Sherrington-Kirkpatrick model near $T=T_c$: expanding around the Replica Symmetric Solution

0.25 MB·English

by  A. Crisanti

Sherrington-Kirkpatrick model near T = T :
c
0 expanding around the Replica Symmetric Solution
1
0
2
A Crisanti1 and C De Dominicis2

n
1 DipartimentodiFisica,Universit`adiRomaLa SapienzaandSMC,P.leAldo
a
Moro2,I-00185Roma,Italy.
J
2 Institut de Physique Th´eorique,CEA-Saclay-OrmedesMerisiers,91191Gif

8 surYvette, France
1
E-mail: [email protected], [email protected]

]
n
Abstract. An expansion for the free energy functional of the Sherrington-
n
Kirkpatrick(SK)model,aroundtheReplicaSymmetric(RS)SKsolutionQ(RS)=
- ab
s δab+q(1−δab)isinvestigated. Inparticular,whentheexpansionistruncatedto
di fourthorderin. Qab−Q(aRbS). TheFullReplicaSymmetryBroken(FRSB)solution
is explicitly found but it turns out to exist only in the range of temperature
.
t 0.549... ≤ T ≤ Tc = 1, not including T = 0. On the other hand an expansion
a
m aroundtheparamagneticsolutionQ(aPbM)=δab uptofourthorderyieldsaFRSB
solutionthatexistsinalimitedtemperaturerange0.915...≤T ≤Tc=1.
-
d
n
o PACSnumbers: 75.10.Nr,64.70.Pf
c
[

2 Submitted to: J. Phys. A: Math. Gen.
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Sherrington-Kirkpatrick model near T =T 2
c
1. Introduction:
The Sherrington-Kirkpatrick(SK) model is defined by the Hamiltonian [1]:
1,N
1
= J σ σ (1)
ij i j
H −2
i6=j
X
wheretheσ are 1IsingspinsandthecouplingsJ areindependentGaussianrandom
i ij
±
variables of zero mean and variance equal to 1/N.
The thermodynamic properties of the model are described by the free energy
(density) f averaged over the quenched disorder. To overcame the difficulties of
averaging a logarithm, the average over the disorder is computed using the so called
replica trick:
Zn 1
βNf = lim − (2)
− n→0 n
where β = 1/T is the inverse temperature and, as usual, ( ) denotes the average
···
over the disorder. For n integer Zn is the partition functions of n identical, non
interacting, replicas of the system. The average over disorder couples the different
replicas. Performing this average, and introducing the auxiliary symmetric replica
overlap matrix Q = 1 σ σ , with a = b, the disorder averaged replicated
ab N i ia ib 6
partition functions can be written as [1]:
P
Nβ2
Zn = dQ eNL[Q] (3)
ab
2π
Z a<br
Y
with the effective Lagrangian(density):
β2 β2
[Q] = Q2 +Ω[Q] n (4)
L − 4 ab − 4
ab
X
β2
Ω[Q]= lnTr exp Q σ σ (5)
σa 2 ab a b
(cid:18) ab (cid:19)
X
The last term in (4) follows from the definition Q =1. The normalizationfactor in
aa
(3) gives a sub-leading contributions for N and is omitted in the following.
→∞
In the thermodynamic limit, N , the value of the integral in (3) is given by
→∞
the stationary point value, and the replica free energy density reads:
nβf = [Q] (6)
− L
with Q evaluated from the stationary condition
ab
∂
[Q]=0, a<b (7)
∂Q L
ab
that is from the self-consistent equation
Trσσaσb exp β22 abQabσaσb
Q = = σ σ , a=b. (8)
ab Trσexp β2(cid:16)2 aPbQabσaσb (cid:17) h a bi 6
(cid:16) (cid:17)
To solve the self-consistentstationarypPointequationwe haveto specify the structure
of the matrix Q . This is not straightforward since the symmetry of the replicated
ab
partition function under replica permutation is broken in the low temperature phase.
