Statistical mechanics of glasses and jamming systems: the replica method and itsapplications

Hajime Yoshino1, 2

1Cybermedia Center, Osaka University, Toyonaka, Osaka 560-0043, Japan2Graduate School of Science, Osaka University, Toyonaka, Osaka 560-0043, Japan

2

CONTENTS

I. Introduction 4A. Edwards-Anderson model 4B. Edwards-Anderson order parameter: the glass order parameter 4C. Replicas and the glass order parameter 5

1. Two replicas 52. n replicas and the replica symmetry 6

II. Prototypes: mean field spinglass models 7A. p-spin models 7B. Random energy model 8

1. p→∞ limit of the p-spin model 82. Exact analysis of the thermodynamics 9

C. Replicated free-energy functional 9D. Use of Parisi’s ansatz 11E. Replica symmetric solution 11

1. p = 2 case: Sherrington-Kirkpatrick’s analysis 122. p→∞ limit 12

F. 1 step replica symmetry breaking (1RSB) 131. p→∞ limit 142. Distribution of overlap P (q) 153. System with many metastable states 154. Complexity and Monasson’s observation 165. Finite p model: explicit computations on the p = 3 model 176. Disentangling hierarchy of responses 177. Glass state following: Franz-Parisi potential 19

G. d’Almeida-Thouless-Gardner instability 201. Stability of the RS symmetric solution 202. Instability of the 1RSB solution:the Gardner’s transition 21

H. More RSB: full replica symmetry breaking 211. RS case 222. 1-RSB, 2-RSB and k-RSB 223. Variational equations 234. Continuous RSB: k →∞ limit 26

I. Hierarchical energy landscape and linear response with 2, 3, . . . , k RSB 27J. Overlap distribution function P (q) and the order parameter function q(x) 28

III. Replicated liquid theory 30A. Our system 30B. Liquid theory 30

1. Mayer function 302. Density functional 30

C. Cloned liquid: basic ideas 311. Glass order parameter 312. Playing with the number of replicas 32

D. Density functional of the replicated free-energy 32E. Construction of the gaussian ansatz 33

1. General form 332. Hierarchical gaussian ansatz 343. Another representation of the hierarchical Gaussian 34

F. Free-energy 351. Entropic part of the replicated free-energy with the hierarchical gaussian ansatz 352. Interaction part of the replicated free-energy with the hierarchical gaussian ansatz 363. Variational equations 374. Effective potential and g(r) 385. Continuous RSB: k →∞ limit 396. d→∞ limit 39

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IV. Hardspheres in d→∞ limit 42A. 1 RSB: dynamical transition, Kauzmann transition, Gardner transition,... 43

1. Free energy 432. Effective potential 433. Another representation 444. Thermodynamics of the liquid 445. Dynamical glass transition point in the liquid phase 446. Cage size in glassy phases 447. Thermodynamics in glassy phases 458. Jamming 469. Gardner transition 47

B. Full RSB: Jamming critically and beyond 471. Vanishing cage size 472. Anomalous features in the contact region 473. Complex responses to shear in the Gardner phase 49

C. Glass state following under compression and shear 50

V. Outlook 52

Acknowledgments 52

References 52

A. Properties of low lying states in the glass phase of the random energy model 531. Distribution of energy gaps 532. Two level system 53

B. Clustering property 54

C. p-spin spherical model 541. RS ansatz 552. 1RSB ansatz 55

D. Useful formulae of Gaussian integrals 55

E. ”Spin” density functional approach 561. RS Gaussian ansatz 562. 1RSB Gaussian ansatz 573. kRSB Gaussian ansatz 58

F. Glass state following: p-spin Ising model 59

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I. INTRODUCTION

A. Edwards-Anderson model

A standard theoretical model for spin glasses is the so called Edwards-Anderson (EA) model [1] which is definedas follows.The Hamiltonian is given by,

H[{S}] = −∑

〈i,j〉JijSiSj − h

N∑

i=1

Si, (1)

where Si is the spin variable on site i(= 1, 2, . . . , 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 = ±1, 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,∑Ni=1 S

2i = N .

The matrix Jij represents the interaction between the spins which can be ferromagnetic Jij > 0 or antiferromagneticJij < 0. A standard choice is a Gaussian distribution with zero mean and variance J2/z with z being the coordinationnumber on the lattice (z = 2d for the hypercubic lattice).

P (Jij) =1√

2πJ2/zexp

[−z

2

(JijJ

)2]. (2)

The last term on the r. h. s. of Eq. (1) is the Zeeman energy term with the external magnetic field h. If h = 0,the Hamiltonian is invariant under Si → −Si ∀i, i. e. time reversal symmetry. In the EA model, with zero externalfield h = 0, the spinglass transition must break this symmetry much as the ferromagnetic phase transition. As soonas h 6= 0, this symmetry is lost from the beginning. The long standing problem in the spinglass physics is whether ornot any phase transition is present in the case h 6= 0. The mean-field theory, which is exact in the large dimensionallimit, 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.

B. Edwards-Anderson order parameter: the glass order parameter

Edwards-Anderson (EA) order parameter [1] is defined as follows. One way to defined it is the dynamical one whichsignals ergodicity breaking:

qEA = limt→∞

C(t) (3)

where C(t) is the spin auto-correlation function

C(t) = limN→∞

1

N

N∑

i=1

〈Si(t)Si(0)〉eq (4)

where 〈. . .〉eq represents the thermal average within an equilibrium state. For the following discussion it is useful todecompose the spin auto-correlation function as

C(t) = C(t) + limN→∞

1

N

N∑

i=1

〈Si〉2eq C(t) ≡ limN→∞

1

N

N∑

i=1

〈(Si(t)− 〈Si〉)(Si(0)− 〈Si〉)〉eq (5)

We generally expect that the “connected part” C(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 limt→∞ C(t) = 0. Based on this expectationwe find a second definition of the EA order parameter,

qEA = limN→∞

1

N

N∑

i=1

〈Si〉2eq, (6)

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 = 0. We expect at high enough temperatures the equilibrium state is paramagnetic so that limN→∞〈Si〉eq = 0because of the the time-reversal symmetry Si → −Si(∀i). Then we conclude that qEA = 0 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 EAorder parameter, as defined above, becomes finite qEA > 0 in the spinglass phase.

Now what happens if the external magnetic field is finite |h| > 0? In this case, the time reversal symmetry is alwaysbroken such that the magnetization,

m =1

N

N∑

i=1

〈Si〉 (7)

should be always finite |m| > 0. In this case we have to redefine the EA order parameter by replacing Si by Si −m:

qEA =1

N

N∑

i=1

〈δSi〉2 δSi = Si −m (8)

As mentioned before, whether or not there remains any spinglass transition which does not involve the spontaneoustime reversal symmetry breaking is an intriguing and long-debated issue.

C. Replicas and the glass order parameter

1. Two replicas

Let us consider two replicas a and b. Each of them has its own spin variables Sai (i = 1, 2, . . . , N) and Sbi(i = 1, 2, . . . , N) but obey exactly the same Hamiltonian H[{S}](that is the realization of Jij). Then we may considera composite system,

HA+B = H[{Sa}] +H[{Sa}]− εN∑

i=1

qi qi = Sai Sbi (9)

where we added an artificial coupling between the two replicas. In the following we assume h = 0 for simplicity. Forthe case |h| > 0, we will just need to replace qi by qi = δSai δS

bi similarly to Eq. (8)).

If the coupling is attractive, i.e. ε > 0, the two replicas would take somewhat similar spin patterns. In other words,the overlap between the two replicas,

Q(N, ε) ≡ 1

N

N∑

i=1

〈qi〉N,ε, (10)

will be larger for larger ε > 0. Here 〈. . .〉N,ε is the canonical ensemble average of the composite system with couplingε and system size N . By defining the free-energy of the composite system Fa+b(N, ε),

− βFa+b(N, ε) = ln(TrSaTrSbe

−β(H[{Sa}]+H[{Sb}])+βε∑Ni=1 S

ai S

bi

)(11)

we find

Q(N, ε) = − 1

N

∂Fa+b(N, ε)

∂ε. (12)

Particularly interesting is the comparison of the two following two limits,

limε→0+

limN→∞

Q(N, ε) limN→∞

limε→0+

Q(N, ε) (13)

• In the paramagnetic phase (liquids) both of the limits should give 0.

• On the contrarily, in spin glass phase the two should give different results because there must be more than oneequilibrium spinglass states. Thus it is natural to expect,

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– By taking N →∞ first, in the presence of the attractive coupling, we bring the two replicas into the samespinglass state. Thus it is natural to expect that we find the EA order parameter qEA,

limε→0+

limN→∞

Q(N, ε) = qEA > 0. (14)

– Now in the other case, which takes ε → 0+ first, we are left with two decoupled replicas. In this case thetwo replicas may well end up staying in different states so that we expect,

limN→∞

limε→0+

Q(N, ε) < qEA. (15)

Based on the above observations, it becomes natural to define the so called spinglass susceptibility,

χSG =1

N

∂2Fa+b(N, ε)

∂ε2

∣∣∣∣∣ε=0

=1

N

∑

ij

(〈qiqj〉 − 〈qi〉〈qj〉 =1

N

∑

ij

(〈SiSj〉 − 〈Si〉〈Sj〉

)2. (16)

If the spinglass transition takes place as a 2nd order phase transition, accompanying divergence of the correlationlength of the the thermal fluctuation of the local overlap qi, the spinglass susceptibility would diverge approachingthe transition point.

2. 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 replicasa = 1, 2, . . . , n and consider a composite system which is described by the replicated Hamiltonian,

Hn =

n∑

a=1

H[{Sa}]−∑

a<b

εab

N∑

i=1

Sai Sbi (17)

for which we naturally define

Qab(N, {ε}) =1

N

N∑

i=1

〈Sai Sbi 〉N,{ε} (18)

Then we become naturally interested with Qab = lim{ε→0+} limN→∞Qab(N, {ε}).An obvious but important symmetry property of the replicated Hamiltonian Eq. (17) with εab = 0 ∀εab is that

the system is invariant under permutations of the replica indexes. This permutation symmetry is called as replicasymmetry. Parisi considered the possibility of spontaneous breaking of this symmetry [2] proposing the followingansatz for the structure of the glass order parameter in the spin glass phase.

Qab = q0 +

k∑

i=1

(qi − qi−1)Imiab =

k∑

i=0

qi(Imiab − I

mi+1

ab ) (19)

where Imab is a kind of generalized (’fat’) identity matrix of size n × n composed of blocks of size m × m. (seeFig. /reffig:parisi-matrix) The matrix elements in the diagonal blocks are 1 while those in the off-diagonal blocks areall 0. The Parisi’s matrix has a hierarchical structure such that

1 = mk+1 < mk < . . . < m1 < m1 < m0 = n (20)

which becomes

0 = m0 < m1 < . . . < mk < mk+1 = 1 (21)

in the n→ 0 limit.

• Replica symmetric (RS) ansatz:

The so called replica symmetric (RS) ansatz corresponds to the case k = 0. In this ansatz permutations of thereplica indexes does not affect the matrix structure. In order words, in the RS ansatz the permutation symmetryof the replicated Hamiltonian is respected.

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0

1

mi

mi

n

n

qi

mi+1

mi+1 qi+1

1

1

0

0

1

1

mi

mi

n

n

q1n1

mk m1m2 m0mk+1

n 1m1 m2

q1

0 mk

q1

qk

q(x)

q2

a) b) c)

d)

FIG. 1. Parametrization of the Parisi’s matrix a) the ’fat’ identity matrix Imiab b) Parisi’s order parameter matrix Eq. (19) c)the hierarchy of the sizes mi of the sub-matrices d) the q(x) function with 0 < n < 1.

• Replica symmetry breaking (RSB) ansatz:

For k ≥ 1, the replica symmetry is broken (RSB), i.e. the permutation symmetry of the replicas is violated. Inthe k-th RSB, the permutation symmetry holds only within a sub-block of size mk.

By selecting any pair of replicas among the n replicas and exchanging them we will obtain another Parisi matrixQ′ab. By considering all such permutations we find a family of RSB ansatz. We will see that they have exactlythe same equilibrium measure. Then choosing one out of all of them means to consider a spontaneous breakingof 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. TheSherrington-Kirkpatrick (SK) model [3] is just a special case p = 2 of this, which corresponds to the d = ∞ limit ofthe EA model.

H = −∑

1≤i1<i2···ip≤NJi1i2···ipSi1Si2 · · ·Sip − h

∑

i

Si (22)

P (Ji1i2···ip) =

√Np−1

πJ2p!exp

[−N

p−1

p!

(Ji1,i2,...,ip

J

)2]

(23)

We denote the average over the quenched disorder as

· · ·J ≡∫ ∏

1≤i1<i2···ip≤NdJi1i2···ipP (Ji1i2···ip) · · · (24)

In the following we mainly consider the p-spin model with Ising spins Si → σi = ±1. A great advantage for ourpedagogical purposes is that it becomes identical to the random energy model (REM)[4] in the p→∞ limit [5], whichcan be analyzed exactly without using replicas. Another standard version is the spherical model which we discussbriefly in the appendix C. The latter is also quite instructive and useful especially to analyze connection between thestatics and dynamics [6].

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B. Random energy model

We consider the random energy model (REM) [4]: to each of the micro-states α = 1, 2, . . . , 2N a random energyEα = NeαJ is assigne which is drawn from a Gaussian distribution with zero mean and variance NJ2/2,

p(E) =e−E

2/NJ2

√πNJ2

(25)

1. 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. Tosee this, let us examine the distribution of the energy of generic spin configurations.

First we examine the distribution of the energy EJ(σ) of a spin configuration σ = (σ1, σ2, . . . , σN ), among differentrealization of the quenched disorder,

p(E) ≡ δ(E − EJ(σ))J

(26)

It is easy to evaluate this,

p(E) =

∫dκ

2πeiκEe−iκEJ (σ)

J=

1√NπJ2

e−E2/NJ2

. (27)

Next we look at the joint distribution of the energy of two spin configurations σ(1) and σ(2),

p(E1, E2) ≡ δ(E1 − EJ(σ(1)))δ(E2 − EJ(σ(2)))J

(28)

which is found to be

p(E1, E2) =1

2

1√NπJ2A+

1√NπJ2A−

exp

(− E2

+

NJ2A+− E2

−NJ2A−

)(29)

where E± = E1±E2

2 and

A± =1

2[1± qp] q =

1

N

N∑

i=1

σ(1)i σ

(2)i . (30)

By noting that limp→∞A± = 1/2 for |q| < 1 we find,

p(E1, E2)→ p(E1)p(E2) (p→∞, |q| < 1). (31)

This means that the energies of two arbitrarily chosen configurations are statistically independent from each other inthe p→∞ limit: the random energy model. From the above discussion we also find a corollary that in p→∞ limit,

p(E1, E2, q) = p(E1)p(E2)P (q). (32)

The overlap between two arbitrary chosen configurations σ(1) and σ(2) are easily obtained as,

P (q) =1

2NN !

(N((1 + q)/2))!(N((1− q)/2))!−−−−→N→∞

δ(q). (33)

The last equation follows from the observation that limN→∞ lnP (q) = −N2 q2 + O(q4) which can be obtained usingthe Stirling’s formula.

9

2. Exact analysis of the thermodynamics

The partition function of the REM is computed as

Z =

2N∑

α=1

e−βEα = 2N∫de

1

2N

2N∑

α=1

δ(e− eα)e−NβJeα =

∫deeN(Σ(e)−βJe) (34)

with

Σ(e) = ln 2− e2 emin < e < emax (35)

with emin = −√

ln 2 and emax =√

ln 2.In the N →∞ limit, the integral can be evaluated by the saddle point method yielding,

−N−1βF = N−1 lnZ = Σ(e∗(T ))− βJe∗(T )dΣ(e)

de

∣∣∣∣e=e∗(T )

= βJ (36)

From the saddle point equation (the 2nd equation) we find the internal energy as,

e∗(T ) =

{−βJ/2 T > Tc

emin T < Tc

where Tc is the critical temperature

kBTc/J = − 1

2emin=

1

2√

ln 2(37)

below which the saddle point is stuck at emin. Thus we find the free-energy and entropy as,

− βF/N =

{ln 2 + (βJ)2/4 T > Tc

−βemin T < Tc

S/N = Σ(e∗(T )) =

{ln 2− (βJ)2/4 T > Tc

0 T < Tc

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,

− βFJ = lnZJ ZJ = Trσ exp

β

∑

i<j<k

Jijkσiσjσk + βh∑

i

σi

(38)

where Trσ ≡∏Ni=1

∑σi=±1. Apparently the free-energy depends on how the random couplings Jijk are chosen so that

we wrote it as FJ . However we expect that the free-energy/spin converges to a unique value in the thermodynamicslimit, i.e. self-averaging. And by the same token this should be the same as the disorder-averaged one. Thus weexpect

f ≡ limN→∞

FJJ

N= limN→∞

FJN

(39)

To compute FJJ

, we rely on the replica trick: assuming that ZnJ can be expanded as ZnJ = 1 + n lnZJ + O(n2) forsmall n we obtain,

f = limN→∞

1

N∂nZnJ |n=0. (40)

10

In the following we proceed as follows. First we evaluateZnJ assuming that n is an integer. Then we make an analyticcontinuation of the resultant formula to real values of n.

Now we consider the partition function of a replicated system a = 1, 2, . . . , n and take its disorder average. Aftersome algebra we find,

ZnJ =

∫ ∏

i<j<k

dJijkP (Jijk)Tr{σ} exp

β

∑

i<j<k

Jijk

n∑

a=1

σai σaj σ

ak + βh

∑

i,a

σai

= Tr{σ} exp

(βJ)2

4N

n+

∑

a6=b

N−1

N∑

i=1

σai σbi

p+ βh

∑

i,a

σai

(41)

where Tr{σ} =∏na=1 Tr{σa}. Now let us introduce the overlap matrix Qab,

Qab ≡1

N

N∑

i=1

σai σbi (42)

with which we find

ZnJ =

∫ ∏

a<b

dQabe−Nnβf [{Qab}] (43)

e−Nnβf [{Qab}] = Tr{σ}∏

a<b

Nδ(NQab −

N∑

i=1

σai σbi )

exp

(βJ)2

4N

n+

∑

a 6=b

N−1

N∑

i=1

σai σbi

p+ βh

∑

i,a

σai

= Tr{σ}

∫ ∏

a<b

Nδ(NQab −

N∑

i=1

σai σbi )

exp

(βJ)2

4N

n+

∑

a 6=bQpab

+ βh

∑

i,a

σai

=

∫ ∏

a<b

{dλab

4πi/N

}exp

(βJ)2

4N

n+

∑

a6=bQpab

− N

2

∑

a6=bλabQab

Trσ exp

1

2

∑

a 6=bλabσ

aσb + βh∑

a

σa

N

=

∫ ∏

a<b

{dλab

4πi/N

}exp

[−NG({Qab, λab})

](44)

G({Qab, λab}) = − (βJ)2

4n− (βJ)2

4

∑

a 6=bQpab +

∑

a6=b

1

2λabQab − ln Trσ exp

1

2

∑

a6=bλabσ

aσb + βh∑

a

σa

(45)

The integrals can be evaluated by the saddle point method. The saddle points Q∗ab and λ∗ab must verify the saddlepoint equations which read

0 =∂G

∂Qab→ λab =

(βJ)2

2pQp−1

ab 0 =∂G

∂λab→ Qab = 〈σaσb〉eff (46)

where

〈. . .〉eff ≡∑σ=±1 e

−βHeff . . .∑σ=±1 e

−βHeff− βHeff ≡

1

2

∑

a6=bλ∗abσ

aσb + βh∑

a

σa (47)

Using the 1st equation of Eq. (46) in Eq. (45) we find

G({Q∗ab}) = − (βJ)2

4n+

(βJ)2

4(p− 1)

∑

a 6=b(Q∗)pab − ln Trσ exp

1

2

∑

a6=b

(βJ)2

2p(Q∗)p−1

ab σaσb + βh∑

a

σa

(48)

11

Once we find all the relevant saddle points we can evaluate the free-energy as,

f = − 1

βlimN→∞

1

N∂n

∫ ∏

a<b

dQabe−NβG[{Qab}]

∣∣∣∣∣∣n=0

= minimizeSP

∂nG[{Q∗ab}] (49)

In the last equation, minimizeSP

means to choose a saddle point Q∗ab which minimizes the free-energy out of all the

locally stable saddle points.

