p D q EA Figure 3: F as a function of p d plotted for p d ' q EA, t = 0:05 and y =?0:01.

Transcription

1 F p D q EA Figure 3: F as a function of p d plotted for p d ' q EA, t = :5 and y =?:. 5

2 F p A Figure : F as a function of p d plotted for < p d < 2p A, t = :5 and y =?:. F p A p max p p p D q EA Figure 2: F as a function of p d plotted for < p d < q EA, t = :5 and y =?:. 4

3 Table of gures F as a function of p d plotted for < p d < 2p A, t = :5 and y =?:. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 4 2 F as a function of p d plotted for < p d < q EA, t = :5 and y =?:. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 4 3 F as a function of p d plotted for p d ' q EA, t = :5 and y =?:. 5 3

4 [4] S. Franz, G. Parisi, and M.A. Virasoro, The replica method on and o equilibrium, Journal de Physique (France), 2:69, 993. [5] J. Kurchan, G. Parisi, and M.A. Virasoro, arriers and Metastable States as Saddle Points in the Replica Approach, Journal de Physique, 3:9, 993. [6] D.J. Gross, I.Kanter, and H.Sompolinsky, Mean-eld theory of the potts glass, Physical Review, 55(3):34{37, 95. [7]. de Dominicis and I. Kondor, Ultrametricity and zero modes in the shortrange Ising spin glass, Europhysics Letters, 2():67{624, 96. [] We thank I. Kondor for pointing out to us that there was an apparent inconcistency in ref.4 because this distinction in the path of integration was overlooked. 2

5 2RS. We conjecture that with more constraints one will be able to stabilise solutions with more levels of breaking. Therefore the linear solution, which is the limit for an innite number of breakings, will reappear. The Parisi parameter q(x) plays two dierent roles in the SK model. On the one hand it appears in the Gibbs measure and determines the P eq (q). In this case q(x) must be necessarily monotonous. On the other hand x(q) also appears in the avity Method where monotonicity does not seem to be required. It remains as an intriguing open question whether in systems as the present a non monotonous x(q) could be relevant. Perhaps solving the R real replicas by a generalisation of the avity Method could elucidate this issue. Acknowledgements We are grateful to S. Franz, I. Kondor, J. Kurchan and T. Temesvari for useful discussions. References [] For a recent exposition on mean eld theory of Spin Glasses see for example: M. Mezard, G. Parisi and M.A. Virasoro, "Spin Glass Theory and eyond" (97) (Singapore:World Scientic). [2]. de Dominicis and I. Kondor, Eigenvalues of the stability matrix for Parisi solution of the long-range spin-glass, Physical Review, 27():66{6, 92. [3] A. risanti and H.J. Sommers, The spherical p-spin interaction spin glass model: the statics, Z. Phys., 7:34{354, 992.

6 The free energy, expressed in term of = p d? q EA y, reads (g.3) and the minimum eigenvalue is = 4?2yt2 3. F Dq EA = (?22 t 2 y? 3 ) + O(y 4 ) (23) learly for positive y these solutions are just approximations to the full solution. In particular we found (2) to have the same instability than the 2RS free solution ( =? yt2 6 ). 4 Discussions Let us analyse the previous results for y <. For p d = the two real replicas lie on congurations belonging to dierent equilibrium states. As the multiplicity of pure states give no contributions to the entropy, the free energy per replica is equal to that of the unconstrained system and q(x) = q(x) unconstr and p(x) =. IF p d increases we force the overlap between couple of replicas to be out of equilibrium. The free energy increases until the maximum is reached at p max while if p d > p max the free energy falls until p d of order of q EA, where we are choosing congurations inside the same state. Since the overwhelming majority of pairs of congurations inside a state are at distance q EA from each other, if p d = q EA we have twice the energy and twice the intra-state entropy. At p d = p max we have F A p d = df A dp d =. The constraining force is then zero (on the S.P.) and the 2 real replicas are unconstrained. In fact the solution (2) is related through a replica permutation to the solution (3), because m 2RR = m RR =2. Then F 2RR max = F RR 2RS. However the constraint still acts on the uctuations and prevents the 2 replicas from sliding along the unstable direction. It is remarkable that in this way we have been able to interpret an unstable

