I have a question about Bose-Einstein condensation. Namely, people say that if we go from the summation over the number of particles to an integral using the density of states, we make a flaw in the calculation if the temperature is below the critical temperature and therefore we write:

$N = N_0 + N_{ex}$.

Where $N$ is the total number of particles, $N_0$ is the number of particles in the ground-state and $N_{ex}$ is the number of particles in the excited states (given by the integral that contains the density of states).

Now my question is: why do some references like Yoshioka statistical physics say that we miss the particles in the ground state since $D(\epsilon)=0$ for the density of states while this expression is inside an integral? Could somebody give a more rigorous proof or reference for this?


1 Answer 1


If for instance $D(\epsilon)\propto \sqrt{\epsilon}$ then $\int_0^{\Delta\epsilon} d\epsilon D(\epsilon)\approx0$ for any small $\Delta\epsilon$. But what you have to keep in mind is that the proper expression is \begin{equation} N= \sum_{i=1}^\infty n_i \neq \int_0^\infty d\epsilon D(\epsilon) \end{equation} Approximating the sum by the integral does not hold if $n_1$ is $O(N)$, because the density assigns no weight to the ground state (at $\epsilon=0$). If you count the occupations in the interval $[0,\Delta \epsilon]$ discretely you always have $n_1=O(N)$in the sum, no matter how small $\Delta \epsilon$ is. But using the integral on the right you get $\int_0^{\Delta\epsilon} d\epsilon D(\epsilon)\propto \Delta \epsilon^{3/2}\rightarrow0$ as $\epsilon\rightarrow0$.

  • $\begingroup$ How do you define $\Delta \epsilon$ 'small'? Does this mean that it is between the ground state and the first excited state? Thanks already for the answer. $\endgroup$ Sep 13, 2018 at 12:49
  • $\begingroup$ @dani for approximating the sum by a continuous density, you have to assume that there are many energy levels within the interval $\Delta\epsilon$. Otherwise $D(\epsilon)$ would never be smooth. I updated the answer. $\endgroup$
    – jkds
    Sep 14, 2018 at 7:01

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