量子力學(xué)英文課件：Chapt4 The Radial Schreodinger Equation

QuantumMechanicsChapter4.TheRadialSchreodingerEquation§4.0TheRadialSchroedingerEquationEquation(6.36),thetime-independentSchreodingerequationinthreedimensionsmayberewrittentoeliminate(消除）theangulardependence,yielding

?Considerthespecialcasel=0first.Equation(8.1)canthenbewritten

Thisequationisidenticaltothetime-independentone-dimensionalSchreodingerequation[Eq.(3.5)],exceptthatthevariableisrratherthanxandtheeigenfunctionisrR(r)ratherthanu(x).Therefore,theeigenfunctionsarealsoidenticalwhenxisreplacedbyrandu(x)isreplacedbyrR(r).——————Ascomparison,Eq.(3.5)isasfollows:§4.1SolutionsforaFreeParticle

ForafreeparticlewecansetV(r)=0,andifl=0wehavewhosesolutionscanbeexpressedasThecompletesolution,includingtimedependence,isthereforeψ(r,θ,φ,t)=R(r)eiωt=Ne±i(kr-ωt)/r(8.5)whereNisanormalizingconstant;ψisindependentofθandφ,becausel=0.Theradialfunctionforl=0mayalsobewrittenintheformR(r)=(Acoskr+Bsinkr)/r(8.6)

BoundaryConditionattheOrigin

Equation(8.6)describesastandingwave（駐波）thatcannotexistattheorigin,becauseofthe1/rfactor.HoweverwecanmakethiswaveacceptablebysettingAequaltozero,because(sinkr)/risfiniteatr=0.Thus,althoughEq.(8.3)hasthesameformastheone-dimensionalSchreodingerequation,the

eigenfunctionsrR(r)thatreplaceu(x)inthatequationmustbezeroattheorigin,becausetheycontainthefactorr.?Thismakesthesolutionforasphericalwellsignificantlydifferentfromthesolutionforaone-dimensionalwell,evenwhentheangularmomentumiszero.?TheCentrifugalPotential（離心力勢）

Evenifl≠0,wecanwriteEq.(8.1)inaformthatresemblesEq.(8.2)bydefininganeffectivepotentialVeffgivenby

wherethesecondtermiscalledthecentrifugalpotential.Thisisnotactuallypotentialenergy,butratherthekineticenergyassociatedwithangularmotion.Equation(8.l)cannowbewrittenThequantityE-Veffistheenergythatremainsafterwesubtractthepotentialenergyandtheenergyassociatedwithangularmotion.Thusthisexpressionistheenergyassociatedwithradialmotion,justasinonedimensiontheexpressionE-Visthekineticenergyassociatedwithmotionalongoneaxis.TheRadialProbabilityDensityTheparallelbetweenEq.(8.8)andtheone-dimensionalSchreodingerequation[Eq.(3.5)]canbestrengthenedbyconsideringprobabilitydensities.InChapter2wesawthatwherer2sinθdθdφdristhevolumeelementinsphericalcoordinates,andwealsosawthat

WecanuseEq.(6.54)toeliminatetheangularpartandobtainthenormalizationconditionfortheradialpart:?TheproductrR(r)istheprobabilityamplitude（幾率幅）fortheradialcoordinate,justasu(x)istheprobabilityamplitudeforthexcoordinate.?ThatmeansthattheprobabilityP(r1,r2)offindingthercoordinateoftheparticletobeintheranger1<r<r2isgivenby

Equation(8.8)cannowbeexpressedinwordsasoperator×probabilityamplitude=kineticenergy×probabilityamplitude

wherethekineticenergyistobeunderstoodasthepartoftheenergythatresultsfromthecomponentofthevelocityalongtheraxis.Thisexpressionappliestomotionalongthex,y,orzaxisinrectangularcoordinates（直角坐標(biāo)）.

