144 Chemical thermodynamics
whichmakeupthegrandcanonicalensemble corresponds to a particular terminthe
extended product obtained by multiplying out (13.9). And the statistical weight of
eachmicrostateissimplythe magnitudeofthis term, as postulatedabove. We need
to examine therole of the( 1 +γ)factorforwhichj=iinthisexpansion. Ifour state
iis full(ni=1), then we have taken theγγγifrom the bracket; if the state is empty we
have taken the1.Thethermalaverage requiredisthus equalto the sum ofthe terms
in whichni= 1 dividedbythe totalsumZZZGofallthe terms.Hence
ni=γγγi
( 1 +γγγi)×
∏
j∏
∏= i(^1 +γγγj)
j∏
= i(^1 +γγγj)The contribution for the ‘other’states(those withj= i)conveniently factors out. We
areleftthereforewith
ni=γγγi
1 +γγγi=
1
γγγi−^1 + 1=
1
exp[(εi−μ)/kkkBT]+ 1as expected and hoped, in agreement with ( 5 .13). This is the Fermi–Dirac distribution.
Bose–Einstein. Here the same sort of technique works, in that the same cancellation
ofthefactorsfrom thej= istates takes place. However we must now allowfor all
occupation numbers in the statei.Lookingat the bracket forj=iin(13.10a),we
recall that the term 1 corresponds toni=0, the termγγγitoni=1, the termγγγi^2 to
ni=2andso on. Therefore theexpressionfor thedistributionis
ni=γγγi+ 2 γγγi^2 +···
1 +γγγi+γγγi^2 +···This expression looks somewhat intractable until we take a hint from the summation
of the denominator, as in the transition from(13.10a)to(13.10b). We writeF =
( 1 −γ)−^1 = 1 +γ+γ^2 +...asbefore. Differentiatingbothforms ofF,we obtain
dF
dγ=( 1 −γ)−^2 = 1 + 2 γ+ 3 γ^2 +...Hence the expression for the distribution becomes
ni=γγγidF/dγγγi
F=
γγγi
1 −γγγi=
1
γγγi−^1 − 1=
1
exp[(εi−μ)/kkkBT]− 1This is the Bose–Einstein distribution of (5.13).
Maxwell–Boltzmann limit. Finally we may note without further ado that, since in
theMBlimitallγγγi 1 ,either ofthe twodistributions tendto the even simpler result:
ni=γγγi=exp[(μ−εi)/kkkBT], the Maxwell–Boltzmann distribution as expected.