Derivation from the microcanonical ensemble In the
microcanonical ensemble, one considers a system with fixed energy, volume, and number of particles. We take a system composed of N = \sum_i n_i identical bosons, n_i of which have energy \varepsilon_i and are distributed over g_i levels or states with the same energy \varepsilon_i, i.e. g_i is the degeneracy associated with energy \varepsilon_i. The total energy of the system is E = \sum_i n_i \varepsilon_i. Calculation of the number of arrangements of n_i particles distributed among g_i states is a problem of
combinatorics. Since particles are indistinguishable in the quantum mechanical context here, the number of ways for arranging n_i particles in g_i boxes (for the ith energy level), where each box is capable of containing an infinite number of bosons (because for bosons the
Pauli exclusion principle does not apply), would be (see image): w_{i,\text{BE}} = \frac{(n_i+g_i-1)!}{n_i! (g_i-1)!} = C^{n_i+g_i-1}_{n_i}, where C^m_k is the
k-combination of a set with
m elements (Note also that w_{i,\text{BE}} represents the absolute non-normalized probability of an energy state with n_i bosons and a degeneracy of g_i, it is not the same as the w_i associated with the Gibbs formulation of entropy). The total number of arrangements in an ensemble of bosons is simply the product of the binomial coefficients C^{n_i+g_i-1}_{n_i} above over all the energy levels, i.e. W_\text{BE} =\prod_i w_{i,\text{BE}}=\prod_i\frac{(n_i +g_i -1)!}{(g_i-1)! n_i!}, which for very large n_i and g_i can be simplified using
Stirling's approximation to W_\text{BE} = \prod_i \frac{(\frac{n_i+g_i-1}{e})^{n_i+g_i-1}}{(\frac{g_i-1}{e})^{g_i-1}(\frac{n_i}{e})^{n_i}}. The entropy of the system can then be expressed as S_\text{BE} = k_B\text{ln}W_\text{BE} = k_B \sum_i [(n_i+g_i-1)(\text{ln}(n_i+g_i-1)-1)-(g_i-1)(\text{ln}(g_i-1)-1)-n_i(\text{ln}n_i-1)]. The three constraints we can impose on the system can be expressed as \sum_i \delta n_i = 0 (conservation of N), \sum_i \epsilon_i \delta n_i = 0 (conservation of E), and \delta S_\text{BE} = 0 (
second law of thermodynamics for a system at equilibrium). This final constraint can be expanded to be in terms of n_i : \delta S_\text{BE} = \frac{\partial}{\partial n_i} S_\text{BE} \delta n_i = k_B \sum_i[\text{ln}(n_i+g_i-1)-\text{ln}n_i]\delta n_i = 0. Now we can write \sum_i[\text{ln}(n_i+g_i-1)-\text{ln}n_i]\delta n_i + C\sum_i \delta n_i -\beta \sum_i \epsilon_i \delta n_i = 0, for which to be true, it must be the case that for any i \text{ln}(n_i+g_i-1)-\text{ln}n_i + C -\beta \epsilon_i = 0. By solving for n_i and simplifying we obtain n_i = \frac{g_i-1}{\alpha e^{\beta \epsilon_i} - 1}, which for sufficiently large g_i reduces to n_i = \frac{g_i}{\alpha e^{\beta \epsilon_i} - 1}, which is the form of the Bose-Einstein distribution. Note that this form holds even for a system of interacting bosons.
