I am wondering about the partition function of the classical microcanonical ensemble. It contains Planck's constant and also an indistinguishability argument about the particles I am looking at and I find this confusing for the following reasons:

If I describe for example an ideal gas using classical mechanics, there is no reason why I should assume that there is a problem to distinguish the gas molecules or why $\hbar$ should occur there.

I mean, how could people like Gibbs etc. derive this equation if they did not know about QM?

In some sense, there has to be a reason why we have these two QM properties in there and maybe it is just to get the limit to QM right, but is this really the only reason for this?

I would love to receive a few ideas about this strange definition by this community, in order to get impressions, why we model a (NVE) ensemble this way?


A footnote on Wikipedia actually explains this.

The $h$ appearing in the partitioning of the phase space is not Planck's constant, it is simply the size of a phase space cell in which we do not distinguish single states anymore. It's initially arbitrary. Since Planck's constant provides a natural scale below which we should not apply classical thinking, it is nowadays often taken to be that constant, but nothing in the classical formalism necessitates this.

You are correct that it influences the entropy, since that is

$$ S = k\ln(W)$$


$$ W = \frac{1}{h^n C} \int_{\text{phase space}}f(\frac{H - E}{\omega})$$

but the logarithm property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$ means this drops out of all entropy differences, meaning the choice of $h$ has no physical impact. The overcounting factor $C$ depends on whether swapping individual particles/phase space coordinates will make a difference in the macrostate, and is $C = N!$ for $N$ identical (not really indistiguishable, classically) particles.

1Here, $n$ is the number of generalized coordinates (the number of particles in the usual settings), $f(\frac{H - E}{\omega})$ is an "indicator function" of the Hamiltonian $H$ that peaks at $H = E$ with width $\omega$ (for any suitable notion of width, e.g. it could be taken to be a Gaussian or rectangle function. Often, one takes $\omega \to 0$, leaving $f$ to be the Dirac delta $\delta(H - E)$, which means the integral then collapses to the surface of the object traced out by the equation $H = E$ in phase space, which, for free particles, is a sphere.

  • $\begingroup$ This is not really an answer, especially it does not explain where this N! comes from. And honestly, if this was completely arbitrary, then people would not have used it. I mean, using Planck's constant and N! chanes completely the magnitude of the entropy, so well yeah... $\endgroup$ – Xin Wang Nov 5 '14 at 19:10
  • $\begingroup$ @XinWang: The "classical indistinguishability" of the molecules is not fundamental, but rather like "There are $\sim 10^{23}$ of them, and we don't care which of them does what, since they're all the same, leading to the same macrostate if their individual states are simply swapped". You seem to jump to conclusions about QM without seeing that it's really just the difference between micro- and macrostate that is at play here. $\endgroup$ – ACuriousMind Nov 5 '14 at 19:15
  • $\begingroup$ but I mean when you calculate an entropy and compare this with the entropy in Thermodynamics, then don't you think that it makes a difference if you put there $h=1$ or $h=$Planck constant? I mean, there are a certain limits that should hold and I have some doubts that it should still somehow matter. $\endgroup$ – Xin Wang Nov 5 '14 at 19:18
  • 2
    $\begingroup$ @XinWang: Yep, the value of $h$ influences the absolute value of the entropy. But if you fix $h$, it doesn't matter since the absolute value of the entropy has no meaning, and it will drop out of entropy differences - essentially, it is just an offset. $\endgroup$ – ACuriousMind Nov 5 '14 at 19:27
  • $\begingroup$ @ACuriousMind I was under the impression that even though classical theory does not predict the numerical value of $h$, it is not arbitrary: Sackur and Tetrode used experimental vapor pressure data to show that $h$ has to be the same as Planck's constant (see my answer to this question). Do you know how these two perspectives can be reconciled? $\endgroup$ – user8153 Sep 10 '18 at 4:55

You can make the analogy with quantum field theory where due to a lack of a well defined microscopic theory, you are forced to regularize the theory and hope to be able to eliminate the regularization parameters by absorbing them in a few physical quantities (mass, charge etc.).

Now, if you start with the correct partition function that takes into account the correct statistics (Fermi Dirac of Bose Einstein), you end up with the Maxwell Boltzmann statistics in the dilute limit, where you take overcounting into account by the N! in the denominator. This is a universal feature in the dilute limit, because the probability of two particles occupying the same state vanishes in this limit.

So, you could say that Gibbs was able to succeed because there exists a universal theory in the dilute scaling limit. You then only have one scaling parameter left (h). In terms of h some quantities like entropy are infinite, but as the ACuriousMind points out in his answer, entropy differences will be independent of h. Now this is not true if you consider changes where in one case you have effectively frozen degrees of freedom (e.g. atoms in a solid) and in the other case you don't (e.g. a gas).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.