§4.2TheSphericalPotentialWell

Letuscomparethesphericalpotentialwellwiththeone-dimensionalsquarewelltreatedinSection3.2.A"square"sphericalwellcanbedescribedbyapotentialV(r)thathasasharpstep(Figure8.1):V(r)=-V0forr<aV(r)=0forr>a(8.11)Whenl=0,thesituationisverymuchlikethatoftheone-dimensionalwell.Forr<a,theradialSchreodingerequationisEq.(8.3)andthesolutionisgivenbyEq.(8.6)withA=0.Thatis,U(r,θ,φ)=B(sinkr)/r(8.12)where,asbefore,thekineticenergyisButinthiscasethekineticenergywithinthewellisequaltoE+V0.Forr>a,theradialequationis,fromEq.(8.2)andthesolutionforaboundstatemustgotozeroasr→∞.Thusforr>a,wehaveadecayingexponential（衰減的指數(shù)函數(shù)）:HereαisrealwhenEisnegative,justasintheone-dimensionalwell.

FIGURE8.1Theeffectivepotentialwellthatappearsintheradialequationforthesquare-wellpotentialofEq.(8.11).Forr<atheeffectivepotentialissimplythecentrifugalpotential.Wenowapplythecontinuityconditionatr=atofindtheallowedvaluesofEaswedidbefore.Becausethemathematicalfunctionsareexactlythesame,theenergylevelsmustbethesameasbefore,withoneimportantexception:TheeigenfunctionrR(r)mustbezeroatr=0.Noneoftheeven（偶）functionsfromSection3.2meetsthisrequirement.Thusallofthesolutionsforthesphericalwellareodd（奇）functionsofr.———————Theoddwavefunctioninonedimensionalsquarewell:Thereforetheresultforl=0musthavethesameformastheresultfoundinSection3.2fortheodd-paritysolution;theconditionthatrR=0atr=0eliminatestheevensolution.TheenergylevelsarethusfoundfromEq.(3.39):Sinka=±ka/βa(β2=k2+α2andcotka<0) (3.39b)Becausecotka<0,thelowest-energysolutionhasπ/2<ka<π(likethelowestoddsolutionintheone-dimensionalcase).Butka≤βa.Therefore,ifβa<π/2(thatis,if2mV0a2/h2<(π/2)2),thereisnoboundstate.

Thereisoneboundstateifπ/2<βa<3π/2,therearetwoif3π/2<βa<5π/2.andsoon.TheallowedenergiesarefoundbythesamemethodfollowedinSection3.2.

Example:EnergyLeveloftheDeuteron（氘

核）Thedeuteron,thenucleusofthe2Hatom,isaboundstateofaneutronandaproton.Ithasonlyoneenergylevel,atanenergyof-2.2MeV.(Thismeansthatanenergyofatleast2.2MeVisrequiredtoseparatetheneutronfromtheprotoninthisnucleus.)ExperimentsshowthatthepotentialenergyV(r)canbeapproximatedbyEq.(8.11)withthevalueofaequalto2fm.FromthisinformationyoucanverifythatthewelldepthV0isabout37MeV.Inspiteofitsdepth,thisnarrowwellhasnoexcitedstate;thevalueofβaisnotmuchlargerthanπ/2.Curiously,thereisno"dineutron"(aboundstateoftwoneutronsintheabsenceofprotons),eventhoughthereisastrongattractiveforcebetweenneutrons.

ThePauliexclusionprinciple(tobediscussedinChapters10and11)providestheexplanationforthat.Inadineutron'sgroundstate,bothneutronswouldbeinthestateoflowestenergy.Theexclusionprincipledoesnotpermitthistohappenforneutrons(andmanyotherparticles).andnoexcitedstateisbound.Thusthereisnodineutron.SolutionsforNonzeroAngularMomentumWhenl>0,theradialequationforthesphericalsquarewellbecomesWiththesubstitutionsthesebecomeSolutionsofEq.(8.17)arecalledsphericalBesselfunctions[jl(kr)]andsphericalNeumannfunctions[n--l(kr)].ForEq.(8.18)theargumentofthefunctionsisofcourseiαrratherthankr.Foreachvalueofltherearetwolinearlyindependentsolutions-asphericalBesselfunctionandasphericalNeumannfunction.Thefirstthreeofeachare

Wehaveseenthatj0(kr)andn0(kr),asjustgiven,aresolutionsoftheradialequation(8.17)whenl=0.YoumayverifyyourselfthattheotherfunctionsgivenabovearesolutionsofEq.(8.17)withthegivenvaluesofl.Figure8.2showsradialprobabilitydensities whereR(r)isthesphericalBesselfunction.forvariousvaluesofl.Noticehowis