7 For p d greater than p the stable solution becomes q(x) = p(x) = q = < x < m q = p d? 9 y(?p2 d + 4p d t? 4t 2 ) m < x < m 2 q 2 = 2t + y 3 t2? p dy 6 (p d? 4t) m 2 < x < p = < x < m p = p d? 9 y(p2 d? 2p d t + 2t 2 ) m < x < m 2 p 2 = p d m 2 < x < m = 4 + y p d 4 m 2 = 2 + y(t + p d 2 ) (2) The free energy of this solution (F ), which diers from () by terms of O(y 3 ), is decrasing in p d (g2). Again at p d = p the two solutions crosses each other. The minimum eigenvalue is = y(p 2 d? p d t + t 2 )=6. We have p? q =?2y(p 2 d? pdt + t 2 )=9 and the free energy dierence between () and (2), expressed in terms of (2), reads (p? q) 3?y 3. We found (2) to be stable for p < p d < p ' 3=2t. In the range p < p d < p D < q EA we did'nt nd any 2RS stable solution. In particular (2) is unstable with respect to replicon uctuations. We expect that additional RS lead to the stable solution. Thus, let us consider p d > p D = 2t + 23yt2 6. The stable solution is < x < m q(x) = 2 (p 3 d + t) + y (p d + t) 2 m < x < p(x) = q(x) m = 4 + y 6 (p d + t) (22) 9

8 p(x) = p = < x < m p = p d? 2 3 yt(p d? t) m < x < m 2 p 2 = p d m 2 < x < m = 4 + y p d 4 m 2 = 2 + y(t + p d 2 ) () The free energy is F A = y2 (2p dt 4? 2p 2 d t3 + p 3 d t2? 4 p4 d t) + O(y3 ) (9) This solution crosses, at p d = p A, the previous one. For p d?y the minimum eigenvalue is = p d 2 + yt2 3. As can be seen in (g.) and (g.2), F A is an increasing function until p d = p max = t yt2, where df dp d we rewrite () on this specic value = F A p d =. For future reference q(x) = p(x) = q = < x < m q = t yt2 m < x < m 2 q 2 = 2t yt2 m 2 < x < p = < x < m p = t yt2 m < x < m = 4 + y t 4 m 2 = 2 + y 3t 2 (2) The minimum eigenvalue is =? yt2 6. As p d increases further, F A begins to decrease while all eigenvalues read positive until the second critical value p. The minimum eigenvalue for p d ' p is =?y(p 2 d? p d t + t 2 )=6 and it corresponds to uctuations of q and p. p is dened by j p =.

9 3.2 2RR constrained saddle points We investigate the possible 2RS solutions in the range < p d < q EA. We try to follow the solution by continuity when p d varies. At dierent points in the interval a stable solution becomes unstable and we have to search for the new stable solution. Let us consider y <. We nd four critical p d values < p A < p < p < p D < q EA. At each critical p d value, two dierent solutions coincide and zero modes appear to modify their stability. We start at p d '. The stable solution is q(x) = p(x) = < x < m 2 q EA = 2t + 3 yt2 m 2 < x < < x < m 2 p d m 2 < x < m 2 = 2 + yt (6) and the free energy density is F A? F unconstr RS = F A = 2 (?2yt2 p 2 d? p 3 d) + O(y 4 ) (7) If p d = the free energy is the same as the equilibrium one and F A p d = df A dp d =. For small p d the free energy is an increasing function of p d (g.). The minimum eigenvalue is =? 2 3 yt2? p d. Thus when p d > p A =? 2 3 yt2 this solution becomes unstable and the new stable solution is: q(x) = q = < x < m q = p d? 2 3 yt(p d? t) m < x < m 2 q 2 = 2t + 3 yt2? 6 yp d(p d? 4t) m 2 < x < 7

