Thermodynamics – Why Gibbs’ Factor Appears in the Partition Function of Maxwell-Boltzmann Statistics

partition functionstatistical mechanicsthermodynamics

In the assumption of Maxwell-Boltzmann statistics, particles are treated as distinguishable. So for identical particles with discrete energy levels, there are degeneracies for any specific occupation state: $|n_1,n_2,…\rangle$

$$
\frac{N!}{n_1!n_2!n_3!…}
$$

The partition function in the Maxwell-Boltzmann statistics is:

$$
\begin{split}
Z=tr(\exp{-(\beta H)})&=\frac{1}{N!}\sum^{N}_{n_1,n_2,…=0}'\frac{N!}{n_1!n_2!n_3!…}\langle n_1,n_2,…|\exp{-(\beta H)}|n_1,n_2,…\rangle\\
&=\frac{1}{N!}\sum^{N}_{n_1,n_2,…=0}'\frac{N!}{n_1!n_2!n_3!…}\exp{(-\beta\sum^{\infty}_{k=1}n_kE_k)}\\
&=\frac{1}{N!}\Big(\sum^{\infty}_{k=1}\exp{(-\beta E_k)}\Big)^N\\
&=\frac{1}{N!}(Z(T,V,1))^N,
\end{split}
$$
thereof prime on the summation means we calculate the partition function when the total number of particles is fixed, that is $n_1+n_2+…=N.$ $\frac{1}{N!}$ is Gibbs' factor. My question is why do we put a Gibb's factor here? In the precedent, we suppose the particles are distinguishable—the presence of Gibbs' factor here contradicts the assumption of distinguishable particles.

Note: Here I use canonical ensemble.

Sources:

From Thermodynamics and Statistical Mechanics by Walter Greiner, Ludwig Neise, Horst Stöcker, D. Rischke chapter 12, Grand canonical description of ideal quantum system

and

Statistical Mechanics, 4th edition (R.K. Pathria, Paul D. Beale) chapter 6.1-6.2

Best Answer

I know of two arguments:

  1. This is necessary for the result to match the quantum result. Period. (This is my less favorite).

  2. It does not matter that the particles are distinguishable in principle. It matters whether we actually care about them being distinguishable. You are not following each particle in the gas, so if two of them swapped places, you wouldn't really mind. Hence, you are not really counting microstates, but rather some sort of "mesostate", which is what really matters to you. Since I don't care about where each individual particle is, but rather only about where there are particles, I treat them as if they were indistinguishable. Because in practice, that is what I actually want to do.

Related Question