1 Cluster expansion and tree graph inequality - CiteSeerX

A Remark on high temperature polymer expansion for lattice systems with in nite range pair interactions. Aldo Procacci and Bernardo N. B. de Lima Dep. Matematica-ICEx, UFMG, CP 702, Belo Horizonte MG 30.161-970, Brazil Benedetto Scoppola Dip. di Matematica, Universita' \La Sapienza" di Roma, P.zza A. Moro 5, 00185, Roma, Italy December 2, 1998

Abstract

We consider the standard high temperature-small activity polymer expansion for lattice spin systems (as presented in [6]) and we show how in many cases using a tree graph equality ([4],[2],[3]) various classical results ([8],[13],[7],[14],[11]) may be in some cases improved, or at least proved in a much simpler way.

1 Cluster expansion and tree graph inequality 1.1 Introduction

The high temperature expansion for lattice system is one of the rst subjects investigated carefully in rigorous statistical mechanics. Althought many of its results are considered \classical", in the sense that they are known since a very long time, it is quite hard to nd in the literature proofs of them reasonably simple and/or optimal In this paper, using the so-called tree graph inequalities, introduced by various authors ([4],[2],[3]) in this contest but then widely used most of all in constructive eld theory, we show that the pressure or the free energy of lattice systems interacting via an in nite range pair potential can be written as an absolutely convergent series if the potential is absolutely summable; a new and simpler proof of the Rota inequality is also given.

1.2 Basic notations

Lattice systems. In general, if X is any nite set, we denote by jX j the number of elements of X . A lattice system is de ned in the following way. Let  be a subset in Z d , the cubic lattice in d dimensions (tipically  is a cube and jj is the number of sites in ); in each site x 2  we place a random variable x , taking values in some probability space ( x ; x ). We assume that

1

( x ; x ) is the same for all x and we denote it ( ; ). We denote, for any nite subset X  ,

X = x2X x and Y Y (1.1) d() = d(x ); d(X ) = d(x ) x2X

x2

We de ne a many-body potential as X :jX j2 (X; ), where (X; ) is a real valued measurable function in X (X is interpreted as a generalized link). Once we give explicitly ( ; ) and the many body potential (chosen in a suitable Banach space, see e.g. [12]), then a lattice system is de ned and its partition function is: P

Z

 ( ) = d( ) expf?

X

X :jX j2

(X; )g

(1.2)

The pressure (or free energy, depending on the system we are considering) of the lattice system is the limit for  % 1 (e.g. as a cube of increasing size) of the following function (1.3) p( ; ) = j1 j log ( ) Connected graphs and trees. Given a nite set A, we de ne a graph g in A as a collection of subsets g = f1 ; 2 ; : : : ; m g with 0  m  jAj(jAj ? 1)=2, such that: i  A; i 6= j for all i; j ; ji j = 2; moreover the graph will be connected if for any pair B; C of subsets of A such that B [ C = A and B \ C = ;, there is a i 2 g such that i \ B 6= ; and i \ C 6= ;. Given a graph g = f1 ; 2 ; : : : ; m g in a set A, 1 ; 2 ; : : : ; m are called links of the graph g and the elements of A are called vertices of g. We denote with GA the set of all connected graphs of A. A connected tree graph  (or simply tree graph) in A is a connected graph in A with jAj ? 1 links. We denote by TA the set of all tree graphs in A. Given a tree graph  in A, the incidence number of the vertex a 2 A, denoted by da , is the number of links i in  such that a 2 i . The number of trees in TA is explicitely computable using the Cayley formulas. Without lost in generality let us consider the case in which A = N = f1; 2; : : : ; ng (hereafter we denote by N the set of the rst n positive integer), then X (1.4) 1 = Qn(n(?d 2)! i=1 i ? 1)!  2TN d1 ;d2 ;:::;dn xed

Observing that, by de nition, 1  di  n for all i, and d1 + d2 + : : : + dn = 2n ? 2, it is easy to obtain from (1.4) that X 1 = nn?2 (1.5)  2TN

1.3 The polymer expansion for lattice systems and the tree graph equality

The high temperature expansion developed in [6] (see also [9]) applies to all lattice systems de ned above. ThePpartition function (1.2) can be rewritten (via a Mayer expansion, i.e. writing Q the factor expf? X (X; )g in (1.2) as X [(expf? (X; )g ? 1) + 1] and then expanding the product, see [6]) as X X (R1 )(R2 ) : : : (Rk )e?U (R1 ;:::;Rk ) (1.6)  ( ) = 1 + k1! R ;:::;R 2  k1 1 k jRi j2 (i=1;2;:::;k)

2

where the distinct subsets Ri , (i = 1; :::; k), with jRi j  2, are identi ed as polymers; the interaction between polymers U (R1 ; : : : ; Rk ) is 0 when k = 1, and for k  2 X

