S. Holzher - Quantum Physics For Dummies

QuantumPhysicsForDummies,®RevisedEditionPublishedbyJohnWiley&Sons,Inc.111RiverSt.Hoboken,NJ07030-5774www.wiley.comCopyright©2013byJohnWiley&Sons,Inc.,Hoboken,NewJerseyPublishedbyJohnWiley&Sons,Inc.,Hoboken,NewJerseyPublishedsimultaneouslyinCanadaNopartofthispublicationmaybereproduced,storedinaretrievalsystem,ortransmittedinanyformorbyanymeans,electronic,mechanical,photocopying,recording,scanning,orotherwise,exceptaspermittedunderSections107or108ofthe1976UnitedStatesCopyrightAct,withouteitherthepriorwrittenpermissionofthePublisher,orauthorizationthroughpaymentoftheappropriateper-copyfeetotheCopyrightClearanceCenter,222RosewoodDrive,Danvers,MA01923,978-750-8400,fax978-646-8600.RequeststothePublisherforpermissionshouldbeaddressedtothePermissionsDepartment,JohnWiley&Sons,Inc.,111RiverStreet,Hoboken,NJ07030,201-748-6011,fax201-748-6008,oronlineatwww.wiley.com/go/permissions.Trademarks:Wiley,theWileyPublishinglogo,ForDummies,theDummiesManlogo,AReferencefortheRestofUs!,TheDummiesWay,DummiesDaily,TheFunandEasyWay,Dummies.comandrelatedtradedressaretrademarksorregisteredtrademarksofJohnWiley&Sons,Inc.and/oritsaffiliatesintheUnitedStatesandothercountries,andmaynotbeusedwithoutwrittenpermission.Allothertrademarksarethepropertyoftheirrespectiveowners.JohnWiley&Sons,Inc.,isnotassociatedwithanyproductorvendormentionedinthisbook.LimitofLiability/DisclaimerofWarranty:Thepublisherandtheauthormakenorepresentationsorwarrantieswithrespecttotheaccuracyorcompletenessofthecontentsofthisworkandspecificallydisclaimallwarranties,includingwithoutlimitationwarrantiesoffitnessforaparticularpurpose.Nowarrantymaybecreatedorextendedbysalesorpromotionalmaterials.Theadviceandstrategiescontainedhereinmaynotbesuitableforeverysituation.Thisworkissoldwiththeunderstandingthatthepublisherisnotengagedinrenderinglegal,accounting,orotherprofessionalservices.Ifprofessionalassistanceisrequired,theservicesofacompetentprofessionalpersonshouldbesought.Neitherthepublishernortheauthorshallbeliablefordamagesarisingherefrom.ThefactthatanorganizationorWebsiteisreferredtointhisworkasacitationand/orapotentialsourceoffurtherinformationdoesnotmeanthattheauthororthepublisherendorsestheinformationtheorganizationorWebsitemayprovideorrecommendationsitmaymake.Further,readersshouldbeawarethatInternetWebsiteslistedinthisworkmayhavechangedordisappearedbetweenwhenthisworkwaswrittenandwhenitisread.Forgeneralinformationonourotherproductsandservices,pleasecontactourCustomerCareDepartmentwithintheU.S.at877-762-2974,outsidetheU.S.at317-572-3993,orfax317-572-4002.Fortechnicalsupport,pleasevisitwww.wiley.com/techsupport.Wileypublishesinavarietyofprintandelectronicformatsandbyprint-ondemand.Somematerialincludedwithstandardprintversionsofthisbookmaynotbeincludedine-booksorinprint-on-demand.IfthisbookreferstomediasuchasaCDorDVDthatisnotincludedintheversionyoupurchased,youmaydownloadthismaterialathttp://booksupport.wiley.com.FormoreinformationaboutWileyproducts,visitwww.wiley.com.LibraryofCongressControlNumber:2012945682ISBN978-1-118-46082-5(pbk);978-1-118-46083-2(ebk);ISBN978-1-11846086-3(ebk);978-1-118-846088-7(ebk)ManufacturedintheUnitedStatesofAmerica10987654321AbouttheAuthorStevenHolznerisanaward-winningauthoroftechnicalandsciencebooks(likePhysicsForDummiesandDifferentialEquationsForDummies).HegraduatedfromMITanddidhisPhDinphysicsatCornellUniversity,wherehewasontheteachingfacultyfor10years.He’salsobeenonthefacultyofMIT.Stevealsoteachescorporategroupsaroundthecountry.Author’sAcknowledgmentsI’dparticularlyliketothankthepeopleatWiley:TracyBoggier,TimGallan,andDanielleVoirol.DedicationToNancy,ofcourse!Publisher’sAcknowledgmentsWeíreproudofthisbook;pleasesendusyourcommentsthroughourDummiesonlineregistrationformlocatedatwww.dummies.com/register.Someofthepeoplewhohelpedbringthisbooktomarketincludethefollowing:Acquisitions,Editorial,andMediaDevelopmentSeniorProjectEditor:TimGallanAcquisitionsEditor:TracyBoggierSeniorCopyEditor:DanielleVoirolAssistantEditor:ErinCalliganMooneyTechnicalEditors:DanFunchWohns,LevKaplan,Ph.D.EditorialManager:MichelleHackerEditorialAssistants:JoeNiesen,JennetteElNaggar,DavidLuttonCartoons:RichTennant(www.the5thwave.com)CoverPhoto:©MEHAUKULYK/GettyImagesCompositionServicesProjectCoordinator:ErinSmithLayoutandGraphics:CarlByers,CarrieA.Cesavice,ShawnFrazier,NikkiGately,MelissaSmith,ChristineWilliamsProofreaders:JoniHeredia,ShannonRamseyIndexer:BroccoliInformationManagementPublishingandEditorialforConsumerDummiesKathleenNebenhaus,VicePresidentandExecutivePublisherDavidPalmer,AssociatePublisherKristinFerguson-Wagstaffe,ProductDevelopmentDirectorPublishingforTechnologyDummiesAndyCummings,VicePresidentandPublisherCompositionServicesGerryFahey,VicePresidentofProductionServicesDebbieStailey,DirectorofCompositionServicesQuantumPhysicsForDummies®TableofContentsIntroductionAboutThisBookConventionsUsedinThisBookFoolishAssumptionsHowThisBookIsOrganizedPartI:SmallWorld,Huh?EssentialQuantumPhysicsPartII:BoundandUndetermined:HandlingParticlesinBoundStatesPartIII:TurningtoAngularMomentumandSpinPartIV:MultipleDimensions:Going3DwithQuantumPhysicsPartV:GroupDynamics:IntroducingMultipleParticlesPartVI:ThePartofTensIconsUsedinThisBookWheretoGofromHerePartI:SmallWorld,Huh?EssentialQuantumPhysicsChapter1:DiscoveriesandEssentialQuantumPhysicsBeingDiscrete:TheTroublewithBlack-BodyRadiationFirstattempt:Wien’sFormulaSecondattempt:Rayleigh-JeansLawAnintuitive(quantum)leap:MaxPlanck’sspectrumTheFirstPieces:SeeingLightasParticlesSolvingthephotoelectriceffectScatteringlightoffelectrons:TheComptoneffectProofpositron?DiracandpairproductionADualIdentity:LookingatParticlesasWavesYouCan’tKnowEverything(ButYouCanFiguretheOdds)TheHeisenberguncertaintyprincipleRollingthedice:QuantumphysicsandprobabilityChapter2:EnteringtheMatrix:WelcometoStateVectorsCreatingYourOwnVectorsinHilbertSpaceMakingLifeEasierwithDiracNotationAbbreviatingstatevectorsasketsWritingtheHermitianconjugateasabraMultiplyingbrasandkets:Aprobabilityof1Coveringallyourbases:Brasandketsasbasis-lessstatevectorsUnderstandingsomerelationshipsusingketsGroovingwithOperatorsHello,operator:HowoperatorsworkIexpectedthat:FindingexpectationvaluesLookingatlinearoperatorsForwardandBackward:FindingtheCommutatorCommutingFindinganti-HermitianoperatorsStartingfromScratchandEndingUpwithHeisenbergEigenvectorsandEigenvalues:They’reNaturallyEigentastic!UnderstandinghowtheyworkFindingeigenvectorsandeigenvaluesPreparingfortheInversion:SimplifyingwithUnitaryOperatorsComparingMatrixandContinuousRepresentationsGoingcontinuouswithcalculusDoingthewavePartII:BoundandUndetermined:HandlingParticlesinBoundStatesChapter3:GettingStuckinEnergyWellsLookingintoaSquareWellTrappingParticlesinPotentialWellsBindingparticlesinpotentialwellsEscapingfrompotentialwellsTrappingParticlesinInfiniteSquarePotentialWellsFindingawave-functionequationDeterminingtheenergylevelsNormalizingthewavefunctionAddingtimedependencetowavefunctionsShiftingtosymmetricsquarewellpotentialsLimitedPotential:TakingaLookatParticlesandPotentialStepsAssumingtheparticlehasplentyofenergyAssumingtheparticledoesn’thaveenoughenergyHittingtheWall:ParticlesandPotentialBarriersGettingthroughpotentialbarrierswhenE>V0Gettingthroughpotentialbarriers,evenwhenE<V0ParticlesUnbound:SolvingtheSchrödingerEquationforFreeParticlesGettingaphysicalparticlewithawavepacketGoingthroughaGaussianexampleChapter4:BackandForthwithHarmonicOscillatorsGrapplingwiththeHarmonicOscillatorHamiltoniansGoingclassicalwithharmonicoscillationUnderstandingtotalenergyinquantumoscillationCreationandAnnihilation:IntroducingtheHarmonicOscillatorOperatorsMindyourp’sandq’s:GettingtheenergystateequationsFindingtheEigenstatesUsingaanda†directlyFindingtheharmonicoscillatorenergyeigenstatesPuttinginsomenumbersLookingatHarmonicOscillatorOperatorsasMatricesAJoltofJava:UsingCodetoSolvetheSchrödingerEquationNumericallyMakingyourapproximationsBuildingtheactualcodeRunningthecodePartIII:TurningtoAngularMomentumandSpinChapter5:WorkingwithAngularMomentumontheQuantumLevelRingingtheOperators:RoundandRoundwithAngularMomentumFindingCommutatorsofLx,Ly,andLzCreatingtheAngularMomentumEigenstatesFindingtheAngularMomentumEigenvaluesDerivingeigenstateequationswithβmaxandβminGettingrotationalenergyofadiatomicmoleculeFindingtheEigenvaluesoftheRaisingandLoweringOperatorsInterpretingAngularMomentumwithMatricesRoundingItOut:SwitchingtotheSphericalCoordinateSystemTheeigenfunctionsofLzinsphericalcoordinatesTheeigenfunctionsofL2insphericalcoordinatesChapter6:GettingDizzywithSpinTheStern-GerlachExperimentandtheCaseoftheMissingSpotGettingDownandDirtywithSpinandEigenstatesHalvesandIntegers:SayingHellotoFermionsandBosonsSpinOperators:RunningAroundwithAngularMomentumWorkingwithSpin1/2andPauliMatricesSpin1/2matricesPaulimatricesPartIV:MultipleDimensions:Going3DwithQuantumPhysicsChapter7:RectangularCoordinates:SolvingProblemsinThreeDimensionsTheSchrödingerEquation:Nowin3D!SolvingThree-DimensionalFreeParticleProblemsThex,y,andzequationsFindingthetotalenergyequationAddingtimedependenceandgettingaphysicalsolutionGettingSquaredAwaywith3DRectangularPotentialsDeterminingtheenergylevelsNormalizingthewavefunctionUsingacubicpotentialSpringinginto3DHarmonicOscillatorsChapter8:SolvingProblemsinThreeDimensions:SphericalCoordinatesANewAngle:ChoosingSphericalCoordinatesInsteadofRectangularTakingaGoodLookatCentralPotentialsin3DBreakingdowntheSchrödingerequationTheangularpartofψ(r,θ,ϕ)Theradialpartofψ(r,θ,ϕ)HandlingFreeParticlesin3DwithSphericalCoordinatesThesphericalBesselandNeumannfunctionsThelimitsforsmallandlargeρHandlingtheSphericalSquareWellPotentialInsidethesquarewell:0<r<aOutsidethesquarewell:r>aGettingtheGoodsonIsotropicHarmonicOscillatorsChapter9:UnderstandingHydrogenAtomsComingtoTerms:TheSchrödingerEquationfortheHydrogenAtomSimplifyingandSplittingtheSchrödingerEquationforHydrogenSolvingforψ(R)Solvingforψ(r)SolvingtheradialSchrödingerequationforsmallrSolvingtheradialSchrödingerequationforlargerYougotthepower:PuttingtogetherthesolutionfortheradialequationFixingf(r)tokeepitfiniteFindingtheallowedenergiesofthehydrogenatomGettingtheformoftheradialsolutionoftheSchrödingerequationSomehydrogenwavefunctionsCalculatingtheEnergyDegeneracyoftheHydrogenAtomQuantumstates:AddingalittlespinOnthelines:GettingtheorbitalsHuntingtheElusiveElectronChapter10:HandlingManyIdenticalParticlesMany-ParticleSystems,GenerallySpeakingConsideringwavefunctionsandHamiltoniansANobelopportunity:Consideringmulti-electronatomsASuper-PowerfulTool:InterchangeSymmetryOrdermatters:SwappingparticleswiththeexchangeoperatorClassifyingsymmetricandantisymmetricwavefunctionsFloatingCars:TacklingSystemsofManyDistinguishableParticlesJugglingManyIdenticalParticlesLosingidentitySymmetryandantisymmetryExchangedegeneracy:ThesteadyHamiltonianNamethatcomposite:GroovingwiththesymmetrizationpostulateBuildingSymmetricandAntisymmetricWaveFunctionsWorkingwithIdenticalNoninteractingParticlesWavefunctionsoftwo-particlesystemsWavefunctionsofthree-or-more-particlesystemsIt’sNotComeOne,ComeAll:ThePauliExclusionPrincipleFiguringoutthePeriodicTablePartV:GroupDynamics:IntroducingMultipleParticlesChapter11:GivingSystemsaPush:PerturbationTheoryIntroducingTime-IndependentPerturbationTheoryWorkingwithPerturbationstoNondegenerateHamiltoniansAlittleexpansion:PerturbingtheequationsMatchingthecoefficientsofλandsimplifyingFindingthefirst-ordercorrectionsFindingthesecond-ordercorrectionsPerturbationTheorytotheTest:HarmonicOscillatorsinElectricFieldsFindingexactsolutionsApplyingperturbationtheoryWorkingwithPerturbationstoDegenerateHamiltoniansTestingDegeneratePerturbationTheory:HydrogeninElectricFieldsChapter12:Wham-Blam!ScatteringTheoryIntroducingParticleScatteringandCrossSectionsTranslatingbetweentheCenter-of-MassandLabFramesFramingthescatteringdiscussionRelatingthescatteringanglesbetweenframesTranslatingcrosssectionsbetweentheframesTryingalab-frameexamplewithparticlesofequalmassTrackingtheScatteringAmplitudeofSpinlessParticlesTheincidentwavefunctionThescatteredwavefunctionRelatingthescatteringamplitudeanddifferentialcrosssectionFindingthescatteringamplitudeTheBornApproximation:RescuingtheWaveEquationExploringthefarlimitsofthewavefunctionUsingthefirstBornapproximationPuttingtheBornapproximationtoworkPartVI:ThePartofTensChapter13:TenQuantumPhysicsTutorialsAnIntroductiontoQuantumMechanicsQuantumMechanicsTutorialGrainsofMystique:QuantumPhysicsfortheLaymanQuantumPhysicsOnlineVersion2.0ToddK.Timberlake’sTutorialPhysics24/7’sTutorialStanZochowski’sPDFTutorialsQuantumAtomTutorialCollegeofSt.Benedict’sTutorialAWeb-BasedQuantumMechanicsCourseChapter14:TenQuantumPhysicsTriumphsWave-ParticleDualityThePhotoelectricEffectPostulatingSpinDifferencesbetweenNewton’sLawsandQuantumPhysicsHeisenbergUncertaintyPrincipleQuantumTunnelingDiscreteSpectraofAtomsHarmonicOscillatorSquareWellsSchrödinger’sCatIntroductionPhysicsasageneraldisciplinehasnolimits,fromtheveryhuge(galaxy-wide)totheverysmall(atomsandsmaller).Thisbookisabouttheverysmallsideofthings—that’sthespecialtyofquantumphysics.Whenyouquantizesomething,youcan’tgoanysmaller;you’redealingwithdiscreteunits.Classicalphysicsisterrificatexplainingthingslikeheatingcupsofcoffeeoracceleratingdownrampsorcarscolliding,aswellasamillionotherthings,butithasproblemswhenthingsgetverysmall.Quantumphysicsusuallydealswiththemicroworld,suchaswhathappenswhenyoulookatindividualelectronszippingaround.Forexample,electronscanexhibitbothparticleandwave-likeproperties,muchtotheconsternationofexperimenters—andittookquantumphysicstofigureoutthefullpicture.Quantumphysicsalsointroducedtheuncertaintyprinciple,whichsaysyoucan’tknowaparticle’sexactpositionandmomentumatthesametime.Andthefieldexplainsthewaythattheenergylevelsoftheelectronsboundinatomswork.Figuringoutthoseideasalltookquantumphysics,asphysicistsprobedeverdeeperforawaytomodelreality.Thosetopicsareallcomingupinthisbook.AboutThisBookBecauseuncertaintyandprobabilityaresoimportantinquantumphysics,youcan’tfullyappreciatethesubjectwithoutgettingintocalculus.Thisbookpresentstheneed-to-knowconcepts,butyoudon’tseemuchinthewayofthoughtexperimentsthatdealwithcatsorparalleluniverses.Ifocusonthemathandhowitdescribesthequantumworld.I’vetaughtphysicstomanythousandsofstudentsattheuniversitylevel,andfromthatexperience,Iknowmostofthemshareonecommontrait:Confusionastowhattheydidtodeservesuchtorture.QuantumPhysicsForDummies,RevisedEditionlargelymapstoacollegecourse,butthisbookisdifferentfromstandardtexts.Insteadofwritingitfromthephysicist’sorprofessor’spointofview,I’vetriedtowriteitfromthereader’spointofview.Inotherwords,I’vedesignedthisbooktobecrammedfullofthegoodstuff—andonlythegoodstuff.Notonlythat,butyoucandiscoverwaysoflookingatthingsthatprofessorsandteachersusetomakefiguringoutproblemssimple.AlthoughIencourageyoutoreadthisbookfromstarttofinish,youcanalsoleafthroughthisbookasyoulike,readingthetopicsthatyoufindinteresting.LikeotherForDummiesbooks,thisoneletsyouskiparoundasyoulikeasmuchaspossible.Youdon’thavetoreadthechaptersinorderifyoudon’twantto.Thisisyourbook,andquantumphysicsisyouroyster.ConventionsUsedinThisBookSomebookshaveadozendizzyingconventionsthatyouneedtoknowbeforeyoucanevenstart.Notthisone.Here’sallyouneedtoknow:Iputnewtermsinitalics,likethis,thefirsttimethey’rediscussed;Ifollowthemwithadefinition.Vectors—thoseitemsthathavebothamagnitudeandadirection—aregiveninbold,likethis:B.Webaddressesappearinmonofont.FoolishAssumptionsIdon’tassumethatyouhaveanyknowledgeofquantumphysicswhenyoustarttoreadthisbook.However,Idomakethefollowingassumptions:You’retakingacollegecourseinquantumphysics,oryou’reinterestedinhowmathdescribesmotionandenergyontheatomicandsubatomicscale.Youhavesomemathprowess.Inparticular,youknowsomecalculus.Youdon’tneedtobeamathpro,butyoushouldknowhowtoperformintegrationanddealwithdifferentialequations.Ideally,youalsohavesomeexperiencewithHilbertspace.Youhavesomephysicsbackgroundaswell.You’vehadayear’sworthofcollege-levelphysics(orunderstandallthat’sinPhysicsForDummies)beforeyoutacklethisone.HowThisBookIsOrganizedQuantumphysics—thestudyofverysmallobjects—isactuallyaverybigtopic.Tohandleit,quantumphysicistsbreaktheworlddownintodifferentparts.Herearethevariouspartsthatarecomingupinthisbook.PartI:SmallWorld,Huh?EssentialQuantumPhysicsPartIiswhereyoustartyourquantumphysicsjourney,andyougetagoodoverviewofthetopichere.Isurveyquantumphysicsandtellyouwhatit’sgoodforandwhatkindsofproblemsitcansolve.Youalsogetagoodfoundationinthemaththatyouneedfortherestofthebook,suchasstatevectorsandquantummatrixmanipulations.Knowingthisstuffpreparesyoutohandletheotherparts.PartII:BoundandUndetermined:HandlingParticlesinBoundStatesParticlescanbetrappedinsidepotentials;forinstance,electronscanbeboundinanatom.Quantumphysicsexcelsatpredictingtheenergylevelsofparticlesboundinvariouspotentials,andthat’swhatPartIIcovers.Youseehowtohandleparticlesboundinsquarewellsandinharmonicoscillators.PartIII:TurningtoAngularMomentumandSpinQuantumphysicsletsyouworkwiththemicroworldintermsoftheangu-larmomentumofparticles,aswellasthespinofelectrons.Manyfamousexperiments—suchastheStern-Gerlachexperiment,inwhichbeamsofparticlessplitinmagneticfields—areunderstandableonlyintermsofquantumphysics,andyougetallthedetailshere.PartIV:MultipleDimensions:Going3DwithQuantumPhysicsInthefirstthreeparts,allthequantumphysicsproblemsareone-dimensionaltomakelifealittleeasierwhileyou’reunderstandinghowtosolvethoseproblems.InPartIV,youbranchouttoworkingwiththree-dimensionalproblemsinbothrectangularandsphericalcoordinatesystems.Takingthingsfrom1Dto3Dgivesyouabetterpictureofwhathappensintherealworld.PartV:GroupDynamics:IntroducingMultipleParticlesInthispart,youworkwithmultiple-particlesystems,suchasatomsandgases.Youseehowtohandlemanyelectronsinatoms,particlesinteractingwithotherparticles,andparticlesthatscatteroffotherparticles.Dealingwithmultipleparticlesisallanotherstepinmodelingreality—afterall,systemswithonlyasingleparticledon’ttakeyouveryfarintherealworld,whichisbuiltofmega,megasystemsofparticles.InPartV,youseehowquantumphysicscanhandlethesituation.PartVI:ThePartofTensYouseethePartoftheTensinallForDummiesbooks.Thispartismadeupoffast-pacedlistsoftenitemseach.Yougettoseesomeofthetenbestonlinetutorialsonquantumphysicsandadiscussionofquantumphysics’tengreatesttriumphs.IconsUsedinThisBookYoufindahandfuloficonsinthisbook,andhere’swhattheymean:Thisiconflagsparticularlygoodadvice,especiallywhenyou’resolvingproblems.Thisiconmarkssomethingtoremember,suchasalawofphysicsoraparticularlyjuicyequation.Thisiconmeansthatwhatfollowsistechnical,insiderstuff.Youdon’thavetoreaditifyoudon’twantto,butifyouwanttobecomeaquantumphysicspro(andwhodoesn’t?),takealook.Thisiconhelpsyouavoidmathematicalorconceptualslip-ups.WheretoGofromHereAllright,you’reallsetandreadytogo.Youcanjumpinanywhereyoulike.Forinstance,ifyou’resureelectronspinisgoingtobeabigtopicofconversationatapartythisweekend,checkoutChapter6.AndifyourupcomingvacationtoGeneva,Switzerland,includesasidetriptoyournewfavoriteparticleaccelerator—theLargeHadronCollider—youcanfliptoChapter12andreaduponscatteringtheory.Butifyouwanttogetthefullstoryfromthebeginning,jumpintoChapter1first—that’swheretheactionstarts.PartISmallWorld,Huh?EssentialQuantumPhysicsInthispart...Thispartisdesignedtogiveyouanintroductiontothewaysofquantumphysics.Youseetheissuesthatgaverisetoquantumphysicsandthekindsofsolutionsitprovides.Ialsointroduceyoutothekindofmaththatquantumphysicsrequires,includingthenotionofstatevectors.Chapter1DiscoveriesandEssentialQuantumPhysicsInThisChapterPuttingforththeoriesofquantizationanddiscreteunitsExperimentingwithwavesactingasparticlesExperimentingwithparticlesactingaswavesEmbracinguncertaintyandprobabilityAccordingtoclassicalphysics,particlesareparticlesandwavesarewaves,andneverthetwainshallmix.Thatis,particleshaveanenergyEandamomentumvectorp,andthat’stheendofit.Andwaves,suchaslightwaves,haveanamplitudeAandawavevectork(wherethemagnitudeofk= ,whereλisthewavelength)thatpointsinthedirectionthewaveistraveling.Andthat’stheendofthat,too,accordingtoclassicalphysics.Buttherealityisdifferent—particlesturnouttoexhibitwave-likeproperties,andwavesexhibitparticle-likepropertiesaswell.Theideathatwaves(likelight)canactasparticles(likeelectrons)andviceversawasthemajorrevelationthatusheredinquantumphysicsassuchanimportantpartoftheworldofphysics.Thischaptertakesalookatthechallengesfacingclassicalphysicsaroundtheturnofthe20thcentury—andhowquantumphysicsgraduallycametotherescue.Uptothatpoint,theclassicalwayoflookingatphysicswasthoughttoexplainjustabouteverything.Butasthosepeskyexperimentalphysicistshaveawayofdoing,theycameupwithabunchofexperimentsthatthetheoreticalphysicistscouldn’texplain.Thatmadethetheoreticalphysicistsmad,andtheygotonthejob.Theproblemherewasthemicroscopicworld—theworldthat’stootinytosee.Onthelargerscale,classicalphysicscouldstillexplainmostofwhatwasgoingon—butwhenitcametoeffectsthatdependedonthemicro-world,classicalphysicsbegantobreakdown.Takingalookathowclassicalphysicscollapsedgivesyouanintroductiontoquantumphysicsthatshowswhypeopleneededit.BeingDiscrete:TheTroublewithBlack-BodyRadiationOneofthemajorideasofquantumphysicsis,well,quantization—measuringquantitiesindiscrete,notcontinuous,units.Theideaofquantizedenergiesarosewithoneoftheearliestchallengestoclassicalphysics:theproblemofblack-bodyradiation.Whenyouheatanobject,itbeginstoglow.Evenbeforetheglowisvisible,it’sradiatingintheinfraredspectrum.Thereasonitglowsisthatasyouheatit,theelectronsonthesurfaceofthematerialareagitatedthermally,andelectronsbeingacceleratedanddeceleratedradiatelight.Physicsinthelate19thandearly20thcenturieswasconcernedwiththespectrumoflightbeingemittedbyblackbodies.Ablackbodyisapieceofmaterialthatradiatescorrespondingtoitstemperature—butitalsoabsorbsandreflectslightfromitssurroundings.Tomakematterseasier,physicspostulatedablackbodythatreflectednothingandabsorbedallthelightfallingonit(hencethetermblackbody,becausetheobjectwouldappearperfectlyblackasitabsorbedalllightfallingonit).Whenyouheatablackbody,itwouldradiate,emittinglight.Well,itwashardtocomeupwithaphysicalblackbody—afterall,whatmaterialabsorbslight100percentanddoesn’treflectanything?Butthephysicistswerecleveraboutthis,andtheycameupwiththehollowcavityyouseeinFigure1-1,withaholeinit.Whenyoushinelightonthehole,allthatlightwouldgoinside,whereitwouldbereflectedagainandagain—untilitgotabsorbed(anegligibleamountoflightwouldescapethroughthehole).Andwhenyouheatedthehollowcavity,theholewouldbegintoglow.Sothereyouhaveit—aprettygoodapproximationofablackbody.Figure1-1:Ablackbody.Youcanseethespectrumofablackbody(andattemptstomodelthatspectrum)inFigure1-2,fortwodifferenttemperatures,T1andT2.Theproblemwasthatnobodywasabletocomeupwithatheoreticalexplanationforthespectrumoflightgeneratedbytheblackbody.Everythingclassicalphysicscouldcomeupwithwentwrong.Figure1-2:Black-bodyradiationspectrum.Firstattempt:Wien’sFormulaThefirstonetotrytoexplainthespectrumofablackbodywasWilhelmWien,in1889.Usingclassicalthermodynamics,hecameupwiththisformula:whereu(υ,T)istheintensitydistributionofthelightspectrumatfrequencyυofablackbodyatthetemperatureT,andAandβareconstantswhichcanbemeasuredinexperiments.