Statistical Mechanics - Physics at Oregon State University

9.2. INTEGRATION OVER THE COUPLING CONSTANT. 193
dG Tr Ve−βH0−βλV
=
dλ Tr e−βH0−βλV (9.8)
It is useful to digress here and show that this result is true in general. We
have
d
dλ e−βH0−βλV =
∞ 1 d
n! dλ [−βH0 − βλV] n
n=0
(9.9)
Because operators do not commute we need to keep track of the order of the
operators and derivatives:
d
dλ e−βH0−βλV =
∞
n=1
1
n!
n
[−βH0 − βλV] m−1 [−βV] [−βH0 − βλV] n−m
m=1
(9.10)
The derivative of the very first term is zero, so the sum starts at n = 1. Next
we realize that we need the trace of this expression, and that the trace obeys
Tr(ABC) = Tr(CAB). This gives
Tr d
dλ e−βH0−βλV =
∞
n=1
1
n!
n
Tr [−βV] [−βH0 − βλV] n−1
m=1
Now we can do the sum on m, it simply gives a factor n and find
Tr d
dλ e−βH0−βλV =
∞
n=1
Finally, we reassemble the sum and obtain
1
(n − 1)! Tr [−βV] [−βH0 − βλV] n−1
(9.11)
(9.12)
Tr d
dλ e−βH0−βλV −βH0−βλV
= Tr [−βV] e (9.13)
leading to the required result. Integrating over the coupling constant λ then
gives the following formula for the free energy
J
G(λ) = G(0) + dλ
0
dG
J
dλ
= G(0) +
dλ 0 λ 〈λV〉λ (9.14)
where it is standard practice to put a factor λ inside the ensemble average.
Remember that G(0) is a known quantity.
The formula above tells us that once we know 〈V〉λ we can calculate the
free energy. That is not surprising, because once we have the pair correlation
function and hence the potential we essentially have solved the problem. The
trick is not how to get the pair correlation function exactly, but how to approximate
it. Using an approximation to the pair correlation function will give us
in general an improved free energy.