., s N = exp[k 2s j (s j1 + s j+1 k 3 σ j s j, (13 P e (σ N,..., σ N, s N,..., s j,..., s N exp[k 2 s j (s j1 + s j+1 + k 3 σ j s j where P e (σ N,..., σ j,..., σ N, s N,..., s N enotes the equilibriu canonical istribution function. However, equations (12 an (13 still o not uniquely eterine the transition probability W j (σ j. The exponential functions, which occur in the ratios (12 an (13, ay be written in the fors: exp(±k 1 σ j (σ j1 + σ j+1 = cosh[k 1 (σ j1 + σ j+1 {1 ± 1 } 2 σ j(σ j1 + σ j+1 tanh 2k 1, exp(±k 2 s j (s j1 + s j+1 = cosh[k 2 (s j1 + s j+1 {1 ± 1 } 2 s j(s j1 + s j+1 tanh 2k 2,

4 556 GAO Xing-Ru an YANG Zhan-Ru Vol. 48 exp(±k 3 σ j s j = (cosh k 3 (1 ± σ j s j tanh k 3. (14 Siilar to what Glauber i we choose W j (σ j an W j (s j as W j (σ j = 1 [1 2 α 1 2 δ 1σ j (σ j1 + σ j+1 3 σ j s j δ 1 3 s j (σ j1 + σ j+1, (15 W j (s j = 1 [1 2 α 1 2 δ 2s j (s j1 + s j+1 3 σ j s j δ 2 3 σ j (s j1 + s j+1 (16 with δ 1 = tanh(2k 1, δ 2 = tanh(2k 2, an 3 = tanh(k 3, where we have use the equality (14, which is vali special for Ising spin an α is an arbitrary constant inicating relaxation tie of non-interacting spin syste. We are intereste in the evolution of local agnetization (local orer paraeter. It is efine as σ k (t = σ k (tp (σ N (t,..., σ N (t, s N (t,..., s N (t, (17 {σ,s} s k (t = s k (tp (σ N (t,..., σ N (t, s N (t,..., s N (t. (18 {σ,s} Accoring to the above efinition an the aster equation (11, we can erive the tie-evolution equations of σ k (t an s k (t (see Appenix A, (αt σ k(t = σ k (t δ 1(σ k1 (t + σ k+1 (t+ 3 s k (t 1 2 δ 1 3 ( σ k (ts k (tσ k1(t + σ k (ts k (tσ k+1 (t, (19 (αt s k(t = s k (t δ 2(s k1 (t + s k+1 (t+ 3 σ k (t 1 2 δ 2 3 ( σ k (ts k (ts k1 (t + σ k (ts k (ts k+1 (t. (20 They are a set of two coupling equations. Noting that there are three-spin correlation functions in Eqs. (19 an (20. For getting the solution, we have to eploy soe approxiate technique. Here we use ecoupling technique as [11 σ k s k σ k1 σ k s k σ k1 = r k,k σ k1, σ k s k σ k+1 σ k s k σ k+1 = r k,k σ k+1, σ k s k s k1 σ k s k s k1 = r k,k s k1, σ k s k s k+1 σ k s k s k+1 = r k,k s k+1. (21 In which r k,k σ k s k is two-spin, locate at two ifferent spin chains, correlation function. For siplicity, we