Localized harmonic oscillators 393 .2.2 The extreme quantum and the classical limits
The assembly of localized oscillators forms a good example of the use ofkkkBTas the
thermalenergy scale.
Firstlyconsider the oscillators at an extremely low temperature,i.e.Tθor
kkkBThν. The oscillators are frozen into the ground state, in that virtually none is
thermally excitedat any one time. WehaveU=U( 0 ),C= 0 ,S= 0 .Thissituation
should bethoughtofas the ‘extreme quantumlimit’. Thediscrete (i.e. quantum)
nature of the states given by (3.11) is totally dominating the properties.
However, at the opposite extreme oftemperature,i.e.TθorkkkBThν,we
reach a ‘classical limit’. In this limit the simple expressions forUandCinvolve
onlykkkBTandnot hν. Planck’s constant, i.e. the scale of the energy level split-
ting,isirrelevant now. Oscillators ofanyfrequencyhave the same average energy
kkkBT.This dependence ofUonkkkBTalone is associated with the old classical ‘law
of equipartition of energy’, which states that each so-called degree of freedom
ofasystem contributes^12 kkkBTtotheinternalenergy.Inthis case eachoscillator
displays two degrees of freedom (one for its kinetic energyand one for its poten-
tial energy), and the old law gives the resultU = NkkkBTfor theNoscillators.
But the questionleft unansweredbytheclassicallaw aloneis: Whenisadegree
of freedom excited and when is it frozen out? The answer is in the energy (or
temperature) scales! We shallreturn to theseideasin ourdiscussions ofgasesin
Chapters 6 and 7.
Before leaving this topic, let us examine the entropyS. One can see from(3.16)
that evenintheclassicallimit, the expressionforSinvolvesh,Planck’s constant.
Entropyisfundamentallya quantum effect. From theclassicalregion alone one can
deduce thatS=S 0 +NkkkBlnT, but there is no way of finding the constantS 0 without
integrating thethermalproperties upwardsfromT=0,i.e. throughthe quantum
region. This problemlies at theheart ofa number ofthehistoricalparadoxes and
controversies of classical statistical mechanics in the pre-quantum age.
This quantum nature ofSiswell illustratedfromamicroscopicinterpretation of
(3.1 6 ). If one has an assemblyof particles, each of which maybe inGstates with
equal probability, thenSis readily evaluated.(G = 2 is the spin-^12 solid at high
temperatures,for example, andwe usedG= 4 for Cu nucleihavingspin^32 .) Since
=GNfor thissituation, we maywriteS=kkkBln=NkkkBlnG.Ifwe compare this
expression with the high temperature limit of (3.1 6 ), we see that the twogive the same
answerifG=kkkBT/hν, theratioofthetwoenergyscales. Thisisapleasinglyplausible
result. It means that the entropyofthe oscillatorsisthe same athightemperatures
as that of a hypothetical equal-occupation system in whichGstates are occupied. As
kkkBTrises,Gcontinues to rise ashigherlevels(there are aninfinite numberinall!)
comeinto play. Hence thelnTterm. But the actualvalue ofGdependsonthe ratio
ofkkkBTtohν. It involves the quantum energy-level scale of the oscillators, so thatG
andSareinevitably quantum properties.