Statistical mechanics of glasses and jamming systems: the replica method and its applications

Transcription

1 Statistical mechanics of glasses and jamming systems: the replica method and its applications Hajime Yoshino, Cybermedia Center, Osaka University, Toyonaka, Osaka 56-3, Japan Graduate School of Science, Osaka University, Toyonaka, Osaka 56-3, Japan

2 CONTENTS I. Introduction A. Edwards-Anderson model B. Edwards-Anderson order parameter: the glass order parameter C. Replicas and the glass order parameter 5. Two replicas 5. n replicas and the replica symmetry 6 II. Prototypes: mean field spinglass models 7 A. p-spin models 7 B. Random energy model 8. p limit of the p-spin model 8. Exact analysis of the thermodynamics 9 C. Replicated free-energy functional 9 D. Use of Parisi s ansatz E. Replica symmetric solution. p = case: Sherrington-Kirkpatrick s analysis. p limit F. step replica symmetry breaking (RSB) 3. p limit. Distribution of overlap P (q) 5 3. System with many metastable states 5. Complexity and Monasson s observation 6 5. Finite p model: explicit computations on the p = 3 model 7 6. Disentangling hierarchy of responses 7 7. Glass state following: Franz-Parisi potential 9 G. d Almeida-Thouless-Gardner instability. Stability of the RS symmetric solution. Instability of the RSB solution:the Gardner s transition H. More RSB: full replica symmetry breaking. RS case. -RSB, -RSB and k-rsb 3. Variational equations 3. Continuous RSB: k limit 6 I. Hierarchical energy landscape and linear response with, 3,..., k RSB 7 J. Overlap distribution function P (q) and the order parameter function q(x) 8 III. Replicated liquid theory 3 A. Our system 3 B. Liquid theory 3. Mayer function 3. Density functional 3 C. Cloned liquid: basic ideas 3. Glass order parameter 3. Playing with the number of replicas 3 D. Density functional of the replicated free-energy 3 E. Construction of the gaussian ansatz 33. General form 33. Hierarchical gaussian ansatz 3 3. Another representation of the hierarchical Gaussian 3 F. Free-energy 35. Entropic part of the replicated free-energy with the hierarchical gaussian ansatz 35. Interaction part of the replicated free-energy with the hierarchical gaussian ansatz Variational equations 37. Effective potential and g(r) Continuous RSB: k limit d limit 39

3 IV. Hardspheres in d limit A. RSB: dynamical transition, Kauzmann transition, Gardner transition, Free energy 3. Effective potential 3 3. Another representation. Thermodynamics of the liquid 5. Dynamical glass transition point in the liquid phase 6. Cage size in glassy phases 7. Thermodynamics in glassy phases 5 8. Jamming 6 9. Gardner transition 7 B. Full RSB: Jamming critically and beyond 7. Vanishing cage size 7. Anomalous features in the contact region 7 3. Complex responses to shear in the Gardner phase 9 C. Glass state following under compression and shear 5 V. Outlook 5 Acknowledgments 5 References 5 A. Properties of low lying states in the glass phase of the random energy model 53. Distribution of energy gaps 53. Two level system 53 B. Clustering property 5 C. p-spin spherical model 5. RS ansatz 55. RSB ansatz 55 D. Useful formulae of Gaussian integrals 55 E. Spin density functional approach 56. RS Gaussian ansatz 56. RSB Gaussian ansatz krsb Gaussian ansatz 58 F. Glass state following: p-spin Ising model 59 3

4 I. INTRODUCTION A. Edwards-Anderson model A standard theoretical model for spin glasses is the so called Edwards-Anderson (EA) model [] which is defined as follows.the Hamiltonian is given by, H[{S}] = i,j J ij S i S j h N S i, () where S i is the spin variable on site i(=,,..., N). The summation i,j is took over pairs of nearest neighbors on a lattice, say a hyper-cubic lattice in a d-dimensional space. The spin variable can be an Ising variable S = σ = ±, vectorial variable with two components (XY model) or three components (Heisenberg model). We may also consider spherical models in which the spins are scalar variables which can take any real numbers but subjected to a global constraint such that, N i= S i = N. The matrix J ij represents the interaction between the spins which can be ferromagnetic J ij > or antiferromagnetic J ij <. A standard choice is a Gaussian distribution with zero mean and variance J /z with z being the coordination number on the lattice (z = d for the hypercubic lattice). [ P (J ij ) = πj /z exp z ( ) ] Jij. () J The last term on the r. h. s. of Eq. () is the Zeeman energy term with the external magnetic field h. If h =, the Hamiltonian is invariant under S i S i i, i. e. time reversal symmetry. In the EA model, with zero external field h =, the spinglass transition must break this symmetry much as the ferromagnetic phase transition. As soon as h, this symmetry is lost from the beginning. The long standing problem in the spinglass physics is whether or not any phase transition is present in the case h. The mean-field theory, which is exact in the large dimensional limit, predicts that there still remains a phase transition because of the so called replica symmetry breaking (RSB) which we will discuss. The questions remains if it is the case also for finite dimensional systems. i= B. Edwards-Anderson order parameter: the glass order parameter Edwards-Anderson (EA) order parameter [] is defined as follows. One way to defined it is the dynamical one which signals ergodicity breaking: where C(t) is the spin auto-correlation function C(t) = lim N N q EA = lim t C(t) (3) N S i (t)s i () eq () where... eq represents the thermal average within an equilibrium state. For the following discussion it is useful to decompose the spin auto-correlation function as i= C(t) = Ĉ(t) + lim N N N S i eq i= Ĉ(t) lim N N N (S i (t) S i )(S i () S i ) eq (5) i= We generally expect that the connected part Ĉ(t) which represents the correlation of the fluctuation of the spin in a given equilibrium state at different times vanishes in the large time limit lim t Ĉ(t) =. Based on this expectation we find a second definition of the EA order parameter, q EA = lim N N N S i eq, (6) i=

