How many ways can two particles be arranged in three phase cells according to Bose Einstein?

First of all, the distributions are actually the macroscopical result from certain microsocopical assumptions. As your system is microscopic, I assume you refer to these assumptions, instead of the actual distributions. (The word distributions makes (in physics) only sense for a large number of particles.)

(If particles A is in state 1 and particle B in state 2, I'll write it in the following as A1B2 etc.)

These assumtions are: a) Maxwell-Boltzman: Particles are distinguishable. (As your question sounds a lot like it's from a problem sheet, I guess you understand the meaning of this word in the context of statistical physics.)

For a a 2 particle system with two states each, the possible states are: A1B1, A1B2, A2B1, A2B2

b) Bose-Einstein: Particles are indistinguishable and an exchange of particles gives us a + in the wave function. This implies that the wave function needs to be symmetric.

For a a 2 particle system with two states each, the possible states are: A1B1, A1B2 + A2B1, A2B2 (I didn't put normailzation factors.)

c) Fermi-Dirac: Particles are indistinguishable and an exchange of particles gives us a - in the wave function. This implies that the wave function needs to be anti-symmetric.

For a a 2 particle system with two states each, the possible states are: A1B2 - A2B1 (I didn't put normailzation factors.)

This should help you to solve you excercise. I didn't fully answer the question on purpose, as I assume you would like to actually underrstand it yourself. I hope this answer brings you on the way. If you need more details, just write a comment with a precise question and I'll edit the answer.

I get #165# using an equation I derived (which is correct), but when I do it the manual way I'm short by #36#... This question is too much to ask of you to do the manual way, particularly on an exam.

DISCLAIMER: LONG ANSWER! (obviously)

METHOD ONE: DRAWING OUT MACROSTATES

I'm going to do this a bit out of order, because having one particle per box is the most confusing one.

#1)# All particles in the same box

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

With #9# boxes, you get #bb9# ways to have three-particle boxes.

#3)# Different boxes in the same row

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

With two particles in the same box, you have #6# configurations of the remaining particle in its own box within the same row. With #3# rows, that gives #bb18# configurations.

#4)# Different boxes in the same column

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

Rotate the box #90^@#, then repeat #(3)# for #bb18# more, except we would have technically done it column-wise because the boxes are distinguishable.

#5)# Different boxes in different rows AND columns (at the same time)

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

You should get #4# configurations with one two-particle box in the upper left, times three columns for the two-particle box within the same row equals #12# configurations. Multiply by the three rows to get #bb36# configurations.

#2i)# Each particle in its own box, row-wise

Play around with this. You should get:

#a)# All particles in the same row: #3# configurations

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

#b)# Two particles in the same row: #2# configurations if two particles are in the first two columns, #2# configurations if the particles are in columns #1# and #3#, #2# configurations if the particles are in the last two columns, times #3# rows, for #18# total.

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

#c)# All particles in different rows: #2# diagonal configurations, #4# configurations with two particles on the off-diagonal, for a total of #6#.

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

Apparently, I get #3 + 18 + 6 = bb27# here.

#2ii)# Each particle in its own box, column-wise

Play around with this. You should get:

#a)# All particles in the same column: #3# configurations

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

#b)# Two particles in the same column: #2# configurations if two particles are in the first two rows, #2# configurations if the particles are in rows #1# and #3#, #2# configurations if the particles are in the last two rows, multiplied by the #3# columns, for #18# total.

Here is the representative macrostate:

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

#c)# All particles in different columns: We don't count these, because they are redundant.

Apparently, I get #3 + 18 = bb21# here.

However, that only accounts for #129# microstates. That's not enough. Maybe I missed #36# in #(5)#?

METHOD TWO: DERIVING AN EQUATION

We begin with the assumption of #N# distinguishable particles in #g# distinguishable boxes. If we arrange the boxes linearly, it doesn't change the number of microstates.

They would each be separated by #g - 1# box walls, shown as #|#:

#"x x" cdots | "x x x" cdots | "x x" cdots#

The walls are distinguishable, and thus could be swapped to obtain distinctly different configurations. With some amount of distinguishable things, there are a factorial number of ways to arrange them.

So, there are #(g - 1 + N)!# arrangements if the particles and boxes are both distinguishable. However, we wish to keep the box walls fixed and the particles are indistinguishable.

To avoid double counting, we then divide by #N!# and #(g-1)!# to account for redundant wall configurations and to ignore identical configurations of indistinguishable particles.

This gives

#bb(t = ((g + N - 1)!)/(N!(g-1)!)" microstates")#

when the #N# particles are indistinguishable and the #g# boxes are distinguishable. This matches the equation shown in Statistical Mechanics by Norman Davidson (pg. 66).

