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Stability of a two-sublattice spin-glass model Carlos S. O. Yokoi∗ 6 Instituto de F´ısica, Universidade de São Paulo, 0 0 Caixa Postal 66318, 05315-970 São Paulo, SP, Brazil 2 n a Francisco A. da Costa† J 0 Departamento de F´ısica Teórica e Experimental, 3 Universidade Federal do Rio Grande do Norte, ] n Caixa Postal 1641, 59072-970 Natal, RN, Brazil n - s (Dated: February 6, 2008) i d . t Abstract a m We study the stability of the replica-symmetric solution of a two-sublattice infinite-range spin- - d n glass model, which can describe the transition from antiferromagnetic to spin glass state. The o c eigenvalues associated with replica-symmetric perturbations are in general complex. The natural [ generalization of the usual stability condition is to require the real part of these eigenvalues to be 1 v positive. The necessary and sufficient conditions for all the roots of the secular equation to have 3 4 6 positive real parts is given by the Hurwitz criterion. The generalized stability condition allows a 1 0 consistent analysis of the phase diagram within the replica-symmetric approximation. 6 0 t/ PACS numbers: 05.50.+q,64.60.-i,75.10.Nr,75.50.Lk a m - d n o c : v i X r a ∗Electronic address: [email protected] †Electronic address: [email protected] 1 I. INTRODUCTION The infinite-range Sherrington-Kirkpatrick (SK) model [1] for a spin glass has attracted considerable attention over the past decades [2, 3, 4]. These investigations have revealed highlynon-trivialpropertiessuchastheinstabilityofreplica-symmetric (RS)solution[5]and the replica-symmetry-breaking scheme to produce a stable solution [6, 7, 8, 9]. Most studies have concentrated on situations where the exchange distributions are either symmetric or with an additional ferromagnetic interaction. More recently a two-sublattice version of the SK model was introduced [10, 11, 12, 13] to allow for antiferromagnetic interactions between different sublattices. Such extension is quite natural in view of the existence of many experimental systems such as Fe Mg Cl [14, 15, 16] and Fe Mn TiO [17, 18], x 1−x 2 x 1−x 3 which exhibit transition from and Ising antiferromagnetic into an Ising spin glass state for certain range of x values. In contrast to the standard SK model, in the two-sublattice SK model with antiferromagnetic intersublattice interactions, the ordered (antiferromagnetic) phase extends to finite fields and the de Almeida-Thouless instability line [5] has distinct branches in the paramagnetic and antiferromagnetic phases, which do not meet at a first- order transition [10, 11, 12, 13]. Experimental determination of the field-temperature phase diagram in Fe Mn TiO , as well as the de Almeida-Thouless instability line [19], are in x 1−x 3 qualitative agreement with mean-field results [13]. In the previous studies of this model the stability of the RS solution against transversal fluctuations, i.e., outside the RS space, has already been investigated [10, 11, 12, 13], and the stability against longitudinal fluctuations, i.e., inside the RS space, was also briefly considered [12]. The stability of the RS solution against transversal fluctuations is important to establish whether replica symmetry breaking is necessary. The stability against longitudinal fluctuations, however, is also necessary to ensure the validity of RS solution. For certain parameter values of the two-sublattice SK model there may be up to three RS solutions, all of them stable against transversal fluctuations. In such a situation the analysis of the stability against longitudinal fluctuations is important for a consistent study of the phase diagram by eliminating unstable solutions. In