Derivation from the grand canonical ensemble The Bose–Einstein distribution, which applies only to a quantum system of non-interacting bosons, is naturally derived from the
grand canonical ensemble without any approximations. In this ensemble, the system is able to exchange energy and exchange particles with a reservoir (temperature
T and chemical potential
μ fixed by the reservoir). Due to the non-interacting quality, each available single-particle level (with energy level
ϵ) forms a separate thermodynamic system in contact with the reservoir. That is, the number of particles within the overall system
that occupy a given single particle state form a sub-ensemble that is also grand canonical ensemble; hence, it may be analysed through the construction of a
grand partition function. Every single-particle state is of a fixed energy, \varepsilon. As the sub-ensemble associated with a single-particle state varies by the number of particles only, it is clear that the total energy of the sub-ensemble is also directly proportional to the number of particles in the single-particle state; where N is the number of particles, the total energy of the sub-ensemble will then be N\varepsilon. Beginning with the standard expression for a grand partition function and replacing E with N \varepsilon, the grand partition function takes the form \mathcal Z = \sum_N \exp((N\mu - N\varepsilon)/k_\text{B} T) = \sum_N \exp(N(\mu - \varepsilon)/k_\text{B} T) This formula applies to fermionic systems as well as bosonic systems. Fermi–Dirac statistics arises when considering the effect of the
Pauli exclusion principle: whilst the number of fermions occupying the same single-particle state can only be either 1 or 0, the number of bosons occupying a single particle state may be any integer. Thus, the grand partition function for bosons can be considered a
geometric series and may be evaluated as such: \begin{align}\mathcal Z & = \sum_{N=0}^\infty \exp(N(\mu - \varepsilon)/k_\text{B} T) = \sum_{N=0}^\infty [\exp((\mu - \varepsilon)/k_\text{B}T)]^N \\ & = \frac{1}{1 - \exp((\mu - \varepsilon)/k_\text{B} T)}.\end{align} Note that the geometric series is convergent only if e^{(\mu - \varepsilon)/k_\text{B}T}, including the case where \varepsilon = 0. This implies that the chemical potential for the Bose gas must be negative, i.e., \mu, whereas the Fermi gas is allowed to take both positive and negative values for the chemical potential. The average particle number for that single-particle substate is given by \langle N\rangle = k_\text{B} T \frac{1}{\mathcal Z} \left(\frac{\partial \mathcal Z}{\partial \mu}\right)_{V,T} = \frac{1}{\exp((\varepsilon-\mu)/k_\text{B} T)-1} This result applies for each single-particle level and thus forms the Bose–Einstein distribution for the entire state of the system. The
variance in particle number, \sigma_N^2 = \langle N^2 \rangle - \langle N \rangle^2, is: \sigma_N^2 = k_\text{B} T \left(\frac{d\langle N\rangle}{d\mu}\right)_{V,T} = \frac{\exp((\varepsilon-\mu)/k_\text{B} T)}{(\exp((\varepsilon-\mu)/k_\text{B} T)-1)^2} = \langle N\rangle(1 + \langle N\rangle). As a result, for highly occupied states the
standard deviation of the particle number of an energy level is very large, slightly larger than the particle number itself: \sigma_N \approx \langle N\rangle. This large uncertainty is due to the fact that the
probability distribution for the number of bosons in a given energy level is a
geometric distribution; somewhat counterintuitively, the most probable value for
N is always 0. (In contrast,
classical particles have instead a
Poisson distribution in particle number for a given state, with a much smaller uncertainty of \sigma_{N,{\rm classical}} = \sqrt{\langle N\rangle}, and with the most-probable
N value being near \langle N \rangle.)
