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SJP QM 3220 3D 1

Page H-1 M. Dubson, (typeset by J. Anderson) Mods by S. Pollock Fall 2008

AngularMomentum(warm‐upforH‐atom)Classically,angularmomentumdefinedas(fora1‐particlesystem)

Note: definedw.r.t.anoriginofcoords.

(InQM,theoperatorcorrespondingtoLxis

accordingtoprescriptionofPostulate2,part3.)

Classically,torquedefinedas and (rotationalversionof )

Iftheforceisradial(centralforce),then H‐atom:Inamulti‐particlesystem,totalaveragemomentum:

isconservedforsystemisolatedfromexternaltorques.

sumoverparticlesInternaltorquescancauseexchangeofaveragemomentumamongparticles,but

remainsconstant.Inclassicalandquantummechanics,only4thingsareconserved:

energy linearmomentum angularmomentum electriccharge

Ox

ym

protonatorigin

electron

(Coulombforce)

SJP QM 3220 3D 1


BacktoQM.Definevectoroperator operatorunitvector

Recall

Claim:foracentralforcesuchasinH‐atom

(willshowthislater)

Thisimplies (justlikeinclassicalmechanics)

AngularmomentumofelectronisH‐atomisconstant,solongasitdoesnotabsorboremitphoton.Throughoutpresentdiscussion,weignoreinteractionofH‐atomw/photons.WillshowthatforH‐atomorforanyatom,molecule,solid–anycollectionofatoms–theangularmomentumisquantizedinunitsofħ. canonlychangebyintegernumberofħ's.

Claim:

and (i,j,kcyclic: xyzor yzxor zxy)

SJP QM 3220 3D 1


Toprove,needtwoveryusefulidentities:

Proof:

(Haveused

I'mdroppingtheˆoveroperatorswhennodangerofconfusion.Since[Lx,Ly]≠0,cannothavesimultaneouseigenstatesof

However, doescommutewithLz.

Claim:

,i=x,y,orz

Proof:

=0(Notecancellations)[L2,Lz]=0=>canhavesimultaneouseigenstatesof

allothertermslike[y,px]=0

SJP QM 3220 3D 1


LookingforwardtoH‐atom:

Wewillshowthat

=>simultaneouseigenstatesof

WhenwesolvetheTISE�ψ=EψfortheH‐atom,thenaturalcoordinatestousewillbesphericalcoordinates:r,θ,φ(notx,y,z) x=rsinθcosφ y=rsinθsinφ z=rcosθ

Justrewriting insphericalcoordinatesisgawd‐awful.But

separationofvariableswillgivespecialsolutions,energyeigenstates,offormTheangularpartofthesolutionY(θ,φ)willturnouttobeeigenstatesofL2,LzandwillhaveformcompletelyindependentofthepotentialV(r). *

energyq‐nbr

Lzq‐nbr

L2q‐nbr

y

x

z

φ

θ

SJP QM 3220 3D 1


Givenonly[L2,Lz]=0and hermiteanweknowtheremustexistsimultaneouseigenstatesf(whichwillturnouttobetheY(θ,φ)mentionedabove)suchthat (λwillberelatedtol,andμwillberelatedtom)Wewillshowthatfwilldependonquantum‐numbersl,m,sowewriteitasflm,andthat willbedeterminedlater.NoticemaxeigenvalueofLz(=lħ)issmallerthansquarerootofeigenvalueof

So,inQM,Lz<|L|Odd!Alsonoticel=0,m=0statehaszeroangularmomentum(L2=0,Lz=0)so,unlikeBohrmodel,canhaveelectroninstatethatis"justsittingthere"ratherthanrevolvingaboutprotoninH‐atom.Proofofboxedformulae:(Thisprooftakes2½pages!) DefineL+=Lx+iLy="raisingoperator"L‐=Lx‐iLy="loweringoperator"(NoteL+†=L‐,L‐†=L+,A†=hermiteanadjointofA)NeitherL+orL‐arehermitean(self‐adjoint).

