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I Partition Function and Degeneracy

  1. Dec 15, 2016 #1

    stevendaryl

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    There doesn't seem to be a forum that is specifically about statistical mechanics, so I'm posting this question here. I apologize for the long-winded introduction, but I think it's needed to provide context for my question:

    If you have a discrete collection of single-particle energy levels [itex]\epsilon_i[/itex], then the grand canonical partition function (for noninteracting particles) is defined by:

    [itex]\mathcal{Z}(\mu, \beta) = \sum_i \sum_{N_i} exp(N_i \beta(\mu - \epsilon_i))[/itex]
    [itex]= \sum_i \sum_{N_i} exp(\beta(\mu - \epsilon_i))^{N_i} [/itex]

    where [itex]N_i[/itex] is the occupancy number: the number of particles in state [itex]i[/itex]. To get Bose statistics, the allowable values for [itex]N_i[/itex] are [itex]N_i = 0, 1, 2, ...[/itex], leading to

    [itex]\mathcal{Z}(\mu, \beta) = \sum_i \dfrac{1}{1 - exp(\beta(\mu - \epsilon_i))}[/itex]

    (because [itex]1+x+x^2 + ... = \frac{1}{1-x}[/itex])

    For Fermi statistics, the only possible values for [itex]N_i[/itex] are [itex]N_i = 0, 1[/itex], leading to:

    [itex]\mathcal{Z}(\mu, \beta) = \sum_i (1 + exp(\beta(\mu - \epsilon_i)))[/itex]

    Here's the question: Suppose that the energy levels for an electron are independent of spin direction. That means that for every single-particle state, there is a second state with the same energy and opposite spin state. Then it seems to me that there are two different ways to take this degeneracy into account:

    (1) Replace [itex]\sum_i ...[/itex] by [itex]\sum_i g_i ...[/itex], where [itex]g_i[/itex] is the degeneracy of energy level [itex]i[/itex], and where the index [itex]i[/itex] ranges only over states with distinct energies. In this case, [itex]g_i = 2[/itex], so the result is just to multiple [itex]\mathcal{Z}[/itex] by 2.

    (2) Modify the allowable occupancy number [itex]N_i[/itex] to range from [itex]0[/itex] to [itex]g_i[/itex], rather than just 0 or 1.

    These two approaches give different answers, but I don't understand, physically, why. It seems that the only thing that should be important is how many electrons can have energy [itex]\epsilon_i[/itex].
     
  2. jcsd
  3. Dec 15, 2016 #2

    mfb

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    The second approach would require the states with the same number of electrons in such a state to be indistinguishable, but "electron up" and "electron down" are different things.
     
  4. Dec 16, 2016 #3

    stevendaryl

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    Thank you.
     
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