5 In order to evaluate this, we will use replicas as we discuss later. Let us discuss what can happen in the EA model. First let us consider the case of zero external magnetic field h =. We expect at high enough temperatures the equilibrium state is paramagnetic so that lim N S i eq = because of the the time-reversal symmetry S i S i ( i). Then we conclude that q EA = in the paramagnetic phase. On the other hand, in a spin glass phase state, the ergodicity should be broken. From this we conclude that the EA order parameter, as defined above, becomes finite q EA > in the spinglass phase. Now what happens if the external magnetic field is finite h >? In this case, the time reversal symmetry is always broken such that the magnetization, 5 m = N N S i (7) i= should be always finite m >. In this case we have to redefine the EA order parameter by replacing S i by S i m: q EA = N N δs i δs i = S i m (8) i= As mentioned before, whether or not there remains any spinglass transition which does not involve the spontaneous time reversal symmetry breaking is an intriguing and long-debated issue. C. Replicas and the glass order parameter. Two replicas Let us consider two replicas a and b. Each of them has its own spin variables S a i (i =,,..., N) and S b i (i =,,..., N) but obey exactly the same Hamiltonian H[{S}](that is the realization of J ij ). Then we may consider a composite system, H A+B = H[{S a }] + H[{S a }] ɛ N q i q i = Si a Si b (9) where we added an artificial coupling between the two replicas. In the following we assume h = for simplicity. For the case h >, we will just need to replace q i by q i = δsi aδsb i similarly to Eq. (8)). If the coupling is attractive, i.e. ɛ >, the two replicas would take somewhat similar spin patterns. In other words, the overlap between the two replicas, i= Q(N, ɛ) N N q i N,ɛ, () i= will be larger for larger ɛ >. Here... N,ɛ is the canonical ensemble average of the composite system with coupling ɛ and system size N. By defining the free-energy of the composite system F a+b (N, ɛ), βf a+b (N, ɛ) = ln (T r S atr S be β(h[{sa}]+h[{s b}])+βɛ ) N i= Sa i Sb i () we find Q(N, ɛ) = N Particularly interesting is the comparison of the two following two limits, F a+b (N, ɛ). () ɛ lim lim ɛ + N Q(N, ɛ) lim lim N ɛ + Q(N, ɛ) (3) In the paramagnetic phase (liquids) both of the limits should give. On the contrarily, in spin glass phase the two should give different results because there must be more than one equilibrium spinglass states. Thus it is natural to expect,

6 By taking N first, in the presence of the attractive coupling, we bring the two replicas into the same spinglass state. Thus it is natural to expect that we find the EA order parameter q EA, lim ɛ + lim Q(N, ɛ) = q EA >. () N Now in the other case, which takes ɛ + first, we are left with two decoupled replicas. In this case the two replicas may well end up staying in different states so that we expect, lim lim Q(N, ɛ) < q EA. (5) N ɛ + Based on the above observations, it becomes natural to define the so called spinglass susceptibility, χ SG = F a+b (N, ɛ) N ɛ = ( q i q j q i q j = ( Si S j S i S j ). (6) N N ɛ= ij If the spinglass transition takes place as a nd order phase transition, accompanying divergence of the correlation length of the the thermal fluctuation of the local overlap q i, the spinglass susceptibility would diverge approaching the transition point. ij 6. n replicas and the replica symmetry There is no reason to restrict ourselves to just the two replicas. We now extend the above discussion to n replicas a =,,..., n and consider a composite system which is described by the replicated Hamiltonian, for which we naturally define H n = n a= H[{S a }] a<b N ɛ ab i= S a i S b i (7) Q ab (N, {ɛ}) = N N Si a Si b N,{ɛ} (8) i= Then we become naturally interested with Q ab = lim {ɛ + } lim N Q ab (N, {ɛ}). An obvious but important symmetry property of the replicated Hamiltonian Eq. (7) with ɛ ab = ɛ ab is that the system is invariant under permutations of the replica indexes. This permutation symmetry is called as replica symmetry. Parisi considered the possibility of spontaneous breaking of this symmetry [] proposing the following ansatz for the structure of the glass order parameter in the spin glass phase. Q ab = q + k k (q i q i )I mi ab = q i (I mi ab i= i= Imi+ ab ) (9) where Iab m is a kind of generalized ( fat ) identity matrix of size n n composed of blocks of size m m. (see Fig. /reffig:parisi-matrix) The matrix elements in the diagonal blocks are while those in the off-diagonal blocks are all. The Parisi s matrix has a hierarchical structure such that which becomes in the n limit. Replica symmetric (RS) ansatz: = m k+ < m k <... < m < m < m = n () = m < m <... < m k < m k+ = () The so called replica symmetric (RS) ansatz corresponds to the case k =. In this ansatz permutations of the replica indexes does not affect the matrix structure. In order words, in the RS ansatz the permutation symmetry of the replicated Hamiltonian is respected.

7 7 a) b) c) m i m i n m i+ m i+ q i+ m i m i q i n m k+ d) q(x) q k m k m m m n q q q q n m m m k n n FIG.. Parametrization of the Parisi s matrix a) the fat identity matrix I m i ab b) Parisi s order parameter matrix Eq. (9) c) the hierarchy of the sizes m i of the sub-matrices d) the q(x) function with < n <. Replica symmetry breaking (RSB) ansatz: For k, the replica symmetry is broken (RSB), i.e. the permutation symmetry of the replicas is violated. In the k-th RSB, the permutation symmetry holds only within a sub-block of size m k. By selecting any pair of replicas among the n replicas and exchanging them we will obtain another Parisi matrix Q ab. By considering all such permutations we find a family of RSB ansatz. We will see that they have exactly the same equilibrium measure. Then choosing one out of all of them means to consider a spontaneous breaking of the replica symmetry. We will come back to this point later. II. PROTOTYPES: MEAN FIELD SPINGLASS MODELS A. p-spin models Let us consider a class of mean-field spin glass models, called as p-spin model with p-body spin interactions. The Sherrington-Kirkpatrick (SK) model [3] is just a special case p = of this, which corresponds to the d = limit of the EA model. H = i <i i p N J ii i p S i S i S ip h i S i () P (J ii i p ) = [ N p πj p! exp N p p! ( Ji,i,...,i p J ) ] (3) We denote the average over the quenched disorder as J i <i i p N dj ii i p P (J ii i p ) () In the following we mainly consider the p-spin model with Ising spins S i σ i = ±. A great advantage for our pedagogical purposes is that it becomes identical to the random energy model (REM)[] in the p limit [5], which can be analyzed exactly without using replicas. Another standard version is the spherical model which we discuss briefly in the appendix C. The latter is also quite instructive and useful especially to analyze connection between the statics and dynamics [6].

8 8 B. Random energy model We consider the random energy model (REM) []: to each of the micro-states α =,,..., N a random energy E α = Ne α J is assigne which is drawn from a Gaussian distribution with zero mean and variance NJ /, p(e) = e E /NJ πnj (5). p limit of the p-spin model In the limit p, the p-spin Ising mean-field spinglass model becomes identical to the random energy model. To see this, let us examine the distribution of the energy of generic spin configurations. First we examine the distribution of the energy E J (σ) of a spin configuration σ = (σ, σ,..., σ N ), among different realization of the quenched disorder, It is easy to evaluate this, p(e) = p(e) δ(e E J (σ)) J (6) dκ π eiκe e iκe J J (σ) = /NJ NπJ e E. (7) Next we look at the joint distribution of the energy of two spin configurations σ () and σ (), p(e, E ) δ(e E J (σ () ))δ(e E J (σ () )) J (8) which is found to be ( ) p(e, E ) = exp E + NπJ A + NπJ A NJ E A + NJ A (9) where E ± = E±E and A ± = [ ± qp ] By noting that lim p A ± = / for q < we find, q = N N i= σ () i σ () i. (3) p(e, E ) p(e )p(e ) (p, q < ). (3) This means that the energies of two arbitrarily chosen configurations are statistically independent from each other in the p limit: the random energy model. From the above discussion we also find a corollary that in p limit, p(e, E, q) = p(e )p(e )P (q). (3) The overlap between two arbitrary chosen configurations σ () and σ () are easily obtained as, P (q) = N! N (N(( + q)/))!(n(( q)/))! δ(q). (33) N The last equation follows from the observation that lim N ln P (q) = N q + O(q ) which can be obtained using the Stirling s formula.

