Fermi and Bose Distributions

Identical particles cannot be told apart

So far we have counted as though the particles carried labels. The very first piece did, when it counted the ways for n of them to be on the left: which particle was on the left made it a different state.

But particles of the same kind carry no labels. Swap one electron for another and there is no telling that anything happened. The upper row of the figure counts with labels, the lower row without. The two in the middle have collapsed into one below.

Change the counting and the answer changes. From here on we count the collapsed way.

Fermions admit one at most

Particles that cannot be told apart come in two kinds: those that admit at most one to a level, and those that admit any number. The first are called fermions, and electrons and protons are among them.

Write the grand partition function for one level. The number in it is either 0 or 1, so \Xi_i = 1 + e^{-\beta(\varepsilon - \mu)}, and that is all. It has come out in exactly the form of the two-level system we handled before.

The mean occupation is \langle n \rangle = \dfrac{1}{e^{\beta(\varepsilon - \mu)} + 1}. The figure warms the system up from near absolute zero. When cold, everything below \mu is packed full and everything above is empty. Raise the temperature and only that boundary blurs, over a width of about kT.

Bosons admit any number

The other kind are called bosons, and photons and helium 4 atoms are among them. Any number of them may sit in one level.

The grand partition function for one level now runs over 0, 1, 2 and onwards. That is a geometric series of ratio e^{-\beta(\varepsilon - \mu)}, exactly the calculation we did for the oscillator. It gives \Xi_i = \dfrac{1}{1 - e^{-\beta(\varepsilon - \mu)}}.

The mean occupation is \langle n \rangle = \dfrac{1}{e^{\beta(\varepsilon - \mu)} - 1}. As the figure shows, the lower levels stack up several deep. As \varepsilon approaches \mu the occupation grows without limit, so \mu can never rise above the lowest level. That restriction is what gives rise to Bose condensation later.

A single sign is all that differs

Set the two answers side by side. A +1 in the denominator is Fermi, a -1 is Bose. That is the whole of the difference. Whether a level admits only 0 and 1 or any number at all, the whole of that difference in counting is folded into one sign.

The three curves in the figure are laid out against \varepsilon - \mu. The faint middle one has the sign taken away, e^{-\beta(\varepsilon - \mu)}, which is the Boltzmann distribution, the one from counting with labels.

On the left, where \varepsilon is close to \mu, the three part company widely. Fermi stops at 1, Bose has no ceiling at all, and Boltzmann lies between.

Where it is thin, all three agree

Look to the right, where \varepsilon - \mu is much larger than kT. e^{\beta(\varepsilon - \mu)} is far larger than 1, so the \pm 1 in the denominator makes no difference at all. The three lie exactly on one another.

This is also to say that the occupation is far below 1. If every level is nearly empty, a rule about how many may share one has nothing to bite on. In a room with seats to spare, a rule about sharing seats is beside the point.

The two panels in the figure are the same three curves, cold and hot. On the right they cannot be told apart. Air at ordinary temperature and pressure was handled well by classical physics because it sits on this thin side. Electrons in a metal and a gas cooled right down sit on the dense side, and those are what the pieces ahead are about.

Probability and large numbers
The microcanonical ensemble
The canonical ensemble
Applications
Phase transitions
Quantum statistics
Time and fluctuations