Statistical Mechanics - Physics at Oregon State University

258 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
since the matrices commute. The value of κ is small, hence:
Z = T re −βH0
1 − βκV + 1
2 β2κ 2 V 2
Using the definition of thermodynamic average we have
and hence
Z = T re −βH0
Using log(Z) = −βF we find
< X > T re −βH0 = T rXe −βH0
1 − βκ < V > + 1
2 β2κ 2 < V 2
>
−βFκ = −βF0 + log(1 − βκ < V > + 1
2 β2 κ 2 < V 2 >)
Expanding the log in second order gives
or
Problem 5.
−βFκ = −βF0 − βκ < V > + 1
2 β2 κ 2 < V 2 > − 1
2 β2 κ 2 < V > 2
Fκ − F0 = κ < V > − 1
2 βκ2 < V 2 > − < V > 2
In a two-dimensional Hilbert space the density operator is given by its matrix
elements:
x R
ρ =
R∗
1 − x
This form is clearly Hermitian and has trace one. Calculate the entropy as a
function of x and R, and find the values of x and R that make the entropy
maximal. Note that you still need to check the condition that the matrix is
positive! Also, show that it is a maximum!
The entropy is given by:
Use
S = −kBT rρ log(ρ)
∂
∂ρ
∂ρ
T rρ log(ρ) = T r log(ρ) + T r
∂X ∂X
∂X