Bose–Einstein statistics

Suppose we have a number of energy levels, labeled by index ${\displaystyle \displaystyle i}$, each level having energy ${\displaystyle \displaystyle \varepsilon _{i}}$ and containing a total of ${\displaystyle \displaystyle n_{i}}$ particles. Suppose each level contains ${\displaystyle \displaystyle g_{i}}$ distinct sublevels, all of which have the same energy, and which are distinguishable. For example, two particles may have different momenta, in which case they are distinguishable from each other, yet they can still have the same energy. The value of ${\displaystyle \displaystyle g_{i}}$ associated with level ${\displaystyle \displaystyle i}$ is called the "degeneracy" of that energy level. Any number of bosons can occupy the same sublevel.

Let ${\displaystyle \displaystyle w(n,g)}$ be the number of ways of distributing ${\displaystyle \displaystyle n}$ particles among the ${\displaystyle \displaystyle g}$ sublevels of an energy level. There is only one way of distributing ${\displaystyle \displaystyle n}$ particles with one sublevel, therefore ${\displaystyle \displaystyle w(n,1)=1}$. It is easy to see that there are ${\displaystyle \displaystyle (n+1)}$ ways of distributing ${\displaystyle \displaystyle n}$ particles in two sublevels which we will write as:

${\displaystyle w(n,2)={\frac {(n+1)!}{n!1!}}.}$

With a little thought (see Notes below) it can be seen that the number of ways of distributing ${\displaystyle \displaystyle n}$ particles in three sublevels is

${\displaystyle w(n,3)=w(n,2)+w(n-1,2)+\cdots +w(1,2)+w(0,2)}$

so that

${\displaystyle w(n,3)=\sum _{k=0}^{n}w(n-k,2)=\sum _{k=0}^{n}{\frac {(n-k+1)!}{(n-k)!1!}}={\frac {(n+2)!}{n!2!}}}$

where we have used the following theorem involving binomial coefficients:

${\displaystyle \sum _{k=0}^{n}{\frac {(k+a)!}{k!a!}}={\frac {(n+a+1)!}{n!(a+1)!}}.}$

Continuing this process, we can see that ${\displaystyle \displaystyle w(n,g)}$ is just a binomial coefficient (See Notes below)

${\displaystyle w(n,g)={\frac {(n+g-1)!}{n!(g-1)!}}.}$

For example, the population numbers for two particles in three sublevels are 200, 110, 101, 020, 011, or 002 for a total of six which equals 4!/(2!2!). The number of ways that a set of occupation numbers ${\displaystyle \displaystyle n_{i}}$ can be realized is the product of the ways that each individual energy level can be populated:

${\displaystyle W=\prod _{i}w(n_{i},g_{i})=\prod _{i}{\frac {(n_{i}+g_{i}-1)!}{n_{i}!(g_{i}-1)!}}\approx \prod _{i}{\frac {(n_{i}+g_{i})!}{n_{i}!(g_{i})!}}}$

where the approximation assumes that ${\displaystyle n_{i}\gg 1}$.

Following the same procedure used in deriving the Maxwell–Boltzmann statistics, we wish to find the set of ${\displaystyle \displaystyle n_{i}}$ for which W is maximised, subject to the constraint that there be a fixed total number of particles, and a fixed total energy. The maxima of ${\displaystyle \displaystyle W}$ and ${\displaystyle \displaystyle \ln(W)}$ occur at the same value of ${\displaystyle \displaystyle n_{i}}$ and, since it is easier to accomplish mathematically, we will maximise the latter function instead. We constrain our solution using Lagrange multipliers forming the function:

${\displaystyle f(n_{i})=\ln(W)+\alpha (N-\sum n_{i})+\beta (E-\sum n_{i}\varepsilon _{i})}$

Using the ${\displaystyle n_{i}\gg 1}$ approximation and using Stirling's approximation for the factorials ${\displaystyle \left(x!\approx x^{x}\,e^{-x}\,{\sqrt {2\pi x}}\right)}$ gives

${\displaystyle f(n_{i})=\sum _{i}(n_{i}+g_{i})\ln(n_{i}+g_{i})-n_{i}\ln(n_{i})+\alpha \left(N-\sum n_{i}\right)+\beta \left(E-\sum n_{i}\varepsilon _{i}\right)+K.}$

