QuantumMechanicsChapter6.Spin§6.1TheSpinOperatorsFollowingthemethodofSection2.3,werepresenttheeigenstatesofthespinoperatorastwo-dimensionalvectors.Inthecaseofelectronspin,thereareonlytwoeigenfunctionsinsteadofthreeormore.TheHilbertspinspaceisthereforetwo-dimensional

andwecanwritetheeigenfunctionsascolumnvectorsjustaswedidfororbitalangularmomentum;theunitvectorsthatareeigenfunctionsofSzarethetwo-componentvectors

ComparewithEq.(7.18)forthethreeeigenstatesofLzwithl=1.Followingthereasoninggiventhere,wesaythatanyelectron-spinfunctionFscanbearepresentedbyasuperpositionof│+>zand│->z:

Continuingthislineofreasoning,wecanrepresentthespinoperatorsSx,Sy,andSzasthefollowing2×2matricesthatoperateonthesevectors:Thesemaybewrittenwherethedimensionlessoperatorsσx,σy,andσz(thePaulispinmatrices)areexpressedbythematricesinEq.(10.3).ThesematricescanbederivedbyusingrulesanalogoustothoseofEqs.(7.17):Itisleftasanexercisetoverifythattheseequationsareindeedsatisfiedbythematricesgiven.Youcanalsoverifytheanticommutationrelationσxσy+σyσx=0,andthecorrespondingrelationsinvolvingσz.ExampleProblem10.1FindvaluesoftheconstantsaandbthatyieldnormalizedeigenfunctionsofSx,byapplyingthisoperatortothegeneralexpressionforaspinstate:a│+>z+b│->z.

Solution.WeknowthattheeigenvaluesofSx,mustbethesameasthoseofSz.ThestatementthatisaneigenfunctionofSxwitheigenvalueisexpressedbythenotationSubstitutingthematrixforSxandperformingtheindicatedmatrixoperation.wehaveThereforeorb=a.Normalizationrequiresthatsothata=b=l/(2)1/2.Similarly,ifisaneigenfunctionwiththeeigenvalue-?/2,thenInthiscaseorb=-a=-1/(2)1/2.Inconclusion,theeigenvectorsareandrespectively.§6.2FineStructureinTheHydrogenSpectrumEffectofSpinonEnergyLevelsTheintrinsicmagneticmomentoftheelectron,whichresultsfromitsspin,leadstoasplittingoftheenergylevelsofhydrogenintosublevels,providedthatthereisamagneticfieldintheatom,becauseineachofthestatesdeterminedbytheSchreodingerequationtheelectronwouldthenhavetwopossibleva1uesforitsmagneticpotentialenergy,accordingtothevalueofthezcomponentofitsmagneticmoment.Butwhyshouldtherebeamagneticfieldinthehydrogenatom?Theonlyotherparticlepresentistheproton.Onemightexpecttheproton,liketheelectrontohaveinternalspinandmagneticmoment,anditdoes,butitturnsoutthattheproton'sintrinsicmagneticmomentproducesafieldwhichistoosmalltoaccountfortheobservedfinestructureinthehydrogenspectrum.(Theproton'smagneticmomentleadstohyperfinestructure,tobediscussedlaterinthissection.)However,theproton'selectricchargeleadstotheexistenceofamagneticfieldofsufficientstrength,becauseintheelectron'srestframetheprotonmovesinanorbitaroundtheelectron.Toputinanotherway,abodymovingthroughanelectricfieldalso“sees”amagneticfield,accordingtothetheoryofrelativity.EnergyoftheSpin-OrbitInteraction

Letususetheprecedingideatoestimatethemagnitudeofthesplittingofthehydrogenlevels.Firstwemustknowthemagneticmomentoftheelectron,whichmaybededucedfromthemagnitudeofthesplittingoftheatomicbeamintheStern-Gerlachexperiment.Thismomentisveryclosetoμ=-(e/me)S.(orμ=-(e/mec)Sincgsunits) (10.6)whereSisthespinangularmomentumoftheelectron,andeisthemagnitudeoftheelectroniccharge.SinceSz=±?/2,thezcomponentofμis(inmksunits)μz=±e?/2me=±1Bohrmagneton

