H.2 Euler-Maclaur in Sum Formula
19.1.3 Derivation III: Most Probable Distribution
In this section, we give yet another derivation of the canonical ensemble but from the point of view of the most probable distribution. We consider a large numberNens of identical systems, each having the same volume V and the same number of particles N, and each in a stationary quantum state.4These systems constitute the ensemble and theysharea constant total energy NensE, where¯ E¯ is the average energy per system.Ni
members of the ensemble are in an eigenstate having energyEisuch that the probability of occurrence of that eigenstate isPi=Ni/Nens. The set{Ni} =N1,N2,. . .,Nris such that r
i=1Ni=Nenswhich is equivalent to5 r i=1
Pi=1. (19.24)
Sincer
i=1NiEi=NensE, we also have¯ r
i=1
PiEi= ¯E. (19.25)
The number of ways of constructing such an ensemble is Wens{Ni}:= Nens!
N1!N2! ã ã ãNr!. (19.26) We would like to choose the set{Ni}to maximizeWens{Ni}subject to the constraints above to give the most probable distribution. Since d lnx=(1/x)dx, the maximum of lnxoccurs at the same value ofx as the maximum ofx. Therefore, for convenience, we maximize lnWens. With the aid of Stirling’s approximation we have
lnWens=NenslnNens− r i=1
NilnNi= −Nens
r i=1
PilnPi. (19.27) SinceNensis a constant, we can just maximize the function
D{Pi} = − r i=1
PilnPi, (19.28)
4If other extensive variables are necessary to specify our system of interest, they are also the same for all members of the ensemble.
5If a system has a number of eigenstatesr, we certainly needNi>rto represent the ensemble. But ultimately we can take the limitNens→ ∞in such a way thatNi→ ∞but the ratioPi=NiNensremains finite. Thus, there is essentially no problem even ifr→ ∞.
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subject to the constraints Eqs. (19.24) and (19.25). Once the set{Ni}is determined,Wensis the number of microstates of the whole ensemble, sokBlnWens represents the entropy of the whole ensemble. Thus, S=(1/Nens)kBlnWens is the entropy per system of the ensemble. It therefore plays the role of the thermodynamic entropy of the system that the ensemble represents. We therefore have
S= −kB
r i=1
PilnPi=kBD{Pi}, (19.29) where D{Pi}is seen to be a dimensionless measure of the entropy. We note thatD{Pi} is the same as the disorder function of Section 15.1, where we have shown (see the first example problem) from its form that it is additive for a composite system. But here we are maximizingD{Pi}subject to the additional constraint Eq. (19.25) on theaverageenergy of members of an ensemble that have different energies. For the microcanonical ensemble, all members of the ensemble have the same energy.
In the maximization process, we handle the constraints by means of Lagrange multi- pliersβandαand solve the problem
∂
∂Pj
− r
i=1
PilnPi−β r i=1
PiEi−α r i=1
Pi
=0 (19.30)
with eachPjnow (temporarily) considered to be an independent variable. We obtain
−lnPj−1−βEj−α=0, (19.31)
which may be exponentiated to give
Pj=e−α−1e−βEj. (19.32)
Summing Eq. (19.32) over all values ofjand applying the constraint Eq. (19.24) allows us to determine that exp(−α−1)=1/Z, whereZ=
jexp(−βEj)is the partition function as given by Eq. (19.5). Therefore, Eq. (19.32) becomes
Pj=exp(−βEj)
Z . (19.33)
It remains to determine the Lagrange multiplier β. Formally, this could be done in terms of E¯ by satisfying the constraint Eq. (19.25) but this would lead to a difficult transcendental equation forβ. Therefore, one takes instead an alternative approach by appealing to thermodynamics which allows β to be identified as a physical quantity.
To strengthen this identification, we recognize that the energies Ei of the eigenstates depend on the volumeVof the system6and its number of particlesN. Then
6This is convenient but not essential to the identification ofβ. It simply allows the system to do reversible workδW=pdV. IfEiwere to depend on a set of extensive mechanical parametersYjinstead of justV, one could write the reversible work in the form
jfjdYj, where thefjare generalized forces. Thenfj= −
iPi∂Ei/∂Yj.
dE¯ = r i=1
EidPi+ r i=1
PidEi=
i
EidPi+ r i=1
Pi∂Ei
∂V dV+ r i=1
Pi∂Ei
∂N dN. (19.34) From Eq. (19.29), the differential of the entropy is
dS= −kB
r i=1
(1+lnPi)dPi= −kB
r i=1
lnPidPi=kBβ r i=1
EidPi, (19.35) where
i dPi=0 has been used in the second and third steps. Substitution of Eq. (19.35) into Eq. (19.34) then gives
dE¯ =(kBβ)−1dS+ r i=1
Pi∂Ei
∂V dV+ r i=1
Pi∂Ei
∂N dN. (19.36)
Comparison of Eq. (19.36) with dU = TdS−pdV +μdN and the identificationE¯=U shows thatβ=1/(kBT)as expected. We also deduce
p= − r i=1
Pi∂Ei
∂V; μ= r i=1
Pi∂Ei
∂N. (19.37)
According to Eq. (19.37), the pressurepcan be interpreted heuristically as if∂Ei/∂Vwere a force per unit area associated with each state and∂Ei/∂N were an energy per particle associated with each state. From the forms of Eqs. (19.35) and (19.36), we see that a change dSin entropy results from a change in populationsPiat fixedEi; however, reversible work results from a change dE¯ of energy at constantS, and therefore from a change ofEi at constant populationPiand fixed particle numberN. Similarly, the chemical potentialμ results from a change inEiwithN at constant populationPiand fixedV.
Recognizing that the philosophy of this ensemble is to specifyT and take whatever E¯ corresponds, we return to the notation of thermodynamics and write Eq. (19.25) in the form
U= r i=1
PiEi, (19.38)
where the summation is over all states. From Eqs. (19.29) and (19.33) we deduce that TS= −kBT
r i=1
Pi(−βEi−lnZ)=U+kBTlnZ. (19.39) Thus the Helmholtz free energy
F=U−TS= −kBTlnZ (19.40)
in agreement with Eq. (19.14) or (19.23).
As an alternative procedure, we could have identifiedβ by comparing dU with dSat constantV andN, in which case dU=TdS. Then we could calculateSfrom Eq. (19.39) and the results in Eq. (19.37) could be obtained from p= −∂F/∂V, μ= −∂F/∂N and Eq. (19.33).
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Before leaving this section, we remark that instead of the most probable values of Pj
or, equivalentlyNj =NensPj, we could deal with themeanvaluesNjwith respect to the quantitiesWens{Ni}given by Eq. (19.26). Specifically,
Nj =
{Ni}NjWens{Ni}
{Ni}Wens{Ni} , (19.41)
where the sums are to be taken overallvalues of the set{Ni}that are compatible with the constraint Eqs. (19.24) and (19.25), written in terms of theNi. By means of a somewhat technical and lengthy calculation (e.g., see Schrửdinger [99, p. 27] or Pathria [8, p. 46]), it can be shown thatNjandNjcalculated for the most probable distribution are the same in the limitNens→ ∞.