Note

=>Considerf:

SJP QM 3220 3D 1


Claim:g=L+fisaneigenfunctionofLzwitheigenvalue=(μ+ħ).SoL+operatorraiseseigenvalueofLzby1ħ.Proof: ToproveLzg=(μ+ħ)g,needtoshowthat[Lz,L+]=ħL+

Now

So,operatingonfwithraisingoperatorL+raiseseigenvaluesofLZby1ħbutkeepseigenvalueofL2unchanged.(Similarly,L‐lowerseigenvalueofLzby1ħ.)OperatingrepeatedlywithL+raiseseigenvalueofLzbyħeachtime:L+(L+f)has(μ+2ħ)etc.ButeigenvalueofLzcannotincreasewithoutlimitsince cannotexceed

Thereisonlyonewayout.Theremustbeforagivenλa"topstate"ftforwhichL+ft=0.Likewise,theremustbeforagivenλa"bottomstate"fbforwhichL‐fb=0.

Lz

fb

ftL‐

L‐L+

L+

ħ

allwithsameλ=eigenvalueofL2

SJP QM 3220 3D 1


WriteLzf=mħ∙f,mchangesbyintegersonlyLzft=ℓħ∙ft,ℓ=maxvalueofmL2ft=?WanttowriteL2intermsofL+,Lz:

=> (Also, )=>

So, whereℓ=maxm,sameλforallm's.

Repeatfor minvalueofm.

(tryit!)

Sommin=‐mmaxandmchangesonlyinunitsof1.=>m=‐ℓ,‐ℓ+1,...ℓ‐2,ℓ‐1,ℓ Nintegersteps=>2ℓ=N,ℓ=N/2=>ℓ=0,1/2,1,3/2,2,5/2,...Endofproofof

SJP QM 3220 3D 1


We'llseelaterthatthereare2flavorsofangularmomentum:

m

0ℓ

1 2 3

2

‐2

‐1

0

1

‐1

0

1

0

1.OrbitalAng.Mom.(integerℓonly)

2.SpinAng.Mom.(integeror½integerOK)

SJP QM 3220 3D 1


mp>>me=>proton(nearly) stationary

Hamiltonianofelectron

TISE: specialsolutions(stationarystates).

GeneralSolutiontoTDSE:

SphericalCoordinateSystem: z=rcosθ x=rsinθcosφ y=rsinθsinφ ψ=ψ(r,θ,φ)Normalization:Need insphericalcoordinates

HardWay:

TheHatom

mp≈1840me

me

z

φ

θ

r

volume

SJP QM 3220 3D 1


Alsoneed9derivatives:

EasierWay:Curvilinearcoordinates(SeeBoas)pathelement:

Sphericalcoordinates:

*InClassicalMechanics(CM),KE=p2/2m=KE=(radialmotionKE)+(angular,axialmotionKE)O

SJP QM 3220 3D 1


*SamesplittinginQM:

(Notice dependsonlyonθ,φandnotr.)

SeparationofVariables!(asusual)Seekspecialsolutionofform:

Normalization:∫dV|ψ|2=(Convention:normalizeradial,angularpartsindividually)Plugψ=R∙YintoTISE=>

Multiplythruby :

=>f(r)=g(θ,φ)=constantC=ℓ(ℓ+1)

SJP QM 3220 3D 1


HaveseparatedTISEintoradialpartf(r)=ℓ(ℓ+1),involvingV(r),andangularpartg(θ,φ)=ℓ(ℓ+1)whichisindependentofV(r).=>Allproblemswithsphericallysymmetricpotential(V=V(r))haveexactlysameangularpartofsolution:Y=Y(θ,φ)called"sphericalharmonics".We'lllookatangularpartlater.Now,let'sexamine

RadialSE:

Changeofvariable:u(r)=r∙R(r)

Canshowthat

Notice:identicalto1DTISE:

except

r:0‐>∞insteadofx:‐∞‐>+∞and

V(x)replacedwith

Veff="effectivepotential"Boundaryconditions:

same!

SJP QM 3220 3D 1


Seek bound state solutions E < 0

E > 0 solutions are unbound states, scattering solutions

u(r=∞)=0fromnormalization∫dr|u|2=1

u(r=0)=0,otherwise blowsupatr=0(subtle!)