9 9. Exact analysis of the thermodynamics with The partition function of the REM is computed as Z = N e βeα α= = N de N N α= δ(e e α )e NβJeα = dee N(Σ(e) βje) (3) Σ(e) = ln e e min < e < e max (35) with e min = ln and e max = ln. In the N limit, the integral can be evaluated by the saddle point method yielding, N βf = N ln Z = Σ(e (T )) βje dσ(e) (T ) de = βj (36) e=e (T ) From the saddle point equation (the nd equation) we find the internal energy as, { e βj/ T > Tc (T ) = e min T < T c where T c is the critical temperature k B T c /J = e min = ln below which the saddle point is stuck at e min. Thus we find the free-energy and entropy as, (37) { ln + (βj) βf/n = / T > T c βe min T < T c { S/N = Σ(e ln (βj) (T )) = / T > T c T < T c C. Replicated free-energy functional For simplicity we consider the p = 3 model in the following but generalizations to general p case is straightforward. So we may sometimes write down equations for general p without explanations. We wish to compute the free-energy of the system, βf J = ln Z J Z J = Tr σ exp β σ i (38) i<j<k J ijk σ i σ j σ k + βh i where Tr σ N i= σ. Apparently the free-energy depends on how the random couplings J i=± ijk are chosen so that we wrote it as F J. However we expect that the free-energy/spin converges to a unique value in the thermodynamics limit, i.e. self-averaging. And by the same token this should be the same as the disorder-averaged one. Thus we expect J F J f lim N N = lim F J N N (39) J To compute F J, we rely on the replica trick: assuming that Z n J can be expanded as ZJ n = + n ln Z J + O(n ) for small n we obtain, f = lim N N nzj n n=. ()

10 In the following we proceed as follows. First we evaluatezj n assuming that n is an integer. Then we make an analytic continuation of the resultant formula to real values of n. Now we consider the partition function of a replicated system a =,,..., n and take its disorder average. After some algebra we find, ZJ n = dj ijk P (J ijk )Tr {σ} exp β n i<j<k = Tr {σ} exp (βj) N n + a b N i<j<k where Tr {σ} = n a= Tr {σ a }. Now let us introduce the overlap matrix Q ab, N i= J ijk σ a i σi a σj a σk a + βh a= i,a p σi a σi b + βh σi a () i,a with which we find Q ab N N σi a σi b () i= Z n J = a<b dq ab e Nnβf[{Q ab}] (3) N e Nnβf[{Qab}] = Tr {σ} Nδ(NQ ab σi a σi b ) exp (βj) N n + N p N σi a σ b i + βh σ a i a<b i= a b i= i,a N = Tr {σ} Nδ(NQ ab σi a σi b ) exp (βj) N n + Q p ab + βh σ a i a<b i= a b i,a N { } dλab = exp (βj) N n + Q p ab N λ ab Q ab Tr σ exp λ ab σ a σ b + βh σ a πi/n a<b a b a b a b a { } dλab = exp [ NG({Q ab, λ ab }) ] () πi/n a<b G({Q ab, λ ab }) = (βj) n (βj) Q p ab + λ abq ab ln Tr σ exp λ ab σ a σ b + βh a b a b a b a σ a (5) The integrals can be evaluated by the saddle point method. The saddle points Q ab and λ ab must verify the saddle point equations which read where = G λ ab = (βj) Q ab... eff σ=± e βh eff... σ=± e βh eff pq p ab = G λ ab Q ab = σ a σ b eff (6) βh eff λ abσ a σ b + βh a b a σ a (7) Using the st equation of Eq. (6) in Eq. (5) we find G({Q ab}) = (βj) n + (βj) (p ) a b(q ) p ab ln Tr σ exp (βj) p(q ) p ab σa σ b + βh a a b σ a (8)

11 Once we find all the relevant saddle points we can evaluate the free-energy as, f = β lim N N n dq ab e NβG[{Q ab}] = minimize SP n= a<b n G[{Q ab}] (9) In the last equation, minimize SP locally stable saddle points. means to choose a saddle point Q ab which minimizes the free-energy out of all the D. Use of Parisi s ansatz In the following we look for the saddle points assuming the Parisi s ansatz Eq. (9). First we analyze the replica symmetric case (k = ) and then study the RSB ansatz (k = ). Note that a family of equivalent saddle points other than the one Eq. (9) can be obtained just by making permutations of the replica indicies. All of them take exactly the same value of the free-energy functional f[{q ab }] so that if we take a sum over all of them, the replica (permutation) symmetry is restored. Choosing one out of them, like the one parametrized as Eq. (9), means to break the replica symmetry. Physically we justify this by saying that it is realized as a spontaneous symmetry breaking, like the ferromagnetic phase transition of Ising model. The coupling parameter ɛ ab discussed previously can be regarded as the symmetry breaking field. E. Replica symmetric solution We assume k =, i.e. λ ab = λ and Q ab = q for all replica pairs a b. With this ansatz we find, λ ab σ a σ b = λ σ a σ b = λ n [( σ a ) n] (5) a b a b a= Then by the HubbardStratonovich (HS) transformation we obtain e a b λ abσ a σ b = e n λ dz e z e z λ n a= σa = e n λ e z λ n a= σa z (5) π where we introduced a shorthand notation... z With this we find, Dz.... dz π e z /.... (5) Tr σ e λ σ a bσ σ b +βh a σa = e nλ/ cosh(ξ) n z (53) where we introduced Using the above expression we find, so that G RS (q, λ) n = (βj) + (n ) Ξ βh + z λ (5) ( ) (βj) q p + λq + λ n ( cosh(ξ))n z (55) βf RS (q, λ) = (βj) ( q p ) + λ( q) ln( cosh(ξ)) z (56) where and q and λ must verify the saddle point equations = f RS q λ = p (βj) q p (57)

12 and = f RS λ q = tanh Ξ z (58) To sum up we find the equation of state within the RS ansatz q = ( ) dz e z tanh p βh + z π (βj) q p (59) This would be regarded as the Edwards-Anderson (EA) order parameter. Physically the nd term in (...) would be interpreted as the effective field (cavity field) produced by the surrounding spins. The same equation can be derived by analyzing the Thouless-Anderson-Parlmer (TAP) equations [7] or equivalently by the cavity method [8].. p = case: Sherrington-Kirkpatrick s analysis The RS solution p = was analyzed by Sherrington-Kirkpatrick [3]. The equation of state Eq. (59) reads in this case as dz ( q = e z tanh βh + z ) λ λ = (β) q (6) π Assuming q, h are small and using tanh(x) = x x 3 / and thus tanh (x) = x /3x +... for small x and using z z =, z z = 3, we find, q = (βh) + λ z z 3 (βh + z λ) z +... = (βh) + (βj) q (βj) q +... (6) Under zero magnetic field h = we find the continuous emergence of the spinglass order parameter: with the spinglass transition temperature, Under small magnetic field we find, q = (βj) (βj) T T c (6) q = χ SG h χ SG = T c /J =. (63) β (βj) (6) suggesting a spin-lass transition signaled by divergence of the spin-glass susceptibility χ SG approaching the spinglass transition temperature T T + c. These observations suggest a spinglass transition as a second order phase transition.. p limit Assuming h =, we find q =,λ = is always a solution: paramagnetic. The free-energy is obtained as, βf RS (q =, λ = ) = ln + (βj) (65) which is independent of p. Note that this agrees with the free-energy of the random energy model in the paramagnetic phase T > T c.

