Statistical Mechanics - Physics at Oregon State University

2.8. A SIMPLE EXAMPLE. 37
2.8 A simple example.
In order to illustrate the ideas developed in this chapter, we consider a simple
example, even though simple examples are dangerous. The energy eigenstates
of a system are given by two quantum numbers, n and l. The possible values
for n are 1, 2, 3, · · · , ∞, but the values for l are limited by l = 1, 2, · · · , n. The
energy eigenvalues depend only on n:
ɛnl = nω (2.58)
where ω is a positive number. This problem is essentially a harmonic oscillator
in two dimensions. The partition function follows from
Z(T ) =
∞
n=1
ω n
− k n e B T
(2.59)
and since the factor in parenthesis is always between zero and one, this can be
summed to give
The Helmholtz free energy is
Z(T ) =
ω − k e B T
ω − k (1 − e B T ) 2
(2.60)
ω − k F (T ) = ω + 2kBT log(1 − e B T ) (2.61)
The derivative with respect to temperature yields the entropy
S(T ) = ω 2
T
1
e ω
k B T − 1
The internal energy is F + T S and is given by
ω − k − 2kB log(1 − e B T ) (2.62)
U(T ) = ω coth( ω
) (2.63)
2kBT
Note that the term with the logarithm disappeared in the formula for the energy
U. This is characteristic of many problems. The Helmholtz free energy is
F = −kBT log(Z) and hence the entropy is S = kB log Z + kBT 1
∂Z
Z ∂T V,N .
Therefore the terms with the logarithms cancel in F + T S.
At this point it is always useful to consider the entropy and energy in two
limiting cases, small and large temperatures. In our problem the natural energy
scale is ω and the temperature is small if the thermal energy kBT is small compared
with the spacing between the energy levels, kBT ≪ ω. The temperature
is large if the thermal energy kBT covers many energy levels, kBT ≫ ω. For
small values of the temperature we find
T → 0 : U ≈ ω, S ≈ 0 (2.64)