FullsolutionofradialSEisverymessy,eventhoughitiseffectivelya1Dproblem(differentproblemforeachℓ)Powerseriessolution(seetextfordetails).Solutionsdependon2quantumnumbers:nandℓ(foreacheffectivepotentialℓ=0,1,2,…haveasetofsolutionslabeledbyindexn.)Solutions:n=1,2,3,… forgivenn ℓ=0,1,…(n‐1) ℓmax=(n–1)n="principalquantumnumber"energyeigenvaluesdependonnonly(itturnsout)

(independentofℓ)

•sameasBohrmodel,agreeswithexperiment!

Notice that energy eigenvalues given by solution to radial equation alone.

SJP QM 3220 3D 1


Firstfewsolutions:Rnℓ(r)normalization"Bohrradius"

NOTE:•forℓ=0(sstates),R(r=0)≠0=>wavefunctionψ"touches"nucleus.•forℓ≠0,R(r=0)=0=>ψdoesnottouchnucleus.ℓ≠0=>electronhasangularmomentum.Sameasclassicalbehavior,particlewithnon‐zeroLcannotpassthruorigin CanalsoseethisinQM:forℓ≠0,Veffhasinfinitebarrieratorigin=>u(r)mustdecaytozeroatr=0exponentially.

=>exponentialdecayin

aswell.

SJP QM 3220 3D 1


Backtoangularequation: Wanttosolveforthe ‐"sphericalharmonics".Before,startedwithcommutationrelations,

and,usingoperatoralgebra,solvedfortheeigenvaluesofL2,Lz.Wefound whereℓ=0,½,1,3/2,… m=‐ℓ,‐ℓ+1…+ℓIntheprocess,wedefinedraisingandloweringoperators:

(cmissomeconstant)

So,ifwecanfind(foragivenℓ)asingleeigenstate ,thenwecangenerateallthe

others(otherm's)byrepeatedapplicationof .

It'seasytofindtheφ‐dependence;don'tneedthe businessyet.

€

ˆ L z =

i∂∂ϕ

(showed in HW)

ˆ L zY =

i∂Y∂ϕ

= mY (and you can cancel the )

Assume

Ifweassume(postulate)thatψissingle‐valuedthan=>m=0,±1,±2,…Butm=‐ℓ,…+ℓSofororbitalangularmomentum,ℓmustbeintegeronly:ℓ=0,1,2,…(throwout½integervalues)

SJP QM 3220 3D 1


* (algebra!)

Candeduce from

=>

Solution:(un‐normalized)Checks:Plugbackin.

Now,cangetother byrepeatedapplicationof Somewhatmessy(HW!)

Normalizationfrom Noticecaseℓ=0 :Example:Conventionon±sign:

SJP QM 3220 3D 1


Thesphericalharmonicsformacomplete,orthonormalset(sinceeigenfunctionsofhermiteanoperators)Anyfunctionofanglesf=f(θ,φ)canbewrittenaslinearcomboof :Likewise:

=>H‐atomenergyeigenstatesare n=1,2,…;ℓ=0,1…(n‐1);m=‐ℓ…+ℓArbitrary(bound)stateis (c'sareanycomplexconstants)energyofstate(n,ℓ,m)dependsonlyonn.En=‐constant/n2(statesℓ,mwithsamenaredegenerate)

Degeneracyofnthlevelisn2(2•n2ifyouincludespin)

ℓ=

n=

2

3d

1

4d(1)4s

4f

3

4

0 1 2 3

3s2s1s

4p

3p

2p

(3) (5) (7)

SJP QM 3220 3D 1


Prob(findparticleindVabout )=

Ifℓ=0,ψ=ψ(r)then Prob(findinr�r+dr)=P(r)=radialprobabilitydensity

Groundstate:

NoticeP(r)verydifferentfromψ(r):

