414 Quantum ideal gases
For example, if there were only two states, then the occupation numbers aref 1 and
f 2 wheref 1 +f 2 =N. The above formula gives
g(f 1 ,f 2 ) =N!
f 1 !f 2!=
N!
f 1 !(N−f 1 )!, (11.3.4)
which is the expected binomial coefficient.
Substituting eqn. (11.3.3) into eqn. (11.3.2) gives
Q(N,V,T) =∑
{f}N!
∏
nfn!∏
ne−βfnεn, (11.3.5)which is just a multinomial expansion for
Q(N,V,T) =
(
∑
ne−βεn)N
. (11.3.6)
Again, if there were two states, then the partition function would be
(e−βε^1 + e−βε^2 )N=∑
f 1 ,f 2 ,f 1 +f 2 =NN!
f 1 !f 2!
e−f^1 βε^1 e−f^2 βε^2 (11.3.7)from the binomial theorem. Therefore, in order to evaluate the partition function, we
just need to perform the sum
∑
ne−βεn=∑
ne−^2 π(^2) β ̄h (^2) |n| (^2) /mL 2
. (11.3.8)
Ultimately, we are interested in the thermodynamic limit, whereL→∞. In this limit,
the spacing between the single-particle energy levels becomes quitesmall, and the
discrete sum overncan, to a very good approximation, be replaced by an integral
over a continuous variable (which we also denote asn):
∑
ne−^2 π(^2) β ̄h (^2) |n| (^2) /mL 2
−→
∫
dne−^2 π(^2) βh ̄ (^2) |n| (^2) /mL 2
. (11.3.9)
Since the single-particle eigenvalues only depend on the magnitude ofn, we can trans-
form the integral overnx,ny, andnzinto spherical polar coordinates (n,θ,φ), where
n=|n|, andθandφretain their usual meaning. Thus, the integral becomes
4 π∫∞
0dn n^2 e−^2 π(^2) β ̄h (^2) |n| (^2) /mL 2
=V
(
m
2 πβ ̄h^2) 3 / 2
=
(
V
λ^3)
, (11.3.10)
whereλis the thermal wavelength. The partition function now becomes
Q(N,V,T) =
(
V
λ^3)N
, (11.3.11)
which is just the classical canonical partition function for an ideal gas. Therefore, we see
that an ideal gas of distinguishable particles, even when treated quantum mechanically,
has precisely the same properties as a classical ideal gas. Thus, weconclude that all
the quantum effects are contained in the particle spin statistics, which we will now
consider.