13 3 We assume k =, i.e. F. step replica symmetry breaking (RSB) λ ab = λ + λ I m ab and Q ab = q + q I m ab where m = m. With this ansatz we find, a b λ ab σ a σ b = λ We find doing the HS transformation,s with which we get where we introduced e Tre a b a= σ a σ b + λ λ = λ n [( σ a ) n] + λ λ σ a σ b C=,...,n/m a b a,b C C=,,...,n/m[( a C a b λ abσ aσ b = e n λ e λ ( a σa) + λ λ C ( a C σa) = e n λ e λ z a b λ abσ aσ b +βh a σa = e n λ n/m a σa z C= Dz { σ a ) m] (66) e λ λ z C a C σa zc (67) Dz [ cosh(ξ)] m } n/m (68) Ξ βh + λ z + λ λ z (69) Observing also the following we obtain G RSB (q, q, λ, λ, m) n Q p ab = n(m )qp + n(n m)qp a b λ ab Q ab = n(m )λ q + n(n m)λ q a b = (βj) + (βj) [ (m n)q p + ( ] m)qp [(m n)λ q + ( m)λ q ] + λ n log Dz { Dz [ cosh(ξ)] m } n/m (7) so that βf RSB (q, q, λ, λ, m) = (βj) + (βj) mq p + (βj) ( m)q p mλ q λ ( m)q + λ Dz log m Dz [ cosh(ξ)] m (7) where q and λ must verify the saddle point equations = f RSB q i λ i = p (βj) q p i (i =, ) (7) and after some algebra we find = f RSB q = λ ( ) Dz cosh m (Ξ) tanh(ξ) Dz Dz cosh m (73) (Ξ)) and = f RSB q = λ Dz Dz cosh m (Ξ) tanh (Ξ) Dz cosh m (Ξ)) (7)

14 Using Eq. (7) in Eq. (7) we find, βf RSB (q, q, m) = (βj) + p (βj) q p m (p ) (βj) mq p (p )(βj) ( m)q p Dz log Dz [ cosh(ξ)] m (75) Finally the parameter m must be extremized. Thus we find, = f RSB m = (p )(βj) (q p qp )+ m Dz ln Dz [ cosh(ξ)] m m The internal energy is obtained as Dz ln Dz [ cosh(ξ)] m ln[ cosh(ξ)] (76) e J = βf RSB = βj β + βj mqp + (βj) ( m)qp (77) Assuming h =, we easily find that q = (and thus λ = ) is a solution. We limit ourselves to this case in the following. In this case the calculation simplifies a lot. For clarity let us display the results below. The free-energy becomes βf RSB (q = q, q =, λ = λ, λ =, m) = (βj) [ ( m)q p ] + λ[ ( m)q] m log Dz[ cosh(z λ)] m (78) where q and λ are given by q = Dz cosh m (z λ) tanh (z λ) Dz cosh m (z λ) λ = p (βj) q p (79). p limit In particular let us examine the case p to compare with the random energy model. We notice that in the limit p, if we assume < q < the nd equation of Eq. (79) implies λ = which cannot satisfy the st equation. The only possibilities are (q, λ) = (, ) and (, ). Let us evaluate the free-energy for the solution q =, λ =. To this end we have to examine the integral, I = Dz[ cosh(zλ)] m (8) in λ limit. By noting that cosh(z λ) e z λsgn(z) we find With this we find Finally we require = m βf RSB which yields λ lim I = e m (8) λ βf RSB (q =, q =, λ =, λ =, m) = (βj) m + ln m (8) m = k BT J ln (83) which continuously decreases with the temperature. Comparing with the RS free-energy we find the RSB free-energy becomes lower than the RS energy for m < suggesting a critical temperature T c such that m(t c ) =. Thus we find k B T c = J/( ln ) (8)

15 5 with which we find m = T T c. (85) As the result the RSB free-energy becomes independent of the temperature below T c ; f RSB = J ln (86) The internal energy can be obtained from Eq. (77) as well. From the above results we realize that the exact results of the REM is recovered by the RSB solution.. Distribution of overlap P (q) Now we have to ask : what is the physical meaning of the replica symmetry breaking?. Let us consider two real replicas, say and and examine the probability distribution P (q) of the overlap q between them. More precisely we consider the disorder average of this distribution function, P (q) = lim N δ(q N N σi σ i (87) We wish to evaluate this using the RSB solution. To this end we just need to add n more replicas so that we have n replicas as a whole. We apply the RSB ansatz for this n replica system as discussed above. At this point we have to remember that there is a family of equivalent solutions which can be obtained by permuting the replica indicies of the Parisi s matrix and that all of them has the same statistical weights as discussed in sec I C and sec II D. The overlap between the two replicas Q can take either q or q depending on the permutations. Then assuming equal weights of all of these permuted solutions we find, ( ) P (q) = m(m ) n m δ(q q ) + n(n ) m(m ) n m n(n ) i= δ(q q ) n ( m)δ(q q ) + mδ(q q ) (88) To be specific let us consider the case of REM for which we know q =, q = and m = T/T c at < T < T c. In the limit T, m so that P (q) = δ(q ) which is expected since the two replicas should stay in the ground state at zero temperature. In the other limit T T c, m so that P (q) = δ(q). This could also be expected because down to T c the two replicas are allowed to stay in arbitrarily different states which happen to have almost the same energy. As discussed in sec II B (see Eq. (3) and Eq. (33)) the overlap between such states are typically. More generally, in the spinglass phase, < T < T c, we find an intriguing situation: the two distinct delta peaks one at q = q = and the other at q = q = coexists in P (q) whose relative weights evolves as m = T/T c increases with the temperature. See Fig. a) for a schematic picture. Actually this result can be obtained without using the replicas but directly developing a low temperature expansion of the REM as we discuss in Appendix A. With this approach the picture becomes clearer. In the low temperature limit, the peak at q = accounts for the case that both of the replicas remain in the ground state and the peak at q = accounts for the situation that one of the two replicas stay at the st excited state while the other one remains in the ground state. Note that typical overlap between the ground state and the excited states are. 3. System with many metastable states Based on the above observation in the p limit (REM), we may interpret the RSB solution in more general cases in the following way. In the RSB phase, there are multiple metastable spinglass states α =,,... whose mutual overlap is typically q. On the other hand q would be interpreted as the EA order parameter q EA, which can be regarded as self-overlap of metastable states. Then in REM (p ) it happens that q EA = suggesting that there is no thermal fluctuation within a metastable state of REM. This means that every spin configuration becomes a metastable state in the p limit. Then similarly to the case of REM Eq. (3), we naturally expect that the partition function of a generic p-spin model can be represented as a summation over the metastable states, Z = α e βnfα(t ) (89)