Ifℓ≠0,ψ=ψ(r,θ,φ)=R(r)Y(θ,φ),then

Prob(findinr�r+dr)=r2|R|2dr evenifℓ≠0Note: ifH‐atomandemission/absorptionofradiation:IfH‐atomisinexcitedstate(n=2,ℓ=1,m=0)thenitisinenergyeigenstate=stationarystate.Ifatomisisolated,thenatomshouldremaininstateψ210forever,sincestationarystatehassimpletimedependence:

RadialProbabilityDensity

"solidangle"

P(r)=r2|R|2

SJP QM 3220 3D 1


But,experimentally,wefindthatH‐atomemitsphotonandde‐excites:ψ210‐>ψ100in≈10‐7s‐>10‐9s

Thereasonthattheatomdoesnotremaininstationarystateisthatitisnottrulyisolated.TheatomfeelsafluctuatingEMfielddueto"vacuumfluctuations".QuantumElectrodynamicsisarelativistictheoryoftheQMinteractionofmatterandlight.Itpredictsthatthe"vacuum"isnot"empty"or"nothing"aspreviouslysupposed,butisinsteadaseethingfoamofvirtualphotonsandotherparticles.ThesevacuumfluctuationsinteractwiththeelectronintheH‐atomandslightlyalterthepotentialV(r).Soeigenstatesofthecoulombpotentialarenoteigenstatesoftheactualpotential:Vcoulomb+VvacuumPhotonspossessanintrinsicangularmomentum(spin)of1ħ,meaningSowhenanatomabsorbsoremitsasinglephoton,itsangularmomentummustchangeby1ħ,byConservationofAngularMomentum,sotheorbitalangularmomentumquantumnumberℓmustchangeby1."SelectionRule":∆ℓ=±1inanyprocessinvolvingemissionorabsorptionof1photon=>allowedtransitionsare:

IfanH‐atomisinstate2s(n=2,ℓ=0)thenitcannotde‐excitetogroundstatebyemissionofaphoton.(sincethiswouldviolatetheselectionrule).Itcanonlyloseitsenergy(de‐excite)bycollisionwithanotheratomorviaarare2‐photonprocess.

ps

n=1

2

3

4

5d

2p

1s

E

Eγ=hf=∆E

SJP QM 3220 3D 1


completeorthonormalset

€

ψ = cnn∑ n , ψ → cn{ } =

c1

c2

c3

cn

€

u1 =

100

u2 =

010

Ifket'sarerepresentedbycolumnvectors,thenbra'sarerepresentedbythetransposeconjugateofcolumn=row,complexconjugate.

Operatorscanberepresentedbymatrices: nohatonmatrixelement

where{|n>}issomecompleteorthonormalset.

MatrixFormulationofQM

SJP QM 3220 3D 1


Whyisthat?Wheredoesthatmatrixcomefrom?Considertheoperator and2statevectors relatedby

() Inbasis{|n>},

€

ψ = cnn∑ n = n

n∑ n ψ

cn

ϕ = dnn∑ n = n

n∑ n ϕ

dn

Nowprojectequationonto|m>byactingwithbra:

€

m ϕ = m ˆ A ψ = cnn∑ m ˆ A n

dn = Amnn∑ cn

But,thisissimplytheruleformultiplicationofmatrixcolumn.

€

d1

d2

d3

=

A11 A12 …

A21 A22

c1

c2

Sothereyouhaveit,that'swhytheoperatorisdefinedasthismatrix,inthisbasis!Now,suppose areenergyeigenstates,then

Amatrixoperator isdiagonalwhenrepresentedinthebasisofitsowneigenstates,andthediagonalelementsaretheeigenvalues.Noticethatingeneraloperatorsdon'tcommute .SamegoesforMatrixMultiplication:A B≠B A

SJP QM 3220 3D 1


Claim:Thematrixofahermitianoperatorisequaltoitstransposeconjugate:Proof:

€

m ˆ A n = ˆ A m n = n ˆ A m*

⇒ Amn = Anm*

Similarly,adjoint(or"Hermitianconjugate") Proof:

