Hi I have a doubt (I'm not very expert in statistical mechanics, so sorry for this question). We consider a gas of hydrogen atoms with no interactions between them. The partition function is: $$ Z=\frac{Z_s^N}{N!} $$ where $Z_s$ is the partition function of one atom. So we write $$ Z_s =Tr\{e^{-\beta \hat H}\} $$ and we must consider sum to discrete and continous spectrum. So I'd write for the contribute of the discrete spectrum: $$ Z_{s_{disc}}=\sum_{n=1}^{+\infty}n^2 e^{\beta \frac{E_0}{n^2}} $$ but this serie doesn't converge. For continuum spectrum, I'm not be able to write the contribute to the sum, because I have infinite degeneration, so where's my mistake?

I have thought that the spectrum tha I considered for energy values is for free atom in the space, and partition function maybe is defined for systems with finite volume.

So, this doubt however would me ask a problem. Can whe prove that operator $e^{-\beta \hat H}$ always has got finite trace for all physical systems?

  • $\begingroup$ I remember my stat mech prof. - the incredible Milton Cole - mentioning this problem in class. He said that Fermi has solved this problem by putting the "atom in a box". This box serves as an (infra-red) regulator. $\endgroup$
    – user346
    Feb 2, 2011 at 6:27
  • $\begingroup$ You mean $e^{-\beta H}$? $\endgroup$
    – QGR
    Feb 2, 2011 at 9:50

3 Answers 3


This is quite a subtle problem:

  • Notice that you have the same problem if you have only a single hydrogen atom.
  • You've ignored all the unbound states of the hydrogen atom, which make thing even worse (they are higher in energy, and have even larger degeneracies).
  • Divergent partition function is actually not really a problem --- all physical properties will depend on derivatives of the logarithm of it.
  • However, you will find that at any finite temperature things like entropy (which are physical) does indeed diverge.
  • This is again not so bad if you realise that you've effectively got an infinite system --- the entropy per volume should still be finite.

Concretely, you should regularise things by considering a finite box of volume V, with N particles. This will modify the relevant states (and their energies) in correct (but incredibly hard to compute) ways to render all physical properties finite.

Intuitively the effect is due to the n'th bound state having a radius $O(n^2)$. Thus you can approximate the effects of a finite boundary by simply removing all states above a certain size --- the divergences are so slow that in practice physical quantities will only depend very weakly on the cut-off.


  1. http://scienceblogs.com/builtonfacts/2011/01/the_hydrogen_partition_functio.php
  2. http://pubs.acs.org/doi/abs/10.1021/ed043p364

For a canonical ensemble consisting of a single hydrogen atom in infinite space, the electron will ionize sooner or later. This reduces the free energy. The actual energy will go up, but the infinite increase in entropy available from infinite space will more than compensate for it.

But when we go over to a gas of hydrogen atoms, even a very dilute gas, things change drastically. The atoms are no longer independent of each other; if one atom ionizes, the probability the electron of another atom becomes bound to the original atom goes up.

Interestingly enough, if we start off with a nonzero temperature less than the Bohr temperature, but adjust the density so that the average volume per atom is sufficiently large exponentially, the free energy is actually minimized for a plasma.


The contribution of the continuum states to the partition function is not really a problem. Actually it solves a problem, canceling the divergence on the sum over the discrete spectrum and leaving as result the so called Planck-Larkin partition function

$$Z_{discrete}=\sum_{n=1}^\infty n^2 [\exp (-\beta E_n)-1+\beta E_n]\; , $$

where $E_n=R(1-1/n^2)$ (the ground state is at zero energy) and $R$ is the ionization energy.

A good paper on this subject is F.J.Rogers, The Astrophysical Journal 310, p. 723.

  • $\begingroup$ Hm I don't understand very well...where does that expression come from? And degeneration of continuum states is infinite, how can you calculate Z with this thing? I can't have access at your article, sorry... $\endgroup$
    – Boy S
    Feb 2, 2011 at 13:14
  • 1
    $\begingroup$ @Boy: Sorry, I should have mentioned that the article is freely available at the NASA-ADS site, at dx.doi.org/10.1086/164725 . You will find all the gory details there. $\endgroup$
    – Carlos
    Feb 2, 2011 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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