16 where f α (T ) is the free-energy/spin of the metastable state α. The above representation is just the same as that of the REM Eq. (3) but here the energy e α is replaced by the free-energy f α. This reflects the fact that q = q EA, which is in the REM (p ), is smaller than for general p suggesting thermal fluctuations within metastable states. In order words, for generic p, a free-energy minimum consists of not only an energy minimum but nearby spin configurations which defines a basin around the minimum in the phase space. Then the thermal averages would be decomposed as 6... = α w α... α w α = e Nβfα α e Nβfα (9) where... α is the thermal average with respect to the thermal fluctuation within α. For example the P (q) function can be rewritten as follows, P (q) = lim N δ(q N = α N σi σi = w α w α δ(q N σi σi ) α,α = w α w α δ(q q αα ) N i= α α i= α α w αδ(q q αα ) + α α w α w α δ(q q αα ) (9) where q α,α N N m i,α m i,α m i,α σ i α (9) i= is the mutual overlap between the metastable states α and α. For α = α, it is the self-overlap, which should be identified as the Edwards-Anderson order parameter q EA. To derive Eq. (9) we assumed the clustering property. (See Appendix B). In the case of RSB we have found Eq. (88) which implies q α,α = q, q α,α = q for α α, α w α = m and α α w αw α = m. We discuss later general connection between the Parisi s ansatz Eq. (9) and the overlap distribution function P (q).. Complexity and Monasson s observation Similarly to the case of REM we may rewrite the partition function Eq. (89) as Z = e βnfα(t ) = dfe N(Σ(f,T ) βf) (93) α where we introduced the so called structural entropy or complexity Σ(f, T ) N log α δ(f f α (T )). (9) The above representations Eq. (89), Eq. (93) and Eq. (9) are justified by studying the solutions of the equations of state of the p-spin model, which is the so called Thouless-Anderson-Palmer (TAP) equations. ion Monasson[9] proposed to introduce an extra parameter m into the game, Nβmφ m ln Z m Z m e Nmβfα(T ) = dfe N(Σ(f,T ) βmf). (95) α Formally evaluating the integral by the saddle point method we find, βmφ m = Σ(f, T ) βmf Σ(f, T ) f f=f (T,m) = m k B T (96) From this we find f = m (mφ m ) Σ = βm m φ m (97)

17 The complexity function Σ(f, T ) can be extracted from the above result varying m as a parameter. Now the above construction can be implemented using the replicas as follows: consider m replicas which are forced to stay together such that at any instance their configuration belong to a common metastable state. In the case of n replicas this amount to consider a situation such that n replicas are divided into n/m groups and that replicas belonging to a common group move around different states together while different groups move around independently from each other; Z n = Z n/m m = ( ) n/m e Nmβfα(T ) α Actually this is readily realized precisely by the RSB ansatz : the n replicas: are divided into n/m groups of size m and the replicas belonging to each of the groups have mutual overlap which is equal to q EA while mutual overlap q between different groups. This implies Thus we find Nβf RSB (q, q, m) = n (Z m ) n/m n= = ln Z m /m. (99) 7 (98) φ m = f RSB (q, q, m) () An important observation made by Monasson [9] is that the condition = m f RSB Eq. (76) for T T c is equivalent to require vanishing complexity Σ = because of Eq. (97). Let us check these explicitly in the case of REM. From Eq. (8) we readily find, βφ m = βf RSB = (βj) m + ln m. () Then using the prescription Eq. (97) we find ( ) ( ) Σ(f ) = m m (βj) m ln = ln (βj) m βf = m m (βj) ln m from which we find the complexity which agrees with Eq. (35) = (βj) m () Σ(f) = ln (f/j) (3) RSB solution continues to be present above T c up to T d (which is in the case of REM) where it finally disappears... Energy landscape changes at T d... The above ideas play key roles to develop the cloned liquid theory (replicated liquid theory) for structural glasses.[, ] 5. Finite p model: explicit computations on the p = 3 model Let us present here some results of the finite p models taking p = 3 as an example. See Fig. 6. Disentangling hierarchy of responses Let us discuss a generic feature of linear response in the system with many metastable state captured by the RSB ansatz following [], [3] and []. We consider a perturbing field h which is conjugated to a physical observable O. We denote the value of the observable/spin at state α as o α and that the associated linear susceptibility as χ α, o α = f α h χ α = o α h = f α h ()

18 8 a) b) c) complexity vs m (RSB saddle point) T= m_s(t) m_g(t) complexity T f m FIG.. RSB solution of the p = 3 Ising MFSG model: (left panel) complexity Σ(f, T =.5) vs f (center panel) The replicon eigenvalue λ R(m, T ) and complexity Σ (f (m, T ), T ) vs m at T =.5 and T =.6 (right panel) the static glass transition line ( Kauzmann line ) m s(t ) above which the complexity is positive Σ > and Gardner line m G(T ) below which the replicon eivgen value is positive. In other words o α is the value of the observable/spin o averaged over the intra-state fluctuation, i. fluctuations within the state α. Similarly χ α is related to the fluctuation of o within the state α. In the RSB ansatz we have e. the thermal βnf RSB (T, h) = m ln α e Nmβfα(T,h) (5) Then we find o(t, h) = f RSB(T, h) h = ln α e Nmβfα(T,h) = h βmn α w α o α = o α (6) where... α w α... is the average over different metastable states with the statistical weight w α e Nmβfα(T,h)... α e Nmβfα(T,h) (7) Similarly we find the susceptibility as, with χ RSB (T, h) = o(t, h) h = α w α h o α + α ( h w α )o α = χ + m χ (8) χ = χ α χ = Nβ( o α o α ) (9) Physically we can interpret χ as the response within a metastable state and χ as the response due to transition between different states. The total response is the mixture. One may wish to disentangle the two qualitatively different susceptibilities. A trick to disentangle the two susceptibilities was noticed in [] as follows. Let us apply different probing fields on each of the replicas, χ a,b (T, h) m f RSB (T, {h a }) h a h b = χ δ ab + χ {ha=h} We also observe that the total response is recovered as, = βn h a h b ln α e Nβ n a= fα(ha) {ha=h} m χ ab = χ + m χ = χ RSB (T, h). () b=