D. Use of Parisi’s ansatz

In the following we look for the saddle points assuming the Parisi’s ansatz Eq. (19). First we analyze the replicasymmetric case (k = 0) and then study the 1RSB ansatz (k = 1).

Note that a family of equivalent saddle points other than the one Eq. (19) can be obtained just by making permu-tations of the replica indicies. All of them take exactly the same value of the free-energy functional f [{Q∗ab}] so thatif we take a sum over all of them, the replica (permutation) symmetry is restored. Choosing one out of them, likethe one parametrized as Eq. (19), means to break the replica symmetry. Physically we justify this by saying that itis realized as a spontaneous symmetry breaking, like the ferromagnetic phase transition of Ising model. The couplingparameter εab discussed previously can be regarded as the symmetry breaking field.

E. Replica symmetric solution

We assume k = 0, i.e. λab = λ and Qab = q for all replica pairs a 6= b. With this ansatz we find,

1

2

∑

a 6=bλabσ

aσb =λ

2

∑

a6=bσaσb =

λ

2[(

n∑

a=1

σa)2 − n] (50)

Then by the HubbardStratonovich (HS) transformation we obtain

e12

∑a 6=b λabσ

aσb = e−n2 λ

∫dz√2πe−

z2

2 ez√λ∑na=1 σ

a

= e−n2 λ〈ez

√λ∑na=1 σ

a〉z (51)

where we introduced a shorthand notation

〈. . .〉z ≡∫Dz . . . . ≡

∫dz√2πe−z

2/2 . . . . (52)

With this we find,

Trσeλ2 σa6=bσ

2σb+βh∑a σ

a

= e−nλ/2〈2 cosh(Ξ)n〉z (53)

where we introduced

Ξ ≡ βh+ z√λ (54)

Using the above expression we find,

GRS(q, λ)

n= − (βJ)2

4+ (n− 1)

(− (βJ)2

4qp +

1

2λq

)+λ

2− 1

n〈(2 cosh(Ξ))n〉z (55)

so that

βfRS(q, λ) = − (βJ)2

4(1− qp) +

1

2λ(1− q)− 〈ln(2 cosh(Ξ))〉z (56)

where and q and λ must verify the saddle point equations

0 =∂fRS

∂q→ λ =

p

2(βJ)2qp−1 (57)

12

and

0 =∂fRS

∂λ→ q = 〈tanh2 Ξ〉z (58)

To sum up we find the equation of state within the RS ansatz

q =

∫dz√2πe−

z2

2 tanh2

(βh+ z

√p

2(βJ)2qp−1

)(59)

This would be regarded as the Edwards-Anderson (EA) order parameter. Physically the 2nd term in (. . .) would beinterpreted as the effective field (cavity field) produced by the surrounding spins. The same equation can be derivedby analyzing the Thouless-Anderson-Parlmer (TAP) equations [7] or equivalently by the cavity method [8].

1. p = 2 case: Sherrington-Kirkpatrick’s analysis

The RS solution p = 2 was analyzed by Sherrington-Kirkpatrick [3]. The equation of state Eq. (59) reads in thiscase as

q =

∫dz√2πe−

z2

2 tanh2(βh+ z

√λ)

λ = (β)2q (60)

Assuming q, h are small and using tanh(x) = x − x3/3 + . . . and thus tanh2(x) = x2 − 2/3x4 + . . . for small x andusing 〈z2〉z = 1, 〈z4〉z = 3, we find,

q = (βh)2 + λ〈z2〉z −2

3〈(βh+ z

√λ)4〉z + . . . = (βh)2 + (βJ)2q − 2(βJ)4q2 + . . . (61)

Under zero magnetic field h = 0 we find the continuous emergence of the spinglass order parameter:

q =(βJ)2 − 1

2(βJ)4' 1− T

Tc(62)

with the spinglass transition temperature,

Tc/J = 1. (63)

Under small magnetic field we find,

q = χSGh2 χSG =

β2

1− (βJ)2(64)

suggesting a spin-lass transition signaled by divergence of the spin-glass susceptibility χSG approaching the spinglasstransition temperature T → T+

c . These observations suggest a spinglass transition as a second order phase transition.

2. p→ ∞ limit

Assuming h = 0, we find q = 0,λ = 0 is always a solution: paramagnetic. The free-energy is obtained as,

− βfRS(q = 0, λ = 0) = ln 2 +(βJ)2

4(65)

which is independent of p. Note that this agrees with the free-energy of the random energy model in the paramagneticphase T > Tc.

13

F. 1 step replica symmetry breaking (1RSB)

We assume k = 1, i.e. λab = λ0 + λ1Imab and Qab = q0 + q1I

mab where m = m1. With this ansatz we find,

1

2

∑

a6=bλabσ

aσb =λ0

2

∑

a6=bσaσb +

λ1 − λ0

2

∑

C=1,...,n/m

∑

a6=ba,b∈C

σaσb

=λ0

2[(

n∑

a=1

σa)2 − n] +λ1 − λ0

2

∑

C=1,2,...,n/m

[(∑

a∈Cσa)2 −m] (66)

We find doing the HS transformation,s

e12

∑a6=b λabσaσb = e−

n2 λ1e

λ02 (

∑a σa)2+

λ1−λ02

∑C(

∑a ∈C σa)2

= e−n2 λ1〈e−

√λ0z0

∑a σa〉z0

n/m∏

C=1

〈e−√λ1−λ0z1C

∑a∈C σa〉z1C (67)

with which we get

Tre−12

∑a 6=b λabσaσb+βh

∑a σa = e−

n2 λ1

∫Dz0

{∫Dz1[2 cosh(Ξ)]m

}n/m(68)

where we introduced

Ξ ≡ βh+√λ0z0 +

√λ1 − λ0z1 (69)

Observing also the following∑

a6=bQpab = n(m− 1)qp1 + n(n−m)qp0

∑

a 6=bλabQab = n(m− 1)λ1q1 + n(n−m)λ0q0

we obtain

G1RSB(q1, q0, λ1, λ0,m)

n= − (βJ)2

4+

(βJ)2

4

[(m− n)qp0 + (1−m)qp1

]

− 1

2[(m− n)λ0q0 + (1−m)λ1q1] +

1

2λ1 −

1

nlog

∫Dz0

{∫Dz1[2 cosh(Ξ)]m

}n/m(70)

so that

βf1RSB(q1, q0, λ1, λ0,m) = − (βJ)2

4+

(βJ)2

4mqp0 +

(βJ)2

4(1−m)qp1

− 1

2mλ0q0 −

1

2λ1(1−m)q1 +

1

2λ1 −

1

m

∫Dz0 log

∫Dz1[2 cosh(Ξ)]m (71)

where q and λ must verify the saddle point equations

0 =∂f1RSB

∂qi→ λi =

p

2(βJ)2qp−1

i (i = 0, 1) (72)

and after some algebra we find

0 =∂f1RSB

∂λ0→ q0 =

∫Dz0

(∫Dz1 coshm(Ξ) tanh(Ξ)∫

Dz1 coshm(Ξ))

)2

(73)

and

0 =∂f1RSB

∂λ1→ q1 =

∫Dz0

∫Dz1 coshm(Ξ) tanh2(Ξ)∫

Dz1 coshm(Ξ))(74)

14

Using Eq. (72) in Eq. (71) we find,

βf1RSB(q1, q0,m) = − (βJ)2

4− (p− 1)

(βJ)2

4mqp0 − (p− 1)

(βJ)2

4(1−m)qp1

+p

4(βJ)2qp−1

1 − 1

m

∫Dz0 log

∫Dz1[2 cosh(Ξ)]m (75)

Finally the parameter m must be extremized. Thus we find,

0 =∂f1RSB

∂m→ 0 = −(p−1)

(βJ)2

4(qp0−qp1)+

1

m2

∫Dz0 ln

∫Dz1[2 cosh(Ξ)]m− 1

m

∫Dz0 ln

∫Dz1[2 cosh(Ξ)]m ln[2 cosh(Ξ)]

(76)The internal energy is obtained as

e

J=∂βf1RSB

∂β= −βJ

2+βJ

2mqp0 +

(βJ)

2(1−m)qp1 (77)

Assuming h = 0, we easily find that q0 = 0 (and thus λ0 = 0) is a solution. We limit ourselves to this case in thefollowing. In this case the calculation simplifies a lot. For clarity let us display the results below. The free-energybecomes

βf1RSB(q1 = q, q0 = 0, λ1 = λ, λ0 = 0,m) = − (βJ)2

4[1− (1−m)qp]

+1

2λ[1− (1−m)q]− 1

mlog

∫Dz[2 cosh(z

√λ)]m (78)

where q and λ are given by

q =

∫Dz coshm(z

√λ) tanh2(z

√λ)∫

Dz coshm(z√λ)

λ =p

2(βJ)2qp−1 (79)

1. p→ ∞ limit

In particular let us examine the case p→∞ to compare with the random energy model. We notice that in the limitp → ∞, if we assume 0 < q < 1 the 2nd equation of Eq. (79) implies λ = 0 which cannot satisfy the 1st equation.The only possibilities are (q, λ) = (0, 0) and (1,∞).

Let us evaluate the free-energy for the solution q = 1, λ =∞. To this end we have to examine the integral,

I =

∫Dz[2 cosh(zλ)]m (80)

in λ→∞ limit. By noting that 2 cosh(z√λ) ' ez

√λsgn(z) we find

limλ→∞

I = 2em2λ

2 (81)

With this we find

− βf1RSB(q1 = 1, q0 = 0, λ1 =∞, λ0 = 0,m) =(βJ)2

4m+

ln 2

m(82)

Finally we require 0 = ∂mβF1RSB which yields

m =kBT

J2√

ln 2 (83)

which continuously decreases with the temperature.Comparing with the RS free-energy we find the 1RSB free-energy becomes lower than the RS energy for m < 1

suggesting a critical temperature Tc such that m(Tc) = 1. Thus we find

kBTc = J/(2√

ln 2) (84)

15

with which we find

m =T

Tc. (85)

As the result the 1RSB free-energy becomes independent of the temperature below Tc;

f1RSB = −J√

ln 2 (86)

The internal energy can be obtained from Eq. (77) as well. From the above results we realize that the exact resultsof the REM is recovered by the 1RSB solution.

2. Distribution of overlap P (q)

Now we have to ask : what is the physical meaning of the replica symmetry breaking?. Let us consider two realreplicas, say 1 and 2 and examine the probability distribution P (q) of the overlap q between them. More precisely weconsider the disorder average of this distribution function,

P (q) = limN→∞

〈δ(q − 1

N

N∑

i=1

σ1i σ

2i 〉 (87)

We wish to evaluate this using the 1RSB solution. To this end we just need to add n − 2 more replicas so thatwe have n replicas as a whole. We apply the 1RSB ansatz for this n replica system as discussed above. At thispoint we have to remember that there is a family of equivalent solutions which can be obtained by permuting thereplica indicies of the Parisi’s matrix and that all of them has the same statistical weights as discussed in sec I C 2and sec II D. The overlap between the two replicas Q12 can take either q0 or q1 depending on the permutations. Thenassuming equal weights of all of these permuted solutions we find,

P (q) =m(m− 1) nmn(n− 1)

δ(q − q1) +

(1− m(m− 1) nm

n(n− 1)

)δ(q − q0) −−−→

n→0(1−m)δ(q − q1) +mδ(q − q0) (88)

To be specific let us consider the case of REM for which we know q1 = 1, q0 = 0 and m = T/Tc at 0 < T < Tc. In

the limit T → 0, m → 0 so that P (q) = δ(q − 1) which is expected since the two replicas should stay in the groundstate at zero temperature.

In the other limit T → T−c , m → 1 so that P (q) = δ(q). This could also be expected because down to Tc the tworeplicas are allowed to stay in arbitrarily different states which happen to have almost the same energy. As discussedin sec II B 1 (see Eq. (32) and Eq. (33)) the overlap between such states are typically 0.

More generally, in the spinglass phase, 0 < T < Tc, we find an intriguing situation: the two distinct delta peaksone at q = q1 = 1 and the other at q = q0 = 0 coexists in P (q) whose relative weights evolves as m = T/Tc increaseswith the temperature. See Fig. 4 a) for a schematic picture. Actually this result can be obtained without using thereplicas but directly developing a low temperature expansion of the REM as we discuss in Appendix A. With thisapproach the picture becomes clearer. In the low temperature limit, the peak at q = 1 accounts for the case that bothof the replicas remain in the ground state and the peak at q = 0 accounts for the situation that one of the two replicasstay at the 1st excited state while the other one remains in the ground state. Note that typical overlap between theground state and the excited states are 0.

3. System with many metastable states

Based on the above observation in the p → ∞ limit (REM), we may interpret the 1RSB solution in more generalcases in the following way. In the 1RSB phase, there are multiple metastable spinglass states α = 1, 2, . . . whosemutual overlap is typically q0. On the other hand q1 would be interpreted as the EA order parameter qEA, which canbe regarded as self-overlap of metastable states. Then in REM (p → ∞) it happens that qEA = 1 suggesting thatthere is no thermal fluctuation within a metastable state of REM. This means that every spin configuration becomesa metastable state in the p→∞ limit.

Then similarly to the case of REM Eq. (34), we naturally expect that the partition function of a generic p-spinmodel can be represented as a summation over the metastable states,

Z =∑

α

e−βNfα(T ) (89)

16

where fα(T ) is the free-energy/spin of the metastable state α. The above representation is just the same as that ofthe REM Eq. (34) but here the energy eα is replaced by the free-energy fα. This reflects the fact that q1 = qEA,which is 1 in the REM (p → ∞), is smaller than 1 for general p suggesting thermal fluctuations within metastablestates. In order words, for generic p, a free-energy minimum consists of not only an energy minimum but nearby spinconfigurations which defines a basin around the minimum in the phase space. Then the thermal averages would bedecomposed as

〈. . .〉 =∑

α

wα〈. . .〉α wα =e−Nβfα∑α e−Nβfα (90)

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) = limN→∞

〈δ(q − 1

N

N∑

i=1

σ1i σ

2i 〉 =

∑

α

∑

α′

wαwα′〈δ(q −1

N

N∑

i=1

σ1i σ

2i )〉α,α′ =

∑

α

∑

α′

wαwα′δ(q − qαα′)

=∑

α

w2αδ(q − qαα) +

∑

α6=α′wαwα′δ(q − qαα′) (91)

where

qα,α′ ≡1

N

N∑

i=1

mi,αmi,α′ mi,α ≡ 〈σi〉α (92)

is the mutual overlap between the metastable states α and α′. For α′ = α, it is the self-overlap, which should beidentified as the Edwards-Anderson order parameter qEA. To derive Eq. (91) we assumed the clustering property.(See Appendix B).

In the case of 1RSB we have found Eq. (88) which implies qα,α = q1, qα,α′ = q0 for α 6= α′,∑α w

2α = 1 − m

and∑α 6=α′ wαwα′ = m. We discuss later general connection between the Parisi’s ansatz Eq. (19) and the overlap

distribution function P (q).

4. Complexity and Monasson’s observation

Similarly to the case of REM we may rewrite the partition function Eq. (89) as

Z =∑

α

e−βNfα(T ) =

∫dfeN(Σ(f,T )−βf) (93)

where we introduced the so called structural entropy or complexity

Σ(f, T ) ≡ 1

Nlog∑

α

δ(f − fα(T )). (94)

The above representations Eq. (89), Eq. (93) and Eq. (94) are justified by studying the solutions of the equations ofstate 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,

−Nβmφm ≡ lnZm Zm ≡∑

α

e−Nmβfα(T ) =

∫dfeN(Σ(f,T )−βmf). (95)

Formally evaluating the integral by the saddle point method we find,

− βmφm = Σ(f∗, T )− βmf∗ ∂Σ(f, T )

∂f

∣∣∣∣f=f∗(T,m)

=m

kBT(96)

From this we find

f∗ = ∂m(mφm) Σ∗ = βm2∂mφm (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 caseof n replicas this amount to consider a situation such that n replicas are divided into n/m groups and that replicasbelonging to a common group move around different states together while different groups move around independentlyfrom each other;

Zn = Zn/mm =

(∑

α

e−Nmβfα(T )

)n/m(98)

Actually this is readily realized precisely by the 1RSB ansatz : the n replicas: are divided into n/m groups of size mand the replicas belonging to each of the groups have mutual overlap which is equal to qEA while mutual overlap q0

between different groups. This implies

−Nβf1RSB(q1, q0,m) = ∂n(Zm)n/m∣∣∣n=0

= lnZm/m. (99)

Thus we find

φm = f1RSB(q1, q0,m) (100)

An important observation made by Monasson [9] is that the condition 0 = ∂mf1RSB Eq. (76) for T ≤ Tc is equivalentto require vanishing complexity Σ = 0 because of Eq. (97).

Let us check these explicitly in the case of REM. From Eq. (82) we readily find,

− βφm = −βf1RSB =(βJ)2

4m+

ln 2

m. (101)

Then using the prescription Eq. (97) we find

Σ(f∗) = m2∂m

(− (βJ)2

4m− ln 2

m

)= ln 2− (βJ)2

4m2 βf∗ = ∂m

(−m2 (βJ)2

4− ln 2

)= − (βJ)2

2m (102)

from which we find the complexity

Σ(f) = ln 2− (f/J)2 (103)

which agrees with Eq. (35)

• 1RSB solution continues to be present above Tc up to Td (which is ∞ in the case of REM) where it finallydisappears....

• Energy landscape changes at Td...

• The above ideas play key roles to develop the cloned liquid theory (replicated liquid theory) for structuralglasses.[10, 11]

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. 2

6. Disentangling hierarchy of responses

Let us discuss a generic feature of linear response in the system with many metastable state captured by the 1RSBansatz following [12], [13] and [14].

We consider a perturbing field h which is conjugated to a physical observable O. We denote the value of theobservable/spin at state α as oα and that the associated linear susceptibility as χα,

oα = −∂fα∂h

χα =∂oα∂h

= −∂2fα∂h2

(104)

18

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.2 0.4 0.6 0.8 1

T

m

m-T phase diagram

m_s(T)m_G(T)

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

-0.83 -0.825 -0.82 -0.815 -0.81 -0.805 -0.8 -0.795 0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

com

plex

ity

com

plex

ity

f

complexity vs m (1RSB saddle point)

T=0.5

a) b) c)

FIG. 2. 1RSB solution of the p = 3 Ising MFSG model: (left panel) complexity Σ(f, T = 0.5) vs f (center panel) The repliconeigenvalue λR(m,T ) and complexity Σ∗(f∗(m,T ), T ) vs m at T = 0.5 and T = 0.6 (right panel) the static glass transition line(’Kauzmann line’) ms(T ) above which the complexity is positive Σ > 0 and ’Gardner line’ mG(T ) below which the repliconeivgen value is positive.

In other words oα is the value of the observable/spin o averaged over the intra-state fluctuation, i. e. the thermalfluctuations within the state α. Similarly χα is related to the fluctuation of o within the state α.

In the 1RSB ansatz we have

− βNf1RSB(T, h) =1

mln∑

α

e−Nmβfα(T,h) (105)

Then we find

o(T, h) = −∂f1RSB(T, h)

∂h=

∂

∂h

ln∑α e−Nmβfα(T,h)

βmN=∑

α

wαoα = 〈oα〉0 (106)

where 〈. . .〉0 ≡∑α wα . . . is the average over different metastable states with the statistical weight

wα ≡e−Nmβfα(T,h) . . .∑α e−Nmβfα(T,h)

(107)

Similarly we find the susceptibility as,

χ1RSB(T, h) =∂o(T, h)

∂h=∑

α

wα∂hoα +∑

α

(∂hwα)oα = χ1 +mχ0 (108)

with

χ1 = 〈χα〉0 χ0 = Nβ(〈o2α〉0 − 〈oα〉20) (109)

Physically we can interpret χ1 as the response within a metastable state and χ0 as the response due to transitionbetween different states. The total response is the mixture. One may wish to disentangle the two qualitativelydifferent susceptibilities.