Sherrington-Kirkpatrick model near T =T 3
c
TheReplicaSymmetric(RS)AnsatzQ =δ +q(1 δ )ofSherrington-Kirkpatrick
ab ab ab
−
[1],thatassumesthesameoverlapforanypairofreplicas,indeedyieldsanunphysical
negative entropy at zero temperature. Following the parameterization introduced
by Parisi [2, 3], the overlap matrix Q for R breaking in the replica permutation
ab
symmetry is divided into successive boxes of decreasing size p , with p = n and
r 0
p =1,alongthediagonal,andtheelementsQ oftheoverlapmatrixareassigned
R+1 ab
so that
Q q =Q , r =0, ,R+1 (9)
ab a∩b=r r
≡ ···
with 1 = Q Q Q > Q . The notation a b = r means that a and b
R+1 R 1 0
≥ ≥ ··· ≥ ∩
belongtothesameboxofsizep buttotwodistinctboxesofsizep <p . Thecase
r r+1 r
R = 0 gives back the RS solution, while the opposit limit R describes a state
→ ∞
with an infinite, continuum, number of possible spontaneous breaking of the replica
permutation symmetry. It turns out that a physical solution is obtained only in the
lattercase. Using this structureforQ ,Parisiandothers[2,3,4]haveshownhowto
ab
obtain solutions with R steps of replica symmetry breaking (RSB) and in particular
withR (FRSB),andhowtoconstructequationssatisfiedbyQ(x),thecontinuous
→∞
limit of the order parameter Q for R [5]. These equations can be solved in
ab
→ ∞
the full low temperature phase [6, 7, 8, 9, 10, 11]. However working directly with
Q(x) makes it difficult to keep track, for instance, of the Hessian, and hence of the
stabilityofthe solution,sincethe matrixstructureofthe overlapmatrixQ islostin
ab
the continuous limit. The study of the Hessian of the fluctuations around the RSB
‡
solution with an arbitrary R from the Lagrangean (4)-(5) is a very hard task. As a
resultstabilityanalysishasbeenmostlyinvestigatednearthecriticaltemperatureand
withthehelpofasimplifiedmodel[12,13],thesocalledTruncatedModel[2,14],that
similarly to the Landau Lagrangian retains only the main mathematical structure of
the expansionofthe replicatedfreeenergyinpowersofQ nearT ,where Q 1.
ab c ab
| |≪
In the present work, we take a different viewpoint and consider the expansion
of the Lagrangean (4)-(5) around the Replica Symmetric ansatz of Sherrington and
Kirkpatrick. The main motivation for such an expansion is to obtain a simpler
Lagrangean which, while retaining the replica symmetry breaking properties of the
original model, is a priori valid in the whole low temperature phase. Anticipating
our conclusions, we find that the model obtained by truncating the expansion to the
fourth order, the minimum order required to have a FRSB solution, while improving
theresultsobtainedfromtheexpansionsnearT isvalidinatemperaturerangewhich
c
does not reach zero temperature.
Theoutlineofthepaperisasfollows: inSection2weconstructtheapproximation
of Ω[Q] obtained expanding it around the Replica Symmetric SK solution Q(RS) = q
ab
(a = b) up to fourth order in Q q. The stationarity equation and its solutions
ab
6 −
are discussed in Section 3. The Truncated Model was obtained considering the
main features of the mathematical structure of the expansion of Ω[Q] around the
paramagnetic solution Q(PM) = 0 (a = b) to fourth order in Q . The parameters
ab 6 ab
entering in the model are, however, usually arbitrary and so it is difficult to make
contactwiththeoriginalSKmodel. ByusingtheresultsofSection2wecandetermine
the coefficientsoftheexpansionandstudythepropertiesofthesolution. Thisisdone
in Section 4. Discussion and conclusions are deferred to Section 5.
‡ Paradoxically,itisthiscontinuous limitR→∞,thatimposestheexistenceofzeromodes(atthe
bottom of the replicon bands). Indeed, this limitis necessary to transform the replicapermutation
invarianceintoa(broken) continuous groupthusgeneratingGoldstonezeromodes.
