# What about the grand canonical ensemble allows you to use it to describe quantum gases as compared to the canonical ensemble?

Probably an obvious question but this is my understanding so far: Ideal classical gas - gas where occupancy of an SPS <<1. Ideal quantum gas - cold dense gas where occupancy of SPS can be greater than 1.

In my lecture notes, derivation of the gibbs distribution/ grand canonical distribution preludes the whole fermi dirac and bose einstein distribution derivations.

I am looking back at these notes and I remember the reason needing to introduce the grand canonical ensemble was because the occupancy of SPS can be greater than 1.

What about the grand canonical ensemble allows for occupancies greater than 1? Equivalently, what about the canonical ensemble limits the occupancy to 1?

• You may find my answer here helpful Jan 20, 2020 at 17:20
• Take into account that even if the answer cited by @BySymmetry was in connection with a question on BEC, his argument is completely general and valid for fermions as well as for bosons. Jan 20, 2020 at 19:18

What about the grand canonical ensemble allows for occupancies greater than 1? Equivalently, what about the canonical ensemble limits the occupancy to 1?

I believe you misunderstood your instructor. There is nothing about the framework of the canonical ensemble which limits the occupancy numbers of the single-particle states to be $$\ll 1$$.

The correct statement is that we can get away with using classical statistics in that limit, but if the occupancy numbers are of order $$1$$ (or greater), the classical prescription no longer works.

The key is the way we enumerate states. Consider a simple system of $$N$$ particles which have allowed energies $$0,\epsilon,2\epsilon,$$ etc. I now ask for the number of (micro)states of the system which correspond to a fixed total energy $$E$$. For concreteness, let $$N=3$$ and $$E=6\epsilon$$.

There are a few (possibly naive) approaches we could take. The first is to define a microstate as a list of $$N=3$$ energies, corresponding to the state inhabited by each of the three particles. Under this prescription, we have the following possibilities:

This yields a total of $$21$$ microstates, and a corresponding entropy of $$\log(21)$$.

This approach is not correct, because it treats the particles as being distinguishable. It turns out that the resulting entropy is not an extensive quantity (see Gibbs' Paradox), which is not acceptable. The solution to the problem is to treat the particles as indistinguishable. Note that the state $$(0,1,5)$$ is identical to the microstates $$(0,5,1), (1,0,5), (1,5,0), (5,1,0),$$ and $$(5,0,1)$$ if the particles are indistinguishable. By treating the microstates as distinct, we are overcounting by a factor of $$3!=6$$.

How do we fix this? The simplest possible solution is to simply divide our previous count by $$3!$$. However, this is not really correct. Note that the state $$(0,0,6)$$ has not been overcounted by a factor of $$3!$$, but only by a factor of $$3$$ because two of the particles are in the same state. Likewise, the state $$(2,2,2)$$ has not been overcounted at all.

Dividing by $$N!$$ is not correct because it assumes that none of the particles occupy the same state. However, if the average occupancy of each state is very low (that is, if the number of microstates with more than one particle per state is small compared to the total number of microstates under consideration), the $$N!$$ prescription is actually a fairly good approximation. This corresponds to what we call classical statistics.

Quantum statistics is obtained by being more accurate with the correction factor. It's possible to do this within the canonical ensemble picture, but it is generally quite complicated because the total particle number is taken to be fixed. Moving to the grand canonical ensemble picture makes things vastly simpler, so that's normally what we do, but it is not necessary in principle.