A trick to disentangle the two susceptibilities was noticed in [12] as follows. Let us apply different probing fieldson each of the replicas,

χa,b(T, h) ≡ − m∂2f1RSB(T, {ha})∂ha∂hb

∣∣∣∣∣{ha=h}

=1

βN

∂2

∂ha∂hbln∑

α

e−Nβ∑na=1 fα(ha)

∣∣∣∣∣{ha=h}

= χ1δab + χ0

We also observe that the total response is recovered as,

m∑

b=1

χab = χ1 +mχ0 = χ1RSB(T, h). (110)

19

As an example let us consider the magnetic response of the p-spin Ising model. To this end we just need to applydifferent fields on ha = h+ δha (a = 1, 2, . . . ,m) on the replicas as outlined above. Then Eq. (71) becomes

βf1RSB({h1, . . . , hm}) = − (βJ)2

4+

(βJ)2

4mqp0 +

(βJ)2

4(1−m)qp1 −

1

2mλ0q0 −

1

2λ1(1−m)q1 +

1

2λ1

− 1

m

∫Dz0 log

∫Dz1

m∏

a=1

[2 cosh(βha +√λ1 − λ0z1 +

√λ0z0)] (111)

from this we find the following susceptibility matrix,

χab = − ∂2f1RSB({h1, . . . , hm})∂ha∂hb

∣∣∣∣∣ha=h

= χ1δab + χ0 (112)

with

χ1 =

∫Dz0

∫Dz1[2 cosh Ξ]mβ(1− tanh2 Ξ)∫

Dz1[2 cosh Ξ]m= β(1− q1) (113)

where Ξ = βh+√λ1 − λ0z1 +

√λ0z0 and

χ0 = β

∫Dz0

∫Dz1[2 cosh Ξ]m tanh2 Ξ)∫Dz1[2 cosh Ξ]m

−(∫Dz1[2 cosh Ξ]m tanh2 Ξ)∫Dz1[2 cosh Ξ]m

) = β(q1 − q0) (114)

The total susceptibility becomes

χ1RSB = β[1− q1 +m(q1 − q0)] (115)

which of course includes the RS result with m = 1, χRS = β(1− q0).

7. Glass state following: Franz-Parisi potential

Following the idea of Franz-Parisi [15], let us consider m+s replicas which are forced to stay in the same metastablestate where we regard the m replicas as reference system and the s replicas as slave system on which we put aperturbation (e.g. temperature change) parametrized by the strength η. The partition function of the whole system

can be written as,

−βFm+s = ln∑

α

e−Nβ[mfα+sfα(η)]

= ln

[∑

α

e−Nβmfα[1 + s(−Nβfα(η)) +O(s2

]]

= −βmφm −NβVFP(η)s+O(s2) (116)

Here we expanded in power series of s around s = 0. In the last equation the 1st term is the free-energy of thereference system, which is nothing but Eq. (95) considered by Monasson [9],

− βmφm = ln∑

α

e−Nβmfα (117)

while the 2nd term gives the free-energy of the slave system considered by Franz and Parisi [16],

VFP(η) =

∑α e−Nβmfαfα(η)∑α e−Nβmfα = 〈f(η)〉reference (118)

where 〈. . .〉reference represents the averaging over the configurations of the references system,

〈. . .〉reference ≡∑α e−Nβmfα . . .∑

α e−Nβmfα (119)

As discussed in sec II F 4, 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 eigenmodes of the Hessian matrix defined as

Ha6=b,c 6=d =∂G[{Qab}]∂Qa 6=bQc 6=d

(120)

with

G[{Qab}] = − (βJ)2

4n− (βJ)2

4(p− 1)

∑

a 6=bQpab − ln Trσ exp

(βJ)2

4p∑

a 6=bQp−1ab σaσb + βh

∑

a

σa

(121)

which follows Eq. (45) and the 1st equation of Eq. (46). After some algebra, using the 2nd equation of Eq. (46), onefinds,

Ha6=b,c 6=d =(βJ)2

4p(p− 1)Qp−2

ab (δacδbd + δadδbc)−(

(βJ)2

4p(p− 1)

)2

Qp−2ab Qp−2

cd

[〈σaσbσcσd〉eff − 〈σaσb〉eff〈σcσd〉eff

]

(122)where 〈. . .〉eff is evaluated using Eq. (47).

1. Stability of the RS symmetric solution

Let us first examine the stability of the replica symmetric solution [17]. Because of the replica symmetry onegenerally expects the following structure.

Ha 6=b,c 6=d = M1δacδbd + δadδbc

2+M2

δac + δad + δbc + δbd4

+M3 (123)

Then one finds the three group of eigen values

λL = M1 + (n− 1)(M2 + nM3) λA = M1 +1

2(n− 2)M2 λR = M1 (124)

where λL,λA,λR are called as longitudinal, anomalous and replicon eigenvalues.Now we have to find M1,M2 and M3 from Eq. (122). Within the RS ansatz Qab = Qcd = q and 〈σaσb〉eff =

〈σcσd〉eff = q. We also need to evaluate 〈σaσbσcσd〉eff which is obtained as the following,

〈σaσbσcσd〉eff = (δacδbd + δabδbc) +[δac(1− δbd) + δad(1− δbc) + δbc(1− δad) + δbd(1− δac)

]q

+[(1− δac)(1− δbd) + (1− δad)(1− δbc)

]r

= (δacδbd + δadδbc)(1− 2q + r) (125)

where

q =

∫Dz tanh2 Ξ r =

∫Dz tanh4 Ξ (126)

with

Ξ = z√λ+ βh λ =

p

2(βJ)2qp−1 (127)

In particular, the replicon eigenvalue is obtained as

M3 = (βJ)2p(p− 1)qp−2

{1− (βJ)2

2p(p− 1)qp−2(1− 2q + r)

}(128)

where

1− 2q + r =

∫Dz(1− tanh2 Ξ)2 =

∫Dz sech4 Ξ (129)

Then positivity of the replicon eigenvalue λR = M3 > 0 implies(kBT

J

)22

p(p− 1)

1

qp−2>

∫Dz sech4 Ξ (130)

This is the so called d’Almeida-Thouless condition [17].

21

2. Instability of the 1RSB solution:the Gardner’s transition

Here we limit ourselves with h = 0 for which q0 = 0 is a solution. Then positivity of the replicon eigenvalueλR = M3 > 0 implies

(kBT

J

)22

p(p− 1)

1

qp−2>

∫Dz coshm Ξ sech4 Ξ∫

Dz coshm ΞΞ = z

√λ (131)

with λ = p2 (βJ)2qp−1. This was studied first by Gardner [18].

H. More RSB: full replica symmetry breaking

It is straightforward to consider generalization to k-RSB and eventually take k → ∞ [2]. We plug in the Parisi’sansatz Eq. (19) into the general expression of the free-energy functional Eq. (45) which reads,

G({Qab, λab}) = − (βJ)2

4n− (βJ)2

4

∑

a6=bQpab +

∑

a 6=b

1

2λabQab − ln Trσ exp

1

2

∑

a6=bλabσ

aσb + βh∑

a

σa

(132)

Given the Parisi’s ansatz Eq. (19) (See Fig. 1 ) which reads,

Qab = q0 +

k∑

i=1

(qi − qi−1)Imiab =

k∑

i=0

qi(Imiab − I

mi+1

ab ) (133)

we easily obtain

1

n

∑

a 6=bQpab =

∑

b(6=a)

k∑

i=0

qpi (Imiab −Imi+1

ab ) =

k∑

i=0

qpi ((mi−1)−(mi+1−1)) =

k∑

i=0

qpi (mi−mi+1) −−−−→κ→∞

−∫ 1

n

dxq(x)p (134)

In the last equation dmi = mi −mi+1 → −dx (note mi decreases with increasing i when n becomes smaller than 1(See Fig. 1 d)). Similarly we find

1

n

∑

a6=bλabqab −−−−→

k→∞−∫ 1

n

dxλ(x)q(x) (135)

Next step is to evaluate

Ak = Tr exp

1

2

∑

a 6=bλabσ

aσb + βh∑

a

σa

= e−

n2 λkTr exp

1

2

∑

a,b

λabσaσb

∏

a

exp [βhσa] (136)

= e−n2 λk exp

1

2

∑

a,b

λab∂2

∂ha∂hb

∏

a

2 cosh(ha)

∣∣∣∣∣∣ha=βh

(137)

= e−n2 λk exp

1

2λ0

∑

a,b

∂2

∂ha∂hb

k∏

i=1

exp

1

2(λi − λi−1)

∑

a,b

Imiab

∂2

∂ha∂hb

∏

a

2 cosh(ha)

∣∣∣∣∣∣ha=βh

(138)

= e−n2 λk exp

1

2λ0

∑

a,b

∂2

∂ha∂hb

k∏

i=1

exp

1

2(λi − λi−1)

∑

a,b

Imiab

∂2

∂ha∂hb

∏

a

2 cosh(ha)

∣∣∣∣∣∣ha=βh

(139)

22

1. RS case

For an exercise let us consider k = 0 (RS) case:

A0 = e−n2 λ0 exp

λ0

2

∑

a,b

∂2

∂ha∂hb

∏

a

(2 cosh(βha))

∣∣∣∣∣ha=h

= e−n2 λ0 exp

[λ0

2

∂2

∂h2

](2 cosh(βh))n

= e−n2 λ0

∫dy√2πλ0

e−y2

2λ0 (2 cosh(βh))n = e−n2 λ0

∫Dz(2 cosh(βh+

√λ0z))

n (140)

In the 2nd equation we used a simple identity like,

(∂

∂x1+

∂

∂x2

)f(x1)f(x2)

∣∣x1=x2=x

=∂

∂xf(x)2 (141)

We also used the identity

exp

[λ

2

∂2

dx2

]f(x) =

∫dy√2πλ

e−y2

2λ f(x− y) ≡ γλ ⊗ f(x) =

∫Dze− z

2

2 f(x−√λz). (142)

This can be checked by formally expanding f(x−y) around y = 0. Here γa represents a Gaussian with zero mean andvariance a. the symbol ⊗ represents a convolution and Collecting the above result and noting that m0 = n, m1 = 1for the RS (k = 0) case, we recover Eq. (55),

GRS(q, λ)

n= − (βJ)2

4+ (n− 1)

(− (βJ)2

4qp0 +

1

2λ0q0

)+λ0

2− 1

nln

∫Dz(2 cosh(βh+

√λ0z))

n (143)

from which we find

βfRS(q, λ) = − (βJ)2

4(1− qp) +

1

2λ(1− q)− 〈ln(2 cosh(Ξ))〉z (144)

2. 1-RSB, 2-RSB and k-RSB

One easily finds,

A1 = e−n2 λ1 exp

1

2λ0

∑

a,b

∂2

∂ha∂hb

exp

1

2(λ1 − λ0)

∑

a,b

Im1

ab

∂2

∂ha∂hb

∏

a

2 cosh(βha)

= e−n2 λ1 exp

1

2λ0

∑

a,b

∂2

∂ha∂hb

n/m1∏

C=1

exp

1

2(λ1 − λ0)

∑

a,b∈C

∂2

∂ha∂hb

∏

a∈C2 cosh(βha)

= e−n2 λ1 exp

1

2λ0

∑

a,b

∂2

∂ha∂hb

[∫

dz√2πe−

z2

2 (2 cosh(βh+√λ1 − λ0z))

m1

]n/m1

= e−n2 λ1

∫Dz0

[∫Dz1(2 cosh(Ξ1))m1

]n/m1

Ξ1 ≡ βh+√λ1 − λ0z1 +

√λ0z0 (145)

Similarly one can find,

A2 = e−n2 λ2

∫Dz0

[∫Dz1

[∫Dz2

[2 cosh(Ξ2)

]m2

]m1/m2]n/m1

Ξ2 = Ξ1 +√λ2 − λ1z2 (146)

23

and

A3 = e−n2 λ3

∫Dz0

∫Dz1

[∫Dz2

[∫Dz3

[(2 cosh(Ξ2))m3

]]m2/m3]m1/m2

n/m1

Ξ3 = Ξ2 +√λ3 − λ2z3 (147)

and so on.Now we are naturally lead to define a family of functions g(mi, y) for i = 1, 2, . . . , k which satisfy a recursion

relation,

g(mi, y) = eλi−λi−1

2∂2

∂y2 g(mi+1, y)mi/mi+1 = γλi−λi−1 ⊗ g(mi+1, y)mi/mi+1 =

∫Dzig(mi+1, y +

√λi − λi−1zi)

mi/mi+1

(148)with ’the boundary condition’,

g(mk+1 = 1, y) = 2 cosh(y). (149)

With these we find,

Ak = e−n2 λk

∫Dz0g(m1, βh+

√λ0z0)n/m1 . (150)

which is valid for k = 0, 1, 2, . . . ,∞. The we find

GkRSB({q, λ})n

= − (βJ)2

4− (βJ)2

4

k∑

i=0

qpi (mi−mi+1)+1

2

k∑

i=0

λiqi(mi−mi+1)+λk2− 1

nln

∫Dz0g(m1, βh+

√λ0z0)n/m1

(151)and thus

βfkRSB({q, λ}) = − (βJ)2

4− (βJ)2

4

k∑

i=0

qpi (mi−mi+1)+1

2

k∑

i=0

λiqi(mi−mi+1)+λk2−∫Dz0f(m1, βh+

√λ0z0) (152)

In the last equation we introduced,

f(mi, y) =1

miln g(mi, y]) (153)

whose recursion relation can be obtained as,

f(mi, y) = ln[γλi−λi−1

⊗ emif(mi+1,y)]

=1

miln

∫Dzie

mif(mi+1,y+√λi−λi−1zi) (154)

with ’the boundary condition’,

f(mk+1 = 1, y) = ln 2 cosh(y). (155)

3. Variational equations

Now we have to obtain the self-consistent equations for λi (i = 0, 1, 2, . . . , k) and qi (i = 0, 1, 2, . . . , k),

0 =∂βfkRSB

∂qi→ λi =

(βJ)2

2pqp−1i (156)

0 =∂βfkRSB

∂λi→ qi =

2

mi −mi+1

{∫Dz0

∂

∂λif(m1, β +

√λ0z0)− 1

2δi,k

}(157)

For the time being we focus on Eq. (157) for the case i 6= k, i. e. i = 1, 2, . . . , k− 1. We come back later to consideri = k case. By a repeated use of the recursion formula Eq. (148) we find

∂λig(mi, y) =mi

mi+1

∫Dzig(mi+1,Ξi)

mi/mi+1−1∂λig(mi+1,Ξi) (158)

24

where Ξi = y +√λi − λi−1zi and for i ≤ k − 1,

∂λig(mi+1, y) =mi+1

mi+2

∫Dzi+1g(mi+2,Ξi+1)mi+1/mi+2−1∂λig(mi+2,Ξi+1) (159)

with Ξi+1 = Ξi +√λi+1 − λizi+1 = y +

√λi − λi−1zi +

√λi+1 − λizi+1 thus for i ≤ k − 1,

∂λig(mi+2,Ξi) = g′(mi+2,Ξi+1)

(−1

2

zi+1√λi+1 − λi

+1

2

zi√λi − λi−1

). (160)

Here and in the following we use the following shorthand notation

A(x, y) ≡ ∂A(x, y)

∂xA′(x, y) ≡ ∂A(x, y)

∂y(161)

Collecting the above results and doing integrations by parts one finds after some algebra for i ≤ k − 1,

∂λig(mi, y) =1

2mi(mi −mi+1)

∫Dzig(mi+1,Ξi)

mi/mi+1(∂yf(mi+1,Ξi)

)2

=1

2(mi −mi+1)

∫dh

δg(mi, y)

δf(mi+1, h)

(f ′(mi+1, h)

)2(162)

Similarly one can show that for i ≤ j ≤ k − 1,

∂λjg(mi, y) =1

2(mj −mj+1)

∫dh

δg(mi, y)

δf(mj+1, h)

(f ′(mj+1, h)

)2(163)

and thus equivalently for i ≤ j ≤ k − 1,

∂λjf(mi, y) =1

2(mj −mj+1)

∫dhPij+1(y, h)

(f ′(mj+1, h)

)2(164)

where we introduced

Pij(y, z) ≡δf(mi, y)

δf(mj , z). (165)

Now collecting the above results we can rewrite Eq. (157) as

qi =

∫Dz0

∫dhP1,i+1(βh+

√λ0z0, h)(f ′(mi+1, h))2 =

∫dhPi(h)(f ′(mi+1, h))2 (166)

where we introduced

Pi(h) =

∫Dz0P1,i+1(βh+

√λ0z0, h). (167)

The function Pi(h) can be understood physically as distribution function of internal field ( at level i of the hierarchy).And f ′(mi+1, j) as the magnetization which is thermalized at level i+ 1 and subjected to slower random field at leveli.

One can find from Eq. (154) that f ′(mi, h) obeys a recursion formula.

f ′(mi, h) =

∫Dziemif(mi+1,y+

√λi−λi−1zi)f ′(mi+1, h)

∫Dzie

mif(mi+1,y+√λi−λi−1zi)

(168)

with the ’boundary condition’ f ′(mk+1, h) = tanh(h).The remaining task is to find a way to compute Pij(y, z). To this end let us begin with the following object,

Rij(y, z) ≡δg(mi, y)

δg(mj , z)(169)

25

which can be rewritten by the chain rule,for j ≥ i+ 1, as,

Rij(y, z) =

∫dxRi,j−1(y, x)Rj−1,j(x, z) Rj−1,j(x, z) =

mj−1

mj

e− (x−z)2

2(λj−1−λj−2)

√2π(λj−1 − λj−2)

gmj−1/mj−1(mj , z) (170)

where the 2nd equation can be obtained using the recursion formula Eq. (148). Then we obtain the recursion formulafor Rij ,

Rij(y, z) =mj−1

mjgmj−1/mj−1(mj , z)γλj−1−λj−2

⊗z Ri,j−1(y, z) (171)

The two objects Pij(y, z) and Rij(y, z) are simply related by Pij(y, z) = (mj/mi)(g(mj , z)/g(mi, y))Rij(y, z). Thuswe find

Pij(y, z) = gmj−1/mj (mj , z)γλj−1−λj−2⊗z

Pi,j−1(y, z)

g(mj−1, z)(172)

where A⊗zB(y, z) stands for a convolution with respect to z. This recursion formula can be solved with the ’boundarycondition’

Pii(y, z) = δ(y − z). (173)

Similarly the internal field distributions Pi(z) follows the same recursion formula

Pi(z) = gmi/mi+1(mi+1, z)γλi−λi−1⊗ Pi−1(z)

g(mi, z)(174)

for i = 1, 2, . . . , k with the ’boundary condition’,

P0(z) =

∫Dz0δ(βh+

√λ0z0 − z) =

e−(z−βh)2

2λ0√2πλ0

(175)

One can check easily that∫dzPi(z) = 1 is satisfied for i = 0 and the recursion relation Eq. (174) preserves the

normalization.Finally let us come back to case i = k of Eq. (157). To this end we begin by observing the following, which follows

from the recursion relation Eq. (148) and an integration by parts,

∂g(mk, y)

∂λk=

∂

∂λk

∫Dzkg(mk+1, y +

√λk − λk−1zk)mk/mk+1

=mk

2

∫Dzkg(mk+1,Ξk)mk

[(mk − 1)(f ′(mk+1,Ξk))2 + 1

]

=1

2

∫dh∂g(mk+1, h)

∂f(mk+1, y)

[(mk − 1)(f ′(mk+1, y))2 + 1

](176)

Here we also used the fact that g(mk+1, y) = g′′(mk+1, y) = 2 cosh(y). More generally we find,

∂f(mi, y)

∂λk=

1

2

∫dhPi,k+1(h, y)

[(mk − 1)(f ′(mk+1, y))2 + 1

](177)

Then Eq. (157) for i = k becomes

qk =2

mk − 1

(∫Dz0

∂

∂λkf(m1, β +

√λ0z0)− 1

2

)=

∫dhPk(h)(f ′(mk+1, h))2 (178)

Thus this is formally the same as those i 6= k given in Eq. (166).It is a simple exercise to check that the above results reproduce the results of k = 0 RSB, i.e. Eq. (59) and k = 1

RSB, i.e. Eq. (73) and Eq. (74).The values of qi = 0, 1, 2, . . . , k in the k(≥ 1) RSB ansatz with mi’s fixed as 0 = m0 < m1 < . . .mk < mk+1 = 1

can be solved numerically by iterating the following steps:

26

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

T=0.10.20.30.40.5 0

0.05

0.1

0.15

0.2

-60 -40 -20 0 20 40 60

T=0.50.40.30.20.1

hx

q(x) P (h)

FIG. 3. Numerical solution of the k-RSB ansatz in the case of the SK model (p = 2) whose critical temperature is Tc/J = 1.Here k = 60. (Left panel) The order parameter function q(x) and (Right panel) the internal field distribution P (mk+1 = 1, h)

1. make an initial guess on qi for i = 0, 1, 2, . . . , k by which we also obtain λi for i = 0, 1, 2, . . . , k via Eq. (156):

λi = (βJ)2

2 pqp−1i .