(Thespectrumisgivenbyu[υ,T],whichistheenergydensityoftheemittedlightasafunctionoffrequencyandtemperature.)Thisequation,Wien’sformula,workedfineforhighfrequencies,asyoucanseeinFigure1-2;however,itfailedforlowfrequencies.Secondattempt:Rayleigh-JeansLawNextupintheattempttoexplaintheblack-bodyspectrumwastheRayleighJeansLaw,introducedaround1900.ThislawpredictedthatthespectrumofablackbodywaswherekisBoltzmann’sconstant(approximately1.3807×10–23J·K–1).However,theRayleigh-JeansLawhadtheoppositeproblemofWien’slaw:Althoughitworkedwellatlowfrequencies(seeFigure1-2),itdidn’tmatchthehigher-frequencydataatall—infact,itdivergedathigherfrequencies.Thiswascalledtheultravioletcatastrophebecausethebestpredictionsavailabledivergedathighfrequencies(correspondingtoultravioletlight).Itwastimeforquantumphysicstotakeover.Anintuitive(quantum)leap:MaxPlanck’sspectrumTheblack-bodyproblemwasatoughonetosolve,andwithitcamethefirstbegin-ningsofquantumphysics.MaxPlanckcameupwitharadicalsuggestion—whatiftheamountofenergythatalightwavecanexchangewithmatterwasn’tcontinuous,aspostulatedbyclassicalphysics,butdiscrete?Inotherwords,Planckpostulatedthattheenergyofthelightemittedfromthewallsoftheblack-bodycavitycameonlyinintegermultipleslikethis,wherehisauniversalconstant:Withthistheory,crazyasitsoundedintheearly1900s,PlanckconvertedthecontinuousintegralsusedbyRayleigh-Jeanstodiscretesumsoveraninfinitenumberofterms.MakingthatsimplechangegavePlanckthefollowingequationforthespectrumofblack-bodyradiation:Thisequationgotitright—itexactlydescribestheblack-bodyspectrum,bothatlowandhigh(andmedium,forthatmatter)frequencies.Thisideawasquitenew.WhatPlanckwassayingwasthattheenergyoftheradiatingoscillatorsintheblackbodycouldn’ttakeonjustanylevelofenergy,asclassicalphysicsallows;itcouldtakeononlyspecific,quantizedenergies.Infact,Planckhypothesizedthatthatwastrueforanyoscillator—thatitsenergywasanintegralmultipleofhυ.AndsoPlanck’sequationcametobeknownasPlanck’squantizationrule,andhbecamePlanck’sconstant:h=6.626×10–34Joule-seconds.Sayingthattheenergyofalloscillatorswasquantizedwasthebirthofquantumphysics.OnehastowonderhowPlanckcameupwithhistheory,becauseit’snotanobvioushypothesis.Oscillatorscanoscillateonlyatdiscreteenergies?Wheredidthatcomefrom?Inanycase,therevolutionwason—andtherewasnostoppingit.TheFirstPieces:SeeingLightasParticlesLightasparticles?Isn’tlightmadeupofwaves?Light,itturnsout,exhibitspropertiesofbothwavesandparticles.Thissectionshowsyousomeoftheevidence.SolvingthephotoelectriceffectThephotoelectriceffectwasoneofmanyexperimentalresultsthatmadeupacrisisforclassicalphysicsaroundtheturnofthe20thcentury.ItwasalsooneofEinstein’sfirstsuccesses,anditprovidesproofofthequantizationoflight.Here’swhathappened.Whenyoushinelightontometal,asFigure1-3shows,yougetemittedelectrons.Theelectronsabsorbthelightyoushine,andiftheygetenoughenergy,they’reabletobreakfreeofthemetal’ssurface.Accordingtoclassicalphysics,lightisjustawave,anditcanexchangeanyamountofenergywiththemetal.Whenyoubeamlightonapieceofmetal,theelectronsinthemetalshouldabsorbthelightandslowlygetupenoughenergytobeemittedfromthemetal.Theideawasthatifyouweretoshinemorelightontothemetal,theelectronsshouldbeemittedwithahigherkineticenergy.Andveryweaklightshouldn’tbeabletoemitelectronsatall,exceptinamatterofhours.Butthat’snotwhathappened—electronswereemittedassoonassomeoneshonelightonthemetal.Infact,nomatterhowweaktheintensityoftheincidentlight(andresearcherstriedexperimentswithsuchweaklightthatitshouldhavetakenhourstogetanyelectronsemitted),electronswereemitted.Immediately.Figure1-3:Thephoto-electriceffect.Experimentswiththephotoelectriceffectshowedthatthekineticenergy,K,oftheemittedelectronsdependedonlyonthefrequency—nottheintensity—oftheincidentlight,asyoucanseeinFigure1-4.Figure1-4:Kineticenergyofemittedelectronsversusfrequencyoftheincidentlight.InFigure1-4,υ0iscalledthethresholdfrequency,andifyoushinelightwithafrequencybelowthisthresholdonthemetal,noelectronsareemitted.Theemittedelectronscomefromthepooloffreeelectronsinthemetal(allmetalshaveapooloffreeelectrons),andyouneedtosupplytheseelectronswithanenergyequivalenttothemetal’sworkfunction,W,toemittheelectronfromthemetal’ssurface.Theresultswerehardtoexplainclassically,soenterEinstein.Thiswasthebeginningofhisheyday,around1905.EncouragedbyPlanck’ssuccess(seetheprecedingsection),Einsteinpostulatedthatnotonlywereoscillatorsquantizedbutsowaslight—intodiscreteunitscalledphotons.Light,hesuggested,actedlikeparticlesaswellaswaves.Sointhisscheme,whenlighthitsametalsurface,photonshitthefreeelectrons,andanelectroncompletelyabsorbseachphoton.Whentheenergy,hυ,ofthephotonisgreaterthantheworkfunctionofthemetal,theelectronisemitted.Thatis,hυ=W+KwhereWisthemetal’sworkfunctionandKisthekineticenergyoftheemittedelectron.SolvingforKgivesyouthefollowing:K=hυ–WYoucanalsowritethisintermsofthethresholdfrequencythisway:K=h(υ–υ0)Soapparently,lightisn’tjustawave;youcanalsoviewitasaparticle,thephoton.Inotherwords,lightisquantized.ThatwasalsoquiteanunexpectedpieceofworkbyEinstein,althoughitwasbasedontheearlierworkofPlanck.Lightquantized?Lightcomingindiscreteenergypackets?Whatnext?Scatteringlightoffelectrons:TheComptoneffectToaworldthatstillhadtroublecomprehendinglightasparticles(seetheprecedingsection),ArthurComptonsuppliedthefinalblowwiththeComptoneffect.Hisexperimentinvolvedscatteringphotonsoffelectrons,asFigure1-5shows.Figure1-5:Lightincidentonanelectronatrest.Incidentlightcomesinwithawavelengthofλandhitstheelectronatrest.Afterthathappens,thelightisscattered,asyouseeinFigure1-6.Figure1-6:Photonscatteringoffanelectron.Classically,here’swhatshould’vehappened:Theelectronshould’veabsorbedtheincidentlight,oscillated,andemittedit—withthesamewavelengthbutwithanintensitydependingontheintensityoftheincidentlight.Butthat’snotwhathappened—infact,thewavelengthofthelightisactuallychangedbyΔλ,calledthewavelengthshift.Thescatteredlighthasawavelengthofλ+Δλ—inotherwords,itswavelengthhasincreased,whichmeansthelighthaslostenergy.AndΔλdependsonthescatteringangle,θ,notontheintensityoftheincidentlight.ArthurComptoncouldexplaintheresultsofhisexperimentonlybymakingtheassumptionthathewasactuallydealingwithtwoparticles—aphotonandanelectron.Thatis,hetreatedlightasadiscreteparticle,notawave.Andhemadetheassumptionthatthephotonandtheelectroncollidedelastically—thatis,thatbothtotalenergyandmomentumwereconserved.Makingtheassumptionthatboththelightandtheelectronwereparticles,Comptonthenderivedthisformulaforthewavelengthshift(it’saneasycalculationifyouassumethatthelightisrepresentedbyaphotonwithenergyE=hυandthatitsmomentumisp=E/c):wherehisPlanck’sconstant,meisthemassofanelectron,cisthespeedoflight,andθisthescatteringangleofthelight.Youalsoseethisequationintheequivalentform:whereλcistheComptonwavelengthofanelectron,confirmsthisrelation—bothequations..AndexperimentNotethattoderivethewavelengthshift,Comptonhadtomaketheassumptionthathere,lightwasactingasaparticle,notasawave.Thatis,theparticlenatureoflightwastheaspectofthelightthatwaspredominant.Proofpositron?DiracandpairproductionIn1928,thephysicistPaulDiracpositedtheexistenceofapositivelychargedanti-electron,thepositron.Hedidthisbytakingthenewlyevolvingfieldofquantumphysicstonewterritorybycombiningrelativitywithquantummechanicstocreaterelativisticquantummechanics—andthatwasthetheorythatpredicted,throughaplus/minus-signinterchange—theexistenceofthepositron.Itwasaboldprediction—ananti-particleoftheelectron?Butjustfouryearslater,physicistsactuallysawthepositron.Today’shigh-poweredelementaryparticlephysicshasallkindsofsynchrotronsandotherparticleacceleratorstocreatealltheelementaryparticlestheyneed,butintheearly20thcentury,thiswasn’talwaysso.Inthosedays,physicistsreliedoncosmicrays—thoseparticlesandhighpoweredphotons(calledgammarays)thatstriketheEarthfromouterspace—astheirsourceofparticles.Theyusedcloud-chambers,whichwerefilledwithvaporfromdryice,toseethetrailssuchparticlesleft.Theyputtheirchambersintomagneticfieldstobeabletomeasurethemomentumoftheparticlesastheycurvedinthosefields.In1932,aphysicistnoticedasurprisingevent.Apairofparticles,oppositelycharged(whichcouldbedeterminedfromthewaytheycurvedinthemagneticfield)appearedfromapparentlynowhere.Noparticletrailledtotheoriginofthetwoparticlesthatappeared.Thatwaspair-production—theconversionofahigh-poweredphotonintoanelectronandpositron,whichcanhappenwhenthephotonpassesnearaheavyatomicnucleus.Soexperimentally,physicistshadnowseenaphotonturningintoapairofparticles.Wow.Asifeveryoneneededmoreevidenceoftheparticlenatureoflight.Lateron,researchersalsosawpairannihilation:theconversionofanelectronandpositronintopurelight.PairproductionandannihilationturnedouttobegovernedbyEinstein’snewlyintroducedtheoryofrelativity—inparticular,hismostfamousformula,E=mc2,whichgivesthepureenergyequivalentofmass.Atthispoint,therewasanabundanceofevidenceoftheparticle-likeaspectsoflight.ADualIdentity:LookingatParticlesasWavesIn1923,thephysicistLouisdeBrogliesuggestedthatnotonlydidwavesexhibitparticle-likeaspectsbutthereversewasalsotrue—allmaterialparticlesshoulddisplaywave-likeproperties.Howdoesthiswork?Foraphoton,momentum,whereυisthephoton’sfrequencyandλisitswavelength.Andthewavevector,k,isequaltok=p/ℏ,whereℏ=h/2π.DeBrogliesaidthatthesamerelationshouldholdforallmaterialparticles.Thatis,DeBrogliepresentedtheseapparentlysurprisingsuggestionsinhisPh.D.thesis.Researchersputthesesuggestionstothetestbysendingabeamthroughadual-slitapparatustoseewhethertheelectronbeamwouldactlikeitwasmadeupofparticlesorwaves.InFigure1-7,youcanseethesetupandtheresults.Figure1-7:Anelectronbeamgoingthroughtwoslits.InFigure1-7a,youcanseeabeamofelectronspassingthroughasingleslitandtheresultingpatternonascreen.InFigure1-7b,theelectronsarepassingthroughasecondslit.Classically,you’dexpecttheintensitiesofFigure1-7aand1-7bsimplytoaddwhenbothslitsareopen:I=I1+I2Butthat’snotwhathappened.Whatactuallyappearedwasaninterferencepatternwhenbothslitswereopen(Figure1-7c),notjustasumofthetwoslits’electronintensities.TheresultwasavalidationofdeBroglie’sinventionofmatterwaves.Experimentboreouttherelationthat,anddeBrogliewasasuccess.Theideaofmatterwavesisabigpartofwhat’scomingupintherestofthebook.Inparticular,theexistenceofmatterwavessaysthatyouaddthewaves’amplitude,ψ1(r,t)andψ2(r,t),nottheirintensities,tosumthem:ψ(r,t)=ψ1(r,t)+ψ2(r,t)Yousquaretheamplitudetogettheintensity,andthephasedifferencebetweenψ1(r,t)andψ2(r,t)iswhatactuallycreatestheinterferencepatternthat’sobserved.YouCan’tKnowEverything(ButYouCanFiguretheOdds)Soparticlesapparentlyexhibitwave-likeproperties,andwavesexhibitparticle-likeproperties.Butifyouhaveanelectron,whichisit—awaveoraparticle?Thetruthisthatphysically,anelectronisjustanelectron,andyoucan’tactuallysaywhetherit’sawaveoraparticle.Theactofmeasurementiswhatbringsoutthewaveorparticleproperties.Youseemoreaboutthisideathroughoutthebook.Quantummechanicsliveswithanuncertainpicturequitehappily.Thatviewoffendedmanyeminentphysicistsofthetime—notablyAlbertEinstein,whosaid,famously,“Goddoesnotplaydice.”Inthissection,Idiscusstheideaofuncertaintyandhowquantumphysicistsworkinprobabilitiesinstead.TheHeisenberguncertaintyprincipleThefactthatmatterexhibitswave-likepropertiesgivesrisetomoretrouble—wavesaren’tlocalizedinspace.AndknowingthatinspiredWernerHeisenberg,in1927,tocomeupwithhiscelebrateduncertaintyprinciple.Youcancompletelydescribeobjectsinclassicalphysicsbytheirmomentumandposition,bothofwhichyoucanmeasureexactly.Inotherwords,classicalphysicsiscompletelydeterministic.Ontheatomiclevel,however,quantumphysicspaintsadifferentpicture.Here,theHeisenberguncertaintyprinciplesaysthatthere’saninherentuncertaintyintherelationbetweenpositionandmomentum.Inthexdirection,forexample,thatlookslikethis:whereΔxisthemeasurementuncertaintyintheparticle’sxposition, isitsmeasurementuncertaintyinitsmomentuminthexdirectionandℏ=h/2π.Thatistosay,themoreaccuratelyyouknowthepositionofaparticle,thelessaccuratelyyouknowthemomentum,andviceversa.Thisrelationholdsforallthreedimensions:AndtheHeisenberguncertaintyprincipleisadirectconsequenceofthewavelikenatureofmatter,becauseyoucan’tcompletelypindownawave.Quantumphysics,unlikeclassicalphysics,iscompletelyundeterministic.Youcanneverknowtheprecisepositionandmomentumofaparticleatanyonetime.Youcangiveonlyprobabilitiesfortheselinkedmeasurements.Rollingthedice:QuantumphysicsandprobabilityInquantumphysics,thestateofaparticleisdescribedbyawavefunction,ψ(r,t).ThewavefunctiondescribesthedeBrogliewaveofaparticle,givingitsamplitudeasafunctionofpositionandtime.(Seetheearliersection“ADualIdentity:LookingatParticlesasWaves”formoreondeBroglie.)Notethatthewavefunctiongivesaparticle’samplitude,notintensity;ifyouwanttofindtheintensityofthewavefunction,youhavetosquareit:|ψ(r,t)|2.Theintensityofawaveiswhat’sequaltotheprobabilitythattheparticlewillbeatthatpositionatthattime.That’showquantumphysicsconvertsissuesofmomentumandpositionintoprobabilities:byusingawavefunction,whosesquaretellsyoutheprobabilitydensitythataparticlewilloccupyaparticularpositionorhaveaparticularmomentum.Inotherwords,|ψ(r,t)|2d3ristheprobabilitythattheparticlewillbefoundinthevolumeelementd3r,locatedatpositionrattimet.Besidestheposition-spacewavefunctionψ(r,t),there’salsoamomentumspaceversionofthewavefunction:ϕ(p,t).Thisbookislargelyastudyofthewavefunction—thewavefunctionsoffreeparticles,thewavefunctionsofparticlestrappedinsidepotentials,ofidenticalparticleshittingeachother,ofparticlesinharmonicoscillation,oflightscatteringfromparticles,andmore.Usingthiskindofphysics,youcanpredictthebehaviorofallkindsofphysicalsystems.Chapter2EnteringtheMatrix:WelcometoStateVectorsInThisChapterCreatingstatevectorsUsingDiracnotationforstatevectorsWorkingwithbrasandketsUnderstandingmatrixmechanicsGettingtowavemechanicsQuantumphysicsisn’tjustaboutplayingaroundwithyourparticleacceleratorwhiletryingnottodestroytheuniverse.Sometimes,yougettodothingsthatarealittlemoremundane,liketurnlightsoffandon,performabitofcalculus,orplaywithdice.Ifyou’reactuallydoingphysicswiththosedice(beyondhurlingthemacrosstheroom),thelabdirectorwon’tevengetmadatyou.Inquantumphysics,absolutemeasurementsarereplacedbyprobabilities,soyoumayusedicetocalculatetheprobabilitiesthatvariousnumberswillcomeup.Youcanthenassemblethosevaluesintoavector(single-columnmatrix)inHilbertspace(atypeofinfinitelydimensionalvectorspacewithsomepropertiesthatareespeciallyvaluableinquantumphysics).Thischapterintroduceshowyoudealwithprobabilitiesinquantumphysics,startingbyviewingthevariouspossiblestatesaparticlecanoccupyasavector—avectorofprobabilitystates.Fromthere,Ihelpyoufamiliarizeyourselfwithsomemathematicalnotationscommoninquantumphysics,includingbras,kets,matrices,andwavefunctions.Alongtheway,youalsogettoworkwithsomeimportantoperators.CreatingYourOwnVectorsinHilbertSpaceInquantumphysics,probabilitiestaketheplaceofabsolutemeasurements.Sayyou’vebeenexperimentingwithrollingapairofdiceandaretryingtofiguretherelativeprobabilitythatthedicewillshowvariousvalues.Youcomeupwithalistindicatingtherelativeprobabilityofrollinga2,3,4,andsoon,allthewayupto12:SumoftheDiceRelativeProbability(NumberofWaysofRollingaParticularTotal)213243566576859410 311 212 1Inotherwords,you’retwiceaslikelytorolla3thana2,you’refourtimesaslikelytorolla5thana2,andsoon.Youcanassembletheserelativeprobabilitiesintoavector(ifyou’rethinkingofa“vector”fromphysics,thinkintermsofacolumnofthevector’scomponents,notamagnitudeanddirection)tokeeptrackofthemeasily:Okay,nowyou’regettingclosertothewayquantumphysicsworks.Youhaveavectoroftheprobabilitiesthatthedicewilloccupyvariousstates.However,quantumphysicsdoesn’tdealdirectlywithprobabilitiesbutratherwithprobabilityamplitudes,whicharethesquarerootsoftheprobabilities.Tofindtheactualprobabilitythataparticlewillbeinacertainstate,youaddwavefunctions—whicharegoingtoberepresentedbythesevectors—andthensquarethem(seeChapter1forinfoonwhy).Sotakethesquarerootofalltheseentriestogettheprobabilityamplitudes:That’sbetter,butaddingthesquaresofalltheseshouldadduptoatotalprobabilityof1;asitisnow,thesumofthesquaresofthesenumbersis36,sodivideeachentryby361/2,or6:Sonowyoucangettheprobabilityamplitudeofrollinganycombinationfrom2to12byreadingdownthevector—theprobabilityamplitudeofrollinga2is1/6,ofrollinga3isandsoon.MakingLifeEasierwithDiracNotationWhenyouhaveastatevectorthatgivestheprobabilityamplitudethatapairofdicewillbeintheirvariouspossiblestates,youbasicallyhaveavectorindicespace—allthepossiblestatesthatapairofdicecantake,whichisan11dimensionalspace.(Seetheprecedingsectionformoreonstatevectors.)Butinmostquantumphysicsproblems,thevectorscanbeinfinitelylarge—forexample,amovingparticlecanbeinaninfinitenumberofstates.Handlinglargearraysofstatesisn’teasyusingvectornotation,soinsteadofexplicitlywritingoutthewholevectoreachtime,quantumphysicsusuallyusesthenotationdevelopedbyphysicistPaulDirac—theDiracorbra-ketnotation.AbbreviatingstatevectorsasketsDiracnotationabbreviatesthestatevectorasaket,likethis:|ψ>.Sointhediceexample,youcanwritethestatevectorasaketthisway:Here,thecomponentsofthestatevectorarerepresentedbynumbersin11dimensionaldicespace.Morecommonly,however,eachcomponentrepresentsafunctionofpositionandtime,somethinglikethis:Ingeneral,asetofvectorsϕNinHilbertspaceislinearlyindependentiftheonlysolutiontothefollowingequationisthatallthecoefficientsan=0:Thatis,aslongasyoucan’twriteanyonevectorasalinearcombinationoftheothers,thevectorsarelinearlyindependentandsoformavalidbasisinHilbertspace.WritingtheHermitianconjugateasabraForeveryket,there’sacorrespondingbra.(Thetermscomefrombra-ket,orbracket,whichshouldbeclearerintheupcomingsectiontitled“GroovingwithOperators.”)AbraistheHermitianconjugateofthecorrespondingket.Supposeyoustartwiththisket:The*symbolmeansthecomplexconjugate.(Acomplexconjugateflipsthesignconnectingtherealandimaginarypartsofacomplexnumber.)Sothecorrespondingbra,whichyouwriteas<ψ|,equals|ψ>T*.Thebraisthisrowvector:Notethatifanyoftheelementsoftheketarecomplexnumbers,youhavetotaketheircomplexconjugatewhencreatingtheassociatedbra.Forinstance,ifyourcomplexnumberintheketisa+bi,itscomplexconjugateinthebraisa–bi.Multiplyingbrasandkets:Aprobabilityof1Youcantaketheproductofyourketandbra,denotedas<ψ|ψ>,likethis:Thisisjustmatrixmultiplication,andtheresultisthesameastakingthesumofthesquaresoftheelements:Andthat’sthewayitshouldbe,becausethetotalprobabilityshouldaddupto1.Therefore,ingeneral,theproductofthebraandketequals1:Ifthisrelationholds,theket|ψ>issaidtobenormalized.Coveringallyourbases:BrasandketsasbasislessstatevectorsThereasonketnotation,|ψ>,issopopularinquantumphysicsisthatitallowsyoutoworkwithstatevectorsinabasis-freeway.Inotherwords,you’renotstuckinthepositionbasis,themomentumbasis,ortheenergybasis.That’shelpful,becausemostoftheworkinquantumphysicstakesplaceinabstractcalculations,andyoudon’twanttohavetodragallthecomponentsofstatevectorsthroughthosecalculations(oftenyoucan’t—theremaybeinfinitelymanypossiblestatesintheproblemyou’redealingwith).Forexample,saythatyou’rerepresentingyourstatesusingpositionvectorsinathree-dimensionalHilbertspace—thatis,youhavex,y,andzaxes,formingapositionbasisforyourspace.That’sfine,butnotallyourcalculationshavetobedoneusingthatpositionbasis.Youmaywantto,forexample,representyourstatesinathree-dimensionalmomentumspace,withthreeaxesinHilbertspace,px,py,andpz.Nowyou’dhavetochangeallyourpositionvectorstomomentumvectors,adjustingeachcomponent,andkeeptrackofwhathappenstoeverycomponentthroughallyourcalculations.SoDirac’sbra-ketnotationcomestotherescuehere—youuseittoperformallthemathandthenpluginthevariouscomponentsofyourstatevectorsasneededattheend.Thatis,youcanperformyourcalculationsinpurelysymbolicterms,withoutbeingtiedtoabasis.Andwhenyouneedtodealwiththecomponentsofaket,suchaswhenyouwanttogetphysicalanswers,youcanalsoconvertketstoadifferentbasisbytakingtheket’scomponentsalongtheaxesofthatbasis.Generally,whenyouhaveavector|ψ>,youcanexpressitasasumoverNbasisvectors,|ϕi,likeso:whereNisthedimensionoftheHilbertspace,andiisanintegerthatlabelsthebasisvectors.UnderstandingsomerelationshipsusingketsKetnotationmakesthematheasierthanitisinmatrixformbecauseyoucantakeadvantageofafewmathematicalrelationships.Forexample,here’sthesocalledSchwarzinequalityforstatevectors:Thissaysthatthesquareoftheabsolutevalueoftheproductoftwostatevectors,|<ψ|ϕ>|2,islessthanorequalto<ψ|ψ><ϕ|ϕ>.Thisturnsoutthebetheanalogofthevectorinequality:SowhyistheSchwarzinequalitysouseful?ItturnsoutthatyoucanderivetheHeisenberguncertaintyprinciplefromit(seeChapter1formoreonthisprinciple).Otherketrelationshipscanalsosimplifyyourcalculations.Forinstance,twokets,|ψ>and|ϕ>,aresaidtobeorthogonalifAndtwoketsaresaidtobeorthonormaliftheymeetthefollowingconditions:Withthisinformationinmind,you’renowreadytostartworkingwithoperators.GroovingwithOperatorsWhataboutallthecalculationsthatyou’resupposedtobeabletoperformwithkets?Takingtheproductofabraandaket,<ψ|ϕ>,isfineasfarasitgoes,butwhataboutextractingsomephysicalquantitiesyoucanmeasure?That’swhereoperatorscomein.Hello,operator:HowoperatorsworkHere’sthegeneraldefinitionofanoperatorAinquantumphysics:Anoperatorisamathematicalrulethat,whenoperatingonaket,|ψ>,transformsthatketintoanewket,|ψ'>inthesamespace(whichcouldjustbetheoldketmultipliedbyascalar).SowhenyouhaveanoperatorA,ittransformsketslikethis:Forthatmatter,thesameoperatorcanalsotransformbras:Hereareseveralexamplesofthekindsofoperatorsyou’llsee:Hamiltonian(H):ApplyingtheHamiltonianoperator(whichlooksdifferentforeverydifferentphysicalsituation)givesyouE,theenergyoftheparticlerepresentedbytheket|ψ>;Eisascalarquantity:Unityoridentity(I):Theunityoridentityoperatorleavesketsunchanged:Gradient(∇):Thegradientoperatorworkslikethis:Linearmomentum(P):Thelinearmomentumoperatorlookslikethisinquantummechanics:Laplacian :YouusetheLaplacianoperator,whichismuchlikeasecondordergradient,tocreatetheenergy-findingHamiltonianoperator:Ingeneral,multiplyingoperatorstogetherisnotthesameindependentoforder,sofortheoperatorsAandB,AB≠BA.AndanoperatorAissaidtobelinearifIexpectedthat:FindingexpectationvaluesGiventhateverythinginquantumphysicsisdoneintermsofprobabilities,makingpredictionsbecomesveryimportant.Andthebiggestsuchpredictionistheexpectationvalue.Theexpectationvalueofanoperatoristheaveragevaluethatyouwouldmeasureifyouperformedthemeasurementmanytimes.Forexample,theexpectationvalueoftheHamiltonianoperator(seetheprecedingsection)istheaverageenergyofthesystemyou’restudying.Theexpectationvalueisaweightedaverageoftheprobabilitiesofthesystem’sbeinginitsvariouspossiblestates.Here’showyoufindtheexpectationvalueofanoperatorA:Notethatbecauseyoucanexpress<ψ|asarowoperatorand|ψ>asacolumnvector,youcanexpresstheoperatorAasasquarematrix.Forexample,supposeyou’reworkingwithapairofdiceandtheprobabilitiesofallthepossiblesums(seetheearliersection“CreatingYourOwnVectorsinHilbertSpace”).Inthisdiceexample,theexpectationvalueisasumofterms,andeachtermisavaluethatcanbedisplayedbythedice,multipliedbytheprobabilitythatthatvaluewillappear.Thebraandketwillhandletheprobabilities,soit’suptotheoperatorthatyoucreateforthis—callittheRolloperator,R—tostorethedicevalues(2through12)foreachprobability.Therefore,theoperatorRlookslikethis:SotofindtheexpectationvalueofR,youneedtocalculate<ψ|R|ψ>.Spellingthatoutintermsofcomponentsgivesyouthefollowing:Doingthemath,yougetSotheexpectationvalueofarollofthediceis7.Nowyoucanseewherethetermsbraandketcomefrom—they“bracket”anoperatortogiveyouexpectationvalues.Infact,theexpectationvalueissuchacommonthingtofindthatyou’lloftenfind<ψ|R|ψ>abbreviatedas<R>,soLookingatlinearoperatorsAnoperatorAissaidtobelinearifitmeetsthefollowingcondition:Forinstance,theexpression|ϕ><χ|isactuallyalinearoperator.Inorderforustoseethisweshallneedtoknowjustalittlemoreaboutwhathappenswhenwetaketheproductsofbrasandkets.Firstly,ifwetaketheproductofthebra,<χ|,withtheket,c|ψ>,wherecisacomplexnumber,thenwegettheanswer,Secondly,ifwetaketheproductofthebra,<χ|,withthesumoftwokets,|ψ1>+|ψ2>,thenwegettheanswer,Nowthatweknowthiswecantesttoseeif|ϕ><χ|isactuallyalinearoperator.OKthen,let’sapply|ϕ><χ|toalinearcombinationofbras,likeso,wherec1andc2arecomplexnumbers.Nowthatyouknowhowtheproductofabrawithasumoftwoketsgoes,youcansay,Then,asyouknow,<χ|c|ψ>=c<χ|ψ>,youcanfinallywritethisas,Thisisexactlywhatalinearoperatorshoulddo—ifyoureplaceAintheaboveequationdefiningalinearoperator,with|ϕ><χ|,thentheresultisthesameastheoneyoujustfound.So|ϕ><χ|isindeedalinearoperator—thoughIwouldagree,it’saprettyfunnylookingone!GoingHermitianwithHermitianOperatorsandAdjointsTheHermitianadjoint—alsocalledtheadjointorHermitianconjugate—ofanoperatorAisdenotedA†.TofindtheHermitianadjoint,followthesesteps:1.Replacecomplexconstantswiththeircomplexconjugates.TheHermitianadjointofacomplexnumberisthecomplexconjugateofthatnumber:2.Replaceketswiththeircorrespondingbras,andreplacebraswiththeircorrespondingkets.YouhavetoexchangethebrasandketswhenfindingtheHermitianadjointofanoperator,sofindingtheHermitianadjointofanoperatorisnotjustthesameasmathematicallyfindingitscomplexconjugate.3.ReplaceoperatorswiththeirHermitianadjoints.Inquantummechanics,operatorsthatareequaltotheirHermitianadjointsarecalledHermitianoperators.Inotherwords,anoperatorisHermitianifHermitianoperatorsappearthroughoutthebook,andtheyhavespecialproperties.Forinstance,thematrixthatrepresentsthemmaybediagonalized—thatis,writtensothattheonlynonzeroelementsappearalongthematrix’sdiagonal.Also,theexpectationvalueofaHermitianoperatorisguaranteedtobearealnumber,notcomplex(seetheearliersection“Iexpectedthat:Findingexpectationvalues”).4.Writeyourfinalequation.HerearesomerelationshipsconcerningHermitianadjoints:ForwardandBackward:FindingtheCommutatorThemeasureofhowdifferentitistoapplyoperatorAandthenB,versusBandthenA,iscalledtheoperators’commutator.Here’showyoudefinethecommutatorofoperatorsAandB:[A,B]=AB–BACommutingTwooperatorscommutewitheachotheriftheircommutatorisequaltozero.Thatis,itdoesn’tmakeanydifferenceinwhatorderyouapplythem:[A,B]=0Noteinparticularthatanyoperatorcommuteswithitself:[A,A]=0Andit’seasytoshowthatthecommutatorofA,BisthenegativeofthecommutatorofB,A:[A,B]=–[B,A]It’salsotruethatcommutatorsarelinear:[A,B+C+D+...]