10 f = +f +f x +f x +f A + +f z +f y?f y?f z A () where the f ;x;z are n n symmetrical matrices and f y antisymmetrical. The eigenvalue equations are now decoupled and read, for a 6= b, and (2t + q ab + yq 2 ab + )f ab + fq; f g ab + ff x ; P g ab = (2t + p ab + yp 2 ab + )f x ab + fp; f g ab + ff x ; Qg ab = (2) (2t + q ab + yq 2 ab + )f z ab + fq; f z g ab + [f y ; P ] ab = (2t + p ab + yp 2 ab + )f y ab + fq; f y g ab + [f z ; P ] ab = (3) The eigenvectors of (2) and (3) are then respectively of the form f = +f +f x +f x +f A f = +f +f x +f x +f A f = +f +f x +f x +f A (4) f = +f z +f x?f x?f z A f = +f z +f y?f y?f z A f = +f z +f y?f y?f z A (5) where f x, f y and f y are antisymmetrical n n matrices breaking the hierarchical subgroups of permutation group (as the RR ones). 6

11 In the following we consider the particular case R = 2 and we make for Q and P a Parisi ansatz. efore analysing the solution, let us explain how we studied the stability. 3. 2RR Stability From (9) we see that the saddle point for a, which originally is integrated along the imaginary axis, lies on the real axis. The integration path can be deformed to go through the saddle point but the integration path will be perpendicular to the real axis. As usual this S.P. has to be a maximum when compared to values along the integration path and therefore a minimum when compared to values along the real axis. Notice that this distinguishes a from all other matrix terms which are integrated along the real axis []. The stability with respect to a uctuations around the saddle point depends on the quadratic term () that is not proportional to t. Therefore these uctuations are much smaller and we can substitute everywhere a by its S.P. value which we will denote p d. The equation for the hessian of F 2 is then (2t + q + yq 2 + )f + fq; fg = () where = ; :::; 2n but? 6= n. In order to reduce this problem to an ultrametric one we can observe that the symmetry group of the Hamiltonian without constraint is S 2n (all replicas are equivalent). After imposing the constraint it becomes S n (S 2 ) n (to permute simultaneasly replicas in both systems or to permute equivalent replicas between systems). Since all solutions found are symmetrical under the action of S 2 on all the replicas, we divide the eigenvectors in two superfamilies, the symmetrical and the antisymmetrical one. We then parametrise the uctuation matrix f around the Q S:P: in terms of 5

12 tion (to negative y) of the continuum solution (slope = ), which is the only one y marginally stable for positive y. It has obviously a free energy greater than the other solutions. 3 The constrained model Let us consider R Real Replicas and introduce in the partition function R (R? ) 2 complex Lagrangean multipliers ( rs ) to keep the mutual overlaps constrained (q rs c ). The eective averaged free energy has two contributions [4] F R (Q; q rs c ) = F (Q) + F constr ( rs a ; qrs c ) (6) The functional F (Q) has the same form as for a single real replica with Q a Rn Rn matrix Q = Q P 2 ::: P R P 2T Q 22 ::: P 2R ::: ::: ::: ::: P RT P 2RT ::: Q RR A (7) where (P rs ) aa = rs a. The "interaction" term F constr ( rs a ; qrs c ) reads F constr ( rs a ; qrs c ) =? 2 2 The stationarity conditions are then X (q rs c r;s;a? rs a ) 2 () F q rr ab = F p rs ab = F rs a = 2 ( rs a? qrs c ) (9) 4

