As an illustration of the use of occupation numbers in the evaluation of the quantum partition function, let us consider the simple case of Boltzmann statistics (ignoring spin statistics or treating the particles as distinguishable). The canonical partition function Q(N,V,T) can be expressed as a sum over the quantum numbers for each particle:
In terms of occupation numbers, the partition is
where is a factor that tells how many different physical states can be represented by a given set of occupation numbers . For Boltzmann particles, exchanging the momentum labels of two particles leads to a different physical state but leaves the occupation numbers unchanged. In fact the counting problem is merely one of determining how many different ways can N particles be placed in the different physical states. This is just
For example, if there are just two states, then the occupation numbers must be and where . The above formula gives
which is the expected binomial coefficient.
The partition function therefore becomes
which is just the multinomial expansion for
Again, if there were two states, then the partition function would be
using the binomial theorem.
Therefore, we just need to be able to evaluate the sum
But we are interested in the thermodynamic limit, where . In this limit, the spacing between the single-particle energy levels becomes quite small, and the discrete sum over can, to a very good approximation, be replaced by an integral over a continuous variable:
Since the single-particle eigenvalues only depend on the magnitude of , this becomes
where is the thermal deBroglie wavelength.
which is just the classical result. Therefore, we see that an ideal gas of distinguishable particles, even when treated fully quantum mechanically, will have precisely the same properties as a classical ideal gas. Clearly, all of the quantum effects are contained in the particle spin statistics. In the next few lectures we will see just how profound an effect the spin statistics can have on the equilibrium properties.