approxiately use r e k,k θ 0, an equilibriu correlation function, instea of r k,k, then equations (19 an (20 are siplifie as (αt σ k(t = σ k (t δ 1(σ k1 (t + σ k+1 (t + 3 s k (t, (22 where 3.2 Solutions of Master Equations (αt s k(t = s k (t δ 2(s k1 (t + s k+1 (t + 3 σ k (t, (23 δ 1 = δ 1 (1 θ 0 3, δ 2 = δ 2 (1 θ 0 3. (24 To solve the coupling aster equations (22 an (23, let us construct two generating functions, F (λ, t = λ k σ k (t, G(λ, t = λ k s k (t. (25 k= Accoring to Eqs. (22 an (23, the two generating functions F (λ, t an G(λ, t satisfy the ifferential equations, (αt F (λ, t = F (λ, t δ 1(λ 1 + λf (λ, t + 3 G(λ, t, Let us efine k= (αt G(λ, t = G(λ, t δ 2(λ 1 + λg(λ, t + 3 F (λ, t. (26 1 = δ 1(λ 1 + λ, 2 = δ 2(λ 1 + λ. (27 The ifferential equations (26 can be siplifie as (αt F (λ, t = 1F (λ, t + 3 G(λ, t, (αt G(λ, t = 2G(λ, t + 3 F (λ, t, (28 which are typical linear an hoogeneous ifferential equations. Eviently, we can obtain their solutions as follows: F (λ, t = F (λ, 0 β 2 1 exp(β 1 αt + G(λ, 0 exp(β 1 αt β 2 β 1 β 1 β 2 3

5 No. 3 Ising Moel on an Infinite Laer Lattice F (λ, 0 β 1 1 exp(β 2 αt + G(λ, 0 exp(β 2 αt, β 1 β 2 β 2 β 1 G(λ, t = F (λ, 0 (β 1 1 (β 2 1 (β 2 β 1 3 exp(β 1 αt + G(λ, 0 β 1 1 β 1 β 2 exp(β 1 αt + F (λ, 0 (β 1 1 (β 2 1 exp(β 2 αt + G(λ, 0 β 2 1 exp(β 2 αt, (29 (β 1 β 2 3 β 2 β 1 where F (λ, 0 an G(λ, 0 are respectively the initial values of F (λ, t an G(λ, t, an β 1 an β 2 are respectively as follows: β 1 = [(1 2 2 /43 2, β 2 = [(1 2 2 /43 2. ( If we consier the case that k 1 is close to k 2, an thus ( 1 2 /2 3 1, then the expression (30 can be expresse approxiately as β 1 = ( 1 2 2, β 2 = ( ( We note that each ter on the right-han sie of expression (29 can be expresse in virtue of the Bessel functions of the iaginary arguent. The etaile process is given in Appenix B. Then the four ters of F (λ, t in the expression (29 can be written respectively as follows: F (λ, 0 β 2 1 exp(β 1 αt = 1 β 2 β 1 2 F (λ, 0 exp[( ηαt λ k I (ηαt I k2 ( + αt + } [ I k2+1 ( + αt + I k21 ( + αt, ( G(λ, 0 exp(β 1 αt = 1 β 1 β 2 2 G(λ, 0 exp[( ηαt λ k I (ηαt (1 η I k2 ( + αt η } 2 [ I k2+2( + αt + I k22 ( + αt, (33 F (λ, 0 β 1 1 exp(β 2 αt = 1 β 1 β 2 2 F (λ, 0 exp[(1 3 ηαt λ k I (ηαt I k2 ( + αt + } [ I k2+1 ( + αt + I k21 ( + αt, ( G(λ, 0 exp(β 2 αt = 1 β 2 β 1 2 G(λ, 0 exp[(1 3 ηαt λ k I (ηαt (1 + η I k2 ( + αt + η } 2 [ I k2+2( + αt + I k22 ( + αt, (35 where I (x is the oifie Bessel function, at the sae tie +, an η are respectively + = 1 2 [tanh(2k 2 + tanh(2k 1 (1 θ 0 tanh k 3, = 1 2 [tanh(2k 2 tanh(2k 1 (1 θ 0 tanh k 3, η = 2. (36 In the sae way, the four ters of G(λ, t in the expression (29 can be written respectively as follows: F (λ, 0 (β 1 1 (β 2 1 (β 2 β 1 3 exp(β 1 αt = 1 4 F (λ, 0 exp[( ηαt η λ k I (ηαt 2( 1 I k2 ( + αt + η } [ I k2+2 ( + αt + I k22 ( + αt, ( G(λ, 0 β 1 1 β 1 β 2 exp(β 1 αt = 1 2 G(λ, 0 exp[( ηαt λ k I (ηαt I k2 ( + αt + } [ I k2+1 ( + αt + I k21 ( + αt, ( F (λ, 0 (β 1 1 (β 2 1 (β 1 β 2 3 exp(β 2 αt

6 558 GAO Xing-Ru an YANG Zhan-Ru Vol. 48 = 1 4 F (λ, 0 exp[(1 3 ηαt η λ k I (ηαt 2( 1 I k2 ( + αt + η } [ I k2+2 ( + αt + I k22 ( + αt, ( G(λ, 0 β 2 1 β 2 β 1 exp(β 2 αt = 1 2 G(λ, 0 exp[(1 3 ηαt λ k I (ηαt I k2 ( + αt } [ I k2+1 ( + αt + I k21 ( + αt. ( We consier the case in which all of the spin expectations σ k (t an s k (t vanish initially except for one, which we ay choose to be the one at the origin, σ k (0 = δ k,0, s k (0 = δ k,0. (41 Fro the forula (25, the initial value of the generating function is just one (F (λ, 0 = G(λ, 0 = 1. For convenience, let us assue that the spin expectation is σ k (t = A(1 + A(2 + A(3 + A(4, s k (t = B(1 + B(2 + B(3 + B(4. (42 By coparing with the forula (25, we conclue that A(1 = 1 2 exp[( ηαt I (ηαt I k2 ( + αt + } [ I k2+1 ( + αt + I k21 ( + αt, 2 3 A(2 = 1 2 exp[( ηαt I (ηαt {(1 η I k2 ( + αt η } [ I k2+2 ( + αt + I k22 ( + αt, A(3 = 1 2 exp[(1 3 ηαt I (ηαt I k2 ( + αt + } [ I k2+1 ( + αt + I k21 ( + αt, 2 3 an A(4 = 1 2 exp[(1 3 ηαt I (ηαt {(1 η I k2 ( + αt η } [ I k2+2 ( + αt + I k22 ( + αt ; ( B(1 = 1 4 exp[( ηαt η I (ηαt 2( 1 I k2 ( + αt + η } [ I k2+2 ( + αt + I k22 ( + αt, 3 3 B(2 = 1 2 exp[( ηαt I (ηαt I k2 ( + αt + } [ I k2+1 ( + αt + I k21 ( + αt, 2 3 B(3 = 1 4 exp[(1 3 ηαt η I (ηαt 2( 1 I k2 ( + αt + η } [ I k2+2 ( + αt + I k22 ( + αt, 3 3 B(4 = 1 2 exp[(1 3 ηαt I (ηαt I k2 ( + αt } [ I k2+1 ( + αt + I k21 ( + αt. ( The expressions (42 (44 enote tie evolutions of the spin expectations σ k (t an s k (t.