"pushedaway"fromtheoriginwhenl>0.Inthatcaseismaximumnearkr=landisquitesmallforkr<l.AclassicalparticlewithmomentumPandangularmomentumLcannotbeclosertotheoriginthanr=L/p.Because

Usingthevalueswefindthatkr1.FIGURE8.2Radialprobabilitydensitiesforl=0.l=2.andl=5,CalculationofEnergyLevelsfortheSphericalWellTofindtheenergylevels,wenowmustapplythecontinuityconditionsatr=atothesolutionsforr<aandr>a.Forr<a,thesolutionmustbejl(kr),becauseitistheonlysolutionthatsatisfiestheconditionthatrR(r)mustbezeroatr=0.Forr>athereisnosuchcondition,andjl(kr)doesnotgotozeroasr→∞.ThereforeweneedtousetheNeumannfunctionnl.

Thegeneralsolutionisalinearcombinationofjlandnl,andthecombinationthathasthecorrectasymptotic(漸近）behaviorasr→∞iscalledasphericalHankelfunctionofthefirstkind,definedasTousethisfunctionasasolutionofEq.(8.18)weletxequaliαr.Then,forl=1andl=2,thesolutionsareThesefunctionsclearlyhavetherequiredbehaviorasr→∞,soweusethemfortheregionr>a.Thecompletesolutionforl=1isthereforewhereAandBareconstantstobedeterminedbytheboundaryconditionsatr=aandbynormalization.WemayeliminatebothAandBandsolvefortheenergylevel;weusetherequirementthattheratio(dR/dr)/Rbecontinuousatr=a,asfollows:Takingthederivativesandperformingthedivisiononeachsideleads(withsomerearrangementofterms)to

Thisequation,withthedefinitions

maybesolvednumericallytofindthepossiblevaluesofEforl=1.

ExampleProblem8.1UseEqs.(8.26)and(8.27)toshowthatthereisaboundstateforl=1onlyif?

Solution.Theright-handsideofEq.(8.26)isnevernegative,buttheleft-handsideisnegativewhencotka<l/ka,whichistruewhenka<π.Thuswemusthaveka≥πifEq.(8.26)istohaveasolution.Foraboundstateinthiswell,Emustbenegative;thereforefromEq.(8.27),withka≥π,wehaveNoticethattherequiredvalueofVa2forl=1isfourtimesthevaluerequiredforaboundstatewithl=0.Tohaveaboundstateforl=1requiresthat§4.3Example:TheSphericallySymmetricHarmonicOscillatorGiventhesphericallysymmetricharmonicoscillatorpotential[Eq.(6.16)]:V(r)=Kr2/2(8.28)WemaywritetheradialSchreodingerequation[Eq.(8.8)]asForl=0,Eq.(8.29)isidenticaltotheone-dimensionalequation,exceptthatu(x)isreplacedbyrR(r).Thereforethereisasolutionwhoseenergy

eigenvalueisequaltoasinonedimension.?However,wefoundpreviously(Section2.1)that,forthesphericallysymmetricharmonicoscillator,theenergylevelsaregivenbywherenisapositiveorzero.Howcantheseresultsbereconciled?Obviously,thefunctionthatgivesanenergyeigenvalueofmustnotbeanacceptablesolution.ThereasonisclearintheexpressionforrR(r):whereNisanormalizationfactor (8.30)?ThisfailstosatisfytherequiredboundaryconditionthatrR(r)mustbezeroforr=0.Ontheotherhand,thesolutionforenergylevelor(8.31)Thiseigenfunction,havingnoangulardependence,mustrepresentastatewithzeroangularmomentum.(WemightalsosaythatitsangulardependencecanbeexpressedasthesphericalharmonicY0,0,forwhichl=0andm=0.)ByapplyingtheraisingoperatortoR0(r),wecangeneratetheeigenfunctionSimilarly,wecangeneratetwoothereigenfunctionswiththesameeigenvalue:Foreachofthesefunctions,n=1andtheenergyisConnectionwiththeSphericalHarmonicsWehavealreadyseethat,inasphericallysymmetricpotential,eacheigenfunctionoftheSchreodingerequationistheproductofapurelyradialfunctionandapurelyangularfunctionandthattheangularfunctionmustbeasphericalharmonicoralinearsuperpositionofsphericalharmonics.