2. evaluate the functions f(mi, h) for an appropriate range of h doing ’Gaussian’ convolutions recursively in theorder (i = k → k − 1 → . . . 2 → 1) using the recursion formula Eq. (154) with the boundary conditionf(mk+1, h) = ln(2 cosh(h)). Simultaneously compute also f ′(mi, h) using the recursion formula Eq. (168) withthe boundary condition f(mk+1, h) = tanh(h).

3. evaluate the functions Pi(h) by doing another series of ’Gaussian’ convolutions recursively in the order (i = 1→2 → . . . k − 1 → k) using the recursion formula Eq. (174) with the boundary condition P0(z) = e−z

2/2/√

2π.Then qi for i = 0, 1, 2, . . . , k can be determined using Eq. (166). Go back to the initial step.

In Fig. 3 we show an example of the numerical analysis done for the SK model (p = 2).

4. Continuous RSB: k → ∞ limit

Finally let us consider k → ∞ limit [2, 18]. Writing mi = x − δx,mi = x and λi − λi−1 = λ(x)δx, the recursionrelation Eq. (148) becomes a partial differential equation, called as Parisi’s equation.

− ∂g(x, y)

dx=λ(x)

2

∂2

dy2g(x, y)− 1

xg(x, y) ln g(x, y) (179)

which must be solved with the boundary condition,

g(1, y) = 2 cosh(y). (180)

Equivalently for

f(x, y) ≡ 1

xln g(x, y) (181)

we find

∂f

∂x= − λ(x)

2

[∂2f

dx2+ x

(∂f

dx

)2]

(182)

which must be solved with the boundary condition,

f(1, y) = ln 2 cosh(y). (183)

27

Similarly the recursion relation for the distribution of the internal field Eq. (174) becomes, after some algebra, thefollowing partial differential equation in the limit k →∞,

P (x, y) =λ(x)

2x

(P ′′(x, y)− 2x(P (x, y)f ′(x, y))′

)(184)

where the dots and dashes are the shorthand notations introduced before Eq. (III F 3). The initial condition is givenby Eq. (175) which reads as,

P (0, y) =e−

(y−βh)2

2λ(0)

√2πλ(0)

. (185)

With these the free-energy becomes

βf∞RSB = − (βJ)2

4+

(βJ)2

4

∫ 1

0

dxqp(x)− 1

2

∫ 1

0

dxλ(x)q(x) +1

2λ(1)−

∫Dz0f(m1, βh+

√λ(0)z0) (186)

where

λ(x) =(βJ)2

2pq(x)p−1 q(x) =

∫dhP (x, h)(f ′(x, h))2. (187)

In Fig. 3 an interesting feature emerges at low temperature on the distribution of the internal field P (1, h): openingof a sudo-gap. This reflects the marginal stability of the full RSB spinglass phase where one finds massless modes, e.g. λR = 0.

I. Hierarchical energy landscape and linear response with 2, 3, . . . , k RSB

In sec. II F 4 we have argued that the 1RSB ansatz is equivalent to assume factorization of the replicated partitionfunction as,

− βNf1RSB = ∂nZn|n=0 Zn =

∑

α1

exp [−Nβm1fα1]

n/m1

. (188)

Here n replicas are group into n/m1 clusters of size m1 and all the replicas belonging to a given cluster are forced tostay in the same state.

Then it is natural to expect that the 2RSB ansatz is equivalent to make a further decomposition,

Zα1= exp [−Nβm2fα1

] =∑

α2∈α1

exp [−Nβm2fα2] (189)

using this in Eq. (188) we find

− βNf2RSB = ∂nZn|n=0 Zn ≡

∑

α1

∑

α2∈α1

exp [−Nβm2fα2 ]

m1/m2

n/m1

(190)

Here we recognize that each clusters is decomposed into m1/m2 sub-clusters of size m2 such that the replicas belongingto a given sub-cluster are forced to stay in the same state. Moreover we see that in the 2RSB ansatz the states labeledby α1 are grouped into meta-states or metabasins labeled by α2 such that the sub-clusters belonging to a given clusterare forced to stay in the same meta-basin, which is a union of states.

Now we wish to extend II F 6 [12], [13] to 2-RSB (and thus to k-RSB) so that we can disentangle the linear responsesin the hierarchical energy landscape [14]. One has

o(T, h) = −∂f2RSB(T, h)

∂h= 〈〈oα2〉1〉0 =

∑

α1

wα1

∑

α2∈α1

wα1|α2oα2 (191)

28

where oα = −∂hfα is the value of the physical observable o averaged over the thermal fluctuation within the state α.We also introduced

〈. . .〉0 ≡∑

α1

wα1. . . 〈. . .〉1 ≡

∑

α2

wα2|α1. . . (192)

with wα1and wα2|α1

, which can be interpreted as the statistical weights of the metabasins and states,

wα1=

Zm1/m2α1∑

α1Zm1/m2α1

wα2|α1=e−Nm2βfα2

Zα1

Zα1=∑

α2∈α1

e−Nm2βfα2 . (193)

Now one can find,

χ2RSB =∑

α1

wα1

∑

α2∈α1

wα2|α1∂hoα2

+∑

α1

wα1

∑

α2∈α1

∂hwα2|α1oα2

+∑

α1

∂hwα1

∑

α2∈α1

wα2|α1oα2

= χ2 +m2χ1 +m1χ0 (194)

with

χ2 = 〈〈χα2〉1〉0 χ1 = Nβ[〈〈o2

α2〉1 − 〈〈oα2

〉21〉0 χ0 = Nβ[〈〈oα2〉21〉0 − 〈〈oα2

〉1〉20 (195)

Physically we can interpret χ2 as the response within a state, χ1 as the response due to transition between differentstates within a common metabasin and finally χ0 as the response due to transition between different metabasins. Thetotal response is the mixture. Just as we did for the 1RSB case we wish to disentangle the three qualitatively differentsusceptibilities.

Again let us suppose that each of the replica is subjected to difference probing fields. Then one easily finds,

χa,b(T, h) ≡ − m∂2f2RSB(T, {ha})∂ha∂hb

∣∣∣∣∣{ha=h}

= χ2δab + χ1Im1

ab + χ0

andm∑

b=1

χab = χ2 +m2χ1 +m1χ0 = χ2RSB(T, h). (196)

Generalization to k-RSB is now straightforward but we do not attempt it here because it would require heaviernotation 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 1RSB. Here we discuss the connectionbetween P (q) and the Parisi’s ansatz Eq. (19) for the more general cases. The structure of the matrix is shown inFig. 1.

We can repeat the argument in sec II F 2 for the general case. The probability to observe the overlap q within thesection δqi = qi+1 − qi will be given by,

P (q)δqi =

∑a6=b δ(Qab − q)δqi

n(n− 1)=mi(mi − 1) n

mi−mi+1(mi+1 − 1) n

mi+1

n(n− 1)−−−→n→0

δmi (197)

where δmi = mi+1 −mi. Thus

P (q) =δmi

δqi−−−−→k→∞

dx(q)

dq(198)

where we took the continuous limit by k →∞ and recalled m as x. The above result establishes the connection betweenthe order parameter function q(x) which parametrized the Parisi’s ansatz and the overlap distribution function P (q).In Fig. 4 we display the two important examples: 1RSB and continuous RSB cases.

Now considering three replicas, one can ask distribution of mutual overlaps, q12, q23 and q31. By doing the sameargument as above one finds,

P (q12, q23, q31) = limn→0

1

n(n− 1)(n− 2)

∑

a 6=b6=cδ(Qab − q12)δ(Qac − q23)δ(Qbc − q31) (199)

Then, just from the structure of the matrix, we find, the smallest two among q12, q23, q31 are equal, which meansultrametricity [8].

29

q(1)q(0) 10

P (q)

q

10 xm

q(x)

x

q(0)

q(1)

1

xp

q1q0 10

P (q)

q

m

1 � m

10 m

q(x)

x

q0

q1

1

a)

b)

c)

d)

FIG. 4. Schematic pictures of the overlap distribution function P (q) and the order parameter function q(x). a) and b) arefor the 1RSB case. c) and d) are for the full RSB case. (For simplicity we assumed flat distribution of P (q) in the continuouspart.)

30

III. REPLICATED LIQUID THEORY

A. Our system

We consider a simple system of N particles in d-dimensional space, with 2-body interactions,

H =

N∑

i=1

|p|2i2m

+∑

i<j

v(rij) rij = |xi − xj |

where xi (i = 1, 2, . . . , N) are d-dimensional vectors which represent the positions of the particles and pi’s are theirmomenta.

B. Liquid theory

1. Mayer function

The free-energy of the system is given by

− βF = lnZ (200)

with the partition function

Z = Z id

∫ N∏

i=1

ddxiV

∏

i<j

(1 + f(rij)) Z id =(V/λdth)N

N !(201)

where Z id is the partition function of the ideal gas, and f(r) is the Mayer function

f(r) = e−βv(r) − 1 (202)

and λth = h/√

2πmkBT is the de Broglie wave length.

Expansion of the free-energy in power series of the Mayer function converges in liquids [19]. We can naturallydecompose the free-energy into the ideal gas part, which is purely entropic, and the excessive part due to interactions,

F = F id + F ex (203)

where the ideal gas part is obtained as,

− βF id = ρV [1− ln(ρλdth)] (204)

with the number density ρ = N/V and the volume of the container V . The excessive part due to interactions reads,

−βF ex =1

2!

∑

ij

∫ddxiV

∫ddyiV

f(xi − yi)

+1

3!

∑

i,j,k

∫ddxiV

∫ddyiV

∫ddziV

f(xi − yj)f(yj − zk)f(zk − xi) + . . . (205)

2. Density functional

We now express the free-energy as a functional of the density field ρ(x) =∑Ni=1 δ

d(x − xi). (See Hansen-Mcdonald[20]) The free-energy consists of the entropic part (ideal gas) term and the interaction part, which can

31

be expanded in terms of the Mayer function:

F [ρ(x)] = F id[ρ(x)] + F ex[ρ(x)]

−βF id[ρ(x)] =

∫ddxρ(x)[1− ln(ρ(x)λdth)]

−βF ex[ρ(x)] =1

2!

∫ddxddyρ(x)ρ(y)f(x− y)

+1

3!

∫ddxddyddzρ(x)ρ(y)ρ(z)

×f(x− y)f(y − z)f(z− y) + . . .

Equilibrium free-energy: minimum of F [ρ(x)] with respect variation of ρ[x].A useful identity to compute the radial distribution function g(r) [20];

δ(−βF ex)

∂δ(−βv(x12))=

1

2ρ(x1)ρ(x2)g(x12) (206)

2

II. THE METHOD

In the replica method one considers a liquid made of m copies of the original system, with the constraint that eachatom of a given replica must be close to an atom of the other m − 1 replicas, i.e. that the replicated liquid mustbe made of molecules of m atoms, each one belonging to a different replica. It was shown that this trick allows tocompute all the properties of the glass, including the size of the cages, the vibrational entropy, etc., provided one isable to perform the analytical continuation to real m ≤ 1, see [14, 15, 16, 25] for a detailed discussion. Thus theproblem is to compute the free energy of the replicated system for integer m and to continue the expression for realm.

This strategy has been first tested on mean field spin glass models [15, 25], and then applied to systems of particlesinteracting through a Lennard-Jones like potential [14, 16, 17]. To compute the replicated free energy for a system ofparticles, the idea was to start from the standard HNC free energy for a molecular liquid [21] and expand it in a powerseries of the “cage radius”, that represents the amplitude of the vibrations of the particles in the glass state [15]. Themethod was successful but could not be extended straightforwardly to hard spheres, because at some stage it wasassumed that vibrations were harmonic, an approximation that clearly breaks down for hard core potentials.

The small cage expansion of the replicated HNC free energy was worked out in [19]. It was shown that the theory isin reasonable quantitative agreement with numerical data even if the HNC approximation is a very poor descriptionof the hard spheres liquid. In next section we will show that the result of [19] for the replicated free energy, obtainedstarting from the HNC free energy functional, can be derived from the full diagrammatic expansion, without the needof neglecting any class of diagrams.

III. SMALL CAGE EXPANSION

Our approach will be based on the standard diagrammatic approach (virial expansion). It seems to us that similarresults could also be obtained by a direct computation, however we think that it is more instructive to use the morefamiliar diagrammatic approach.

We start from the diagrammatic expansion of the canonical free energy as a function of the single-molecule densityρ(x) and of the interaction potential between molecules. It can be obtained from the grancanonical partition functionof the system via a Legendre transform [21]. Calling x ≡ (x1, . . . , xm) the coordinate of a molecule, the expressionfor the replicated free energy functional is [20, 21] (setting from now on β = 1):

Ψ[ρ(x), f(x,y)] =

∫dxρ(x)[log ρ(x) − 1] −

[+ + + + + · · ·

](1)

where the black circles represent a function ρ(x) and the lines represent a function f(x,y) = e−ϕ(x,y) − 1, andϕ(x,y) =

∑a ϕ(|xa − ya|) is the interaction potential between two molecules, with ϕ(r) = ∞ for r < D and 0

otherwise, the usual hard sphere interaction. The function f(x,y) is equal to −1 if |xa − ya| < D for at least one paira of spheres, and vanishes otherwise. The information on the inter-replica interaction that is used to build moleculesis encoded in the function ρ(x).

The generic diagram of the series expansion of Ψ represents an integral of the form

=1

S

∫dx1dx2dx3ρ(x1)ρ(x2)ρ(x3)f(x1,x2)f(x2,x3)f(x3,x1) . (2)

where S is the symmetry number of the diagram [20].For the one-molecule density we make the ansatz

ρ(x) = ρ

∫dX

∏

a

e− (xa−X)2

2A

(√

2πA)d= ρ

∫dX ρ(u) , ρ(u) =

∏

a

e− (ua)2

2A

(√

2πA)d(3)

where ρ = N/V is the number density of molecules, X is the center of mass of the molecule and ua ≡ xa − X . Thisallows to compute exactly the ideal gas term of the free energy:

1

N

∫dxρ(x)[log ρ(x) − 1] = log ρ − 1 +

d

2(1 − m) log(2πA) +

d

2(1 − m − log m) . (4)

We want to perform a power series expansion in√

A of the interaction term.

FIG. 5. Diagrams of the Mayer expansion

C. Cloned liquid: basic ideas

Here we sketch the idea of replicated or cloned liquid [9, 16, 21].

1. Glass order parameter

Replicas a = 1, 2, . . . ,m obeying exactly the same form of Hamiltonian, plus fictitious attractive coupling betweenthe replicas:

H =

m∑

a=1

H[{xai }] +ε

2

N∑

i=1

m∑

a,b=1

〈(xai − xbi )2〉 (ε > 0)

Because of the attractive forces, the replicas would develop a molecular states as shown in Fig. 6.The size of the molecule, the cage size, can be regarded as the glass order parameter,

A = limε→0+

limN→∞

〈(xai − xbi )2〉/2 =1

Nm(m− 1)

∂ lnZm(T, ε)

∂(−βε)

∣∣∣∣ε=0

which is A =∞ in liquids A <∞ in solids. Note that the idea is the same as the Edwards-Anderson order parameterqEA for spinglasses (See sec. I B, sec. I C 1 I C 2).

�A

uai

xai = Xi + ua

i Xi =1

m

mX

a=1

xai

FIG. 6. Schematic picture of molecular liquid. Each molecule consists of replicas a = 1, 2, . . . ,m confined to each other at scaleof order

√A.

32

2. Playing with the number of replicas

Basic assumption is to assume that there are exponentially large number of metastable states so that the complexityor the structural entropy is finite in the supercooled liquid state. Then using the replicas, we may be able to realizethe molecular liquid such that all the replicas behave as a group. This is the situation of 1RSB discussed in sec II F 4.

−βmFm(T ) = lnZm Zm =∑

α

e−Nmβfα(T ) =

∫dfe−N(mβf−Σ(f,T ))

Mezard and Parisi noticed that by choosing m small enough, compared with 1 [21], such a molecule feels effectivelyhigher temperature Teff = T/m than the physical temperature T such that they remain in the liquid state (See Fig.7, right panel). Then we can use the methods of the liquid theory [20] to compute the free-energy of the system.

0

slope m/T

f

�(f)/kB

f⇤(T, m) 0m

1

T

TK

Tc

Glass

Liquid

m⇤

T

FIG. 7. Complexity and the phase diagram of the replicated liquid

Complexity can be extracted using the Monasson’s prescription as [9],

NΣ(f∗, T ) = m2β∂Fm(T )

∂mNf∗ = m

∂Fm(T )

∂m+ Fm(T )

Free-energy of the glass can be obtained as [21],

limm→1−

(Fm(T )/N) =ln∑α e−Nm∗(T )βfα(T )

(−βm∗(T ))N= f∗(T,m∗)− Σ(f∗, T )

βm∗

D. Density functional of the replicated free-energy

For the replicated liquid of m replicas we write the coordinates as

x = {x1 · · ·xm}

. Let us introduce a shorthand notation

∫dx . . . =

∫

m∏

a=1

ddxa

. . . (207)

The free-energy functional is given by

Fm[ρ(x)] = F idm [ρ(x)] + F ex

m [ρ(x)]

−βmF idm [ρ(x)] =

∫dxρ(x)[1− ln(ρ(x)λdth)]

−βmF exm [ρ(x)] =

1

2

∫dxdyρ(x)ρ(y)f(x− y) + . . .

f(x− y) = −1 +

m∏

a=1

e−βv(|(xa−ya)|)

33

E. Construction of the gaussian ansatz

1. General form

Here we construct a general form of the Gaussian ansatz for the replicated density field. We first decompose thecoordinates of the particles xa belonging to a molecule into the center of mass XCM and relative displacement uawith respect to it.

xa = XCM + ua∑

a

ua = 0 (208)

including also the coordinate of the center of mass we can rewrite the integrals as,

∫ddx . . . =

∫ddXCMDu

∫Du . . . ≡

∫mdδ(

m∑

a=1

ua)ddu . . . (209)

We assume Gaussian distribution for ua such that

〈ua · ub〉G = dAab 〈ua · ub〉G = dDab = d(Aaa +Abb − 2Aab) (210)

where 〈. . .〉G represents averaging over the Gaussian distribution which we specify below

〈. . .〉G =1

ρ

∫Duρ(u) (211)

Note that the matrix A must be trace-less;

∑

a

Aab =∑

b

Aab = 0. (212)

Thus the matrix A can be parametrized as

Aab∈(1,2,...,m−1) = Amma,b Aam = −m−1∑

b=1

Ammab Amm =

m−1∑

a,b=1

Ammab (213)

Now we can write down explicitly the Gaussian ansatz for the replicated density as,

ρA(u) =ρm−d

(2π)(m−1)d/2(det(Amm))d/2e−

12

∑m−1a,b=1(Amm)−1ua·ub (214)

which is normalized such that∫DuρA(u) = ρ.

The entropic part of the free-energy functional can be easily evaluated within this ansatz.

1

N

∫dxρ(x)[1− ln ρ(x)] = 1− 〈ln ρ(x)〉G (215)

where

〈ln ρ(x)〉G = ln ρ− d lnm− (m− 1)d

2ln(2π)− d

2ln detAmm −

1

2

m−1∑

a,b=1

(Amm)−1〈ua · ub〉 (216)

To sum up we obtain,

−βmF idm (A)

N= 1− ln(ρλdth) + d lnm+ (m− 1)

d

2ln(2πe) +

d

2ln detAmm

34

1RSB 2RSB

FIG. 8. Schematic picture of molecule made of replicas in 1RSB and 2RSB ansatz

2. Hierarchical gaussian ansatz

To proceed further we assume the Parisi-type hierarchical structure for the matrix Aab similarly to the spinglasscase. Let us explain the parametrization. In the following we switch to the convention used in [22, 23] : mk = 1 >mk > . . . > m2 > m0 ≥ 0 and we identify m0 with m. Note that this is shifted with respect to the convention weused for the spinglass case Eq. (21) where mk+1 = 1 > mk > . . . > m2 > m1 ≥ 0 and we identified m1 with m. Thenwe assume the matrix A is parametrized as

Aab =

k∑

i=1

(Imi−1

ab − Imiab )(−Ai) = (−A1)Im0

ab +

k−1∑

i=1

((−A)i+1 − (−A)i)Imiab − (−Ak)Imkab a 6= b (217)

The reason for the − sign for the off diagonal elements is that fluctuations of different replicas are anti-correlatedwithin a molecule whose center of the mass is fixed. In the k →∞ limit the matrix A is parametrized by −A(x) withm < x < 1.