Sherrington-Kirkpatrick model near T =T 4
c
2. Expansion of the free energy functional around the SK solution:
To expandthe functionalΩ[Q]aroundthe SK solutionQ =q for a=b,we consider
ab
6
an overlapmatrix Q of the form
ab
Q =δ +q(1 δ )+q (10)
ab ab ab ab
−
where q is given by the SK Replica Symmetric solution (see below) and q the
ab
deviation from the Replica Symmetric solution. Inserting this form of Q into the
ab
free energy functional (6) yields:
β2 β2 β2 β2
nβf =n q2 n q q q q2
− 4 − 2 − 2 ab− 4 ab
ab ab
X X
β2 2 β2 1,n
+lnTrσ exp q σa + qabσaσb +O((n121))
 2 2 
!
ab ab
X X
Setting q =0 the above expressionleads to the Sherrington-Kirkpatrickfree energy
ab
β2 β2
βf = q2 q+lncosh(βz)+ln2+O(n) (12)
SK
− 4 − 2
where the overbar denotes the averageover the Gaussian variable z:
g(z)= +∞ dz e−z2/2qg(z). (13)
√2πq
Z−∞
Stationarity of f with respect to q leads to SK Replica Symmetric solution:
SK
q =θ2, θ tanh(βz). (14)
≡
For q =0 the free energy functional f can be written, expanding the last term
ab
6
in (11) in powers of q , as:
ab
β2 β2
nβf = nβf q q q2
− − SK− 2 ab− 4 ab
ab ab
X X
1 β2 k k
+ q σ σ (15)
ab a b
k! 2
Xk≥1 (cid:18) (cid:19) * Xab ! +c
where the subscript “c” indicates that only connected contributions, i.e., only those
terms that cannot be written as the product of two or more independent sums, must
be considered. The angular brackets denote the average
n
g(σ) = eβzσag(σ)+O(n). (16)
h i
a=1
Y
Since σ2 = 1, the last term in (15) contains only averages of products of spins with
a
different replica index. These are easily evaluated yielding
n h
σ σ = eβzσa σ
h a1··· ahi l
a=1 l=1
Y Y
=[2cosh(βz)]n−h[2sinh(βz)]h+O(n)
=θh+O(n), a = =a . (17)
1 h
6 ···6
Form the study of the truncated model it is known that terms of order O(q4 ) must
ab
be included into the free energyto break the replica symmetry. Thus in the following
we shall consider the first four terms of the expansion.
Sherrington-Kirkpatrick model near T =T 5
c
2.0.1. Term O(q ): The term of order O(q ) is
ab ab
q σ σ = q σ σ =θ2 q (18)
ab a b ab a b ab
* + h i
ab ab ab
X X X
Thechoiceq =θ2,see(14),cancelsthelinearterminthe expansion(15)andremoves
the tad-poles.
2.0.2. Terms O(q2 ): The term of order O(q2 ) reads
ab ab
2
q σ σ = q q σ σ σ σ (19)
ab a b ab cd a b c d
* ! + h i
ab ab
X Xcd
To evaluate this term we have to find all different possible ways of equating the ab
indexes to cd indexes with the constraint, imposed by q =0, that a=b and c=d.
