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316 Appendix F: The statistical mechanics of Bose–Einstein condensation

to this summation when the system occupies many levels to give an

almost continuous distribution. The density of states in the integral

represents a summation over the number of levels within each frequency

(or energy) interval dω(≡dε/
).

F.2 Bose–Einstein condensation

The crucial difference between photons and particles in statistical me-

chanics is that photons simply disappear as the temperature tends to

zero, as we can see from eqn F.2. In contrast, for a system of particles,

such as a gas of atoms or molecules in a box, the number remains con-

stant. This introduces a second constraint on the distribution function,

namely that the sum of the populations in all the energy levels equals

the total numberN. Note that the symbol ‘N’ is used here because ‘N’

was used in Chapter 7 to represent the number density (the convention

usually adopted in laser physics). Here, number density is denoted by

‘n’, as is usual in statistical mechanics. Hence

N=

∑

i

f(εi). (F.4)

This equation for conservation of number, and eqn F.3 for the total

energy, apply to any system of particles. We shall consider particles

with integer spin that follow the Bose–Einstein distribution function

fBE(ε)=

1

eβ(ε−μ)− 1

. (F.5)

This function has two parametersβandμthat can be determined by the

two constraints. For the particular case of bosons at low temperatures,

the chemical potentialμhas little consequenceexceptfor atoms in the

lowest energy state, so that for the higher-lying levelsfBE(ε) closely

resembles the distribution of photons! We can justify this assertion by

considering the properties of a system with a significant populationN 0 in

the ground state.^3 The number of atoms in the lowest level with energy

(^3) This may appear to be a circular argu-
ment since the large occupation of the
lowest level is the signature of the Bose–
Einstein condensation that we want to
investigate! This section shows that
this is a consistent solution of the equa-
tions for bosons, and the validity of
this treatment can be appreciated more
readily after deriving the equations.
ε 0 is given by
N 0 =

1

eβ(ε^0 −μ)− 1

. (F.6)

Hence

ε 0 −μ

kBT

=ln

(

1+

1

N 0

)



1

N 0

. (F.7)

Einstein originally considered gases with a large total number of atoms,

N∼ 1023 ,so, even whenN 0 is only a small fraction ofN,the difference

ε 0 −μis negligible in comparison to the thermal energykBT.(Thisther-

modynamic, or large number, approximation works even for samples of

106 magnetically-trapped atoms.^4 ) The equation shows that the chem-

(^4) We assume a thermal energy kBT
much greater than the spacing be-
tween the energy levels. Otherwise,
particles sit in the ground state sim-
ply because of the Boltzmann factor
exp{−β(ε 1 −ε 0 )}1, whereε 1 is the
energy of the first excited level (Sec-
tion 12.9). ical potential is lower thanε
5 0 , the lowest energy level in the system.^5
Otherwise there would be an energy
for whichε−μ=0;thiswouldmakethe
denominator in eqn F.5 equal to zero so
thatf(ε)→∞.
For the first excited level we find that
ε 1 −μ=(ε 1 −ε 0 )+(ε 0 −μ)
ω+
kBT
N 0

ω.