To check this equation, we put #bb3# indistinguishable particles in #bb4# distinguishable boxes to get:

#color(blue)(t) = ((4 + 3 - 1)!)/(3! (4 - 1)!) = (6!)/(6 cdot 6) = (5! cdot cancel6)/(cancel6 cdot 6)#

#= (120)/6 = color(blue)(20)# microstates #color(blue)(sqrt"")#,

just like the example shows.

In your case, with #bb3# indistinguishable particles and #bb9# distinguishable boxes:

#color(blue)(t) = ((9 + 3 - 1)!)/(3! (9 - 1)!) = (11!)/(6 cdot 8!) = (cancel(8!) cdot 9 cdot 10 cdot 11)/(6 cdot cancel(8!))#

#= color(blue)(165)# microstates

So we were short by #36# in our manual guess...

Description of the behavior of bosons

In quantum statistics, Bose–Einstein statistics (B–E statistics) describes one of two possible ways in which a collection of non-interacting, indistinguishable particles may occupy a set of available discrete energy states at thermodynamic equilibrium. The aggregation of particles in the same state, which is a characteristic of particles obeying Bose–Einstein statistics, accounts for the cohesive streaming of laser light and the frictionless creeping of superfluid helium. The theory of this behaviour was developed (1924–25) by Satyendra Nath Bose, who recognized that a collection of identical and indistinguishable particles can be distributed in this way. The idea was later adopted and extended by Albert Einstein in collaboration with Bose.

The Bose–Einstein statistics applies only to the particles not limited to single occupancy of the same state – that is, particles that do not obey the Pauli exclusion principle restrictions. Such particles have integer values of spin and are named bosons.

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

Comparison of average occupancy of the ground state for three statistics

Bose–Einstein distribution

At low temperatures, bosons behave differently from fermions (which obey the Fermi–Dirac statistics) in a way that an unlimited number of them can "condense" into the same energy state. This apparently unusual property also gives rise to the special state of matter – the Bose–Einstein condensate. Fermi–Dirac and Bose–Einstein statistics apply when quantum effects are important and the particles are "indistinguishable". Quantum effects appear if the concentration of particles satisfies

N V ≥ n q , {\displaystyle {\frac {N}{V}}\geq n_{q},}

where N is the number of particles, V is the volume, and nq is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength, so that the wavefunctions of the particles are barely overlapping.

Fermi–Dirac statistics applies to fermions (particles that obey the Pauli exclusion principle), and Bose–Einstein statistics applies to bosons. As the quantum concentration depends on temperature, most systems at high temperatures obey the classical (Maxwell–Boltzmann) limit, unless they also have a very high density, as for a white dwarf. Both Fermi–Dirac and Bose–Einstein become Maxwell–Boltzmann statistics at high temperature or at low concentration.

B–E statistics was introduced for photons in 1924 by Bose and generalized to atoms by Einstein in 1924–25.

The expected number of particles in an energy state i for B–E statistics is:

n ¯ i = g i e ( ε i − μ ) / k B T − 1 {\displaystyle {\bar {n}}_{i}={\frac {g_{i}}{e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}-1}}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

with εi > μ and where ni is the occupation number (the number of particles) in state i, g i {\displaystyle g_{i}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is the degeneracy of energy level i, εi is the energy of the i-th state, μ is the chemical potential, kB is the Boltzmann constant, and T is absolute temperature.

The variance of this distribution V ( n ) {\displaystyle V(n)}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is calculated directly from the expression above for the average number.[1]

V ( n ) = k T ∂ ∂ μ n ¯ i = ⟨ n ⟩ ( 1 + ⟨ n ⟩ ) = n ¯ + n ¯ 2 {\displaystyle V(n)=kT{\frac {\partial }{\partial \mu }}{\bar {n}}_{i}=\langle n\rangle (1+\langle n\rangle )={\bar {n}}+{\bar {n}}^{2}}

For comparison, the average number of fermions with energy ε i {\displaystyle \varepsilon _{i}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
given by Fermi–Dirac particle-energy distribution has a similar form:

n ¯ i ( ε i ) = g i e ( ε i − μ ) / k B T + 1 . {\displaystyle {\bar {n}}_{i}(\varepsilon _{i})={\frac {g_{i}}{e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}+1}}.}

As mentioned above, both the Bose–Einstein distribution and the Fermi–Dirac distribution approaches the Maxwell–Boltzmann distribution in the limit of high temperature and low particle density, without the need for any ad hoc assumptions:

  • In the limit of low particle density, n ¯ i = g i e ( ε i − μ ) / k B T ± 1 ≪ 1 {\displaystyle {\bar {n}}_{i}={\frac {g_{i}}{e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}\pm 1}}\ll 1}
    How many ways can two particles be arranged in three phase cells according to Bose Einstein?
    , therefore e ( ε i − μ ) / k B T ± 1 ≫ 1 {\displaystyle e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}\pm 1\gg 1}
    How many ways can two particles be arranged in three phase cells according to Bose Einstein?
    or equivalently e ( ε i − μ ) / k B T ≫ 1 {\displaystyle e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}\gg 1}
    How many ways can two particles be arranged in three phase cells according to Bose Einstein?
    . In that case, n ¯ i ≈ g i e ( ε i − μ ) / k B T = 1 Z e − ε i / k B T {\displaystyle {\bar {n}}_{i}\approx {\frac {g_{i}}{e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}}}={\frac {1}{Z}}e^{-\varepsilon _{i}/k_{\text{B}}T}}
    How many ways can two particles be arranged in three phase cells according to Bose Einstein?
    , which is the result from Maxwell–Boltzmann statistics.
  • In the limit of high temperature, the particles are distributed over a large range of energy values, therefore the occupancy on each state (especially the high energy ones with ε i − μ ≫ k B T {\displaystyle \varepsilon _{i}-\mu \gg k_{\text{B}}T}
    How many ways can two particles be arranged in three phase cells according to Bose Einstein?
    ) is again very small, n ¯ i = g i e ( ε i − μ ) / k B T ± 1 ≪ 1 {\displaystyle {\bar {n}}_{i}={\frac {g_{i}}{e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}\pm 1}}\ll 1} . This again reduces to Maxwell–Boltzmann statistics.

In addition to reducing to the Maxwell–Boltzmann distribution in the limit of high T {\displaystyle T}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
and low density, B–E statistics also reduces to Rayleigh–Jeans law distribution for low energy states with
ε i − μ ≪ k B T {\displaystyle \varepsilon _{i}-\mu \ll k_{\text{B}}T}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, namely

n ¯ i = g i e ( ε i − μ ) / k B T − 1 ≈ g i ( ε i − μ ) / k B T = g i k B T ε i − μ . {\displaystyle {\begin{aligned}{\bar {n}}_{i}&={\frac {g_{i}}{e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}-1}}\\&\approx {\frac {g_{i}}{(\varepsilon _{i}-\mu )/k_{\text{B}}T}}={\frac {g_{i}k_{\text{B}}T}{\varepsilon _{i}-\mu }}.\end{aligned}}}

History

Władysław Natanson in 1911 concluded that Planck's law requires indistinguishability of "units of energy", although he did not frame this in terms of Einstein's light quanta.[2][3]

While presenting a lecture at the University of Dhaka (in what was then British India and is now Bangladesh) on the theory of radiation and the ultraviolet catastrophe, Satyendra Nath Bose intended to show his students that the contemporary theory was inadequate, because it predicted results not in accordance with experimental results. During this lecture, Bose committed an error in applying the theory, which unexpectedly gave a prediction that agreed with the experiment. The error was a simple mistake—similar to arguing that flipping two fair coins will produce two heads one-third of the time—that would appear obviously wrong to anyone with a basic understanding of statistics (remarkably, this error resembled the famous blunder by d'Alembert known from his Croix ou Pile article[4][5]). However, the results it predicted agreed with experiment, and Bose realized it might not be a mistake after all. For the first time, he took the position that the Maxwell–Boltzmann distribution would not be true for all microscopic particles at all scales. Thus, he studied the probability of finding particles in various states in phase space, where each state is a little patch having phase volume of h3, and the position and momentum of the particles are not kept particularly separate but are considered as one variable.

Bose adapted this lecture into a short article called "Planck's law and the hypothesis of light quanta"[6][7] and submitted it to the Philosophical Magazine. However, the referee's report was negative, and the paper was rejected. Undaunted, he sent the manuscript to Albert Einstein requesting publication in the Zeitschrift für Physik. Einstein immediately agreed, personally translated the article from English into German (Bose had earlier translated Einstein's article on the general theory of relativity from German to English), and saw to it that it was published. Bose's theory achieved respect when Einstein sent his own paper in support of Bose's to Zeitschrift für Physik, asking that they be published together. The paper came out in 1924.[8]

The reason Bose produced accurate results was that since photons are indistinguishable from each other, one cannot treat any two photons having equal quantum numbers (e.g., polarization and momentum vector) as being two distinct identifiable photons. By analogy, if in an alternate universe coins were to behave like photons and other bosons, the probability of producing two heads would indeed be one-third, and so is the probability of getting a head and a tail which equals one-half for the conventional (classical, distinguishable) coins. Bose's "error" leads to what is now called Bose–Einstein statistics.