U (R1 ; : : : ; Rk ) =

1i 1, the polymer activity de ned in (1.9) satis es the following condition  n X j(R)j  61 (1.12) sup x2 R:jRj=n g2GN

Rx

then, uniformely in , the series (1.10) is absolutely convergent An indirect proof of (1.12), which make use of the Kirkwood-Salsburg equations, is known since a long time (see e.g. [8] and references therein, and [9])). In [6] a \direct" proof is given in the sense that the explicit expression of the terms of the free energy series and an explicit bound for each of such terms are provided (see also [13],[7],[14]). Inequality (1.12) is a condition on allowed many body potentials  and on the temperature (remark that (R) goes to zero as goes to zero).

3

In the proof of theorem 1 the following combinatorial inequality is used:

j(f )j = j

X

gf g2GN

(?1)jgj j  N (f )

(1.13)

where f 2 GN is a connected graph in N , g 2 GN is a connected subgraph of f , jgj denotes the number of links of g and N (f ) is the number of tree graphs contained in f . The number (f ) P is the Ursell coecent of the polymer gas Gibbs factor expf? 1i 1 V (X ) = X V . We denote now by X1 = f1g, X2 ; X3 ; : : : Xl (l < n) a sequence of subsets of N such that X1  X2  : : :  Xl and jXij = i (1  i  l) and de ne for any X  N such that Xl  X

WX (X1 ; X2 ; : : : ; Xl ; t1 ; t2 ; : : : tl ) = where Observe that

X

X

t1()t2 () : : : tl ()V

(A.2)

ti() = ti 2 [0; 1] if  crosses Xi 

1

otherwise

WX (X1 : : : ; Xl ; t1 ; : : : tl?1 ; tl = 0) = WXl (X1 ; : : : ; Xl?1 ; t1 ; : : : ; tl?1 ) + V (X nXl ) WX (X1 ; : : : ; Xl ; t1 ; : : : ; tl?1 ; tl = 1) = WX (X1 ; : : : ; Xl?1 ; t1 ; : : : tl?1 ) We now give an iteration procedure to rewrite e?V (N ) . Consider thus 2

WN (X1 ; t1 ) = 4

3 X

crossesX1

11

t1 V5 + V (NnX1 )

(A.3) (A.4)

then we can rewrite, by (A.4) 1

Z

e?V (N ) =

0

=

dt1 @t@ e?WN (X1 ;t1 ) + e?WN (X1 ;t1=0) 1

X

1 crossesX1

(?V )

1

Z

(A.5)

dt1 e?WN (X1 ;t1 ) + e?WN (X1 ;t1 =0)

0

(A.6)

Now, for each 1 in the sum above, take X2 = X1 [ 1 and

WN (X1 ; X2 ; t1 ; t2 ) =

X

2 crosses X2

t1 (2 )t2 (2 )V + V (NnX2 )

Thus, again using that WN (X1 ; X2 ; t1 ; t2 = 1) = WN (X1 ; t1 )

e?V (N ) = + = +

1

Z

Z

1

 @ ? W ( X ;X ; t ;t ) 1 2 1 2 N dt2 @t e + 2

(?V1 ) dt1 0 0 1 crosses X1 o e?WN (X1 ;X2;t1 ;t2 =0) + e?WN (X1 ;t1 =0) X

X

X

1 crosses X1 1 crosses X2 X

1 crosses X1

(?V1 )

1

Z

0

(?V1 )(?V2 )

Z

1

0

dt1

Z

1

0

dt2 t1 (2 )e?WN (X1 ;X2 ;t1 ;t2 ) +

dt1 e?WN (X1 ;X2 ;t1 ;t2 =0) + e?WN (X1 ;t1 =0)

We can now iterate the procedure for n ? 1 steps (recall that by de nition Xn = N ) and also using (A.3) we obtain n

X e?V (N ) =

X

l=1 1 crosses X1

lY ?2 j =1

:::

lY ?1

X

l?1 crosses Xl?1 j =1

(?Vj )

Z

0

1

dt1




Z

0

t1 (l?1 )


tj (j+1 )e?WXl (X1 ;:::;Xl?1 ;t1 ;:::;tl?1 ) e?V (NnXl )

1

dtl?1 (A.7)

Now recall that the sum 1 crosses X1 : : : l?1 crosses Xl?1 has been constructed with the restriction that Xj +1 = Xj [ j . This means that f1 ; : : : ; l?1 g is necessarely a tree graph in TXl . So the sum can be reorganized in the following way P

P

X

1 crosses X1

:::

X

l?1 crosses Xl?1

=

X

Xl N : jXl j=l

X

X

 2TXl X1 ;Xcomp 2 ;:::Xl?1 :

