Statistical Mechanics - Physics at Oregon State University

174 CHAPTER 8. MEAN FIELD THEORY: CRITICAL TEMPERATURE.
Since the first parts are independent of a we can write this as
G(h, T ) min
m
and we find the value of a from
− 1
2 JNqm2
− hNm + NkBT min {Tr ρ log(ρ)}
a
This number a is now determined by requiring
(8.82)
min
a {Tr ρ log(ρ)} (8.83)
d
Tr ρ log(ρ) = 0 (8.84)
da
This expression is similar to the one we discussed in the chapter on operator
methods, and here, too, we can write
d
dρ
dρ
Tr ρ log(ρ) = Tr log(ρ) + Tr
da da da
(8.85)
Since Tr ρ has to remain equal to one, the second term is zero. The first term
is [log ρ]12. Similarly, we find that the derivative with respect to a ∗ is [log ρ]21.
In order to vary a and a ∗ we can either vary the real and imaginary part of a
independently, but also can vary a and a ∗ independently. Both procedures give
[log ρ]12 = [log ρ]21 = 0 (8.86)
If log ρ is diagonal, ρ must be diagonal and hence a = 0. This follows from
A = elog(A) = 1
n! [log(A)]n , and because log A is diagonal, all powers of log A
are diagonal. Therefore we find that
1
ρ = 2 (1 + m)
0
0
(1 − m)
(8.87)
Note that ρ is Hermitian with trace one, and that the condition of being positive
definite requires |m| 1, which is also expected.
There is one more detail, however. We found an extremum for the expression
Tr ρ log(ρ), but is it a minimum? Since there is only one extremum, independent
of the value of m, we can look at one other easy case and check if the results
are smaller or larger. Let us consider a real, positive, and small, and m = 0.
ρ =
1
2
1
2 a
1 a 2
(8.88)
Perturbation theory tells us that the eigenvalues of this matrix are 1
2 ±a. Hence
the trace is now equal to T (a) = ( 1
1 1
1
2 +a) log( 2 +a)+( 2 −a) log( 2 −a). If we take
the derivative d
dT 1
1
da of this expression we find da = log( 2 + a) − log( 2 − a). This
expression is indeed zero for a = 0, as needed, and is also positive for a > 0,as
). Hence in this
can be seen after combining the logarithms to dT
da
= log( 1+2a
1−2a