19 As an example let us consider the magnetic response of the p-spin Ising model. To this end we just need to apply different fields on h a = h + δh a (a =,,..., m) on the replicas as outlined above. Then Eq. (7) becomes βf RSB ({h,..., h m }) = (βj) + (βj) mq p + (βj) ( m)q p mλ q λ ( m)q + λ m Dz log Dz [ cosh(βh a + λ λ z + λ z )] () m from this we find the following susceptibility matrix, a= χ ab = f RSB ({h,..., h m }) h a h b ha=h 9 = χ δ ab + χ () with χ = Dz Dz [ cosh Ξ] m β( tanh Ξ) Dz [ cosh Ξ] m = β( q ) (3) where Ξ = βh + λ λ z + λ z and χ = β Dz [ cosh Ξ] m tanh Ξ) Dz Dz [ cosh Ξ] m ( Dz [ cosh Ξ] m tanh Ξ) Dz [ cosh Ξ] m ) = β(q q ) () The total susceptibility becomes which of course includes the RS result with m =, χ RS = β( q ). χ RSB = β[ q + m(q q )] (5) 7. Glass state following: Franz-Parisi potential Following the idea of Franz-Parisi [5], let us consider m+s replicas which are forced to stay in the same metastable state where we regard the m replicas as reference system and the s replicas as slave system on which we put a perturbation (e.g. temperature change) parametrized by the strength η. The partition function of the whole system can be written as, βf m+s = ln α e Nβ[mfα+sfα(η)] = ln [ α e Nβmfα [ + s( Nβf α (η)) + O(s ]] = βmφ m NβV FP (η)s + O(s ) (6) Here we expanded in power series of s around s =. In the last equation the st term is the free-energy of the reference system, which is nothing but Eq. (95) considered by Monasson [9], βmφ m = ln α e Nβmfα (7) while the nd term gives the free-energy of the slave system considered by Franz and Parisi [6], α V FP (η) = e Nβmfα f α (η) = f(η) reference (8) α e Nβmfα where... reference represents the averaging over the configurations of the references system, α reference α e Nβmfα (9) e Nβmfα As discussed in sec II F, the parameter m can be varied to select a group of states with specific value of free-energy (per spin). In Appendix F we discuss the implementation of this scheme to the p-spin Ising model.

20 G. d Almeida-Thouless-Gardner instability Let us discuss stability of the saddle points with respect to small fluctuations. To this end we examine the eigen modes of the Hessian matrix defined as H a b,c d = G[{Q ab}] Q a b Q c d () with G[{Q ab }] = (βj) n (βj) (p ) a b Q p ab ln Tr σ exp (βj) p Q p ab σa σ b + βh a b a σ a () which follows Eq. (5) and the st equation of Eq. (6). After some algebra, using the nd equation of Eq. (6), one finds, ( ) H a b,c d = (βj) p(p )Q p ab (δ (βj) ] acδ bd + δ ad δ bc ) p(p ) Q p ab Qp cd [ σ a σ b σ c σ d eff σ a σ b eff σ c σ d eff () where... eff is evaluated using Eq. (7).. Stability of the RS symmetric solution Let us first examine the stability of the replica symmetric solution [7]. generally expects the following structure. Because of the replica symmetry one δ ac δ bd + δ ad δ bc H a b,c d = M Then one finds the three group of eigen values + M δ ac + δ ad + δ bc + δ bd + M 3 (3) λ L = M + (n )(M + nm 3 ) λ A = M + (n )M λ R = M () where λ L,λ A,λ R are called as longitudinal, anomalous and replicon eigenvalues. Now we have to find M,M and M 3 from Eq. (). Within the RS ansatz Q ab = Q cd = q and σ a σ b eff = σ c σ d eff = q. We also need to evaluate σ a σ b σ c σ d eff which is obtained as the following, where σ a σ b σ c σ d eff = (δ ac δ bd + δ ab δ bc ) + [ δ ac ( δ bd ) + δ ad ( δ bc ) + δ bc ( δ ad ) + δ bd ( δ ac ) ] q + [ ( δ ac )( δ bd ) + ( δ ad )( δ bc ) ] r = (δ ac δ bd + δ ad δ bc )( q + r) (5) q = Dz tanh Ξ r = Dz tanh Ξ (6) with Ξ = z λ + βh λ = p (βj) q p (7) In particular, the replicon eigenvalue is obtained as M 3 = (βj) p(p )q p { where (βj) p(p )q p ( q + r) } (8) q + r = Dz( tanh Ξ) = Dz sech Ξ (9) Then positivity of the replicon eigenvalue λ R = M 3 > implies ( ) kb T J p(p ) q p > This is the so called d Almeida-Thouless condition [7]. Dz sech Ξ (3)

21 . Instability of the RSB solution:the Gardner s transition Here we limit ourselves with h = for which q = is a solution. λ R = M 3 > implies Then positivity of the replicon eigenvalue ( ) kb T J p(p ) q p > Dz cosh m Ξ sech Ξ Dz cosh m Ξ Ξ = z λ (3) with λ = p (βj) q p. This was studied first by Gardner [8]. H. More RSB: full replica symmetry breaking It is straightforward to consider generalization to k-rsb and eventually take k []. We plug in the Parisi s ansatz Eq. (9) into the general expression of the free-energy functional Eq. (5) which reads, G({Q ab, λ ab }) = (βj) n (βj) Q p ab + λ abq ab ln Tr σ exp λ ab σ a σ b + βh a b a b a b a Given the Parisi s ansatz Eq. (9) (See Fig. ) which reads, we easily obtain n Q p ab = a b k b( a) i= Q ab = q + q p i (Imi ab Imi+ ab ) = k k (q i q i )I mi ab = q i (I mi ab i= i= k q p i ((m i ) (m i+ )) = i= σ a (3) Imi+ ab ) (33) k q p i (m i m i+ ) dxq(x) p (3) κ In the last equation dm i = m i m i+ dx (note m i decreases with increasing i when n becomes smaller than (See Fig. d)). Similarly we find Next step is to evaluate n a b λ ab q ab k n i= dxλ(x)q(x) (35) A k = Tr exp λ ab σ a σ b + βh σ a = e n λ k Tr exp λ ab σ a σ b exp [βhσ a ] (36) a b a a,b a = e n λ k exp λ ab cosh(h a ) h a h b (37) a,b a ha=βh = e n λ k exp λ k exp h a h b (λ i λ i ) I mi ab cosh(h a ) h a,b i= a h b (38) a,b a ha=βh = e n λ k exp λ k exp h a h b (λ i λ i ) I mi ab cosh(h a ) h a,b i= a h b (39) a,b a n ha=βh

22 . RS case For an exercise let us consider k = (RS) case: A = e n λ exp λ ( cosh(βh a )) h a h b a,b a [ ] = e n λ λ exp h ( cosh(βh)) n = e n dy λ e y λ ( cosh(βh)) n = e n λ πλ ha=h Dz( cosh(βh + λ z)) n () In the nd equation we used a simple identity like, ( + ) f(x )f(x ) x x = x=x =x x f(x) () We also used the identity [ ] λ exp dx f(x) = dy e y λ f(x y) γλ f(x) = πλ Dze z f(x λz). () This can be checked by formally expanding f(x y) around y =. Here γ a represents a Gaussian with zero mean and variance a. the symbol represents a convolution and Collecting the above result and noting that m = n, m = for the RS (k = ) case, we recover Eq. (55), G RS (q, λ) n from which we find = (βj) + (n ) ( ) (βj) q p + λ q + λ n ln Dz( cosh(βh + λ z)) n (3) βf RS (q, λ) = (βj) ( q p ) + λ( q) ln( cosh(ξ)) z (). -RSB, -RSB and k-rsb One easily finds, A = e n λ exp λ exp h a h b (λ λ ) I m ab cosh(βh a ) h a h b a,b a,b a = e n λ exp λ n/m h a h b exp (λ λ ) cosh(βh a ) h a h b a,b C= a,b C a C = e n λ exp λ [ dz ] n/m e z ( cosh(βh + λ λ z)) m h a h b π = e n λ Similarly one can find, Dz [ a,b Dz ( cosh(ξ )) m ] n/m Ξ βh + λ λ z + λ z (5) A = e n λ Dz [ Dz [ Dz [ cosh(ξ ) ] m ] m/m ] n/m Ξ = Ξ + λ λ z (6)