LetusshowthatthisistrueforthefunctionsofEqs.(8.32).Thefactorx[inEq.(8.32a)]maybewrittenintermsofthesumofthesphericalharmonicsY1,1andY1,-1,because[asshowninExampleProblem6.3]Y1,1-

Y1,-1=(3/2π)1/2x/rorx/r=(2π/3)1/2(Y1,1-Y1,-1)andthereforewhereNisanormalizationconstant.Thusu100isaneigenfunctionofL2withl=1,butitisamixtureofm=+1andm=-1withequalamplitudes.AmeasurementofLzwouldyield+?and–?withequalprobability.FromEqs.(8.32)wecanalsodeducethatu100isalsoanequalmixtureofm=1andm=-1,whichcanbewrittenwhereasu001isaneigenfunctionofLzwithm=0:ItisoftenconvenienttousecombinationsofharmonicoscillatorfunctionswhicharelinearlyindependentandareeigenfunctionsofLz.Thesemaybewrittenwithasanormalizingfactor,asThesefunctionsaresimultaneouseigenfunctionsofenergy(n=1),ofL2(l=1),andofLz(withm=+1,0and–1,respectively).Itisinterestingthat,foranygivenvalueofn,asetofsimilarequationscanbewritten.Byapplicationoftheraisingoperator,youcanverifythattherearesixlinearlyindependentharmonic-oscillatorfunctionsforn=2,containingtherespectivefactorsx2,y2,z2,xy,xz,andyz.Wecanconstructsixindependentcombinationsofthesefactorsbycombiningthefivesphericalharmonicsforl=2withtheharmonicforl=0.Forexample,thecombinationu200+u020+u002containsx,y,andzonlyinthecombinationx2+y2=z2;thusitissphericallysymmetricandisproportionaltothesphericalharmonicY0,0withl=0.Similarly,forn=3,theraisingoperatorsyieldtendifferentfactorswithacombinedexponentof3:X3,y3,z3,x2y,xy2,x2z,xz2,y2z,yz2andxyz.Thesecanbewrittenaslinearcombinationsofthesevensphericalharmonicsforl=3plusthethreesphericalharmonicsforl=1.Theprocessisvalidforanyvalueofn.(SeeProblems5and6.)§4.4ScatteringofParticlesfromaSphericallySymmetricPotential

Letusnowconsiderthethree-dimensionalcounterpartofthetransmissionofparticlespastapotentialbarrier(Section5.3).Inthiscasethesituationisobviouslymorecomplicated,becausetheparticlescanemergeinanydirection(be"scattered")insteadofsimplybeingtransmittedorreflected.Supposethata"beam"ofparticles-aplanewaveoftheformΨin=Aei(kz-ωt)－encountersapotentialwell(a"scatteringcenter")wherethepotentialenergyisnonzerooveralimitedregion(r<a)surroundingthisscatteringcenter.(SeeFigure8.3.)Thedensityofparticlesinthebeamisandtheintensityofthebeam(thenumberofparticlescrossingaunitareainaunitoftime)istheproductofparticledensityandparticlevelocity,or asdiscussedinSection5.2.?Somefractionoftheparticleswillinteractwiththescatteringcentertoproduceawavethattravelsoutwardfromthecenterwithanamplitudethatingeneralisaproductoftwofunctions:(1)afunctionf(θ,φ)oftheangularcoordinatesθandφand(2)afunctionR(r)oftheradialcoordinaterFIGURE8.3Scatteringofaplanewavefromascatteringcenter,producingasphericalscatteredwave.Theinteractionthatproducesthescatteredwaveoccursonlyintheregionr<a.Thefunctionf(θ,φ)enablesustofindtheprobabilitythataparticlewillbescattered,asafunctionofthescatteringangle.Butbeforesolvingawaveequation,itmaybehelpfultoinvestigatethescatteringofclassicalparticles.ScatteringCrossSection,Classical

Considerthescatteringofclassicalparticlesfromasphere.Insteadofawave,wemighthaveastreamoftinypellets(小球)aimedtowardthesphere.Therearethreepossibilities;apelletcould1.Bedeflectedbythesphere2.Missthespherecompletely3.Gothroughthespherewithoutdeviation(ifthesphereisporous(多孔的）)-ahighlyimprobableclassicalresult,butcommoninquantummechanics.Ifthebeamintensity(thenumberofpelletsperunitareapersecondinthebeam)isIandthesphereisnotporous,thenthenumberNscofpelletsthatarescatteredpersecondmustbeequaltoIσ,whereσisthesphere'scross-sectionalarea,or