In order to satisfy Eq. (213) with the matrix structure Eq. (217) the diagonal part Ad = A11 = A22 . . . = Amm isfixed automatically as,

Ad = −∑

b( 6=a)

Aab =

k∑

i=1

(mi−1 −mi)Ai −−−−→k→∞

−∫ 1

m

dxA(x) (218)

The matrix Dab = Aaa +Abb − 2Aab takes the same form with off-diagonal elements parametrized by,

Dab =

k∑

i=1

(Imi−1

ab − Imiab )Di (219)

Di = 2(Ad +Ai) = 2(

k∑

j=1

(mj−1 −mj)Aj +Ai) −−−−→k→∞

2

(−∫ 1

m

dyA(y) +A(x)

)(220)

and the diagonal element Dd = 0.

3. Another representation of the hierarchical Gaussian

As we see below it is convenient to introduce an alternative representation of the Gaussian ansatz [22, 23],

ρ(x) = ρ

∫ddX1

m∏

a=1

γdD1/2(X1 − xa) (221)

where we introduced γdA(x) which is a d-dimensional Gaussian,

γdA(x) =e−|x|22A

(√

2πA)d(222)

With this one can confirm that

〈(ua − ub)2〉G = 〈(xa − xb)2〉G = dD1 (223)

35

This is equivalent to the general Gaussian form Eq. (214) at 1RSB level since they give the same first two moments,which are sufficient to fix a Gaussian distribution. We just need to choose D1 as D1 = 2(Ad +A1.

Similarly 2 RSB ansatz can be written as,

ρ(x) = ρ

∫ddX1

m/m2∏

C=1

ddX2Cγd(D1−D2)/2(X1 −X2C)∏

a∈CγdD2/2

(X2c − xa) (224)

Let us check if this correctly describes the fluctuation of the distance between replicas xa − xb. To this end let usdecompose it as,

xa − xb = (xa −X2C(a))− (xb −X2C(b)) + (X2C(a) −X1)− (X2C(b) −X1) (225)

where C(a) and C(b) represents the cluster at level 2 where the replicas a and b belong to. Then it is easily to see that

〈(xa − xb)2〉G =

{2D2/2 = D2 (C(a) = C(b))2D2/2 + 2(D1 −D2)/2 = D1 (C(a) 6= C(b)) (226)

Thus we see that the fluctuation is correctly captured. Generalization to k-RSB is straightforward. A hierarchy ofGaussians can be constructed by noting,

Di =

k−1∑

j=i

(Dj −Dj+1) +Dk. (227)

F. Free-energy

1. Entropic part of the replicated free-energy with the hierarchical gaussian ansatz

Here we discuss how to evaluate the entropic part of the free-energy F idm in Eq. (208) within the hierarchical Gaussian

ansatz. To this end we have to compute the determinant of the matrix Amm of size m− 1×m− 1.Let us start from the 1RSB ansatz which amounts to assume,

Aab = A(δab − 1/m) (228)

in Eq. (214). Here the cage size A is related to A1 by A = mA1. The determinant is readily computed as,

detAmm = det[A(δab − 1/m)] = Am−1det(δab − 1/m) = Am−1/m (229)

using the identity det(δab + uavb) = 1 +∑m−1a=1 uava with ua = 1 ∀a and va = −1/m ∀a. Thus we obtain

−βmF idm

N= 1− ln(ρλdth) +

d

2lnm+

d

2(m− 1) ln(2πemA1) (230)

For the k-RSB structure, let us just quote the result obtained in [23] (See Appendix II of [21]),

ln detAmm = − lnm+

k∑

i=1

(m

mi− m

mi−1

)ln(Ei/2) −−−−→

k→∞− lnm−m

∫ 1

m

dy

y2ln[E(y)/2] (231)

with

Ei = miDi +

k∑

j=i+1

(mj −mj−1)Dj −−−−→k→∞

E(x) = xD(x) +

∫ 1

x

dyD(y) (232)

Let us make a quick check for the 1RSB case. We see that Ad = −(m− 1)(−A1 = (m− 1)A1 to meet the sum ruleEq. (212). Then D1 = 2(Ad − (−A1)) = 2mA1. We have m1 = 1 and m0 = m. Using these in the above equationswe obtain ln detAmm = − lnm+ (m− 1) ln(mA1). This is consistent with Eq. (229).

To sum up we obtained the entropic part of the free-energy at k-RSB level as,

−βmF idm

N= 1− ln(ρλdth) +

d

2lnm+ (m− 1)

d

2ln(πe) +

d

2

k∑

i=1

(m

mi− m

mi−1

)lnEi (233)

where we used∑ki=1(m/mi −m/mi−1) = m− 1.

36

2. Interaction part of the replicated free-energy with the hierarchical gaussian ansatz

Let us discuss how to evaluate the interaction part of the free-energy F exm within the Gaussian ansatz. In the

following we limit ourselves to the 1st virial correction,

− βF exm =

1

2N=

∫dxdyρ(x)ρ(y)f(x− y) f(x− y) = −1 +

m∏

a=1

e−βv(xa−ya) (234)

Let us first consider 1RSB case (k = 1). Using Eq. (222) in the above equation we find,

−βF exm =

ρ2

2N

∫ddX1

∫ddY 1

m∏

a=1

∫ddxa

∫ddyaγ

dD1/2

(X1 − xa)γdD1/2(Y 1 − ya)

−1 +

m∏

a=1

e−βv(xa−ya)

=ρ2

2N

∫ddX1ddY 1

[g(m1, X

1 − Y 1)m/m1 − 1]

=ρ

2

∫ddr

[g(m1, r)

m/m1 − 1]

(235)

where we introduced

g(m1, X − Y ) ≡∫ddxddyγdD1/2

(X − x)γdD1/2(Y − y)e−βv(x−y)

=

∫ddrγdD1

(r)e−βv(X−Y−r) =(γdD1⊗ e−βv

)(X − Y ) =

∫Ddz e−βv(X−Y+

√D1z) (236)

In the last equation we introduced a short hand notation for the d-dimensional integral with the Gaussian weight ofunit variance,

∫Ddz · · · =

∫ddze−z

2/2 · · · . (237)

Generalization to k-RSB is straightforward. We find again that the interaction part of the free-energy is give byEq. (235) but g(m1, r) replaced by the one which is obtained by solving the following recursive relation,

g(mi, r) = γdDi−Di+1⊗ g(mi+1, r)

mi/mi+1 =

∫Ddzig(mi+1, r +

√Di −Di+1zi)

mi/mi+1 (238)

for i = 1, 2, . . . , k, assuming Dk+1 = 0 and mk+1 = mk = 1, given the initial condition,

g(mk+1, r) = e−βv(r). (239)

The function g(mi, r) would be interpreted as the statistical weight or the partition function associated with thelink between a pair of molecules at level i separated by a given distance r. We may also define its associated free-energy/replica,

f(mi, r) =1

miln g(r,mi) (240)

which obeys

f(mi, r) =1

miln γdDi−Di+1

⊗ emif(mi+1,r) (241)

for i = 1, 2, . . . , k given the initial condition

f(mk+1, r) = −βv(r) (242)

Note: since we have chosen the convention mk+1 = mk = 1, it is problematic to write f(mk+1) and f(mk) whichare actually different. To avoid the confusion we could instead wrote them as fk+1 and fk. In spite of this problemlet us use this notation f(m) which is convenient to take the continuous limit.

37

3. Variational equations

Next let us extract variational equations to find the saddle points. Up to the 1st virial correction we have obtainedthe total free-energy as,

−βmFN

= 1− ln(ρλdth) + d lnm+ (m− 1)d

2ln(πe) +

d

2

k∑

i=1

(m

mi− m

mi−1

)lnEi

+ρ

2

∫ddr

[g(m1, r)

m/m1 − 1]

(243)

with

Ei = miDi +

k∑

j=i+1

(mj −mj−1)Dj (244)

which can be inverted to find

Di =Eimi

+

k∑

j=i+1

(1

mj− 1

mj−1

)Ej (245)

We consider below the variational equations

0 =∂

∂El

−βmFN

=d

2

(m

ml− m

ml−1

)1

El+ρ

2

∫ddr

1

ml

∂g(m1, r)m/m1

∂Dl+

(1

ml− 1

ml−1

) l−1∑

i=1

∂g(m1, r)m/m1

∂Di

(246)

where we used

∂ElDi =δilml

+

(1

ml− 1

ml−1

)θ(l − (i+ 1)) (247)

for l = 1, 2, . . . , k.Now we need to study objects like ∂Djg(mi, r). This can be done similarly to what we did in the spinglass model

in sec II H 3. Again we again use the convention of : dash is to mean a partial derivative with respecrt to the secondvariable. Using Eq. (238) and Eq. (240) we find after some algebra

∂Dlg(mi, r) =1

2(mj −mj−1)

∫dhδg(mi, h)

δf(ml, y)(f ′(ml, h))2 (248)

or equivalently

∂Dlf(mi, r) =1

2(ml −ml−1)

∫dhPi,l+1(f ′(ml, h))2 (249)

where we introduced

Pij(y, z) =δf(mi, y)

δf(mj , z)(250)

which obeys the recursion formula

Pij(y, z) = gmj−1/mj (mj , z)γdDj−1−Dj ⊗z

Pi,j−1(y, z)

g(mj−1, z)(251)

for j = i+ 1, . . . , k, k + 1 with the initial condition,

Pii(y, z) = δ(y − z). (252)

38

For convenience, let us introduce the a pair of integrated quantities,

P (ml, s) =

∫drrd−1emf(m1,r)P1,l(r, s) P (ml, s) =

∫dremf(m1,r)P1,l(r, s) (253)

both of which follows the formally same recursion formula as Pij ,

P (mj , s) = gmj−1/mj (mj , s)γdD1−D2

⊗ P (mj−1, s)

g(mj−1, s)P (mj , s) = gmj−1/mj (mj , s)γ

dD1−D2

⊗ P (mj−1, s)

g(mj−1, s)(254)

for j = 1, 2, . . . , k + 1 with the initial condition,

P (m1, s) = sd−1emf(m1,s) P (m1, s) = emf(m1,s) (255)

Then we see that the relation

P (mj , s) = sd−1P (mj , s) (256)

should hold for j = 1, 2, . . . , k + 1.Using the above result we can write

∂Dlg(m1, r)m/m1 =

1

2m(ml −ml−1)emf(m1,r)

∫dsP1,l(r, s)(f

′(ml, s))2. (257)

and find the desired variational equation as,

1

El= ml−1κ(ml)−

l−1∑

i=1

(mi −mi−1)κ(mi) κ(mi) =ρ

2Vd

∫dsP (ml, s)(f

′(ml, s))2 (258)

where Vd is the volume of unit sphere in d-dimension so that dVd is the solid angle.

4. Effective potential and g(r)

Let us discuss the physical meaning of P (mi, h), which was distribution of internal field in the spinglass problem.Let us introduce the effective potential φeff(s) through,

e−φeff (s) ≡ P (mk+1, s) = gmk/mk+1(mk+1, s)γdDk−Dk+1

⊗ P (mk, s)

g(mk, s)

= e−βv(s)γdDk ⊗P (mk, s)

g(mk, s)(259)

using mk = mk+1 = 1, Dk+1 = 0 and g(mk+1, r) = e−βv(r). In particular, in the 1RSB case, we find,

e−φeff (s) = e−βv(s)

∫ddy

(√

2πD1)de−

(y−s)22D1 gm−1(m1, y) g(m1, y) =

∫ddy′

(√

2πD1)de−

(y′−y)2

2D1 e−βv(y′) (260)

using P (m1, s) = emf(m1,s) = g(m1, s)m/m1 with m1 = 1.

Physically the potential φeff(r) describes effective inter-molecular interaction obtained by integrating out fluctuationwithin the molecules [22, 24]. Within the 1st order virial expansion, that we are considering, the effective potential isintimately related to the radial distribution function g(r) in the glass [22]. From the identity Eq. (206) we have

δ

δ(−βv(r))

−βF exm

N= m

ρ

2dVdr

d−1g(r) (261)

We readily fnd

δ

δ(−βv(r))

−βF exm

N=ρ

2dVd

∫dr′(r′)d−1 δg(r′)m/m1

δ(−βv(r))= m

ρ

2dVdP (mk+1, r) = m

ρ

2dVdr

d−1P (mk+1, r) (262)

in the last equation we used Eq. (256). Thus we find,

g(r) = P (mk+1, r) = e−φeff (r) (263)

This relation is particularly useful information to study the jamming transition since the behavior of g(r) become oneof the main interests.

39

5. Continuous RSB: k → ∞ limit

Let us consider k →∞ limit. This can be done just as we did for the spinglass discussed in sec. II H 4. We readilyfind the following partial different equations corresponding to the recursion relations for g(mi, r), f(mi, r) and Pi(r).

The only needed changes are λ(x)→ −D(x) and the boundary conditions.

∂g

dx=D

2

∂2

∂r2g +

1

xg ln g (264)

∂f

∂x=D

2

[∂2f

∂r2+ x

(∂f

∂r

)2]

(265)

∂P

∂x= −D

2

[∂2P

∂r2− 2x

∂

∂r

(P∂f

∂r

)](266)

with

g(1, r) = γdD(1) ⊗ e−βv(r) (267)

f(1, r) = ln[γdD(1) ⊗ e−βv(r)

](268)

P (m, r) = rd−1emf(m,r) (269)

Note: in the k-RSB ansatz we used the convention mk+1 = mk = 1, ∆k+1 = 0. But to work with x coordinate it isbetter to choose the initial condition at mk.

The saddle point equation Eq. (258) becomes in the limit k →∞,

1

E(x)= xκ(x)−

∫ x

m

dyκ(y) κ(x) =ρ

2Vd

∫dsP (ml, s)(f

′(ml, s))2 (270)

from which we can determine D(x) through

D(x) =E(x)

x−∫ 1

x

dy

y2E(y). (271)

In terms of these we can write the k →∞ limit of the free-energy Eq. (243).

−βmFN

= 1− ln(ρλdth) + d lnm+ (m− 1)d

2ln(πe)− d

2m

∫ 1

m

dx

x2lnE(x)

+ρ

2

∫ddr

[g(m1, r)

m/m1 − 1]

(272)

6. d→ ∞ limit

In order to study the problem in the limit d → ∞, we have to introduce properly rescaled quantities. Let us lookat the interaction term discussed in sec. III F 2. The interaction part of the free-energy in Eq. (243) is

−βF ex

N=ρ

2

∫ddr

[g(m1, r)

m/m1 − 1]

=ρ

2dVd

∫ ∞

0

drrd−1[g(m1, r)

m/m1 − 1]

(273)

Here Vd is the volume of unit sphere in d-dimension so that dVd is the solid angle. The function g(mi, r) fori = 1, 2, . . . , k follows the recursion relations Eq. (238) (with Dk+1 = 0,mk+1 = mk = 1,gmk+1

(r) = e−βv(r)).Let us first examine the equation for i = k,

g(mk, r) = γdDk ⊗ e−βv(r) =

∫ddu

(√

2πDk)de− (r−u)2

2Dk e−βv(u) =

∫ ∞

0

du

(u

r

)(d−1)/2e− (r−u)2

2Dk√2πDk

e−Dk

2d2ru e−βv(u) (274)

in the last we used the polar coordinate (see [22] appendix C). Now let us examine the limit d→∞. In order to havenon-trivial limit we can choose

u = D(1 + z/d) r = D(1 + h/d) Di =D2

d2∆i Ai =

D2

d2αi (i = 1, 2, . . . , k) (275)

40

where D is an arbitrary chosen scale of length, which we may take as the diameter of spheres, with which we find

g(mk, h) =

∫dze(z−h)/2 e

− (z−h)2

2∆k√2π∆k

e−∆k8 e−βv(D(1+z/d)) (276)

Here we see that, in order to have non-trivial d→∞ limit, the potential must have the form,

v(r) = v(d(r/D − 1)) (277)

The hardphsere potential meets this requirement. Now we find a simple expression,

g(mk, h) =

∫dze− (z−(h+∆k/2))2

2∆k√2π∆k

e−βv(z) = (γ∆k⊗ e−βv(z))(h+ ∆k/2). (278)

which is expressed by ’one dimensional’ convolution. The recursion relation Eq. (238) for i 6= k can be analyzedsimilarly and one finds

g(mi, h) ≡ g(mi, h−∆i/2) (279)

for i = 1, 2, . . . , k obeys ’one dimensional’ recursion formula,

g(mi, h) = γ∆i−∆i+1⊗ g(mi+1, h)mi/mi+1 =

∫Dzig(mi+1, h+

√∆i −∆i+1zi)

mi/mi+1 (280)

with the initial condition (note that ∆k+1 = 0),

g(mk+1, h) = e−βv(h) (281)

Naturally we also introduce,

f(mi, h) ≡ f(mi, h−∆i/2). (282)

which obeys the recursion relation,

f(mi, h) =1

miln γ∆i−∆i+1 ⊗ emif(mi+1,h) (283)

for i = 1, 2, . . . , k with the initial condition,

f(mk+1, h) = −βv(h). (284)

By the same token ’one dimensional’ version of the recursion formula Eq. (251)

Pij(y, z) = gmj−1/mj (mj , z)γ∆j−1−∆j⊗z

Pi,j−1(y, z)

g(mj−1, z)(285)

is satisfied by

Pij(y, z) ≡ Pij(y −∆i/2, z −∆j/2) (286)

for j = i+ 1, . . . , k, k + 1 with the initial condition,

Pii(y, z) = δ(y − z). (287)

For convenience let us introduce again a pair of integrated quantities corresponding to Eq. (253) which read,

P (ml, h) =

∫dh′eh

′emf(m1,h

′)P1,l(h′, h) = e−∆1/2

∫dh′eh

′+∆1/2emf(m1,h′+∆1/2)P1,l(h

′ + ∆1/2, h+ ∆i/2)

P (ml, h) =

∫dh′emf(m1,h

′)P1,l(h′, h) =

∫dh′emf(m1,h

′+∆1/2)P1,l(h′ + ∆1/2, h+ ∆i/2) (288)

41

where we mean P1,l(h′, h) = P1,l(r = D(1 + h′/d), h). Clearly it is natural to introduce,

P (ml, h) = e∆1/2P (ml, h−∆i/2) (289)

We see that both P (ml, h) and P (ml, h) obey the same recursion relation as Eq. (285),

P (mj , z) = gmj−1/mj (mj , z)γ∆j−1−∆j ⊗zP (mj−1, z)

g(mj−1, z)P (mj , z) = gmj−1/mj (mj , z)γ∆j−1−∆j ⊗z

P (mj−1, z)

g(mj−1, z)(290)

for j = 1, 2, . . . , i with the initial condition,

P (m1, h) = ehemf(m1,h) P (m1, h) = emf(m1,h) (291)

Then corresponding to Eq. (256) we find,

P (mj , s) = esP (mj , s) (292)

should hold for j = 1, 2, . . . , k + 1.Turning to the entropic part of the free-energy Eq. (233), we may rewrite it as,

−βmF idm

N= 1− ln(ρλdth) +

d

2lnm+ (m− 1)

d

2ln((D/d)2πe) +

d

2

k∑

i=1

(m

mi− m

mi−1

)lnGi (293)

where we introduced another scaled quantity Gi through,

Ei =D2

d2Gi (i = 1, 2, . . . , k). (294)

Thus we see Eq. (244) and Eq. (245) becomes

Gi = mi∆i +

k∑

j=i+1

(mj −mj−1)∆j ∆i =Gimi

+

k∑

j=i+1

(1

mj− 1

mj−1

)Gj (295)

Let us then come back to the interaction part of the free-energy Eq. (273). Let us examine the integral,

ρ

2dVd

∫ ∞

0

drrd−1 . . . =d

2ϕ

∫ ∞

−∞dheh . . . (296)

where we introduced,

ϕ = 2dφ/d φ = ρVd(D/2)d (297)

where ϕ is the usual volume fraction. The reason for the scaled volume fraction can be understood from the fact thatthe leading term of the entropic part of the free-energy Eq. (233) is proportional to d/2. In the integrant in Eq. (273),we can use g(m1, r) = g(m1, h+ ∆1/2) thus we find the interaction part of the free-energy as,

−βmF exm

N=d

2ϕe−∆1/2

∫dheh

[g(m1, h)m/m1 − 1

](298)

where g(m1, h) should be obtained by solving the recursion relation Eq. (280) for i = 1, 2, . . . , k with the initialcondition g(mk+1, h) = e−βv(h).

