A different BE statitics

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In summary, the conversation discusses the calculation of the number of ways to arrange N_i particles in energy levels with degeneracies g_i, assuming the particles are indistinguishable. The formula for this is \prod_i \frac{g_i^{N_i}}{N_i!}. However, this result is different from the expected result for Bose-Einstein statistics, which is \prod_i \frac{(N_i+g_i-1)!}{N_i!(g_i-1)!}. This is due to overcounting states when dividing by N_i!.
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ShayanJ
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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
 
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  • #2
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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