13 where f = Q? Q SP. (2t + q ab + yq 2 ab + )f ab + fq; fg ab = (4) This equation, analysed in details in the context of the SK model [7], can be solved through the explicit construction of the eigenvectors exploiting the symmetry of the equation. In the notation of [7] the eigenvectors are classied as longitudinal (f ab ), anomalous (f ab) and replicon (f ab ). For positive y the minimum eigenvalue is =? yt2 6 and it is well known that unstable directions belong also to the replicon family. Let us discuss the analytic continuation to negative y of this solution. It leads to m 2 < m, a result which does not allow the usual probabilistic interpretation. In this case we expect new unstable directions. We found the minimum eigenvalue to be negative ( = 5yt2 ). The unstable 6 eigenvectors belong to the longitudinal and anomalous families. They are longitudinal! f ab = anomalous! f ab = f () = if a \ b = f () 6= if a \ b = f (2) 6= if a \ b = 2 f () = if a \ b = f () 6= if a \ b = max(a \ ; b \ ) = f (2) 6= if a \ b = 2 max(a \ ; b \ ) = f ()(? n m ) if a \ b = max(a \ ; b \ ) > f (2)(? n m ) if a \ b = 2 max(a \ ; b \ ) > (5) where, as in [7], a \ b = k if q ab = q k. For future reference, we anticipate that by considering additional RS the new unstable S.P. will have m i+? m i y. There is, of course, the analytic continua- 3

14 appears in the p-spin spherical model and in the Potts model [6]. The y parameter will play the role of a control parameter. We will assume it to be small (y << t) but positive or negative (in the SK model it is positive). In [6] it was shown that for y < the stable solution is simply a matrix with one replica symmetry breaking (RS). The probability distribution of the overlap between equilibrium congurations is then P eq (q) = m (q) + (? m )(q? q EA ) (2) where q EA = 2t + 3 yt2 and m = + yt are obtained by stationarity conditions. 2 In the solution we will consistently truncate the O(y 2 ) terms. As a consequence of the stationarity conditions, the O(y 2 ) terms in the solution give O(y 4 ) contribution to free energy. In the following we will then consider free energy up to O(y 3 ). 2.2 Unstable saddle points For positive y we know that a better approximation to the full solution is given by the 2RS saddle point q(x) = q = < x < m q = t yt2 m < x < m 2 q 2 = 2t yt2 m 2 < x < m = yt m 2 = yt (3) 2 and free energy F 2RS > F RS (F 2RS? F RS y 2 ). To study uctuations around this 2RS S.P. we consider the eigenvalue equation 2

15 Introduction In spin glass models, replica symmetry breaking implies the existence of many pure states of equilibrium. Their hierarchical organisation is described by the order parameter Q = fq ab g []. In the SK model q ab has a continuum form and it is a marginally stable saddle point of the replica free energy surface in the limit n! [2]. In other models, like the p-spin spherical model, the stable solution is just one-step, yielding a simpler equilibrium state organisation [3]. In this paper we investigate this dierent feature from the stability point of view. We use the generalisation of the replica method recently developed to study non equilibrium states [4] [5]. The general idea is to consider R identical copies of the model (real replicas) constrained to have specic mutual overlaps. Then, by forcing the overlaps out of equilibrium, one can probe the phase-space structure of the original model and obtain information on some of its non equilibrium states. y performing the stability analysis in the replica approach o equilibrium, we uncover the physical meaning for some saddle points other than the one yielding the Gibbs-measure results. 2 The unconstrained model and its saddle points 2. The model Let us consider the following truncated replica free energy density F (Q) =? n (t T rq2 + 3 T rq3 + 6 X a6=b q 3 ab + y 2 X a6=b q 4 ab) () where we consider t T c? T to be small. The cubic term P q 3 ab, absent in the SK-model, gives a y independent replica symmetry breaking. A term of this kind

16 Saddle Points Stability in the Replica Approach O Equilibrium M.E. Ferrero and M.A. Virasoro October 26, 994 Dipartimento di Fisica, Universita di Roma, La Sapienza, I-5 Roma, Italy and INFN Sezione di Roma I, Roma, Italy. Abstract We study the replica free energy surface for a spin glass model near the glassy temperature. In this model the simplicity of the equilibrium solution hides non trivial metastable saddle points. y means of the stability analysis performed for one and two real replicas constrained, an interpretation for some of them is achieved.