Exact Solution of Ising Model on a Small-World Network

Transcription

1 1/2 Exact Solution of Ising Model on a Small-World Network J. Viana Lopes, Yu. G. Pogorelov, J.M.B. Lopes dos Santos CFP, Departamento de Física, Faculdade de Ciências, Universidade do Porto and Raul Toral IMEDEA, CSIC-UIB cond-mat/

2 2/2 Outline The model The solution Thermodynamics Finite size scaling behavior Discussion and conclusions

3 3/2 Small World Network Watts e Strogatz 1998 N ln(n) Shortest distance, averaged over pairs: O(N) for ordered networks O(lnN) for networks with pn shortcuts (p > 0)

4 4/2 Building the network Start with ring of N connected points (periodic bc). For each site, with a probability p, we make a connection with a randomly chosen site.

5 4/2 Building the network Start with ring of N connected points (periodic bc). For each site, with a probability p, we make a connection with a randomly chosen site.

6 4/2 Building the network Start with ring of N connected points (periodic bc). For each site, with a probability p, we make a connection with a randomly chosen site.

7 4/2 Building the network Start with ring of N connected points (periodic bc). For each site, with a probability p, we make a connection with a randomly chosen site.

8 4/2 Building the network Start with ring of N connected points (periodic bc). For each site, with a probability p, we make a connection with a randomly chosen site.

9 5/2 The model H = J N 1 i=0 σ i σ i+1 I σ i σ j = J i j σ i σ j (i j) S (i, j) Short range links: interaction strength J. Long range shortcuts: interaction strength I. S is the set of pairs with long shortcuts (#S = pn).

10 6/2 Previous results Simulations (Hong et. al 2002, Herrero 2002) T c finite for any p > 0 ; Mean-field critical behavior; Analytical results (Barratt and Weigt 2000, Gitterman 2000) Agreement over mean-field behavior; Gitterman predicts no order for p < 1/2. Bethe lattice representation (Dorogvtsev, Goltsev and Mendes, 2002)

11 7/2 Motivation If it is mean-field, does it have an analytical solution?

12 8/2 Solving... (please wait) Spin Bond transformation Z = Tr {σ} exp(β J i j σ i σ j ) = Tr {σ} exp(βj i j σ i σ j ) (i, j) (i, j) Identity (σ i = ±1) exp(βj i j σ i σ j ) cosh(βj i j ) = 1 + σ i σ j tanh(βj i j ) = ( ) bi tanh(βji j )σ i σ j j b i j =0,1

13 9/2... Z = ( ) coshβj i j (σ i σ j tanhβj i j ) b i j (i, j) b σ (i, j) Sum over σ k, for a given bond configuration: factor 2, if j b k j is even; 0, if j b k j is odd;

14 10/2... Z = Z 0 Tr {b} (tanh(βj i j )) b i j (i, j) Z 0 = 2 N (i, j) coshβj i j Tr : Trace restricted to the configurations of b s with j b k j even, for all k.

15 11/2 Example: Ising 1D Each site i has 2 links: surviving bond configurations must have b i 1,i = b i,i+1. Only two possible bond configurations contribute to Z: (a) : b i j = 0 i, j (b) : b i j = 1 i, j Z = 2 N cosh N (βj) ( 1 + tanh N (βj) ) (a) (b)

16 12/2 This model Restriction: at most 1 shortcut per site. b =0 12 b = b =1 b = Fix the b values of shortcuts and the b value of one short range link and all the b s are determined. 12 b = b 12=0 b =0 b =1 23 b = b =0 24 4

17 13/2 Partition function Z = Z 0 c N b I b 0 {b 1,...b pn } t L[b] J t M[b] I With c I = cosh(βi), t J = tanh(βj), t I = tanh(βi) and L[b] Number of short range links with b = 1; M[b] Number of shortcuts with b = 1. Z = Z 0 c N b I b 0 e lnω(m,l)+llnt J+M lnt I M,L

18 14/2 Number of states Two models solved A B

19 15/2 Number of states 0 N 1 l 1 l 2 l 3 l 4=2s l 5 Number of shortcuts, N b = pn Choose 0 2M 2N b points (filled). L/d = l 2 + l l 2M (l i, integer, d = 1/2p) (N L)/d = l 1 + l l 2M+1 C L/d 1 M 1 C (N L)/d M

20 16/2 Number of states Pick bonds, not sites! Normalizing factor: C 2pN 2M C pn M number of ways to pick 2M sites; number of ways to pick M bonds; Ω(M,L) = CpN M C 2pN 2M C 2p(N L) M C 2pL 1 M 1

21 17/2 Thermodynamical behaviour Intensive variables n = M/pN l = L/N lnω(m,l) + Llnt J + M lnt I = N (s(l,n) + l lnt J + pnlnt I ) + O(ln N) lim N + = steepest descent!

22 18/2 Maximization n l lnz = N lnz 0 + pn lnc I + pn f 1

23 19/2 Critical temperature (T c ) Maximum exits at a temperature given by (2t I + 1)t d J = 1 T c p=0.25 p=0.1 p=0.01 p=0.001 P F Log(I)

24 20/2 Asymptotic behaviour I/T c 1 I/T c 1 T c = T c = ( ln 2J ( ln 1 pln3 2J ) J pi ln(j/(pi)) )

25 21/2 Partition Function lnz = N lnz 0 + pn lnc I + pn f 1 f 1 = ln ( ) 4(1 l) 2 (1 n) (2 2l n) 2 T < T c 0 T > T c n T C T l T C T

26 22/2 Specific heat C C N (T)/C (T) (T/T c 1)N 1/ T

27 23/2 Finite size scaling (FSS) Does this solution give the FSS behaviour? 2 MC Simulation Sum Analytic 1.5 c ζ C N (T ) C (T) vs ζ tn 1/2

28 24/2 Analytical Result Z N = exp[ βn( f 0 + f a )]Z FSS Z FSS = N 2πp Z + 0 dl Z 2 0 dug(l,u)exp(pn h(l,u)) h(l,u) c 1 (l l ) 2 c 2 l(u u ) 2 Z F SS (ζ) πn 1/4 k 1 Z 0 e (x+k 2ζ) 2 dx x.

29 25/2 Bethe Lattice Approach Network is locally a Bethe lattice (loops O(logN)) σ 0 σ (L,M) N s (L,M) = e κ(x,t)(l +M), x L L + M κ min (x) 0 at T c!

30 26/2 Bethe Lattice Approach σ 0 σ(l,m) exp( L + M ξ 1D ) L + M lnn σ 0 σ(l,m) N 1/ξ 1D

31 27/2 Conclusions Mean-field ferromagnetic transition; Finite size scaling not governed by ξ (violation of hyperscaling); Lacking order parameter, χ, finite field; any way of having lcd < d e f f < ucd.