=[A,B]+[A,C]+[A,D]+...AndtheHermitianadjointofacommutatorworksthisway:Youcanalsofindtheanticommutator,{A,B}:{A,B}=AB+BAFindinganti-HermitianoperatorsHere’sanotherone:WhatcanyousayabouttheHermitianadjointofthecommutatoroftwoHermitianoperators?Here’stheanswer.First,writetheadjoint:Thedefinitionofcommutatorstellsyouthefollowing:Youknow(AB)†=B†A†(seetheearliersection“GoingHermitianwithHermitianOperatorsandAdjoints”forpropertiesofadjoints).Therefore,ButforHermitianoperators,A=A†,soremovethe†symbols:ButBA–ABisjust–[A,B],soyouhavethefollowing:AandBhereareHermitianoperators.WhenyoutaketheHermitianadjointofanexpressionandgetthesamethingbackwithanegativesigninfrontofit,theexpressioniscalledanti-Hermitian,sothecommutatoroftwoHermitianoperatorsisanti-Hermitian.(Andbytheway,theexpectationvalueofananti-Hermitianoperatorisguaranteedtobecompletelyimaginary.)StartingfromScratchandEndingUpwithHeisenbergIfyou’vereadthroughthelastfewsections,you’renowarmedwithallthisnewtechnology:Hermitianoperatorsandcommutators.Howcanyouputittowork?YoucancomeupwiththeHeisenberguncertaintyrelationstartingvirtuallyfromscratch.Here’sacalculationthattakesyoufromafewbasicdefinitionstotheHeisenberguncertaintyrelation.Thiskindofcalculationshowshowmucheasieritistousethebasis-lessbraandketnotationthanthefullmatrixversionofstatevectors.Thisisn’tthekindofcalculationthatyou’llneedtodoinclass,butfollowitthrough—knowinghowtousekets,bras,commutators,andHermitianoperatorsisvitalinthecomingchapters.TheuncertaintyinameasurementoftheHermitianoperatornamedAisformallygivenbyThatis,ΔAisequaltothesquarerootoftheexpectationvalueofA2minusthesquaredexpectationvalueofA.Ifyou’vetakenanymathclassesthatdealtwithstatistics,thisformulamaybefamiliartoyou.Similarly,theuncertaintyinameasurementusingHermitianoperatorBisNowconsidertheoperatorsΔAandΔB(nottheuncertaintiesΔAandΔBanymore),andassumethatapplyingΔAandΔBasoperatorsgivesyoumeasurementvalueslikethis:Likeanyoperator,usingΔAandΔBcanresultinnewkets:Here’sthekey:TheSchwarzinequaility(fromtheearliersection“Understandingsomerelationshipsusingkets”)givesyouSoyoucanseethattheinequalitysign,≥,whichplaysabigpartintheHeisenberguncertaintyrelation,hasalreadycreptintothecalculation.BecauseΔAandΔBareHermitian,<χ|χ>isequalto<ψ|ΔA2|ψ>and<ϕ|ϕ>isequalto<ψ|ΔB2|ψ>.BecauseΔA†=ΔA(thedefinitionofaHermitianoperator),youcanseethatThismeansthatThatis,<χ|χ>isequalto<ΔA2>and<ϕ|ϕ>isequalto<ΔB2>.SoyoucanrewritetheSchwarzinequalitylikethis:Okay,wherehasthisgottenyou?It’stimetobeclever.NotethatyoucanwriteΔAΔBasHere,{ΔA,ΔB}=ΔAΔB+ΔBΔAistheanticommutatoroftheoperatorsΔAandΔB.Because[ΔA,ΔB]=[A,B](theconstants<A>and<B>subtractout),youcanrewritethisequation:Here’swherethemathgetsintense.Takealookatwhatyouknowsofar:ThecommutatoroftwoHermitianoperators,[A,B],isanti-Hermitian.Theexpectationvalueofananti-Hermitianisimaginary.{ΔA,ΔB}isHermitian.TheexpectationvalueofaHermitianisreal.Allthismeansthatyoucanviewtheexpectationvalueoftheequationasthesumofreal({ΔA,ΔB})andimaginary([A,B])parts,soAndbecausethesecondtermontherightispositiveorzero,youcansaythatthefollowingistrue:Whew!ButnowcomparethisequationtotherelationshipfromtheearlieruseoftheSchwarzinequality:Combiningthetwoequationsgivesyouthis:ThishasthelookoftheHeisenberguncertaintyrelation,exceptforthepeskyexpectationvaluebrackets,<>,andthefactthatΔAandΔBappearsquaredhere.YouwanttoreproducetheHeisenberguncertaintyrelationhere,whichlookslikethis:Okay,sohowdoyougettheleftsideoftheequationfrom<ΔA2><ΔB2>toΔAΔB?BecauseanearlierequationtellsyouthatΔA=A–<A>,youknowthefollowing:Takingtheexpectationvalueofthelastterminthisequation,yougetthisresult:SquaretheearlierequationΔA=(<A2>–<A>2)1/2togetthefollowing:Andcomparingthatequationtothebeforeit,youconcludethatCool.ThatresultmeansthatbecomesThisinequalityatlastmeansthatWell,well,well.Sotheproductoftwouncertaintiesisgreaterthanorequalto1/2theabsolutevalueofthecommutatoroftheirrespectiveoperators?Wow.IsthattheHeisenberguncertaintyrelation?Well,takealook.Inquantummechanics,themomentumoperatorlookslikethis:AndtheoperatorforthemomentuminthexdirectionisSowhat’sthecommutatoroftheXoperator(whichjustreturnsthexpositionofaparticle)andPx?[X,Px]=iℏ,sofromyougetthisnextequation(remember,ΔxandΔpxherearetheuncertaintiesinxandΔpx,nottheoperators):Hotdog!ThatistheHeisenberguncertaintyrelation.(Noticethatbyderivingitfromscratch,however,youhaven’tactuallyconstrainedthephysicalworldthroughtheuseofabstractmathematics—you’vemerelyproved,usingafewbasicassumptions,thatyoucan’tmeasurethephysicalworldwithperfectaccuracy.)EigenvectorsandEigenvalues:They’reNaturallyEigentastic!Asyouknowifyou’vebeenfollowingalonginthischapter,applyinganoperatortoaketcanresultinanewket:Tomakethingseasier,youcanworkwitheigenvectorsandeigenvalues(eigenisGermanfor“innate”or“natural”).Forexample,|ψ>isaneigenvectoroftheoperatorAifThenumberaisacomplexconstantNotewhat’shappeninghere:ApplyingAtooneofitseigenvectors,|ψ>,givesyou|ψ>back,multipliedbythateigenvector’seigenvalue,a.Althoughacanbeacomplexconstant,theeigenvaluesofHermitianoperatorsarerealnumbers,andtheireigenvectorsareorthogonal(thatis,<ψ|ϕ>=0).Castingaproblemintermsofeigenvectorsandeigenvaluescanmakelifealoteasierbecauseapplyingtheoperatortoitseigenvectorsmerelygivesyouthesameeigenvectorbackagain,multipliedbyitseigenvalue—there’snopeskychangeofstate,soyoudon’thavetodealwithadifferentstatevector.Takealookatthisidea,usingtheRoperatorfromrollingthedice,whichisexpressedthiswayinmatrixform(seetheearliersection“Iexpectedthat:Findingexpectationvalues”formoreonthismatrix):TheRoperatorworksin11-dimensionalspaceandisHermitian,sothere’llbe11orthogonaleigenvectorsand11correspondingeigenvalues.BecauseRisadiagonalmatrix,findingtheeigenvectorsiseasy.Youcantakeunitvectorsintheelevendifferentdirectionsastheeigenvectors.Here’swhatthefirsteigenvector,ξ1,wouldlooklike:Andhere’swhatthesecondeigenvector,ξ2,wouldlooklike:Andsoon,uptoξ11:Notethatalltheeigenvectorsareorthogonal.Andtheeigenvalues?They’rethenumbersyougetwhenyouapplytheRoperatortoaneigenvector.Becausetheeigenvectorsarejustunitvectorsinall11dimensions,theeigenvaluesarethenumbersonthediagonaloftheRmatrix:2,3,4,andsoon,upto12.UnderstandinghowtheyworkTheeigenvectorsofaHermitianoperatordefineacompletesetoforthonormalvectors—thatis,acompletebasisforthestatespace.Whenviewedinthis“eigenbasis,”whichisbuiltoftheeigenvectors,theoperatorinmatrixformatisdiagonalandtheelementsalongthediagonalofthematrixaretheeigenvalues.Thisarrangementisoneofthemainreasonsworkingwitheigenvectorsissouseful;youroriginaloperatormayhavelookedsomethinglikethis(Note:Bearinmindthattheelementsinanoperatorcanalsobefunctions,notjustnumbers):Byswitchingtothebasisofeigenvectorsfortheoperator,youdiagonalizethematrixintosomethingmorelikewhatyou’veseen,whichismucheasiertoworkwith:Youcanseewhythetermeigenisappliedtoeigenvectors—theyformanaturalbasisfortheoperator.Iftwoormoreoftheeigenvaluesarethesame,thateigenvalueissaidtobedegenerate.Soforexample,ifthreeeigenvaluesareequalto6,thentheeigenvalue6isthreefolddegenerate.Here’sanothercoolthing:IftwoHermitianoperators,AandB,commute,andifAdoesn’thaveanydegenerateeigenvalues,theneacheigenvectorofAisalsoaneigenvectorofB.(Seetheearliersection“ForwardandBackward:FindingtheCommutator”formoreoncommuting.)FindingeigenvectorsandeigenvaluesSogivenanoperatorinmatrixform,howdoyoufinditseigenvectorsandeigenvalues?Thisistheequationyouwanttosolve:Andyoucanrewritethisequationasthefollowing:Irepresentstheidentitymatrix,with1salongitsdiagonaland0sotherwise:Thesolutionto(A–aI)|ψ>=0existsonlyifthedeterminantofthematrixA–aIis0:det(A–aI)=0FindingeigenvaluesAnyvaluesofathatsatisfytheequationdet(A–aI)=0areeigenvaluesoftheoriginalequation.Trytofindtheeigenvaluesandeigenvectorsofthefollowingmatrix:First,convertthematrixintotheformA–aI:Next,findthedeterminant:det(A–aI)=(–1–a)(–4–a)+2det(A–aI)=a2+5a+6Andthiscanbefactoredasfollows:det(A–aI)=a2+5a+6=(a+2)(a+3)Youknowthatdet(A–aI)=0,sotheeigenvaluesofAaretherootsofthisequation;namely,a1=–2anda2=–3.FindingeigenvectorsHowaboutfindingtheeigenvectors?Tofindtheeigenvectorcorrespondingtoa1(seetheprecedingsection),substitutea1—thefirsteigenvalue,–2—intothematrixintheformA–aI:SoyouhaveBecauseeveryrowofthismatrixequationmustbetrue,youknowthatψ1=ψ2.Andthatmeansthat,uptoanarbitraryconstant,theeigenvectorcorrespondingtoa1isthefollowing:Dropthearbitraryconstant,andjustwritethisasamatrix:Howabouttheeigenvectorcorrespondingtoa2?Plugginga2,–3,intothematrixinA–aIform,yougetthefollowing:ThenyouhaveSo2ψ1–ψ2=0,andψ1=ψ2÷2.Andthatmeansthat,uptoanarbitraryconstant,theeigenvectorcorrespondingtoa2isDropthearbitraryconstant:Sotheeigenvaluesofthisnextmatrixoperatorarea1=–2anda2=–3.Andtheeigenvectorcorrespondingtoa1isTheeigenvectorcorrespondingtoa2isPreparingfortheInversion:SimplifyingwithUnitaryOperatorsApplyingtheinverseofanoperatorundoestheworktheoperatordid:A–1A=AA–1=ISometimes,findingtheinverseofanoperatorishelpful,suchaswhenyouwanttosolveequationslikeAx=y.SolvingforxiseasyifyoucanfindtheinverseofA:x=A–1y.However,findingtheinverseofalargematrixoftenisn’teasy,soquantumphysicscalculationsaresometimeslimitedtoworkingwithunitaryoperators,U,wheretheoperator’sinverseisequaltoitsadjoint,U–1=U†.(Tofindtheadjointofanoperator,A,youfindthetransposebyinterchangingtherowsandcolumns,AT.Thentakethecomplexconjugate,AT*=A†.)Thisgivesyouthefollowingequation:Theproductoftwounitaryoperators,UandV,isalsounitarybecauseWhenyouuseunitaryoperators,ketsandbrastransformthisway:Andyoucantransformotheroperatorsusingunitaryoperatorslikethis:Notethattheprecedingequationsalsomeanthefollowing:Herearesomepropertiesofunitarytransformations:IfanoperatorisHermitian,thenitsunitarytransformedversion,A'=UAU†,isalsoHermitian.TheeigenvaluesofAanditsunitarytransformedversion,A'=UAU†,arethesame.Commutatorsthatareequaltocomplexnumbersareunchangedbyunitarytransformations:[A',B']=[A,B].ComparingMatrixandContinuousRepresentationsWernerHeisenbergdevelopedthematrix-orientedviewofquantumphysicsthatyou’vebeenusingsofarinthischapter.It’ssometimescalledmatrixmechanics.Thematrixrepresentationisfineformanyproblems,butsometimesyouhavetogopastit,asyou’reabouttosee.Oneofthecentralproblemsofquantummechanicsistocalculatetheenergylevelsofasystem.TheenergyoperatoriscalledtheHamilitonian,H,andfindingtheenergylevelsofasystembreaksdowntofindingtheeigenvaluesoftheproblem:Here,EisaneigenvalueoftheHoperator.Here’sthesameequationinmatrixterms:TheallowableenergylevelsofthephysicalsystemaretheeigenvaluesE,whichsatisfythisequation.Thesecanbefoundbysolvingthecharacteristicpolynomial,whichderivesfromsettingthedeterminantoftheabovematrixtozero,likesoThat’sfineifyouhaveadiscretebasisofeigenvectors—ifthenumberofenergystatesisfinite.Butwhatifthenumberofenergystatesisinfinite?Inthatcase,youcannolongeruseadiscretebasisforyouroperatorsandbrasandkets—youuseacontinuousbasis.GoingcontinuouswithcalculusRepresentingquantummechanicsinacontinuousbasisisaninventionofthephysicistErwinSchrödinger.Inthecontinuousbasis,summationsbecomeintegrals.Forexample,takethefollowingrelation,whereIistheidentitymatrix:Itbecomesthefollowing:Andeveryket|ψ>canbeexpandedinabasisofotherkets,|ϕn>,likethis:DoingthewaveTakealookatthepositionoperator,R,inacontinuousbasis.Applyingthisoperatorgivesyour,thepositionvector:Inthisequation,applyingthepositionoperatortoastatevectorreturnsthelocations,r,thataparticlemaybefoundat.Youcanexpandanyketinthepositionbasislikethis:AndthisbecomesHere’saveryimportantthingtounderstand:ψ(r)=<r|ψ>isthewavefunctionforthestatevector|ψ>—it’stheket’srepresentationinthepositionbasis.Orincommonterms,it’sjustafunctionwherethequantity|ψ(r)|2d3rrepresentstheprobabilitythattheparticlewillbefoundintheregiond3ratr.Thewavefunctionisthefoundationofwhat’scalledwavemechanics,asopposedtomatrixmechanics.What’simportanttorealizeisthatwhenyoutalkaboutrepresentingphysicalsystemsinwavemechanics,youdon’tusethebasis-lessbrasandketsofmatrixmechanics;rather,youusuallyusethewavefunction—thatis,brasandketsinthepositionbasis.Therefore,yougofromtalkingabout|ψ>to<r|ψ>,whichequalsψ(r).Thiswavefunctionappearsalotinthecomingchapters,andit’sjustaketinthepositionbasis.Soinwavemechanics,H|ψ>=E|ψ>becomesthefollowing:Youcanwritethisasthefollowing:Butwhatis<r|H|ψ>?It’sequaltoHψ(r).TheHamiltonianoperator,H,isthetotalenergyofthesystem,kinetic(p2/2m)pluspotential(V(r))soyougetthefollowingequation:ButthemomentumoperatorisTherefore,substitutingthemomentumoperatorforpgivesyouthis:UsingtheLaplacianoperator,yougetthisequation:Youcanrewritethisequationasthefollowing(calledtheSchrödingerequation):Sointhewavemechanicsviewofquantumphysics,you’renowworkingwithadifferentialequationinsteadofmultiplematricesofelements.Thisallcamefromworkinginthepositionbasis,ψ(r)=<r|ψ>insteadofjust|ψ>.Thequantumphysicsintherestofthebookislargelyaboutsolvingthisdifferentialequationforavarietyofpotentials,V(r).Thatis,yourfocusisonfindingthewavefunctionthatsatisfiestheSchrödingerequationforvariousphysicalsystems.WhenyousolvetheSchrödingerequationforψ(r),youcanfindtheallowedenergystatesforaphysicalsystem,aswellastheprobabilitythatthesystemwillbeinacertainpositionstate.Notethat,besideswavefunctionsinthepositionbasis,youcanalsogiveawavefunctioninthemomentumbasis,ψ(p),orinanynumberofotherbases.TheHeisenbergtechniqueofmatrixmechanicsisonewayofworkingwithquantumphysics,andit’sbestusedforphysicalsystemswithwelldefinedenergystates,suchasharmonicoscillators.TheSchrödingerwayoflookingatthings,wavemechanics,useswavefunctions,mostlyinthepositionbasis,toreducequestionsinquantumphysicstoadifferentialequation.PartIIBoundandUndetermined:HandlingParticlesinBoundStatesInthispart...Thispartiswhereyougetthelowdownononeofquantumphysics’favoritetopics:solvingtheenergylevelsandwavefunctionsforparticlestrappedinvariousboundstates.Forexample,youmayhaveaparticletrappedinasquarewell,whichismuchlikehavingapeainabox.Oryoumayhaveaparticleinharmonicoscillation.Quantumphysicsisexpertathandlingthosekindsofsituations.Chapter3GettingStuckinEnergyWellsInThisChapterUnderstandingpotentialwellsWorkingwithinfinitesquarewellsDeterminingenergylevelsTrappingparticleswithpotentialbarriersHandlingfreeparticlesWhat’sthat,Lassie?Stuckinanenergywell?Gogethelp!Inthischapter,yougettoseequantumphysicsatwork,solvingproblemsinonedimension.Youseeparticlestrappedinpotentialwellsandsolvefortheallowableenergystatesusingquantumphysics.Thatgoesagainstthegraininclassicalphysics,whichdoesn’trestricttrappedparticlestoanyparticularenergyspectrum.Butasyouknow,whentheworldgetsmicroscopic,quantumphysicstakesover.TheequationofthemomentistheSchrödingerequation(derivedinChapter2),whichletsyousolveforthewavefunction,ψ(x),andtheenergylevels,E:LookingintoaSquareWellAsquarewellisapotential(thatis,apotentialenergywell)thatformsasquareshape,asyoucanseeinFigure3-1.Figure3-1:Asquarewell.Thepotential,orV(x),goestoinfinityatx<0andx>a(wherexisdistance),likethis:V(x)=∞,wherex<0V(x)=0,where0≤x≤aV(x)=∞,wherex>aUsingsquarewells,youcantrapparticles.Ifyouputaparticleintoasquarewellwithalimitedamountofenergy,it’llbetrappedbecauseitcan’tovercometheinfinitepotentialateithersideofthesquarewell.Therefore,theparticlehastomoveinsidethesquarewell.Sodoestheparticlejustsortofrollaroundonthebottomofthesquarewell?Notexactly.Theparticleisinaboundstate,anditswavefunctiondependsonitsenergy.Thewavefunctionisn’tcomplicated:Soyouhavetheallowedwavefunctionsforthestatesn=1,2,3,andsoon.Theenergyoftheallowableboundstatesaregivenbythefollowingequation:Therestofthischaptershowsyouhowtosolveproblemslikethisone.TrappingParticlesinPotentialWellsTakealookatthepotentialinFigure3-2.Noticethedip,orwell,inthepotential,whichmeansthatparticlescanbetrappedinitiftheydon’thavetoomuchenergy.Theparticle’skineticenergysummedwithitspotentialenergyisaconstant,equaltoitstotalenergy:IfitstotalenergyislessthanV1,theparticlewillbetrappedinthepotentialwell,youseeinFigure3-2;togetoutofthewell,theparticle’skineticenergywouldhavetobecomenegativetosatisfytheequation,whichisimpossible.Figure3-2:Apotentialwell.Inthissection,youtakealookatthevariouspossiblestatesthataparticlewithenergyEcantakeinthepotentialgivenbyFigure3-2.Quantum-mechanicallyspeaking,thosestatesareoftwokinds—boundandunbound.Thissectionlooksattheminoverview.BindingparticlesinpotentialwellsBoundstateshappenwhentheparticleisn’tfreetotraveltoinfinity—it’sassimpleasthat.Inotherwords,theparticleisconfinedtothepotentialwell.AparticletravelinginthepotentialwellyouseeinFigure3-2isboundifitsenergy,E,islessthanbothV1andV2.Inthatcase,theparticlemoves(inaclassicalapproximation)betweenx1andx2.Aparticletrappedinsuchawellisrepresentedbyawavefunction,andyoucansolvetheSchrödingerequationfortheallowedwavefunctionsandtheallowedenergystates.Youneedtousetwoboundaryconditions(theSchrödingerequationisasecond-orderdifferentialequation)tosolvetheproblemcompletely.Boundstatesarediscrete—thatis,theyformanenergyspectrumofdiscreteenergylevels.TheSchrödingerequationgivesyouthosestates.Inaddition,inone-dimensionalproblems,theenergylevelsofaboundstatearenot​degenerate—thatis,notwoenergylevelsarethesameintheentireenergyspectrum.EscapingfrompotentialwellsIfaparticle’senergy,E,isgreaterthanthepotentialV1inFigure3-2,theparticlecanescapefromthepotentialwell.Therearetwopossiblecases:V1<E<V2andE>V2.Thissectionlooksatthemseparately.Case1:Energybetweenthetwopotentials(V1<E<V2)IfV1<E<V2,theparticleinthepotentialwellhasenoughenergytoovercomethebarrierontheleftbutnotontheright.Theparticleisthusfreetomovetonegativeinfinity,soitsallowedxregionisbetween–∞andx1.Here,theallowedenergyvaluesarecontinuous,notdiscrete,becausetheparticleisn’tcompletelybound.Theenergyeigenvaluesarenotdegenerate—thatis,notwoenergyeigenvaluesarethesame(seeChapter2formoreoneigenvalues).TheSchrödingerequationisasecond-orderdifferentialequation,soithastwolinearlyindependentsolutions;however,inthiscase,onlyoneofthosesolutionsisphysicalanddoesn’tdiverge.Thewaveequationinthiscaseturnsouttooscillateforx<x2andtodecayrapidlyforx>x2.Case2:Energygreaterthanthehigherpotential(E>V2)IfE>V2,theparticleisn’tboundatallandisfreetotravelfromnegativeinfinitytopositiveinfinity.Theenergyspectrumiscontinuousandthewavefunctionturnsouttobeasumofafunctionmovingtotherightandonemovingtotheleft.Theenergylevelsoftheallowedspectrumarethereforedoublydegenerate.That’salltheoverviewyouneed—timetostartsolvingtheSchrödingerequationforvariousdifferentpotentials,startingwiththeeasiestofall:infinitesquarewells.TrappingParticlesinInfiniteSquarePotentialWellsInfinitesquarewells,inwhichthewallsgotoinfinity,areafavoriteinphysicsproblems.Youexplorethequantumphysicstakeontheseproblemsinthissection.Findingawave-functionequationTakealookattheinfinitesquarewellthatappearsbackinFigure3-1.Here’swhatthatsquarewelllookslike:V(x)=∞,wherex<0V(x)=0,where0≤x≤aV(x)=∞,wherex>aTheSchrödingerequationlookslikethisinthreedimensions:WritingouttheSchrödingerequationgivesyouthefollowing:You’reinterestedinonlyonedimension—x(distance)—inthischapter,sotheSchrödingerequationlookslikeBecauseV(x)=0insidethewell,theequationbecomesAndinproblemsofthissort,theequationisusuallywrittenaswhere(kisthewavenumber).Sonowyouhaveasecond-orderdifferentialequationtosolveforthewavefunctionofaparticletrappedinaninfinitesquarewell.Yougettwoindependentsolutionsbecausethisequationisasecond-orderdifferentialequation:ψ1(x)=Asin(kx)ψ2(x)=Bcos(kx)AandBareconstantsthatareyettobedetermined.Thegeneralsolutionofisthesumofψ(x)=Asin(kx)+Bcos(kx)DeterminingtheenergylevelsTheequationψ(x)=Asin(kx)+Bcos(kx)tellsyouthatyouhavetousetheboundaryconditionstofindtheconstantsAandB(theprecedingsectionexplainshowtoderivetheequation).Whataretheboundaryconditions?Thewavefunctionmustdisappearattheboundariesofaninfinitesquarewell,soψ(0)=0ψ(a)=0Thefactthatψ(0)=0tellsyourightawaythatBmustbezerobecausecos(0)=1.Andthefactthatψ(a)=0tellsyouthatψ(a)=Asin(ka)=0.Becausesineiszerowhenitsargumentisamultipleofπ,thismeansthatka=nπn=1,2,3...Notethatalthoughn=0istechnicallyasolution,ityieldsψ(x)=0forallx,whichisnotnormalizable,soit’snotaphysicalsolution—thephysicalsolutionsbeginwithn=1.ThisequationcanalsobewrittenasAndbecausek2=2mE/ℏ2,youhavethefollowingequation,wheren=1,2,3,...