[TheBohrmagnetonisoftendenotedbythesymbolμbandsimplycalledthemagneticmomentoftheelectron.Strictlyspeaking,μbisnottheelectron'smagneticmoment;itisonlyonecomponentofthatmoment.However,onecomponentisallthatisobservedinanexperimentalmeasurementofthemagneticmomentoftheelectron.]Todeterminetheenergyattributabletotheexistenceoftheelectron'smagneticmoment,wewritetheenergyofamagneticmomentμinamagneticfieldBasW=-μ?B,andfromEq.(10.6)wehaveW=-(e/me)S?B(10.7)Bmayberelatedtotheelectron'sorbitalangularmomentuminthefollowingway:Ifthevelocityoftheelectronisv,thevelocityoftheprotonrelativetotheelectronis-v.ThemagneticfieldBproducedbytheproton'schargeeistherefore

Thismaybewrittenintermsoftheelectron'sangularmomentumL=r×mvasTheinteractionenergyisthen,intheelectron'srestframeExpression(10.10)isnotthecorrectonetouseforW,however,becausetheenergysplittingseeninthelabisnotthesameasthatseenintheelectron'srestframe.Thereasonforthisisrathercomplicated,involvingaphenomenonknownastheThomasprecession,

whichisrelatedtothefactthatthetimescaleisnotthesameinthelaboratoryasintherestframe,oftheelectron.Theproperrelativistictreatmentshowsthattheenergyshiftseeninthelaboratoryisjustone-halfofthevaluegiveninEq.(10.10),orWecannowestimatethevalueofW.Weset∣L∣≌?.ForS,weknowthatthecomponentinanydirection,suchasthedirectionofLmustbe?/2inmagnitude.Ifwethenletr≌a0,wefindWtobeoftheorderof10-4

eV,whichisoftheorderofmagnituderequiredbytheobservedfinestructure.WeshallseeinChapter8thateffectsofthissizemaybecalculatedbythemethodsofperturbationtheory.ItshouldcomeasnosurprisetofindthatasafirstapproximationthisenergycanbeassumedtobetheexpectationvalueofS·L,computingthisvaluebyusingtheeigenfunctionsalreadyfoundforthehydrogenatom;thatiswhereuisaneigenfunctionofEq.(9.10).Weshouldthereforebeabletoexpressthisenergyintermsofthequantumnumbersn,l,andmfortheseeigenfunctions,ifwecanevaluatetheS·Loperator;thatis,weneedtoknowtheresultoftheoperation(S·L)opu.CombiningSpinandOrbitalAngularMomentumTheS·LoperatorcanbeincludedintheoperatorforthetotalenergyasitappearsintheSchreodingerequation.TheenergyisthenfoundbyfindingtheeigenvaluesoftheSchreodingerequationinthisform,withthevariablesSandLrepresentedasoperators.Thuswherethefunctionislabeledustoshowthatitisafunctionofthespincoordinateaswellasthethreespacecoordinates,becauseS·Lmustoperateonit.Wemayaccountforthisspindependencebyassumingthatusissomesuperpositionoftwowavefunctionsψ1andψ2thatdifferinthevalueofthespincoordinate,asfollows:ψ1=e-iωtu1(x,y,z)∣+>s(10.14a)ψ2=e-iωtu1(x,y,z)∣->s(10.14b)wherethestatevectors∣+>sand∣->sarethetwoeigenfunctionsintroducedinEqs.(10.1).TherewesawthatthespinoperatorSzoperatesonthesefunctionstoyieldthetwopossibleeigenvaluesofSz:NowwemustexaminetheeffectofallthreecomponentsofS,astheyappearintheS?Loperator.Toproceed,wemustintroducethetotalangularmomentumvectorJ=L+S.Becausetherearenoexternaltorquesonthesystemthetotalangularmomentumisconstantduringthemotion,butneitherLnorSisnecessarilyconstant.TheBfieldcausedbytheelectron'sorbitalmotionexertsatorqueonthespinmagneticmoment;whenthismagneticmomentchangesdirection,Lmustalsochangedirection,inorderthatL+Sremainconstant.Inquantummechanicalterms,thismeansthatthewavefunctioncannotbeeitherofthefunctionsofEqs.(10.14),whichhavefixedvaluesofS.Rather,thewavefunctionmustbeasuperpositionofthesetwofunctions.ThevectorS,ofcourse,remainsconstantinmagnitude,becausethisisafundamentalpropertyoftheelectron,butthedirectionofSisnotdetermined.HavingintroducedJ,wenowcanexpressS?LintermsofJ,L,andS,usingtheequation