Bose and Einstein extended the idea to atoms and this led to the prediction of the existence of phenomena which became known as Bose–Einstein condensate, a dense collection of bosons (which are particles with integer spin, named after Bose), which was demonstrated to exist by experiment in 1995.

Derivation

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 = ∑ i n i {\textstyle N=\sum _{i}n_{i}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
identical bosons, n i {\displaystyle n_{i}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
of which have energy ε i {\displaystyle \varepsilon _{i}} and are distributed over g i {\displaystyle g_{i}} levels or states with the same energy ε i {\displaystyle \varepsilon _{i}} , i.e. g i {\displaystyle g_{i}} is the degeneracy associated with energy ε i {\displaystyle \varepsilon _{i}} of total energy E = ∑ i n i ε i {\textstyle E=\sum _{i}n_{i}\varepsilon _{i}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. Calculation of the number of arrangements of n i {\displaystyle n_{i}} particles distributed among g i {\displaystyle 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 {\displaystyle n_{i}} particles in g i {\displaystyle g_{i}} boxes (for the i {\displaystyle i}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
th energy level) would be (see image).

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

The image represents one possible distribution of bosonic particles in different boxes. The box partitions (green) can be moved around to change the size of the boxes and as a result of the number of bosons each box can contain.

w i , BE = ( n i + g i − 1 ) ! n i ! ( g i − 1 ) ! = C n i n i + g i − 1 , {\displaystyle w_{i,{\text{BE}}}={\frac {(n_{i}+g_{i}-1)!}{n_{i}!(g_{i}-1)!}}=C_{n_{i}}^{n_{i}+g_{i}-1},}

where C k m {\displaystyle C_{k}^{m}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is the k-combination of a set with m elements. The total number of arrangements in an ensemble of bosons is simply the product of the binomial coefficients C n i n i + g i − 1 {\displaystyle C_{n_{i}}^{n_{i}+g_{i}-1}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
above over all the energy levels, i.e.

W BE = ∏ i w i , BE = ∏ i ( n i + g i − 1 ) ! ( g i − 1 ) ! n i ! , {\displaystyle W_{\text{BE}}=\prod _{i}w_{i,{\text{BE}}}=\prod _{i}{\frac {(n_{i}+g_{i}-1)!}{(g_{i}-1)!n_{i}!}},}

The maximum number of arrangements determining the corresponding occupation number n i {\displaystyle n_{i}} is obtained by maximizing the entropy, or equivalently, setting d ( ln ⁡ W BE ) = 0 {\displaystyle \mathrm {d} (\ln W_{\text{BE}})=0}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
and taking the subsidiary conditions N = ∑ n i , E = ∑ i n i ε i {\textstyle N=\sum n_{i},E=\sum _{i}n_{i}\varepsilon _{i}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
into account (as Lagrange multipliers).[9] The result for n i ≫ 1 {\displaystyle n_{i}\gg 1}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, g i ≫ 1 {\displaystyle g_{i}\gg 1}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, n i / g i = O ( 1 ) {\displaystyle n_{i}/g_{i}=O(1)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is the Bose–Einstein distribution.

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.[10] 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, ε {\displaystyle \varepsilon }

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. 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 {\displaystyle N}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is the number of particles, the total energy of the sub-ensemble will then be N ε {\displaystyle N\varepsilon }
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. Beginning with the standard expression for a grand partition function and replacing E {\displaystyle E}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
with N ε {\displaystyle N\varepsilon } , the grand partition function takes the form

Z = ∑ N exp ⁡ ( ( N μ − N ε ) / k B T ) = ∑ N exp ⁡ ( N ( μ − ε ) / k B T ) {\displaystyle {\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:

Z = ∑ N = 0 ∞ exp ⁡ ( N ( μ − ε ) / k B T ) = ∑ N = 0 ∞ [ exp ⁡ ( ( μ − ε ) / k B T ) ] N = 1 1 − exp ⁡ ( ( μ − ε ) / k B T ) . {\displaystyle {\begin{aligned}{\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{aligned}}}

Note that the geometric series is convergent only if e ( μ − ε ) / k B T < 1 {\displaystyle e^{(\mu -\varepsilon )/k_{\text{B}}T}<1}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, including the case where ϵ = 0 {\displaystyle \epsilon =0}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. This implies that the chemical potential for the Bose gas must be negative, i.e., μ < 0 {\displaystyle \mu <0}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, whereas the Fermi gas is allowed to take both positive and negative values for the chemical potential.[11]

The average particle number for that single-particle substate is given by

⟨ N ⟩ = k B T 1 Z ( ∂ Z ∂ μ ) V , T = 1 exp ⁡ ( ( ε − μ ) / k B T ) − 1 {\displaystyle \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.[12][13]