Finally let us discuss the variational equation Eq. (258) in the d→∞ limit.Going back to Eq. (246) and using the scaled quantities introduced above we readily find,

1

Gl=D2

d2

∫dhϕ

∫dh′eh

′ emf(m1,h′)

2

ml−1P1,l(h

′, h)(∂rf(ml, h))2 −l−1∑

i=1

(mi −mi−1)P (h′, h)(∂rf(mi, h)2)

=ϕ

2

∫dh

P (ml, h)(f ′(ml, h))2 −

l−1∑

i=1

(mi −mi−1)P (mi, h)(f ′(mi, h))2

=ϕ

2e−∆1/2

∫dh

P (ml, h)(f ′(ml, h))2 −

l−1∑

i=1

(mi −mi−1)P (mi, h)(f ′(mi, h))2

(299)

42

Note that by f(mi, h) we mean f(mi, r = D(1 + h/d)). So in the first equation we used that ∂rf(mi, h) =(d/D)∂hf(mi, h).

Now let us look at the effective potential in the large-d limit. Using Eq. (289), Eq. (292) in Eq. (259) we find,

e−φeff (h) = e−βv(h)γdDk ⊗e−z−∆1/2P (mk, h+ ∆k/2)

g(mk, h+ ∆k/2)

= e−βv(h)

∫dze

z−h2e− (z−h)2

2∆k√2π∆k

e−∆k8e−z−∆1/2P (mk, z + ∆k/2)

g(mk, z + ∆k/2)

= e−βv(h)

∫dzez−hγ∆k

(h− z)⊗ e−z−∆1/2P (mk, z)

g(mk, z)(300)

In the continuous limit k →∞, we have the following partial differential equations corresponding to the recursionrelations for g(mi, h), g(mi, h) and P (mi, h), which are just the same as Eq. (266),

∂g

dx=

∆

2

∂2

∂h2g +

1

xg ln g (301)

∂f

∂x=

∆

2

∂

2f

∂h2+ x

(∂f

∂h

)2 (302)

∂P

∂x= −∆

2

∂

2P

∂h2− 2x

∂

∂h

(P∂f

∂h

) (303)

with

g(1, h) = γ∆(1) ⊗ e−βv(h) (304)

f(1, h) = ln[γ∆(1) ⊗ e−βv(h)

](305)

P (m,h) = ehemf(m,h) (306)

The saddle point equation Eq. (258) becomes in the limit k →∞,

1

G(x)= xκ(x)−

∫ x

m

dyκ(y) κ(x) =ϕ

2e−∆(m)/2

∫dhP (x, s)(f ′(x, s))2 (307)

Given the function G(x) we can determine ∆(x) through

∆(x) =G(x)

x−∫ 1

x

dy

y2G(y). (308)

In terms of these we can write the k →∞ limit of the free-energy.

−βmFN

= 1− ln(ρλdth) + d lnm+ (m− 1)d

2ln(πe)− d

2m

∫ 1

m

dx

x2lnG(x)

+d

2ϕe−∆1/2

∫dheh

[emf(m,h) − 1

](309)

IV. HARDSPHERES IN d→ ∞ LIMIT

We look at hardspheres in large d limit. So we assume,

e−βv(h) = θ(h) (310)

43

A. 1 RSB: dynamical transition, Kauzmann transition, Gardner transition,...

1. Free energy

Let us first look at the 1RSB ansatz. We readily find the entropic part as,

−βmF id

N= 1− ln(ρλdth) +

d

2lnm+ (m− 1)

d

2ln(πe(D/d)2) +

d

2(m− 1) ln ∆1

= 1− ln(ρλdth) +d

2lnm+ (m− 1)

d

2ln(2πe(D/d)2) +

d

2(m− 1) lnα (311)

Here we used m1 = 1, m0 = m, G1 = m1∆1 = ∆1. In the last equation we used ∆1 = 2mα1 = 2α introducingα = mα1 such that we can write αab = α(δab − 1/m). The interaction term becomes

−βmF ex

N=d

2ϕe−∆1/2

∫dheh

[g(m1, h)m − 1

](312)

where

g(m1, h) =

∫Dz1g(m2, h+

√∆1z1) =

∫dy′√2π∆1

e−(y′−h)2

2∆1 θ(y′) = Θ(h/√

2∆1) (313)

with ∆1 = 2α. Here we introduced

Θ(x) =

∫ ∞

−x

dz√πe−z

2

=1 + erf(z)

2efc(z) =

2√π

∫ z

0

dte−t2

(314)

To sum up we find the free-energy as

−βmF1RSB

N= 1− ln(ρλdth) +

d

2lnm+ (m− 1)

d

2ln(2πe(D/d)2)

+d

2

[(m− 1) lnα+ ϕe−∆1/2

∫dheh

[Θ(h/

√2∆1)m − 1

]](315)

with ∆1 = 2α.The variational equation for ∆1 = 2α reads as,

1

∆1=ϕ

2e−∆1/2

∫dhP (m1, h)(f ′(m1, h))2 (316)

where

P (m1, h) = ehemh(m1,h) = ehΘm(h/√

2∆1) (317)

and

f ′(m1, h) =mΘm−1(h/

√2∆1)

Θm(h/√

2∆1)

e−h2

2∆1√2π∆1

(318)

To sum up we find

1

∆1=ϕ

2e−∆1/2

∫dhehΘm(h/

√2∆1)

[mΘm−1(h/

√2∆1)

Θm(h/√

2∆1)

]2

(319)

2. Effective potential

Let us look at the effective potential Eq. (300), which is identical to the radial distribution function g(r) in thed→∞ limit where only the 1st virial contributes. With h = (r −D)/D, and e−βv(h) = θ(h) for hardspheres,

e−φeff (h) = θ(h)

∫dzez−h

e−(z−h)2

2∆1√2π∆1

e−z−∆1/2ezΘm(z/√

2∆1)

Θ(z/√

2∆1)

= θ(h)

∫dze−

(z−(h+∆1))2

2∆1√2π∆1

Θm−1(z/√

2∆1) (320)

where ∆1 = 2α1m.

44

3. Another representation

The above expressions can also be written in the following equivalent form [22]. Doing an integration by parts onecan show

e−∆1/2

∫dheh

[Θ(h/

√2∆1)m − 1

]= −m

∫DλΘm−1

(√2α− λ√

2

)(321)

Then the free-energy can be written as,

−βmF1RSB

N= 1− ln(ρλdth) +

d

2lnm+ (m− 1)

d

2ln(2πe(D/d)2)

+d

2

[(m− 1) lnα− ϕ(1− Gm(α))

](322)

with

Gm(α) = 1−m∫DλΘm−1

(√2α− λ√

2

)

= 2√α

∫dte2t

√α

Θm

(t+

√α

2

)− θ(t)

(323)

We obtain the variational equation for α from Eq. (322) as,

1

ϕ= ζ(α) ζ(α) =

α

1−m∂Gm(α)

∂α(324)

4. Thermodynamics of the liquid

In the m→ 1 limit we find the the fee-energy of liquid,

limm→1

−βmF1RSB

N= 1− ln(ρλdth)− d

2ϕ (325)

The reduced pressure is found as

βP

ρ= ϕ

∂

∂ϕlimm→1

(βF1RSB

N

)=d

2ϕ (326)

which increases with increasing density ϕ. This is the equation of the state of the Hardsphere liquid in d→∞ limit.

5. Dynamical glass transition point in the liquid phase

In Fig. 9 we show a plot of the function ζ(α) computed in the limit m → 1. By increasing the scaled volumefraction, we meet the dynamical transition point with ϕd ∼ 4.8 and αd ∼ 0.576 [22]. At higher density we find asolution with the cage size α decreases, which is the physical solution,

α = αd − const

√ϕ− ϕd

ϕd(327)

6. Cage size in glassy phases

At higher densities we expect smaller cage sizes. Assuming small cage size one can show that ,

Gm(α) ∼ 2√α

∫ ∞

−∞dt(Θm(t)− θ(t)

)+O(α)

45

0

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5A

m=1

⇣m(↵)

↵

1/'

m = 1

FIG. 9. The function ζ(α) = α1−m

∂Gm(α)∂α

here m = 1 is assumed.

Using this in Eq. (324) we indeed find that the saddle point value of the cage size as,

α∗ ∼ (cϕ)−2 c =1−m∫∞

−∞ dt(Θm(t)−Θ(t))2. (328)

7. Thermodynamics in glassy phases

The saddle point value of the free-energy and the complexity can be computed using the Monasson’s scheme Eq. (97),

− βf∗ = ∂m

(−βmFmN

)Σ∗ = m2∂m

(βmFmNm

)= −βmFm

N− (−βmf∗) (329)

The result is the following [22]. Dropping off subleading terms in d→∞ we find,

−βf∗ =d

2

1

m+d

2ln(2παe(D/d)2) +

d

2ϕ∂mGm(α)

Σ∗ =d

2

{ln d− [1 + ln(α/m)]− ϕ(1− Gm(α) +m∂mGm(α))

}(330)

where saddle point value α = α∗ must be used. Here we used the relations ϕ = 2dϕ/d, ϕ = ρ(D/2)dVd, Vd =πd/2/(Γ(1 + d/2)) and the Stirling’s formula ln Γ(X) = X lnX −X for large X.

Since the cage size scales as α∗ ∝ ϕ−2 we find the complexity,

Σ(α∗) =d

2ln d− d

2[ϕ+O(ln ϕ)]. (331)

This implies the static glass transition density, where the Kauzmann transition takes place, scales with the dimensiond as,

ϕK ∝ ln d. (332)

The pressure is obtained as,

βP

ρ= ϕ

∂

∂ϕ

(βFm(α∗)

N

)=

1

m

[1 +

d

2ϕ(1− G(α∗)

]∼ 1

m

d

2ϕ (333)

which diverges deeper in the glass phase m→ 0.

46

substantiate this picture for hard spheres in the d → ∞ limit. Insection III, we provide a proof of the correctness of theGaussian ansatz for a generic form of the overlap matrix,extending the main result of ref 11; in section IV, we recall afew important results of refs 9 and 11 that are directly neededhere; in section V, we present our main results for the Hessianmatrix of the 1RSB solution and in particular its so-calledreplicon eigenvalue, that is responsible for the instability of thissolution (i.e., the Gardner transition); in section VI, we discussthe cubic terms in the expansion around the 1RSB solution andfrom them we extract the dynamical exponents that characterizethe glass transition; in section VII, we present an approximatecalculation to obtain an order of magnitude for the Gardnertransition pressure in finite d; in section VIII, we summarizeand draw our conclusions.

II. A GENERAL RFOT SCENARIO FOR THE GLASS ANDJAMMING TRANSITIONSA. The Generic Phase Diagram of RFOT Models. As is

by now well-known, Kirkpatrick, Thirumalai, and Wolynes’scenario for the liquid−glass transition involves a first pointat which the equilibrium state fractures into an exponentialnumber of ergodic components: this is the dynamicaltemperature Td (or pressure Pd, see Figure 1), also called the

mode-coupling temperature because in low dimensions it canbe computed using mode-coupling theory. The ergodic com-ponents are only truly dynamically separated in the mean-field limit, while in a realistic short-range finite-dimensionalsituation the system is still ergodic, although the dynamicsslows down. As the temperature is lowered, or the pressureincreased, the number of metastable states contributing toequilibrium diminishes, until a point is reached where theequilibrium system is left with only the deepest amorphousstates: this is the Kauzmann point, beyond which thethermodynamics stays dominated by (or “frozen in”) thosestates. From a purely equilibrium point of view, one maypicture the situation at P > PK (or T < TK) as in the sketch ofFigure 2, with widely separated states of “size” q, defined, forexample, as

∑= · −qN

k r r1 cos[ ( )]i

ia

ib

(1)

with a and b being two copies (replicas) of the system and k avector of length comparable to the inverse of the interparticledistance (alternative definitions of q are possible; see ref 15 fora review). A more precise way of stating the same thing isto introduce the effective potential V(q),16 which counts thelogarithm of the number of configurations having correlationexactly q with a “reference” equilibrium configuration. One ob-tains a picture as in Figure 2, where one sees that configurationsare either close, with overlap q = qEA (with probability 1 − m),or very far awayin other states, with overlap q = 0withprobability m, corresponding to the two minima in the effectivepotential and to the two peaks in the Parisi order parameterP(q). Note that P(q) may in general only be nonzero whereV(q) takes the minimal value, and that qEA plays the role of theEdwards−Anderson order parameter.6

This construction concerns equilibrium configurations butmay be generalized to describe metastable states17,18 by choos-ing the “reference” configuration, rather than from equilibrium,from a system perturbed by a small “pinning field”, itself ther-malized at a higher temperature T′ = T/m. Technically speak-ing, following Monasson,18 this amounts to the followingcalculation: one considers m weakly coupled replicas at temper-ature T, takes an equilibrium configuration of one of thereplicas as the reference configuration, and couples to it anadditional replica that is forced to stay at distance q from it.Then, one computes the free energy of the additional replica,and averages it over the other m. This amounts to fixing theParisi parameter m, rather than choosing the value thatmaximizes the free energy. In this way, one obtains a bistableform for the effective potential, up to a threshold value Tth′ =T/mth at which the minimum close to qEA disappears (Figure 3),and at precisely the threshold level, the stability matrix cor-responding to the minimum at qEA develops zero modes,signaling the fact that the states close to the threshold level aremarginal. This shows up within the replica scenario as thevanishing of the “replicon” eigenvalue,19 and within the Thouless−Anderson−Palmer approach6 as the free-energy Hessian develop-ing zero eigenvalues. A crucial result of ref 20 is that the out-of-equilibrium aging dynamics happens exactly at this threshold

Figure 1. A sketch of the phase diagram.

Figure 2. A sketch for equilibrium TG ≤ T ≤ Tk or PG ≥ P ≥ PK. Top:a cartoon of the free energy and its minima of width qEA. Bottom: theParisi order parameter P(q) and the effective potential V(q).

The Journal of Physical Chemistry B Article

dx.doi.org/10.1021/jp402235d | J. Phys. Chem. B XXXX, XXX, XXX−XXXB

FIG. 10. Schematic phase diagram of hardspheres in large-d limit Fig 1. of [25]

8. Jamming

We have seen that pressure diverges in the limit m→ 0 which signals jamming. By remembering α = α1m, we canwrite,

Gm(α) = 2√α

∫dte2t

√α

[Θm(t+

√α

2)− θ(t)

]=

∫dyey

[Θm(

y + α1m

2√α1m

)− θ(y)

]−−−→m→0

∫ ∞

0

dye−ye−y2

4α1 (334)

where we used

limm→0

θm(s/√m) =

{e−s

2

x < 01 x > 0

(335)

The the saddle point equation Eq. (324) becomes in the limit m→ 0,

1

ϕ= α1∂α1

limm→0

G(mα1) =1

4α

∫ ∞

0

dyy2e−y2

4α−y (336)

from which we find ’lowest jamming density’ ϕth ' 6.26, the density below which the jammed state is absent.Since the cage size is α = α1m and P ∝ 1/m, we see that the cage size vanishes approaching the jamming density

as 1/P within the 1RSB ansatz.Finally let us consider the behavior of the effective potential Eq. (320) in the jamming limit m→ 0. It is important

to note that ∆1 = 2mα1 vanishes in this limit. We find

limm→0

e−φeff (h) = θ(h)

∫dze−

(z−(h+2mα1))2

2·2mα1√2π · 2mα1

Θm−1(z/√

2 · 2mα1) (337)

If we send m→ 0 with fixed h(> 0) we find it becomes featureless

limm→0

e−βφeff (h) = 1 (338)

In stead let uslook at h = O(m), using h = λm and fix λ, we find only finite contribution from the z < 0 region but

large contribution from z > 0. Using limy→−∞Θ(y) = e−y2

2√π|y| we find,

limm→0

e−βφeff (η=λm) ' limm→0

∫ 0

−∞dze−

(z−(λ+2α1)m)2

2·2mα1√2π · 2mα1

2√π|z|√

4mα1e

z2

4mα1 ' 1

m

1

2α1F1(λ/2α1) (339)

with

F1(x) =

∫ ∞

0

dzze−z(x+1) (340)

47

For large λ we find

F1(λ/2α1) ∝ λ−2 (341)

So in the jamming limit m→ 0, we find a delta peak in the contact region r ∼ D.[22, 24]

9. Gardner transition

In [25], the 1RSB solution was found to become unstable exhibiting the Gardner transition, which is analogous tothe one in the p-spin spinglass model discussed in sec. II G 2. Then we have to try the full RSB ansatz [23, 26].

B. Full RSB: Jamming critically and beyond

1. Vanishing cage size

Here we sketch the very interesting observations on the critical properties of hardspheres approaching the jammingm → 0 revealed by the full RSB ansatz.[23, 26]. In order to study jamming m → 0, it become convenient to workwith

y ≡ 1/m (342)

instead of x, and parametrized x dependent (i.e. hierarchical) quantities in terms of y. Note that 1 < y < 1/mcorresponding to m < x < 1 and y ∼ 1/m describes physics on the most fine scale of the hierarchical energylandscapes, i.e. inherent structures. Note that the maximum value of y diverges in the jamming limit m→ 0.

By numerically solving the k-RSB problem with large k [23], it was realized that the order parameter ∆(x) exhibita non-trivial scaling feature as shown in the left panel of Fig. 12),

∆(y) = ∆∞y−κ y � 1 (343)

with κ ' 1.4. Since pressure scales as P ∝ 1/m, this implies that the cage size vanishing approaching the jammingas,

∆EA = ∆(1) ∝ p−κ. (344)

Numerical data shown in the right panel of Fig. 12 suggests that this holds also in finite dimensional systems.This implies

γ(y) ≡ G(x)/m = ∆∞y−c c = κ− 1 (345)

because of the relation

γ(y) = y∆(y) +

∫ 1/m

y

dz∆(z) (346)

2. Anomalous features in the contact region

The function P (y, h) is closely related to the effective potential and thus the radial distribution function g(r). In[23] the following scaling features was found in the scaling region y � 1 (see Fig. 12),

P (y, h) =

ycp0(hyc) h ∼ −y−cyap1(hyb) |h| ∼ y−bp2(h) h� y−b

(347)

Here the scaling functions must satisfy matching conditions so that they behave as,

p0(|z| � 1) ∼ |z|θ p1(z → −∞) ∼ |z|θ (348)

p1(z → +∞) ∼ z−α p2 = O(1) (349)

4845

10−5

10−4

10−3

10−2

10−5

10−4

10−3

10−2

∆(y

)

1 2 5 10 2 5 1022

y1 2 5 10 2 5 102

2

k = 20, ymax = 100

k = 50, ymax = 100

k = 100, ymax = 100

∆∞y−κ

10−7

10−6

10−5

10−4

10−3

10−2

10−7

10−6

10−5

10−4

10−3

10−2

∆(y

)1 2 5 10 2 5 102

2 5 1032

y1 2 5 10 2 5 102

2 5 1032

k = 100, ymax = 50

k = 100, ymax = 100

k = 100, ymax = 200

k = 100, ymax = 500

k = 100, ymax = 1000

k = 100, ymax = 10000

∆∞y−κ

FIG. 9: The function ∆(y) at m = 0 and ϕ = 10. In left panel we report results for fixed ymax = 100 and k = 20, 50, 100,which show that k = 100 is large enough to be considered infinite. In right panel we report results for fixed k = 100 andymax = 50, 100, 200, which show that for large ymax the cutoff at large y disappears. The power law regime ∆(y) ∼ ∆∞y−κ

with ∆∞ ≈ 0.023 and κ given in Eq. (200) is approached at large y.

0.0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

γ

0.0 0.05 0.1 0.15 0.2 0.25 0.3 0.35y−c

0.0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

k = 100, ymax = 10000

γ∞x

0·100

1·10−5

2·10−5

3·10−5

4·10−5

5·10−5

0·100

1·10−5

2·10−5

3·10−5

4·10−5

5·10−5

∆

0.0 0.0005 0.001 0.0015 0.002 0.0025y−1−c

0.0 0.0005 0.001 0.0015 0.002 0.0025

k = 100, ymax = 10000

∆∞x

FIG. 10: Plot of γ versus y−c (left), and of ∆ versus y−1−c (right), with c given in Eq. (200). The linear fits give γ∞ ≈ 0.080and ∆∞ = c

c+1γ∞ ≈ 0.023.

the plots are linear at large y and provide an estimate of γ∞ ≈ 0.080 and ∆∞ = cc+1γ∞ ≈ 0.023, which is also

perfectly compatible with Fig. 9.Having an estimate of γ∞, we can test the scaling of Eq. (187), see Fig. 11, and the scaling in Eq. (167), see Fig. 12.