7 No. 3 Ising Moel on an Infinite Laer Lattice Discussion of the Result Now we iscuss our result. At first we investigate the asyptotic behavior of σ k (t an s k (t as t by the asyptotic expressions of I (x. Then we iscuss two special cases: (i k 1 = k 2 = k an k 3 0, which eans the two sae Ising chains are epenent; (ii k 1 = k 2 = k an k 3 = 0, which eans the two sae Ising chains are inepenent. Since the asyptotic expressions of I (x as x is I (x = 1 2 x exp(x, I (x = (1 1 2 exp(x. (45 x The asyptotic expressions of σ k (t an s k (t as t are ( σ k (t exp(( η + + αt = exp t (, s k (t exp(( η + + αt = exp t τ τ where τ =, (46 1 α(1 3 2η +, (47 which is the relaxation tie of evolution of the syste. Fro the above expression an Eq. (36 we can fin that the relaxation tie epens on the teperature of the syste. For our oel, in Sec. 2, we have known that the critical teperature is T c = 0. At the critical point (T c = 0, k i (i = 1, 2, 3, then 3 = tanh(k 3 = 1, + = (1 θ 0 an η = 0, therefore the relaxation tie is 1 τ = α(θ 0 1. (48 We can prove that the equilibriu value θ 0 = 1 (See Appenix C. Fro the expression (48, we get the relaxation tie is infinite at the critical point. It iplies that the syste will very slowly approach the equilibriu state, thus the syste will exhibit the critical slowing own phenoenon. Now we iscuss a special case in which the interactions are k 1 = k 2 = k an k 3 0. The paraeters in the expression (36 are as follows: + = tanh(2k(1 tanh k 3, = 0, η = 0. (49 We apply the relation of the oifie Bessel function The expression (43 reuces to an expression (44 reuces to I (0 = δ,0. (50 A(1 = 1 2 exp[(1 + 3αt I k ( + αt, A(2 = 1 2 exp[(1 + 3αt I k ( + αt, A(3 = 1 2 exp[(1 3αt I k ( + αt, A(4 = 1 2 exp[(1 3αt I k ( + αt, (51 B(1 = 1 2 exp[(1 + 3αt I k ( + αt, B(2 = 1 2 exp[(1 + 3αt I k ( + αt, B(3 = 1 2 exp[(1 3αt I k ( + αt, B(4 = 1 2 exp[(1 3αt I k ( + αt. (52 Therefore the expression of the spin expectation σ k (t an s k (t is copletely the sae an very siple. σ k (t = exp[(1 + 3 αt I k ( + αt, s k (t = exp[(1 + 3 αt I k ( + αt, (53 As t, using the asyptotic expressions of I k ( + αt, we obtain the asyptotic expressions of σ k (t an s k (t as follows: σ k (t = s k (t exp[( αt. (54 So we can get the relaxation tie 1 τ = α(1 3 + = 1 α(1 tanh 2k(1 tanh k 3. (55 At the critical point (T c = 0, k, an k 3, the relaxation tie is also infinite. The syste exhibits the critical slowing own phenoenon again. When k 3 = 0, then 3 = 0, it iplies that there is no interaction between the two infinite chains of the laer. The syste becoes two inepenent chains. The expression (54 reprouces precisely Glauber s one-iensional chain result. [6 4 Conclusions In this paper, we stuie the equilibriu phase transition an ynaical properties of Ising oel on an infinite laer lattice. In the first part, we investigate phase transition property of the syste by using the transfer atrix