aa
6 6
There are three possible cases: all indexes different, a pair of equal indexes, and two
pairsofequalindexes. Bynoticingthatthe spinproductaveragesdependonlyonthe
number of different indexes, and not on the value of the indexes, and that the matrix
q is symmetric, these yield
ab
2
′ ′ ′
q σ σ =θ4 q q +4θ2 q q +2 q2 , (20)
ab a b ab cd ac cb ab
* ! +
ab abcd abc ab
X X X X
since there are 4 possible ways of equating one index in ab with one index in cd and
2 was of equating the pair of indexes ab to the pair cd. All sums are restricted to
different indexes, this is denoted by the prime “′” over the sum sign. Transforming
the restricted sums into unrestricted ones, i.e., sums over free index, one finally ends
up with:
2
q σ σ =θ4 q q +4θ2(1 θ2) q q
ab a b ab cd ac cb
* ! + −
ab abcd abc
X X X
+2(1 θ2)2 q2 (21)
− ab
ab
X
This equation has a simple diagrammatic expression. Indeed denoting q by a line
ab
and the vertex where two (or more) indexes are equal by a “dot”, the above equation
can be written as
2
q σ σ =θ4 +4θ2(1 θ2)
ab a b
* ! + −
ab
X
+2(1 θ2)2 (22)
−
More details can be found in Appendix B. From this form we easily see that the first
term is a disconnected contribution and hence it does not appears in the free energy
(15), therefore to order O(q2 ) the free energy reads
ab
β4 β4
nβf = nβf + M q q + N q2 +O(n2,q3 ) (23)
− − SK 4 ac cb 4 ab ab
abc ab
X X
where
M =2θ2(1 θ2), N =(1 θ2)2 T2. (24)
− − −
Notice that the coefficient N is (minus) the Replicon eigenvalue of the Replica
Symmetric solution [15]. The q =0 solution is hence unstable below T =1.
ab
Sherrington-Kirkpatrick model near T =T 6
c
2.0.3. Terms O(q3 ) and O(q4 ): These are evaluated as done for the O(q2 ) by
ab ab ab
computing all connected contributions that follows from the expansion of the k = 3
and k = 4 terms in (15). By using a self-explanatory diagrammatic representation
these are given by:
3
q σ σ =P +Q +R +J +K (25)
ab a b
* ! +
Xab c
where
P =24θ2(1 θ2)2, Q= 16θ4(1 θ2), R= 48θ2(1 θ2)2, (26)
− − − − −
J =16θ2(1 θ2)2, K =8(1 θ2)3 (27)
− −
and
4
q σ σ = A +B B
ab a b
* ! + − −
Xab c
+C C +4D
−
3D +E 2E
− −
+F +G H (28)
−
with
A=32θ4(1 3θ2)(1 θ2), B =384θ4(1 θ2)2, C =384θ2(1 θ2)3, (29)
− − − −
D =64θ2(1 3θ2)(1 θ2)2, E =192θ2(1 θ2)2, F =48(1 θ2)4, (30)
− − − −
G=32(1 3θ2)2(1 θ2)2, H =96(1 3θ2)(1 θ2)3. (31)
− − − −
CollectingallcontributionsuptoorderO(q4 ),thereplicafreeenergyfunctionalreads:
ab
1 1
nβf = nβf + M q q +N q2 + P q q q
− − SK 4T4 " ac cb ab# 6(2T2)3 " ac cd db
abc ab abcd
X X X
+Q q q q +R q2 q +J q3 +K q q q
ad bd cd ac cb ab ac cb ba
#
abcd abc ab abc
X X X X
1
+ A q q q q +B q q2 q (32)
24(2T2)4 "− ae be ce de ac cd db
abcde abcd
X X
B q q q q +C q q2 q C q q q q
− ac dc ce eb ac cb ba− ac ad dc cb
abcde abc abcd
X X X
+4D q3 q 3D q2 q q +E q q q q
ac cb− ab bc bd ab bc cd de
abc abcd abcde
X X X