Nsc=Iσorσ=Nsc/I (8.34)Andσ=πR2,whereRistheradiusofthesphere.Ingeneral,theratioNsc/Iiscalledthescatteringcrosssectionofthespherefortheseparticles,anditisdenotedbythesymbolσ.Butwhatifthesphereisporous?Inthatcase,σisnotdefinedasthegeometricalcrosssectionofthesphere;rather,itisdefinedastheratioNsc/I,usingEq.(8.34).DifferentialCrossSection,Classical

Weareofteninterestedinthenumberofparticlesthatarescatteredintoaspecificrangeofangles.Classically,allpelletsinaparallelbeamwillbescatteredatthesameangleθiftheyhavethesameimpactparameter(碰撞參數(shù))b.Bydefinition,bisthedistancebywhichthepelletwouldmissthecenterofthesphereifitpassedthroughthesphereinastraightline.Ifthebeamisparalleltothezaxisandthecenterofthesphereisattheorigin,thenwecanrelatebtoRandthescatteringangleθ.FromFigure8.4weseethatb=Rsinθi,whereθiistheanglebetweenthevectorRandthezaxisinFigure8.4.Whenapelletstrikesasolidsphereandreboundselastically,wecanseefromFigure8.4thatthescatteringangleθ,whichistheanglebetweenthepellet'soriginaldirectionanditsfinaldirection,isgivenbyθ=π-2θi=π-2sin-l(b/R)(8.35)whereRisthesphere'sradius.FIGUIEE8.4Classicalelasticscatteringofapelletbyahardsphere.Eachpelletisdeflectedthroughanangleofθ=π-2θi.

Thenumberofpelletsthatscatterintoananglebetweenθandθ+dθisproportionaltothedifferentialcrosssectiondσ/dθ,whichisdependentontheangleθ:dσissimplythesizeoftheareathroughwhichapelletmustgoinordertobescatteredintoanangleinthisrange.Intermsoftheimpactparameterb,thisistheareaAofaringofradiusbandthicknessdb.WecanwriteAintermsofθanddθbysolvingEq.(8.35)forb,thendifferentiating,asfollows:b=Rsin[(π/2)-(θ/2)]=Rcos(θ/2)(8.36)db=-R/2sin(θ/2)dθ(8.37)

Thereforedσ=2πbdb=2πRcos(θ/2)(-R/2)sin(θ/2)dθ (8.38)whichcanbewrittendσ=-(πR2/2)sinθdθ(8.39)Nowthetotalcrosssectionσcanbefoundbyintegrationfromθ=πto0(becauseθ=πwhenb=0andθ=0whenb=R):asitshouldbe.ScatteringCrossSection,QuantumMechanical

Thescatteredwavecanbewrittenasψsc=Af(θ,ф)ei(kr-ωt)/r,awavetravelingoutwardfromtheorigininthedirectionofincreasingr(justasafunctionofkx-ωttravelsinthe+xdirection).?Thefactorl/rgivestheproper1/r2dependenceintheintensityofthewave,andthefactorAexpressesthefactthatthescatteredwaveamplitudeshouldbeproportionaltotheamplitudeoftheincidentwave[writtenbeforeasψin=Aei(kz-ωt)].?Wecanrelatef(θ,φ)tothescatteringcrosssectionasfollows.IfaperfectlyefficientparticledetectorwereplacedatpointP(see

Figure8.3),thenumberNdofparticlesobservedperunittimewouldbetheproductoftheintensityofthescatteredwaveandtheareadAofthedetector.Theintensityiswhereυistheparticlevelocity;thereforeButdA/r2isthesolidangledΩsubtendedbythedetectoratthescatteringcenter,istheintensityIincoftheincidentbeam,andEq.(8.41)gives

wheredΩ=sinθdθdφ.WecannowuseEq.(8.34)tointroducethescatteringcrosssection;byanalogtothatequation,

wemusthave

or

Noticethatdσ/dΩ,beingacrosssection,hasthedimensionsofanarea.Thereforef(θ,φ)musthavethedimensionsofalength.Weshallnowdemonstratethatitisproportionaltothewavelengthoftheincomingwave.

