# What is the probability that a single-particle bosonic quantum state is occupied?

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 $\bar n_r$?

• $P(E)=Z^{-1}\exp(-\beta E)$? – AccidentalFourierTransform Jun 9 '18 at 18:02
• @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? – SRS Jun 9 '18 at 18:17
• – AccidentalFourierTransform Jun 9 '18 at 18:19
• 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 – Aleksey Druggist Jun 10 '18 at 7:09
• or of course $p(n_r=0)=\frac{1}{1+\bar n_r}$ – Aleksey Druggist Jun 10 '18 at 7:43

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 $$p_0 = \frac{1}{1 + \langle n \rangle}.$$
• As far as I can see the question was to express the probability through $\bar n_r$ – Aleksey Druggist Jun 13 '18 at 16:21
• It is interesting that in the case of fermions, the result is completely different: $p_0=1-<n>$ – Aleksey Druggist Jun 13 '18 at 16:47
• @AlekseyDruggist Yeah, but also, note that the two match at first order in $\langle n \rangle$, because that is the classical limit! – knzhou Jun 13 '18 at 16:49
• 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 – Aleksey Druggist Jun 19 '18 at 8:37