2E q q q2 +F q q q q +G q4 H q2 q2
− ab bc cd ab bc cd da ab− ac cb#
abcd abcd ab abc
X X X X
+O(n2,q5 )
ab
Sherrington-Kirkpatrick model near T =T 7
c
3. Stationarity equation:
The equation for q follows from the stationarity condition (∂/∂q )f =0 applied to
ab ab
the replica free energy functional (32). In the limit R this yields
→∞
1 1
[MS +Nq(x)]+ 3(P +Q)S2+R S +2S q(x) +3Jq(x)2
2T4 1 6(2T2)3 1 2 1
"
(cid:16) (cid:17)
x 1
+6K dyq˙(y)q(y)+S q(0) + 4AS3
(cid:18)Z0 1 (cid:19)# 24(2T2)4"− 1
+B 2S S 4S3+b2S2q(x) +C∆(x)
1 2− 1 1
(cid:16) (cid:17)
+D 4S 6S S +12S q(x)2 6S2q(x)
3− 1 2 1 − 1
(cid:16) (cid:17)
+E 4S3 4S S 4S2q(x)
1 − 1 2− 1
(cid:16) x (cid:17)
+12F dyq˙(y)q(y)2+S2q(0)
1
(cid:18)Z0 (cid:19)
+4Gq(x)3 4HS qb(x) =0, 0 x x , (33)
2 c
− # ≤ ≤
where
x d
∆(x)=2 dy q(y)2q(y)+q2(y)q˙(y) +S q(0)2+S q(0)
1 2
dy
(cid:20)Z0 (cid:18) (cid:19) (cid:21)
x
b
+ 4q(x) 6bS dyq˙(y)q(y)+S q(0)
1 1
−
(cid:16) x (cid:17)(cid:20)Z0 (cid:21)
3 dyq(y)q(y)2 3S2q(0b)+q(x)3 (34)
− − 1
Z0
and
b
1 xc
S = dxq(x)n = dxq(x)n (1 x )q(x )n (35)
n c c
− − − −
Z0 Z0
The “dot” indicates the derivative, q˙(x) = (d/dx)q(x), while the “hat” the Replica
Fourier Transform (RFT), that for R reads [17]:
→∞ §
xc d
q(x)= dyy q(y) q(x ), RFT (36)
c
dy −
Zx
b x 1 d
q(x)= dy q(y)+q(0) inverse RFT (37)
− ydy
Z0
where q(0) = q(x = 0), and we habve neglected the surface term at x = 1 since
q(x=1)=q =0.
aa
3.1. Solution of the Stationarity equation
The complicate integro-differential stationarity equation (33) can be solved reducing
it to an ordinary differential equations using the differential operator O =
§ TheRFTwasfirstintroduced,directlyinthecontinuumlimit(R→∞)byMezardandParisi[18]
b
Sherrington-Kirkpatrick model near T =T 8
c
(1/q˙(x))(d/dx) to eliminate integrals. Application of O to (33) leads to
N 1 1
2T2 + 3(2T2)3 RS1+3Jq(x)+3Kq(x) + 12(2T2)4bBS12
(cid:20) (cid:21) (cid:20)
x
+C 2 dyq˙(y)q(yb)+q2(x)+4q(x)q(x) 3S q(x)
1
−
(cid:18) Z0
+2S q(0) +D 1b2S q(x)b 3S2 2EbS2+6Fq(bx)2
1 1 − 1 − 1
(cid:19)
(cid:0) (cid:1)
+6Gq(x)2 2HS =0. b (38)
2
−
(cid:21)
The equation is not yet simple enough to be solved. A second application of O, and
a rearrangementof terms, yields
8T2X(x)+Y(x)q(x)+U(x)q(x)+Z(x)S =0 b (39)
1
where
b
X(x)=J Kx, Y(x)=2C 4Fx, U(x)=4G 2Cx, Z(x)=4D+Cx (40)
− − −
Theintegralequation(39)cannowbetransformedintoadifferentialequationdividing
it by Y(x) and taking the derivative with respect to x. This leads to the first order
differential equation
Y(x)[U(x) Y(x)x]q˙(x)+µq(x)+8T2λ+νS =0 (41)
1
−
with coefficients
λ=X˙Y XY˙ = 2CK+4FJ (42)
− −
µ=U˙Y UY˙ = 4C2+16FG (43)
− −
ν =Z˙Y ZY˙ =2C2+16DF. (44)
−
The solution of equation (41) reads:
x s
q(x)=Γ − a bS , 0 x x , (45)
1 c
(x s)2+∆ − − ≤ ≤
−
where p
λ ν C G
a=8T2 b= s= ∆= s2, (46)
µ µ 2F F −
and we have absorbed a factor µ into the definition of the integration constant Γ.