In both case we find excellent agreement between the analytical results of Sec. IX and the numerical solution of theequations.

XI. CRITICAL SCALING OF PHYSICAL OBSERVABLES

We have confirmed that the asymptotic solution of Eqs. (113) and (116) found in Sec. IX is realized by the numericalsolution of the fullRSB equations. We now start to investigate the physical consequences, in particular to identifyobservables that display critical scaling controlled by the exponents in Eq. (200).

In Sec. XI A we show that the exponent κ is related to the scaling of the cage radius with pressure. In Sec. XI Bwe discuss the scaling of the pair correlation function of the glass on approaching jamming and we show that it isdetermined by the exponents α and θ. In Sec. XI C we show that the distribution of forces in the packing can beobtained from the pair correlation function and that it is characterized by the exponent θ. Finally, in Sec. XI D weshow that the fullRSB solution predicts that jammed packings are isostatic, i.e. particles touch on average 2d otherparticles.

51

10−16

10−12

10−8

10−4

1

⟨∆r(

t)2⟩

10−8 10−6 10−4 10−2 1 102

t10−8 10−6 10−4 10−2 1 102

10−12

10−9

10−6

10−3

10−12

10−9

10−6

10−3

∆E

A

102 104 106 108

p102 104 106 108

d = 3d = 4d = 6d = 8∼ p−κ

∼ p−3/2

FIG. 13: (Left) Time evolution of the mean-square displacement in the equilibrated liquid at ϕ = 0.405 as well as in the glassfor different p = 102–108 . (Right) Pressure evolution of the Debye-Waller factor ∆EA in d = 3, 4, 6, 8. The black line is a

power-law with exponent κ given by Eq. (200). For comparison, we also show a power-law p−3/2 as a grey line.

XIII. COMPARISON WITH RESULTS OF MOLECULAR DYNAMICS SIMULATIONS IN FINITEDIMENSIONS

The scaling of the fullRSB solution provides a number of predictions for the critical scaling at the jamming transition.It predicts that packings are isostatic with average contact number z = 2d, that the cage radius scales as ∆EA ∼ p−κ,that the correlation function at jamming diverges on approaching contact as (r − D)−α, that the contact peak of thecorrelation function, on approaching jamming, has a scaling form given by Eq. (210) (see also [11]) with a scalingfunction F∞(λ) ∼ λ−2−θ at large λ, and that the force distribution is characterized by P∞(f) ∼ fθ for small f . Theexponents κ, α, θ are given in Eq. (200).

Some of these predictions have already been verified in the past using molecular dynamics simulation. In particularisostaticity is a well-known property of jammed packings (see [11, 45, 71] for reviews). Also, the exponent α ing(r) ∼ (r − D)−α has been independently measured with good precision by a number of groups [27, 28, 43, 72, 73]and its mostly accepted value α ≈ 0.42 is perfectly compatible with the prediction in Eq. (200).

The scaling of ∆EA ∼ p−κ has been studied in [21–23, 26] and the data were found to be compatible with κ = 3/2,which is slightly different from our prediction, and had been proposed in [20, 23] using a scaling argument based onmarginal stability. To check whether the numerical data are also compatible with our prediction for κ, we performedadditional molecular dynamics simulations. Hard-sphere systems in d=3, 4, 6, and 8 with N=8000 particles aresimulated under periodic boundary conditions using a modified version of the event-driven molecular dynamics codedescribed in Refs. [12, 13, 72]. We consider monodisperse spheres with unit mass and unit diameter D and unit

mass m, hence time is expressed in units of√

βmD2 at fixed unit inverse temperature β. Hard sphere glasses areobtained using a Lubachevski-Stillinger algorithm initiated in the low-density fluid state with a slow growth rateγ = 3 × 10−4. The fluid then falls out of equilibrium near the dynamical transition. Using these configurations, themean-square displacement ⟨∆r2(t)⟩ = ⟨1/N

∑i[ri(t)−ri(0)]2⟩ is obtained, and reported in Fig. 13. The rattlers, which

are identified as the particles having fewer than d + 1 contacts at p = 1010 (following Ref. [27]), are removed whenanalyzing systems at p ! 105. The long-time mean square displacement plateau provides ⟨∆r2(t → ∞)⟩ = d ∆EA,where the Debye-Waller factor ∆EA is an estimate of the average cage size in the glass [13]. Using this approach, wefind that in all dimensions the exponent κ is close to 3/2, but the data are better described by our predicted value of

κ, see Fig. 13. Note that the theoretical prediction is that ∆EA = ∆EA/d2 should decrease as d2 (at fixed ϕ), while inFig. 13 we see that ∆EA is roughly independent of dimension in the range of d we investigated. This discrepancy isprobably due to the fact that the numerically investigated dimensions are quite far from the asymptotic d → ∞ limit(as far as prefactors are concerned), as it has been already noted in [12, 13]. An approximate analytical computationof the prefactor ∆∞ of the p−κ scaling in finite d is certainly possible and would shed light on this issue.

The measure of the exponent θ is more problematic. In fact, previous attempts at measuring θ using differenttechniques reported results in the range 0.2 ÷ 0.45 [27, 28]. In [28] it has been shown that the behavior of the forcedistribution P (f) at small forces is dominated by two different ways in which the force network responds to an externalperturbation: extended modes and local buckling modes. According to the results of [28, 42], extended modes givean exponent θe = 1/α − 2, which perfectly agrees with our results given in Eq. (200). Buckling modes, instead, give

FIG. 11. The behavior of order parameters. (Left panel) ∆(y) plotted against y = x/m. This figure is took from Fig.9 of[23]obtained by numerically solving the k-RSB problem. The straight line represents the power law ∆(y) = ∆∞y

−κ with theκ extracted from theoretical analysis. (Right panel) The scaling property of ∆EA extracted from the simulation data of thehardspheres in finite dimensional systems. The cage size ∆EA is associated with the plateau value of the MSD. This figure istook from Fig. 13 of [23].

38

a < c < b

P (y, h) ∼ ycp0(hyc)

y−c

yc

y−b

ya

h

P (y, h)

∼ 1yc−b

Q(y, z)

z

Q(y, z) ∼ p0(z)

y−c

ya−c

FIG. 7: Scaling of P (y, h) and Q(y, z).

B. Complete scaling of P (y, h)

Let us conjecture that the scaling of Eq. (166) holds for all h < 0. This means that for h < 0 and large y,

P (y, h) ∼ yc diverges, while we know from Eq. (111) that for large h > 0, P (y, h) ∼ exp(−∆(y)/2), therefore it

remains finite for large y. Combining this information we expect that, on increasing h from −∞, P (y, h) increasesup to a value ≈ yc on a scale |h| ≈ y−c, it reaches a peak and then it decreases fast around h ≈ 0 to approach itsasymptotic limit exp(−∆(y)/2) ≈ 1 at h → ∞.

It is natural therefore to conjecture that the decrease from the peak down to values of order 1 happens aroundh ∼ 0 on another scale |h| ≈ y−b with b > c, which matches between the behavior at h < 0 and h > 0. We pose that

in this regime P (y, h ∼ 0) ∼ ya with a < c. In summary, we have

P (y, h) ∼

⎧⎪⎨⎪⎩

ycp0(hyc) for h ∼ −y−c

yap1(hyb) for |h| ∼ y−b

p2(h) for h ≫ y−b

(167)

with the condition a < c < b, and this scaling is illustrated in Fig. 7.Assuming this scaling, we can match the different regimes. Note first that obviously the scaling (167) requires

that p0(z = 0) = 0. We can assume that p0(z) ∼ |z|θ for small z. Then, to match with p1(z), we must assume thatp1(z → −∞) ∼ |z|θ too. Matching requires that yc|hyc|θ ∼ ya|hyb|θ, therefore c(1+θ) = a+bθ, which implies θ = c−a

b−c .

Similarly, in order to match with the regime of positive h, we must have p1(z → ∞) ∼ z−α, and ya(hyb)−α ∼ O(1),from which we obtain that a − bα = 0, hence α = a/b, and p2(h) ∼ h−α for h → 0. In summary, we obtain thefollowing scaling relations between exponents6:

α =a

bθ =

c − a

b − cκ = c + 1 . (168)

We will see later that the exponents α, θ, κ are directly related to the scaling of physical observables (the cage radiusand the pair correlation function).

Eq. (167) suggests to define the scaled variable z = hyc and the functions

H(yi, z) = yij(yi, zy−ci ) , ⇔ j(yi, h) =

1

yiH(yi, hyc

i ) ,

Q(yi, z) = y−ci P (yi, zy−c

i ) , ⇔ P (yi, h) = yci Q(yi, hyc

i ) .

(169)

6 The reader should not confuse the exponent α introduced here with the previously used matrix α.

FIG. 12. Scaling of P (y, h). Took from Fig. 7 of [23].

which implies

θ =c− ab− c α =

a

b(350)

These scaling features are reflected on the effective potential and thus the radial distribution function g(r). In thetail part one finds,

g(r) ∝ (r −D)−α r > D (351)

which was just a constant 1 at 1RSB level. And deeper into the contact region, by defining λ = hp/d = p(r −D)/Dwith p being the pressure one finds,

g(r)

g(D)= F∞(λ) F∞(λ) ∼ λ−2+θ λ� 1 (352)

The anomalous exponent θ was absent at the 1RSB level.Closer theoretical analysis revealed the following features by which the exponents were determined [23],

• analysis of p1(t): revealed a scaling relation

b =1 + c

2(353)

49

and that the exponent a is a function of c in order to have non-trivial scaling,

a = a(c) (354)

At this stage we are left with the exponent c undetermined.

• Exact identity which follows from the structure of the full RSB ansatz:

1 =ϕ

2

∫ ∞

−∞dhehP (y, h)[(γ(y)f(y, h))′′]2 (355)

With this the remaining uncertainty is removed.

The exponents were found to be

a = 0.29213... b = 0.70787... c = 0.41574... (356)

which implies

α = 0.41269... θ = 0.42311... κ = 1.41574... (357)

3. Complex responses to shear in the Gardner phase

It was also shown explicitly in [23] that the full RSB solution is marginally stable in the sense that the repliconeigenvalue remains λR = 0 throughout the Gardner’s phase. This is the same as the spinglass phases with full RSB,like the one in the SK model or the Gardner’s phase in the p-spin models. This means that under compression, theenergy landscape becomes progressively more complex [26]. Now it becomes very interesting to study consequencesof this in glassy systems. Recently some very suggestive evidences were reported [27, 28].

As an example, let us discuss response to shear [14] which is analogous to study response of a spinglass systemagainst external magnetic field. The replicated free-energy under shear can be written as

−βF ({γa}) =

∫dxρ(x)[1− log ρ(x)] +

1

2

∫dxdyρ(x)ρ(y)f{γa}(x, y) (358)

with the modified Mayer function

f{γa}(x, y) = −1 +

m∏

a=1

e−βv(|S(γa)(xa−ya)|) (359)

with

S(γ)µν = δµν + γδν,1δµ,2 (360)

Note that we are imposing different shear strain γa on different replicas. The aim is to disentangle the responses inthe hierarchical energy landscape along the idea discussed in sec. II F 6 and sec. II I.

One finds that in the d→∞ limit, the free-energy can be expressed exactly as the functional of the order parameterplus the shear strain γ,

−βmFm(∆, {γa})/N = 1− log ρ+ d logm+ d2 (m− 1) log(2πeD2/d2) + d

2 log det(αm,m)

− d2 ϕ∫

dλ√2πF(

∆ab + λ2

2 (γa − γb)2)

�1 �2...

FIG. 13. Replicas under shear

50

where F represents the interaction part of the free-energy. This can be used to study not only the linear response toshear but also non-linear responses including yielding [29]. In the following let us limit our selves to linear responsesfor the moment.

The replicated free-energy would be expanded in power series of the strain γa as

F ({γa})/N = F ({0})/N +

m∑

a=1

σaγa +1

2

1,m∑

a,b

µabγaγb + · · ·

from which it is possible to extract the shear-modulus matrix µab as,

βµab =d

2ϕ

δab

∑

c6=a

∂F∂∆ac

− (1− δab)∂F∂∆ab

(361)

It is interesting to note that the following sum rule holds∑

b

µab = 0

The physical meaning is that if we include all kinds of thermal fluctuations, including the α relaxations, i.e. transitionbetween different metabasins, we end up with a liquid so that the total shearmodulus is 0. However it does notnecessarily mean that the shearmodulus is 0 at all levels in the energy landscape.

It turned out that the shear-modulus associated with the energy landscape at level y, is given by,

βµ(y) ∝ pκ (362)

which follows from p ∝ 1/m,γ(y) ∝ y−c and the scaling relation κ = 1 + c.This means in particular that the rigidity of the inherent structures scales as,

βµEA =1

mγ(1/m)∝ p1+c=κ (363)

while that of metabasins scales differently, and much smaller,

βµ(y) =1

mγ(y)∝ p (364)

Recent numerical simulation of a 3 dim system of softspheres [28] have found an interesting evidence which isconsistent with the above result. Similarly to the difference of the field cooled susceptibility χFC and zero-field cooledsusceptibility χZFC studied in spinglass [30], zero-field compressed shear modulus µZFC and field compressed shear

modulus µFC was observed. It was found that µZFC ∝√P while µFC ∝ P .

C. Glass state following under compression and shear

The glass state following idea discussed in sec. II F 7 was explicitely formulated and analyzed in the hardspheresystem in d→∞ limit in [29] to study the evolution of glass states under shear and compression.

The Franz-Parisi potential under compression/shear is obtained as, Decomposition of the total free-energy of them+ s replica system,

−βFm+s[α] = −βFm[α] + s(−βVFP(γ)) +O(s2)

Franz-Parisi potential under compression/shear is obtained as,

−βVFP

N=d

2+d

2

∆g +m∆f

mD+d

2log(πD/d2) +

dϕg2

∫dy ey Θ

(y + ∆g/2√

2∆g

)m

×∫Dζ Dx log

Θ

(√∆γ(ζ)x+ ∆γ(ζ)/2 + y − η + D/2√

2D

) .

51

Here the volume fraction ϕslave of the slave system is parametrized as ϕslave = ϕgeη and γ is the shear-strain. We

also introduced ∆γ(ζ) = ∆f + ζ2γ2 , ∆f = 2∆r −∆g −∆ and Dx ≡∫dxe−x

2/2.

βP/ρ

d= −∂βVFP/(Nd)

∂η=ϕg2

∫dyDxDζ ey

Θ(y+∆g/2√

2∆g

)m

Θ(

ξ√2D

) e−ξ2

2D√2πD

βσ

d= −∂βVFP/(Nd)

∂γ= −γ ϕg

2

∫dyDxDζ ey

Θ(y+∆g/2√

2∆g

)m

Θ(

ξ√2D

) e−ξ2

2D√2πD

(1 +

x√∆γ(ζ)

)ζ2 .

0.0

0.2

0.4

0.6

0.8

1.0

3 4 5 6 7 8 9 10

Liquid EOS

= 8

= 7

= 6.667

= 6

= 5.5

= 5.25

= 5

= 4.9

d/p

ϕ

ϕG

ϕd

ϕg

ϕg

ϕg

ϕg

ϕg

ϕg

ϕg

ϕg

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0.0 0.1 0.2 0.3 0.4 0.5

2

3

4

5

6

0.0 0.1 0.2 0.3 0.4 0.5

βσ/d

γ

ϕg =7ϕg =6.5ϕg =6ϕg =5.5ϕg =5

p/d

γ

FIG. 14. Evolution of pressure and shear stress of the glassy states extracted by the state following computation Took from[29].

52

V. OUTLOOK

ACKNOWLEDGMENTS

This work was supported by KAKENHI (No. 25103005 “Fluctuation & Structure” and No. 50335337) from MEXT,Japan, by JPS Core-to-Core Program “Non-equilibrium dynamics of soft matter and informations”.

[1] S. F. Edwards and P. W. Anderson, Journal of Physics F: Metal Physics 5, 965 (1975).[2] G. Parisi, Physical Review Letters 43, 1754 (1979).[3] D. Sherrington and S. Kirkpatrick, Physical review letters 35, 1792 (1975).[4] B. Derrida, Physical Review Letters 45, 79 (1980).[5] D. J. Gross and M. Mezard, Nuclear Physics B 240, 431 (1984).[6] T. Castellani and A. Cavagna, Journal of Statistical Mechanics: Theory and Experiment 2005, P05012 (2005).[7] D. J. Thouless, P. W. Anderson, and R. G. Palmer, Philosophical Magazine 35, 593 (1977).[8] M. Mezard, G. Parisi, and M. A. Virasoo (1987).[9] R. Monasson, Physical review letters 75, 2847 (1995).

[10] M. Mezard and G. Parisi, The Journal of Chemical Physics 111, 1076 (1999).[11] M. Mezard and G. Parisi, Phys. Rev. Lett. 82, 747 (1999).[12] H. Yoshino and M. Mezard, Physical review letters 105, 015504 (2010).[13] H. Yoshino, The Journal of Chemical Physics 136, 214108 (2012).[14] H. Yoshino and F. Zamponi, Physical Review E 90, 022302 (2014).[15] S. Franz and G. Parisi, Journal de Physique I 5, 1401 (1995).[16] S. Franz and G. Parisi, Journal de Physique I 5, 1401 (1995).[17] J. De Almeida and D. J. Thouless, Journal of Physics A: Mathematical and General 11, 983 (1978).[18] E. Gardner, Nuclear Physics B 257, 747 (1985).[19] J.-P. Hansen and I. R. McDonald, Theory of simple liquids (Elsevier, 1990).[20] J.-P. Hansen and I. R. McDonald, Theory of simple liquids (Academic Press, London, 1986).[21] M. Mezard and G. Parisi, Journal de Physique I 1, 809 (1991).[22] G. Parisi and F. Zamponi, Rev. Mod. Phys. 82, 789 (2010).[23] P. Charbonneau, J. Kurchan, G. Parisi, P. Urbani, and F. Zamponi, Journal of Statistical Mechanics: Theory and Exper-

iment 2014, P10009 (2014).[24] L. Berthier, H. Jacquin, and F. Zamponi, Phys. Rev. E 84, 051103 (2011).[25] J. Kurchan, G. Parisi, P. Urbani, and F. Zamponi, J. Phys. Chem. B 117, 12979 (2013).[26] P. Charbonneau, J. Kurchan, G. Parisi, P. Urbani, and F. Zamponi, Nature Communications 5, 3725 (2014).[27] L. Berthier, P. Charbonneau, Y. Jin, G. Parisi, B. Seoane, and F. Zamponi, arXiv preprint arXiv:1511.04201 (2015).[28] D. Nakayama, H. Yoshino, and F. Zamponi, arXiv preprint arXiv:1512.06544 (2015).[29] C. Rainone, P. Urbani, H. Yoshino, and F. Zamponi, Phys. Rev. Lett. 114, 015701 (2015).[30] S. Nagata, P. Keesom, and H. Harrison, Physical Review B 19, 1633 (1979).[31] E. Bertin, Physical review letters 95, 170601 (2005).[32] R. Monasson, Journal of Physics A: Mathematical and General 31, 513 (1998).

53

Appendix A: Properties of low lying states in the glass phase of the random energy model

1. Distribution of energy gaps

Let Eg be the energy of the ground state among the 2N micro-states of a given realization of the REM. Then wecan write,

E2 = (Eg + E)2 = E2g + 2EgE + E2 E = E − Eg (A1)

In the following let us measure the energy in the shifted energy scale E which is 0 at the ground state. Then fromEq. (25) we find distribution of the shifted energy E of the low lying states can be written as

p(E) ' A(κ)eκE (A2)

with

κ = −2Eg/NJ2 = 1/(kBTc) ≡ βc (A3)

, which follows from Eq. (37) and A(κ) being a normalization constant.Suppose that M samples of random energies E (we omitˆin the following) are drawn from the distribution Eq. (A2)

in the range, say 0 < E < Emax so that A(κ) = κ/(eκEmax − 1) and put then order such that E1 < E2 . . . EM . Letus introduce the spacing between the adjacent energy levels yn = En+1 − En (1 ≤ n ≤ M − 1) and yM = EM . It isstraightforward to show that the distribution of the energy spacing becomes[31],

pM ({yn}) ≡M !AM (κ)

∫ Emax

0

dEMeκEM

∫ EM

0

dEM−1eκEM−1 · · ·

∫ E2

0

dE1eκE1δ(yM − EM )

M−1∏

n=1

δ[yn − (En+1 − En)]

= M(A(κ)/κ)MeMκyM

N−1∏

m=1

pm(ym) pm(ym) = mκe−mκym . (A4)

Here we see that the statistics of the energy spacing at different levels are independent from each other and that eachof them follows a level dependent exponential distribution.

