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When Bose gas it cooled below critical temperature some of it condenses into Bose-Einstein condensate, resulting in seemingly infinite occupation of 0th state because $\mu = 0$. In reality, the 0th state should be excluded from the calculation and then the number of particles in states of energy >0 is given by an appropriate integral (Landau and Lifshitz, chapter 56). But what if I want to find how many of them are in the first excited state? The Bose-Einstein Statistic, with $\mu = 0$, is $$ n_i = \frac{1}{e^{\beta\epsilon_i}-1} $$ So do I just use it for $i=1$ to find the mean number of particles in the first excited state? But then I need to know what $\epsilon_0$ is. Or do I do the same integral as in Landau but with upper limit of $\epsilon_0$? I know the anwser is supposed to be proportional to $N^{2/3}$

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So I think I have managed to figure it out in the end. So we have Bose-Einstein Distribution: $$ n_i = \frac{1}{e^{\beta(e_i-\mu)}-1}$$ Now, going off an assumption that the gas is degenerate and that chemical potential is zero, this becomes" $$ n_i = \frac{1}{e^{\beta e}-1}$$ Now $\beta = \frac{1}{k_bT}$ I will make another assumption that $T= T_{crtitical} \propto N^{2/3} $. Expanding the exponential as a Taylor series, get $$n_i \propto N^{2/3}$$ I guess this is a very rough and not very rigorous, but this gives the right idea about occupation of first excited level.

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  • $\begingroup$ So is your prediction that $\sim N^{2/3}$ of the particles, i.e. most of them, are in the first excited state? I don't think this is the right idea. In the thermodynamic limit, the number of particles in any individual state (except the ground state) is going to tend to zero. $\endgroup$ – Mark Mitchison May 10 '16 at 19:31
  • $\begingroup$ @MarkMitchison I think the point is that I am assuming a degenerate gas near critical temperature. I imagine there should be a way to generalise this result to give occupation number of the first state for any temperature, which should give the right limit for zero temperature... I will give it a little bit more thought $\endgroup$ – Ilya Lapan May 10 '16 at 20:05

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