The quantity S is function of Γ and x (and temperature), see (35). Introducing the
1 c
auxiliary function
z x s z s
h(z)= dx − +(1 z) −
(x s)2+∆ − (z s)2+∆
Z0 − −
(z s)(1 s)+∆
= − p− s2+∆ p (47)
(z s)2+∆ −
− p
this reads p
Γh(x ) a
c
S = − (48)
1
b 1
−
The value of Γ, and x , is determined from equations (38) and (39). Replacing
c
in equation(39) q(x) with the expression(45) yields alinear equationfor Γ. This can
Sherrington-Kirkpatrick model near T =T 9
c
be readily solved noticing that since Γ does not depend on x we can just set x = 0
and use the identity q(0)=S . This leads to
1
Γ
0
Γ= (49)
bΓ +Γ h(x )
1 2 c
where
Γ =4T2J(b 1)+a(2G C 2D),
0
− − −
s
Γ =2G(b 1) , (50)
1
− √s2+∆
Γ =2Gb C 2D.
2
− −
Finally the value of x , for a given the temperature T, is determined from (38).
c
Again we can take advantage of the fact that x does not depend to x and choose
c
in (38) a suitable value for x, e.g., x = x or x = 0. Setting x = 0 into (38) a
c
straightforwardalgebra leads to the equation
1
2N + 3Jq(0)+(R+3K)S
6T2 1
(cid:2) 1 (cid:3)
+ 6Gq(0)2+(6C+12D)S q(0)+
48T4 1
(B 3C(cid:2) 3D 3E+6F)S2+(C 2H)S =0 (51)
− − − 1 − 2
where (cid:3)
∆ b(b 2)
S = Γ2 (1 x ) +I (x ) I (0)+1 − h(x )2
2 − − − c (x s)2+∆ 2 c − 2 − (b 1)2 c
(cid:20) c− − (cid:21)
2
a a
+2 Γh(x ) (52)
(b 1)2 c − b 1
− (cid:18) − (cid:19)
and
∆ x s
I (x)= dx = √∆ tan−1 − , ∆>0 (53)
2 − (x s)2+∆ − √∆
Z − (cid:18) (cid:19)
Solvingequation(51)forx atfixedT yields the value ofx (T),thatsubstituted
c c
back gives the solution q(x) as function of temperature. In figures 1 and 2 we show
the solutions Q(x)=q+q(x) for two different temperatures.
From figures one clearly sees that Q(x = 0) = 0. It grows as the temperature
6
decreases, and overcomes q for T < 0.618..., see also figure 3. Retaining in the
expansionofΩ[Q]onlytermsuptoorderO(q4 )breaksthereplicasymmetry,however,
ab
this approximation is not good enough to change the SK result Q(x = 0)= q = 0 to
6
the expected one Q(x=0)=0. To recover the latter one has to add more terms in
k
the expansion, probably all terms.
Below temperature T =0.549...equation(38) ceasesto have a physicalsolution
and only the SK solution Q(x) = q survives. In figure 3 we show the values of Q(0),
Q(x ) and x as function of temperature.
c c
k Q(x=0)mustvanishinabsenceofexternalfieldsthatbreaktheup/downsymmetry. Forinstance
intheq≥4PottsmodelthesymmetryisbrokenandindeedQ(0)6=0
Sherrington-Kirkpatrick model near T =T 10
c
0.12
Q(x)
0.08
0.04
0
0 0.2 0.4 0.6 0.8 1
x
Figure 1. Q(x) versus x at temperature T = 0.9. The horizontal dashed line
showstheSKsolutionQ(x)=q. Forthistemperaturewehavexc=0.168846...,
Q(xc)=0.109238...Q(0)=0.013570...andq=0.102701....
0.5
0.4
0.3
Q(x)
0.2
0.1
0
0 0.2 0.4 0.6 0.8 1
x
Figure 2. Q(x) versus x at temperature T = 0.7. The full line is the result
from the expansion around T =1 to order O(τ13), whilethe circleare obtained
from the numerical solution of equation (38). The horizontal dashed line shows
the SK solution Q(x) = q. For this temperature we have xc = 0.3920(6),
Q(xc)=0.3879(1)...Q(0)=0.2232(5)...andq=0.3166(5)....