When calculating the entropy of an classical gas we consider the Gibbs correction factor, when dealing with quantum gases we consider the appropriate Fermi/Bose statistics.

But in my introductory notes to statistical mechanics one calculates the canonical partition sum of $N$ harmonic oscillators without such corrections. Isn't this assuming 1) we can stick a label on each and every oscillator 2) their wave functions are not correlated (large distance between oscillators)? Does this make sense?


1 Answer 1

  1. when you multiply the partition function of each harmonic oscillator to get the partition function of $N$ such harmonic oscillators (or equivalently sum the free energy of each individual harmonic oscillator) you already, implicitly, included the Gibbs correction term. The point of the Gibbs correction is exactly to get extensive quantities to behave properly, which they do when you add up the free energy. That is - you already treat the extensive properties as extensive. If you wanted to treat the Gibbs correction explicitly you have to start with $N$ independent harmonic oscillators and consider all the ways the energy can be distributed between them and then you will get a different result, which will converge to the correct one (i.e. the one you get when you add up the individual free energies) when you explicitly put in the Gibbs correction factor.

  2. Their wave function is not correlated because in this scenario they are not interacting. If you'll put in interactions - for example consider them to have electric charge - you will get an extra term in the Hamiltonian, and the calculation of the partition function can be complicated. Notice that if the interactions are long-ranged you might not get a proper ensemble (that is - extensive properties might not behave properly). Usually in condensed matter when we consider interactions we consider short-range ones. This is enough, however, to get very different behavior than the noninteracting setup in certain cases.

  • $\begingroup$ I really don't get your first point @yu-v. Isn't this what I'm doing? I assume the energy spectrum $H=\sum_i(n_i+1/2)\omega$ and then take the partition sum $\sum_{\{n_1..n_N\}}\exp(-\beta H)$. How is this different from the ideal gas, in which case I need the gibbs factor? $\endgroup$
    – user224659
    Commented Jun 18, 2020 at 13:21
  • $\begingroup$ how do you use the Gibbs factor in an ideal gas? $\endgroup$
    – user245141
    Commented Jun 18, 2020 at 13:33
  • $\begingroup$ I think ths paper answers my questions lajpe.org/dec14/4302_Udayanandan.pdf thanks though! $\endgroup$
    – user224659
    Commented Jun 18, 2020 at 13:35
  • $\begingroup$ Abstract: Boltzmann Correction Factor (BCF) N! is used in micro canonical ensemble and canonical ensemble as a dividing term to avoid Gibbs paradox while finding the number of states and partition function for ideal gas. For harmonic oscillators this factor does not come since they are considered to be distinguishable. We here show that BCF comes twice for harmonic oscillators in grand canonical ensemble for entropy to be extensive in classical statistics. Then we extent this observation for all distinguishable systems. $\endgroup$
    – user224659
    Commented Jun 18, 2020 at 13:36
  • $\begingroup$ The oscillators are distinguishable. $\endgroup$
    – mike stone
    Commented Apr 28 at 0:14

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