this work we remedy the lack of such investigation by a detailed numerical and ana- lytical study of the eigenvalues associated with longitudinal fluctuations. Surprisingly, these eigenvalues are in general complex. It is natural to assume that stability condition should 2 require the real part of these eigenvalues to be positive. The necessary and sufficient condition for allthe roots of the secular equation to have positive real part is given by the Hurwitz criterion. We show that this generalized stability condition allows a consistent study of the phase diagram within the RS approximation. II. THE MODEL We consider a system of 2N Ising spins S = 1 located at the sites of two identical i ± sublattices A and B. The interactions are described by the Hamiltonian = J S S J′ S S J′ S S H S , (1) H − ij i j − ij i j − ij i j − i i∈AX,j∈B (iXj)∈A (iXj)∈B Xi where the first sum is over all distinct pairs of spins belonging to different sublattices, the secondandthirdonesrefertoalldistinct pairsofspinsbelongingtothesamesublattices, and the last sum is over all spins in the two sublattices. J is the exchange interaction between ij spins in different sublattices, J′ is the exchange interaction between spins in the same ij sublattice, and H is the applied magnetic field. The exchange interactions are independent, quenched, Gaussian random variables with mean values J J′ J = 0, J′ = 0, (2) h ijiJ N h ijiJ N and variances J2 J′2 J2 J 2 = , J′2 J′ 2 = . (3) h ijiJ −h ijiJ N h ij iJ −h ijiJ N The mean intrasublattice interactions will always assumed to be ferromagnetic (J′ > 0), 0 whereas the mean intersublattice interactions may be ferromagnetic (J > 0) or antiferro- 0 magnetic (J < 0). 0 Thestandardapproachtocomputethequenchedaverageistointroducennon-interacting replicas α = 1,2,...,nofthesystem, calculatetheannealed averages andthen takethelimit n 0 [2, 4]. In this replica method the free energy per spin f is given by → 1 1 1 f = lim φ, φ = lim ln Zn , (4) 2β n→0 n −N→∞ N h iJ where β = 1/k T and Zn is the partition function of n replicas of the system. Performing B the average of Zn over the random couplings we find β2J2n n β2J′2n n Zn = Tr exp N +βJ′ 1 βH (mα +mα) h iJ − (− 2 0N − 2 (cid:18) − N(cid:19)− α A B X 3 βJ′ βJ mαmα 0 (mα)2 +(mα)2 β2J2 qαβqαβ − 0 A B − 2 A B − A B Xα Xα h i (Xαβ) β2J′2 2 2 qαβ + qαβ , (5) − 2 A B  (Xαβ)(cid:20)(cid:16) (cid:17) (cid:16) (cid:17) (cid:21) where (αβ) denotes distinct pairs of replicas and we have introduced the sublattice magne- tization and sublattice overlap function of the replicas, 1 1 mα = Sα, qαβ = SαSβ, (X = A,B). (6) X N i X N i i i∈X i∈X X X The trace over the spin variables in (5) can be performed by taking into account the con- straints (6) by means of the identities ∞ i∞ Ndλα 1 1 = dmα X exp Nλα mα Sα (X = A,B), (7) Z−∞ X Z−i∞ 2πi "− X X − N i∈X i !# X and ∞ i∞ Ndλαβ 1 1 = dqαβ X exp Nλαβ qαβ SαSβ (X = A,B). (8) Z−∞ X Z−i∞ 2πi "− X X − N i∈X i i !# X We then obtain ∞ i∞ Ndλα ∞ i∞ Ndλα ∞ i∞ Ndλαβ Zn = dmα A dmα B dqαβ A h iJ Yα Z−∞ AZ−i∞ 2πi Z−∞ BZ−i∞ 2πi (Yαβ)Z−∞ A Z−i∞ 2πi ∞ i∞ Ndλαβ dqαβ B exp Nφ(mα,mα,qαβ,qαβ;λα,λα,λαβ,λαβ) , (9) ×Z−∞ B Z−i∞ 2πi h− A B A B A B A B i where β2J2n n β2J′2n n φ = +βJ′ 1 βH (mα +mα) βJ mαmα − 2 0N − 2 − N − A B − 0 A B (cid:18) (cid:19) α α X X βJ0′ (mα)2 +(mα)2 β2J2 qαβqαβ β2J′2 qαβ 2 + qαβ 2 − 2 A B − A B − 2 A B Xα h i (Xαβ) (Xαβ)(cid:20)(cid:16) (cid:17) (cid:16) (cid:17) (cid:21) + (λαmα +λαmα)+ λαβqαβ +λαβqαβ lnTr exp lnTr exp (,10) A A B B A A B B − HA − HB Xα (Xαβ)(cid:16) (cid:17) with and denoting the “effective sublattice Hamiltonians” A B H H = λαSα + λαβSαSβ (X = A,B). (11) HX X X Xα (Xαβ) In the limit of large N the integrations over the λ variables in (9) can be