€

ˆ A m n = m ˆ A tn = n ˆ A m*

Ofcourse,it'sdifficulttodocalculationsifthematricesandcolumnsareinfinitedimensional.ButthereareHilbertsubspacesthatarefinitedimensional.Forinstance,intheH‐atom,thefullspaceofboundstatesisspannedbythefullset{n,ℓ,m}(=|nℓm>).Thesub‐set{n=2,ℓ=1,m=+1,0,‐1}formsavectorspacecalledasubspace.Subspace?InordinaryEuclideanspace,anyplaneisasubspaceofthefullvolume.Ifweconsiderjustthexycomponentsofavector ,thenwehaveaperfectlyvalid2Dvectorspace,eventhoughthe"true"vectoris3D.Likewise,inHilbertspace,wecanrestrictourattentiontoasubspacespannedbyasmallnumberofbasisstates.Example:H‐atomsubspace{n=2,ℓ=1,m=+1,0,‐1}Basisstatesare (candropn=2,ℓ=1inlabelsincetheyarefixed.)

€

ˆ L z m = m m

ˆ L 2 m = 2( +1) m

( =1)= 22 m (for all m)

€

⇒ Lz( )mn= m ˆ L z n =

+1 0 00 0 00 0 −1

SJP QM 3220 3D 1


€

L2mn = m ˆ L 2 n = 2

1 0 00 1 00 0 1

(WhataboutLx?Ly?)Beforeseeingwhatallthismatrixstuffisgoodfor,let'sexaminespinbecauseit'sveryimportantphysicallyandbecauseitwillleadto2DHilbertspacewithsimple2x2matrices.bracketorinner‐product:

Whichintegralyoudodependsontheconfigurationspaceofproblem.Keydefiningpropertiesofbracket:

<f|g>*=<g|f> c=constant

<f|c•g>=c<f|g>,<c•f|g>=c*<f|g>

<α|(b|β>+c|γ>)=b<α|β>+c<α|γ>Diracproclaims:<g|f>=<g|nextto|f> bracket="bra"and"ket"Ket|f>representsvectorinH‐space(HilbertSpace)"ket""wavefunction"

Bothψandψ(x)describesamestate,but|ψ>ismoregeneral:

ReviewofDiracBraKetNotation

Different"representations"ofsameH=spacevector|ψ>

SJP QM 3220 3D 1


�position‐representation,momentum‐rep,energy‐rep.

SJP QM 3220 3D 1


Whatisa"bra"?<g|isanewkindofmathematicalobject,calleda"functional".

insertstatefunctionhere input outputfunction: number numberoperator: function functionfunctional: function numbe<g|wantstobindwith|f>toproduceinnerproduct<g|f>Foreveryket|f>thereisacorrespondingbra<f|.Likethekets,thebra'sformavectorspace.

|cf>�<cf|=c*<f| (?)

|αf+βg>�<αf+βg|=α*<f|+β*<g|

<αf+βg|h>=α*<f|h>+β*<g|h>

Complexnumberbra=anotherbra =>bra'sformanylinearcomboofbra's=anotherbra vectorspaceThevectorspaceofbrasiscalleda"dualspace".It'sthedualoftheketvectorspace.

isaket.Whatisthecorrespondingbra?

Definition:hermiteanconjugateoradjoint

€

ˆ A f g ≡ f Atg forallf,g.

(If ishermiteanorself‐adjoint.)Someproperties:

€

Proof : f ˆ A t( )tg = ˆ A t f g =

g ˆ A t f*

= ˆ A g f*

= f ˆ A f

Def'nofA†

SJP QM 3220 3D 1


Theadjointofanoperatorisanalogoustocomplexconjugateofacomplexnumber:

The"ket‐bra"|f><g|isanoperator.Itturnsaket(function)intoanotherket(function):

€

f g ( ) h = f g h ProjectionOperators

€

ψ(x) = cnun (x) = un ψ n∑

n∑ un (x)→

ψ = cn nn∑ = n ψ

n∑ n = n

n∑ n ψ

=> "Completenessrelation" (discretespectrumcase)