J2≡J·J=(L+S)·(L+S)=S2+L2+(S·L)+(L·S)(10.16)ButtheLoperatormustcommutewiththeSoperator,becauseLopandSopoperateondifferentcoordinates.Therefore(S·L)=(L·S),andwecansolveforS·L,obtaining

S·L=(J2

–S2

–L2)/2 (10.17)Theproblemofevaluating(S·L)isnowreducedtoevaluating(J2

–S2

–L2).Wehavealreadyfound[Eq.(6.58)]thattheeigenvaluesofL2canbewrittenintheform

L2=l(l+1)?2 (10.18)

wherelisdefinedasthemaximumvalueofm,orofLz/?(Section2.3).ByanalogwemightsuspectthatwecouldalsowriteS2=s(s+1)?2(10.19)andJ2=j(j+l)?2(10.20)wheresisdefinedasthemaximumvalueofSz/?andjisdefinedasthemaximumvalueofJz/?.BecauseSz=±?/2,thevalueofsmustbel/2.ThisleadstoavalueforS2thatisconsistentwithwhatweknowaboutorbitalangularmomentum:

Fromtheisotropyofspace,Sx2=Sy2=Sz2=?2/4,andbydefinition,S2=Sx2+Sy2+Sz2=3?2/4(10.21)thesameresultthatwefindfromEq.(10.19).NoticethesimilarityofthisdiscussiontotheonethatledtoEq.(6.58).TheonlydifferenceisthatEq.(6.58)involvedexpectationvalues(e.g.,<Lx2>).Inthecaseofspin,Sx2hasonlyonepossiblevalue:<Sx2>=Sx2inallcases.§6.3SteppingOperatorsandEigenvaluesofJ2

TheeigenvaluesofJ2followasimilarpattern;weonlyhavetodeterminethemaximumvalueofJz.WeknowthatJz=Lz+SzandthatSz=±?/2.ThusitwouldseemclearthatthemaximumvalueofJzmustequalthemaximumvalueofLzplusthemaximumvalueofSz.Apparently,therefore,j=(l?+?/2)/?=l+1/2,bydefinition.Thisisapossiblevalueforj,butthereismoretoit.

ThevalueofjisrelatedtotheeigenvalueofJ2.whichinturnisrelatedtothevalueofL·S.Foragivenvalueofl,theremustbetwopossiblevaluesofL·S,becausethespinvectoralwayshastwopossibledirectionsrelativetoanyotherdirection(includingthedirectionoftheLvector).OneofthesedirectionsyieldsapositivevalueforL·S,theotheranegativevalueofequalmagnitude.Thistellsusthattheremustbetwopossiblevaluesofj.SteppingOperatorsforAngularMomentumSates

Toputthediscussiononafirmerbasis,wenowgeneratetheeigenvaluesofJ2byusingthestepping-operatormethodthatwasintroducedin

Section4.2forharmonic-oscillatoreigenfunctions.Indoingthis,itisconvenienttointroducethesymbol[Lx,Ly]torepresenttheso-calledcommutatorofLxandLy:theoperatorLxLy

–

LyLx.Itiseasytoshow(Exercise6,Chapter6)thatthethreeoperatorsofthistypeobeythecommutationrelations[Lx,Ly]=ihLz,[Ly,Lz]=ihLx,[Lz,Lx]=ihLy

(10.22)Eachrelationcanbegeneratedfromthepreviousonebythereplacementofxbyy,ybyz,andzbyx.WecannowderivetheeigenvaluesofJ2bymakingtworeasonableassumptions:1.Similarcommutationrelationsapplytoallformsofangularmomentum,sothat[Sx,Sy]=i?Sz,[Jx,Jy]=i?Jz,andsoon.2.WecanfindafunctionthatisaneigenfunctionofbothJ2andJz,withtheeigenvaluea?2forJ2andeigenvalueb?forJz,whereaandbaredimensionless.Denotingthisfunctionbythesymboluab,wewriteJ2uab=a?2uab(10.23)

Jzuab=b?uab

(10.24)LetusnowintroducetheoperatorJ+=Jx+iJyanddemonstratethatthefunctionJ+uabisalsoaneigenfunctionofJz.ApplyingJztoJ+uabgivesus

