I am reading material related to statistical physics, and I am having a problem understanding why we care about the notions of distinguishably and indistinguishability. I have found What are distinguishable and indistinguishable particles in statistical mechanics? on here, but it doesn't answer my question since it doesn't really discuss why we care. I have also seen Classical, identical particles which are distinguishable, which explains that we can't 'watch' quantum mechanical particles, but this doesn't really fix my confusion either since I don't really understand mathematically what 'watch' actually means. I will explain my confusion here.

My understanding is that we say particles are indistinguishable if there is no way, even in principle, to tell them apart, and distinguishable otherwise. I can understand how we might not be able to tell particles apart, but I can't see why this would ever be important. We can derive the partition function for (distinguishable, I think) particles basically by making a combinatoric argument about how many ways there are to partition $N$ particles into $k$ groups (states or energy levels), with these groups having sizes $n_1, \ldots, n_k$, finding that there are $\displaystyle\frac{N!}{\Pi_i n_i !}$ ways of doing this.

I don't see how not being able to tell the particles apart actually changes anything. As far as I can see, counting arguments like this don't rely on the ''indistinguishable'' and ''particle'' having any meaning whatsoever: $\displaystyle\frac{N!}{\Pi_i n_i !}$ is the number of ways of partitioning a set of size $N$ into $k$ sets with sizes $n_i$, regardless of what the elements of that set are, and whether we can ''tell them apart''.

I hope I've explained my confusion sufficiently: in a nutshell, it could be summarised as "given that counting arguments don't depend on distinguishability, why does having indistinguishable particles change anything at all?". My suspicion is that I am mistaken in believing this: certainly if we say a set has size $N$ we are assuming that it has $N$ 'distinct' elements, in some sense. I really can't see how this is the same notion of distinguishability as applies to particles, though, so I am quite stuck.

If anyone could help me out here, that would be great! Just to restate what I'm after: I feel that I understand what people mean by "indistinguishable", I just can't see why it would matter in terms of counting arguments.

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    $\begingroup$ My go-to recommendation for this point is E. T. Jaynes' paper 'the Gibbs Paradox.' You might like it. He argues very clearly that the distinguishability that matters is the ability to practically distinguish the particles experimentally, rather than whether they could be distinguished in principle. bayes.wustl.edu/etj/articles/gibbs.paradox.pdf $\endgroup$
    – N. Virgo
    Jul 18, 2020 at 14:45

2 Answers 2


Let's say you have a system of two distinguishable, non-interating particles (say, a proton and a neutron), each of which can inhabit one of two available energy levels, $E=0$ and $E=\epsilon$. The macrostates of this system can be characterized by the total energy, which means there are three possible macrostates corresponding to total energy $\mathcal E_0=0,\mathcal E_1=\epsilon$, and $\mathcal E_2 = 2\epsilon$.

How many microstates correspond to each macrostate?

  1. $\Omega(\mathcal E_0)=1$ because there's only one way for both particles to have zero energy
  2. $\Omega(\mathcal E_1)=2$ because the proton could have energy $\epsilon$ and the neutron could have energy $0$, or the proton could have energy $0$ and the neutron could have energy $\epsilon$
  3. $\Omega(\mathcal E_2)=1$ because there's only one way for both particles to have energy $\epsilon$

Now repeat the same analysis for two indistinguishable electrons, and there's a big difference - $\Omega(\mathcal E_1)=1$, not $2$. The state with electron A having energy $0$ and electron B having energy $\epsilon$ is the same state as the one with electron A having energy $\epsilon$ and electron B having energy $0$. In fact, it doesn't even make sense to label the electrons as A and B.

This is the essence of indistinguishability - for a system of indistinguishable particles, we identify states which are related to each other via permutation as being the same state. For a system of distinguishable particles, this is not so.