performed by the saddle-point method. The saddle point is given by Tr Sαexp Tr SαSβexp mα = HX, qαβ = HX, (X = A,B). (12) X Tr exp X Tr exp X X H H 4 These equations determine λ variables in terms of m and q variables. The remaining integrations over the m and q variables in (9) can be performed by the Laplace method in the limit of large N. The stationary-point equations are given by λα = βH +βJ′mα +βJ mα, λαβ = β2J′2qαβ +β2J2qαβ, (X = A,B), (13) X 0 X 0 X X X X where X is the sublattice complementary to X, i.e., if X = A then X = B, and vice versa. Substituting these results in the expression of φ given by Eq. (10) we find β2n βJ′ φ = (J2 +J′2)+βJ mαmα + 0 (mα)2 +(mα)2 +β2J2 qαβqαβ − 2 0 A B 2 A B A B Xα Xα h i (Xαβ) β2J′2 2 2 + qαβ + qαβ lnTr exp lnTr exp , (14) 2 A B − HA − HB (Xαβ)(cid:20)(cid:16) (cid:17) (cid:16) (cid:17) (cid:21) where we have discarded terms that vanish in the limit of large N. Analogously, the effective sublattice Hamiltonians (11) become = β H +J′mα +J mα Sα +β2 J′2qαβ +J2qαβ SαSβ, (X = A,B). (15) HX 0 X 0 X X X Xα (cid:16) (cid:17) (Xαβ)(cid:16) (cid:17) To evaluate the general expressions obtained thus far it is necessary to impose some structureonmandq variables. Thesimplest assumptioncorrespondstotheRSsolution[2, 4] obtained by assuming order parameters independent of replica indices, mα = m , qαβ = q , (X = A,B). (16) X X X X Proceeding in the usual way [2, 4], one finds that the saddle-point equations (12) and stationary-point equations (13) give the equations of state m = tanhH , q = tanh2H , (X = A,B), (17) X X X X h i D E where H = β H +J′m +J m + J′2q +J2q x , (X = A,B), (18) X 0 X 0 X X X (cid:18) q (cid:19) and the brackets without subscript denote Gaussian averages, h···i ∞ dx = e−x2/2( ). (19) h···i −∞ √2π ··· Z Analogously, the free energy per spin (4) becomes βJ2 βJ′2 J J′ f = (1 q )(1 q ) (1 q )2 +(1 q )2 + 0m m + 0 m2 +m2 − 4 − A − B − 8 − A − B 2 A B 4 A B 1 1 h i (cid:16) (cid:17) ln2coshH ln2coshH . (20) A B − 2β h i− 2β h i 5 III. THE STABILITY OF REPLICA-SYMMETRIC SOLUTION The validity of the RS solution (17) rests on the applicability of Laplace method used to perform the integrations over m and q variables for large N. The integral converges only if the stationary point (13) is a minimum of φ, i.e., only if the eigenvalues of the Hessian matrixformedbythesecondderivatives ofthefunctionφgivenbyequation(10)withrespect to the m and q variables are all positive. We can equivalently consider φ as a function of λ variables, related to m and q variables by means of Eq. (12). We will follow the latter approach because it leads to simpler calculations. The Hessian is a [n(n+1)/2] [n(n+1)/2] × matrix whose elements are 2 2 matrices given by × Gαβ Gαβ Gα(βγ) Gα(βγ) G(αβ)(γδ) G(αβ)(γδ) Gαβ = AA AB ,Gα(βγ) = AA AB ,G(αβ)(γδ) = AA AB ,       Gαβ Gαβ Gα(βγ) Gα(βγ) G(αβ)(γδ) G(αβ)(γδ) BA BB BA BB BA BB            (21) where ∂2φ ∂2φ ∂2φ Gαβ = , Gα(βγ) = , G(αβ)(γδ) = (X,Y = A,B). (22) XY ∂λα∂λβ XY ∂λα∂λ(βγ) XY ∂λαβ∂λγδ X Y X Y X Y At the stationary point of the RS solution (17) there are seven different types of 2 2 × elements of the Hessian matrix. We denote these elements by [5] Gαα = A, Gαβ = B, Gα(αβ) = C, G(αβ)α = C, Gα(βγ) = D, G(αβ)γ = D, G(αβ)(αβ) = P, G(αβ)(αγ) = Q, G(αβ)(γδ) = R, e (23) f where the indices α, β, γ and δ are all distinct and the tilde denotes the transpose of the matrix. We do not quote the lengthy expressions for these elements because only their linear combinations are needed in the calculation of the eigenvalues. The eigenvalues of the Hessian matrix can now be determined by finding the eigenvectors that divide the space into orthogonal subspaces closed to the permutation operation. The procedure are analogous to the case of the SK model [5] except that now the elements of the Hessian matrix are 2 2 matrices (23). These eigenvectors are [20]: n(n 3) transversal × − or replicon eigenvectors depending on two replica indices, 4(n 1) anomalous eigenvalues − depending on a single replica index, and 4 longitudinal eigenvectors independent of replica indices. 