16.2. THEFREEELECTRONGAS 259
While,attheboundaries,weimposethe”box”conditions
ψ(0,y,z) = ψ(L,y,z)= 0
ψ(x, 0 ,z) = ψ(x,L,z)= 0
ψ(x,y,0) = ψ(x,y,L)= 0 (16.28)
Thesolutionoftheparticleinaboxisasimplegeneralizationoftheparticleina
tube(Lecture5),andtheparticleinasquare(Lecture10).Theeigenstatesare
ψn 1 n 2 n 3 (x,y,z)=
( 2
L
) 3 / 2
sin(
πn 1
L
x)sin(
πn 2
L
y)sin(
πn 3
L
z) (16.29)
withenergyeigenvalues
En 1 n 2 n 3 =
π^2 ̄h^2
2 mL^2
(n^21 +n^22 +n^23 ) (16.30)
SupposethesolidisattemperatureT=0. Weaskthequestion: (i)whatisthe
totalenergyET oftheelectronsinthesolid;and(ii)whatistheenergyEF ofthe
mostenergeticelectroninthesolid? Theenergyofthemostenergeticelectronina
coldsolid,EF,isknownastheFermiEnergy.
StartwithEF. ThemethodforcalculatingtheFermienergyistofirstsuppose
thatweknow it,and,assumingthateveryenergyEn 1 n 2 n 3 <EF isfilledwithtwo
electrons(spinup/down),figureoutthetotalnumberofelectronsinthesolid. By
settingthisnumbertoN,wecandetermineEF.
Toeacheigenstatetherecorrespondsasetofintegersn 1 ,n 2 ,n 3 ,soeachpossible
energystatecanberepresentedapointinathree-dimensionalspace,withpositive
integercoordinates.Denotethemaximumvalueofn^21 +n^22 +n^23 ,forthemostenergetic
state,byR^2 .TheFermiEnergyistherefore
EF =
π^2 ̄h^2
2 mL^2
R^2 (16.31)
NowallofthestateswithE≤EF areoccupied. Soweneedtocountthenumberof
pointswithintegercoordinates(n 1 n 2 n 3 )suchthat
n^21 +n^22 +n^23 ≤R^2 (16.32)
Butsincethereisonesitewithintegercoordinatesperunitvolume,thenumberof
sitessatisfying(16.32)issimplythevolumeofaoneoctantofasphereofradiusR
(seeFig.16.4).Sincetherecanbenomorethantwoelectronspersite(n 1 n 2 n 3 ),the
totalnumberofelectronswithenergieslessthanEF,withalllevelsfilled,is
N= 2 ×
1
8
×
4
3
πR^3 =
1
3
πR^3 (16.33)