JzJ+uab=Jz(Jx+iJy)uab=(JzJx+iJzJy)uab

(10.25)WecanapplyJzdirectlytouabbyusingthecommutationrelationsintheform

JzJx=JxJz+i?Jy

andJzJy=JyJz-i?Jx

(10.26)SubstitutingfromEq.(10.26)intoEq.(10.25)produces

JzJ+uab=(JxJz+i?Jy+iJyJz+?Jx)uabv

(10.27)ThesubstitutionofJzuab=b?uab

fromEq.(10.24)nowyields

JzJ+uab=(Jxb?+i?Jy+iJyb?+?Jx)uab

(10.28)whichreducesto

JzJ+uab=(Jx+iJy)(b?+?)uab=(b+1)?J+uab

(10.29)ThuswehaveproventhatthefunctionJ+uabisaneigenfunctionofJz,witheigenvalue(b+l)?.Followingthesameprocedure,wecouldprovethatJ-uab,whereJ-=Jx

–

Jy,isaneigenfunctionofJzwitheigenvalue(b-l)?.Thereforewecangeneratefromuabasetofeigenfunctionswhoseeigenvaluesare?multipliedbyb,b±1,b±2,……EachofthesefunctionsisalsoaneigenfunctionofJ2,asyoumayveritybyapplyingtheJ2operatortothefunctionJ+uab,thenapplyingthecommutationrelationsandEqs.(10.23)and(10.24).

Thissetofeigenfunctionscannotbeinfinite,becauseweknowthatnoeigenvalueofJz2canbegreaterthantheeigenvalueofJ2,whichequalsa?2foreveryeigenfunctionintheset.ThustheremustbealargesteigenvalueofJz,whichwedenoteasj?.Wewritetheeigenfunctioncorrespondingtothiseigenvalueasuaj;thatmeansthatJzuaj=j?uaj.ItisclearthatJ+uaj=0 (10.30)forotherwiseJ+uajwouldbeaneigenfunctionofJzwitheigenvalue(j+1)?,incontradictiontothedefinitionofj?asthelargesteigenvalue.

Letusnowapplytheoperator(Jx-iJy)tothisfunctiontoobtain(Jx-iJy)(Jx+iJy)uaj=0(10.31)or[Jx2+Jy2+i(JxJy

–

JyJx)]uaj=0(10.32)ThecommutationruleforJxandJyshowsthatthiscanbewrittenas(Jx2+Jy2-?Jz)uaj=0(10.33)or,sinceJzuaj=j?uaj

(Jx2+Jy2-j?2)Uaj=0(10.34)Adding(Jz2+j?2)uajtobothsides,wehave(Jx2+Jy2+Jz2)uaj=(Jz2+j?2)uaj

(10.35)whichreducesto J2uaj=j(j+1)?2uaj (10.36)Insummary,wehaveshowndirectlyfromtheassumedcommutationrelationsthattheeigenvaluesofJ2arej(j+1)?2andthatthestateswithagivenvalueofjformasetforwhichtheeigenvalueofJzisoneofthesequence+j?,+(j-1)?,...,-j?.ThuswemaywriteJz=mj?,wheremjhasamaximumvalueofjandaminimumvalueof-j.Sincethedifferencebetweenanytwovaluesofmjmustbeanintegerandthedifferencebetweenjand-jis2j,thevalueofjmustbeeitheranintegerorahalf-integer.

Thisderivationisbasedsolelyuponthecommutationrelations;nothinginthederivationdependsuponthenatureoftheparticleinvolved.Forasingleelectron,thefactthats=1/2makesjahalf-integer;foraparticlewhosespinisaninteger,jwouldalsobeaninteger.Nowwemustdeterminethegeneralrelationbetweenthevalueofjandthevalueofs.WedothisbyreferringtothezcomponentsofthevectorsJ,L,andS.SinceJ=L+SweknowthatJz=Lz+Sz.Theeigenvaluerelationismj?=ml?+ms?,sothatmj=ml+ms,or

mj=ml±1/2(10.37)Themaximumvalueofmjisj.so

mj≤l+1/2 (10.38)andtheremustbeasetofstatesforwhichmjtakesonthevalues

mj

=l+1/2,l–1/2,l-3/2,...,-(l+1/2)(j=l+?) (10.39)andforeverystateinthissetthevalueofjisl+1/2,ThismeansthatthemagnitudeofJisgreaterthanthemagnitudeofLforeachofthesestates,andwededucethatS·Lispositiveforeachofthesestates.TheremustalsobestatesforwhichS·Lisnegative;theseformasetforwhichjequalsl–1/2,andinthisset