="projectionoperator"

picksoutportionofvector|ψ>thatliesalong|n>

€

ˆ P n ψ = n n ψ = cn n

|ψ>

u2=|2>

u1=|1>

|2><2|ψ>

u1<u1|ψ>=|1><1|ψ>

SJP QM 3220 3D 1


€

ψ = nn∑ n ψ like

R = ˆ x x ⋅

R ( ) + ˆ y ˆ y ⋅

R ( )

€

ˆ 1 = nn∑ n like 1 = ˆ x x ⋅ _ ( ) + ˆ y ˆ y ⋅ _ ( )

Anywherethereisaverticalbarinthebracket,oraketorabra,wecanreplacethebarwith

Example:

=>

Ifeigenvaluespectrumiscontinuous(asfor )thenmustuseintegral,ratherthansum,overstates.

CompletenessRelation(continuousspectrum)

Example:

€

Φ(p) = f p ψ = dx∫ f p x x ψ =1

2πdx∫ e− ipx ψ(x)

TheMeasurementPostulates3and4canberestatedintermsoftheprojectionoperator:Startingwithstate ,

wheresum{n}isoveranycompletesetofstates,ifwemeasureobservableassociatedwithn,thenwewillfindvaluen0withprobability

Probabilityoffindingeigenvaluen0=expectationvalueofprojectionoperator .Andasresultofmeasurementstate|ψ>collapsestostate . (apartfromnormalization)

SJP QM 3220 3D 1


Wecannowgeneralizetocaseofstatesdescribedbymorethanoneeigenvalue,suchasH‐atom.

Ifwemeasureenergy(butnotalso ),findn0,thenweareprojectingonto

subspacespannedby{ℓ,m}withsomen0.

Statecollapsesto

mustrenormalize

SJP QM 3220 3D 1


Spin½RecallthatintheH‐atomsolution,weshowedthatthefactthatthewavefunction

Ψ(r)issingle‐valuedrequiresthattheangularmomentumquantum#beinteger:l=

0,1,2..However,operatoralgebraallowedsolutionsl=0,1/2,1,3/2,2…

Experimentshowsthattheelectronpossessesanintrinsicangularmomentum

calledspinwithl=½.Byconvention,weusethelettersinsteadofl forthespin

angularmomentumquantumnumber:s=½.

Theexistenceofspinisnotderivablefromnon‐relativisticQM.Itisnotaformof

orbitalangularmomentum;itcannotbederivedfrom .

(Theelectronisapointparticlewithradiusr=0.)

Electrons,protons,neutrons,andquarksallpossessspins=½.Electronsand

quarksareelementarypointparticles(asfaraswecantell)andhavenointernal

structure.However,protonsandneutronsaremadeof3quarkseach.The3half‐

spinsofthequarksaddtoproduceatotalspinof½forthecompositeparticle(ina

sense,↑↑↓makesasingle↑).Photonshavespin1,mesonshavespin0,thedelta‐

particlehasspin3/2.Thegravitonhasspin2.(Gravitonshavenotbeendetected

experimentally,sothislaststatementisatheoreticalprediction.)

SJP QM 3220 3D 1


SpinandMagneticMoment

Wecandetectandmeasurespinexperimentallybecausethespinofa

chargedparticleisalwaysassociatedwithamagneticmoment.

Classically,amagneticmomentisdefinedasavectorµ associatedwith

aloopofcurrent.Thedirectionofµ isperpendiculartotheplaneof

thecurrentloop(right‐hand‐rule),andthemagnitudeis

.

Theconnectionbetweenorbitalangularmomentum(notspin)andmagnetic

momentcanbeseeninthefollowingclassicalmodel:Consideraparticlewithmass

m,chargeqincircularorbitofradiusr,speedv,periodT.

|angularmomentum|=L=pr=mvr,sovr=L/m,and .

Soforaclassicalsystem,themagneticmomentisproportionaltotheorbital

angularmomentum: .

Thesamerelationholdsinaquantumsystem.

InamagneticfieldB,theenergyofamagneticmomentisgivenby

(assuming ).InQM, .

Writingelectronmassasme(toavoidconfusionwiththemagneticquantumnumber

m)andq=–ewehave ,wherem=−l..+l.Thequantity

ri

µ

r

i

m,q

SJP QM 3220 3D 1


iscalledtheBohrmagneton.Thepossibleenergiesofthemagnetic

momentin isgivenby .