8 560 GAO Xing-Ru an YANG Zhan-Ru Vol. 48 etho. It is foun that there is no nonzero teperature phase transition as expecte. In the special case of k 1 = k, k 2 = 0, k 3 = 0, our oel reprouces precisely the result of one-iensional Ising chain. In the secon part, we stuy the tie evolution of the syste fro the nonequilibriu state to the equilibriu state. Base on Glauber assuption, we solve a set of ecoupling aster equations an obtain approxiation solution of the aster equations uner a special conition (( 1 2 / We foun the syste will exhibit the critical slowing own phenoenon. Finally, let us point out that the result about ynaics in this paper is obtaine uner the conition ( 1 2 /2 3 1, which iplies that the conclusion is vali, only when k 1 is close to k 2 or k 3 = J 3 /k β T is large enough, i.e., J 3 large enough or teperature low enough. Therefore for a laer spin syste with strong coupling between both spin chains or a low-teperature laer spin syste, our result an conclusion are vali, especially, at zero teperature the conclusion is true an creible. Appenix A: Derivation of Eqs. (19 an (20 Let us ultiply both sies of the aster equation (11 by σ k (t an su over all values of the σ an s variables, we obtain σ k (tp (σ 1 (t,..., σ N (t, s 1 (t,..., s N (t = 2 σ k (tw k (σ k (tp (σ 1 (t,..., σ N (t, s 1 (t,..., s N (t. (A1 t {σ,s} {σ,s} Base on the efinition (17 of σ k (t, we have t σ k(t = 2 σ k (tw k (σ k (t. (A2 Substituting the for (15 for the transition probabilities into Eq. (A2 an consiering σk 2 = 1, we obtain a ifferential equation for the expectation values σ k (t, (αt σ k(t = σ k (t δ 1(σ k1 (t+σ k+1 (t+ 3 s k (t 1 2 δ 1 3 ( σ k (ts k (tσ k1 (t + σ k (ts k (tσ k+1 (t. (A3 Siilarly we can get a ifferential equation for the expectation values s k (t. So we erive a set of coupling equations (19 an (20 fro the aster equation (11. Appenix B: Solutions for Generating Functions Fro the expressions (31 an (27, we have (β 1 1 (β 2 1 = 1 [ 2 3 (β 1 β (λ 2 + λ , (A4 β 1 1 = 1 [ 1 + (λ 1 + λ, (A5 β 1 β (β 2 1 (β 2 β 1 = 1 [ 1 (λ 1 + λ, (A where = (1/2[tanh(2k 2 tanh(2k 1 [1 θ 0 tanh(k 3. We also have [( exp(β 1 αt = exp [ 2 αt exp[ 2 (λ1 + λαt exp (λ 2 + λ 2 αt, (A7 8 3 [( exp(β 2 αt = exp [ αt exp[ 2 (λ1 + λαt exp 2 (λ 2 + λ 2 αt, (A8 8 3 where + = (1/2[tanh(2k 2 + tanh(2k 1 [1 θ 0 tanh(k 3. Noting the generating function of the Bessel function of iaginary arguent, [6 an we obtain an x exp[ 2 (λ1 + λ = x exp[ 2 (λ2 + λ 2 = [( exp(β 1 αt = exp αt [( exp(β 2 αt = exp αt k= k= k= k= n, n, λ k I k (x, λ 2k I k (x, (A9 (A10 ( λ n+2 2 I n ( + αt I αt, (A11 ( λ n+2 I n ( + αt I 2 αt. (A12

9 No. 3 Ising Moel on an Infinite Laer Lattice 561 Fro the forulas (A6 an (A11, we obtain the first ter of F (λ, t, F (λ, 0 β 2 1 β 2 β 1 exp(β 1 αt = 1 2 F (λ, 0 exp [( αt {[ = 1 [( 2 F (λ, 0 exp αt n, (λ 1 + λ { n, ( λ n+2 2 I n ( + αt I αt ( λ n+2 2 } I n ( + αt I αt [ ( λ n+21 2 ( I n ( + αt I αt + λ n } I n ( + αt I αt. (A Consier the following property of the Bessel function of iaginary arguent: λ k I k2 (x I (y, λ k I k21 (x I (y, λ n+2 I n (x I (y = n, n, λ n+21 I n (x I (y = λ k I k2+1 (x I (y, n, λ n+2+1 I n (x I (y = siplifying the forula (A13, we can