When getting the overcounting factor in statistical mechanics, how does one compute it?

Let's say each property is unique in one aspect (a string with an unique address in pc memory for example).

Should I count the number of ways it is possible to exchange subsets of properties of one particle with subsets of properties of another particle such that they are described with the same (standardized or semantically same) string?

Ex: {charge=1, spin=1/2, likesteddybears=true} is indistinguishable from {charge=-1, spin=1/2} because "spin=1/2" from the first listed particle can be exchanged with "spin=1/2" from the second listed particle?

Should I count the number of ways it is possible to change the set of all properties of one particle with the set of all properties of another such that they have matching properties (semantically)?

Ex: {spin=1/2,charge=-1} is indistinguishable from {charge=-1,spin=1/2}?

Should I count the number of ways the function I want to calculate the overcounting factor for gives the same result when I permute the input arguments' order (how should the permutation be restricted? only arguments with all properties equal? only arguments with one property equal? should it not be restricted at all and be super hard to implement in a type based programming language? should properties be exchanged simultaneously with argument permutation? should particles be converted into different argument types when necessary, say position or time?)

TL;DR: please tell me how to count the overcounting factor, in a way so that even a computer would understand it.


2 Answers 2


I will address the stat mech part.

Over-counting refers only to indistinguishable particles, i.e. those with all the same properties. It also only comes into play when you are performing calculations by "labeling" or "tracking" indistinguishable particles. This is not the same as using the number of particles in each energy state. In other words, if you treat the state "particle 1 with energy $E_0$ and particle 2 with energy $E_1$" as different from the state "particle 1 with energy $E_1$ and particle 2 with energy $E_0$", and your particles are indistinguishable, you are over-counting states by a factor of 2, and must divide that out. This is not the case if you are using particle numbers, since then your state is described by "1 particle in $E_0$ and one in $E_1$," and you are not over-counting.

The factor is then, as you say, calculated by permuting all identical particles in a each state (configuration). This is usually just $N!$ for $N$ identical particles. You can verify this easily by using a 3-particle labeling scheme following the reasoning from above.


Classical particles are always "distinguishable" , even if "identical". This is because every particle has a differnt trajectory in phase space. For atoms, this is experimentally impossibile to determine, but for, shall we say, fluorescent proteins or fluorescent multiprotein complexes, the trajectories are experimentally acessible

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    $\begingroup$ This might explain the difference between distinguishable and indistinguishable, but it doesn't seem to discuss the overcounting factor OP requested. $\endgroup$
    – Kyle Kanos
    Mar 9, 2018 at 11:02

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