QMGreensite_merged

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)