Statistical Mechanics - Physics at Oregon State University
Statistical Mechanics - Physics at Oregon State University
Statistical Mechanics - Physics at Oregon State University
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192 CHAPTER 9. GENERAL METHODS: CRITICAL EXPONENTS.<br />
9.2 Integr<strong>at</strong>ion over the coupling constant.<br />
Since exact answers are hard to get for most realistic models, it is useful to know<br />
a variety of tricks to obtain good approxim<strong>at</strong>ions. The procedure of integr<strong>at</strong>ion<br />
over the coupling constant is one of those tricks. We will not investig<strong>at</strong>e the<br />
rel<strong>at</strong>ions between this method and others, but simply remark th<strong>at</strong> this method<br />
is another way of obtaining a correl<strong>at</strong>ed density m<strong>at</strong>rix. We use the Ising model<br />
again for illustr<strong>at</strong>ion. We will break up the Hamiltonian into two parts:<br />
V = − <br />
{σ1,···,σN }<br />
H0 = −h <br />
{σ1,···,σN }<br />
and study the general Hamiltonian<br />
|σ1, · · · , σN > <br />
σiσj < σ1, · · · , σN | (9.1)<br />
<br />
|σ1, · · · , σN > <br />
σi < σ1, · · · , σN| (9.2)<br />
i<br />
H = H0 + λV (9.3)<br />
In this division the term H0 contains all the easy terms. In other systems it<br />
would include the kinetic energy in many cases. The important point is th<strong>at</strong><br />
the oper<strong>at</strong>or H0 can be diagonalized, we can find eigenvalues and eigenvectors.<br />
The oper<strong>at</strong>or V contains all the hard terms, the ones th<strong>at</strong> we do not know how<br />
to deal with, and have approxim<strong>at</strong>ed before. We are obviously interested in the<br />
case λ = J. We will, however, tre<strong>at</strong> λ as a variable parameter. The division<br />
is made in such a way th<strong>at</strong> the problem is simple for λ = 0. Again, we always<br />
make sure th<strong>at</strong> for H0 the solutions are known.<br />
The free energy as a function of λ is<br />
and the deriv<strong>at</strong>ive with respect to λ is<br />
G(λ) = −kBT log Tr e −βH0−βλV<br />
(9.4)<br />
dG kBT d<br />
= −<br />
Tr<br />
dλ Tr e−βH0−βλV dλ e−βH0−βλV<br />
(9.5)<br />
One always has to be careful when taking deriv<strong>at</strong>ives of oper<strong>at</strong>ors. In this<br />
case H0 and V commute and the exponent of the sum is the product of the<br />
single exponents.<br />
[H0, V] = 0 ⇒ e −βH0−βλV = e −βH0 e −βλV<br />
The deriv<strong>at</strong>ive with respect to λ affects the second term only, and we have<br />
d<br />
dλ e−βλV = d<br />
dλ<br />
which leads to<br />
∞<br />
n=0<br />
1<br />
n! (−βλ)n V n =<br />
∞<br />
n=1<br />
(9.6)<br />
1<br />
(n − 1)! λn−1 (−β) n V n = −βVe −βλV<br />
(9.7)