Where K is the sum of a number of terms which are not functions of the ${\displaystyle n_{i}}$. Taking the derivative with respect to ${\displaystyle \displaystyle n_{i}}$, and setting the result to zero and solving for ${\displaystyle \displaystyle n_{i}}$, yields the Bose–Einstein population numbers:

${\displaystyle n_{i}={\frac {g_{i}}{e^{\alpha +\beta \varepsilon _{i}}-1}}.}$

By a process similar to that outlined in the Maxwell–Boltzmann statistics article, it can be seen that:

${\displaystyle d\ln W=\alpha \,dN+\beta \,dE}$

which, using Boltzmann's famous relationship ${\displaystyle S=k_{\text{B}}\,\ln W}$ becomes a statement of the second law of thermodynamics at constant volume, and it follows that ${\displaystyle \beta ={\frac {1}{k_{\text{B}}T}}}$ and ${\displaystyle \alpha =-{\frac {\mu }{k_{\text{B}}T}}}$ where S is the entropy, ${\displaystyle \mu }$ is the chemical potential, k_B is Boltzmann's constant and T is the temperature, so that finally:

${\displaystyle n_{i}={\frac {g_{i}}{e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}-1}}.}$

Note that the above formula is sometimes written:

${\displaystyle n_{i}={\frac {g_{i}}{e^{\varepsilon _{i}/k_{\text{B}}T}/z-1}},}$

where ${\displaystyle \displaystyle z=\exp(\mu /k_{\text{B}}T)}$ is the absolute activity, as noted by McQuarrie.[12]

Also note that when the particle numbers are not conserved, removing the conservation of particle numbers constraint is equivalent to setting ${\displaystyle \alpha }$ and therefore the chemical potential ${\displaystyle \mu }$ to zero. This will be the case for photons and massive particles in mutual equilibrium and the resulting distribution will be the Planck distribution.

A much simpler way to think of Bose–Einstein distribution function is to consider that n particles are denoted by identical balls and g shells are marked by g-1 line partitions. It is clear that the permutations of these n balls and g − 1 partitions will give different ways of arranging bosons in different energy levels. Say, for 3 (= n) particles and 3 (= g) shells, therefore (g − 1) = 2, the arrangement might be |●●|●, or ||●●●, or |●|●● , etc. Hence the number of distinct permutations of n + (g-1) objects which have n identical items and (g − 1) identical items will be:

${\displaystyle {\frac {(g-1+n)!}{(g-1)!n!}}}$

OR

The purpose of these notes is to clarify some aspects of the derivation of the Bose–Einstein (B–E) distribution for beginners. The enumeration of cases (or ways) in the B–E distribution can be recast as follows. Consider a game of dice throwing in which there are ${\displaystyle \displaystyle n}$ dice, with each die taking values in the set ${\displaystyle \displaystyle \{1,\dots ,g\}}$, for ${\displaystyle g\geq 1}$. The constraints of the game are that the value of a die ${\displaystyle \displaystyle i}$, denoted by ${\displaystyle \displaystyle m_{i}}$, has to be greater than or equal to the value of die ${\displaystyle \displaystyle (i-1)}$, denoted by ${\displaystyle \displaystyle m_{i-1}}$, in the previous throw, i.e., ${\displaystyle m_{i}\geq m_{i-1}}$. Thus a valid sequence of die throws can be described by an n-tuple ${\displaystyle \displaystyle (m_{1},m_{2},\dots ,m_{n})}$, such that ${\displaystyle m_{i}\geq m_{i-1}}$. Let ${\displaystyle \displaystyle S(n,g)}$ denote the set of these valid n-tuples:

${\displaystyle S(n,g)={\Big \{}(m_{1},m_{2},\dots ,m_{n}){\Big |}{\Big .}m_{i}\geq m_{i-1},m_{i}\in \left\{1,\ldots ,g\right\},\forall i=1,\dots ,n{\Big \}}.}$

(1)

Then the quantity ${\displaystyle \displaystyle w(n,g)}$ (defined above as the number of ways to distribute ${\displaystyle \displaystyle n}$ particles among the ${\displaystyle \displaystyle g}$ sublevels of an energy level) is the cardinality of ${\displaystyle \displaystyle S(n,g)}$, i.e., the number of elements (or valid n-tuples) in ${\displaystyle \displaystyle S(n,g)}$. Thus the problem of finding an expression for ${\displaystyle \displaystyle w(n,g)}$ becomes the problem of counting the elements in ${\displaystyle \displaystyle S(n,g)}$.