mj

=l–1/2,l–3/2,...,-(l–1/2)(j=l–1/2) (10.40)Inbothofthesesetswefindvaluesofmjintherangel–1/2≥mj≥-(l–1/2).Thusthereisonlyonestateforwhichmj=l+1/2,onlyonestateforwhichmj=-(l+?),andtwostatesforeachoftheotherpossiblevaluesofmj.Letusnowexamineindetailthedistinctionbetweentwostateswithdifferentjandthesamemj.Considerthecasemj=l–1/2.Thisvalueofmjcanbeproducedintwoways;eitherml=l-1andthespinispositive,orml=landthespinisnegative.InDiracnotation(aspresentedinSection2.4)wewouldwritethesestatesas∣l,l–1>∣+>sand∣l,l>∣->s,respectively.Thesetwostatesareofcourseorthogonaltoeachother.NeitherofthemisaneigenstateoftheJ2operator,butwecanformtwolinearcombinationsofthem,suchthatonecombinationisaneigenstateoftheJ2witheigenvaluej=l+1/2,andtheothercombinationisaneigenstateoftheJ2witheigenvaluej=l–1/2.Asillustratedinthefollowingexample,S·Lispositivewhenj=l+1/2,andS·Lisnegativewhenj=l–1/2.

ExampleProblem10.2FindtheeigenvalueofJ2foreachofthetwoindependentlinearcombinationsof∣1,0>∣+>sand∣1,1>∣->sthatareeigenfunctionsofJ2.Solution.ApplyingtheS·LoperatortothegeneralexpressionF=a∣1,0>∣+>s+b∣1,1>∣->sgivesustheexpression∣SxLx+SyLy+SzLz[a∣1,0>∣+>s+b∣1,1>∣->s].Usingthespinoperators[Eqs.(10.3)]combinedwiththeoperatorsforthecomponentsofL[Eqs.(7.19)]wefindthattheindividualtermsare

SxLx[a∣1,0>∣+>s+b∣1,1>∣->s]=?2/2(2)1/2[a∣1,1>∣->s+a∣1,-1>∣->s

+b∣1,0>∣+>s]

SyLy[a∣1,0>∣+>s+b∣1,1>∣->s]=?2/2(2)1/2[a∣1,1>∣->s-a∣1,-1>∣->s

+b∣1,0>∣+>s]

SzLz[a∣1,0>∣+>s+b∣1,1>∣->s=?2/2[-b∣1,1>∣->s]Thisbecomes,afteraddingandgroupingterms:S·LF=?2/(2)1/2[a∣1,1>∣->s+b∣1,0>∣+>s] –?2/2[b∣1,1>∣->s]=A?2[a∣1,0>∣+>s+b∣1,1>∣->s]whereAh2istheassumedeigenvalue,Abeingadimensionlessnumber.Equatingcorrespondingcoefficientswehavea/(2)1/2

–b/2=Abandb/(2)1/2=Aa(10.41)WecansolvethesetofindtworootsforA:A=+1/2orA=-l.ThereforetheeigenvaluesofS·Lare+?2/2and-?2.SinceL2=2?2andS2=3?2/4,J2=L2+S2+2S·L=15?2/4(whenS·L=+?2/2),andJ2=3?2/4(whenS·L=-?2).

NoticethatthesevaluesagreewiththeformulaJ2=j(j+1)?2withj=3/2inthefirstcaseandj=l/2inthesecondcase.Tosummarize:1.Therearetwosetsofstatesforagivenvalueofl,whens=1/2;thesehavethevaluesofmlgiveninEqs.(10.39)and(10.40).2.Therearetwoindependentstatesforeachvalueofmjfrommj=l-(l/2)downtomj=-[l-(l/2)].3.IfoneofthesestatesisaneigenstateofJ2,thatstateisnotaneigenstateofeitherLzorSz,butitisalinearcombinationofthetwoeigenstates∣l,ml>∣->sand∣l,ml

–1>∣+>s(asillustratedintheprecedingexample).4.Ingeneral.thequantitiesLz,Sz,andJ2arenotsimultaneouslyobservable,becausetheJ2operatordoesnotcommutewiththeLzandSzoperators.