2. Two level system

For simplicity let us consider a two level system: the ground state + 1st excited state with the energy gap ∆E;

Z = 1 + x x ≡ e−β∆E (A5)

From Eq. (??) and Eq. (A3) we find that the spacing between the ground state and the 1st exited state is given by,

p(∆E) = βceβc∆E (A6)

We assume that the self-overlap of the two states is 1 and the mutual overlap is 0. Then we find

P (q = 1) =1 + x2

(1 + x)2= 1− 2x

(1 + x)2. (A7)

so that

P (q = 0) = 1− P (q = 1) = βc

∫ ∞

0

d∆Ee−βc∆E2x

(1 + x)2=

T

Tc

∫ 1

0

dxx(T/Tc)−1 2x

(1 + x)2=

T

Tc+O(T/Tc)2 (A8)

So we see that this agrees with the result of the replica computation at the lowest order. Now it is possible to showthat using three-level model, we find the O(T/Tc)2 term actually vanishes!

54

Appendix B: Clustering property

Let us note the instantaneous value of the overlap between the replica 1 and 2 as,

q(σ1, σ2) =1

N

N∑

i=1

σ1i σ

2i (B1)

Let us consider the following object.

〈δ(q − q(σ1, σ2))〉α,α′ =

∫dλ

2πeiλ(q−qα,α′ )〈e−iλδq〉α,α′ δq ≡ q(σ1, σ2)− qα,α′ (B2)

where

δq =

N∑

i=1

δqi δqi = σ1i σ

2i − qα,α′ (B3)

〈e−iλδq〉 = 1− iλ〈δq〉α,α′ +λ2

2〈(δq)2〉α,α′ + . . . (B4)

where 〈δq〉α,α′ = 0 and

〈(δq)2〉α,α′ =1

N2

N∑

i=1

N∑

j=1

〈δqiδqj〉 (B5)

In equilibrium states the clustering property holds, which reads as,

〈δqiδqj〉 −−−−−−→|i−j|→∞〈qi〉〈qj〉 (B6)

which implies

N∑

i=1

〈δqiδqj〉 ∼ O(N0) (B7)

so that

〈(δq)2〉α,α′ ∼ O(N−1) (B8)

Thus

limN→∞

〈e−iλδq〉α,α′ = 1 (B9)

To sum up

〈δ(q − q(σ1, σ2))〉α,α′ =

∫dλ

2πeiλ(q−qα,α′ ) = δ(q − qα,α′) (B10)

Appendix C: p-spin spherical model

In the spherical model the spins are normalized such that

N∑

i=1

S2i = N (C1)

55

The disorder-averaged replicated partition function reads,

ZnJ =

∫ n∏

a=1

Ndλaa4πi

∫ ∏

a<b

dQabNdλab4πi

exp

N (βJ)2

4

∑

a,b

Qpab −1

2

∑

ab

λabQab

∫ n∏

a=1

dSa exp

1

2

∑

a,b

λa,bSaSb

N

=

∫ n∏

a=1

Ndλaa4πi

∫ ∏

a<b

dQabNdλab4πi

exp(−NG({Qab, λab})

)(C2)

The integral can be performed simply (see Eq. (D1)). Here we defined the diagonal part of the matrix Qab asqaa = 1 ∀a to enforce the spherical constraint. The effective action is obtained as,

G({Qab, λab}) = − (βJ)2

4

∑

ab

Qpab +1

2

∑

ab

λabQab −1

2ln det(−λab)−

n

2

√2π (C3)

The saddle point equations are obtained as the following.

0 =∂G

∂λab=

1

2Qab −

1

2λ−1ab → λab = Q−1

ab (C4)

Here we used the identity ∂detA/∂Aij = A−1ij . Another equation is,

0 =∂G

∂Qab= − (βJ)2

4Qp−1ab −

1

2Q−1ab . (C5)

1. RS ansatz

For the RS ansatz we assume Qab = (1 − q)δab + q. This can be easily inverted to find Q−1ab . (hint: assume

Q−1ab = 1/(1 − q)δab + A and find A.) It will be easy to verify that q = 0 is a solution, which describes the

paramagnetic phase.

2. 1RSB ansatz

For the 1RSB ansatz we assume Qab = [(1− q1)δab + q1 − q0]Imab + q0. Let us assume that q0 = 0. Then the matrixQab can be easily inverted as the RS case. Find the saddle point equation of q1. Considering the limit m → 1, itwill turn out to become identical to the self-consistent equation for the plateau value Cd obtained by the dynamicalmean-field theory (T > Td).

Appendix D: Useful formulae of Gaussian integrals

∫ M∏

i=1

dxi exp

−1

2

∑

i,j

xiKijxj

=

√2π

M/√

detK (D1)

1√

2πM/√

detK

∫ M∏

i=1

dxi exp

−1

2

∑

i,j

xiKijxj +∑

i

hixi

= exp

∑

ij

hiK−1ij hj

(D2)

56

Appendix E: ”Spin” density functional approach

Frollowing [32], let us introduce replicated “spin” density ρ(σ) as,

Nρ(σ) =

N∑

i=1

n∏

a=1

δσai ,σa (E1)

where σ = (σ1, σ2, . . . , σn). With this we can write

1

N

N∑

i=1

A(σ1i , σ

2i , . . . , σ

ni ) = Trσρ(σ)A(σ1, σ2, . . . , σn) (E2)

for any quantity A. We also note that the trance over the microscopic configurations of the replicated system as,

N∏

i=1

n∏

a=1

∑

σai =±1

=

(∏

σ

∫dρ(σ)

)N !∏

σ (Nρ(σ))!=

∫Dρ[σ]e−NTrσρ(σ) ln ρ(σ) (E3)

where we used the Stirling’s formula assuming N � 1. We also introduced a compact notation,

∫Dρ[σ] =

∏

σ

∫dρ(σ) (E4)

Now we can express the disorder averaged replicated partition of the p-spin Ising spinglass model as,

ZnJ =

N∏

i=1

n∏

a=1

∑

σai =±1

exp

β

∑

i<j<k...

Jijk...

n∑

a=1

σai σaj σ

ak . . .

=

N∏

i=1

n∏

a=1

∑

σai =±1

e

(βJ)2

4 Nn exp

(βJ)2

4N∑

a 6=b(Trσρ(σ)σaσb)p

=

∫Dρ(σ)e−NGn[ρ(σ)] (E5)

with the effective action,

Gn[ρ(σ)] = Trσρ(σ) ln ρ(σ)− (βJ)2

4n− (βJ)2

4

∑

a6=b(Trσρ(σ)σaσb)p (E6)

which consists of the entropic part (1st term) and interaction part (2nd term).The functional integral can be done by saddle point integration. Then the free-energy can be obtained as

βf = minimizeρ∗

∂nGn[ρ∗(σ)] (E7)

where ρ∗(σ) is the replica spin density which locally extremizes the action.

1. RS Gaussian ansatz

Let us try the following Gaussian ansatz:

ρ(σ) = C−1

∫dh√2π∆2

e−h2

2∆2 eβh∑na=1 σa = C−1e

(β∆)2

2

∑a,b σ

aσb (E8)

with the normalization factor defined as

C =

∫dh√2π∆2

e−h2

2∆2 (2 cosh(βh))n (E9)

57

such that Trσρ(σ) = 1.Then the entropic part can be easily evaluated as,

Trσρ(σ) ln ρ(σ) =(β∆)2

2

n+

∑

a6=b〈σaσb〉

− ln

∫dh√2π∆2

e−h2

2∆2 (2 cosh(βh))n (E10)

Then we need to evaluate

〈σaσb〉 = Trσρ(σ)σaσb =∑

σa=±1

∑

σb=±1

ρ(σa, σb)σaσb (E11)

where we introduced an effective 2-replica spin density

ρ(σa, σb) =

∏

c 6=a 6=b

∑

σc

ρ(σ). (E12)

In the RS ansatz we easily find the 2-replica spin density as,

ρ0(σa, σb) =

∫dh√2π∆2

e−h2

2∆2 (2 cosh(βh))n−2eβhσaeβhσb −−−→n→0

∫dh√2π∆2

e−h2

2∆2eβhσa

2 cosh(βh)

eβhσb

2 cosh(βh)(E13)

then we find

〈σaσb〉 =

∫dh√2π∆2

e−h2

2∆2 tanh2(βh) ≡ q (E14)

Then the whole action is evaluated as

Gn =(β∆)2

2(n+ n(n− 1)q)− ln

∫dh√2π∆2

e−h2

2∆2 (2 cosh(βh))n − (βJ)2

4n− (βJ)2

4n(n− 1)qp (E15)

thus

βf = ∂nGn =(β∆)2

2(1− q)−

∫dh√2π∆2

e−h2

2∆2 ln[2 cosh(βh)]− (βJ)2

4+

(βJ)2

4qp (E16)

By substituting (β∆)2 = λ, one can easily verify that the above results are identical to those exact results discussedin sec. II E.

2. 1RSB Gaussian ansatz

A natural extension of the RS Gaussian ansatz is the following, which corresponds to 1RSB,

ρ(σ) = C−1

∫dh0√2π∆2

0

e− h2

02∆2

0+βh0

∑na=1 σa

n/m∏

C=1

∫dh1,C√2π∆2

1

e−h2

1,C2∆2

1+βh1,C

∑a∈C σa

= C−1 exp

(β∆0)2

2

∑

a,b

σaσb +(β∆1)2

2

n/m∑

C=1

∑

a,b∈Cσaσb

(E17)

C =

∫dh0√2π∆2

0

e− h2

02∆2

0

{∫dh1√2π∆2

1

e− h2

12∆2

1 [2 cosh(β(h0 + h1))]m

}n/m(E18)

The entropic part is evaluated as

Trσρ(σ) ln ρ(σ) = − lnC +(β∆0)2

2

∑

a,b

〈σaσb〉+(β∆1)2

2

n/m∑

C=1

∑

a,b∈C〈σaσb〉

= − lnC +(β∆0)2 + (β∆1)2

2n+

(β∆0)2

2n(n−m)q0 +

(β∆0)2 + (β∆1)2

2n(m− 1)q1 (E19)

58

We now need the 2-replica spin density Eq. (E12) in the 1 RSB ansatz. We have two cases: in one case the tworeplicas a and b belong to the same group so that they join at level 1 in the hierarchy. In the other case they belongto different groups and join only at the level 0 of the hierarchy. It turns out that the 2-replica spin density dependson the level i of the hierarchy where the two replicas join. In the 1 RSB ansatz it is easy to find,

ρ0(σa, σb) =

∫Dz0

∫Dz1 coshm(Ξ) eΞσa

2 cosh(Ξ)∫Dz1 coshm(Ξ)

∫Dz1 coshm(Ξ) eΞσb

2 cosh(Ξ)∫Dz1 coshm(Ξ)

(E20)

ρ1(σa, σb) =

∫Dz0

∫Dz1 coshm(Ξ) eΞσa

2 cosh(Ξ)eΞσb

2 cosh(Ξ)∫Dz1 coshm(Ξ)

(E21)

where

Ξ = β∆0z0 + β∆1z1 (E22)

from these we find

q0 =∑

σa

∑

σb

ρ0(σa, σb)σaσb =

∫Dz0

(∫Dz1 coshm(Ξ) tanh(Ξ)∫

Dz1 coshm(Ξ))

)2

(E23)

q1 =∑

σa

∑

σb

ρ1(σa, σb)σaσb =

∫Dz0

∫Dz1 coshm(Ξ) tanh2(Ξ)∫

Dz1 coshm(Ξ)(E24)

The action is evaluated as

Gn = − lnC +(β∆0)2 + (β∆1)2

2n+

(β∆0)2

2n(n−m)q0 +

(β∆0)2 + (β∆1)2

2n(m− 1)q1

+(βJ)2

4n+

(βJ)2

4

[n(n−m)qp0 + n(m− 1)qp1

](E25)

and

βf = ∂nGn = −∫Dz0

1

mln

∫Dz1[2 cosh(β(∆0z0 + ∆1z1))]m

− (β∆0)2

2mq0 +

(β∆0)2 + (β∆1)2

2(m− 1)q1 −

(βJ)2

4− (βJ)2

4

[−mqp0 + (m− 1)qp1

](E26)

These results agree with the exact results Eq. (73), Eq. (74) and Eq. (71) by substitutions (β∆0)2 = λ0 and (β∆0)2 +(β∆1)2 = λ1.

3. kRSB Gaussian ansatz

It is straightforward to generalized the formulation to the k-RSB. We consider

ρ(σ) = C−1 exp

1

2

k∑

i=0

(β∆i)2∑

ab

Imiab σaσb

(E27)

where (we consider only zero external field cases)

C = g(m0, 0) (E28)

which is related to a family of functions g(m1, y) for i = 1, 2, . . . , k + 1 by the recursion formula

g(mi, y) =

∫Dzig(mi+1, y + β∆izi)

mi/mi+1 (E29)

with the boundary condition,

g(mk+1 = 1, y) = 2 cosh(y). (E30)

59

Note that the above recursion relation is identical to Eq. (148) obtained in the exact computation, with the identifi-cation,

(β∆)2i = λi − λi−1 (E31)

Using the above results we find the entropic term as,

1

nTrρ({σ})ρ(σ) ln ρ(σ) = − 1

nlnC +

1

n

k∑

j=0

(β∆j)2

2

∑

ab

Imiab 〈σaσb〉

−−−→n→0

− 1

m1

∫Dz0 ln g(m1,Ξ0) +

k∑

i=0

1

2(

i∑

j=0

(β∆j)2)(mi −mi+1)qi (E32)

while the interaction term becomes

1

n

∑

a 6=b(Trρ(bfσ)σaσb)

p =1

n

∑

a6=bQpab −−−→n→0

k∑

i=0

qpi (mi −mi+1) (E33)

Collecting the above results we find the k-RSB free-energy as

βfk−RSB = − (βJ)2

4− (βJ)2

4

k∑

i=0

qpi (mi −mi+1) +

k∑

i=0

1

2(

i∑

j=0

(β∆j)2)(mi −mi+1)qi −

1

m1

∫Dz0 ln g(m1,Ξ0) (E34)

which agrees with the exact result Eq. (152) using Eq. (E31) and Eq. (153).We still need to know how to evaluate 〈σaσb〉 In principle we need to repeat the procedure explained II H for this.

But given the Gaussian ansatz, we may proceed more intuitively (see [23] (sec V B) for the same type of argumentsfor hardspheres in large-d limit). Suppose that replica a and b meets first in the replica cluster at level l. Then theybelong to different subclusters at level l + 1. Then the hierarchical construction of the Gaussian ansatz for the spindensity function Eq. (E27) implies, for l = 0, 1, 2, . . . , k,

ql = Trσρ(σ)σaσb =

∫dhPl(h)〈σa〉h,l+1〈σb〉h,l+1 =

∫dhPl(h)(f ′(ml+1, h))2. (E35)

Here Pl(h) represents the distribution of effective fields h at level l which are applied to subclusters at level l+ 1. Wealso introduced a notation 〈. . .〉h,l+1 which represents the thermal average within a cluster at level l + 1 subjectedto ’external field’ h. The 3rd equation simply follows by observing that 〈σa〉h,l+1 = 〈σb〉h,l+1 = f ′(ml+1, h) wheref(ml+1, h) = (1/ml+1) ln g(ml+1, h) is the ’free-energy’/replica of the cluster at level l + 1.

The distribution of the internal field Pi(h) obeys the following recursion relation

Pi(z) = gmi/mi+1(mi+1, z)γ(β∆i)2 ⊗zPi−1(z)

g(mi, z)(E36)

for i = 1, 2, . . . , k with the ’boundary condition’,

P0(z) =e− z2

2(β∆0)2

√2π(β∆0)2

. (E37)

The meaning of the recursion formula could be given as follows. The 1st factor is the statistical weight of the clusterat level l where the two replicas a and b meet with given value of the ’external’ field z. The 2nd factor is the statisticalweight of the rest of the system at given value of the ’external’ field z.

Appendix F: Glass state following: p-spin Ising model

Let us consider the implementation of the glass state following scheme discussed in sec II F 7 in the p-spin Isingmodel. The structure of the parisi matrix may be took as shown in Fig. 15. Then we just need to slightly modify thecalculations in secII F as the following.

60

1

2

∑

a 6=bλabσ

aσb =λ0

2

∑

a6=bσaσb +

λ1r − λ0

2

∑

(a 6=b)∈refrence

σaσb +λ1s − λ0

2

∑

(a 6=b)∈slave

σaσb

= −m2λ1r −

s

2λ1s −

λ0

2

n∑

a=1

σa

2

− λ1r − λ0

2

n∑

a=1

σa

2

− λ1s − λ0

2

n∑

a=1

σa

2

(F1)

then we find

e12

∑a 6=b λabσ

aσb = e−m2 λ1r− s2λ1s〈e−

√λ0z0

∑na=1 σ

a〉z0〈e−√λ1r−λ0z1r

∑a∈reference σ

a〉z1r 〈e−√λ1r−λ0z1s

∑a∈slave σ

a〉z1s (F2)

with this we get

Tre12

∑a 6=b λabσ

aσb = e−m2 λ1r− s2λ1s

∫Dz0

[∫Dz1r[2 cosh(Ξr)]

m

∫Dz1r[2 cosh(Ξs)]

m

](F3)

with

Ξr =√λ0z0 +

√λ1r − λ0z1r Ξs =

√λ0z0 +

√λ1s − λ0z1s (F4)

∑

a 6=bQpab = m(m− 1)qp1r + s(s− 1)qp1s + 2msqp0

∑

a 6=bλabQab = m(m− 1)λ1rq1r + s(s− 1)λ1sq

p1s + 2msλ0q0 (F5)

Let us consider to make variation of the temperature of the slave system. We denote the temperature of thereference system as T0 and that of the slave system as T0/η. The effective action is obtained as,

Gm+s = − (β0J)2

4(m+ η2s)− (βJ)2

4{m(m− 1)qp1r + s(s− 1)qp1s + 2msqp0}

+1

2{m(m− 1)λ1rq1r + s(s− 1)λ1sq

p1s + 2msλ0q0}+

m

2λ1r +

s

2λ1s

− ln

∫Dz0

[∫Dz1r[2 cosh(Ξr)]

m

∫Dz1r[2 cosh(Ξs)]

m

](F6)

where β0 = 1/T0.In the s → 0 limit, which we consider, the reference system decouples with the slave system. Then we see that

the saddle point values of λ1r, q1r becomes identical to those of the standard 1RSB ansatz discussed in sec II F attemperature T0(= 1/β0).

Now the Franz-Parisi potential is obtained as,

βVFP = ∂sGm+s|s=0 = − (β0J)2

4η2 − (β0J)2

4

(−η2qp1s + η2mqp0

)+

1

2(−λ1sq1s + 2mλ0q0) +

1

2λ1s

−∫Dz0

∫Dz1r[cosh Ξr]

m∫Dz1s ln Ξs∫

Dz0

∫Dz1r[cosh Ξr]m

(F7)

The saddle point values of q0, q1s, λ0,λ1s must obey

0 =∂Gm+s

∂q0→ λ0 = ηp

(β0J)2

2qp−10 (F8)

0 =∂Gm+s

∂q1s→ λ1s = η2p

(β0J)2

2qp−11s (F9)

0 =∂Gm+s

∂λ0→ q0 =

∫Dz0

∫Dz1r[cosh Ξr]

m∫Dz1s tanh2 Ξs∫

Dz0

∫Dz1r[cosh Ξr]m

(F10)

0 =∂Gm+s

∂λ0→ q1s =

∫Dz0

∫Dz1r[cosh Ξr]

m tanh Ξr∫Dz1s tanh Ξs∫

Dz0

∫Dz1r[cosh Ξr]m

(F11)

In the limit η = 1, it is easy to see that q0 = q1s = q1r solves the equations as expected.

61

q1ss

s

n

n

q0m

m

q1r

q1r

q1sq0

FIG. 15. Simplest ansatz for the Parisi’s matrix for the m+ s system.

In particular, it will be interesting to examine the evolution of the internal energy of the slave system

eslave

J=

1

β0J

∂βVFP

∂η=β0J

2

(ηqp1s −mqp0 − η

)(F12)

In the limit η = 1, using q0 = q1s = q1r, this becomes identical that of the refrence system ( Eq. (77) with q0 = 0 andβ = β0) as it should.