Forspinangularmomentum,itisfoundexperimentallythattheassociatedmagnetic

momentistwiceasbigasfortheorbitalcase:

(WeuseSinsteadofLwhenreferringtospinangularmomentum.)

Thiscanbewritten .

Theenergyofaspininafieldis (m=±1/2)afactwhichhasbeen

verifiedexperimentally.

Theexistenceofspin(s=½)andthestrangefactorof2inthegyromagneticratio

(ratioof )wasfirstdeducedfromspectrographicevidencebyGoudsmitand

Uhlenbeckin1925.

Another,evenmoredirectwaytoexperimentallydeterminespiniswithaStern‐

Gerlachdevice,nextpage

SJP QM 3220 3D 1


(ThispagefromQMnotesofProf.RogerTobin,PhysicsDept,TuftsU.)

Stern-Gerlach Experiment (W. Gerlach & O. Stern, Z. Physik 9, 349-252 (1922).

€

F = ∇ µ • B ( ) = ( µ •

∇ ) B (in current free regions), or here,

F = ˆ z (µz

∂Bz

∂z)(thisisalittle

crude‐seeGriffithsExample4.4forabettertreatment,butthisgivesthemainidea)

Deflectionofatomsinz‐directionisproportionaltoz‐componentofmagnetic

momentµz,whichinturnisproportionaltoLz.Thefactthattherearetwobeamsis

proofthatl=s=½.Thetwobeamscorrespondtom=+1/2andm=–1/2.Ifl=1,

thentherewouldbethreebeams,correspondingtom=–1,0,1.Theseparationof

thebeamsisadirectmeasureofµz,whichprovidesproofthat

Theextrafactorof2intheexpressionforthemagneticmomentoftheelectronis

oftencalledthe"g‐factor"andthemagneticmomentisoftenwrittenas

.Asmentionedbefore,thiscannotbededucedfromnon‐relativistic

QM;itisknownfromexperimentandisinserted"byhand"intothetheory.

SJP QM 3220 3D 1


However,arelativisticversionofQMduetoDirac(1928,the"DiracEquation")

predictstheexistenceofspin(s=½)andfurthermorethetheorypredictsthevalue

g=2.Alater,betterversionofrelativisticQM,calledQuantumElectrodynamics

(QED)predictsthatgisalittlelargerthan2.Theg‐factorhasbeencarefully

measuredwithfantasticprecisionandthelatestexperimentsgiveg=

2.0023193043718(±76inthelasttwoplaces).ComputingginQEDrequires

computationofabinfiniteseriesoftermsthatinvolveprogressivelymoremessy

integrals,thatcanonlybesolvedwithapproximatenumericalmethods.The

computedvalueofgisnotknownquiteaspreciselyasexperiment,neverthelessthe

agreementisgoodtoabout12places.QEDisoneofourmostwell‐verified

theories.

SpinMath

Recallthattheangularmomentumcommutationrelations

werederivedfromthedefinitionoftheorbitalangularmomentumoperator:

.

Thespinoperator doesnotexistinEuclideanspace(itdoesn'thaveapositionor

momentumvectorassociatedwithit),sowecannotderiveitscommutation

relationsinasimilarway.Insteadweboldlypostulatethatthesamecommutation

relationsholdforspinangularmomentum:

.Fromthese,wederive,justabefore,that

SJP QM 3220 3D 1


(sinces=½)

(sincems=−s,+s=−1/2,+1/2)

Notation:sinces=½always,wecandropthisquantumnumber,andspecifythe

eigenstatesofL2,Lzbygivingonlythemsquantumnumber.Therearevariousways

towritethis:

Thesestatesexistina2DsubsetofthefullHilbertSpacecalledspinspace.Since

thesetwostatesareeigenstatesofahermitianoperator,theyformacomplete

orthonormalset(withintheirpartofHilbertspace)andany,arbitrarystateinspin

spacecanalwaysbewrittenas (Griffiths'notationis

)

Matrixnotation: .Notethat

IfwewereworkinginthefullHilbertSpaceof,say,theH‐atomproblem,thenour

basisstateswouldbe .Spinisanotherdegreeoffreedom,sothatthe

fullspecificationofabasisstaterequires4quantumnumbers.(Moreonthe

connectionbetweenspinandspacepartsofthestatelater.)