Derivation in the canonical approach It is also possible to derive approximate Bose–Einstein statistics in the
canonical ensemble. These derivations are lengthy and only yield the above results in the asymptotic limit of a large number of particles. The reason is that the total number of bosons is fixed in the canonical ensemble. The Bose–Einstein distribution in this case can be derived as in most texts by maximization, but the mathematically best derivation is by the
Darwin–Fowler method of mean values as emphasized by Dingle. See also Müller-Kirsten. The fluctuations of the ground state in the condensed region are however markedly different in the canonical and grand-canonical ensembles. Suppose we have a number of energy levels, labeled by index i, each level having energy \varepsilon_i and containing a total of n_i particles. Suppose each level contains 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 g_i associated with level i is called the "degeneracy" of that energy level. Any number of bosons can occupy the same sublevel. Let w(n,g) be the number of ways of distributing n particles among the g sublevels of an energy level. There is only one way of distributing n particles with one sublevel, therefore w(n,1)=1. It is easy to see that there are (n+1) ways of distributing n particles in two sublevels which we will write as: 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 n particles in three sublevels is w(n,3) = w(n,2) + w(n-1,2) + \cdots + w(1,2) + w(0,2) so that 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: \sum_{k=0}^n\frac{(k+a)!}{k!a!}=\frac{(n+a+1)!}{n!(a+1)!}. Continuing this process, we can see that w(n,g) is just a binomial coefficient (See
Notes below) 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 n_i can be realized is the product of the ways that each individual energy level can be populated: 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 n_i \gg 1. Following the same procedure used in deriving the
Maxwell–Boltzmann statistics, we wish to find the set of 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 W and \ln(W) occur at the same value of 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: f(n_i)=\ln(W)+\alpha(N-\sum n_i)+ \beta(E-\sum n_i \varepsilon_i) Using the n_i \gg 1 approximation and using
Stirling's approximation for the factorials \left(x!\approx x^x\,e^{-x}\,\sqrt{2\pi x}\right) gives 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 n_i. Taking the derivative with respect to n_i, and setting the result to zero and solving for n_i, yields the Bose–Einstein population numbers: 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: d\ln W = \alpha\,dN + \beta\,dE which, using Boltzmann's famous relationship S=k_\text{B}\,\ln W becomes a statement of the
second law of thermodynamics at constant volume, and it follows that \beta = \frac{1}{k_\text{B}T} and \alpha = - \frac{\mu}{k_\text{B}T} where
S is the
entropy, \mu is the
chemical potential,
kB is the
Boltzmann constant and
T is the
temperature, so that finally: n_i = \frac{g_i}{e^{(\varepsilon_i-\mu)/k_\text{B}T}-1}. Note that the above formula is sometimes written: n_i = \frac{g_i}{e^{\varepsilon_i/k_\text{B}T}/z-1}, where z=\exp(\mu/k_\text{B}T) is the absolute
activity, as noted by McQuarrie. Also note that when the particle numbers are not conserved, removing the conservation of particle numbers constraint is equivalent to setting \alpha and therefore the chemical potential \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 , the arrangement might be
|●●|●, or
||●●●, or
|●|●● , etc. Hence the number of distinct permutations of objects which have
n identical items and (
g − 1) identical items will be: \frac{(g-1 + n)!}{(g-1)! n!} See the image for a visual representation of one such distribution of
n particles in
g boxes that can be represented as partitions.
OR The purpose of these notes is to clarify some aspects of the derivation of the Bose–Einstein distribution for beginners. The enumeration of cases (or ways) in the Bose–Einstein distribution can be recast as follows. Consider a game of dice throwing in which there are n dice, with each die taking values in the set \{ 1, \dots, g \}, for g \ge 1. The constraints of the game are that the value of a die i, denoted by m_i, has to be