PartialWaveAnalysis

WemustnowrelatethescatteringphenomenontotheSchreodingerequationsolutionsthatwehaveseen.First,letusassumethatthescatteringpotentialissphericallysymmetric,whichimpliesthatthescat-

teredfunctionhassymmetrywithrespecttorotationaboutthezaxis.?Inthatcase,thereisnoφdependence,andf(θ,φ)becomessimplyf(θ).?Next,weapplytheconservationofangularmomentum;wedecomposetheincomingwaveintocomponentscalledpartialwaves,eachwithadifferentvalueofl.?Wecanthencalculatethescatteringofeachcomponentindependently.Inthelimitr→∞,thecompletewavefunctionapproachesψ→A(eikz+f(θ)eikr/r)(8.45)ThisistobecomparedwiththegeneralsolutionoftheSchreodingerequation,alinearcombinationofsphericalharmonics,eachmultipliedbytheappropriateradialfactorRl(r).Withnoφdependence,wehave

wherePl(cosθ)istheLegendrepolynomialoforderl.Inthisandthefollowingequations,thesummationrunsfroml=0toinfinity.

Ifthescatteringpotentialgoestozeroforr>a,thefunctionRlcanbewrittenasasuperpositionofjl(kr)andnl(kr)inthatregion:

Rl(r)=cosδljl(kr)-sinδlnl(kr)(r>a)(8.47)wherethecoefficientsarewrittenascosineandsinetopreservethenormalizationofthesolution(becausecos2δl+sin2δl=1regardlessofthevalueofδl).Atthispointweneedonlyfindthevaluesofthephaseanglesδl(calledphaseshifts)inordertodeterminef(θ)andhencethescatteringcrosssection.TodothiswematchEq.(8.45)tothelimitofEq.(8.46)forr→∞,eventuallyexpressingf(θ)intermsofaseriesofLegendrepolynomialwithcoefficientsthataredeterminedbythescatteringpotential.Itisknownthatthelimitsofthefunctionsjl(kr)andnl(kr)are,respectively,

Consequentlywemaywrite

Thustheonlyeffectofthescatteringcenteroneachcomponentoftheradialfunction(eachpartialwave)istoshiftitsphasebytheangleδl.Wenowcanfindanexpressionforthecrosssectionσintermsofthephaseshiftsδl.Weequatetheright-handsideof(8.45)tothelimit(asr→∞)oftheright-handsideof(8.46),using(8.50)toeliminateRlobtainingToeliminatezasavariable,wecanexpandeikzasaseriesofsphericalharmonics,aswecoulddoforanyfunctioninthreedimensions.(Thisisoftenausefulwaytoexpandaplanewave,expressingitasasumofcomponentwaves,eachwithadefiniteangularmomentum.SeeFigure8.5)Sinceeikzhasnoфdependence(zbeingequaltorcosθ).wewrite

andweevaluatethecoefficientsalinthesamewaythatwefindcoefficientsinaFourierseries,byusingthefactthatthePlareorthogonalandnormalized.TheresultisFIGURE8.5Analysisofaparticlebeamintodistinctannularbeams.Aparticlegoingthroughtheshadedareahasangularmomentumbetweenl?and(l+1)?.SubstitutionofthisexpressionintoEq.(8.51),andusingthelimitofjl(kr)asr→∞,yieldsor,writingthesinefunctionsinexponentialform[usingtheidentitysinx≡(eix

–e-ix)/2i],

Byequatingthecoefficientsofe-ikronthetwosideswefindthattheconstantcoefficientsblaregivenby

bl=(2l+1)ileiδ(8.56)Byequatingthecoefficientsofeikrwethenfindanexpressionforf(θ):ThetotalcrosssectionisthereforeYoucanverifyEq.(8.58)yourself.NoticethataftersquaringtheseriesofEq.(8.57),productsoftermsinvolvingdifferentvaluesofldonotappearintheintegral,becauseoftheorthogonalityoftheLegendrepolynomials.Thereforewecanexchangethesumandintegralsigns.ThentheintegralscanbeevaluatedbyuseofthenormalizationoftheLegendrepolynomials,withthefinalresultthat

Again.thesummationisoverallvaluesofl—allpositiveintegerspluszero.Theproblemnowbecomesthecalculationofthephaseshiftsδl—andthereareinfinitelymanyofthem!Fortunately,therearemanysituationsinwhichwecandeducethatonlythewaveswithl=0