Statistical Mechanics - Physics at Oregon State University

132 CHAPTER 6. DENSITY MATRIX FORMALISM.
intensive state variables their correct value, then to take the thermodynamic
limit N → ∞, and finally consider the results as a function of ɛ and take the
limit ɛ → 0. In practice, one should make N large and choose ɛ as small as
possible, but large compared to the average spacing of the energy levels. We
will discuss this question by looking at a simple example.
A quantum system in one dimension has single-particle eigenstates with
energy nω, with n = 1, 2, 3, · · ·. The energy of N particles is given by ( N i ni)ω.
Hence the entropy is equal to
S(U, N) = −kB
n1,n2,.,nN
C 1
ɛ √ π e− (U−( N i ni )ω)2
ɛ2
log C 1
ɛ √ π e− (U−( N i ni )ω)2
ɛ2
(6.76)
We assume that the particles are not identical. Next we set xi = ni ω
Nɛ and
assume that ɛ ≫ ω
N , in agreement with the statement that we can only take
the limit ɛ → 0 after taking the limit N → ∞. The quantity ω
N is the spacing
between the possible energy values per particle and we have to choose ɛ much
larger than this value. The summations are replaced by integrations and we
have
S(U, N) = −kB
N
Nɛ
ω
· · ·
C 1
ɛ √ U
e−( ɛ
π −Nx1−···−NxN ) 2
log C 1
ɛ √
− (
π
U
dx1 · · · dxN
ɛ − Nx1 − · · · − NxN) 2
(6.77)
Now we define x =
i xi and replace the integration variables by x, x2, · · · , xN.
Since all the variables xi have to be positive, we know that for a given value
of x all other variables are limited by N i=2 xi x. Since the integrand only
depends on the value of x, the integration over x2, · · · , xN is just a geometrical
factor g(x). Therefore, the entropy is equal to
N
ɛN
S(U, N) = −kB
dxg(x)
ω
− ( U
− Nx)2
ɛ
C 1
ɛ √ U
e−( ɛ
π −Nx)2
log C 1
ɛ √ π
The coefficient C follows from Tr ρ = 1:
or
C −1 = Tr
C −1 =
n1,n2,.,nN
(6.78)
1
ɛ √ (U−H)2
e− ɛ
π 2 (6.79)
1
ɛ √ π e− (U−( N i n i )ω)2
ɛ 2 (6.80)