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Random-energy model in random fields Luiz O de Oliveira Filho,1,∗ Francisco Alexandre da Costa,2 and Carlos S O Yokoi1 6 0 1Instituto de F´ısica, Universidade de São Paulo, 0 2 Caixa Postal 66318, 05315-970 São Paulo, SP, Brazil n a 2Departamento de F´ısica Teórica e Experimental, J 7 Universidade Federal do Rio Grande do Norte, 2 Caixa Postal 1641, 59072-970 Natal, RN, Brazil ] h c (Dated: February 6, 2008) e m Abstract - t a The random-energy model is studied in the presence of random fields. The problem is solved t s . exactly both in the microcanonical ensemble, without recourse to the replica method, and in the t a m canonical ensemble using the replica formalism. The phase diagrams for bimodal and Gaussian - d random fields are investigated in detail. In contrast to the Gaussian case, the bimodal random n o field may lead to a tricritical point and a first-order transition. An interesting feature of the phase c [ diagram is the possibility of a first-order transition from paramagnetic to mixed phase. 1 v 0 PACS numbers: 05.50.+q,75.10.Nr,64.60.-i 4 6 1 0 6 0 / t a m - d n o c : v i X r a ∗ Permanent Adress: Coordenaçaõ de F´ısica, Universidade Estadual Vale do Acarau´, Av. Doutor Guarani 317, 62040-730Sobral, CE, Brazil 1 I. INTRODUCTION Spin-glass [1, 2] and random-field models [3] have played prominent roles in the study of disordered systems in the last few decades. Although the random-exchange and random- field effects are usually considered separately, it has been argued that in proton glasses such as Rb (NH ) H PO [4] it is necessary to take into account the effect of random 1−x 4 x 2 4 fields generated by the presence of impurities. Another example where the spin-glass and random-field effects are present simultaneously is the diluted antiferromagnets Fe Zn F x 1−x 2 [5, 6]. The effect of random fields on the well known Sherrington-Kirkpatrick (SK) model for spin glass [7] has been investigated for Gaussian random field [8], bimodal random field [9] and trimodal random field [10]. Since the low temperature properties of the SK model is rather difficult to work out explicitly [1, 2], it seems worthwhile to consider a simpler spin-glass model where the effect of random fields can be investigated thoroughly. The random-energy model (REM) [11, 12] is probably the simplest spin-glass model [13] retaining some important properties of the SK model. The REM is related to the gener- alization of the SK model to include interaction between every set of p-spins [13]. In the p limit the energies of the spin configurations become independent random variables → ∞ and the model reduces to the REM. In this paper we investigate the effect of random fields on the REM. The model is given as the p limit of the Hamiltonian → ∞ = J S S J S S H S , (1) H − i1...ip i1 ··· ip − 0 i j − i i i1<X···<ip Xi<j Xi where S = 1 are Ising spins, J are independent quenched Gaussian random couplings i ± i1...ip with zero mean and variance p!J2/2Np−1, J 0 are ferromagnetic couplings and H are 0 i ≥ independent identically distributed quenched random fields. The Hamiltonian (1) for p = 2 is the SK model in a random field, whereas for p → ∞ it reduces to the REM model in a random field. We have solved the problem exactly by two complementary approaches. In section 2 we employ the microcanonical formalism [12] to obtain the thermodynamic quantities directly. In section 3 we employ the replica formalism [13] to determine the spin-glass order parameters. In section 4 we study the phase diagram for bimodal and Gaussian distribution of random fields. Finally, in section 5 we 2 compare our results with the previous studies on related models and make some concluding