In the r.h.s. we have rst the sum over all sequences of subests of Xl compatible with a given tree graph  (the compatibility is in the sense speci ed in the enunciate of theorem 2), then we sum over all treeQ graphs in Xl and nally we sum over all possible subset Xl . The factor lj?=12 t1 (j +1 )


tj (j +1 ) in these notations depends now from the tree  and from the sequence X1 ; X2 ; : : : ; Xl?1 compatible with  and it can be rewritten for a xed  and a xed sequence X1 ; X2 ; : : : Xl?1 (i.e. for a given choice of the links 1 ; : : : ; l?1 ) as lY ?2 j =1

t1(j+1 )


tj (j+1 ) = tb11?1


tlb?l?11 ?1 12

where bi is the number of links in  which cross Xi . Thus

e?V (X ) =

n

X

X

X

(?V )

Y

l=1 Xl N :  2TXl 2 jXl j=l

Z

X

X1 ;X2 ;:::Xl?1 0 comp: 

1

dt1




Z

0

1

dtl?1 tb11 ?1


tlb?l?11 ?1

e?WXl (X1 ;:::;Xl?1 ;t1 ;:::;tl?1 ) e?V (X nXl )

De ning now Q(Xl ) = 1 if jXl j = 1, and

Q(Xl ) =

X

Y

 2TXl 2

(?V )

if jXl j  2, we have

Z

X

X1 ;X2 ;:::Xl?1 comp: 

1

0

e?V (X ) =

Iterating this formula we obtain

e?V (X ) =

dt1


n

X

Z

0

X

l=1 Xl N

n

X

1

(A.8)

dtl?1 tb11 ?1


tlb?l?11 ?1 e?WXl (X1 ;:::;Xl?1 ;t1 ;:::;tl?1 )

Q(Xl )e?V (X nXl )

X

l=r fY1 ;Y2 ;:::;Yr g2(N )

Q(Y1)Q(Y2 )


Q(Yr )

Note that no double countings arise in the expression above because in Y1 we set X1 = f1g, while in the subsequent sets Yj we set as X1 the rst point in the lexicographic order, and therefore any sequence of Y 's can be ordered in a unique way. Comparing formula above with (A.1) we obtain K (N ) = Q(N ) which is the tree equality (1.14), and the theorem 2 is proven. Proof of Lemma 3. Let us denote as, for any X  N , X X Vij V~ (X ) = Vii + i2X

fi;j g2X

The stability condition is thus V~ (X )  0 for all X  N . De ne also X W~ Xl (X1 ; : : : ; Xl?1 ; t1 ; : : : ; tl?1 ) = Vii + WXl (X1 ; : : : ; Xl?1 ; t1 ; : : : ; tl?1 ) i2Xl

Thus lemma 3 is equivalent to show that W~ Xl (X1 ; : : : ; Xl?1 ; t1 ; : : : ; tl?1 )  0

(A.9)

(A.10)

We now prove the following proposition.

Proposition. For any s > 0, and with the convention t0 = ts+1 = 0 and X0 = ;, the following identity holds

W~ Xs+1 (X1 ; : : : ; Xs ; t1 ; : : : ; ts) =

s s?r

(1 ? tk )tk+1 tk+2 : : : tk+r (1 ? tk+r+1)V~ (Xk+r+1 nXk ) r=1 k=0 (A.11) XX

13

Obviously lemma 3 follows trivially from this proposition. Just observe that the r.h.s. of (A.11) is non negative since is the sum of non negative terms. The proof of the proposition can be obtained by induction. It is easy to check the formula for s = 1 (which is a trivial case) and s = 2. Assuming that it is true for s ? 1 we will now show that it is true for s. In fact, by de nition W~ Xs+1 (X1 ; : : : ; Xs ; t1; : : : ; ts ) = V (X1 ) + t1 It2 ;:::;ts (X1 ; Xs+1 nX1 )+ W~ Xs+1 nX1 (X2 nX1 ; : : : ; XsnX1 ; t2 ; : : : ; ts) where It2 ;:::;ts (X1 ; Xs+1 nX1 ) represents the interaction energy between X1 and Xs+1 nX1 , e.g. if X1 = f1g, X2 = f1; 2g : : : Xs = f1; 2; : : : ; sg. then

It2 ;:::;ts (X1 ; Xs+1 nX1 ) = V12 + t2 V13 + t2 t3 V14 + : : : + t2 : : : tsV1s+1 Now note that

V (X1 ) + It2 ;:::;ts (X1 ; Xs+1 nX1 ) =

sX +1 j =2

t2 : : : tj?1 (1 ? tj ) [V (Xj ) ? V (Xj nX1 )]

where conventionally t2 : : : tj ?1 (1 ? tj ) = (1 ? t2 ) when j = 2. Thus W~ Xs+1 (X1 ; : : : ; Xs ; t1 ; : : : ; ts ) = 8