For systems of identical femions and identical bosons, an
exchange of particles does not change the physical state.
Therefore the factor 
is just 1 for
both of kinds of systems. Moreover, the occupation number
of a state characterized by 
for a system of
identical bosons can be any number between 0 and N:
For fermions, the Pauli exclusion principle forbids two identical particles from occupying the same quantum state. This restricts the occupation numbers to be either 0 or 1:
Given these possibilities for the occupation numbers, the canonical partition function can be formulated:
Note that the sum over occupation numbers must be performed subject to the restriction
a condition that makes the evaluation of Q(N,V,T) extremely difficult.
Therefore, it seems that the canonical ensemble is not the best choice
for carrying out the calculation. No worry, there are other ensembles
from which to choose, and of these, it turns out that the grand canonical
ensemble is significantly easier to work with. Recall that in the
grand canonical ensemble, 
, V and T are the control variables and
the partition function is given by
Note that the inner sum over occupation numbers is still subject to the
restriction 
. However, there is a final sum over
all possible values that N, the number that restricts the sum over
occupation numbers, can take on. Therefore, if we let the sum over
occupation numbers be unrestricted, then they could sum to any value they
liked. This would be equivalent to performing an unrestricted sum
over occupation numbers without performing the final sum over N,
since in the course of summing, unrestricted, over occupation numbers,
we would obtain every possible value of N as required by the
final sum over N. This is the main advantage of using this ensemble
for bosonic and fermonic systems. Thus, the grand canonical partition function
becomes
Note also that the sum of products is just
For bosons, each individual sum is just the sum of a geometric series. Hence,
whereas, for fermions, each individual sum contains only two terms
corresponding to 
and 
. Thus, for fermions:
Note that the summands are independent of the quantum number m so that we may perform the product over m values trivially with the result
for bosons and
for fermions, where g=(2s+1) is the number of eigenstates of 
(also known as the spin degeneracy).
At this point, let us recall the procedure for calculating the equation of state in the grand canonical ensemble. The free energy in this ensemble is PV/kT given by
and the average particle number is given by
The fugacity 
must be eliminated in favor of 
using the
second equation and substituted into the first equation to yield the
equation of state. Recall that, for the classical ideal gas,
Eliminating in favor is trivial in this case, leading
to the classical ideal gas equation
For the ideal gas of identical fermions, the equations one must solve are
and for bosons, they are
It is not difficult to see that the problem of solving for
in terms of is highly non-trivial for both systems.
The next two lectures will be devoted to just this problem and
exploring the rich behavior that the quantum ideal gases exhibit.