greater than or equal to the value of die (i-1), denoted by m_{i-1}, in the previous throw, i.e., m_i \ge m_{i-1}. Thus a valid sequence of die throws can be described by an
n-tuple ( m_1 , m_2 , \dots , m_n), such that m_i \ge m_{i-1}. Let S(n,g) denote the set of these valid
n-tuples: {{NumBlk|| S(n,g) = \left\{ ( m_1 , m_2 , \dots , m_n ) \, \Big| \, m_i \ge m_{i-1}, m_i \in \left\{ 1, \ldots, g \right\}, \forall i = 1, \dots , n \right\}. |}} Then the quantity w(n,g) (
defined above as the number of ways to distribute n particles among the g sublevels of an energy level) is the cardinality of S(n,g), i.e., the number of elements (or valid
n-tuples) in S(n,g). Thus the problem of finding an expression for w(n,g) becomes the problem of counting the elements in S(n,g). '
Example n
= 4, g
= 3:' S(4,3) = \left\{ \underbrace{(1111), (1112), (1113)}_{(a)}, \underbrace{(1122), (1123), (1133)}_{(b)}, \underbrace{(1222), (1223), (1233), (1333)}_{(c)}, \underbrace{(2222), (2223), (2233), (2333), (3333)}_{(d)} \right\} w(4,3) = 15 (there are 15 elements in S(4,3)) S(4,3)) --> Subset (a) is obtained by fixing all indices m_i to 1, except for the last index, m_n, which is incremented from 1 to g=3. Subset (b) is obtained by fixing m_1 = m_2 = 1, and incrementing m_3 from 2 to g=3. Due to the constraint m_i \ge m_{i-1} on the indices in S(n,g), the index m_4 must automatically take values in \left\{ 2, 3 \right\}. The construction of subsets (c) and (d) follows in the same manner. Each element of S(4,3) can be thought of as a
multiset of cardinality n=4; the elements of such multiset are taken from the set \left\{ 1, 2, 3 \right\} of cardinality g=3, and the number of such multisets is the
multiset coefficient \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 S(n,g) is a
multiset of cardinality n (number of dice) with elements taken from the set \left\{ 1, \dots, g \right\} of cardinality g (number of possible values of each die), and the number of such multisets, i.e., w(n,g) is the
multiset coefficient {{NumBlk|| 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)!} }} which is exactly the same as the
formula for w(n,g), as derived above with the aid of a
theorem involving binomial coefficients, namely {{NumBlk|| \sum_{k=0}^n\frac{(k+a)!}{k!a!}=\frac{(n+a+1)!}{n!(a+1)!}. }} To understand the decomposition {{NumBlk|| w(n,g) = \sum_{k=0}^n w(n-k, g-1) = w(n, g-1) + w(n-1, g-1) + \dots + w(1, g-1) + w(0, g-1) }} or for example, n=4 and g=3 w(4,3) = w(4,2) + w(3,2) + w(2,2) + w(1,2) + w(0,2), let us rearrange the elements of S(4,3) as follows 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)}, \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 (\alpha) of S(4,3) is the same as the set S(4,2) = \left\{ (1111), (1112), (1122), (1222), (2222) \right\}. By deleting the index m_4=3 (shown in red with double underline) in the subset (\beta) of S(4,3), one obtains the set S(3,2) = \left\{ (111), (112), (122), (222) \right\}. In other words, there is a one-to-one correspondence between the subset (\beta) of S(4,3) and the set S(3,2). We write (\beta) \longleftrightarrow S(3,2) . Similarly, it is easy to see that (\gamma) \longleftrightarrow S(2,2) = \left\{ (11), (12), (22) \right\} (\delta) \longleftrightarrow S(1,2) = \left\{ (1), (2) \right\} (\omega) \longleftrightarrow S(0,2) = \{\} = \varnothing. Thus we can write S(4,3) = \bigcup_{k=0}^{4} S(4-k,2) or more generally, {{NumBlk|| S(n,g) = \bigcup_{k=0}^{n} S(n-k,g-1); }} and since the sets S(i,g-1), \text{ for } i = 0, \dots , n are non-intersecting, we thus have {{NumBlk|| w(n,g) = \sum_{k=0}^{n} w(n-k,g-1), }} with the convention that {{NumBlk|| w(0,g) = 1 \ , \forall g, \text{ and } w(n,0) = 1 \ , \forall n. }} Continuing the process, we arrive at the following formula 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 {{NumBlk|| 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, }} keeping in mind that for q and p being constants, we have {{NumBlk||\sum_{k=0}^{q} p = q p.|}} It can then be verified that (8) and (2) give the same result for w(4,3), w(3,3), w(3,2), etc. == Interdisciplinary applications ==