Statistical Mechanics - Physics at Oregon State University

6.3. MAXIMUM ENTROPY PRINCIPLE. 129
which leads to the simple solutions
kB 〈m| log(ρ) |n〉 = (λ − 1)kBδnm
(6.56)
ρ = e λ−1 E (6.57)
We still need to check that this is a Hermitian matrix with eigenvalues less
than one. That is easy, we simply choose λ to be real and at most one.
How do we find the value of λ? By checking the condition Tr ρ = 1. This
gives
e 1−λ = Tr E (6.58)
and the last trace is simply the dimension of the space, or the number of states
with the given energy, the multiplicity function:
e 1−λ = g(U, V, N) (6.59)
Since the multiplicity function is one or larger, λ is one or less. The entropy for
this density matrix is
or
S = −kBTr e λ−1 (λ − 1)E = −kBg −1 (U, V, N) log(g −1 (U, V, N))Tr E (6.60)
S = kB log(g(U, V, N)) (6.61)
which is exactly what we expect! Hence we retrieved our old definition, and the
new procedure is the same as the old one.
The extremum of X we have found is a maximum. In order to test this, we
have to calculate the second order derivatives. Hence we need
2 ∂ X
∂
= −kB (log(ρ)) mn
(6.62)
∂ρij∂ρnm
∂ρij
In order to find the answer, we use again ρ(x) = e τ(x) . As before, we have
∂
∂x eτ(x) =
∞
n=1
n−1
1
τ
n!
m=0
m
∂τ
(x) τ
∂x
n−m−1 (x) (6.63)
In general, as we noted before, this cannot be simplified. Taking the trace
of this expression allows us to make it easier. But here we need all values. But
we only need to calculate the second order derivative at the extremum. Hence
we set ρ = e λ−1 E, or τ = (λ − 1)E, and we have
∂
∂x eτ(x) =
∞
n=1
n−1
1
n!
m=0
(λ − 1) m (x)
∂τ
(λ − 1)
∂x
n−m−1 (x) (6.64)