Example n = 4, g = 3:

${\displaystyle S(4,3)=\left\{\underbrace {(1111),(1112),(1113)} _{(a)},\underbrace {(1122),(1123),(1133)} _{(b)},\underbrace {(1222),(1223),(1233),(1333)} _{(c)},\right.}$

${\displaystyle \left.\underbrace {(2222),(2223),(2233),(2333),(3333)} _{(d)}\right\}}$

${\displaystyle \displaystyle w(4,3)=15}$ (there are ${\displaystyle \displaystyle 15}$ elements in ${\displaystyle \displaystyle S(4,3)}$)

Subset ${\displaystyle \displaystyle (a)}$ is obtained by fixing all indices ${\displaystyle \displaystyle m_{i}}$ to ${\displaystyle \displaystyle 1}$, except for the last index, ${\displaystyle \displaystyle m_{n}}$, which is incremented from ${\displaystyle \displaystyle 1}$ to ${\displaystyle \displaystyle g=3}$. Subset ${\displaystyle \displaystyle (b)}$ is obtained by fixing ${\displaystyle \displaystyle m_{1}=m_{2}=1}$, and incrementing ${\displaystyle \displaystyle m_{3}}$ from ${\displaystyle \displaystyle 2}$ to ${\displaystyle \displaystyle g=3}$. Due to the constraint ${\displaystyle \displaystyle m_{i}\geq m_{i-1}}$ on the indices in ${\displaystyle \displaystyle S(n,g)}$, the index ${\displaystyle \displaystyle m_{4}}$ must automatically take values in ${\displaystyle \displaystyle \left\{2,3\right\}}$. The construction of subsets ${\displaystyle \displaystyle (c)}$ and ${\displaystyle \displaystyle (d)}$ follows in the same manner.

Each element of ${\displaystyle \displaystyle S(4,3)}$ can be thought of as a multiset of cardinality ${\displaystyle \displaystyle n=4}$; the elements of such multiset are taken from the set ${\displaystyle \displaystyle \left\{1,2,3\right\}}$ of cardinality ${\displaystyle \displaystyle g=3}$, and the number of such multisets is the multiset coefficient

${\displaystyle \displaystyle \left\langle {\begin{matrix}3\\4\end{matrix}}\right\rangle ={3+4-1 \choose 3-1}={3+4-1 \choose 4}={\frac {6!}{4!2!}}=15}$

More generally, each element of ${\displaystyle \displaystyle S(n,g)}$ is a multiset of cardinality ${\displaystyle \displaystyle n}$ (number of dice) with elements taken from the set ${\displaystyle \displaystyle \left\{1,\dots ,g\right\}}$ of cardinality ${\displaystyle \displaystyle g}$ (number of possible values of each die), and the number of such multisets, i.e., ${\displaystyle \displaystyle w(n,g)}$ is the multiset coefficient

${\displaystyle \displaystyle w(n,g)=\left\langle {\begin{matrix}g\\n\end{matrix}}\right\rangle ={g+n-1 \choose g-1}={g+n-1 \choose n}={\frac {(g+n-1)!}{n!(g-1)!}}}$

(2)

which is exactly the same as the formula for ${\displaystyle \displaystyle w(n,g)}$, as derived above with the aid of a theorem involving binomial coefficients, namely

${\displaystyle \sum _{k=0}^{n}{\frac {(k+a)!}{k!a!}}={\frac {(n+a+1)!}{n!(a+1)!}}.}$

(3)

To understand the decomposition

${\displaystyle \displaystyle w(n,g)=\sum _{k=0}^{n}w(n-k,g-1)=w(n,g-1)+w(n-1,g-1)+\cdots +w(1,g-1)+w(0,g-1)}$

(4)

or for example, ${\displaystyle \displaystyle n=4}$ and ${\displaystyle \displaystyle g=3}$

${\displaystyle \displaystyle w(4,3)=w(4,2)+w(3,2)+w(2,2)+w(1,2)+w(0,2),}$

let us rearrange the elements of ${\displaystyle \displaystyle S(4,3)}$ as follows