—thosearetheallowedenergystates.Thesearequantizedstates,​‐correspondingtothequantumnumbers1,2,3,andsoon:Notethatthefirstphysicalstatecorrespondston=1,whichgivesyouthisnextequation:Thisisthelowestphysicalstatethattheparticlescanoccupy.Justforkicks,putsomenumbersintothis,assumingthatyouhaveanelectron,mass9.11×10–31kilograms,confinedtoaninfinitesquarewellofwidthoftheorderoftheBohrradius(theaverageradiusofanelectron’sorbitinahydrogenatom);let’ssaya=3.00x10–10meters.givesyouthisenergyforthegroundstate:That’saverysmallamount,about4.2electronvolts(eV—theamountofenergyoneelectrongainsfallingthrough1volt).Evenso,it’salreadyontheorderoftheenergyofthegroundstateofanelectroninthegroundstateofahydrogenatom(13.6eV),soyoucansayyou’recertainlyintherightquantumphysicsballparknow.NormalizingthewavefunctionOkay,youhavethisforthewaveequationforaparticleinaninfinitesquarewell:Thewavefunctionisasinewave,goingtozeroatx=0andx=a.YoucanseethefirsttwowavefunctionsplottedinFigure3-3.Figure3-3:Wavefunctionsinasquarewell.NormalizingthewavefunctionletsyousolvefortheunknownconstantA.Inanormalizedfunction,theprobabilityoffindingtheparticlebetweenxanddx,|ψ(x)|2dx,addsupto1whenyouintegrateoverthewholesquarewell,x=0tox=a:Substitutingforψ(x)givesyouthefollowing:Here’swhattheintegralinthisequationequals:Sofromthepreviousequation,.SolveforA:Therefore,here’sthenormalizedwaveequationwiththevalueofApluggedin:Andthat’sthenormalizedwavefunctionforaparticleinaninfinitesquarewell.AddingtimedependencetowavefunctionsNowhowaboutseeinghowthewavefunctionforaparticleinaninfinitesquarewellevolveswithtime?TheSchrödingerequationlookslikethis:YoucanalsowritetheSchrödingerequationthisway,whereHistheHermitianHamiltonianoperator:Hψ(r)=Eψ(r)That’sactuallythetime-independentSchrödingerequation.ThetimedependentSchrödingerequationlookslikethis:Combiningtheprecedingthreeequationsgivesyouthefollowing,whichisanotherformofthetime-dependentSchrödingerequation:Andbecauseyou’redealingwithonlyonedimension,x,thisequationbecomesThisissimplerthanitlooks,however,becausethepotentialdoesn’tchangewithtime.Infact,becauseEisconstant,youcanrewritetheequationasThatequationmakeslifealotsimpler—it’seasytosolvethetime-dependentSchrödingerequationifyou’redealingwithaconstantpotential.Inthiscase,thesolutionisNeat.Whenthepotentialdoesn’tvarywithtime,thesolutiontothetimedependentSchrödingerequationsimplybecomesψ(x),thespatialpart,multipliedbye–iEt/ℏ,thetime-dependentpart.Sowhenyouaddinthetime-dependentparttothetime-independentwavefunction,yougetthetime-dependentwavefunction,whichlookslikethis:TheenergyofthenthquantumstateisTherefore,theresultisShiftingtosymmetricsquarewellpotentialsThestandardinfinitesquarewelllookslikethis:V(x)=∞,wherex<0V(x)=0,where0≤x≤aV(x)=∞,wherex>aButwhatifyouwanttoshiftthingssothatthesquarewellissymmetricaroundtheorigininstead?Thatis,youmovethesquarewellsothatitextendsfrom–a/2toa/2?Here’swhatthenewinfinitesquarewelllookslikeinthiscase:V(x)=∞,wherex<–a/2V(x)=0,where–a/2≤x≤a/2V(x)=∞,wherex>a/2Youcantranslatefromthisnewsquarewelltotheoldonebyaddinga/2tox,whichmeansthatyoucanwritethewavefunctionforthenewsquarewellinthisequationlikethefollowing:Doingalittletriggivesyouthefollowingequations:Soasyoucansee,theresultisamixofsinesandcosines.Theboundstatesarethese,inincreasingquantumorder:Andsoon.Notethatthecosinesaresymmetricaroundtheorigin:ψ(x)=ψ(–x).Thesinesareanti-symmetric:–ψ(x)=ψ(–x).LimitedPotential:TakingaLookatParticlesandPotentialStepsTrulyinfinitepotentials(whichIdiscussintheprevioussections)arehardtocomeby.Inthissection,youlookatsomereal-worldexamples,wherethepotentialissettosomefiniteV0,notinfinity.Forexample,takealookatthesituationinFigure3-4.There,aparticleistravelingtowardapotentialstep.Currently,theparticleisinaregionwhereV=0,butit’llsoonbeintheregionV=V0.Figure3-4:Apotentialstep,E>V0.TherearetwocasestolookathereintermsofE,theenergyoftheparticle:E>V0:Classically,whenE>V0,youexpecttheparticletobeabletocontinueontotheregionx>0.E<V0:WhenE<V0,you’dexpecttheparticletobouncebackandnotbeabletogettotheregionx>0atall.Inthissection,youstartbytakingalookatthecasewheretheparticle’senergy,E,isgreaterthanthepotentialV0,asshowninFigure3-4;thenyoutakealookatthecasewhereE<V0.AssumingtheparticlehasplentyofenergyStartwiththecasewheretheparticle’senergy,E,isgreaterthanthepotentialV0.Fromaquantumphysicspointofview,here’swhattheSchrödingerequationwouldlooklike:Fortheregionx<0:Here,.Fortheregionx>0:Inthisequation,.Inotherwords,kisgoingtovarybyregion,asyouseeinFigure3-5.Figure3-5:Thevalueofkbyregion,whereE>V0.Treatingthefirstequationasasecond-orderdifferentialequation,youcanseethatthemostgeneralsolutionisthefollowing:ψ1(x)=Aeik1x+Be–ik1x,wherex<0Andfortheregionx>0,solvingthesecondequationgivesyouthis:ψ2(x)=Ceik2x+De–ik2x,wherex>0Notethateikxrepresentsplanewavestravelinginthe+xdirection,ande–ikxrepresentsplanewavestravelinginthe–xdirection.Whatthissolutionmeansisthatwavescanhitthepotentialstepfromtheleftandbeeithertransmittedorreflected.Giventhatwayoflookingattheproblem,youmaynotethatthewavecanbereflectedonlygoingtotheright,nottotheleft,soDmustequalzero.Thatmakesthewaveequationbecomethefollowing:Wherex<0:ψ1(x)=Aeik1x+Be–ik1xWherex>0:ψ2(x)=Ceik2xThetermAeik1xrepresentstheincidentwave,Be–ik1xisthereflectedwave,andCeik2xisthetransmittedwave.CalculatingtheprobabilityofreflectionortransmissionYoucancalculatetheprobabilitythattheparticlewillbereflectedortransmittedthroughthepotentialstepbycalculatingthereflectionandtransmissioncoefficients.ThesearedefinedintermsofsomethingcalledthecurrentdensityJ(x);thisisgivenintermsofthewavefunctionbyIfJristhereflectedcurrentdensity,andJi,istheincidentcurrentdensity,thenR,thereflectioncoefficientisT,thetransmissioncoefficient,isYounowhavetocalculateJr,Ji,andJt.Actually,that’snotsohard—startwithJi.Becausetheincidentpartofthewaveisψi(x)=Aeik1x,theincidentcurrentdensityisAndthisjustequals.JrandJtworkinthesameway:Soyouhavethisforthereflectioncoefficient:T,thetransmissioncoefficient,isFindingA,B,andCSohowdoyoufigureouttheconstantsA,B,andC?Youdothatasyoufigureoutthecoefficientswiththeinfinitesquarewellpotential—withboundaryconditions(seetheearliersection“TrappingParticlesinInfiniteSquareWellPotentials”).However,here,youcan’tnecessarilysaythatψ(x)goestozero,becausethepotentialisnolongerinfinite.Instead,theboundaryconditionsarethatψ(x)anddψ(x)/dxarecontinuousacrossthepotentialstep’sboundary.Inotherwords,ψ1(0)=ψ2(0)Youknowthefollowing:Wherex<0:ψ1(x)=Aeik1x+Be–ik1xWherex>0:ψ2(x)=Ceik2xTherefore,pluggingthesetwoequationsintoψ1(0)=ψ2(0)givesyouA+B=C.Andpluggingthemintogivesyouk1A–k1B=k2CSolvingforBintermsofAgivesyouthisresult:SolvingforCintermsofAgivesyouYoucanthencalculateAfromthenormalizationconditionofthewavefunction:Butyoudon’tactuallyneedA,becauseitdropsoutoftheratiosforthereflectionandtransmissioncoefficients,RandT.Inparticular,Therefore,That’saninterestingresult,anditdisagreeswithclassicalphysics,whichsaysthatthereshouldbenoparticlereflectionatall.Asyoucansee,ifk1≠k2,thentherewillindeedbeparticlereflection.Notethatask1goestok2,Rgoesto0andTgoesto1,whichiswhatyou’dexpect.Soalreadyyouhavearesultthatdiffersfromtheclassical—theparticlecanbereflectedatthepotentialstep.That’sthewave-likebehavioroftheparticlecomingintoplayagain.Assumingtheparticledoesn’thaveenoughenergyOkay,nowtrythecasewhereE<V0whenthere’sapotentialstep,asshowninFigure3-6.Inthiscase,theparticledoesn’thaveenoughenergytomakeitintotheregionx>0,accordingtoclassicalphysics.Seewhatquantumphysicshastosayaboutit.Figure3-6:Apotentialstep,E<V0.Tackletheregionx<0first.There,theSchrödingerequationwouldlooklikethis:where.Youknowthesolutiontothisfromthepreviousdiscussiononpotentialsteps(see“LimitedPotential:TakingaLookatParticlesandPotentialSteps”):Okay,butwhatabouttheregionx>0?That’sadifferentstory.Here’stheSchrödingerequation:where.Buthangon;E–V0islessthanzero,whichwouldmakekimaginary,whichisimpossiblephysically.SochangethesignintheSchrödingerequationfromplustominus:Andusethefollowingfork2(notethatthisispositiveifE<V0):Okay,sonowyouhavetosolvethedifferential.Therearetwolinearlyindependentsolutions:ψ(x)=Ce–k2xψ(x)=Dek2xAndthegeneralsolutiontoisHowever,wavefunctionsmustbefiniteeverywhere,andthesecondtermisclearlynotfiniteasxgoestoinfinity,soDmustequalzero(notethatifxgoestonegativeinfinity,thefirsttermalsodiverges,butbecausethepotentialstepislimitedtox>0,thatisn’taproblem).Therefore,here’sthesolutionforx>0:Soyourwavefunctionsforthetworegionsareψ1(x)=Aeik1x+Be–ik1xx<0ψ2(x)=Ce–k2xx>0Puttingthisintermsoftheincident,reflected,andtransmittedwavefunctions,ψi(x),ψr(x),andψt(x),youhavethefollowing:ψi(x)=Aeik1xψr(x)=Be–ik1xψt(x)=Ce–k2xFindingtransmissionandreflectioncoefficientsNowyoucanfigureoutthereflectionandtransmissioncoefficients,RandT(asyoudoforthecaseE>V0intheearliersection“Assumingtheparticlehasplentyofenergy”):Actually,thisisveryeasyinthiscase;takealookatJt:Butbecauseψt(x)=Ce–k2x,ψt(x)iscompletelyreal,whichmeansthatinthiscase,thefollowingistrue:Andthisequation,ofcourse,isequaltozero.SoJt=0;therefore,T=0.IfT=0,thenRmustequal1.Thatmeansthatyouhaveacompletereflection,justasintheclassicalsolution.Thenonzerosolution:Findingaparticleinx>0Despitethecompletereflection,there’sadifferencebetweenthemathematicalandclassicalsolution:Thereactuallyisanonzerochanceoffindingtheparticleintheregionx>0.Toseethat,takealookattheprobabilitydensityforx>0,whichisP(x)=|ψt(x)|2Plugginginforthewavefunctionψt(x)givesyouP(x)=|ψt(x)|2=|C|2e–2k2xYoucanusethecontinuityconditionstosolveforCintermsofA:ψ1(0)=ψ2(0)Usingthecontinuityconditionsgivesyouthefollowing:Thisdoesfallquicklytozeroasxgetslarge,butnearx=0,ithasanonzerovalue.YoucanseewhattheprobabilitydensitylookslikefortheE<V0caseofapotentialstepinFigure3-7.Figure3-7:Thevalueofkbyregion,E<V0.Okay,you’vetakencareofinfinitesquarewellsandpotentialsteps.Nowwhataboutthecasewherethepotentialstepdoesn’textendouttoinfinitybutisitselfbounded?Thatbringsyoutopotentialbarriers,whichIdiscussinthenextsection.HittingtheWall:ParticlesandPotentialBarriersWhatiftheparticlecouldworkitswaythroughapotentialstep—thatis,thestepwasoflimitedextent?Thenyou’dhaveapotentialbarrier,whichissetupsomethinglikethis:V(x)=0,wherex<0V(x)=V0,where0≤x≤aV(x)=0,wherex>aYoucanseewhatthispotentiallookslikeinFigure3-8.Figure3-8:ApotentialbarrierE>V0.InsolvingtheSchrödingerequationforapotentialbarrier,youhavetoconsidertwocases,correspondingtowhethertheparticlehasmoreorlessenergythanthepotentialbarrier.Inotherwords,ifEistheenergyoftheincidentparticle,thetwocasestoconsiderareE>V0andE<V0.ThissectionstartswithE>V0.GettingthroughpotentialbarrierswhenE>V0InthecasewhereE>V0,theparticlehasenoughenergytopassthroughthepotentialbarrierandendupinthex>aregion.ThisiswhattheSchrödingerequationlookslike:Fortheregionx<0:whereFortheregion0≤x≤a:whereFortheregionx>a:whereThesolutionsforψ1(x),ψ2(x),andψ3(x)arethefollowing:Wherex<0:ψ1(x)=Aeik1x+Be–ik1xWhere0≤x≤a:ψ2(x)=Ceik2x+De–ik2xWherex>a:ψ3(x)=Eeik1x+Fe–ik1xInfact,becausethere’snoleftwardtravelingwaveinthex>aregion,F=0,soψ3(x)=Eeik1x.SohowdoyoudetermineA,B,C,D,andE?Youusethecontinuityconditions,whichworkoutheretobethefollowing:Okay,fromtheseequations,yougetthefollowing:A+B=C+Dik1(A–B)=ik2(C–D)Ceik2a+De–ik2a=Eeik1aik2Ceik2a–ik2De–ik2a=ik1Eeik1aSoputtingalloftheseequationstogether,yougetthisforthecoefficientEintermsofA:Wow.Sowhat’sthetransmissioncoefficient,T?Well,TisAndthisworksouttobeWhew!Notethatask1goestok2,Tgoesto1,whichiswhatyou’dexpect.SohowaboutR,thereflectioncoefficient?I’llspareyouthealgebra;here’swhatRequals:YoucanseewhattheE>V0probabilitydensity,|ψ(x)|2,lookslikeforthepotentialbarrierinFigure3-9.Figure3-9:|ψ(x)|2forapotentialbarrierE>V0.ThatcompletesthepotentialbarrierwhenE>V0.Gettingthroughpotentialbarriers,evenwhenE<V0Whathappensiftheparticledoesn’thaveasmuchenergyasthepotentialofthebarrier?Inotherwords,you’renowfacingthesituationyouseeinFigure310.Figure3-10:ApotentialbarrierE<V0.NowtheSchrödingerequationlookslikethis:Fortheregionx<0:ψ1(x)=Aeik1x+Be–ik1xFortheregion0≤x≤a:where.ButnowE–V0islessthan0,whichwouldmakekimaginary.Andthat’simpossiblephysically.SochangethesignintheSchrödingerequationfromplustominus:Andusethisfork2:.Fortheregionx>a:where.Allthismeansthatthesolutionsforψ1(x),ψ2(x),andψ3(x)arethefollowing:Wherex<0:ψ1(x)=Aeik1x+Be–ik1xWhere0≤x≤a:ψ2(x)=Cek2x+De–k2xWherex>a:ψ3(x)=Eeik1x+Fe–ik1xInfact,there’snoleftwardtravelingwaveintheregionx>a;F=0,soψ3(x)isψ3(x)=Eeik1x.ThissituationissimilartothecasewhereE>V0,exceptfortheregion0≤x≤a.Thewavefunctionoscillatesintheregionswhereithaspositiveenergy,x<0andx>a,butisadecayingexponentialintheregion0≤x≤a.Youcanseewhattheprobabilitydensity,|ψ(x)|2,lookslikeinFigure3-11.Figure3-11:|ψ(x)|2forapotentialbarrierE<V0.FindingthereflectionandtransmissioncoefficientsHowaboutthereflectionandtransmissioncoefficients,RandT?Here’swhattheyequal:Asyoumayexpect,youusethecontinuityconditionstodetermineA,B,andE:ψ1(0)=ψ2(0)ψ2(a)=ψ3(a)AfairbitofalgebraandtrigisinvolvedinsolvingforRandT;here’swhatRandTturnouttobe:Despitetheequation’scomplexity,it’samazingthattheexpressionforTcanbenonzero.Classically,particlescan’tentertheforbiddenzone0≤x≤abecauseE<V0,whereV0isthepotentialinthatregion;theyjustdon’thaveenoughenergytomakeitintothatarea.TunnelingthroughQuantummechanically,thephenomenonwhereparticlescangetthroughregionsthatthey’reclassicallyforbiddentoenteriscalledtunneling.Tunnelingispossiblebecauseinquantummechanics,particlesshowwaveproperties.Tunnelingisoneofthemostexcitingresultsofquantumphysics—itmeansthatparticlescanactuallygetthroughclassicallyforbiddenregionsbecauseofthespreadintheirwavefunctions.Thisis,ofcourse,amicroscopiceffect—don’ttrytowalkthroughanycloseddoors—butit’sasignificantone.Amongothereffects,tunnelingmakestransistorsandintegratedcircuitspossible.Youcancalculatethetransmissioncoefficient,whichtellsyoutheprobabilitythataparticlegetsthrough,givenacertainincidentintensity,whentunnelingisinvolved.Doingsoisrelativelyeasyintheprecedingsectionbecausethebarrierthattheparticlehastogetthroughisasquarebarrier.Butingeneral,calculatingthetransmissioncoefficientisn’tsoeasy.Readon.GettingthetransmissionwiththeWKBapproximationThewayyougenerallycalculatethetransmissioncoefficientistobreakupthepotentialyou’reworkingwithintoasuccessionofsquarebarriersandtosumthem.That’scalledtheWentzel-Kramers-Brillouin(WKB)approximation—treatingageneralpotential,V(x),asasumofsquarepotentialbarriers.TheresultoftheWKBapproximationisthatthetransmissioncoefficientforanarbitrarypotential,V(x),foraparticleofmassmandenergyEisgivenbythisexpression(thatis,aslongasV(x)isasmooth,slowlyvaryingfunction):Sonowyoucanamazeyourfriendsbycalculatingtheprobabilitythataparticlewilltunnelthroughanarbitrarypotential.It’sthestuffsciencefictionismadeof—well,onthemicroscopicscale,anyway.ParticlesUnbound:SolvingtheSchrödingerEquationforFreeParticlesWhataboutparticlesoutsideanysquarewell—thatis,freeparticles?Thereareplentyofparticlesthatactfreelyintheuniverse,andquantumphysicshassomethingtosayaboutthem.Here’stheSchrödingerequation:Whatiftheparticlewereafreeparticle,withV(x)=0?Inthatcase,you’dhavethefollowingequation:Andyoucanrewritethisaswherethewavenumber,k,is.YoucanwritethegeneralsolutiontothisSchrödingerequationasψ(x)=Aeikx+Be–ikxIfyouaddtime-dependencetotheequation,yougetthistime-dependentwavefunction:That’sasolutiontotheSchrödingerequation,butitturnsouttobeunphysical.Toseethis,notethatforeithertermintheequation,youcan’tnormalizetheprobabilitydensity,|ψ(x)|2(seetheearliersectiontitled“Normalizingthewavefunction”formoreonnormalizing):|ψ(x)|2=|A|2or|B|2What’sgoingonhere?Theprobabilitydensityforthepositionoftheparticleisuniformthroughoutallx!Inotherwords,youcan’tpindowntheparticleatall.Thisisaresultoftheformofthetime-dependentwavefunction,whichusesanexactvalueforthewavenumber,k—andp=ℏkandE=ℏk2/2m.SowhatthatequationsaysisthatyouknowEandpexactly.AndifyouknowpandEexactly,thatcausesalargeuncertaintyinxandt—infact,xandtarecompletelyuncertain.Thatdoesn’tcorrespondtophysicalreality.Forthatmatter,thewavefunctionψ(x),asitstands,isn’tsomethingyoucannormalize.Tryingtonormalizethefirstterm,forexample,givesyouthisintegral:Andforthefirsttermofψ(x,t),thisisAndthesameistrueofthesecondterminψ(x,t).Sowhatdoyoudoheretogetaphysicalparticle?Thenextsectionexplains.GettingaphysicalparticlewithawavepacketIfyouhaveanumberofsolutionstotheSchrödingerequation,anylinearcombinationofthosesolutionsisalsoasolution.Sothat’sthekeytogettingaphysicalparticle:Youaddvariouswavefunctionstogethersothatyougetawavepacket,whichisacollectionofwavefunctionsoftheformei(kx–Et/ℏ)suchthatthewavefunctionsinterfereconstructivelyatonelocationandinterferedestructively(gotozero)atallotherlocations:Thisisusuallywrittenasacontinuousintegral:Whatisϕ(k,t)?It’stheamplitudeofeachcomponentwavefunction,andyoucanfindϕ(k,t)fromtheFouriertransformoftheequation:Becausek=p/ℏ,youcanalsowritethewavepacketequationslikethis,intermsofp,notk:Well,youmaybeaskingyourselfjustwhat’sgoingonhere.Itlookslikeψ(x,t)isdefinedintermsofϕ(p,t),butϕ(p,t)isdefinedintermsofψ(x,t).Thatlooksprettycircular.Theansweristhatthetwopreviousequationsaren’tdefinitionsofψ(x,t)orϕ(p,t);they’rejustequationsrelatingthetwo.You’refreetochooseyourownwavepacketshapeyourself—forexample,youmayspecifytheshapeofϕ(p,t),andwouldletyoufindψ(x,t).GoingthroughaGaussianexampleHere’sanexampleinwhichyougetconcrete,selectinganactualwavepacketshape.Chooseaso-calledGaussianwavepacket,whichyoucanseeinFigure312—localizedinoneplace,zerointheothers.Figure3-12:AGaussianwavepacket.Theamplitudeϕ(k)youmaychooseforthiswavepacketisYoustartbynormalizingϕ(k)todeterminewhatAis.Here’showthatworks:Substitutinginϕ(k)givesyouthisequation:Doingtheintegral(thatmeanslookingitupinmathtables)givesyouthefollowing:Therefore,.Sohere’syourwavefunction:Thislittlegemofanintegralcanbeevaluatedtogiveyouthefollowing:Sothat’sthewavefunctionforthisGaussianwavepacket(Note:Theexp[–x2/a2]istheGaussianpartthatgivesthewavepacketthedistinctiveshapethatyouseeinFigure3-12)—andit’salreadynormalized.Nowyoucanusethiswavepacketfunctiontodeterminetheprobabilitythattheparticlewillbein,say,theregion0≤x≤a/2.TheprobabilityisInthiscase,theintegralisAndthisworksouttobeSotheprobabilitythattheparticlewillbeintheregion0≤x≤a/2is1/3.Cool!Chapter4BackandForthwithHarmonicOscillatorsInThisChapterHamiltonians:LookingattotalenergySolvingforenergystateswithcreationandannihilationoperatorsUnderstandingthematrixversionofharmonicoscillatoroperatorsWritingcomputercodetosolvetheSchrödingerequationHarmonicoscillatorsarephysicssetupswithperiodicmotion,suchasthingsbouncingonspringsortick-tockingonpendulums.You’reprobablyalreadyfamiliarwithharmonicoscillatorproblemsinthemacroscopicarena,butnowyou’regoingmicroscopic.Therearemany,manyphysicalcasesthatcanbeapproximatedbyharmonicoscillators,suchasatomsinacrystalstructure.Inthischapter,youseebothexactsolutionstoharmonicoscillatorproblemsaswellascomputationalmethodsforsolvingthem.KnowinghowtosolvetheSchrödingerequationusingcomputersisausefulskillforanyquantumphysicsexpert.GrapplingwiththeHarmonicOscillatorHamiltoniansOkay,timetostarttalkingHamiltonians(andI’mnotreferringtofansoftheU.S.FoundingFatherAlexanderHamilton).TheHamiltonianwillletyoufindtheenergylevelsofasystem.GoingclassicalwithharmonicoscillationInclassicalterms,theforceonanobjectinharmonicoscillationisthefollowing(thisisHooke’slaw):F=–kxInthisequation,kisthespringconstant,measuredinNewtons/meter,andxisdisplacement.Thekeypointhereisthattherestoringforceonwhateverisinharmonicmotionisproportionaltoitsdisplacement.Inotherwords,thefartheryoustretchaspring,theharderit’llpullback.BecauseF=ma,wheremisthemassoftheparticleinharmonicmotionandaisitsinstantaneousacceleration,youcansubstituteforFandwritethisequationasma+kx=0Here’stheequationforinstantaneousacceleration,wherexisdisplacementandtistime:Sosubstitutingfora,youcanrewritetheforceequationasDividingbythemassoftheparticlegivesyouthefollowing:Ifyoutakek/m=ω2(whereωistheangularfrequency),thisbecomesYoucansolvethisequationforx,whereAandBareconstants:x=Asinωt+BcosωtTherefore,thesolutionisanoscillatingonebecauseitinvolvessinesandcosines,whichrepresentperiodicwaveforms.UnderstandingtotalenergyinquantumoscillationNowlookatharmonicoscillatorsinquantumphysicsterms.TheHamiltonian(H)isthesumofkineticandpotentialenergies—thetotalenergyofthesystem:H=KE+PEForaharmonicoscillator,here’swhattheseenergiesareequalto:Thekineticenergyatanyonemomentisthefollowing,wherepistheparticle’smomentumandmisitsmass:Theparticle’spotentialenergyisequaltothefollowing,wherekisthespringconstantandxisdisplacement:(Note:Thekisreplacedbecauseω2=k/m.)Therefore,inquantumphysicsterms,youcanwritetheHamiltonianasH=KE+PE,orwherePandXarethemomentumandpositionoperators.YoucanapplytheHamiltonianoperatortovariouseigenstates(seeChapter2formoreoneigenstates),|ψ>,oftheharmonicoscillatortogetthetotalenergy,E,ofthoseeigenstates:Theproblemnowbecomesoneoffindingtheeigenstatesandeigenvalues.However,thisdoesn’tturnouttobeaneasytask.UnlikethepotentialsV(x)coveredinChapter3,V(x)foraharmonicoscillatorismorecomplex,dependingasitdoesonx2.Soyouhavetobeclever.Thewayyousolveharmonicoscillatorproblemsinquantumphysicsiswithoperatoralgebra—thatis,youintroduceanewsetofoperators.Andthey’recomingupnow.CreationandAnnihilation:IntroducingtheHarmonicOscillatorOperatorsCreationandannihilationmaysoundlikebigmake-or-breaktheuniversekindsofideas,buttheyplayastarringroleinthequantumworldwhenyou’reworkingwithharmonicoscillators.YouusethecreationandannihilationoperatorstosolveharmonicoscillatorproblemsbecausedoingsoisacleverwayofhandlingthetougherHamiltonianequation(seetheprecedingsection).Here’swhatthesetwonewoperatorsdo:Creationoperator:Thecreationoperatorraisestheenergylevelofaneigenstatebyonelevel,soiftheharmonicoscillatorisinthefourthenergylevel,thecreationoperatorraisesittothefifthlevel.Annihilationoperator:Theannihilationoperatordoesthereverse,loweringeigenstatesonelevel.Theseoperatorsmakeiteasiertosolvefortheenergyspectrumwithoutalotofworksolvingfortheactualeigenstates.Inotherwords,youcanunderstandthewholeenergyspectrumbylookingattheenergydifferencebetweeneigenstates.Mindyourp’sandq’s:GettingtheenergystateequationsHere’showpeopleusuallysolvefortheenergyspectrum.First,youintroducetwonewoperators,pandq,whicharedimensionless;theyrelatetotheP(momentum)andX(position)operatorsthisway:Youusethesetwonewoperators,pandq,asthebasisoftheannihilationoperator,a,andthecreationoperator,a†:NowyoucanwritetheharmonicoscillatorHamiltonianlikethis,intermsofaanda†:Asforcreatingnewoperatorshere,thequantumphysicistswentcrazy,evengivinganametoa†a:theNornumberoperator.Sohere’showyoucanwritetheHamiltonian:TheNoperatorreturnsthenumberoftheenergyleveloftheharmonicoscillator.IfyoudenotetheeigenstatesofNas|n>,yougetthis,wherenisthenumberofthenthstate:N|n>=n|n>BecauseH=ℏω(N+1/2),andbecauseH|n>=En|n>,thenbycomparingtheprevioustwoequations,youhaveAmazingly,thatgivesyoutheenergyeigenvaluesofthenthstateofaquantummechanicalharmonicoscillator.Soherearetheenergystates:Thegroundstateenergycorrespondston=0:ThefirstexcitedstateisThesecondexcitedstatehasanenergyofAndsoon.Thatis,theenergylevelsarediscreteandnondegenerate(notsharedbyanytwostates).Thus,theenergyspectrumismadeupofequidistantbands.FindingtheEigenstatesWhenyouhavetheeigenstates(seeChapter2tofindoutallabouteigenstates),youcandeterminetheallowablestatesofasystemandtherelativeprobabilitythatthesystemwillbeinanyofthosestates.ThecommutatorofoperatorsA,Bis[A,B]=AB–BA,sonotethatthecommutatorofaanda†isthefollowing:Thisisequaltothefollowing:Thisequationbreaksdownto[a,a†]=1.AndputtingtogetherthisequationwithFindingtheenergyofa|n>Okay,withthecommutatorrelations,you’rereadytogo.Thefirstquestionisiftheenergyofstate|n>isEn,whatistheenergyofthestatea|n>?Well,tofindthisrearrangethecommutator[a,H]=ℏωatogetHa=aH–ℏωa.Thenusethistowritetheactionofℏona|n>likethis:H(a|n>)=(aH–ℏωa)|n>=(En–ℏω)(a|n>)Soa|n>isalsoaneigenstateoftheharmonicoscillator,withenergyEn–ℏω,notEn.That’swhyaiscalledtheannihilationorloweringoperator:Itlowerstheenergylevelofaharmonicoscillatoreigenstatebyonelevel.Findingtheenergyofa†|n>Sowhat’stheenergylevelofa†|n>?Youcanwritethatcanlikethis:Allthismeansthata†|n>isaneigenstateoftheharmonicoscillator,withenergyEn+ℏω,notjustEn—thatis,thea†operatorraisestheenergylevelofaneigenstateoftheharmonicoscillatorbyonelevel.Usingaanda†directlyIfyou’vebeenfollowingalongfromtheprecedingsection,youknowthatH(a|n>)=(En–ℏω)(a|n>)andH(a†|n>)=(En+ℏω)(a†|n>).Youcanderivethefollowingfromthetheseequations:a|n>=C|n–1>a†|n>=D|n+1>CandDarepositiveconstants,butwhatdotheyequal?Thestates|n–1>and|n+1>havetobenormalized,whichmeansthat<n–1|n–1>=<n+1|n+1>=1.SotakealookatthequantityusingtheCoperator:(<n|a†)(a|n>)=C2<n–1|n–1>Andbecause|n–1>isnormalized,<n–1|n–1>=1:(<n|a†)(a|n>)=C2<n|a†a|n>=C2Butyoualsoknowthata†a=N,theenergyleveloperator,soyougetthefollowingequation:<n|N|n>=C2N|n>=n|n>,wherenistheenergylevel,son<n|n>=C2However,<n|n>=1,son=C2n½=CThisfinallytellsyou,froma|n>=C|n–1>,thata|n>=n½|n–1>That’scool—nowyouknowhowtousetheloweringoperator,a,oneigenstatesoftheharmonicoscillator.Whatabouttheraisingoperator,a†?Firstwerearrangethecommutator[a†,H]=–ℏωa†,togetHa†=a†H+ℏωaThenyoufollowthesamecourseofreasoningyoutakewiththeaoperatortoshowthefollowing:a†|n>=(n+1)1/2|n+1>Soatthispoint,youknowwhattheenergyeigenvaluesareandhowtheraisingandloweringoperatorsaffecttheharmonicoscillatoreigenstates.You’vemadequitealotofprogress,usingtheaanda†operatorsinsteadoftryingtosolvetheSchrödingerequation.FindingtheharmonicoscillatorenergyeigenstatesThecharmofusingtheoperatorsaanda†isthatgiventhegroundstate,|0>,thoseoperatorsletyoufindallsuccessiveenergystates.Ifyouwanttofindanexcitedstateofaharmonicoscillator,youcanstartwiththegroundstate,|0>,andapplytheraisingoperator,a†.Forexample,youcandothis:Andsoon.Ingeneral,youhavethisrelation:WorkinginpositionspaceOkay,isfineasfarasitgoes—butjustwhatis|0>?Can’tyougetaspatialeigenstateofthiseigenvector?Somethinglikeψ0(x),notjust|0>?Yes,youcan.Inotherwords,youwanttofind<x|0>=ψ0(x).Soyouneedtherepresentationsofaanda†inpositionspace.ThepoperatorisdefinedasBecause,youcanwriteAndwritingx0=,thisbecomesOkay,whatabouttheaoperator?YouknowthatAndthatTherefore,YoucanalsowritethisequationasOkay,sothat’sainthepositionrepresentation.What’sa†?Thatturnsouttobethis:Now’sthetimetobeclever.Youwanttosolvefor|0>inthepositionspace,or<x|0>.Here’sthecleverpart—whenyouusetheloweringoperator,a,on|0>,youhavetoget0becausethere’snolowerstatethanthegroundstate,soa|0>=0.Andapplyingthe<x|bragivesyou<x|a|0>=0.That’scleverbecauseit’sgoingtogiveyouahomogeneousdifferentialequation(thatis,onethatequalszero).First,yousubstitutefora:MultiplyingbothsidesbygivesyouthefollowingThesolutiontothiscompactdifferentialequationisThat’sagaussianfunction,sothegroundstateofaquantummechanicalharmonicoscillatorisagaussiancurve,asyouseeinFigure4-1.Figure4-1:Thegroundstateofaquantummechanicalharmonicoscillator.FindingthewavefunctionofthegroundstateAsagaussiancurve,thegroundstateofaquantumoscillatorisψ0(x)=Aexp(–x2/2x02).HowcanyoufigureoutA?Wavefunctionsmustbenormalized,sothefollowinghastobetrue:Substitutingforψ0(x)givesyouthisnextequation:YoucanevaluatethisintegraltobeTherefore,ThismeansthatthewavefunctionforthegroundstateofaquantummechanicalharmonicoscillatorisCool.Nowyou’vegotanexactwavefunction.Alittleexcitement:FindingthefirstexcitedstateOkay,theprecedingsectionshowsyouwhatψ0(x)lookslike.Whataboutthefirstexcitedstate,ψ1(x)?Well,asyouknow,ψ1(x)=<x|1>and|1>=a†|0>,soψ1(x)=<x|a†|0>Andyouknowthata†isthefollowing:Therefore,ψ1(x)=<x|a†|0>becomesAndbecauseψ0(x)=<x|0>,yougetthefollowingequation:Youalsoknowthefollowing:Therefore,becomesWhat’sψ1(x)looklike?Youcanseeagraphofψ1(x)inFigure4-2,whereithasonenode(transitionthroughthexaxis).Figure4-2:Thefirstexcitedstateofaquantummechanicalharmonicoscillator.FindingthesecondexcitedstateAllright,howaboutfindingψ2(x)andsoon?Youcanfindψ2(x)fromthisequation:Substitutingfora†,theequationbecomesUsinghermitepolynomialstofindanyexcitedstateYoucangeneralizethedifferentialequationforψn(x)likethis:Tosolvethisgeneraldifferentialequation,youmakeuseofthefactthatHn(x)isthenthhermitepolynomial,whichisdefinedthisway:Holymackerel!Whatdothehermitepolynomialslooklike?Here’sH0(x),H1(x),andsoon:H0(x)=1H1(x)=2xH2(x)=4x2–2H3(x)=8x3–12xH4(x)=16x4–48x2+12H5(x)=32x5–160x3+120xWhatdoesthisbuyyou?Youcanexpressthewavefunctionsforquantummechanicalharmonicoscillatorslikethis,usingthehermitepolynomialsHn(x):Andthat’swhatthewavefunctionlookslikeforaquantummechanicalharmonicoscillator.Whoknewitwould’veinvolvedhermitepolynomials?Youcanseewhatψ2(x)lookslikeinFigure4-3;notethattherearetwonodeshere—ingeneral,ψn(x)fortheharmonicoscillatorwillhavennodes.Figure4-3:Thesecondexcitedstateofaquantummechanicalharmonicoscillator.PuttinginsomenumbersTheprecedingsectiongivesyouψn(x),andyou’vealreadysolvedforEn,soyou’reontopofharmonicoscillators.Takealookatanexample.Saythatyouhaveaprotonundergoingharmonicoscillationwithω=4.58×1021sec–1,asshowninFigure4-4.Figure4-4:Aprotonundergoingharmonicoscillation.Whataretheenergiesofthevariousenergylevelsoftheproton?Youknowthatingeneral,Soherearetheenergiesoftheproton,inmegaelectronvolts(MeV):Andsoon.Nowwhataboutthewavefunctions?Thegeneralformofψn(x)iswhereConvertalllengthmeasurementsintofemtometers(1fm=1×10–15m),givingyoux0=3.71fm.Here’sψ0(x),wherexismeasuredinfemtometers:Hereareacouplemorewavefunctions:LookingatHarmonicOscillatorOperatorsasMatricesBecausetheharmonicoscillatorhasregularlyspacedenergylevels,peopleoftenviewitintermsofmatrices,whichcanmakethingssimpler.Forexample,thefollowingmaybethegroundstateeigenvector(notethatit’saninfinitevector):Andthismaybethefirstexcitedstate:Andsoon.TheNoperator,whichjustreturnstheenergylevel,wouldthenlooklikethis:SoN|2>givesyouThisisequaltoInotherwords,N|2>=2|2>.Howaboutthea(lowering)operator?Thatlookslikethis:Inthisrepresentation,whatisa|1>?Ingeneral,a|n>=n1/2|n–1>,soa|1>shouldequal|0>.Takealook:Thismatrixmultiplicationequalsthefollowing:Inotherwords,a|1>=|0>,justasexpected.Sohowaboutthea†(raising)operator?Here’showitworksingeneral:a†|n>=(n+1)1/2|n+1>.Inmatrixterms,a†lookslikethis:Forexample,youexpectthata†|1>= |2>.Doesit?ThematrixmultiplicationisThisequalsthefollowing:Soa†|1>= |2>,asitshould.HowabouttakingalookattheHamiltonian,whichreturnstheenergyofaneigenstate,H|n>=En|n>?Inmatrixform,theHamiltonianlookslikethis:Soifyoupreferthematrixwayoflookingatthings,that’showitworksfortheharmonicoscillator.AJoltofJava:UsingCodetoSolvetheSchrödingerEquationNumericallyHere’stheone-dimensionalSchrödingerequation:Andforharmonicoscillators,youcanwritetheequationlikethis,where:Ingeneral,asthepotentialV(x)getsmoreandmorecomplex,usingacomputertosolvetheSchrödingerequationbeginstolookmoreandmoreattractive.Inthissection,IshowyouhowtodojustthatfortheharmonicoscillatorSchrödingerequation.MakingyourapproximationsIncomputerterms,youcanapproximateψ(x)asacollectionofpoints,ψ1,ψ2,ψ3,ψ4,ψ5,andsoon,asyouseeinFigure4-5.Figure4-5:Dividingψ(x)alongthexaxis.Eachpointalongψ(x)—ψ1,ψ2,ψ3,ψ4,ψ5,andsoon—isseparatedfromitsneighborbyadistance,h0,alongthexaxis.Andbecausedψ/dxistheslopeofψ(x),youcanmaketheapproximationthatInotherwords,theslope,dψ/dx,isapproximatelyequaltoΔy/Δx,whichisequaltoψn+1–ψn(=Δy)dividedbyh0(=Δx).Youcanrearrangetheequationtothis:That’sacrudeapproximationforψn+1,givenψn.So,forexample,ifyouknowψ4,youcanfindtheapproximatevalueofψ5,ifyouknowdψ/dxintheregionofψ4.Youcan,ofcourse,findbetterapproximationsforψn+1.Inparticular,physicistsoftenusetheNumerovalgorithmwhensolvingtheSchrödingerequation,andthatalgorithmgivesyouψn+1intermsofψnandψn–1.Here’swhattheNumerovalgorithmsays:Inthisequation,fortheharmonicoscillator,andtheboundaryconditionsareψ(–∞)=ψ(∞)=0.Wow.Imaginehavingtocalculatethisbyhand.Whynotleaveituptothecomputer?Foraprotonundergoingharmonicoscillationwithω=4.58×1021sec–1,theexactgroundstateenergyisYousolvethisproblemexactlyearlierinthischapter.ThefollowingsectionshaveyoutrytogetthissameresultusingtheNumerovalgorithmandacomputer.BuildingtheactualcodeTocalculatethegroundstateenergyoftheharmonicoscillatorusingtheNumerovalgorithm,thissectionusestheJavaprogramminglanguage,whichyoucangetforfreefromjava.sun.com.Here’showyouusetheprogram:Youchooseatrialvalueoftheenergyforthegroundstate,E0,andthencalculateψ(x)at∞,whichshouldbezero—andifit’snot,youcanadjustyourguessforE0andtryagain.Youkeepgoinguntilψ(∞)=0(orifnotactually0,averysmallnumberincomputerterms)—andwhenitdoes,youknowyou’veguessedthecorrectenergy.Approximatingψ(∞)Howdoyoucalculateψ(∞)?Afterall,infinityisaprettybignumber,andthecomputerisgoingtohavetroublewiththat.Inpracticalterms,youhavetouseanumberthatapproximatesinfinity.Inthiscase,youcanusetheclassicalturningpointsoftheproton—thepointswherealltheproton’senergyispotentialenergyandithasstoppedmovinginpreparationforreversingitsdirection.Attheturningpoints,xt,so(thatis,alltheenergyisinpotentialenergy),Andthisisontheorderof±5femtometers(fm),soyouassumethatψ(x)shouldsurelybezeroat,say,±15fm.Here’stheintervaloverwhichyoucalculateψ(x):xmin=–15fmxmax=15fmDividethis30fmintervalinto200segments,makingthewidthofeachsegment,h0,equalto(xmax–xmin)/200=h0=0.15fm.Okay,you’remakingprogress.You’llstartbyassumingthatψ(xmin)=0,guessavalueofE0,andthencalculateψ(xmax)=ψ200(becausethereare200segments,atx=xmax,ψn=ψ200),whichshouldequalzerowhenyougetE0.Here’swhattheresultstellyou:Correct:Ifabs(ψ200)iszero,orinpracticalterms,lessthan,say,yourmaximumallowedvalueofψmax=1×10–8,thenyou’redone—theE0youguessedwascorrect.Toohigh:Ifabs(ψ200)islargerthanyourmaximumallowedψ,ψmax(=1×10–8),andψ200ispositive,theenergyyouchoseforE0wastoohigh.Youhavetosubtractasmallamountofenergy,ΔE—say1×10–7MeV—fromyourguessfortheenergy;thencalculateabs(ψ200)againandseewhetherit’sstillhigherthanyourmaximumallowedψ,ψmax.Ifso,youhavetorepeattheprocessagain.Toolow:Ifabs(ψ200)islargerthanyourmaximumallowedψ,ψmax(=1×10–8),andψ200isnegative,theenergyyouchoseforE0wastoolow.Youhavetoaddasmallamountofenergy,ΔE,toyourguessfortheenergy;thencalculateabs(ψ200)againandseewhetherit’sstillhigherthanyourmaximumallowedψ,ψmax.Ifso,youhavetorepeattheprocess.Sohowdoyoucalculateψ200?Giventwostartingvalues,ψ0andψ1,usetheNumerovalgorithm:Keepcalculatingsuccessivepointsalongψ(x):ψ2,ψ3,ψ4,andsoon.Thelastpointisψ200.Okay,you’reonourway.You’regoingtostartthecodewiththeassumptionthatψ0=0andψ1isaverysmallnumber(youcanchooseanysmallnumberyoulike).Becauseyouknowthattheexactgroundlevelenergyisactually1.50MeV,startthecodewiththeguessthatE0=1.4900000MeVandletthecomputercalculatetheactualvalueusingincrementsofΔE=1×10–7MeV.Notealsothisequationdependsonkn(x)2,kn–1(x)2,andkn+1(x)2.Here’showyoucanfindthesevalues,whereEcurrentisthecurrentguessforthegroundstateenergy(substituten,n–1,andn+1forj):Andyouknowthatω=4.58×1021sec–1,soTherefore,kj2(xj)=0.05Ecurrent–5.63×10–3xj2,wherexjforaparticularsegmentjisxj=jh0+xmin.WritingthecodeOkay,nowI’mgoingtoputtogetheralltheinfofromtheprecedingsectionintosomeJavacode.YoustartwithaJavaclass,se(forSchrödingerEquation),inafileyounamese.java:publicclassse...}Thenyousetupthevariablesandconstantsyou’llneed,includinganarrayforthevaluesyoucalculateforψ(becausetofindψn+1,you’llhavehadtostorethealready-calculatedvaluesofψnandψn–1):publicclassse{doublepsi[];doubleECurrent;doubleEmin=1.490;doublexMin=-15.;doublexMax=15.;doublehZero;doubleEDelta=0.0000001;doublemaxPsi=0.00000001;intnumberDivisions=200;...}Theseclass’sconstructorgetsrunfirst,soyouinitializevaluesthere,includingψ0(=ψ(xmin)=0)andψ1(anysmallnumberyouwant)togetthecalculationgoing.Inthemainmethod,calledaftertheconstructor,youcreateanobjectoftheseclassandcallitcalculatemethodtogetthingsstarted:publicclassse{doublepsi[];doubleECurrent;doubleEmin=1.490;doublexMin=-15.;doublexMax=15.;doublehZero;doubleEDelta=0.0000001;doublemaxPsi=0.00000001;intnumberDivisions=200;publicse(){ECurrent=Emin;psi=newdouble[numberDivisions+1];psi[0]=0;psi[1]=-0.000000001;psi[numberDivisions]=1.0;hZero=(xMax-xMin)/numberDivisions;}publicstaticvoidmain(String[]argv){sede=newse();de.calculate();}...}Therealworktakesplaceinthecalculatemethod,whereyouusethecurrentguessfortheenergyandcalculateψ200:Ifabs(ψ200)islessthanyourmaximumallowedvalueofψ,ψmax,you’vefoundtheanswer—yourcurrentguessfortheenergyisrighton.Ifabs(ψ200)isgreaterthanψmaxandψ200ispositive,youhavetosubtractΔEfromyourcurrentguessfortheenergyandtryagain.Ifabs(ψ200)isgreaterthanψmaxandψ200isnegative,youhavetoaddΔEtoyourcurrentguessfortheenergyandthentryagain.Here’swhatallthislookslikeincode:publicvoidcalculate(){while(Math.abs(psi[numberDivisions])>maxPsi){for(inti=1;i<numberDivisions;i++){psi[i+1]=calculateNextPsi(i);}if(psi[numberDivisions]>0.0){ECurrent=ECurrentñEDelta;}else{ECurrent=ECurrent+EDelta;}System.out.println(ìPsi200:ì+psi[numberDivisions]+ìE:ì+round(ECurrent));}System.out.println(ì\nThegroundstateenergyisì+round(ECurrent)+ìMeV.î);}Notethatthenextvalueofψ(thatis,ψn+1)iscalculatedwithamethodnamedcalculateNextPsi.Here’swhereyouusetheNumerovalgorithm—givenψn,ψn–1,youcancalculateψn+1:publicdoublecalculateNextPsi(intn){doubleKSqNMinusOne=calculateKSquared(n-1);doubleKSqN=calculateKSquared(n);doubleKSqNPlusOne=calculateKSquared(n+1);doublenextPsi=2.0*(1.0-(5.0*hZero*hZero*KSqN/12.0))*psi[n];nextPsi=nextPsi-(1.0+(hZero*hZero*KSqNMinusOne/12.0))*psi[n-1];nextPsi=nextPsi/(1.0+(hZero*hZero*KSqNPlusOne/12.0));returnnextPsi;}Finally,notethattocalculateψn+1,youneedkn,kn–1,andkn+1,whichyoufindwithamethodnamedcalculateKSquared,whichusesthenumericvaluesyou’vealreadyfiguredoutforthisproblem:publicdoublecalculateKSquared(intn){doublex=(hZero*n)+xMin;return(((0.05)*ECurrent)-((x*x)*5.63e-3));}Whew.Here’sthewholeprogram,se.java:publicclassse{doublepsi[];doubleECurrent;doubleEmin=1.490;doublexMin=-15.;doublexMax=15.;doublehZero;doubleEDelta=0.0000001;doublemaxPsi=0.00000001;intnumberDivisions=200;publicse(){ECurrent=Emin;psi=newdouble[numberDivisions+1];psi[0]=0;psi[1]=-0.000000001;psi[numberDivisions]=1.0;hZero=(xMax-xMin)/numberDivisions;}publicstaticvoidmain(String[]argv){sede=newse();de.calculate();}publicvoidcalculate(){while(Math.abs(psi[numberDivisions])>maxPsi){for(inti=1;i<numberDivisions;i++){psi[i+1]=calculateNextPsi(i);}if(psi[numberDivisions]>0.0){ECurrent=ECurrent-EDelta;}else{ECurrent=ECurrent+EDelta;}System.out.println(ìPsi200:ì+psi[numberDivisions]+ìE:ì+round(ECurrent));}System.out.println(ì\nThegroundstateenergyisì+round(ECurrent)+ìMeV.î);}publicdoublecalculateKSquared(intn){doublex=(hZero*n)+xMin;return(((0.05)*ECurrent)-((x*x)*5.63e-3));}publicdoublecalculateNextPsi(intn){doubleKSqNMinusOne=calculateKSquared(n-1);doubleKSqN=calculateKSquared(n);doubleKSqNPlusOne=calculateKSquared(n+1);doublenextPsi=2.0*(1.0-(5.0*hZero*hZero*KSqN/12.0))*psi[n];nextPsi=nextPsi-(1.0+(hZero*hZero*KSqNMinusOne/12.0))*psi[n-1];nextPsi=nextPsi/(1.0+(hZero*hZero*KSqNPlusOne/12.0));returnnextPsi;}publicdoubleround(doubleval){doubledivider=100000;val=val*divider;doubletemp=Math.round(val);return(double)temp/divider;}}Okay,nowyoucancompilethecodewithjavac,theJavacompiler(ifjavacisn'tinyourcomputer'spath,besuretoaddthecorrectpathtoyourcommand-linecommand,suchasC:>C:\java\bin\javacse.java).C:>javacse.javaThiscreatesse.classfromse.java,andyoucanrunse.classwithJavaitself(addingthecorrectpathifneeded):C:>javaseRunningthecodeWhenyourunthejavacodefortheharmonicoscillatorSchrödingerequation,itdisplaysthesuccessivevaluesofψ200asitadjuststhecurrentguessfortheenergyasitnarrowsinontherightanswer—whichitdisplaysattheendoftherun.Here’swhatyousee:C:>javasePSI200:-1.0503644097337778E-4E:1.49PSI200:-1.050354423295303E-4E:1.49PSI200:-1.0503444368533108E-4E:1.49PSI200:-1.0503344504260495E-4E:1.49...PSI200:-6.12820872814324E-8E:1.50066PSI200:-6.031127521356655E-8E:1.50066PSI200:-5.934046348307554E-8E:1.50066PSI200:-5.836965180600015E-8E:1.50066PSI200:-5.739883979461778E-8E:1.50066PSI200:-5.6428029151212084E-8E:1.50066PSI200:-5.5457218252899224E-8E:1.50066PSI200:-5.4486408066519986E-8E:1.50066PSI200:-5.351559702201636E-8E:1.50066PSI200:-5.254478723976338E-8E:1.50066PSI200:-5.157397714326237E-8E:1.50066PSI200:-5.060316801012202E-8E:1.50066PSI200:-4.963235841725704E-8E:1.50066PSI200:-4.866154915227413E-8E:1.50066PSI200:-4.7690740419271214E-8E:1.50066PSI200:-4.6719932089691944E-8E:1.50066PSI200:-4.574912368974434E-8E:1.50066PSI200:-4.4778315322587505E-8E:1.50066PSI200:-4.380750790476514E-8E:1.50066PSI200:-4.28367005783992E-8E:1.50066PSI200:-4.186589345217578E-8E:1.50066PSI200:-4.0895085873184064E-8E:1.50066PSI200:-3.992427935226201E-8E:1.50066PSI200:-3.8953472673066213E-8E:1.50066PSI200:-3.79826665057731E-8E:1.50066PSI200:-3.701186038502826E-8E:1.50066PSI200:-3.604105453620266E-8E:1.50066PSI200:-3.507024949509914E-8E:1.50066PSI200:-3.4099444217875174E-8E:1.50066PSI200:-3.312863911389194E-8E:1.50066PSI200:-3.2157834719961815E-8E:1.50066PSI200:-3.1187030089902856E-8E:1.50066PSI200:-3.021622619594536E-8E:1.50066PSI200:-2.9245421985136167E-8E:1.50066PSI200:-2.8274618172375295E-8E:1.50066PSI200:-2.7303815344369703E-8E:1.50066PSI200:-2.633301196069577E-8E:1.50066PSI200:-2.5362208888510866E-8E:1.50066PSI200:-2.439140632085814E-8E:1.50066PSI200:-2.342060424823075E-8E:1.50066PSI200:-2.244980221960756E-8E:1.50066PSI200:-2.147900005347249E-8E:1.50067PSI200:-2.0508198285622532E-8E:1.50067PSI200:-1.9537397616823192E-8E:1.50067PSI200:-1.8566596602866105E-8E:1.50067PSI200:-1.7595795286272332E-8E:1.50067PSI200:-1.6624994703779555E-8E:1.50067PSI200:-1.565419461892862E-8E:1.50067PSI200:-1.4683394780836424E-8E:1.50067PSI200:-1.3712594592034165E-8E:1.50067PSI200:-1.2741795159638587E-8E:1.50067PSI200:-1.177099622966848E-8E:1.50067PSI200:-1.0800197142733883E-8E:1.50067PSI200:-9.82939798529632E-9E:1.50067Thegroundstateenergyis1.50067MeV.Andthereyouhaveit—theprogramapproximatesthegroundstateenergyas1.50067MeV,prettydarnclosetothevalueyoucalculatedtheoreticallyintheearliersection“Makingyourapproximations”:1.50MeV.Verycool.PartIIITurningtoAngularMomentumandSpinInthispart...Thingsthatspinandrotate—that’sthetopicofthispart.Quantumphysicshasallkindsofthingstosayabouthowangularmomentumandspinarequantized,andyouseeitallinthispart.Chapter5WorkingwithAngularMomentumontheQuantumLevelInThisChapterAngularmomentumAngularmomentumandtheHamiltonianMatrixrepresentationofangularmomentumEigenfunctionsofangularmomentumInclassicalmechanics,youmaymeasureangularmomentumbyattachingagolfballtoastringandwhirlingitoveryourhead.Inquantummechanics,thinkintermsofasinglemoleculemadeupoftwoboundatomsrotatingaroundeachother.That’sthelevelatwhichquantummechanicaleffectsbecomenoticeable.Andatthatlevel,itturnsoutthatangularmomentumisquantized.Andsincethathastangibleresultsinmanycases,suchasthespectrumofexcitedatoms,it’sanimportanttopic.Besideshavingkineticandpotentialenergy,particlescanalsohaverotationalenergy.Here’swhattheHamiltonian(totalenergy—seeChapter4)lookslike:Here,ListheangularmomentumoperatorandIistherotationmomentofinertia.Whataretheeigenstatesofangularmomentum?IfListheangularmomentumoperator,andlisaneigenvalueofL,thenyoucouldwritethe​‐following:Butthatturnsouttobeincompletebecauseangularmomentumisavectorinthree-dimensionalspace—anditcanbepointinganydirection.Angularmomentumistypicallygivenbyamagnitudeandacomponentinonedirection,whichisusuallytheZdirection.Soinadditiontothemagnitudel,youalsospecifythecomponentofLintheZdirection,Lz(thechoiceofZisarbitrary—youcanjustaseasilyusetheXorYdirection).IfthequantumnumberoftheZcomponentoftheangularmomentumisdesignatedbym,thenthecompleteeigenstateisgivenby|l,m>,sotheequationbecomesthefollowing:That’sthekindofdiscussionabouteigenstatesthatIcoverinthischapter,andIbeginwithadiscussionofangularmomentum.RingingtheOperators:RoundandRoundwithAngularMomentumTakealookatFigure5-1,whichdepictsadiskrotatingin3Dspace.Becauseyou’reworkingin3D,youhavetogowithvectorstorepresentbothmagnitudeanddirection.Figure5-1:ArotatingdiskwithangularmomentumvectorL.Asyoucansee,thedisk’sangularmomentumvector,L,pointsperpendiculartotheplaneofrotation.Here,youcanapplytheright-handrule:Ifyouwrapyourrighthandinthedirectionsomethingisrotating,yourthumbpointsinthedirectionoftheLvector.HavingtheLvectorpointoutoftheplaneofrotationhassomeadvantages.Forexample,ifsomethingisrotatingataconstantangularspeed,theLvectorwillbeconstantinmagnitudeanddirection—whichmakesmoresensethanhavingtheLvectorrotatingintheplaneofthedisk’srotationandconstantlychangingdirection.BecauseLisa3Dvector,itcanpointinanydirection,whichmeansthatithasx,y,andzcomponents,Lx,Ly,andLz(whicharen’tvectors,justmagnitudes).YoucanseeLzinFigure5-1.ListhevectorproductofpositionRandlinearmomentumP,so(L=R×P).YoucanalsowriteLx,Ly,andLzatanygivenmomentintermsofoperatorslikethis,wherePx,Py,andPzarethemomentumoperators(whichreturnthemomentuminthex,y,andzdirections)andX,Y,andZarethepositionoperators(whichreturnthepositioninthex,y,andzdirections):Lx=YPz–ZPyLy=ZPx–XPzLz=XPy–YPxYoucanwritethemomentumoperatorsPx,Py,andPzasInthesamewayyoucanrepresentthepositionoperatorsbytheirequivalentcoordinates,i.e.ThenifwesubstitutetheseoperatorrepresentationsintotheequationsforLx,Ly,andLz,youget,FindingCommutatorsofL ,L ,andLxyzFirstexamineLx,Ly,andLzbytakingalookathowtheycommute;iftheycommute(forexample,if[Lx,Ly]=0),thenyoucanmeasureanytwoofthem(LxandLy,forexample)exactly.Ifnot,thenthey’resubjecttotheuncertaintyrelation,andyoucan’tmeasurethemsimultaneouslyexactly.Okay,sowhat’sthecommutatorofLxandLy?UsingLx=YPz–ZPyandLy=ZPx–XPz,youcanwritethefollowing:[Lx,Ly]=[YPz–ZPy,ZPx–XPz]Youcanwritethisequationas[Lx,Ly]=[YPz,ZPx]–[YPz,XPz]–[ZPy,ZPx]+[ZPy,XPz]==iℏ(XPy–YPx)ButXPy–YPx=Lz,so[Lx,Ly]=iℏLz.SoLxandLydon’tcommute,whichmeansthatyoucan’tmeasurethembothsimultaneouslywithcompleteprecision.Youcanalsoshowthat[Ly,Lz]=iℏLxand[Lz,Lx]=iℏLy.Becausenoneofthecomponentsofangularmomentumcommutewitheachother,youcan’tmeasureanytwosimultaneouslywithcompleteprecision.Rats.ThatalsomeansthattheLx,Ly,andLzoperatorscan’tsharethesameeigenstates.Sowhatcanyoudo?HowcanyoufindanoperatorthatshareseigenstateswiththevariouscomponentsofLsothatyoucanwritetheeigenstatesas|l,m>?Theusualtrickhereisthatthesquareoftheangularmomentum,L2,isascalar,notavector,soit’llcommutewiththeLx,Ly,andLzoperators,noproblem:[L2,Lx]=0[L2,Ly]=0[L2,Lz]=0Okay,cool,you’remakingprogress.BecauseLx,Ly,andLzdon’tcommute,youcan’tcreateaneigenstatethatlistsquantumnumbersforanytwoofthem.ButbecauseL2commuteswiththem,youcanconstructeigenstatesthathaveeigenvaluesforL2andanyoneofLx,Ly,andLz.Byconvention,thedirectionthat’susuallychosenisLz.CreatingtheAngularMomentumEigenstatesNow’sthetimetocreatetheactualeigenstates,|l,m>,ofangularmomentumstatesinquantummechanics.Whenyouhavetheeigenstates,you’llalsohavetheeigenvalues,andwhenyouhavetheeigenvalues,youcansolvetheHamiltonianandgettheallowedenergylevelsofanobjectwithangularmomentum.Don’tmaketheassumptionthattheeigenstatesare|l,m>;rather,saythey’re|α,β>,wheretheeigenvalueofL2isL2|α,β>=ℏ2α|α,β>.SotheeigenvalueofL2isℏ2α,whereyouhaveyettosolveforα.Similarly,theeigenvalueofLzisLz|α,β>=ℏβ|α,β>.Toproceedfurther,youhavetointroduceraisingandloweringoperators(asyoudowiththeharmonicoscillatorinChapter4).Thatway,youcansolveforthegroundstateby,forexample,applyingtheloweringoperatortothegroundstateandsettingtheresultequaltozero—andthensolvingforthegroundstateitself.Inthiscase,theraisingoperatorisL+andtheloweringoperatorisL–.TheseoperatorsraiseandlowertheLzquantumnumber.InawayanalogoustotheraisingandloweringoperatorsinChapter4,youcandefinetheraisingandloweringoperatorsthisway:Raising:L+=Lx+iLyLowering:L–=Lx–iLyThesetwoequationsmeanthatYoucanalsoseethatThatmeansthefollowingareallequaltoL2:L2=L+L–+Lz2–ℏL–L2=L–L++Lz2+ℏLzL2=_1(L+L–+L–L+)+Lz22Youcanalsoseethattheseequationsaretrue:[L2,L±]=0[L+,L–]=2ℏLz[Lz,L±]=±ℏL±Okay,nowyoucanputallthistowork.You’regettingtothegoodstuff.TakealookattheoperationofL+on|α,β>:L+|α,β>=?ToseewhatL+|α,β>is,startbyapplyingtheLzoperatoronitlikethis:LzL+|α,β>=?From[Lz,L±]=±ℏL±,youcanseethatLzL+–L+Lz=ℏL+,soLzL+|α,β>=L+Lz|α,β>+ℏL+|α,β>AndbecauseLz|α,β>=ℏβ|α,β>,youhavethefollowing:LzL+|α,β>=ℏ(β+1)L+|α,β>ThisequationmeansthattheeigenstateL+|α,β>isalsoaneigenstateoftheLzoperator,withaneigenvalueof(β+1).Orinamorecomprehensibleway:L+|α,β>=c|α,β+1>wherecisaconstantyoufindlaterin“FindingtheEigenvaluesoftheRaisingandLoweringOperators.”SotheL+operatorhastheeffectofrasingtheβquantumnumberby1.Similarly,theloweringoperatordoesthis:L–|α,β>=d|α,β–1>NowtakealookatwhatL2L+|α,β>equals:L2L+|α,β>=?BecauseL2isascalar,itcommuteswitheverything.L2L+–L+L2=0,sothisistrue:L2L+|α,β>=L+L2|α,β>AndbecauseL2|α,β>=αℏ2|α,β>,youhavethefollowingequation:L2L+|α,β>=αℏ2L+|α,β>Similarly,theloweringoperator,L–,givesyouthis:L2L–|α,β>=αℏ2L–|α,β>SotheresultsoftheseequationsmeanthattheL±operatorsdon’tchangetheαeigenvalueof|α,β>atall.Okay,sojustwhatareαandβ?Readon.FindingtheAngularMomentumEigenvaluesTheeigenvaluesoftheangularmomentumarethepossiblevaluestheangularmomentumcantake,sothey’reworthfinding.Let’stakealookathowtodojustthat.DerivingeigenstateequationswithβmaxandβminNotethatL2–Lz2=Lx2+Ly2,whichisapositivenumber,soL2–Lz2≥0.ThatmeansthatAndsubstitutinginL2|α,β>=αℏ2|α,β>andLz2|α,β>=βℏ|α,β>,andusingthefactthattheeigenstatesarenormalized,givesyouthis:Therefore,α≥β2.Sothere’samaximumpossiblevalueofβ,whichyoucancallβmax.Youcanbeclevernow,becausetherehastobeastate|α,βmax>suchthatyoucan’traiseβanymore.Thus,ifyouapplytheraisingoperator,yougetzero:L+|α,βmax>=0Applyingtheloweringoperatortothisalsogivesyouzero:L–L+|α,βmax>=0AndbecauseL–L+=L2–Lz2–ℏLz,thatmeansthefollowingistrue:(L2–Lz2–ℏLz)|α,βmax>=0PuttinginL2|α,βmax>=αℏ2andgivesyouthis:(α–βmax2–βmax)ℏ2=0α=βmax(βmax+1)=0Cool,nowyouknowwhatαis.Atthispoint,it’susualtorenameβmaxaslandβasm,so|α,β>becomes|l,m>andL2|l,m>=l(l+1)ℏ2|l,m>Lz|l,m>=mℏ|l,m>Youcansayevenmore.Inadditiontoaβmax,theremustalsobeaβminsuchthatwhenyouapplytheloweringoperator,L–,yougetzero,becauseyoucan’tgoanylowerthanβmin:L–|l,βmin>=0AndyoucanapplyL+onthisaswell:L+L–|l,βmin>=0FromL–L+=L2–Lz2+ℏLz,youknowthat(L2–Lz2+ℏLz)|α,βmin>=0whichgivesyouthefollowing:(α–βmin2+βmin)ℏ2=0α–βmin2+βmin=0α=βmin2–βminα=βmin(βmin–1)Andcomparingthisequationtoα=βmax(βmax+1)=0givesyouβmax=–βminNotethatbecauseyoureach|α,βmin>bynsuccessiveapplicationsofL–on|α,βmax>,yougetthefollowing:βmax=βmin+nCouplingthesetwoequationsgivesyouβmax=n/2Therefore,βmaxcanbeeitheranintegerorhalfaninteger(dependingonwhethernisevenorodd).Becausel=βmax,m=β,andnisapositivenumber,youcanfindthat–l≤m≤l.Sonowyouhaveit:Theeigenstatesare|l,m>.Thequantumnumberofthetotalangularmomentumisl.Thequantumnumberoftheangularmomentumalongthezaxisism.L2|l,m>=ℏ2l(l+1)|l,m>,wherel=0,1/2,1,3/2,...Lz|l,m>=ℏm|l,m>,wherem=–l,–(l–1),...,l–1,l.