The variance in particle number, σ N 2 = ⟨ N 2 ⟩ − ⟨ N ⟩ 2 {\textstyle \sigma _{N}^{2}=\langle N^{2}\rangle -\langle N\rangle ^{2}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, is:

σ N 2 = k B T ( d ⟨ N ⟩ d μ ) V , T = exp ⁡ ( ( ε − μ ) / k B T ) ( exp ⁡ ( ( ε − μ ) / k B T ) − 1 ) 2 = ⟨ N ⟩ ( 1 + ⟨ N ⟩ ) . {\displaystyle \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: σ N ≈ ⟨ N ⟩ {\displaystyle \sigma _{N}\approx \langle N\rangle }

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. 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 σ N , c l a s s i c a l = ⟨ N ⟩ {\textstyle \sigma _{N,{\rm {classical}}}={\sqrt {\langle N\rangle }}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, and with the most-probable N value being near ⟨ N ⟩ {\displaystyle \langle N\rangle }
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
.)

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.[14] See also Müller-Kirsten.[9] The fluctuations of the ground state in the condensed region are however markedly different in the canonical and grand-canonical ensembles.[15]

Derivation

Suppose we have a number of energy levels, labeled by index i {\displaystyle i} , each level having energy ε i {\displaystyle \varepsilon _{i}} and containing a total of n i {\displaystyle n_{i}} particles. Suppose each level contains g i {\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 g i {\displaystyle g_{i}} associated with level i {\displaystyle i} is called the "degeneracy" of that energy level. Any number of bosons can occupy the same sublevel.

Let w ( n , g ) {\displaystyle w(n,g)}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
be the number of ways of distributing n {\displaystyle n}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
particles among the g {\displaystyle g}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
sublevels of an energy level. There is only one way of distributing n {\displaystyle n} particles with one sublevel, therefore w ( n , 1 ) = 1 {\displaystyle w(n,1)=1}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. It is easy to see that there are ( n + 1 ) {\displaystyle (n+1)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
ways of distributing n {\displaystyle n} particles in two sublevels which we will write as:

w ( n , 2 ) = ( n + 1 ) ! n ! 1 ! . {\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 n {\displaystyle n} particles in three sublevels is

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

so that

w ( n , 3 ) = ∑ k = 0 n w ( n − k , 2 ) = ∑ k = 0 n ( n − k + 1 ) ! ( n − k ) ! 1 ! = ( n + 2 ) ! n ! 2 ! {\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:

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

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

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

w ( n , g ) = ( n + g − 1 ) ! n ! ( g − 1 ) ! . {\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 n i {\displaystyle n_{i}} can be realized is the product of the ways that each individual energy level can be populated:

W = ∏ i w ( n i , g i ) = ∏ i ( n i + g i − 1 ) ! n i ! ( g i − 1 ) ! ≈ ∏ i ( n i + g i ) ! n i ! ( g i ) ! {\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 n i ≫ 1 {\displaystyle n_{i}\gg 1} .

Following the same procedure used in deriving the Maxwell–Boltzmann statistics, we wish to find the set of n i {\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 W {\displaystyle W}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
and ln ⁡ ( W ) {\displaystyle \ln(W)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
occur at the same value of n i {\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:

f ( n i ) = ln ⁡ ( W ) + α ( N − ∑ n i ) + β ( E − ∑ n i ε i ) {\displaystyle f(n_{i})=\ln(W)+\alpha (N-\sum n_{i})+\beta (E-\sum n_{i}\varepsilon _{i})}

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

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
gives

f ( n i ) = ∑ i ( n i + g i ) ln ⁡ ( n i + g i ) − n i ln ⁡ ( n i ) + α ( N − ∑ n i ) + β ( E − ∑ n i ε i ) + K . {\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 n i {\displaystyle n_{i}} . Taking the derivative with respect to n i {\displaystyle n_{i}} , and setting the result to zero and solving for n i {\displaystyle n_{i}} , yields the Bose–Einstein population numbers:

n i = g i e α + β ε i − 1 . {\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:

d ln ⁡ W = α d N + β d E {\displaystyle d\ln W=\alpha \,dN+\beta \,dE}

which, using Boltzmann's famous relationship S = k B ln ⁡ W {\displaystyle S=k_{\text{B}}\,\ln W}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
becomes a statement of the second law of thermodynamics at constant volume, and it follows that β = 1 k B T {\displaystyle \beta ={\frac {1}{k_{\text{B}}T}}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
and α = − μ k B T {\displaystyle \alpha =-{\frac {\mu }{k_{\text{B}}T}}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
where S is the entropy, μ {\displaystyle \mu }
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is the chemical potential, kB is Boltzmann's constant and T is the temperature, so that finally:

n i = g i e ( ε i − μ ) / k B T − 1 . {\displaystyle n_{i}={\frac {g_{i}}{e^{(\varepsilon _{i}-\mu )/k_{\text{B}}T}-1}}.}