${\displaystyle S(4,3)=\left\{\underbrace {(1111),(1112),(1122),(1222),(2222)} _{(\alpha )},\underbrace {(111{\color {Red}{\underset {=}{3}}}),(112{\color {Red}{\underset {=}{3}}}),(122{\color {Red}{\underset {=}{3}}}),(222{\color {Red}{\underset {=}{3}}})} _{(\beta )},\right.}$

${\displaystyle \left.\underbrace {(11{\color {Red}{\underset {==}{33}}}),(12{\color {Red}{\underset {==}{33}}}),(22{\color {Red}{\underset {==}{33}}})} _{(\gamma )},\underbrace {(1{\color {Red}{\underset {===}{333}}}),(2{\color {Red}{\underset {===}{333}}})} _{(\delta )}\underbrace {({\color {Red}{\underset {====}{3333}}})} _{(\omega )}\right\}.}$

Clearly, the subset ${\displaystyle \displaystyle (\alpha )}$ of ${\displaystyle \displaystyle S(4,3)}$ is the same as the set

${\displaystyle \displaystyle S(4,2)=\left\{(1111),(1112),(1122),(1222),(2222)\right\}}$.

By deleting the index ${\displaystyle \displaystyle m_{4}=3}$ (shown in red with double underline) in the subset ${\displaystyle \displaystyle (\beta )}$ of ${\displaystyle \displaystyle S(4,3)}$, one obtains the set

${\displaystyle \displaystyle S(3,2)=\left\{(111),(112),(122),(222)\right\}}$.

In other words, there is a one-to-one correspondence between the subset ${\displaystyle \displaystyle (\beta )}$ of ${\displaystyle \displaystyle S(4,3)}$ and the set ${\displaystyle \displaystyle S(3,2)}$. We write

${\displaystyle \displaystyle (\beta )\longleftrightarrow S(3,2)}$.

Similarly, it is easy to see that

${\displaystyle \displaystyle (\gamma )\longleftrightarrow S(2,2)=\left\{(11),(12),(22)\right\}}$

${\displaystyle \displaystyle (\delta )\longleftrightarrow S(1,2)=\left\{(1),(2)\right\}}$

${\displaystyle \displaystyle (\omega )\longleftrightarrow S(0,2)=\varnothing }$ (empty set).

Thus we can write

${\displaystyle \displaystyle S(4,3)=\bigcup _{k=0}^{4}S(4-k,2)}$

or more generally,

${\displaystyle \displaystyle S(n,g)=\bigcup _{k=0}^{n}S(n-k,g-1)}$;

(5)

and since the sets

${\displaystyle S(i,g-1),{\text{ for }}i=0,\dots ,n}$

are non-intersecting, we thus have

${\displaystyle \displaystyle w(n,g)=\sum _{k=0}^{n}w(n-k,g-1)}$,

(6)

with the convention that

${\displaystyle w(0,g)=1\ ,\forall g,{\text{ and }}w(n,0)=1\ ,\forall n.}$

(7)

Continuing the process, we arrive at the following formula

${\displaystyle w(n,g)=\sum _{k_{1}=0}^{n}\sum _{k_{2}=0}^{n-k_{1}}w(n-k_{1}-k_{2},g-2)=\sum _{k_{1}=0}^{n}\sum _{k_{2}=0}^{n-k_{1}}\cdots \sum _{k_{g}=0}^{n-\sum _{j=1}^{g-1}k_{j}}w(n-\sum _{i=1}^{g}k_{i},0).}$

Using the convention (7)₂ above, we obtain the formula

${\displaystyle \displaystyle w(n,g)=\sum _{k_{1}=0}^{n}\sum _{k_{2}=0}^{n-k_{1}}\cdots \sum _{k_{g}=0}^{n-\sum _{j=1}^{g-1}k_{j}}1,}$

(8)

keeping in mind that for ${\displaystyle \displaystyle q}$ and ${\displaystyle \displaystyle p}$ being constants, we have

${\displaystyle \displaystyle \sum _{k=0}^{q}p=qp}$.

(9)

It can then be verified that (8) and (2) give the same result for ${\displaystyle \displaystyle w(4,3)}$, ${\displaystyle \displaystyle w(3,3)}$, ${\displaystyle \displaystyle w(3,2)}$, etc.