–l≤m≤l.Foreachl,thereare2l+1valuesofm.Forexample,ifl=2,thenmcanequal–2,–1,0,1,or2.Ifl=5/2,thenmcanequal–5/2,–3/2,–1/2,1/2,3/2,and5/2.YoucanseearepresentativeLandLzinFigure5-2.ListhetotalangularmomentumandLzistheprojectionofthattotalangularmomentumonthezaxis.Figure5-2:LandLz.GettingrotationalenergyofadiatomicmoleculeHere’sanexamplethatinvolvesfindingtherotationalenergyspectrumofadiatomicmolecule.Figure5-3showsthesetup:Arotatingdiatomicmoleculeiscomposedoftwoatomswithmassesm1andm2.Thefirstatomrotatesatr=r1,andthesecondatomrotatesatr=r2.What’sthemolecule’srotationalenergy?Figure5-3:Arotatingdiatomicmolecule.TheHamiltonian(asyoucanseeatthechapterintro)isIistherotationalmomentofinertia,whichisI=m1r12+m2r22=μr2wherer=|r1–r2|andBecauseL=Iω,L=μr2ω.Therefore,theHamiltonianbecomesSoapplyingtheHamiltoniantotheeigenstates,|l,m>,givesyouthefollowing:Andasyouknow,L2|l,m>=l(l+1)ℏ2|l,m>,sothisequationbecomesAndbecauseH|l,m>=E|l,m>,youcanseethatAndthat’stheenergyasafunctionofl,theangularmomentumquantumnumber.FindingtheEigenvaluesoftheRaisingandLoweringOperatorsThissectionlooksatfindingtheeigenvaluesoftheraisingandloweringangularmomentumoperators,whichraiseandlowerastate’szcomponentofangularmomentum.StartbytakingalookatL+,andplantosolveforc:L+|l,m>=c|l,m+1>SoL+|l,m>givesyouanewstate,andmultiplyingthatnewstatebyitstransposeshouldgiveyouc2:(L+|l,m>)†L+|l,m>=c2Toseethisequation,notethat(L+|l,m>)†L+|l,m>=c2<l,m+1|l,m+1>=c2.Ontheotherhand,alsonotethat(L+|l,m>)†L+|l,m>=<l,m|L+L–|l,m>,soyouhave<l,m|L+L–|l,m>=c2WhatdoyoudoaboutL+L–?Well,youseeearlierinthechapter,in“CreatingtheAngularMomentumEigenstates,”thatthisistrue:L+L–=L2–Lz2+ℏLz.Soyourequationbecomesthefollowing:Great!ThatmeansthatcisequaltoSowhatisvalueforc:?ApplyingtheL2andLzoperatorsgivesyouthisc=ℏ[l(l+1)–m(m+1)]1/2Andthat’stheeigenvalueofL+,whichmeansyouhavethisrelation:L+|l,m>=ℏ[l(l+1)–m(m+1)]1/2|l,m+1>Similarly,youcanshowthatL–givesyouthefollowing:L–|l,m>=ℏ[l(l+1)–m(m–1)]1/2|l,m–1>InterpretingAngularMomentumwithMatricesChapter4coversamatrixinterpretationofharmonicoscillatorstatesandoperators,andyoucanhandleangularmomentumthesameway(whichoftenmakesunderstandingwhat’sgoingonwithangularmomentumeasier).Yougettotakealookatthematrixrepresentationofangularmomentumonaquantumlevelnow.Considerasystemwithangularmomentum,withthetotalangularmomentumquantumnumberl=1.Thatmeansthatmcantakethevalues–1,0,and1.Soyoucanrepresentthethreepossibleangularmomentumstateslikethis:Okay,sowhataretheoperatorsyou’veseeninthischapterinmatrixrepresentation?Forexample,whatisL2?YoucanwriteL2thiswayinmatrixform:Okay,<1,1|L2|1,1>=l(l+1)ℏ2=2ℏ2;<1,1|L2|1,0>=0;<1,0|L2|1,0>=2ℏ2;andsoon;Therefore,theprecedingmatrixbecomesthefollowing:AndyoucanalsowritethisasSoinmatrixform,theequationL2|1,1>=2ℏ2|1,1>becomesHowabouttheL+operator?Asyouprobablyknow(fromtheprecedingsection),L+|l,m>=ℏ[l(l+1)–m(m+1)]1/2|l,m+1>.Inthisexample,l=1andm=1,0,and–1.Soyouhavethefollowing:L+|1,1>=0SotheL+operatorlookslikethisinmatrixform:Therefore,L+|1,0>wouldbeAndthisequalsInotherwords,.Okay,whataboutL–?YouknowthatL–|l,m>=ℏ[l(l+1)–m(m–1)]1/2|l,m–1>.Inthisexample,l=1andm=1,0,and–1.Sothatmeansthefollowing:L—|1,–1>=0SotheL–operatorlookslikethisinmatrixform:ThatmeansthatL–|1,1>wouldbeThisequalsWhichtellsyouthatJustasyou’dexpect.Okay,you’vefoundL2,L+,andL–.FindingthematrixrepresentationofLzissimplebecauseℏ|1,1>=Lz|1,1>0=Lz|1,0>–ℏ|1,1>=Lz|1,–1>SoyouhavethatThusLz|1,–1>equalsAndthisequalsSoLz|1,–1>=–ℏ|1,–1>.NowwhataboutfindingtheLxandLyoperators?That’snotashardasyoumaythink,becauseandTakealookatLxfirst.L+equalsAndL–equalsSoLxequals:Okay,nowwhataboutLy?,so:Cool.Thisisgoingprettywell—howaboutcalculating[Lx,Ly]?Todothat,youneedtocalculate[Lx,Ly]=LxLy–LyLx.FirstfindLxLy:ThisequalsAndsimilarly,LyLxequalsAndthisequalsSoAndthisequalsButbecauseYoucanwritethecommutator,[Lx,Ly]likethis:Thisisjusttheoldresultthatweknowandlove,soitallchecksout!RoundingItOut:SwitchingtotheSphericalCoordinateSystemSofar,thischapterhasbeendealingwithangularmomentumintermsofbrasandkets,suchas:Thecharmofbrasandketsisthattheydon’tlimityoutoanyspecificsystemofrepresentation(seeChapter2).Soyouhavethegeneraleigenstates,butwhataretheactualeigenfunctionsofLzandL2?Thatis,you’regoingtotrytofindtheactualfunctionsthatyoucanusewiththeangularmomentumoperatorslikeL2andLz.Tofindtheactualeigenfunctions(notjusttheeigenstates),youturnfromrectangularcoordinates,x,y,andz,tosphericalcoordinatesbecauseit’llmakethemathmuchsimpler(afterall,angularmomentumisaboutthingsgoingaroundincircles).Figure5-4showsthesphericalcoordinatesystem.Figure5-4:Thesphericalcoordinatesystem.Intherectangular(Cartesian)coordinatesystem,youusex,y,andztoorientyourself.Inthesphericalcoordinatesystem,youalsousethreequantities:r,θ,andϕ,asFigure5-4shows.Youcantranslatebetweenthesphericalcoordinatesystemandtherectangularonethisway:Thervectoristhevectortotheparticlethathasangularmomentum,θistheangleofrfromthezaxis,andϕistheangleofrfromthexaxis.x=rsinθcosϕy=rsinθsinϕz=rcosθConsidertheequationsforangularmomentum:Whenyoutaketheangularmomentumequationswiththesphericalcoordinate-systemconversionequations,youcanderivethefollowing:Okay,theseequationslookprettyinvolved.Butthere’sonethingtonotice:Theydependonlyonθandϕ,whichmeanstheireigenstatesdependonlyonθandϕ,notonr.Sotheeigenfunctionsoftheoperatorsintheprecedinglistcanbedenotedlikethis:<θ,ϕ|l,m>Traditionally,yougivethenameYlm(θ,ϕ)totheeigenfunctionsofangularmomentuminsphericalcoordinates,soyouhavethefollowing:Ylm(θ,ϕ)=<θ,ϕ|l,m>Allright,timetoworkonfindingtheactualformofYlm(θ,ϕ).YouknowthatwhenyouusetheL2andLzoperatorsonangularmomentumeigenstates,yougetthis:L2|l,m>=l(l+1)ℏ2|l,m>Lz|l,m>=mℏ|l,m>Sothefollowingmustbetrue:L2Ylm(θ,ϕ)=l(l+1)ℏ2Ylm(θ,ϕ)LzYlm(θ,ϕ)=mℏYlm(θ,ϕ)Infact,youcangofurther.NotethatLzdependsonlyonθ,whichsuggeststhatyoucansplitYlm(θ,ϕ)upintoapartthatdependsonθandapartthatdependsonϕ.SplittingYlm(θ,ϕ)upintopartslookslikethis:Ylm(θ,ϕ)=Θlm(θ)Φm(ϕ)That’swhatmakesworkingwithsphericalcoordinatessohelpful—youcansplittheeigenfunctionsupintotwoparts,onethatdependsonlyonθandonepartthatdependsonlyonϕ.TheeigenfunctionsofLzinsphericalcoordinatesStartbyfindingtheeigenfunctionsofLzinsphericalcoordinates.Insphericalcoordinates,theLzoperatorlookslikethis:SoLzYlm(θ,ϕ)=LzΘlm(θ)Φm(ϕ)iswhichisthefollowing:AndbecauseLzYlm(θ,ϕ)=mℏYlm(θ,ϕ),thisequationcanbewritteninthisversion:Cancellingouttermsfromthetwosidesofthisequationgivesyouthisdifferentialequation:Thislookseasytosolve,andthesolutionisjustΦm(ϕ)=CeimϕwhereCisaconstantofintegration.YoucandetermineCbyinsistingthatΦm(ϕ)benormalized—thatis,thatthefollowingholdtrue:whichgivesyouSoΦm(ϕ)isequaltothis:You’remakingprogress—you’vebeenabletodeterminetheformofΦm(ϕ),soYlm(θ,ϕ)=Θlm(θ)Φm(ϕ),whichequalsThat’sgreat—you’rehalfwaythere,butyoustillhavetodeterminetheformofΘlm(θ),theeigenfunctionofL2.That’scomingupnext.TheeigenfunctionsofL2insphericalcoordinatesNowyou’regoingtotackletheeigenfunctionofL2,Θlm(θ).Youalreadyknowthatinsphericalcoordinates,theL2operatorlookslikethis:That’squiteanoperator.AndyouknowthatSoapplyingtheL2operatortoYlm(θ,ϕ)givesyouthefollowing:AndbecauseL2Ylm(θ,ϕ)=l(l+1)ℏ2Ylm(θ,ϕ)=l(l+1)ℏ2Θlm(θ)Φm(ϕ),thisequationbecomesWow,whathaveyougotteninto?Cancellingtermsandsubtractingtherighthandsidefromtheleftfinallygivesyouthisdifferentialequation:Combiningtermsanddividingbyeimϕgivesyouthefollowing:Holycow!Isn’ttheresomeonewho’striedtosolvethiskindofdifferentialequationbefore?Yes,thereis.ThisequationisaLegendredifferentialequation,andthesolutionsarewell-known.(Whew!)Ingeneral,thesolutionstakethisform:Θlm(θ)=ClmPlm(cosθ)wherePlm(cosθ)istheLegendrefunction.SowhataretheLegendrefunctions?Youcanstartbyseparatingoutthemdependence,whichworksthiswaywiththeLegendrefunctions:wherePl(x)iscalledaLegendrepolynomialandisgivenbytheRodrigues​‐formula:YoucanusethisequationtoderivethefirstfewLegendrepolynomialslikethis:P0(x)=1P1(x)=xP2(x)=1/2(3x2–1)P3(x)=1/2(5x3–3x)P4(x)=1/8(35x4–30x2+3)P5(x)=1/8(63x5–70x3+15x)andsoon.That’swhatthefirstfewPl(x)polynomialslooklike.SowhatdotheassociatedLegendrefunctions,Plm(x)looklike?Youcanalsocalculatethem.YoucanstartoffwithPl0(x),wherem=0.ThoseareeasybecausePl0(x)=Pl(x),soP10(x)=xP20(x)=1/2(3x2–1)P30(x)=1/2(5x3–3x)Also,youcanfindthatP11(x)=(1–x2)1/2P21(x)=3x(1–x2)1/2P22(x)=3(1–x2)P32(x)=15x(1–x2)P33(x)=15x(1–x2)3/2TheseequationsgiveyouanoverviewofwhatthePlmfunctionslooklike,whichmeansyou’realmostdone.Asyoumayrecall,Θlm(θ),theθpartofYlm(θ,ϕ),isrelatedtothePlmfunctionslikethis:Θlm(θ)=ClmPlm(cosθ)AndnowyouknowwhatthePlmfunctionslooklike,butwhatdoClm,theconstants,looklike?Assoonasyouhavethose,you’llhavethecompleteangularmomentumeigenfunctions,Ylm(θ,ϕ),becauseYlm(θ,ϕ)=Θlm(θ)Φm(ϕ).YoucangoaboutcalculatingtheconstantsClmthewayyoualwayscalculatesuchconstantsofintegrationinquantumphysics—younormalizetheeigenfunctionsto1.ForYlm(θ,ϕ)=Θlm(θ)Φm(ϕ),thatlookslikethis:Substitutethefollowingthreequantitiesinthisequation:Ylm(θ,ϕ)=Θlm(θ)Φm(ϕ)Θlm(θ)=ClmPlm(cosθ)Yougetthefollowing:Theintegraloverϕgives2π,sothisbecomesYoucanevaluatetheintegraltothis:Soinotherwords:WhichmeansthatSoYlm(θ,ϕ)=Θlm(θ)Φm(ϕ),whichistheangularmomentumeigenfunctioninsphericalcoordinates,isThefunctionsgivenbythisequationarecalledthenormalizedsphericalharmonics.Herearewhatthefirstfewnormalizedsphericalharmonicslooklike:Infact,youcanusetheserelationstoconvertthesphericalharmonicstorectangularcoordinates:Substitutingtheseequationsintogivesyouthesphericalharmonicsinrectangularcoordinates:Chapter6GettingDizzywithSpinInThisChapterDiscoveringspinwiththeStern-GerlachexperimentLookingateigenstatesandspinnotationUnderstandingfermionsandbosonsComparingthespinoperatorswithangularmomentumoperatorsWorkingwithspin1/2andPaulimatricesPhysicistshavesuggestedthatorbitalangularmomentumisnottheonlykindofangularmomentumpresentinanatom—electronscouldalsohaveintrinsicbuilt-inangularmomentum.Thiskindofbuilt-inangularmomentumiscalledspin.Whetherornotelectronsactuallyspinwillneverbeknown—they’reasclosetopoint-likeparticlesasyoucancome,withoutanyapparentinternalstructure.Yetthefactremainsthattheyhaveintrinsicangularmomentum.Andthat’swhatthischapterisabout—theintrinsic,built-inquantummechanicalspinofsubatomicparticles.TheStern-GerlachExperimentandtheCaseoftheMissingSpotTheStern-Gerlachexperimentunexpectedlyrevealedtheexistenceofspinbackin1922.PhysicistsOttoSternandWaltherGerlachsentabeamofsilveratomsthroughthepolesofamagnet—whosemagneticfieldwasinthezdirection—asyoucanseeinFigure6-1.Figure6-1:TheStern-Gerlachexperiment.Because46ofsilver’s47electronsarearrangedinasymmetricalcloud,theycontributenothingtotheorbitalangularmomentumoftheatom.The47thelectroncanbeinThe5sstate,inwhichcaseitsangularmomentumisl=0andthezcomponentofthatangularmomentumis0The5pstate,inwhichcaseitsangularmomentumisl=1,whichmeansthatthezcomponentofitsangularmomentumcanbe–1,0,or1ThatmeansthatSternandGerlachexpectedtoseeoneorthreespotsonthescreenyouseeatrightinFigure6-1,correspondingtothedifferentstatesofthezcomponentofangularmomentum.Butfamously,theysawonlytwospots.Thispuzzledthephysicscommunityforaboutthreeyears.Then,in1925,physicistsSamuelA.GoudsmitandGeorgeE.Uhlenbecksuggestedthatelectronscontainedintrinsicangularmomentum—andthatintrinsicangularmomentumiswhatgavethemamagneticmomentthatinteractedwiththemagneticfield.Afterall,itwasapparentthatsomeangularmomentumotherthanorbitalangularmomentumwasatworkhere.Andthatbuilt-inangularmomentumcametobecalledspin.Thebeamofsilveratomsdividesintwo,dependingonthespinofthe47thelectronintheatom,sotherearetwopossiblestatesofspin,whichcametobeknownasupanddown.Spinisapurelyquantummechanicaleffect,andthere’snorealclassicalanalog.TheclosestyoucancomeistolikenspintothespinoftheEarthasitgoesaroundthesun—thatis,theEarthhasbothspin(becauseit’srotatingonitsaxis)andorbitalangularmomentum(becauseit’srevolvingaroundthesun).Buteventhispicturedoesn’twhollyexplainspininclassicalterms,becauseit’sconceivablethatyoucouldstoptheEarthfromspinning.Butyoucan’tstopelectronsfrompossessingspin,andthatalsogoesforothersubatomicparticlesthatpossessspin,suchasprotons.Spindoesn’tdependonspatialdegreesoffreedom;evenifyouweretohaveanelectronatrest(whichviolatestheuncertaintyprinciple),itwouldstillpossessspin.GettingDownandDirtywithSpinandEigenstatesSpinthrowsabitofacurveatyou.Whendealingwithorbitalangularmomentum(seeChapter5),youcanbuildangularmomentumoperatorsbecauseorbitalangularmomentumistheproductofmomentumandradius.Butspinisbuiltin;there’snomomentumoperatorinvolved.Sohere’sthecrux:Youcannotdescribespinwithadifferentialoperator,asyoucanfororbitalangularmomentum.InChapter5,Ishowhoworbitalangularmomentumcanbereducedtothesedifferentialoperators:Andyoucanfindeigenfunctionsforangularmomentum,suchasY20:Butbecauseyoucan’texpressspinusingdifferentialoperators,youcan’tfindeigenfunctionsforspinasyoudoforangularmomentum.Sothatmeansthatyou’releftwiththebraandketwayoflookingatthings(brasandketsaren’ttiedtoanyspecificrepresentationinspatialterms).InChapter5,youalsotakealookatthingsinangularmomentumterms,introducingtheeigenstatesoforbitalangularmomentumlikethis:|l,m>(wherelistheangularmomentumquantumnumberandmisthequantumnumberofthezcomponentofangularmomentum).Youcanusethesamenotationforspineigenstates.Aswithorbitalangularmomentum,youcanuseatotalspinquantumnumberandaquantumnumberthatindicatesthespinalongthezaxis(Note:There’snotruezaxisbuiltinwhenitcomestospin—youintroduceazaxiswhenyouapplyamagneticfield;byconvention,thezaxisistakentobeinthedirectionoftheappliedmagneticfield).Thelettersgiventothetotalspinquantumnumberandthez-axiscomponentofthespinaresandm(yousometimesseethemwrittenassandms).Inotherwords,theeigenstatesofspinarewrittenas|s,m>.Sowhatpossiblevaluescansandmtake?That’scomingupnext.HalvesandIntegers:SayingHellotoFermionsandBosonsInanalogywithorbitalangularmomentum,youcanassumethatm(thez-axiscomponentofthespin)cantakethevalues–s,–s+1,...,s–1,ands,wheresisthetotalspinquantumnumber.Forelectrons,SternandGerlachobservedtwospots,soyouhave2s+1=2,whichmeansthats=1/2.Andtherefore,mcanbe+1/2or–1/2.Soherearethepossibleeigenstatesforelectronsintermsofspin:|1/2,1/2>|1/2,–1/2>Sodoallsubatomicparticleshaves=1/2?Nope.Herearetheiroptions:Fermions:Inphysics,particleswithhalf-integerspinarecalledfermions.Theyincludeelectrons,protons,neutrons,andsoon,evenquarks.Forexample,electrons,protons,andneutronshavespins=1/2,anddeltaparticleshaves=3/2.Bosons:Particleswithintegerspinarecalledbosons.Theyincludephotons,pimesons,andsoon;eventhepostulatedparticlesinvolvedwiththeforceofgravity,gravitons,aresupposedtohaveintegerspin.Forexample,pimesonshavespins=0,photonshaves=1,andsoforth.Soforelectrons,thespineigenstatesare|1/2,1/2>and|1/2,–1/2>.Forphotons,theeigenstatesare|1,1>,|1,0>,and|1,–1>.Therefore,thepossibleeigenstatesdependontheparticleyou’reworkingwith.SpinOperators:RunningAroundwithAngularMomentumBecausespinisatypeofbuilt-inangularmomentum,thespinoperatorshavealotincommonwiththeorbitalangularmomentumoperators.InChapter5,IdiscusstheorbitalangularmomentumoperatorsL2andLz,andasyoumayexpect,thereareanalogousspinoperators,S2andSz.However,theseoperatorsarejustoperators;theydon’thaveadifferentialformliketheorbitalangularmomentumoperatorsdo.Infact,alltheorbitalangularmomentumoperators,suchasLx,Ly,andLz,haveanalogshere:Sx,Sy,andSz.ThecommutationrelationsamongLx,Ly,andLzarethefollowing:[Lx,Ly]=iℏLz[Ly,Lz]=iℏLx[Lz,Lx]=iℏLyAndtheyworkthesamewayforspin:[Sx,Sy]=iℏSz[Sy,Sz]=iℏSx[Sz,Sx]=iℏSyTheL2operatorgivesyouthefollowingresultwhenyouapplyittoanorbitalangularmomentumeigenstate:L2|l,m>=l(l+1)ℏ2|l,m>Andjustasyou’dexpect,theS2operatorworksinananalogousfashion:S2|s,m>=s(s+1)ℏ2|s,m>TheLzoperatorgivesyouthisresultwhenyouapplyittoanorbitalangularmomentumeigenstate(seeChapter5):Lz|l,m>=mℏ|l,m>Andbyanalogy,theSzoperatorworksthisway:Sz|s,m>=mℏ|s,m>Whatabouttheraisingandloweringoperators,L+andL–?Arethereanalogsforspin?Inangularmomentumterms,L+andL–worklikethis:Therearespinraisingandloweringoperatorsaswell,S+andS–,andtheyworklikethis:Inthenextsection,Itakeaspeciallookatparticleswithspin1/2.WorkingwithSpin1/2andPauliMatricesSpin1/2particles(fermions)needalittleextraattention.TheeigenvaluesoftheS2operatorhereareAndtheeigenvaluesoftheSzoperatorareYoucanrepresentthesetwoequationsgraphicallyasshowninFigure6-2,wherethetwospinstateshavedifferentprojectionsalongthezaxis.Figure6-2:Spinmagnitudeandzprojection.Spin1/2matricesTimetotakealookatthespineigenstatesandoperatorsforparticlesofspin1/2intermsofmatrices.Thereareonlytwopossiblestates,spinupandspindown,sothisiseasy.First,youcanrepresenttheeigenstate|1/2,1/2>likethis:Andtheeigenstate|1/2,–1/2>lookslikethis:NowwhataboutspinoperatorslikeS2?TheS2operatorlookslikethisinmatrixterms:Andthisworksouttobethefollowing:Similarly,youcanrepresenttheSzoperatorthisway:ThisworksouttoUsingthematrixversionofSz,forexample,youcanfindthezcomponentofthespinof,say,theeigenstate|1/2,–1/2>.Findingthezcomponentlookslikethis:Sz|1/2,–1/2>Puttingthisinmatrixtermsgivesyouthismatrixproduct:Here’swhatyougetbyperformingthematrixmultiplication:Andputtingthisbackintoketnotation,yougetthefollowing:HowabouttheraisingandloweringoperatorsS+andS–?TheS+operatorlookslikethis:Andtheloweringoperatorlookslikethis:So,forexample,youcanfigureoutwhatS+|1/2,–1/2>is.Hereitisinmatrixterms:Performingthemultiplicationgivesyouthis:Orinketform,it’sS+|1/2,–1/2>=ℏ|1/2,1/2>.Cool.PaulimatricesSometimes,youseetheoperatorsSx,Sy,andSzwrittenintermsofPauli​‐matrices,σx,σy,andσz.Here’swhatthePaulimatriceslooklike:NowyoucanwriteSx,Sy,andSzintermsofthePaulimatriceslikethis:Whoo!Andthatconcludesyourlookatspin.PartIVMultipleDimensions:Going3DwithQuantumPhysicsInthispart...Thepreviouspartsdealmostlywithparticlesinone-dimensionalsystems.Thispartexpandsthatcoveragetothreedimensions,likeintherealworld.Youseehowtohandlequantumphysicsinthree-dimensionalcoordinates—whetherrectangularorspherical—whichlaysthegroundworkforworkingwithelectronsinatoms.Chapter7RectangularCoordinates:SolvingProblemsinThreeDimensionsInThisChapterExploringtheSchrödingerequationinthex,y,andzdimensionsWorkingwithfreeparticlesin3DGettingintorectangularpotentialsSeeingharmonicoscillatorsin3DspaceOne-dimensionalproblemsareallverywellandgood,buttherealworldhasthreedimensions.Thischapterisallaboutleavingone-dimensionalpotentialsbehindandstartingtotakealookatspinlessquantummechanicalparticlesinthreedimensions.Here,youworkwiththreedimensionsinrectangularcoordinates,startingwithalookattheSchrödingerequationinglorious,real-life3D.Youthendelveintofreeparticles,boxpotentials,andharmonicoscillators.Note:Bytheway,thenextchapterusessphericalcoordinatesbecausesomeproblemsarebetterinonesystemthantheother.Problemswithsphericalsymmetryarebesthandledinsphericalcoordinates,forexample.TheSchrödingerEquation:Nowin3D!Inonedimension,thetime-dependentSchrödingerequation(ofthetypeinChapters3and4thatletyoufindthewavefunction)lookslikethis:Andyoucangeneralizethatintothreedimensionslikethis:UsingtheLaplacianoperator,youcanrecastthisintoamorecompactform.Here’swhattheLaplacianlookslike:Andhere’sthe3DSchrödingerequationusingtheLaplacian:Tosolvethisequation,whenthepotentialdoesn’tvarywithtime,breakoutthetime-dependentpartofthewavefunction:Here,ψ(x,y,z)isthesolutionofthetime-independentSchrödingerequation,andEistheenergy:Sofar,sogood.Butnowyou’verunintoawall—theexpressionisingeneralveryhardtodealwith,sothecurrentequationisingeneralveryhardtosolve.Sowhatshouldyoudo?Well,youcanfocusonthecaseinwhichtheequationisseparable—thatis,whereyoucanseparateoutthex,y,andzdependenceandfindthesolutionineachdimensionseparately.Inotherwords,inseparablecases,thepotential,V(x,y,z),isactuallythesumofthex,y,andzpotentials:V(x,y,z)=Vx(x)+Vy(y)+Vz(z)NowyoucanbreaktheHamiltonianinintothreeHamilitonians,Hx,Hy,andHz:(Hx+Hy+Hz)ψ(x,y,z)=Eψ(x,y,z)whereWhenyoudivideuptheHamiltonianasin(Hx+Hy+Hz)ψ(x,y,z)=Eψ(x,y,z),youcanalsodivideupthewavefunctionthatsolvesthatequation.Inparticular,youcanbreakthewavefunctionintothreeparts,oneforx,y,andz:ψ(x,y,z)=X(x)Y(y)Z(z)WhereX(x),Y(y),andZ(z)arefunctionsofthecoordinatesx,y,andzandarenottobeconfusedwiththepositionoperators.Thisseparationofthewavefunctionintothreepartsisgoingtomakelifeconsiderablyeasier,becausenowyoucanbreaktheHamiltonianupintothreeseparateoperatorsaddedtogether:E=Ex+Ey+EzSoyounowhavethreeindependentSchrödingerequationsforthethreedimensions:Thissystemofindependentdifferentialequationslooksaloteasiertosolvethan(Hx+Hy+Hz)ψ(x,y,z)=Eψ(x,y,z).Inessence,you’vebrokenthethreedimensionalSchrödingerequationintothreeone-dimensionalSchrödingerequations.Thatmakessolving3Dproblemstractable.SolvingThree-DimensionalFreeParticleProblemsConsiderthefreeparticleyouseeinthreedimensionsinFigure7-1.Figure7-1:Afreeparticlein3D.Becausetheparticleistravelingfreely,V(x)=V(y)=V(z)=0.SothethreeindependentSchrödingerequationsforthethreedimensionscoveredintheprecedingsectionbecomethefollowing:Ifyourewritetheseequationsintermsofthewavenumber,k,where,thentheseequationsbecomethefollowing:Inthissection,youtakealookatthesolutionstotheseequations,findthetotalenergy,andaddtimedependence.Thex,y,andzequationsTakealookatthexequationforthefreeparticle,itsgeneralsolutionas.YoucanwriteX(x)=AxeikxxY(y)=AyeikyyZ(z)=AzeikzzwhereAx,Ay,andAzareconstants.Becauseψ(x,y,z)=X(x)Y(y)Z(z),yougetthisforψ(x,y,z):whereA=AxAyAz.Thepartintheparenthesesintheexponentisthedotproductofthevectorskandr,k·r.Thatis,ifthevectora=(ax,ay,az)intermsofcomponentsandthevectorb=(bx,by,bz),thenthedotproductofaandbisa·b=(axbx,ayby,azbz).Sohere’showyoucanrewritetheψ(x,y,z)equation:FindingthetotalenergyequationThetotalenergyofthefreeparticleisthesumoftheenergyinthreedimensions:E=Ex+Ey+EzWithafreeparticle,theenergyofthexcomponentofthewavefunctionis.Andthisequationworksthesamewayfortheyandzcomponents,sohere’sthetotalenergyoftheparticle:Notethatkx2+ky2+kz2isthesquareofthemagnitudeofk—thatis,k.k=k2Therefore,youcanwritetheequationforthetotalenergyasNotethatbecauseEisaconstant,nomatterwheretheparticleispointed,alltheeigenfunctionsof,,andareinfinitelydegenerateasyouvarykx,ky,andkz.AddingtimedependenceandgettingaphysicalsolutionYoucanaddtimedependencetothesolutionforψ(x,y,z),givingyouψ(x,y,z,t),ifyourememberthat,forafreeparticle,gives.Thatequationyouthisformforψ(x,y,z,t):Because,theequationturnsintoInfact,nowthattherightsideoftheequationisintermsoftheradiusvectorr,youcanmaketheleftsidematch:That’sthesolutiontotheSchrödingerequation,butit’sunphysical(asIdiscussfortheone-dimensionalSchrödingerequationforafreeparticleinChapter3).Why?Tryingtonormalizethisequationinthreedimensions,forexample,givesyouthefollowing,whereAisaconstant:Thus,theintegraldivergesandyoucan’tnormalizeψ(r,t)asI’vewrittenit.Sowhatdoyoudoheretogetaphysicalparticle?ThekeytosolvingthisproblemisrealizingthatifyouhaveanumberofsolutionstotheSchrödingerequation,thenanylinearcombinationofthosesolutionsisalsoasolution.Inotherwords,youaddvariouswavefunctionstogethersothatyougetawavepacket,whichisacollectionofwavefunctionsoftheformeik·rsuchthatThewavefunctionsinterfereconstructivelyatonelocation.Theyinterferedestructively(gotozero)atallotherlocations.Lookatthetime-independentversion:However,forafreeparticle,theenergystatesarenotseparatedintodistinctbands;thepossibleenergiesarecontinuous,sopeoplewritethissummationasanintegral:Sowhatisϕ(k)?It’sthethree-dimensionalanalogofϕ(k)thatyoufindinChapter3;thatis,it’stheamplitudeofeachcomponentwavefunction.Youcanfindϕ(k)fromtheFouriertransformofψ1(x)=Aeik1x+Be–ik1x(wherex<0)likethis:Inpractice,youchooseϕ(k)yourself.Lookatanexample,usingthefollowingformforϕ(k),whichisforaGaussianwavepacket(Note:TheexponentialpartiswhatmakesthisaGaussianwaveform):whereaandAareconstants.Youcanbeginbynormalizingϕ(k)todeterminewhatAis.Here’showthatworks:Okay.PerformingtheintegralgivesyouwhichmeansthatthewavefunctionisYoucanevaluatethisequationtogiveyouthefollowing,whichiswhatthetimeindependentwavefunctionforaGaussianwavepacketlookslikein3D:Okay,that’showthingslookwhenV(r)=0.Butcan’tyousolvesomeproblemswhenV(r)isnotequaltozero?Yep,yousurecan.Checkoutthenextsection.GettingSquaredAwaywith3DRectangularPotentialsThissectiontakesalookata3Dpotentialthatformsabox,asyouseeinFigure7-2.Youwanttogetthewavefunctionsandtheenergylevelshere.Figure7-2:Aboxpotentialin3D.Insidethebox,saythatV(x,y,z)=0,andoutsidethebox,saythatV(x,y,z)=∞.Soyouhavethefollowing:DividingV(x,y,z)intoVx(