Note that the above formula is sometimes written:

n i = g i e ε i / k B T / z − 1 , {\displaystyle n_{i}={\frac {g_{i}}{e^{\varepsilon _{i}/k_{\text{B}}T}/z-1}},}

where z = exp ⁡ ( μ / k B T ) {\displaystyle z=\exp(\mu /k_{\text{B}}T)}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is the absolute activity, as noted by McQuarrie.[16]

Also note that when the particle numbers are not conserved, removing the conservation of particle numbers constraint is equivalent to setting α {\displaystyle \alpha }

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
and therefore the chemical potential μ {\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.

Notes

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:

( g − 1 + n ) ! ( g − 1 ) ! n ! {\displaystyle {\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 g − 1 partitions.

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

The image represents one possible distribution of bosonic particles in different boxes. The box partitions (green) can be moved around to change the size of the boxes and as a result of the number of bosons each box can contain.

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 n {\displaystyle n} dice, with each die taking values in the set { 1 , … , g } {\displaystyle \{1,\dots ,g\}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, for g ≥ 1 {\displaystyle g\geq 1}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. The constraints of the game are that the value of a die i {\displaystyle i} , denoted by m i {\displaystyle m_{i}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, has to be greater than or equal to the value of die ( i − 1 ) {\displaystyle (i-1)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, denoted by m i − 1 {\displaystyle m_{i-1}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, in the previous throw, i.e., m i ≥ m i − 1 {\displaystyle m_{i}\geq m_{i-1}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. Thus a valid sequence of die throws can be described by an n-tuple ( m 1 , m 2 , … , m n ) {\displaystyle (m_{1},m_{2},\dots ,m_{n})}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, such that m i ≥ m i − 1 {\displaystyle m_{i}\geq m_{i-1}} . Let S ( n , g ) {\displaystyle S(n,g)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
denote the set of these valid n-tuples:

S ( n , g ) = { ( m 1 , m 2 , … , m n ) | m i ≥ m i − 1 , m i ∈ { 1 , … , g } , ∀ i = 1 , … , n } . {\displaystyle S(n,g)=\left\{(m_{1},m_{2},\dots ,m_{n})\,{\Big |}\,m_{i}\geq m_{i-1},m_{i}\in \left\{1,\ldots ,g\right\},\forall i=1,\dots ,n\right\}.}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

(1)

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

Example n = 4, g = 3:

S ( 4 , 3 ) = { ( 1111 ) , ( 1112 ) , ( 1113 ) ⏟ ( a ) , ( 1122 ) , ( 1123 ) , ( 1133 ) ⏟ ( b ) , ( 1222 ) , ( 1223 ) , ( 1233 ) , ( 1333 ) ⏟ ( c ) , ( 2222 ) , ( 2223 ) , ( 2233 ) , ( 2333 ) , ( 3333 ) ⏟ ( d ) } {\displaystyle 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 {\displaystyle w(4,3)=15}

(there are 15 {\displaystyle 15}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
elements in S ( 4 , 3 ) {\displaystyle S(4,3)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
)

Subset ( a ) {\displaystyle (a)}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is obtained by fixing all indices m i {\displaystyle m_{i}} to 1 {\displaystyle 1}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, except for the last index, m n {\displaystyle m_{n}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, which is incremented from 1 {\displaystyle 1} to g = 3 {\displaystyle g=3}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. Subset ( b ) {\displaystyle (b)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
is obtained by fixing m 1 = m 2 = 1 {\displaystyle m_{1}=m_{2}=1}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, and incrementing m 3 {\displaystyle m_{3}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
from 2 {\displaystyle 2}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
to g = 3 {\displaystyle g=3} . Due to the constraint m i ≥ m i − 1 {\displaystyle m_{i}\geq m_{i-1}} on the indices in S ( n , g ) {\displaystyle S(n,g)} , the index m 4 {\displaystyle m_{4}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
must automatically take values in { 2 , 3 } {\displaystyle \left\{2,3\right\}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. The construction of subsets ( c ) {\displaystyle (c)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
and ( d ) {\displaystyle (d)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
follows in the same manner.