get the first ter (32 of F (λ, t. In the sae way, the secon ter (33, the thir ter (34 an the fourth ter (35 can be obtaine. Appenix C: Proof of θ 0 = 1 We efine an equilibriu correlation function an the partition function is We have Fro the efinition of σ i s i, we can obtain Z N = Z N k 3 = {σ i,s i} = {σ i,s i} σ i s i = 1 Z N {σ i,s i} ( N i=n exp r e k,k θ 0, (A14 (A15 N (k 1 σ i σ i+1 + k 2 s i s i+1 + k 3 σ i s i. (A16 i=n σ i s i exp (2N + 1σ i s i exp {σ i,s i} σ i s i exp N (k 1 σ i σ i+1 + k 2 s i s i+1 + k 3 σ i s i i=n N (k 1 σ i σ i+1 + k 2 s i s i+1 + k 3 σ i s i. (A17 i=n N (k 1 σ i σ i+1 + k 2 s i s i+1 + k 3 σ i s i, (A18 i=n Fro the transfer atrix, we have the expression of the partition function 1 Z N Z N k 3 = (2N + 1 σ i s i. (A19 Z N = λ 2N λ 2N λ 2N λ 2N+1 4, (A20 where λ i (i = 1, 2, 3, 4 are the four eigenvalues an λ 1 is the axiu value aong these four eigenvalues. Therefore when N is very large (N, we have Z N = λ 2N+1 1. (A21 Fro the forula (A19 we have where σ i s i = 1 λ 1 λ 1 k 3, λ 1 = 2 cosh(k 1 +k 2 cosh k 3 + cosh(2k 1 +2k 2 (cosh 2k cosh(2k 1 2k 2 + cosh 2k 3 + 1, since cosh 2x = 2 cosh 2 x 1. λ 1 can be rewritten as [ λ 1 = 2 cosh(k 1 +k 2 cosh k 3 + cosh 2 (k 1 +k 2 cosh 2 k 3 cosh 2 (k 1 +k 2 + cosh 2 (k 1 k 2, (A22

10 562 GAO Xing-Ru an YANG Zhan-Ru Vol. 48 fro the above forula, we have λ 1 cosh = 2 cosh(k (k 1 +k 2 cosh k 3 sinh k 1 +k 2 sinh k k 3 cosh 2 (k 1 +k 2 cosh 2 k 3 cosh 2 (k 1 +k 2 + cosh 2 (k 1 k 2 = 2 cosh(k 1 +k 2 sinh k cosh k 3. cosh 2 k cosh 2 (k 1 k 2 / cosh 2 (k 1 +k 2 Because an tanh k i = 1 when k i (i = 1, 2, we have then an Accoring to the forula (A22, we have cosh(k 1 k 2 cosh(k 1 +k 2 = 1 tanh k 1 tanh k tanh k 1 tanh k 2, cosh(k 1 k 2 cosh(k 1 +k 2 = 0, λ 1 k 3 = 2 cosh(k 1 +k 2 sinh k 3 (1 + coth k 3, λ 1 = 2 cosh(k 1 +k 2 cosh k 3 (1 + tanh k 3. θ 0 = σ i s i = tanh k 3 (1 + coth k 3 (1 + tanh k 3 = 1. Acknowlegents One (Gao of us woul like to thank Profs. W.A. Guo an Z. Gao for their goo suggestions. One (Gao of us woul also like to thank Profs. X.M. Kong an J.X. Le for their help an valuable iscussions. One (Gao of us is grateful to Drs. S.H. Li an J.Q. Tao for helpful iscussions. References [1 George A. Baker, Jr., Phys. Rev. 130 ( [2 J.S. Hoye, Phys. Rev. B 6 ( [3 Per Bak, Phys. Rev. Lett. 49 ( [4 M. Barati an A. Raazani, Phys. Rev. B 64 ( [5 Maria Gloria Pini, Phys. Rev. B 48 ( [6 R.J. Glauber, J. Math. Phys. 4 ( [7 K. Kawasaki, Phys. Rev. 145 ( [8 Pei-Qing Tong, Phys. Rev. E 56 ( [9 L.L. Goncalves, Phys. Rev. E 63 ( [10 M. Droz, J. Kaphorst Leal a Silva, an A. Malaspinas, Phys. Lett. A 115 ( [11 Z.R. Yang, Phys. Rev. B 46 ( [12 B. Sutherlan, Phys. Rev. Lett. 31 ( [13 J.Y. Zhu an Z.R. Yang, Phys. Rev. E 61 ( [14 K.-t. Lrung, Phys. Rev. E 63 ( [15 M.C. Yalabik an J.D. Gunton, Phys. Rev. B 25 ( [16 R. Bausch, V. Doh, H.K. Janssen, an R.K.P. Zia, Phys. Rev. Lett. 47 ( [17 S. Wansleben an P. Lanau, J. Appl. Phys. 61 ( [18 S. Tang an D.P. Lanau, Phys. Rev. B 36 ( [19 E. Pytte, Phys. Rev. B 34 ( [20 Z.R. Yang, Progress in Phys. 1(2 (