where . Fermions are identical particles that, for each "box" or one-particle state they can occupy (given e.g. by in the case of the Hydrogen atom-like states), admit either or particles in it. Higher numbers are forbidden by the Pauli exclusion principle. The energies of the multi-particle state with and in a particular one-particle state differ by . Consequently,
where I used the Boltzmann distribution. However, the probabilities that the number of particles in the given one-particle state is equal to or must add to one,
These conditions are obviously solved by
which implies that the expectation value of is equal to the right formula for the Fermi-Dirac distribution:
The calculation for bosons is analogous except that the Pauli exclusion principle doesn't restrict . So the number of particles (indistinguishable bosons) in the given one-particle state may be . For each such number , we have exactly one distinct state (because we can't distinguish the particles). The probability of each such state is called where .
These conditions are solved by
Note that the ratio of the adjacent is what it should be and the denominator was chosen so that all the from sum up to one.
because the number of particles, an integer, must be weighted by the probability of each such possibility. The denominator is still inherited from the denominator of above; it is equal to a geometric series that sums up to
Don't forget that is in the denominator of the denominator, so it is effectively in the numerator.
This result's denominator has a second power. One of the copies gets cancelled with the denominator before and the result is therefore
which is the Bose-Einstein distribution.