23 3 and A 3 = e n λ3 Dz Dz [ Dz [ ] [ Dz 3 ( cosh(ξ )) m3]] m m/m /m 3 n/m Ξ 3 = Ξ + λ 3 λ z 3 (7) and so on. Now we are naturally lead to define a family of functions g(m i, y) for i =,,..., k which satisfy a recursion relation, g(m i, y) = e λ i λ i with the boundary condition, y g(mi+, y) mi/mi+ = γ λi λ i g(m i+, y) mi/mi+ = Dz i g(m i+, y + λ i λ i z i ) mi/mi+ (8) g(m k+ =, y) = cosh(y). (9) With these we find, A k = e n λ k Dz g(m, βh + λ z ) n/m. (5) which is valid for k =,,,...,. The we find G krsb ({q, λ}) n and thus = (βj) (βj) βf krsb ({q, λ}) = (βj) (βj) k q p i (m i m i+ )+ i= In the last equation we introduced, k i= k q p i (m i m i+ )+ i= λ i q i (m i m i+ )+ λ k n ln k i= λ i q i (m i m i+ )+ λ k Dz g(m, βh+ λ z ) n/m (5) Dz f(m, βh+ λ z ) (5) f(m i, y) = m i ln g(m i, y]) (53) whose recursion relation can be obtained as, [ f(m i, y) = ln γ λi λ i e mif(mi+,y)] = ln m i Dz i e mif(mi+,y+ λ i λ i z i) (5) with the boundary condition, f(m k+ =, y) = ln cosh(y). (55) 3. Variational equations Now we have to obtain the self-consistent equations for λ i (i =,,,..., k) and q i (i =,,,..., k), = βf krsb q i = βf krsb λ i q i = λ i = (βj) pq p i (56) { Dz f(m, β + λ z ) } m i m i+ λ i δ i,k (57) For the time being we focus on Eq. (57) for the case i k, i. e. i =,,..., k. We come back later to consider i = k case. By a repeated use of the recursion formula Eq. (8) we find λi g(m i, y) = m i Dz i g(m i+, Ξ i ) mi/mi+ λi g(m i+, Ξ i ) (58) m i+

24 where Ξ i = y + λ i λ i z i and for i k, λi g(m i+, y) = m i+ Dz i+ g(m i+, Ξ i+ ) mi+/mi+ λi g(m i+, Ξ i+ ) (59) m i+ with Ξ i+ = Ξ i + λ i+ λ i z i+ = y + λ i λ i z i + λ i+ λ i z i+ thus for i k, ( ) λi g(m i+, Ξ i ) = g (m i+, Ξ i+ ) z i+ + z i. (6) λi+ λ i λi λ i Here and in the following we use the following shorthand notation A(x, A(x, y) y) x A (x, y) A(x, y) y (6) Collecting the above results and doing integrations by parts one finds after some algebra for i k, λi g(m i, y) = m i(m i m i+ ) Dz i g(m i+, Ξ i ) ( mi/mi+ y f(m i+, Ξ i ) ) = (m i m i+ ) dh δg(m i, y) ( f (m i+, h) ) δf(m i+, h) Similarly one can show that for i j k, λj g(m i, y) = (m j m j+ ) dh δg(m i, y) ( f (m j+, h) ) δf(m j+, h) (6) (63) and thus equivalently for i j k, λj f(m i, y) = (m j m j+ ) dhp ij+ (y, h) ( f (m j+, h) ) (6) where we introduced P ij (y, z) δf(m i, y) δf(m j, z). (65) Now collecting the above results we can rewrite Eq. (57) as q i = Dz dhp,i+ (βh + λ z, h)(f (m i+, h)) = dhp i (h)(f (m i+, h)) (66) where we introduced P i (h) = Dz P,i+ (βh + λ z, h). (67) The function P i (h) can be understood physically as distribution function of internal field ( at level i of the hierarchy). And f (m i+, j) as the magnetization which is thermalized at level i + and subjected to slower random field at level i. One can find from Eq. (5) that f (m i, h) obeys a recursion formula. f (m i, h) = Dzi e mif(mi+,y+ λ i λ i z i) f (m i+, h) Dzi e mif(mi+,y+ λ i λ i z i) (68) with the boundary condition f (m k+, h) = tanh(h). The remaining task is to find a way to compute P ij (y, z). To this end let us begin with the following object, R ij (y, z) δg(m i, y) δg(m j, z) (69)

25 5 which can be rewritten by the chain rule,for j i +, as, R ij (y, z) = dxr i,j (y, x)r j,j (x, z) R j,j (x, z) = m j m j e (x z) (λ j λ j ) π(λj λ j ) gmj /mj (m j, z) (7) where the nd equation can be obtained using the recursion formula Eq. (8). Then we obtain the recursion formula for R ij, R ij (y, z) = m j m j g mj /mj (m j, z)γ λj λ j z R i,j (y, z) (7) The two objects P ij (y, z) and R ij (y, z) are simply related by P ij (y, z) = (m j /m i )(g(m j, z)/g(m i, y))r ij (y, z). Thus we find P ij (y, z) = g mj /mj (m j, z)γ λj λ j z P i,j (y, z) g(m j, z) (7) where A z B(y, z) stands for a convolution with respect to z. This recursion formula can be solved with the boundary condition Similarly the internal field distributions P i (z) follows the same recursion formula P ii (y, z) = δ(y z). (73) P i (z) = g mi/mi+ (m i+, z)γ λi λ i P i (z) g(m i, z) (7) for i =,,..., k with the boundary condition, P (z) = Dz δ(βh + (z βh) λ z z) = e λ (75) πλ One can check easily that dzp i (z) = is satisfied for i = and the recursion relation Eq. (7) preserves the normalization. Finally let us come back to case i = k of Eq. (57). To this end we begin by observing the following, which follows from the recursion relation Eq. (8) and an integration by parts, g(m k, y) = λ k λ k Dz k g(m k+, y + λ k λ k z k ) m k/m k+ = m k Dz k g(m k+, Ξ k ) m k = dh g(m k+, h) [ ] (m k )(f (m k+, y)) + f(m k+, y) [ ] (m k )(f (m k+, Ξ k )) + Here we also used the fact that g(m k+, y) = g (m k+, y) = cosh(y). More generally we find, f(m i, y) = λ k [ ] dhp i,k+ (h, y) (m k )(f (m k+, y)) + (76) (77) Then Eq. (57) for i = k becomes ( q k = m k Dz f(m, β + λ z ) ) = λ k dhp k (h)(f (m k+, h)) (78) Thus this is formally the same as those i k given in Eq. (66). It is a simple exercise to check that the above results reproduce the results of k = RSB, i.e. Eq. (59) and k = RSB, i.e. Eq. (73) and Eq. (7). The values of q i =,,,..., k in the k( ) RSB ansatz with m i s fixed as = m < m <... m k < m k+ = can be solved numerically by iterating the following steps:

