Consider a physical system of r indistinguishable particles.…
Consider a physical system of $r$ indistinguishable particles. In statistical mechanics the phase space is divided into a large number $n$ of cells, and each particle falls into one of the cells. The state of the system is then a distribution of the $r$ particles among the $n$ cells, uniquely determined by the numbers $0\le m_i\le r$ $(i=1,\dots,n)$, where $m_i$ is the number of particles in cell $i$. Photons, atomic nuclei and atoms containing an even number of elementary particles obey Bose–Einstein statistics, in which only the different distributions are considered and each is assigned the same probability. Find this probability.
Explanation of Method
Since all distributions are equally likely, each has probability $1/(\text{number of distributions})$. A distribution is a solution of $m_1+m_2+\dots+m_n=r$ in non-negative integers. We count these with the stars and bars method: write the $r$ particles as stars and separate the $n$ cells by $n-1$ bars; every arrangement of stars and bars is one distribution.
Step 1: What is a distribution?
Because the particles are indistinguishable, a state is described only by how many particles are in each cell:
\[ (m_1, m_2, \dots, m_n), \qquad m_i\ge0, \qquad m_1+m_2+\dots+m_n = r. \]
Step 2: Stars and bars
Write each particle as a star $\ast$ and put $n-1$ bars $|$ between the cells. For example, with $r=4$ particles and $n=4$ cells:
| Picture | Distribution $(m_1,m_2,m_3,m_4)$ |
|---|---|
| $\ast\ast\,|\,\ast\,|\,|\,\ast$ | $(2,1,0,1)$ |
| $|\,\ast\ast\ast\ast\,|\,|$ | $(0,4,0,0)$ |
| $\ast\,|\,\ast\,|\,\ast\,|\,\ast$ | $(1,1,1,1)$ |
Every distribution gives exactly one row of $r$ stars and $n-1$ bars, and every such row gives a distribution. A row has $r+n-1$ symbols; it is determined by choosing which $r$ of the positions are stars:
\[ \text{number of distributions} = \binom{n+r-1}{r} = \binom{n+r-1}{n-1}. \]
Step 3: The probability
All distributions are equally likely, so each has probability
\[ \boxed{P(m_1,\dots,m_n) = \frac{1}{\binom{n+r-1}{r}} = \frac{r!\,(n-1)!}{(n+r-1)!}.} \]
Step 4: Small example — $r=2$ particles, $n=3$ cells
The distributions are $(2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1)$ — six of them, and $\binom{4}{2}=6$. ✓ In Bose–Einstein statistics each has probability $\frac16$.
Step 5: Comparison with the other statistics
| Statistics | Particles | Equally likely are | Count ($r=2$, $n=3$) | P(both in cell 1) |
|---|---|---|---|---|
| Maxwell–Boltzmann | distinguishable | all $n^r$ placements | $3^2=9$ | $1/9$ |
| Bose–Einstein | indistinguishable | all $\binom{n+r-1}{r}$ distributions | $6$ | $1/6$ |
| Fermi–Dirac | indistinguishable, at most one per cell | all $\binom nr$ distributions | $3$ | $0$ |
Bose–Einstein statistics gives crowded cells a larger probability than classical (Maxwell–Boltzmann) counting: bosons tend to "bunch together".
Result
| Answer | |
|---|---|
| Number of distributions | $\binom{n+r-1}{r}$ (stars and bars) |
| Probability of each distribution | $\dfrac{1}{\binom{n+r-1}{r}}=\dfrac{r!\,(n-1)!}{(n+r-1)!}$ |