I've started to learn Quantum mechanics and statistical mechanics on my own. There is a line in my book, which says, that

In an ideal gas, to remove the Gibbs paradox, the gas molecules are treated as indistinguishable. Since the molecules are of the order $10^{23}$, their wave functions overlap, and hence we can treat them as indistinguishable.

Can you please explain, how overlapping of wavefunctions tell us whether particles are distinguishable or not.

  • $\begingroup$ Which textbook is the quote taken from? $\endgroup$
    – user140255
    Jun 8, 2020 at 14:11

1 Answer 1


The quotation you gave perpetuates two old misconceptions:

  1. that classical statistical physics model where molecules are distinguishable by their positions, implies wrong statements about entropy, to be removed by changing our understanding of the gas into a model where particles are not distinguishable by their positions

  2. that Gibbs paradox is something that requires quantum theory to be resolved.

Ad 1) Classical statistical physics does not require that expression $k_B \log W$ where $W$ is number of distinguishable states of a classical system, be extensive (linear in $N$). It is us, based on laws and conventions of thermodynamics (in particular, the fact we can't distinguish molecules of same species in thermodynamic experiments), who require that statistical entropy be extensive (linear in $N$) and we can fulfill this requirement by proper normalization of $W$ in the above formula, without any implication on distinguishability of the particles in the fundamental sense. One possible formula that achieves this is $k_B \log \frac{W}{N!}$, but we could also use $k_B \log \frac{W}{N^N}$. The former is preferred since in quantum theory there is this idea about indistinguishable particles where factor $1/N!$ removes the spurious states , but both formulae are valid formulae for statistical entropy in classical statistical physics, where particles are distinguishable.

Ad 2) The "Gibbs paradox" is not a paradox, or a problem of model/our understanding, at all. Gibbs' observation was that change in the extensive variant of statistical entropy of a macrostate of a system that goes from two spatially separate non-interacting gases in volumes $V$ and $V$ into one mixed gas of volume $2V$, is zero if two gases are the same, and large non-zero if they are not the same. The surprise is in the fact that a minor change in difference between the gases (such as one gas having a different nuclide of the same element) leads to finite jump in the entropy of mixing. But this is not a paradox, but instead reveals important property of statistical entropy: it depends on whether subsets of the system for which we calculate statistical entropy are thermodynamically distinguishable or not. The fact the particles can be distinguished by their positions does not matter - thermodynamically they can't be distinguished. If the particles are the same species, their mixing won't increase extensive statistical entropy. If they are not, mixing changes thermodynamic state and extensive statistical entropy has to increase.

  • $\begingroup$ I liked the explanation. BTW the quote I guess is from pathria. Which book did you refer btw for stat mech $\endgroup$
    – Shashaank
    Jun 8, 2020 at 14:54
  • $\begingroup$ By "extensive variant of statistical entropy", you mean the entropy obtained after normalizing the number of states $W$ by some appropriate normalization factor $f(N)$? $\endgroup$
    – user87745
    Jun 8, 2020 at 17:12
  • 1
    $\begingroup$ @DvijD.C. Indeed, $\log W$ isn't homogeneous function of $U,V,N$, but $\log \frac{W}{N!}$ is (sometimes called "extensive entropy"). $\endgroup$ Jun 8, 2020 at 21:46
  • $\begingroup$ Yes, I just wanted to confirm that you weren't referring to variations in the sense of calculus. Great answer, +1. $\endgroup$
    – user87745
    Jun 8, 2020 at 21:51

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