From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Spin system on lattice
Date: 25 Jan 2000 12:23:01 -0600
Newsgroups: sci.physics.research
Summary: [missing]

In article <85r7bv$iv5$1@nnrp1.deja.com>,
 v_i_smirnov@my-deja.com writes:
>  If you have a spin of either 0 or 1 on each point of the
> two-dimensional lattice Z^2, where Z is the set of all
> integers, and you have a homogenous nearest-neighbour potential
> (in other words, if A and B are adjacent points in the lattice
> then the potential is -AB/T, and if A and B are not adjacent then
> the potential is 0)
> then why is it true that for T (temparature) big enough there is
> only one equilibrium state?

There are a number of proofs.  Mine is in "High-Temperature Analyticity
in Classical Lattice Systems", Commun. math. Phys. 50 (1976) 245-257.
Basically, you can write down an equation of the form K phi + delta = phi
where phi is a linear functional on a certain Banach space of observables,
delta corresponds to the equilibrium state with no interaction,
and K is a certain operator.  The phi corresponding to an equilibrium state
must satisfy this equation.  The result is that there is only one solution
phi = sum_{j=0}^infinity K^j delta when that series converges (in a
certain Banach space), which it does when the temperature is large enough.
This gives you not only uniqueness but also analytic dependence on parameters.


Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2