26 6 q(x).8 P (h)..5 T= T= x h FIG. 3. Numerical solution of the k-rsb ansatz in the case of the SK model (p = ) whose critical temperature is T c/j =. Here k = 6. (Left panel) The order parameter function q(x) and (Right panel) the internal field distribution P (m k+ =, h). make an initial guess on q i for i =,,,..., k by which we also obtain λ i for i =,,,..., k via Eq. (56): λ i = (βj) pq p i.. evaluate the functions f(m i, h) for an appropriate range of h doing Gaussian convolutions recursively in the order (i = k k... ) using the recursion formula Eq. (5) with the boundary condition f(m k+, h) = ln( cosh(h)). Simultaneously compute also f (m i, h) using the recursion formula Eq. (68) with the boundary condition f(m k+, h) = tanh(h). 3. evaluate the functions P i (h) by doing another series of Gaussian convolutions recursively in the order (i =... k k) using the recursion formula Eq. (7) with the boundary condition P (z) = e z / / π. Then q i for i =,,,..., k can be determined using Eq. (66). Go back to the initial step. In Fig. 3 we show an example of the numerical analysis done for the SK model (p = ).. Continuous RSB: k limit Finally let us consider k limit [, 8]. Writing m i = x δx,m i = x and λ i λ i = λ(x)δx, the recursion relation Eq. (8) becomes a partial differential equation, called as Parisi s equation. g(x, y) dx = λ(x) which must be solved with the boundary condition, Equivalently for dy g(x, y) g(x, y) ln g(x, y) (79) x g(, y) = cosh(y). (8) f(x, y) ln g(x, y) (8) x we find f x = λ(x) [ ( ) ] f f dx + x dx (8) which must be solved with the boundary condition, f(, y) = ln cosh(y). (83)

27 Similarly the recursion relation for the distribution of the internal field Eq. (7) becomes, after some algebra, the following partial differential equation in the limit k, P (x, y) = λ(x) ( P (x, y) x(p (x, y)f (x, y)) ) (8) x where the dots and dashes are the shorthand notations introduced before Eq. (III F 3). The initial condition is given by Eq. (75) which reads as, (y βh) P (, y) = e λ(). (85) πλ() 7 With these the free-energy becomes where βf RSB = (βj) + (βj) dxq p (x) λ(x) = (βj) pq(x) p q(x) = dxλ(x)q(x) + λ() Dz f(m, βh + λ()z ) (86) dhp (x, h)(f (x, h)). (87) In Fig. 3 an interesting feature emerges at low temperature on the distribution of the internal field P (, h): opening of a sudo-gap. This reflects the marginal stability of the full RSB spinglass phase where one finds massless modes, e. g. λ R =. I. Hierarchical energy landscape and linear response with, 3,..., k RSB In sec. II F we have argued that the RSB ansatz is equivalent to assume factorization of the replicated partition function as, βnf RSB = n Z n n= Z n = exp [ Nβm f α ] n/m. (88) α Here n replicas are group into n/m clusters of size m and all the replicas belonging to a given cluster are forced to stay in the same state. Then it is natural to expect that the RSB ansatz is equivalent to make a further decomposition, Z α = exp [ Nβm f α ] = exp [ Nβm f α ] (89) α α using this in Eq. (88) we find βnf RSB = n Z n n= Z n exp [ Nβm f α ] α α α m n/m /m (9) Here we recognize that each clusters is decomposed into m /m sub-clusters of size m such that the replicas belonging to a given sub-cluster are forced to stay in the same state. Moreover we see that in the RSB ansatz the states labeled by α are grouped into meta-states or metabasins labeled by α such that the sub-clusters belonging to a given cluster are forced to stay in the same meta-basin, which is a union of states. Now we wish to extend II F 6 [], [3] to -RSB (and thus to k-rsb) so that we can disentangle the linear responses in the hierarchical energy landscape []. One has o(t, h) = f RSB(T, h) h = o α = α w α α α w α α o α (9)

28 where o α = hf α is the value of the physical observable o averaged over the thermal fluctuation within the state α. We also introduced... α w α α w α α... (9) with w α and w α α, which can be interpreted as the statistical weights of the metabasins and states, w α = Zm/m α w α α α Zα m/m = e Nmβfα Z α = e Nmβfα. (93) Z α α α Now one can find, χ RSB = w α w α α h o α + α α α with = χ + m χ + m χ α w α h w α α o α + h w α w α α o α α α α α α 8 (9) χ = χ α χ = Nβ[ o α o α χ = Nβ[ o α o α (95) Physically we can interpret χ as the response within a state, χ as the response due to transition between different states within a common metabasin and finally χ as the response due to transition between different metabasins. The total response is the mixture. Just as we did for the RSB case we wish to disentangle the three qualitatively different susceptibilities. Again let us suppose that each of the replica is subjected to difference probing fields. Then one easily finds, and χ a,b (T, h) m f RSB (T, {h a }) h a h b {ha=h} = χ δ ab + χ I m ab + χ m χ ab = χ + m χ + m χ = χ RSB (T, h). (96) b= Generalization to k-rsb is now straightforward but we do not attempt it here because it would require heavier notation describe the hierarchy of the clusters. J. Overlap distribution function P (q) and the order parameter function q(x) We have already discussed the overlap distribution P (q) for the case of RSB. Here we discuss the connection between P (q) and the Parisi s ansatz Eq. (9) for the more general cases. The structure of the matrix is shown in Fig.. We can repeat the argument in sec II F for the general case. The probability to observe the overlap q within the section δq i = q i+ q i will be given by, P (q)δq i = where δm i = m i+ m i. Thus a b δ(q ab q)δq i n(n ) = m i(m i ) n m i m i+ (m i+ ) n m i+ n(n ) P (q) = δm i δq i k dx(q) dq n δm i (97) where we took the continuous limit by k and recalled m as x. The above result establishes the connection between the order parameter function q(x) which parametrized the Parisi s ansatz and the overlap distribution function P (q). In Fig. we display the two important examples: RSB and continuous RSB cases. Now considering three replicas, one can ask distribution of mutual overlaps, q, q 3 and q 3. By doing the same argument as above one finds, P (q, q 3, q 3 ) = lim n n(n )(n ) a b c (98) δ(q ab q )δ(q ac q 3 )δ(q bc q 3 ) (99) Then, just from the structure of the matrix, we find, the smallest two among q, q 3, q 3 are equal, which means ultrametricity [8].