Statistical Mechanics - Physics at Oregon State University

60 CHAPTER 3. VARIABLE NUMBER OF PARTICLES
1
2
. The remainder of the error converges very rapidly in this case, because of
the special nature of the integrant. Therefore, for T ≫ 10−15K we can replace
the sum by an integral and we find:
Z1 ≈
√ 3 π
2α
(3.90)
Note that in the case α ≪ 1 we have Z1 ≫ 1. Since this sum appears in the
denominator of the probability for a state, the probability of being in a given
state is very small.
The volume V of the box is L 3 and the quantum concentration nQ(T ) is
defined by
Combining these formulas leads to
MkBT
nQ(T ) =
2π2 3
2
(3.91)
Z1(T, V ) = V nQ(T ) (3.92)
Since we have one particle only, the density is n = 1
V and the condition Z1 ≫ 1
is equivalent to n ≪ nQ(T ). For a normal gas (which is very dilute), this
is always obeyed, and we can replace sums by integrals. This is called the
classical limit. Discontinuities due to the discrete energy spectrum in quantum
mechanics are not important, and everything is smooth. On the other hand, if
n nQ(T ) we need to do the sums. This is the case when the density is large
or the temperature is low.
We note that Z1 ∝ T 3
2 . The factor three is due to the presence of three
summations, one for each spatial coordinate. If we would be working in D
dimensions, this would be Z1 ∝ T D
2 . Therefore, the internal energy is
U(T, V, N = 1) = kBT 2
∂Z1
∂T V
= D
2 kBT (3.93)
as expected for one particle.
What do we do with N particles? We assume that the energy states of
the particles are independent, and hence the energy of all particles together
is the sum of all one particle energies. This would suggest that Z(T, V, N) =
Z N 1 (T, V ). If the particles were distinguishable, this would be correct. But we
know from quantum mechanics that the particles are not distinguishable. Hence
we do not know which particle is in which state! Therefore, we propose that the
partition function for N particles in this box is given by
Z(T, V, N) = 1 N
(Z1) (3.94)
N!
The presence of a power of N can be easily argued for, as we did above, but the
factor N! is not clear. If we assume that the particles are indistinguishable, it
makes sense to divide by N!. This factor N! did lead to a lot of discussion, and