For what it's worth, your factor of $\frac{N!}{\prod n_i!}$ is indeed the correct "overcounting factor" which we would need to include. If we assume that the particles are distinguishable, then this factor would be completely absent, but this leads to problem with non-extensive entropy (see the Gibbs Paradox). If we assume that the energy levels are sparsely populated (so the likelihood of any $n_i$ being greater than 1 is low), then we can approximate it by $N!$; this yields what we would usually call classical statistical mechanics.

  • $\begingroup$ excellent exposition, thank you for providing this. $\endgroup$ Jul 17, 2020 at 18:05

Adding to J. Murray's answer.

Quantum mechanics adds a layer of confusion over this, in my opinion. The concept of indistinguishability can be exemplified even in the classical world.

In many instances, the answer lies in your question: "do you care?" Meaning: "is the question you asking (or, in other words, the calculation you are setting up) dependent on particles identity?". Which boils down to asking how you define the state of your system: are you happy with macrostates or do you need microstates?

I don't see how not being able to tell the particles apart actually changes anything.

It does change if the experiment you are "modelling" is able to distinguish between particles. Imagine you are writing the configuration of $N$ (classical) particles as "the vector of number of particles that sit at a given position". This is a macrostate, similar to the example of "how many electrons per energy shell". You will say there are $n_1$ particles at position $x_1$, $n_2$ particles at position $x_2$ and so on. You might like to write this state, borrowing Dirac notation from quantum mechanics, as the vector $|n_1,\,n_2,\,\ldots\rangle$. Implicitly, you are stating that the particles are indistinguishable, as your state representation does not feature particles identity. Here, indistinguishability is a property of your model, independently of whether your particles are indistinguishable or not. Presumably, it is because the experiment you are planning does not involve tracking individual identities.

Obviously, however, you can construct this state vector as a sort of "projection" of the vector specifying the position of each individual particle, that we might instead write $(x_{i_1}, x_{i_2}, \ldots, x_{i_N})$, where the index $i_j$ stands for the position where particle $j$ is sitting. These are your microstates. In the example here, microstates with the the same number of particles per position index contribute to the same macrostate. If you choose this representation of the states (because you are very keen on keeping track of all the identities of your particles), then you are limited by whether your particles actually are indistinguishable $-$ the electrons-vs-nucleons example given above shows this.

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    $\begingroup$ The issue is, what does distinguishability have to do with the probability distributions that we actually see? For example, if the microstates in J. Murray's answer are distributed according to the usual Gibbs distribution, then the probability of energy $\epsilon$ in the first case is $\frac{2}{1+2e^{-\epsilon/(k_B T)}+e^{-2\epsilon/(k_B T)}} e^{-\epsilon/(k_B T)}$ and the probability of energy $\epsilon$ in the second case is $\frac{1}{1+e^{-\epsilon/(k_B T)}+e^{-2\epsilon/(k_B T)}} e^{-\epsilon/(k_B T)}$. $\endgroup$
    – Ian
    Jul 18, 2020 at 0:34
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    $\begingroup$ (Cont.) The former is significantly larger than the latter if $\epsilon$ is not too big compared to $k_B T$. Thus it is not just a matter of "whether we care to pay attention or not", it is a matter of "whether nature cares enough that the probability distribution is affected". $\endgroup$
    – Ian
    Jul 18, 2020 at 0:34
  • $\begingroup$ Just by "not caring" whether two coins are distinguishable or not, it will not suddenly happen that tossing those coins results in two heads with probability $\frac13$ instead of $\frac14$. $\endgroup$ Jul 18, 2020 at 17:14
  • $\begingroup$ Minor amendment to my previous comment: it actually doesn't really matter how big $\epsilon/(k_B T)$ is in J. Murray's answer. The ratio of the probability to see energy $\epsilon$ in the distinguishable case to the probability to see energy $\epsilon$ in the indistinguishable case is always between $1.5$ and $2$. $\endgroup$
    – Ian
    Jul 18, 2020 at 20:22
  • $\begingroup$ @Ian, yes, I was making the point about "what does distinguishability have to do with the probability distributions that we actually see" (or model). $\endgroup$
    – berberto
    Jul 18, 2020 at 22:50

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