Each element of S ( 4 , 3 ) {\displaystyle S(4,3)} can be thought of as a multiset of cardinality n = 4 {\displaystyle n=4}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
; the elements of such multiset are taken from the set { 1 , 2 , 3 } {\displaystyle \left\{1,2,3\right\}}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
of cardinality g = 3 {\displaystyle g=3} , and the number of such multisets is the multiset coefficient

⟨ 3 4 ⟩ = ( 3 + 4 − 1 3 − 1 ) = ( 3 + 4 − 1 4 ) = 6 ! 4 ! 2 ! = 15 {\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 S ( n , g ) {\displaystyle S(n,g)} is a multiset of cardinality n {\displaystyle n} (number of dice) with elements taken from the set { 1 , … , g } {\displaystyle \left\{1,\dots ,g\right\}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
of cardinality g {\displaystyle g} (number of possible values of each die), and the number of such multisets, i.e., w ( n , g ) {\displaystyle w(n,g)} is the multiset coefficient

w ( n , g ) = ⟨ g n ⟩ = ( g + n − 1 g − 1 ) = ( g + n − 1 n ) = ( g + n − 1 ) ! n ! ( g − 1 ) ! {\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)!}}}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

(2)

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

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

(3)

To understand the decomposition

w ( n , g ) = ∑ k = 0 n w ( n − k , g − 1 ) = w ( n , g − 1 ) + w ( n − 1 , g − 1 ) + ⋯ + w ( 1 , g − 1 ) + w ( 0 , g − 1 ) {\displaystyle 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)}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

(4)

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

w ( 4 , 3 ) = w ( 4 , 2 ) + w ( 3 , 2 ) + w ( 2 , 2 ) + w ( 1 , 2 ) + w ( 0 , 2 ) , {\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 S ( 4 , 3 ) {\displaystyle S(4,3)} as follows

S ( 4 , 3 ) = { ( 1111 ) , ( 1112 ) , ( 1122 ) , ( 1222 ) , ( 2222 ) ⏟ ( α ) , ( 111 3 = ) , ( 112 3 = ) , ( 122 3 = ) , ( 222 3 = ) ⏟ ( β ) , ( 11 33 == ) , ( 12 33 == ) , ( 22 33 == ) ⏟ ( γ ) , ( 1 333 === ) , ( 2 333 === ) ⏟ ( δ ) ( 3333 ==== ) ⏟ ( ω ) } . {\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 )},\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 ( α ) {\displaystyle (\alpha )}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
of S ( 4 , 3 ) {\displaystyle S(4,3)} is the same as the set

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

By deleting the index m 4 = 3 {\displaystyle m_{4}=3}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
(shown in red with double underline) in the subset ( β ) {\displaystyle (\beta )}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
of S ( 4 , 3 ) {\displaystyle S(4,3)} , one obtains the set

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

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

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
. We write

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

Similarly, it is easy to see that

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

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

( ω ) ⟷ S ( 0 , 2 ) = { } = ∅ . {\displaystyle (\omega )\longleftrightarrow S(0,2)=\{\}=\varnothing .}

Thus we can write

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

or more generally,

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

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

(5)

and since the sets

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

are non-intersecting, we thus have

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

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

(6)

with the convention that

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

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

(7)

Continuing the process, we arrive at the following formula

w ( n , g ) = ∑ k 1 = 0 n ∑ k 2 = 0 n − k 1 w ( n − k 1 − k 2 , g − 2 ) = ∑ k 1 = 0 n ∑ k 2 = 0 n − k 1 ⋯ ∑ k g = 0 n − ∑ j = 1 g − 1 k j w ( n − ∑ i = 1 g k i , 0 ) . {\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

w ( n , g ) = ∑ k 1 = 0 n ∑ k 2 = 0 n − k 1 ⋯ ∑ k g = 0 n − ∑ j = 1 g − 1 k j 1 , {\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,}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

(8)

keeping in mind that for q {\displaystyle q}

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
and p {\displaystyle p}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
being constants, we have

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

How many ways can two particles be arranged in three phase cells according to Bose Einstein?

(9)

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

How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, w ( 3 , 3 ) {\displaystyle w(3,3)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, w ( 3 , 2 ) {\displaystyle w(3,2)}
How many ways can two particles be arranged in three phase cells according to Bose Einstein?
, etc.

Interdisciplinary applications

Viewed as a pure probability distribution, the Bose–Einstein distribution has found application in other fields:

  • In recent years, Bose–Einstein statistics has also been used as a method for term weighting in information retrieval. The method is one of a collection of DFR ("Divergence From Randomness") models,[17] the basic notion being that Bose–Einstein statistics may be a useful indicator in cases where a particular term and a particular document have a significant relationship that would not have occurred purely by chance. Source code for implementing this model is available from the Terrier project at the University of Glasgow.
  • The evolution of many complex systems, including the World Wide Web, business, and citation networks, is encoded in the dynamic web describing the interactions between the system's constituents. Despite their irreversible and nonequilibrium nature these networks follow Bose statistics and can undergo Bose–Einstein condensation. Addressing the dynamical properties of these nonequilibrium systems within the framework of equilibrium quantum gases predicts that the "first-mover-advantage", "fit-get-rich" (FGR) and "winner-takes-all" phenomena observed in competitive systems are thermodynamically distinct phases of the underlying evolving networks.[18]