6 The eigenvalues associated with the transversal eigenvectors are found to be the eigenvalues of the 2 2 matrix × T = P 2Q+R, (24) − with elements T = (1 2q +r ) (βJ′)2(1 2q +r )2, (25) 11 A A A A − − − T = T = (βJ)2(1 2q +r )(1 2q +r ), (26) 12 21 A A B B − − − T = (1 2q +r ) (βJ′)2(1 2q +r )2, (27) 22 B B B B − − − where t = tanh3H , r = tanh4H , (X = A,B). (28) X X X X D E D E The necessary and sufficient condition for all the eigenvalues to be positive are T +T > 0 and T T T2 > 0, (29) 11 22 11 22 − 12 which are equivalent to the conditions T = 2 (βJ′)2(1 2q +r ) (βJ′)2(1 2q +r ) > 0, (30) 1 A A B B − − − − T = [1 (βJ′)2(1 2q +r )][1 (βJ′)2(1 2q +r )] 2 A A B B − − − − (βJ)4(1 2q +r )(1 2q +r ) > 0, (31) A A B B − − − in agreement with previous studies [10, 13]. A RS solution satisfying these conditions will be called transversally (T) stable, and T unstable otherwise. The eigenvalues associated with anomalous and longitudinal eigenvectors are the same in the limit n 0. They are found to be the eigenvalues of the 4 4 matrix → × A B D C L = − − , (32)   2C 2D P 4Q+3R  − −    where e f L = (1 q ) βJ′(1 q )2 +2(βJ′)2(t m )2, (33) 11 − A − 0 − A A − A L = (1 q ) βJ′(1 q )2 +2(βJ′)2(t m )2, (34) 22 − B − 0 − B B − B L = L = βJ (1 q )(1 q )+2(βJ)2(t m )(t m ), (35) 12 21 0 A B A A B B − − − − − 1 L = L = (t m )[1 βJ′(1 q ) (βJ′)2(1 4q +3r )], (36) 13 −2 31 A − A − 0 − A − − A A 7 1 L = L = (t m )[1 βJ′(1 q ) (βJ′)2(1 4q +3r )], (37) 24 −2 42 B − B − 0 − B − − B B 1 L = L = βJ (t m )(1 q ) (βJ)2(t m )(1 4q +3r ), (38) 14 41 0 B B A A A B B −2 − − − − − − 1 L = L = βJ (t m )(1 q ) (βJ)2(t m )(1 4q +3r ), (39) 23 32 0 A A B B B A A −2 − − − − − − L = (1 4q +3r )[1 (βJ′)2(1 4q +3r )]+2βJ′(t m )2, (40) 33 − A A − − A A 0 A − A L = (1 4q +3r )[1 (βJ′)2(1 4q +3r )]+2βJ′(t m )2, (41) 44 − B B − − B B 0 B − B L = L = (βJ)2(1 4q +3r )(1 4q +3r )+2βJ (t m )(t m ). (42) 34 43 A A B B 0 A A B B − − − − − The characteristic equation has the form λ4 a λ3 +a λ2 a λ+a = 0, (43) 1 2 3 4 − − where the coefficients a are n-th order traces of the matrix L. A numerical study of n equation (43) shows that the eigenvalues are complex for some values of the parameters of themodel,incontrastwithone-sublatticeSKmodelinwhichtheanomalousandlongitudinal eigenvalues never become complex [5]. Even though the Hessian matrix (21) for n > 1 is real and symmetric, in the limit n 0 there is no guarantee that the eigenvalues will → be real. In fact, complex longitudinal and anomalous eigenvalues also arise in the spin 1 one-sublattice infinite-range spin-glass model with crystal-field anisotropy [21, 22]. In general, therefore, the stability condition should require the real part of the eigenvalues to be positive. According to the Hurwitz criterion [23], the necessary and sufficient condition for all the roots of equation (43) to have positive real parts are a a 1 3 D = a > 0, D = = a a a > 0, (44) 1 1 2 (cid:12) (cid:12) 1 2 3 (cid:12) 1 a (cid:12) − (cid:12) 2 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) a a 0 0 (cid:12) (cid:12) 1 3 a a 0 1 3 (cid:12) (cid:12) (cid:12) 1 a a 0 (cid:12) D = (cid:12)(cid:12) 1 a a (cid:12)(cid:12) = a D a2a > 0, D = (cid:12)(cid:12) 2 4 (cid:12)(cid:12) = a D > 0. (45) 3 (cid:12) 2 4 (cid:12) 3 2 − 1 4 4 (cid:12) (cid:12) 4 3 (cid:12) (cid:12) (cid:12) 0 a1 a3 0 (cid:12) (cid:12) 0 a a (cid:12) (cid:12) (cid:12) (cid:12) 1 3 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 0 1 a a (cid:12) (cid:12) (cid:12) (cid:12) 2 4 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) These condition are equivalent to the following four cond(cid:12)itions: (cid:12) (cid:12) (cid:12) L = a > 0, (46) 1 1 L = D = a a a > 0, (47) 2 2 1 2 3 − L = D = a a a a2 a a2 > 0, (48) 3 3 1 2 3 − 3 − 4 1 L = a > 0. (49) 4 4 8 A RS solution satisfying these conditions will be called longitudinally (L) stable, and L unstable otherwise. IV. RESULTS OF THE STABILITY ANALYSIS In this section we present the results of the stability analysis of the RS solution for different values of the parameters of the model. Since the Hamiltonian (1) is invariant under the simultaneous transformations H H, S S , (50) i i → − → − it is sufficient to consider fields H 0. For H = 0 only one of the two solutions related by ≥ the global inversion symmetry has to be considered. A. Zero applied field 1. Ferromagnetic intersublattice interaction In zero applied field (H = 0) and ferromagnetic intersublattice interactions (J > 0) the 0 solutions of the set of equations (17) are of the form m = m = m, q = q = q. (51) A B A B Three types of solutions are possible: Paramagnetic (P) solution: m = 0,q = 0. • Spin Glass (SG) solution: q > 0,m = 0. • Ferromagnetic (F) solution: q > 0,m > 0. • Fig. 1 shows the lines delimiting the regions where different types of solutions can be found in the plane of temperature versus J +J′. 0 0 The P solution is always possible. However it is L stable only above the line (b) and the left portion of line (a), and T stable above line (a). The L instability of P solution occurs due to the violation of the condition (49), which is given in the case of P solution by L = [1 β2(J′2 J2)][1 β2(J′2 +J2)][1 β(J′ J )][1 β(J′ +J )] > 0. (52) 4 − − − − 0 − 0 − 0 0 9 For (J +J′)/√J2 +J′2 1/2, the second factor in (52) becomes negative below line (a). 0 0 ≤ Thus the left portion of line (a) is determined by β2(J′2 +J2) = 1. (53) On the other hand, for (J +J′)/√J2 +J′2 > 1/2 the fourth factor in (52) becomes negative 0 0 below line (b). Thus the equation for line (b) is β(J′ +J ) = 1. (54) 0 0 The T instability of the P solution is due to the violation of the condition (31), which is given in the case of P solution by T = [1 β2(J′2 J2)][1 β2(J′2 +J2)] > 0. (55) 2 − − − The second factor in (55) becomes negative below line (a) for all values of J + J′. Thus 0 0 line (a) is given by equation (53) for all J +J′. The T and L instabilities of the P solution 0 0 occur simultaneously on the line (a) for (J +J′)/√J2 +J′2 1/2. 0 0 ≤ The SG solution is possible only below line (a). It is T unstable throughout this region and L stable to the left of line (c). The L instability of the SG solution occurs due to the violation of the condition (49), which is given in the case of SG solution by L = (1 q)2(1 4q +3r)2 [1 β2(J′2 J2)(1 4q +3r)][1 β2(J′2 +J2)(1 4q+3r)] 4 − − − − − − − [1 β(J′ J )(1 q)][1 β(J′ +J )(1 q)] > 0. (56) × − 0 − 0 − − 0 0 − For (J + J′)/√J2 +J′2 > 1/2 the last factor in (56) becomes negative to the left of line 0 0 (c). Thus the equation determining line (c) is β(J′ +J )(1 q) = 1. (57) 0 0 − The F solution is possible only between lines (b) and (c). It is L stable throughout this region but T stable only above line (d). The T instability of the F solution occurs due to the violation of the condition (31), which is given in the case of F solution by T = [1 β2(J′2 J2)(1 2q+3r)][1 β2(J′2 +J2)(1 2q +3r)] > 0. (58) 2 − − − − − For (J + J′)/√J2 +J′2 > 1/2 the second factor in (58) becomes negative below line (d). 0 0 Thus line (d) is described by equation β2(J′2 +J2)(1 2q +3r) = 1. (59) − 10