remarks. II. MICROCANONICAL APPROACH In this section we solve the model in the microcanonical ensemble [12]. Let S = (S ,...,S ) denote one of 2N spin configurations or the microstates of the system. The 1 N energy of a given microstate is given by E = (S) = J S S + (S), (2) S H − i1···ip i1··· ip H0 i1<X···<ip where denotesthepartoftheHamiltonianwithoutrandomcouplings. SinceE arelinear 0 S H combinations of Gaussian random variables J , they are themselves Gaussian random i1···ip variables with mean E = (S) = E0, (3) h Si H0 S and covariance J2N 1 σSS′ = (ES −ES0)(ES′ −ES0′) = 2 qSpS′ +O N , (4) D E (cid:20) (cid:18) (cid:19)(cid:21) where 1 ′ qSS′ = N SiSi, (5) i X ′ is the overlap between the microstates S and S . In the thermodynamic limit, N , the → ∞ energies ES and ES′ of two macroscopically distinguishable microstates S and S′ become uncorrelated in the p limit, → ∞ J2N σSS′ = 2 !qSpS′ −→ 0, for p → ∞ and |qSS′| < 1. (6) Thus in the p limit the energies E become independent Gaussian random variables. S → ∞ The multivariate probability density is then the product of univariate probability densities given by 1 (E E0)2 f (E) = exp − S . (7) ES √πNJ2 "− NJ2 # Let us consider a given sample, that is, a particular realization of the random couplings J . The entropy of the sample is given by i1···ip S(E) = k lnΩ(E), (8) B 3 where Ω(E) = δ(E E ) (9) S − S X is the density of states. The average density of states is Ω(E) = δ(E E ) = f (E). (10) h i h − S i ES S S X X Due to the statistical independence of E the fluctuations around this average is of order S Ω(E) −1/2, and thus completely negligible [12]. h i We can rewrite the average density of states in the form 1 ∞ (E E )2 Ω(E) = dE exp − 0 δ(E E0). (11) h i √πNJ2 −∞ 0 "− NJ2 # 0 − S Z S X We recognize Ω (E) = δ(E E0) (12) 0 0 − S S X as the density of states of the system described by the Hamiltonian . Therefore 0 H 1 ∞ (E E )2 S (E ) 0 0 0 Ω(E) = dE exp − + , (13) h i √πNJ2 Z−∞ 0 "− NJ2 kB # where S (E ) = k lnΩ (E ), (14) 0 0 B 0 0 is the entropy of the system characterized by the Hamiltonian . In the thermodynamic 0 H limit, N , we have → ∞ (E E )2 S (E ) 0 0 0 ln Ω(E) = max − + . (15) h i E0 "− NJ2 kB # E is determined by 0 1 ∂S (E ) 2k (E E ) 0 0 B 0 = = − , (16) T (E ) ∂E − NJ2 0 0 0 where T (E ) is by definition the temperature of the system described by the Hamiltonian 0 0 . 0 H For energies E such that ln Ω(E) > 0 the average density of states is very large and the h i fluctuation is negligible. Thus we have with probability 1, k (E E )2 B 0 S(E) = k ln Ω(E) = − +S (E ). (17) B h i − NJ2 0 0 For energies E such that ln Ω(E) < 0, the average density of states is very small. Thus h i with probability 1 there are no samples with this energy. 4 The temperature of the system is given by 1 ∂S(E) 2k (E E ) B 0 = = − , (18) T(E) ∂E − NJ2 which coincides with the temperature of the system described by the Hamiltonian , 0 H T(E) = T (E ). (19) 0 0 Therefore the energy of the system as a function of temperature is given by NJ2 E(T) = E (T) , (20) 0 − 2k T B where E (T) is the energy of the system characterized by the Hamiltonian . The entropy 0 0 H as a function of the temperature is NJ2 S(T) = S (T) . (21) 0 − 4k T2 B These results are valid above a critical temperature T determined by c NJ2 S(T ) = S (T ) = 0. (22) c 0 c − 4k T2 B c Below this temperature the system is frozen in its ground state. These results are valid for any Hamiltonian . We now particularize for the case where 0 H the Hamiltonian describes the Ising model with infinite range ferromagnetic interactions 0 H in a random field [14, 15, 16], 2 J J 0 0 = S S H S = S H S , (23) 0 i j i i i i i H −N − −2N ! − i<j i i i X X X X where in the last passage we have dropped the term J /2 that is negligible in the thermo- 0 dynamic limit. The quadratic term can be linearized using the identity λ ∞ eλa2/2 = dxe−λx2/2+λax, (24) s2π −∞ Z and the partition function will be given by βJ N ∞ 1 1 Z = e−βH0 = 0 dmexp N βJ m2 + ln2coshβ(J m+H ) . 