See also

  • Bose–Einstein correlations
  • Bose–Einstein condensate
  • Bose gas
  • Einstein solid
  • Higgs boson
  • Parastatistics
  • Planck's law of black body radiation
  • Superconductivity
  • Fermi–Dirac statistics
  • Maxwell–Boltzmann statistics

Notes

  1. ^ Pearsall, Thomas (2020). Quantum Photonics, 2nd edition. Graduate Texts in Physics. Springer. doi:10.1007/978-3-030-47325-9. ISBN 978-3-030-47324-2.
  2. ^ Jammer, Max (1966). The conceptual development of quantum mechanics. McGraw-Hill. p. 51. ISBN 0-88318-617-9.
  3. ^ Passon, Oliver; Grebe-Ellis, Johannes (2017-05-01). "Planck's radiation law, the light quantum, and the prehistory of indistinguishability in the teaching of quantum mechanics". European Journal of Physics. 38 (3): 035404. arXiv:1703.05635. Bibcode:2017EJPh...38c5404P. doi:10.1088/1361-6404/aa6134. ISSN 0143-0807. S2CID 119091804.
  4. ^ d'Alembert, Jean (1754). "Croix ou pile". L'Encyclopédie (in French). 4.
  5. ^ d'Alembert, Jean (1754). "Croix ou pile" (PDF). Xavier University. Translated by Richard J. Pulskamp. Retrieved 2019-01-14.
  6. ^ See p. 14, note 3, of the thesis: Michelangeli, Alessandro (October 2007). Bose–Einstein condensation: Analysis of problems and rigorous results (PDF) (Ph.D.). International School for Advanced Studies. Archived (PDF) from the original on 3 November 2018. Retrieved 14 February 2019.
  7. ^ Bose (2 July 1924). "Planck's law and the hypothesis of light quanta" (PostScript). University of Oldenburg. Retrieved 30 November 2016.
  8. ^ Bose (1924), "Plancks Gesetz und Lichtquantenhypothese", Zeitschrift für Physik (in German), 26 (1): 178–181, Bibcode:1924ZPhy...26..178B, doi:10.1007/BF01327326, S2CID 186235974
  9. ^ a b H. J. W. Müller-Kirsten, Basics of Statistical Physics, 2nd ed., World Scientific (2013), ISBN 978-981-4449-53-3.
  10. ^ Srivastava, R. K.; Ashok, J. (2005). "Chapter 7". Statistical Mechanics. New Delhi: PHI Learning Pvt. Ltd. ISBN 9788120327825.
  11. ^ Landau, L. D., Lifšic, E. M., Lifshitz, E. M., & Pitaevskii, L. P. (1980). Statistical physics (Vol. 5). Pergamon Press.
  12. ^ "Chapter 6". Statistical Mechanics. January 2005. ISBN 9788120327825.
  13. ^ The BE distribution can be derived also from thermal field theory.
  14. ^ R. B. Dingle, Asymptotic Expansions: Their Derivation and Interpretation, Academic Press (1973), pp. 267–271.
  15. ^ Ziff R. M.; Kac, M.; Uhlenbeck, G. E. (1977). "The ideal Bose–Einstein gas, revisited". Physics Reports 32: 169–248.
  16. ^ See McQuarrie in citations
  17. ^ Amati, G.; C. J. Van Rijsbergen (2002). "Probabilistic models of information retrieval based on measuring the divergence from randomness " ACM TOIS 20(4):357–389.
  18. ^ Bianconi, G.; Barabási, A.-L. (2001). "Bose–Einstein Condensation in Complex Networks". Physical Review Letters 86: 5632–5635.

References

  • Annett, James F. (2004). Superconductivity, Superfluids and Condensates. New York: Oxford University Press. ISBN 0-19-850755-0.
  • Carter, Ashley H. (2001). Classical and Statistical Thermodynamics. Upper Saddle River, New Jersey: Prentice Hall. ISBN 0-13-779208-5.
  • Griffiths, David J. (2005). Introduction to Quantum Mechanics (2nd ed.). Upper Saddle River, New Jersey: Pearson, Prentice Hall. ISBN 0-13-191175-9.
  • McQuarrie, Donald A. (2000). Statistical Mechanics (1st ed.). Sausalito, California 94965: University Science Books. p. 55. ISBN 1-891389-15-7.{{cite book}}: CS1 maint: location (link)

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