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A different BE statitics!

  1. May 21, 2013 #1

    ShayanJ

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    Gold Member

    I want to calculate the number of ways for putting [itex] N_1,N_2,N_3,...,N_i,... [/itex] bosons in energy levels with degeneracies [itex] g_1,g_2,g_3,...,g_i,... [/itex].
    The particles are indistinguishable and there can be any number of particles in a state.
    The level with degeneracy [itex] g_i [/itex] has [itex] N_i [/itex] particles in it.The first of this [itex] N_i [/itex] particles has [itex] g_i [/itex] states to choose.The second,again has [itex] g_i [/itex] choices and the same for all of them.So there are [itex] g_i^{N_i} [/itex] ways for putting [itex] N_i [/itex] particles in [itex] g_i [/itex] sates.But the particles are indistinguishable so their order is not important and so [itex] g_i^{N_i} [/itex] reduces to [itex] \frac{g_i^{N_i}}{N_i!} [/itex]. So the number of ways for putting [itex] N_1,N_2,N_3,...,N_i,... [/itex] bosons in energy levels with degeneracies [itex] g_1,g_2,g_3,...,g_i,... [/itex] is:
    [itex]
    \prod_i \frac{g_i^{N_i}}{N_i!}
    [/itex]

    But the above result will give us sth like Boltzmann distribution,not Bose-Einstein's and we know that the answer should be like below:
    [itex]
    \prod_i \frac{(N_i+g_i-1)!}{N_i!(g_i-1)!}
    [/itex]

    But what is wrong?
    In deriving my formula,I assumed only that the particles are indistinguishable and don't follow Pauli's principle,the same assumptions made for bosons.So what was different?
    Thanks
     
    Last edited: May 21, 2013
  2. jcsd
  3. May 21, 2013 #2

    DrDu

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    Science Advisor

    Dividing by N_i! you are overcounting states. Take just one state with degeneracy g_1=1 and two bosons.
    Obviously dividing by 2! is not correct.
     
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