Unlike the Fermi-Dirac distribution function, the Bose-Einstein distribution function $$f(E)=\bar n_r=\frac{1}{e^{\beta(E-\mu)}-1}$$ can be greater than 1, and therefore, doesn't represent a probability. It represents the average number of particles $\bar n_r$ in a single-particle quantum state $r$. What is the expression for a single-particle quantum state $r$ being occupied or unoccupied? Can we related that to $f(E)$ or $\bar n_r$?

  • $\begingroup$ $P(E)=Z^{-1}\exp(-\beta E)$? $\endgroup$ Jun 9, 2018 at 18:02
  • $\begingroup$ @AccidentalFourierTransform Is there a way the mean number of particles in a given single-particle quantum state be related to the probability of occupation of that state? $\endgroup$
    – SRS
    Jun 9, 2018 at 18:17
  • $\begingroup$ Unoccupied: $p(n_r=0)$ and occupied: $p(n_r\neq0)=1-p(n_r=0)$, $p(n_r=0)=1/Z_r$, $Z_r=\frac{1}{1-e^{-\beta(E-\mu)}}$ (sum of geometric progression), so $p(n_r=0)=\frac{e^{-\beta(E-\mu)}}{\bar n_r}$ - something like this $\endgroup$ Jun 10, 2018 at 7:09
  • $\begingroup$ or of course $p(n_r=0)=\frac{1}{1+\bar n_r}$ $\endgroup$ Jun 10, 2018 at 7:43

1 Answer 1


Let $x = e^{-\beta (E - \mu)}$. Essentially by definition, the probability $p_n$ of having occupancy number $n$ $$p_n = \frac{x^n}{Z}$$ where the probability distribution is normalized by the partition function $$Z = 1 + x + x^2 + \ldots = \frac{1}{1-x}.$$ Then the probability that the occupancy number is zero is $$p_0 = \frac{x^0}{Z} = 1 - x.$$ The probability that the occupancy number is nonzero is $x$. The average occupancy is $$\langle n \rangle = \sum_n n p_n = (1-x) \sum_n n x^n = \frac{1}{x^{-1} - 1}$$ which is the result you quoted. In particular, $p_0$ and $\langle n \rangle$ are related by $$p_0 = \frac{1}{1 + \langle n \rangle}.$$

  • $\begingroup$ As far as I can see the question was to express the probability through $\bar n_r$ $\endgroup$ Jun 13, 2018 at 16:21
  • $\begingroup$ @AlekseyDruggist You're right, I put that in. $\endgroup$
    – knzhou
    Jun 13, 2018 at 16:37
  • $\begingroup$ It is interesting that in the case of fermions, the result is completely different: $p_0=1-<n>$ $\endgroup$ Jun 13, 2018 at 16:47
  • $\begingroup$ @AlekseyDruggist Yeah, but also, note that the two match at first order in $\langle n \rangle$, because that is the classical limit! $\endgroup$
    – knzhou
    Jun 13, 2018 at 16:49
  • $\begingroup$ By the way, the variable $x$ is more suited for the case of parafermions of the order $k$: $p_0=\frac{x-1}{x^{k} - 1}$, to express the probability in terms of $<n>$ would be a problem, I guess $\endgroup$ Jun 19, 2018 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.