FIGURE10.l(a)StateinwhichLzandSzareknownJ2isunknown.becausethedirectionofLrelativetoSisunknown.(b)StateinwhichJ2isknown.thevaluesofLzandSzarenotknown.becauseSandLrotatearoundJ.

Figure10.1illustratesthispoint,Figure10.1ashowsthecaseinwhichL2,Lz,S2,andSzareallknown.EachofthevectorsSandLliesonacone(whoseaxisisthezaxis,butthepositionofthevectoronitsconeiscompletelyundetermined.ThereforethedirectionofSrelativetoLisunknown,S·Lisunknown,andJ2mustbeunknown.Incontrast,Figure10.1bshowsthesituationinwhichJ2,L2,S2,andJzareknown.InthiscaseS·Lmustbeknown;thusweknowthattheanglebetweenSandLmustbefixed,makingSlieonaconewhoseaxisisthedirectionofL.ObservethatthetriangleformedbyS,L,andJiscompletelydeterminedinsizeandshape,butthistrianglerotatesaboutthedirectionofJ.ThusLzandSzareundetermined,butJ2andJzareknown.EvaluationoftheSpin-OrbitEnergyWenowhavethenecessaryinformationtoevaluatethespin-orbitenergy.LetusgeneralizeEq.(10.12)tothecaseofanuclearchargeofZeinaone-electronion.ThenwehaveUsingEq.(1.17)fortheoperator,wehaveIfuisaneigenfunctionofJ2,S2,andL2,wemaywrite(J2-S2-L2)opu=?2[j(j+1)-l(l+l)-s(s+1)]u (10.44)Andaftersettingsequalto1/2,wehaveTheangularpartofthisintegralisequaltol,becauseofthenormalizationofthesphericalharmonics.Itisasimplemattertoevaluatetheradialpartforanyparticularstate,bypluggingintheappropriateradialfunction(seeTable9.1orAppendixG).Evaluatingtheintegralforthegeneralcaseismoredifficult;theresult(forl≠0)iswherea’0isdefinedasa’0=a0(me+M)/M.ItisinstructivetowriteWintermsoftheenergyoftheunperturbedstate[En=-mrc2Z2α2/2n2,whereα=e2/4πε0hc≌1/137(thefinestructureconstant)]:Sincejequalseitherl+1/2orl-1/2,wecanwritethisastwoexpressions:Thus∣W∣isproportionalto∣En∣α2,while∣En∣itselfisproportionaltomc2α2.Thuswecouldsaythatwehavethefirstthreetermsinanexpansionofthetotalenergyinapowerseriesinα2.However,thereisanothereffect,arelativisticcontributiontotheenergy,thatiscomparabletotheeffectjustcomputed.Thiseffectisaconsequenceofthefactthatthekineticenergyofaparticleofmassmandmomentumpisnotaccuratelygivenbytheexpressionp2/2m,ashadbeenassumedinconstructingtheSchreodingerequation.Withbotheffectsincluded,theenergyshiftΔEisgivenbyasingleformulaNoticethatthetotalenergydependsonnandjonly,andnotonl.(ThecalculationinvolvesanapproximationthatwewilldiscussinChapter7.)Forthisreason,thelabelingofstatesincludesasubscriptgivingthevalueofj.Forexample,2s1/2isthestatewithn=2,l=0,andj=1/2,and2p1/2isthestatewithn=2,l=1,andj=1/2.HyperfineStructure

Theprotonandneutronalsohaveintrinsicspinandmagneticmoment.Theseareobservablebecauseoftheenergyshiftresultingfromtheinteractionofthismagneticmomentandtheorbitalandspinmagnetic

momentsoftheelectron.Althoughtheenergyinvolvedistiny,comparedtotheS·Leffect,ithasclearlybeenobserved.Themagnitudeofthis“hyperfine”energysplittingcanbeestimatedtheoreticallybyanextensionofthemethodsusedabove.Ingeneral,anatomhasatotalangularmomentumthatcanbedefinedasthevectorF=J+I,whereIistheinternalangularmomentumofthenucleus(analogoustoSfortheelectron).Inthe

1Hatom,Iissimplythespinangularmomentumoftheproton,whichisequaltothespinangularmomentumoftheelectron,withtwopossiblevalues,±?/2,ofthecomponentalonganyspecifieddirec