0 0 0 i s 2π −∞ ( "−2 N #) S Z i X X (25) In the thermodynamic limit, N , the Laplace method gives → ∞ 1 lnZ = N max βJ m2 + ln2coshβ(J m+H) , (26) 0 0 0 m −2 h i (cid:20) (cid:21) 5 where we have used the law of large numbers to write 1 ln2coshβ(J m+H ) = ln2coshβ(J m+H) , (27) 0 i 0 N h i i X where denotes the expectation value with respect to the random fields H. Thus the h···i free energy is given by 1 1 F = β−1lnZ = N J m2 ln2coshβ(J m+H) , (28) 0 0 0 0 − "2 − β h i# where the magnetization m is determined by the equation m = tanhβ(J m+H) . (29) 0 h i TheinternalenergyE (T)andtheentropyS (T)followfromusualthermodynamicrelations. 0 0 Applying the general results obtained previously for the system described by the full Hamiltonian (1) in the p limit, we obtain for the internal energy → ∞ E βJ2 1 = J m2 Htanhβ(J m+H) , (30) 0 0 N − 2 − 2 −h i and for the entropy S (βJ)2 = βJ m2 β Htanhβ(J m+H) + ln2coshβ(J m+H) . (31) 0 0 0 Nk − 4 − − h i h i B These results are valid for β < β where β is determined by c c (β J)2 c S(β ) = β (H +J m)tanhβ (H +J m) + ln2coshβ (H +J m) = 0 (32) c c 0 c 0 c 0 − 4 − h i h i and m = tanhβ (J m+H) . (33) c 0 h i For β > β the system is frozen in its ground state. Therefore c E(β) = E(β ), S(β) = 0. (34) c III. REPLICA APPROACH In this section we solve the model in the canonical ensemble [13]. We use the replica identity for the free energy Zn 1 βF = lnZ = lim h i− , (35) − h i n→0 n 6 to perform the average over the random couplings J . To evaluate Zn we introduce i1i2···ip h i n replicas of the system α = 1,2,...,n, Zn = Tr e−β nα=1H(Sα) = Tre−βHeff, (36) h i D P E where denotes the effective Hamiltonian that results after taking the average over ran- eff H dom couplings, N(βJ)2 1 p n NβJ 1 2 β = SαSβ + + 0 Sα +β H Sα. − Heff 2  N i i ! 2 2 N i ! i i α<β i α i i α X X X X X X   (37) We have dropped terms that vanish in the thermodynamic limit, N . The nonlinear → ∞ terms can be linearized with the help of the asymptotic relation Nλf′′(a) ∞ eNλf(a) dxeNλ[f(x)−f′(x)(x−a)], (38) ∼ s 2π −∞ Z which canbeproved forλf′′(a) > 0 andN applying the Laplacemethod. Inparticular → ∞ for f(x) = x2/2 the asymptotic relation reduces to the identity (24). Omitting the factors that do not contribute to the free energy in the thermodynamic limit, N , we arrive at → ∞ Zn dq dm e−βFn(qαβ,mα), (39) αβ α h i ∼ Z α<β Z α Y Y where F 1 1 1 n = βJ2n+ βJ2(p 1) qp + J m2 N −4 2 − αβ 2 0 α α<β α X X 1 1 β−1 lnTr exp (βJ)2p qp−1SαSβ +β (H +J m )Sα . (40) − N 2 αβ i 0 α  i α<β α X X X   In the N limit we use the law of large numbers to write the last term as an expectation → ∞ value over the random-field distribution and use Laplace method to evaluate the integral. The free energy is then given by the stationary value of the functional F 1 1 1 J = βJ2 + lim βJ2(p 1) qp + 0 m2 N −4 n→0 n(2 − αβ 2 α α<β α X X 1 β−1 lnTr exp (βJ)2p qp−1SαSβ +β (H +J m )Sα , (41) − * "2 αβ 0 α #+) α<β α X X where denotes the expectation value with respect to the random field H. h···i 7 To compute the free energy we assume m = m, (42) α to be independent of replica indices, and parameterize q following the Parisi’s K-step αβ replica-symmetry-breaking Ansatz [17]. In the n 0 limit the free energy functional be- → comes a function of the magnetization m and the parameters 0 q q q q 1, (43) 0 1 K−1 K ≤ ≤ ≤ ··· ≤ ≤ ≤ 0 = m m m m = 1, (44) 0 1 K K+1 ≤ ≤ ··· ≤ and is given by F βJ2 K J = 1+(p 1) (m m )qp pqp−1 + 0m2 N − 4 " − i+1 − i i − K # 2 i=0 X ∞ − −∞dy Gσ02(y −H −J0m) g0(y), (45) Z D E where g (y) is given recursively by 0 1 ∞ ′ ′ ′ gi−1(y) = βmi ln(Z−∞dy Gσi2(y −y)exp[βmigi(y )]), (46) for i = 1,...,K with the initial condition 1 g (y) = ln(2coshβy). (47) K β Gσ2(y) denotes the Gaussian distribution function 1 y2 Gσ2(y) = Jσ√2π exp −2J2σ2!, (48) where the variances σ2 are given by i p p σ2 = qp−1, σ2 = (qp−1 qp−1) for i = 1,...,K. (49) 0 2 0 i 2 i − i−1 We first assume that all the q’s are less than one, 0 q ... q q < 1. Then 0 K−1 K ≤ ≤ ≤ ≤ σ2 0 when p for i = 0,...,K. Using the expansion i → → ∞ ∞ J2σ2 d2 J2σ2 −∞dy′Gσ2(y′ −y)f(y′) = exp 2 dy2!f(y) = 1+ 2 f′′(y)+O(σ4), (50) Z 8 we obtain F βJ2 K J 1 = 1+(p 1) (m m )qp pqp−1 + 0m2 ln2coshβ(H +J m) N − 4 " − i+1 − i i − K # 2 − βh 0 i i=0 X K β(Jσ )2 i 1 (1 m ) tanh2β(H +J m) +O(σ4,...,σ4 ,σ2σ2,...,σ2σ2 ). − 2 − − i h 0 i 0 K 0 1 0 K Xi=0 h i (51) Stationarity of the free energy with respect to the variational parameters gives, in the limit p , → ∞ m = tanhβ(H +J m) , (52) 0 h i and q = q = = q = tanh2β(H +J m) . (53) 0 1 K 0 ··· h i Thus we arrived at the replica-symmetric solution where all the q’s are identical. The free energy in the p limit is given by → ∞ F βJ2 J 1 = + 0m2 ln2coshβ(H +J m) . (54) 0 N − 4 2 − βh i The entropy is S (βJ)2 = β (J m+H)tanhβ(J m+H) + ln2coshβ(J m+H) . (55) 0 0 0 Nk − 4 − h i h i B This solution corresponds precisely to the high temperature solution found in the microcanonical approach. Since the entropy becomes negative at low temperatures, it is necessary to consider a different solution for low temperatures. We therefore assume that 0 q ... q < q = 1. Then σ2 0 for i = ≤ 0 ≤ ≤ K−1 K i → 0,...,K 1 and σ in the limit p . A simple calculation yields, K − → ∞ → ∞ 1 1 gK−1(y) = βm ln(2coshβmKy)+ 2βmK(JσK)2 +O(e−(βJσKmK)2/2σK−1). (56) K The error is exponentially small and may be safely ignored. The rest of calculation proceeds as before using the expansion (50) and we arrive at F βJ2 K J 1 = 1+(p 1) (m m )qp pqp−1 + 0m2 βm (Jσ )2 N − 4 " − i+1 − i i − K # 2 − 2 K K i=0 X K−1 β(Jσ )2 i [m sech2βm (H +J m) +m tanh2βm (H +J m) ] K K 0 i K 0 − 2 h i h i i=0 X 1 ln2coshβm (H +J m) +O(σ4,...,σ4 ,σ2σ2,...,σ2σ2 ). (57) −βm h K 0 i 0 K−1 0 1 0 K−1 K 9 Stationarity with respect to the variational parameters gives, in the limit p , → ∞ m = tanhβm (H +J m) , (58) K 0 h i q = q = = q = tanh2βm (H +J m) , q = 1, (59) 0 1 K−1 K 0 K ··· h i consistent with initial assumption q = 1, and K (βJ)2 m2 = ln2coshβm (H +J m) βm (H +J m)tanhβm (H +J m) . (60) 4 K h K 0 i− Kh 0 K 0 i These results arethesame forallK 1, showing that noother solutions arepossible beyond ≥ one-step replica symmetry breaking. The free energy in the limit p is given by → ∞ F βJ2 J 1 = m + 0m2 ln2coshβm (H +J m) . (61) K K 0 N − 4 2 − βm h i K The entropy is S (βJ)2 1 = m β (J m+H)tanhβm (J m+H) + ln2coshβm (J m+H) . K 0 K 0 K 0 Nk − 4 − h i m h i B K (62) Taking into account the self-consistency equation (60) we find that the entropy vanishes identically. Thus this solution corresponds to the frozen phase found in the microcanonical approach. The self-consistency equations (58) and (60) imply that βm is independent of temper- K ature. Since this solution is acceptable only for m 1, we have K ≤ βm = β , (63) K c where β is found from the equations c (β J)2 c = ln2coshβ (H +J m) β (H +J m)tanhβ (H +J m) , (64) c 0 c 0 c 0 4 h i− h i and m = tanhβ (H +J m) . (65) c 0 h i Thus we see that β corresponds precisely to the critical temperature for the transition c to the frozen phase found in the microcanonical approach. The Parisi order parameter function q(x) [17] has two flat portions q = m2 and q = 1, with a discontinuous jump at 0 K x = m = T/T , K c T T q(x) = m2θ x +θ x . (66) T − − T (cid:18) c (cid:19) (cid:18) c(cid:19) 10