[Noteonlanguage:throughoutthissectionIwillusethesymbolSz(andSx,etc)to

refertoboththeobservable("themeasuredvalueofSzis ")anditsassociated

operator("theeigenvalueofSzis ").]

SJP QM 3220 3D 1


ThematrixformofS2andSzinthe basiscanbeworkedoutelementby

element.(Recallthatforanyoperator .)

Operatorequationscanbewritteninmatrixform,forinstance,

WearegoingaskwhathappenswhenwemakemeasurementsofSz,aswellasSx

andSy,(usingaStern‐Gerlachapparatus).Willneedtoknow:Whatarethe

matricesfortheoperatorsSxandSy?Thesearederivedfromtheraisingand

loweringoperators:

TogetthematrixformsofS+,S−,weneedaresultfromthehomework:

Forthecases=½,thesquarerootfactorsarealways1or0.Forinstance,s=½,

m=−1/2gives .Consequently,

,leadingto

and

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NoticethatS+,S−arenothermitian.

Using yields

Thesearehermitian,ofcourse.

Oftenwritten: ,where are

calledthePaulispinmatrices.

Nowlet'smakesomemeasurementsonthestate .

Normalization: .

SupposewemeasureSzonasysteminsomestate .

Postulate2saysthatthepossibleresultsofthismeasurementareoneoftheSz

eigenvalues: .Postulate3saystheprobabilityoffinding,say ,

is .

Postulate4saysthat,asaresultofthismeasurement,whichfound ,theinitial

state collapsesto .

ButsupposewemeasureSx?(WhichwecandobyrotatingtheSGapparatus.)

Whatwillwefind?Answer:oneoftheeigenvaluesofSx,whichweshowbeloware

thesameastheeigenvaluesofSz: .(Notsurprising,sincethereis

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nothingspecialaboutthez‐axis.)Whatistheprobabilitythatwefind,say,Sx=

?ToanswerthisweneedtoknowtheeigenstatesoftheSxoperator.Let's

callthese(sofarunknown)eigenstates (Griffithscallsthem

).Howdowefindthese?Wemustsolvetheeigenvalueequation:

,whereλ aretheunknowneigenvalues.Inmatrixformthisis,

whichcanberewritten .In

linearalgebra,thislastequationiscalledthecharacteristicequation.

Thissystemoflinearequationsonlyhasasolutionif

.So

Asexpected,theeigenvaluesofSxarethesameasthoseofSz(orSy).

Nowwecanplugineacheigenvalueandsolvefortheeigenstates:

; .

Sowehave

Nowbacktoourquestion:Supposethesysteminthestate ,andwe

measureSx.Whatistheprobabilitythatwefind,say,Sx= ?Postulate3gives

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therecipefortheanswer:

Questionforthestudent:Supposetheinitialstateisanarbitrarystate

andwemeasureSx.WhataretheprobabilitiesthatwefindSx= and ?

SJP QM 3220 3D 1


Let'sreviewthestrangenessofQuantumMechanics.

SupposeanelectronisintheSx= eigenstate .Ifweask:What

isthevalueofSx?Thenthereisadefiniteanswer: .Butifweask:Whatisthe

valueofSz,thenthisisnoanswer.ThesystemdoesnotpossessavalueofSz.Ifwe

measureSz,thentheactofmeasurementwillproduceadefiniteresultandwillforce

thestateofthesystemtocollapseintoaneigenstateofSz,butthatveryactof

measurementwilldestroythedefinitenessofthevalueofSx.Thesystemcanbein

aneigenstateofeitherSxorSz,butnotboth.

sjp qm 3220 3d 1 - physics€¦ · sjp qm 3220 3d 1 page h-6 m. dubson, (typeset by j. anderson)...

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