x),Vy(y),andVz(z)givesyouOkay,becausethepotentialgoestoinfinityatthewallsofthebox,thewavefunction,ψ(x,y,z),mustgotozeroatthewalls,sothat’syourconstraint.In3D,theSchrödingerequationlookslikethisinthreedimensions:Writingthisoutgivesyouthefollowing:Takethisdimensionbydimension.Becausethepotentialisseparable,youcanwriteψ(x,y,z)asψ(x,y,z)=X(x)Y(y)Z(z).Insidethebox,thepotentialequalszero,sotheSchrödingerequationlookslikethisforx,y,andz:Thenextstepistorewritetheseequationsintermsofthewavenumber,k.Because,youcanwritetheSchrödingerequationsforx,y,andzasthefollowingequations:Startbytakingalookattheequationforx.Nowyouhavesomethingtoworkwith—asecondorderdifferentialequation,.Herearethetwoindependentsolutionstothisequation,whereAandBareyettobedetermined:X1(x)=Asin(kx)X2(x)=Bcos(kx)Sothegeneralsolutionofisthesumofthelasttwoequations:X(x)=Asin(kx)+Bcos(kx)Great.Nowtakealookatdeterminingtheenergylevels.DeterminingtheenergylevelsTobeabletodeterminetheenergylevelsofaparticleinaboxpotential,youneedanexactvalueforX(x)—notjustoneofthetermsoftheconstantsAandB.YouhavetousetheboundaryconditionstofindAandB.Whataretheboundaryconditions?Thewavefunctionmustdisappearattheboundariesofthebox,soX(0)=0X(Lx)=0Sothefactthatψ(0)=0tellsyourightawaythatBmustbe0,becausecos(0)=1.AndthefactthatX(Lx)=0tellsyouthatX(Lx)=Asin(kxLx)=0.Becausethesineis0whenitsargumentisamultipleofπ,thismeansthatAndbecause,itmeansthatThat’stheenergyinthexcomponentofthewavefunction,correspondingtothequantumnumbers1,2,3,andsoon.ThetotalenergyofaparticleofmassminsidetheboxpotentialisE=Ex+Ey+Ez.FollowingforEyandEz:,youhavethisSothetotalenergyoftheparticleisE=Ex+Ey+Ez,whichequalsthis:Andthereyouhavethetotalenergyofaparticleintheboxpotential.NormalizingthewavefunctionNowhowaboutnormalizingthewavefunctionψ(x,y,z)?Inthexdimension,youhavethisforthewaveequation:Sothewavefunctionisasinewave,goingtozeroatx=0andx=Lz.Youcanalsoinsistthatthewavefunctionbenormalized,likethis:Bynormalizingthewavefunction,youcansolvefortheunknownconstantA.SubstitutingforX(x)intheequationgivesyouthefollowing:Therefore,becomes,whichmeansyoucansolveforA:Great,nowyouhavetheconstantA,soyoucangetX(x):Nowgetψ(x,y,z).Youcandividethewavefunctionintothreeparts:ψ(x,y,z)=X(x)Y(y)Z(z)ByanalogywithX(x),youcanfindY(y)andZ(z):Soψ(x,y,z)equalsthefollowing:That’saprettylongwavefunction.Infact,whenyou’redealingwithaboxpotential,theenergylookslikethis:UsingacubicpotentialWhenworkingwithaboxpotential,youcanmakethingssimplerbyassumingthattheboxisactuallyacube.Inotherwords,L=Lx=Ly=Lz.Whentheboxisacube,theequationfortheenergybecomesSo,forexample,theenergyofthegroundstate,wherenx=ny=nz=1,isgivenbythefollowing,whereE111isthegroundstate:Notethatthere’ssomedegeneracyintheenergies;forexample,notethatE211(nx=2,ny=1,nz=1)isE121(nx=1,ny=2,nz=1)isE112(nx=1,ny=1,nz=2)isSoE211=E121=E112,whichmeansthatthefirstexcitedstateisthreefolddegenerate,matchingthethreefoldequivalenceindimensions.Ingeneral,whenyouhavesymmetrybuiltintothephysicallayout(asyoudowhenL=Lx=Ly=Lz),youhavedegeneracy.Thewavefunctionforacubicpotentialisalsoeasiertomanagethanthewavefunctionforageneralboxpotential(wherethesidesaren’tofthesamelength).Here’sthewavefunctionforacubicpotential:So,forexample,here’sthewavefunctionforthegroundstate(nx=1,ny=1,nz=1),ψ111(x,y,z):Andhere’sψ211(x,y,z):Andψ121(x,y,z):Springinginto3DHarmonicOscillatorsInonedimension,thegeneralparticleharmonicoscillator(whichIfirstdescribeinChapter4)lookslikeFigure7-3,wheretheparticleisundertheinfluenceofarestoringforce—hereillustratedasaspring.Figure7-3:Aharmonicoscillator.TherestoringforcehastheformFx=–kxxinonedimension,wherekxistheconstantofproportionalitybetweentheforceontheparticleandthelocationoftheparticle.Thepotentialenergyoftheparticleasafunctionoflocationxiswhere.Thisisalsosometimeswrittenas.Inthissection,youtakealookattheharmonicoscillatorinthreedimensions.Inthreedimensions,thepotentiallookslikethis:Nowthatyouhaveaformforthepotential,youcanstarttalkingintermsofSchrödinger’sequation:Substitutinginforthethree-dimensionpotential,V(x,y,z),givesyouthisequation:Takethisdimensionbydimension.Becauseyoucanseparatethepotentialintothreedimensions,youcanwriteψ(x,y,z)asψ(x,y,z)=X(x)Y(y)Z(z).Therefore,theSchrödingerequationlookslikethisforx:YousolvethatequationinChapter4,whereyougetthisnextsolution:whereandnx=0,1,2,andsoon.TheHnxtermindicatesahermitepolynomial,whichlookslikethis:H0(x)=1H1(x)=2xH2(x)=4x2–2H3(x)=8x3–12xH4(x)=16x4–48x2+12H5(x)=32x5–160x3+120xTherefore,youcanwritethewavefunctionlikethis:That’sarelativelyeasyformforawavefunction,andit’sallmadepossiblebythefactthatyoucanseparatethepotentialintothreedimensions.Whatabouttheenergyoftheharmonicoscillator?Theenergyofaonedimensionalharmonicoscillatoris.Andbyanalogy,theenergyofathree-dimensionalharmonicoscillatorisgivenbyNotethatifyouhaveanisotropicharmonicoscillator,whereωx=ωy=ωz=ω,theenergylookslikethis:Asforthecubicpotential,theenergyofa3Disotropicharmonicoscillatorisdegenerate.Forexample,E112=E121=E211.Infact,it’spossibletohavemorethanthreefolddegeneracyfora3Disotropicharmonicoscillator—forexample,E200=E020=E002=E110=E101=E011.Ingeneral,thedegeneracyofa3Disotropicharmonicoscillatoriswheren=nx+ny+nz.Chapter8SolvingProblemsinThreeDimensions:SphericalCoordinatesInThisChapterProblemsinsphericalcoordinatesFreeparticlesinsphericalcoordinatesSquarewellpotentialsIsotropicharmonicoscillatorsInyourotherlifeasaseacaptain-slash-pilot,you’reprobablyprettyfamiliarwithlatitudeandlongitude—coordinatesthatbasicallynameacoupleofanglesasmeasuredfromthecenteroftheEarth.Puttogethertheangleeastorwest,theanglenorthorsouth,andtheall-importantdistancefromthecenteroftheEarth,andyouhaveavectorthatgivesagooddescriptionoflocationinthreedimensions.Thatvectorispartofasphericalcoordinatesystem.Navigatorstalkmoreaboutthepairofanglesthanthedistance(“Earth’ssurface”isgenerallyspecificenoughforthem),butquantumphysicistsfindbothanglesandradiuslengthimportant.Some3Dquantumphysicsproblemsevenallowyoutobreakdownawavefunctionintotwoparts:anangularpartandaradialpart.Inthischapter,Idiscussthree-dimensionalproblemsthatarebesthandledusingsphericalcoordinates.(For3Dproblemsthatworkbetterinrectangularcoordinatesystems,seeChapter7.)ANewAngle:ChoosingSphericalCoordinatesInsteadofRectangularSayyouhavea3Dboxpotential,andsupposethatthepotentialwellthattheparticleistrappedinlookslikethis,whichissuitedtoworkingwithrectangularcoordinates:Becauseyoucaneasilybreakthispotentialdowninthex,y,andzdirections,youcanbreakthewavefunctiondownthatway,too,asyouseehere:ψ(x,y,z)=X(x)Y(y)Z(z)Solvingforthewavefunctiongivesyouthefollowingnormalizedresultinrectangularcoordinates:Theenergylevelsalsobreakdownintoseparatecontributionsfromallthreerectangularaxes:E=Ex+Ey+EzAndsolvingforEgivesyouthisequation(fromChapter7):Butwhatifthepotentialwellaparticleistrappedinhassphericalsymmetry,notrectangular?Forexample,whatifthepotentialwellweretolooklikethis,whereristheradiusoftheparticle’slocationwithrespecttotheoriginandwhereaisaconstant?Clearly,tryingtostuffthiskindofproblemintoarectangular-coordinateskindofsolutionisonlyaskingfortrouble,becausealthoughyoucandoit,itinvolveslotsofsinesandcosinesandresultsinaprettycomplexsolution.Amuchbettertacticistosolvethiskindofaprobleminthenaturalcoordinatesysteminwhichthepotentialisexpressed:sphericalcoordinates.Figure8-1showsthesphericalcoordinatesystemalongwiththecorrespondingrectangularcoordinates,x,y,andz.Inthesphericalcoordinatesystem,youlocatepointswitharadiusvectornamedr,whichhasthreecomponents:Anrcomponent(thelengthoftheradiusvector)θ(theanglefromzaxistothethervector)ϕ(theanglefromthexaxistothethervector)Figure8-1:Thesphericalcoordinatesystem.TakingaGoodLookatCentralPotentialsin3DThischapterfocusesonproblemsthatinvolvecentralpotentials—thatis,sphericallysymmetricalpotentials,ofthekindwhereV(r)=V(r).Inotherwords,thepotentialisindependentofthevectornatureoftheradiusvector;thepotentialdependsononlythemagnitudeofvectorr(whichisr),notontheangleofr.Whenyouworkonproblemsthathaveacentralpotential,you’reabletoseparatethewavefunctionintoaradialpart(whichdependsontheformofthepotential)andanangularpart,whichisasphericalharmonic.Readon.BreakingdowntheSchrödingerequationTheSchrödingerequationlookslikethisinthreedimensions,whereΔistheLaplacianoperator(seeChapter2formoreonoperators):AndtheLaplacianoperatorlookslikethisinrectangularcoordinates:Insphericalcoordinates,it’salittlemessy,butyoucansimplifylater.CheckoutthesphericalLaplacianoperator:Here,L2isthesquareoftheorbitalangularmomentum:Soinsphericalcoordinates,theSchrödingerequationforacentralpotentiallookslikethiswhenyousubstituteintheterms:Takealookattheprecedingequation.Thefirsttermactuallycorrespondstotheradialkineticenergy—thatis,thekineticenergyoftheparticlemovingintheradialdirection.Thesecondtermcorrespondstotherotationalkineticenergy.Andthethirdtermcorrespondstothepotentialenergy.SowhatcanyousayaboutthesolutionstothisversionoftheSchrödingerequation?Youcannotethatthefirsttermdependsonlyonr,asdoesthethird,andthatthesecondtermdependsonlyonangles.Soyoucanbreakthewavefunction,ψ(r)=ψ(r,θ,ϕ),intotwoparts:AradialpartApartthatdependsontheanglesThisisaspecialpropertyofproblemswithcentralpotentials.Theangularpartofψ(r,θ,ϕ)Whenyouhaveacentralpotential,whatcanyousayabouttheangularpartofψ(r,θ,ϕ)?TheangularpartmustbeaneigenfunctionofL2,andasIshowinChapter5,theeigenfunctionsofL2arethesphericalharmonics,Ylm(θ,ϕ)(wherelisthetotalangularmomentumquantumnumberandmisthezcomponentoftheangularmomentum’squantumnumber).ThesphericalharmonicsequalHerearethefirstseveralnormalizedsphericalharmonics:That’swhattheangularpartofthewavefunctionisgoingtobe:asphericalharmonic.Theradialpartofψ(r,θ,ϕ)YoucangivetheradialpartofthewavefunctionthenameRnl(r),wherenisaquantumnumbercorrespondingtothequantumstateoftheradialpartofthewavefunctionandlisthetotalangularmomentumquantumnumber.Theradialpartissymmetricwithrespecttoangles,soitcan’tdependonm,thequantumnumberofthezcomponentoftheangularmomentum.Inotherwords,thewavefunctionforparticlesincentralpotentialslookslikethefollowingequationinsphericalcoordinates:ψ(r,θ,ϕ)=Rnl(r)Ylm(θ,ϕ)ThenextstepistosolveforRnl(r)ingeneral.Substitutingψ(r,θ,ϕ)fromtheprecedingequationintotheSchrödingerequation,,givesyouOkay,whatcanyoumakeofthis?First,note(fromChapter5)thatthesphericalharmonicsareeigenfunctionsofL2(that’sthewholereasonforusingthem),witheigenvaluel(l+1)ℏ2:Sothelastterminthisequationissimplyl(l+1)ℏ2.Thatmeansthattakestheform,whichequalsTheprecedingequationistheoneyouusetodeterminetheradialpartofthewavefunction,Rnl(r).It’scalledtheradialequationforacentralpotential.WhenyousolvetheradialequationforRnl(r),youcanthenfindψ(r,θ,ϕ)becauseyoualreadyknowYlm(θ,ϕ):ψ(r,θ,ϕ)=Rnl(r)Ylm(θ,ϕ)Thus,thischaptersimplybreaksdowntofindingthesolutiontotheradialequation.Note:Incidentally,theradialequationisreallyadifferentialequationinonedimension:therdimension.Byselectingonlyproblemsthatcontaincentralpotentials,youreducethegeneralproblemoffindingthewavefunctionofparticlestrappedinathree-dimensionalsphericalpotentialtoaonedimensionaldifferentialequation.HandlingFreeParticlesin3DwithSphericalCoordinatesInthissectionandthenext,youtakealookatsomeexamplecentralpotentialstoseehowtosolvetheradialequation(seetheprecedingsectionformoreontheradialpart).Here,youworkwithafreeparticle,inwhichnopotentialatallconstrainstheparticle.Thewavefunctioninsphericalcoordinatestakesthisform:AndyouknowallaboutYlm(θ,ϕ),becauseitgivesyouthesphericalharmonics.Theproblemisnowtosolvefortheradialpart,Rnl(r).Here’stheradialequation:Forafreeparticle,V(r)=0,sotheradialequationbecomesThewayyouusuallyhandlethisequationistosubstituteρforkr,wherek=(2mE)½/ℏ,andbecausewehaveaversionofthesameequationforeachnindexitisconvenienttosimplyremoveit,sothatRnl(r)becomesRl(kr)=Rl(ρ).Thissubstitutionmeansthatbecomesthefollowing:Inthissection,youseehowthesphericalBesselandNeumannfunctionscometotherescuewhenyou’redealingwithfreeparticles.ThesphericalBesselandNeumannfunctionsTheradialpartoftheequation,,lookstough,butthesolutionsturnouttobewell-known—thisequationiscalledthesphericalBesselequation,andthesolutionisacombinationofthesphericalBesselfunctions[jl(ρ)]andthesphericalNeumannfunctions[nl(ρ)]:Rl(ρ)=Aljl(ρ)+Blnl(ρ)whereAlandBlareconstants.SowhatarethesphericalBesselfunctionsandthesphericalNeumannfunctions?ThesphericalBesselfunctionsaregivenbyHere’swhatthefirstfewiterationsofjl(ρ)looklike:HowaboutthesphericalNeumannfunctions?ThesphericalNeumannfunctionsaregivenbyHerearethefirstfewiterationsofnl(ρ):ThelimitsforsmallandlargeρAccordingtothesphericalBesselequation,theradialpartofthewavefunctionforafreeparticlelookslikethis:Rl(ρ)=Aljl(ρ)+Blnl(ρ)TakealookatthesphericalBesselfunctionsandNeumannfunctionsforsmallandlargeρ:Smallρ:TheBesselfunctionsreducetoTheNeumannfunctionsreduceto..Largeρ:TheBesselfunctionsreducetoTheNeumannfunctionsreduceto..NotethattheNeumannfunctionsdivergeforsmallρ.Therefore,anywavefunctionthatincludestheNeumannfunctionsalsodiverges,whichisunphysical.SotheNeumannfunctionsaren’tacceptablefunctionsinthewavefunction.Thatmeansthewavefunctionψ(r,θ,ϕ),whichequalsRnl(r)Ylm(θ,ϕ),equalsthefollowing:ψ(r,θ,ϕ)=Aljl(kr)Ylm(θ,ϕ)wherek=(2mEn)1/2/ℏ.Notethatbecausekcantakeanyvalue,theenergylevelsarecontinuous.HandlingtheSphericalSquareWellPotentialTakealookatasphericalsquarewellpotentialofthekindyoucanseeinFigure8-2(IintroducesquarewellsinChapter3).ThispotentialtrapsparticlesinsideitwhenE<0andscattersparticleswhenE>0.Mathematically,youcanexpressthesquarewellpotentiallikethis:Figure8-2:Thesphericalsquarewellpotential.Notethatthispotentialissphericallysymmetricandvariesonlyinr,notinθorϕ.You’redealingwithacentralpotential,soyoucanbreakthewavefunctionintoanangularpartandaradialpart(seetheearliersection“TakingaGoodLookatCentralPotentialsin3D”).Thissectionhasyoutakealookattheradialequation,handlingthetwocasesof0<r<aandr>aseparately.Insidethesquarewell:0<r<aForasphericalsquarewellpotential,here’swhattheradialequationlookslikefortheregion0<r<a:Inthisregion,V(r)=–V0,soyouhaveTakingtheV0termovertotherightgivesyouthefollowing:Andhere’swhatdividingbyrgivesyou:Then,multiplyingby–2m/ℏ2,yougetNowmakethechangeofvariableρ=kr,wherek=(2m(E+V0))1/2/ℏ,sothatRnl(r)becomesRl(kr)=Rl(ρ).Usingthissubstitutionmeansthattakesthefollowingform:ThisisthesphericalBesselequation(justasyouseeforthefreeparticlein“HandlingFreeParticlesin3DwithSphericalCoordinates”).Thistime,k=[2m(E+V0)]1/2/ℏ,not(2mE)1/2/ℏ.Thatmakessense,becausenowtheparticleistrappedinthesquarewell,soitstotalenergyisE+V0,notjustE.ThesolutiontotheprecedingequationisacombinationofthesphericalBesselfunctions[jl(ρ)]andthesphericalNeumannfunctions[nl(ρ)]:Rl(ρ)=Aljl(ρ)+Blnl(ρ)Youcanapplythesameconstraintherethatyouapplyforafreeparticle:Thewavefunctionmustbefiniteeverywhere.Forsmallρ,theBesselfunctionslooklikethis:Andforsmallρ,theNeumannfunctionsreducetoSotheNeumannfunctionsdivergeforsmallρ,whichmakesthemunacceptableforwavefunctionshere.ThatmeansthattheradialpartofthewavefunctionisjustmadeupofsphericalBesselfunctions,whereAlisaconstant:Rl(ρ)=Aljl(ρ)Thewholewavefunctioninsidethesquarewell,ψinside(r,θ,ϕ),isaproductofradialandangularparts,anditlookslikethis:whereρinside=r(2m(E+V0))1/2/ℏandYlm(θ,ϕ)arethesphericalharmonics.Outsidethesquarewell:r>aOutsidethesquarewell,intheregionr>a,theparticleisjustlikeafreeparticle,sohere’swhattheradialequationlookslike:Yousolvethisequationearlierin“HandlingFreeParticlesin3DwithSphericalCoordinates”:Becauseρ=kr,wherek=(2mE)1/2/ℏ,yousubstituteρforkrsothatRnl(r)becomesRl(kr)=Rl(ρ).Usingthissubstitutionmeansthattheradialequationtakesthefollowingform:ThesolutionisacombinationofsphericalBesselfunctionsandsphericalNeumannfunctions,whereBlandClareconstants:IftheenergyE<0,wemusthaveCl=iBl"sothatthewavefunctiondecaysexponentiallyatlargedistancesr.Sotheradialsolutionoutsidethesquarewelllookslikethis,whereρoutside=r(2mE)1/2/ℏ:Fromtheprecedingsection,youknowthatthewavefunctioninsidethesquarewellisSohowdoyoufindtheconstantsAl,Bl,andCl?Youfindthoseconstantsthroughcontinuityconstraints:Attheinside/outsideboundary,wherer=a,thewavefunctionanditsfirstderivativemustbecontinuous.Sotodeterminetheconstantsyouhavetosolvethesetwoequations:GettingtheGoodsonIsotropicHarmonicOscillatorsThissectiontakesalookatsphericallysymmetricharmonicoscillatorsinthreedimensions.Inonedimension,youwritetheharmonicoscillatorpotentiallikethis:where(here,kisthespringconstant;thatis,therestoringforceoftheharmonicoscillatorisF=–kx).Youcanturnthesetwoequationsintothreedimensionalversionsoftheharmonicpotentialbyreplacingxwithr:where.Becausethispotentialissphericallysymmetric,thewavefunctionisgoingtobeofthefollowingform:whereyouhaveyettosolvefortheradialfunctionRnl(r)andwhereYlm(θ,ϕ)describesthesphericalharmonics.Asyouknow,theradialSchrödingerequationlookslikethis:SubstitutingforV(r)fromgivesyouthefollowing:Well,thesolutiontothisequationisprettydifficulttoobtain,andyou’renotgoingtogainanythingbygoingthroughthemath(pagesandpagesofit),sohere’sthesolution:whereexp(x)=exandAndtheLab(r)functionsarethegeneralizedLaguerrepolynomials:Wow.Aren’tyougladyoudidn’tslogthroughthemath?HerearethefirstfewgeneralizedLaguerrepolynomials:Allright,youhavetheformforRnl(r).Tofindthecompletewavefunction,ψnlm(r,θ,ϕ),youmultiplybythesphericalharmonics,Ylm(θ,ϕ):Nowtakealookatthefirstfewwavefunctionsfortheisotropicharmonicoscillatorinsphericalcoordinates:Asyoucansee,whenyouhaveapotentialthatdependsonr2,aswithharmonicoscillators,thewavefunctiongetsprettycomplexprettyfast.Theenergyofanisotropic3Dharmonicoscillatorisquantized,andyoucanderivethefollowingrelationfortheenergylevels:Sotheenergylevelsstartat3ℏω/2andthengoto5ℏω/2,7ℏω/2,andsoon.Chapter9UnderstandingHydrogenAtomsInThisChapterTheSchrödingerequationforhydrogenTheradialwavefunctionsEnergydegeneracyLocationoftheelectronNotonlyishydrogenthemostcommonelementintheuniverse,butit’salsothesimplest.Andonethingquantumphysicsisgoodatispredictingeverythingaboutsimpleatoms.ThischapterisallaboutthehydrogenatomandsolvingtheSchrödingerequationtofindtheenergylevelsofthehydrogenatom.Forsuchasmalllittleguy,thehydrogenatomcanwhipupalotofmath—andIsolvethatmathinthischapter.UsingtheSchrödingerequationtellsyoujustaboutallyouneedtoknowaboutthehydrogenatom,andit’sallbasedonasingleassumption:thatthewavefunctionmustgotozeroasrgoestoinfinity,whichiswhatmakessolvingtheSchrödingerequationpossible.IstartbyintroducingtheSchrödingerequationforthehydrogenatomandtakeyouthroughcalculatingenergydegeneracyandfiguringouthowfartheelectronisfromtheproton.ComingtoTerms:TheSchrödingerEquationfortheHydrogenAtomHydrogenatomsarecomposedofasingleproton,aroundwhichrotatesasingleelectron.YoucanseehowthatlooksinFigure9-1.Notethattheprotonisn’tattheexactcenteroftheatom—thecenterofmassisattheexactcenter.Infact,theprotonisataradiusofrpfromtheexactcenter,andtheelectronisataradiusofre.Figure9-1:Thehydrogenatom.SowhatdoestheSchrödingerequation,whichwillgiveyouthewaveequationsyouneed,looklike?Well,itincludestermsforthekineticandpotentialenergyoftheprotonandtheelectron.Here’sthetermfortheproton’skineticenergy:where.Here,xpistheproton’sxposition,ypistheproton’syposition,andzpisitszposition.TheSchrödingerequationalsoincludesatermfortheelectron’skineticenergy:where.Here,xeistheelectron’sxposition,yeistheelectron’syposition,andzeisitszposition.Besidesthekineticenergy,youhavetoincludethepotentialenergy,V(r),intheSchrödingerequation,whichmakesthetime-independentSchrödingerequationlooklikethis:whereψ(re,rp)istheelectronandproton’swavefunction.Theelectrostaticpotentialenergy,V(r),foracentralpotentialisgivenbythefollowingformula,whereristheradiusvectorseparatingthetwocharges:Asiscommoninquantummechanics,youuseCGS(centimeter-gramsecond)systemofunits,where.SothepotentialduetotheelectronandprotonchargesinthehydrogenatomisNotethatr=re–rp,sotheprecedingequationbecomeswhichgivesyouthisSchrödingerequation:Okay,sohowdoyouhandlethisequation?Findoutinthenextsection.SimplifyingandSplittingtheSchrödingerEquationforHydrogenHere’stheusualquantummechanicalSchrödingerequationforthehydrogenatom:Theproblemisthatyou’retakingintoaccountthedistancetheprotonisfromthecenterofmassoftheatom,sothemathismessy.Ifyouweretoassumethattheprotonisstationaryandthatrp=0,thisequationwouldbreakdowntothefollowing,whichismucheasiertosolve:Unfortunately,thatequationisn’texactbecauseitignoresthemovementoftheproton,soyouseethemore-completeversionoftheequationinquantummechanicstexts.TosimplifytheusualSchrödingerequation,youswitchtocenter-of-masscoordinates.Thecenterofmassoftheproton/electronsystemisatthislocation:Andthevectorbetweentheelectronandprotonisr=re–rpUsingvectorsRandrinsteadofreandrpmakestheSchrödingerequationeasiertosolve.TheLaplacianforRisforris..AndtheLaplacianHowcanyourelate and totheusualequation’s and ?Afterthealgebrasettles,yougetwhereM=me+mpisthetotalmassandiscalledthereducedmass.Whenyouputtogethertheequationsforthecenterofmass,thevectorbetweentheprotonandtheelectron,thetotalmass,andm,thenthetimeindependentSchrödingerequationbecomesthefollowing:Then,giventhevectors,Randr,thepotentialisgivenby,TheSchrödingerequationthenbecomesThislookseasier—themainimprovementbeingthatyounowhave|r|inthedenominatorofthepotentialenergytermratherthan|re–rp|.BecausetheequationcontainstermsinvolvingeitherRorrbutnotboth,theformofthisequationindicatesthatit’saseparabledifferentialequation.Andthatmeansyoucanlookforasolutionofthefollowingform:ψ(R,r)=ψ(R)ψ(r)Substitutingtheprecedingequationintotheonebeforeitgivesyouthefollowing:Anddividingthisequationbyψ(R)ψ(r)givesyouWell,well,well.Thisequationhastermsthatdependoneitherψ(R)orψ(r)butnotboth.Thatmeansyoucanseparatethisequationintotwoequations,likethis(wherethetotalenergy,E,equalsER+Er):MultiplyingAndmultiplyingbyψ(R)givesyoubyψ(r)givesyouNowyouhavetwoSchrödingerequations.Thenexttwosectionsshowyouhowtosolvethemindependently.Solvingforψ(R)In,howdoyousolveforψ(R),whichisthewavefunctionofthecenterofmassoftheelectron/protonsystem?Thisisastraightforwarddifferentialequation,andthesolutionisψ(R)=Ce–ik⋅rHere,Cisaconstantandkisthewavevector,where.Inpractice,ERissosmallthatpeoplealmostalwaysjustignoreψ(R)—thatis,theyassumeittobe1.Inotherwords,therealactionisinψ(r),notinψ(R);ψ(R)isthewavefunctionforthecenterofmassofthehydrogenatom,andψ(r)isthewavefunctionfora(fictitious)particleofmassm.Solvingforψ(r)TheSchrödingerequationforψ(r)isthewavefunctionforamade-upparticleofmassm(inpractice,m≈meandψ(r)isprettyclosetoψ(re),sotheenergy,Er,isprettyclosetotheelectron’senergy).Here’stheSchrödingerequationforψ(r):Youcanbreakthesolution,ψ(r),intoaradialpartandanangularpart(seeChapter8):ψ(r)=Rnl(r)Ylm(θ,ϕ)Theangularpartofψ(r)ismadeupofsphericalharmonics,Ylm(θ,ϕ),sothatpart’sokay.Nowyouhavetosolvefortheradialpart,Rnl(r).Here’swhattheSchrödingerequationbecomesfortheradialpart:wherer=|r|.Tosolvethisequation,youtakealookattwocases—whererisverysmallandwhererisverylarge.Puttingthemtogethergivesyoutheroughformofthesolution.SolvingtheradialSchrödingerequationforsmallrForsmallr,thetermsand,inthepreviousequation,becomemuchsmallerthantherest,soweneglectthemandwritetheradialSchrödingeras,Andmultiplyingby2m/ℏ2,yougetThesolutiontothisequationisproportionaltoRnl(r)~Arl+Br–l–1Note,however,thatRnl(r)mustvanishasrgoestozero—butther–l–1termgoestoinfinity.AndthatmeansthatBmustbezero,soyouhavethissolutionforsmallr:Rnl(r)~rlThattakescareofsmallr.Thenextsectiontakesalookatverylarger.SolvingtheradialSchrödingerequationforlargerForverylarger,becomesBecausetheelectronisinaboundstateinthehydrogenatom,E<0;thus,thesolutiontotheprecedingequationisproportionaltoRnl(r)~Ae–λr+Beλrwhere.NotethatRnl(r)~Ae–λr+BeλrdivergesasrgoestoinfinitybecauseoftheBeλrterm,soBmustbeequaltozero.ThatmeansthatRnl(r)~e–λr.Inthenextsection,youputthesolutionsforsmallrandlargertogether.Yougotthepower:PuttingtogetherthesolutionfortheradialequationPuttingtogetherthesolutionsforsmallrandlarger(seetheprecedingsections),theSchrödingerequationgivesyouasolutiontotheradialSchrödingerequationofRnl(r)=rlf(r)e–λr,wheref(r)issomeas-yet-undeterminedfunctionofr.Yournexttaskistodeterminef(r),whichyoudobysubstitutingthisequationintotheradialSchrödingerequation,givingyouthefollowing:Performingthesubstitutiongivesyouthefollowingdifferentialequation:Quiteadifferentialequation,eh?Butjustsitbackandrelax—yousolveitwithapowerseries,whichisacommonwayofsolvingdifferentialequations.Here’sthepower-seriesformoff(r)touse:SubstitutingtheprecedingequationintotheonebeforeitgivesyouChangingtheindexofthesecondtermfromktok–1givesyouBecauseeachterminthisserieshastobezero,youhaveDividingbyrk–2givesyouThisequationgivestherecurrencerelationoftheinfiniteseries,.Thatis,ifyouhaveonecoefficient,youcangetthenextoneusingthisequation.Whatdoesthatbuyyou?Well,takealookattheratioofak/ak–1:Here’swhatthisratioapproachesaskgoesto∞:Thisresemblestheexpansionforex,whichisAsfore2x,theratioofsuccessivetermsisAndinthelimitk→∞,theratioofsuccessiveexpansioncoefficientsofe2xapproaches:That’sthecasefore2x.Forf(r),youhaveComparingthesetwoequations,it’sapparentthatTheradialwavefunction,Rnl(r),lookslikethis:where.Pluggingtheformyouhaveforf(r),,intogivesyouthefollowing:Rnl(r)=rle2λre–λr=rleλrOkay,shouldyoubeoverjoyed?Well,no.Here’swhatthewavefunctionψ(r)lookslike:ψ(r)=Rnl(r)Ylm(θ,ϕ).AndsubstitutinginyourformofRnl(r)fromthisequationgivesyouψ(r)=rleλrYlm(θ,ϕ)Thatlooksfine—exceptthatitgoestoinfinityasrgoestoinfinity.Youexpectψ(r)togotozeroasrgoestoinfinity,sothisversionofRnl(r)=rleλrisclearlyunphysical.Inotherwords,somethingwentwrongsomewhere.Howcanyoufixthisversionoff(r)?Fixingf(r)tokeepitfiniteYouneedthesolutionfortheradialequationtogotozeroasrgoestoinfinity.Theproblemofhavingψ(r)gotoinfinityasrgoestoinfinityliesintheformyouassumeforf(r)intheprecedingsection,whichisThesolutionistosaythatthisseriesmustterminateatacertainindex,whichyoucallN.Niscalledtheradialquantumnumber.Sothisequationbecomesthefollowing(notethatthesummationisnowtoN,notinfinity):Forthisseriestoterminate,aN+1,aN+2,aN+3,andsoonmustallbezero.TherecurrencerelationforthecoefficientsakisForaN+1tobezero,thefactormultiplyingak–1mustbezerofork=N+1,whichmeansthatSubstitutingink=N+1givesyou.Anddividingby2givesyou.MakingthesubstitutionN+l+1→n,whereniscalledtheprincipalquantumnumber,givesyouThisisthequantizationconditionthatmustbemetiftheseriesforf(r)istobefinite,whichitmustbe,physically:Because,theequationputsconstraintsontheallowablevaluesoftheenergy.FindingtheallowedenergiesofthehydrogenatomThequantizationconditionforψ(r)toremainfiniteasrgoestoinfinityiswhere.Substitutingλintothequantization-conditionequationgivesyouthefollowing:Nowsolvefortheenergy,E.SquaringbothsidesoftheprecedingequationgivesyouSohere’stheenergy,E(Note:BecauseEdependsontheprincipalquantumnumber,I’verenameditEn):PhysicistsoftenwritethisresultintermsoftheBohrradius—theorbitalradiusthatNielsBohrcalculatedfortheelectroninahydrogenatom,r0.TheBohrradiusis.Andintermsofr0,here’swhatEnequals:Thegroundstate,wheren=1,worksouttobeaboutE=–13.6eV.Noticethatthisenergyisnegativebecausetheelectronisinaboundstate—you’dhavetoaddenergytotheelectrontofreeitfromthehydrogenatom.Herearethefirstandsecondexcitedstates:Firstexcitedstate,n=2:E=–3.4eVSecondexcitedstate,n=3:E=–1.5eVOkay,nowyou’veusedthequantizationcondition,whichistodeterminetheenergylevelsofthehydrogenatom.GettingtheformoftheradialsolutionoftheSchrödingerequationInthissection,youcompletethecalculationofthewavefunctions.GotothecalculationofRnl(r)(seetheearliersectiontitled“Yougotthepower:Puttingtogetherthesolutionfortheradialequation”).Sofar,youknowthat,where.Therefore,Infact,thisisn’tquiteenough;theprecedingequationcomesfromsolvingtheradialSchrödingerequation:Thesolutionisonlygoodtoamultiplicativeconstant,soyouaddsuchaconstant,Anl(whichturnsouttodependontheprincipalquantumnumbernandtheangularmomentumquantumnumberl),likethis:YoufindAnlbynormalizingRnl(r).NowtrytosolveforRnl(r)byjustflat-outdoingthemath.Forexample,trytofindR10(r).Inthiscase,n=1andl=0.Then,becauseN+l+1=n,youhaveN=n–l–1.SoN=0here.ThatmakesRnl(r)looklikethis:Andthesummationinthisequationisequalto,soAndbecausel=0,rl=1,soR10(r)=A10e–λra0,wherecanalsowriteR10(r)=A10e–λra0as.Therefore,youwherer0istheBohrradius.TofindA10anda0,younormalizeψ100(r,θ,ϕ)to1,whichmeansintegrating|ψ100(r,θ,ϕ)|2d3roverallspaceandsettingtheresultto1.Nowd3r=r2sinθdrdθdϕ,andintegratingthesphericalharmonics,suchasY00,overacompletesphere,,givesyou1.Therefore,you’releftwiththeradialparttonormalize:PluggingintogivesyouYoucansolvethiskindofintegralwiththefollowingrelation:Withthisrelation,theequationbecomesTherefore,Thisisafairlysimpleresult.BecauseA10isjusttheretonormalizetheresult,youcansetA10to1(thiswouldn’tbethecaseifmultipleterms).Therefore,involved.That’sfine,anditmakesR10(r),whichisYouknowthatψnlm(r,θ,ϕ)=Rnl(r)Ylm(θ,ϕ).Andsoψ100(r,θ,ϕ)becomesWhew.Ingeneral,here’swhatthewavefunctionψnlm(r,θ,ϕ)lookslikeforhydrogen:whereLn–l–12l+1(2r/nr0)isageneralizedLaguerrepolynomial.HerearethefirstfewgeneralizedLaguerrepolynomials:L0b(r)=1L1b(r)=–r+b+1SomehydrogenwavefunctionsSowhatdothehydrogenwavefunctionslooklike?Intheprecedingsection,youfindthatψ100(r,θ,ϕ)lookslikethis:Herearesomeotherhydrogenwavefunctions:Notethatψnlm(r,θ,ϕ)behaveslikerlforsmallrandthereforegoestozero.Andforlarger,ψnlm(r,θ,ϕ)decaysexponentiallytozero.Soyou’vesolvedtheproblemyouhadearlierofthewavefunctiondivergingasrbecomeslarge—andallbecauseofthequantizationcondition,whichcuttheexpressionforf(r)fromanexponenttoapolynomialoflimitedorder.Notbad.YoucanseetheradialwavefunctionR10(r)inFigure9-2.R20(r)appearsinFigure9-3.AndyoucanseeR21(r)inFigure9-4.Figure9-2:TheradialwavefunctionR10(r).Figure9-3:R20(r).Figure9-4:R21(r).CalculatingtheEnergyDegeneracyoftheHydrogenAtomEachquantumstateofthehydrogenatomisspecifiedwiththreequantumnumbers:n(theprincipalquantumnumber),l(theangularmomentumquantumnumberoftheelectron),andm(thezcomponentoftheelectron’sangularmomentum,ψnlm[r,θ,ϕ]).Howmanyofthesestateshavethesameenergy?Inotherwords,what’stheenergydegeneracyofthehydrogenatomintermsofthequantumnumbersn,l,andm?Well,theactualenergyisjustdependentonn,asyouseeearlierinthesectiontitled“Findingtheallowedenergiesofthehydrogenatom”:wheremisthemass,notthequantumnumber.ThatmeanstheEisindependentoflandm.Sohowmanystates,|n,l,m>,havethesameenergyforaparticularvalueofn?Well,foraparticularvalueofn,lcanrangefromzeroton–1.Andeachlcanhavedifferentvaluesofm,sothetotaldegeneracyisThedegeneracyinmisthenumberofstateswithdifferentvaluesofmthathavethesamevalueofl.Foranyparticularvalueofl,youcanhavemvaluesof–l,–l+1,...,0,...,l–1,l.Andthat’s(2l+1)possiblemstatesforaparticularvalueofl.Soyoucanplugin(2l+1)forthedegeneracyinm:Andthisseriesworksouttobejustn2.Sothedegeneracyoftheenergylevelsofthehydrogenatomisn2.Forexample,thegroundstate,n=1,hasdegeneracy=n2=1(whichmakessensebecausel,andthereforem,canonlyequalzeroforthisstate).Forn=2,youhaveadegeneracyof4:ψ200(r,θ,ϕ)ψ21–1(r,θ,ϕ)ψ210(r,θ,ϕ)ψ211(r,θ,ϕ)Cool.Quantumstates:AddingalittlespinYoumaybeaskingyourself—whataboutthespinoftheelectron?Rightyouare!Thespinoftheelectrondoesprovideadditionalquantumstates.Uptonowinthissection,you’vebeentreatingthewavefunctionofthehydrogenatomasaproductofradialandangularparts:ψnlm(r,θ,ϕ)=Rnl(r)Ylm(θ,ϕ)Nowyoucanaddaspinpart,correspondingtothespinoftheelectron,wheresisthespinoftheelectronandmsisthezcomponentofthespin:Thespinpartoftheequationcantakethefollowingvalues:|1/2,1/2>|1/2,–1/2>Hence,ψnlm(r,θ,ϕ)nowbecomesψnlmms(r,θ,ϕ):Andthiswavefunctioncantaketwodifferentforms,dependingonms,likethis:Infact,youcanusethespinnotation(whichyouuseinChapter6),whereForexample,for|1/2,1/2>,youcanwritethewavefunctionasAndfor|1/2,–1/2>,youcanwritethewavefunctionasWhatdoesthisdototheenergydegeneracy?Ifyouincludethespinoftheelectron,therearetwospinstatesforeverystate|n,l,m>,sothedegeneracybecomesSoifyouincludetheelectron’sspin,theenergydegeneracyofthehydrogenatomis2n2.Infact,youcanevenaddthespinoftheprotontothewavefunction(althoughpeopledon’tusuallydothat,becausetheproton’sspininteractsonlyweaklywithmagneticfieldsappliedtothehydrogenatom).Inthatcase,youhaveawavefunctionthatlookslikethefollowing:whereseisthespinoftheelectron,mseisthezcomponentoftheelectron’sspin,spisthespinoftheproton,andmspisthezcomponentoftheproton’sspin.Ifyouincludetheproton’sspin,thewavefunctioncannowtakefourdifferentforms,dependingonms,likethis:Thedegeneracymustnowincludetheproton’sspin,sothat’safactoroffourforeach|n,l,m>:Onthelines:GettingtheorbitalsWhenyoustudyheatedhydrogeninspectroscopy,yougetaspectrumconsistingofvariouslines,namedthes(forsharp),p(forprincipal),d(fordiffuse),andf(forfundamental)lines.Andother,unnamedlinesarepresentaswell—theg,h,andsoon.Thes,p,d,f,andtherestofthelinesturnouttocorrespondtodifferentangularmomentumstatesoftheelectron,calledorbitals.Thesstatecorrespondstol=0;thepstate,tol=1;thedstate,tol=2;thefstate,tol=3;andsoon.Eachoftheseangularmomentumstateshasadifferentlyshapedelectroncloudaroundtheproton—thatis,adifferentorbital.Threequantumnumbers—n,l,andm—determineorbitals.Forexample,theelectroncloudforthe|1,0,0>state(1s,withm=0)appearsinFigure9-5.Figure9-5:The|1,0,0>state.The|3,2,1>state(3d,withm=2)appearsinFigure9-6.Figure9-6:The|3,2,1>state.The|2,1,1>state(2p,withm=1)appearsinFigure9-7.Figure9-7:The|2,1,1>state.HuntingtheElusiveElectronJustwhereistheelectronatanyonetime?Inotherwords,howfaristheelectronfromtheproton?Youcanfindtheexpectationvalueofr,thatis,<r>,totellyou.Ifthewavefunctionisψnlm(r,θ,ϕ),thenthefollowingexpressionrepresentstheprobabilitythattheelectronwillbefoundinthespatialelementd3r:|ψnlm(r,θ,ϕ)|2d3rInsphericalcoordinates,d3r=r2sinθdrdθdϕ.Soyoucanwrite|ψnlm(r,θ,ϕ)|2d3ras|ψnlm(r,θ,ϕ)|2r2sinθdrdθdϕTheprobabilitythattheelectronisinasphericalshellofradiusrtor+dristhereforeAndbecauseψnlm(r,θ,ϕ)=Rnl(r)Ylm(θ,ϕ),thisequationbecomesthefollowing:TheprecedingequationisequaltoorSphericalharmonicsarenormalized,sothisjustbecomes|Rnl(r)|2r2drOkay,that’stheprobabilitythattheelectronisinsidethesphericalshellfromrtor+dr.Sotheexpectationvalueofr,whichis<r>,iswhichisThisiswherethingsgetmorecomplex,becauseRnl(r)involvestheLaguerrepolynomials.Butafteralotofmath,here’swhatyouget:wherer0istheBohrradius:.TheBohrradiusisabout5.29×10−11meters,sotheexpectationvalueoftheelectron’sdistancefromtheprotonis<r>=[3n2–l(l+1)](2.65×10−11)metersSo,forexample,inthe1sstate(|1,0,0>),theexpectationvalueofrisequalto<r>1s=3(2.65×10−11)=7.95×10−11metersAndinthe4pstate(|4,1,m>),<r>4p=46(2.65×10−11)=1.22×10−9metersAndthatconcludesthischapter,whichhasbeenatriumphfortheSchrödingerequation.Chapter10HandlingManyIdenticalParticlesInThisChapterLookingatwavefunctionsandHamiltoniansinmany-particlesystemsWorkingwithidenticalanddistinguishableparticlesIdentifyingandcreatingsymmetricandantisymmetricwavefunctionsExplainingelectronshellsandtheperiodictableHydrogenatoms(seeChapter9)involveonlyaprotonandanelectron,butallotheratomsinvolvemoreelectronsthanthat.Sohowdoyoudealwithmultipleelectronatoms?Forthatmatter,howdoyoudealwithmulti-particlesystems,suchasevenasimplegas?Ingeneral,youcan’tdealwithproblemslikethis—exactly,anyway.Imaginethecomplexityofjusttwoelectronsmovinginaheliumatom—you’dhavetotakeintoaccounttheinteractionoftheelectronsnotonlywiththenucleusoftheatombutalsowitheachother—andthatdependsontheirrelativepositions.SonotonlydoestheHamiltonianhaveatermin1/r1forthepotentialenergyofthefirstelectronand1/r2forthesecondelectron,butitalsohasaterminforthepotentialenergythatcomesfromtheinteractionofthetwoelectrons.Andthatmakesanexactwavefunctionjustaboutimpossibletofind.However,evenwithoutfindingexactwavefunctions,youcanstilldoasurprisingamountwithmulti-particlesystems,suchasderivingthePauliexclusionprinciple—whichsays,amongotherthings,thatnotwoelectronscanbeintheexactsamequantumstate.Infact,you’llprobablybesurprisedathowmuchyoucanactuallysayaboutmulti-particlesystemsusingquantummechanics.Thischapterstartswithanintroductiontomany-particlesystemsandgoesontodiscussidenticalparticles,symmetry(andanti-symmetry),andelectronshells.Many-ParticleSystems,GenerallySpeakingYoucanseeamulti-particlesysteminFigure10-1,whereanumberofparticlesareidentifiedbytheirposition(ignorespinforthemoment).Thissectionexplainshowtodescribethatsysteminquantumphysicsterms.Figure10-1:Amulti-particlesystem.ConsideringwavefunctionsandHamiltoniansBeginbyworkingwiththewavefunction.Thestateofasystemwithmanyparticles,asshowninFigure10-1,isgivenbyψ(r1,r2,r3,...).Andhere’stheprobabilitythatparticle1isind3r1,particle2isind3r2,particle3isind3r3,andsoon:Thenormalizationofψ(r1,r2,r3,...)demandsthatOkay,sowhatabouttheHamiltonian,whichgivesyoutheenergystates?Thatis,whatisH,whereHψ(r1,r2,r3,...)=Eψ(r1,r2,r3,...)?Whenyou’redealingwithasingleparticle,youcanwritethisasButinamanyparticlesystemtheHamiltonianmustrepresentthetotalenergyofallparticles,notjustone.Thetotalenergyofthesystemisthesumoftheenergyofalltheparticles(omittingspinforthemoment),sohere’showyoucangeneralizetheHamiltonianformulti-particlesystems:This,inturn,equalsthefollowing:Here,miisthemassoftheithparticleandVisthemulti-particlepotential.ANobelopportunity:Consideringmulti-electronatomsThissectiontakesalookathowtheHamiltonianwavefunction(seetheprecedingsection)wouldworkforaneutral,multi-electronatom.Amultielectronatom,whichyouseeinFigure10-2,isthemostcommonmulti-particlesystemthatquantumphysicsconsiders.Here,Risthecoordinateofthenucleus(relativetothecenterofmass),r1isthecoordinateofthefirstelectron(relativetothecenterofmass),r2thecoordinateofthesecondelectron,andsoon.Figure10-2:Amulti-electronatom.IfyouhaveZelectrons,thewavefunctionlookslikeψ(r1,r2,...,rZ,R).Andthekineticenergyoftheelectronsandthenucleuslookslikethis:Andthepotentialenergyofthesystemlookslikethis:Soaddingthetwoprecedingequations,here’swhatyougetforthetotalenergy(E=KE+PE)ofamulti-particleatom:Okay,nowthatlookslikeapropermess.WanttowintheNobelprizeinphysics?Justcomeupwiththegeneralsolutiontotheprecedingequation.Asisalwaysthecasewhenyouhaveamulti-particlesysteminwhichtheparticlesinteractwitheachother,youcan’tsplitthisequationintoasystemofNindependentequations.IncaseswheretheNparticlesofamulti-particlesystemdon’tinteractwitheachother,whereyoucandisconnecttheSchrödingerequationintoasetofNindependentequations,solutionsmaybepossible.ButwhentheparticlesinteractandtheSchrödingerequationdependsonthoseinteractions,youcan’tsolvethatequationforanysignificantnumberofparticles.However,thatdoesn’tmeanallislostbyanymeans.Youcanstillsayplentyaboutequationslikethisoneifyou’reclever—anditallstartswithanexaminationofthesymmetryofthesituation,whichIdiscussnext.ASuper-PowerfulTool:InterchangeSymmetryEventhoughfindinggeneralsolutionsforequationsliketheoneforthetotalenergyofamulti-particleatom(intheprecedingsection)isimpossible,youcanstillseewhathappenswhenyouexchangeparticleswitheachother—andtheresultsareveryrevealing.Thissectioncoverstheideaofinterchangesymmetry.Ordermatters:SwappingparticleswiththeexchangeoperatorYoucandeterminewhathappenstothewavefunctionwhenyouswaptwoparticles.Whetherthewavefunctionissymmetricundersuchoperationsgivesyouinsightintowhethertwoparticlescanoccupythesamequantumstate.Thissectiondiscussesswappingparticlesandlookingatsymmetricandantisymmetricfunctions.TakealookatthegeneralwavefunctionforNparticles:ψ(r1,r2,...,ri,...,rj,...,rN)Note:Inthischapter,Italkaboutsymmetryintermsofthelocationcoordinate,r,tokeepthingssimple,butyoucanalsoconsiderotherquantities,suchasspin,velocity,andsoon.Thatwouldn’tmakethisdiscussionanydifferent,becauseyoucanwrapallofaparticle’squantummeasurements—location,velocity,speed,andsoon—intoasinglequantumstate,whichyoucancallξ.DoingsowouldmakethegeneralwavefunctionforNparticlesintothis:ψ(ξ1,ξ2,...,ξi,...,ξj,...,ξN).ButasIsaid,thissectionjustconsidersthewavefunctionψ(r1,r2,...,ri,...,rj,...,rN)tokeepthingssimple.Nowimaginethatyouhaveanexchangeoperator,Pij,thatexchangesparticlesiandj.Inotherwords,Pijψ(r1,r2,...,ri,...,rj,...,rN)=ψ(r1,r2,...,rj,...,ri,...,rN)AndPij=Pji,soPijψ(r1,r2,...,ri,...,rj,...,rN)=ψ(r1,r2,...,rj,...,ri,...,rN)=Pjiψ(r1,r2,...,ri,...,rj,...,rN)Also,notethatapplyingtheexchangeoperatortwicejustputsthetwoexchangedparticlesbackwheretheywereoriginally,soPij2=1.Here’swhatthatlookslike:PijPijψ(r1,r2,...,ri,...,rj,...,rN)=Pijψ(r1,r2,...,rj,...,ri,...,rN)=ψ(r1,r2,...,ri,...,rj,...,rN)However,ingeneral,PijandPlm(whereij≠lm)donotcommute.Thatis,PijPlm≠PlmPij(ij≠lm).Therefore,[Pij,Plm]≠0(ij≠lm).Forexample,sayyouhavefourparticleswhosewavefunctionisApplytheexchangeoperatorsP12andP14toseewhetherP12P14equalsP14P12.Here’sP14ψ(r1,r2,r3,r4):Andhere’swhatP12P14ψ(r1,r2,r3,r4)lookslike:Okay.NowtakealookatP14P12ψ(r1,r2,r3,r4).Here’sP12ψ(r1,r2,r3,r4):Andhere’swhatP14P12ψ(r1,r2,r3,r4)lookslike:Asyoucanseebycomparingandthislastequation,P12P14ψ(r1,r2,r3,r4)≠P14P12ψ(r1,r2,r3,r4).Inotherwords,theorderinwhichyouapplyexchangeoperatorsmatters.ClassifyingsymmetricandantisymmetricwavefunctionsPij2=1(seetheprecedingsection),sonotethatifawavefunctionisan​‐eigenfunctionofPij,thenthepossibleeigenvaluesare1and–1.Thatis,forψ(r1,r2,...,ri,...,rj,...,rN)aneigenfunctionofPijlookslikePijψ(r1,r2,...,ri,...,rj,...,rN)=ψ(r1,r2,...ri,...,rj,...,rN)or–ψ(r1,r2,...,ri,...,rj,...,rN)Thatmeanstherearetwokindsofeigenfunctionsoftheexchangeoperator:Symmetriceigenfunctions:Pijψs(r1,r2,...,ri,...,rj,...,rN)=ψs(r1,r2,...,ri,...,rj,...,rN)Antisymmetriceigenfunctions:Pijψa(r1,r2,...,ri,...,rj,...,rN)=–ψa(r1,r2,...,ri,...,rj,...,rN)Nowtakealookatsomesymmetricandsomeantisymmetriceigenfunctions.Howaboutthisone—isitsymmetricorantisymmetric?ψ1(r1,r2)=(r1–r2)2YoucanapplytheexchangeoperatorP12:P12ψ1(r1,r2)=(r2–r1)2Notethatbecause(r1–r2)2=(r2–r1)2,ψ1(r1,r2)isasymmetricwavefunction;that’sbecauseP12ψ1(r1,r2)=ψ1(r1,r2).Howaboutthiswavefunction?Again,applytheexchangeoperator,P12:Okay,butbecause,youknowthatP12ψ2(r1,r2)=ψ2(r1,r2),soψ2(r1,r2)issymmetric.Here’sanotherone:NowapplyP12:Howdoesthatequationcomparetotheoriginalone?Well,,soP12ψ3(r1,r2)=–ψ3(r1,r2).Therefore,ψ3(r1,r2)isantisymmetric.Whataboutthisone?Tofindout,applyP12:Allright—how’sthiscomparewiththeoriginalequation?Okay—ψ4(r1,r2)issymmetric.Youmaythinkyouhavethisprocessdownprettywell,butwhataboutthisnextwavefunction?StartbyapplyingP12:Sohowdothesetwoequationscompare?Thatis,ψ5(r1,r2)isneithersymmetricnorantisymmetric.Inotherwords,ψ5(r1,r2)isnotaneigenfunctionoftheP12exchangeoperator.FloatingCars:TacklingSystemsofManyDistinguishableParticlesAllright,ifyou’vebeenreadingthischapterfromthestart,youprettymuchhavetheideaofswappingparticlesdown.Nowyoulookatsystemsofparticlesthatyoucandistinguish—thatis,systemsofidentifiablydifferentparticles.Asyouseeinthissection,youcandecouplesuchsystemsintolinearlyindependentequations.Supposeyouhaveasystemofmanydifferenttypesofcarsfloatingaroundinspace.Youcandistinguishallthosecarsbecausethey’realldifferent—theyhavedifferentmasses,foronething.Nowsaythateachcarinteractswithitsownpotential—thatis,thepotentialthatanyonecarseesdoesn’tdependonanyothercar.Thatmeansthatthepotentialforallcarsisjustthesumoftheindividualpotentialseachcarsees,whichlookslikethis,assumingyouhaveNcars:Beingabletocutthepotentialenergyupintoasumofindependenttermslikethismakeslifealoteasier.Here’swhattheHamiltonianlookslike:NoticehowmuchsimplerthisequationisthantheHamiltonianforthehydrogenatomwhichIgiveyouhere:NotethatyoucanseparatethepreviousequationforthepotentialofallcarsintoNdifferentequations:Andthetotalenergyisjustthesumoftheenergiesoftheindividualcars:Andthewavefunctionisjusttheproductoftheindividualwavefunctions:wheretheΠsymbolisjustlikeΣ,exceptitstandsforaproductofterms,notasum,andnireferstoallthequantumnumbersoftheithparticle.Asyoucansee,whentheparticlesyou’reworkingwitharedistinguishableandsubjecttoindependentpotentials,theproblemofhandlingmanyofthembecomessimpler.YoucanbreakthesystemupintoNindependentone-particlesystems.Thetotalenergyisjustthesumoftheindividualenergiesofeachparticle.TheSchrödingerequationbreaksdownintoNdifferentequations.AndthewavefunctionendsupjustbeingtheproductofthewavefunctionsoftheNdifferentparticles.Takealookatanexample.Sayyouhavefourparticles,eachwithadifferentmass,inasquarewell.Youwanttofindtheenergyandthewavefunctionofthissystem.Here’swhatthepotentialofthesquarewelllookslikethisforeachofthefournoninteractingparticles:Here’swhattheSchrödingerequationlookslike:Youcanseparatetheprecedingequationintofourone-particleequations:I’vealreadysolvedsuchone-dimensionalproblemsinChapter3.TheenergylevelsareAndbecausethetotalenergyisthesumoftheindividualenergiesisenergyingeneralis,theSohere’stheenergyofthegroundstate—whereallparticlesareintheirgroundstates,n1=n2=n3=n4=1:Foraone-dimensionalsystemwithaparticleinasquarewell,thewavefunctionisThewavefunctionforthefour-particlesystemisjusttheproductoftheindividualwavefunctions,soitlookslikethis:Forexample,forthegroundstate,n1=n2=n3=n4=1,youhaveSoasyoucansee,systemsofNindependent,distinguishableparticlesareoftensusceptibletosolution—allyouhavetodoistobreakthemupintoNindependentequations.JugglingManyIdenticalParticlesWhentheparticlesinamulti-particlesystemareallindistinguishable,that’swhentherealadventurebegins.Whenyoucan’ttelltheparticlesapart,howcanyoutellwhichone’swhere?Thissectionexplainswhathappens.LosingidentitySayyouhaveabunchofpoolballsandyouwanttolookatthemclassically.Youcanpainteachpoolballdifferently,andthen,evenastheyhurtlearoundthepooltable,you’reabletodistinguishthem—sevenballinthecornerpocket,andthatsortofthing.Classically,identicalparticlesretaintheirindividuality.Youcanstilltellthemapart.Thesameisn’ttruequantummechanically,becauseidenticalquantumparticlesreallyareidentical—youcan’tpaintthem,asyoucanpoolballs.Forexample,lookatthescenarioinFigure10-3.There,twoelectronsarecollidingandbouncingapart.Seemslikekeepingtrackofthetwoelectronswouldbeeasy.Figure10-3:Anelectroncollidingwithanotherelectron.ButnowlookatthescenarioinFigure10-4—theelectronscould’vebouncedlikethat,notlikethebounceshowninFigure10-3.Andyou’dneverknowit.Figure10-4:Anelectroncollidingwithanotherelectron.Sowhichelectroniswhich?Fromtheexperimenter’spointofview,youcan’ttell.Youcanplacedetectorstocatchtheelectrons,butyoucan’tdeterminewhichoftheincomingelectronsendedupinwhichdetector,becauseofthetwopossiblescenariosinFigures10-3and10-4.Quantummechanically,identicalparticlesdon’tretaintheirindividualityintermsofanymeasurable,observablequantity.Youlosetheindividualityofidenticalparticlesassoonasyoumixthemwithsimilarparticles.ThisideaholdstrueforanyN-particlesystem.AssoonasyouletNidenticalparticlesinteract,youcan’tsaywhichexactoneisatr1orr2orr3orr4andsoon.SymmetryandantisymmetryInpracticalterms,thelossofindividualityamongidenticalparticlesmeansthattheprobabilitydensityremainsunchangedwhenyouexchangeparticles.Forexample,ifyouweretoexchangeelectron10,281withelectron59,830,you’dstillhavethesameprobabilitythatanelectronwouldoccupyd3r10,281andd3r59,830.Here’swhatthisidealookslikemathematically(randsarethelocationandspinsoftheparticles):|ψ(r1s1,r2s2,...,risi,...,rjsj,...,rNsN)|2=|ψ(r1s1,r2s2,...,rjsj,...,risi,...,rNsN)|2Theprecedingequationmeansthatψ(r1s1,r2s2,...,risi,...,rjsj,...,rNsN)=±ψ(r1s1,r2s2,...,rjsj,...,risi,...,rNsN)SothewavefunctionofasystemofNidenticalparticlesmustbeeithersymmetricorantisymmetricwhenyouexchangetwoparticles.Spinturnsouttobethedecidingfactor:Antisymmetricwavefunction:Iftheparticleshavehalf-odd-integralspin(1/2,3/2,andsoon),thenthisishowthewavefunctionlooksunderexchangeofparticles:ψ(r1s1,r2s2,...,risi,...,rjsj,...,rNsN)=–ψ(r1s1,r2s2,...,rjsj,...,risi,...,rNsN)Symmetricwavefunction:Iftheparticleshaveintegralspin(0,1,andsoon),thisishowthewavefunctionlooksunderexchangeofparticles:ψ(r1s1,r2s2,...,risi,...,rjsj,...,rNsN)=ψ(r1s1,r2s2,...,rjsj,...,risi,...,rNsN)Havingsymmetricorantisymmetricwavefunctionsleadstosomedifferentphysicalbehavior,dependingonwhetherthewavefunctionissymmetricorantisymmetric.Inparticular,particleswithintegralspin,suchasphotonsorpimesons,arecalledbosons.Andparticleswithhalf-odd-integralspin,suchaselectrons,protons,andneutrons,arecalledfermions.Thebehaviorofsystemsoffermionsisverydifferentfromthebehaviorofsystemsofbosons.Exchangedegeneracy:ThesteadyHamiltonianTheHamiltonian,whichyoucanrepresentlikethisH(r1s1,r2s2,...,risi​,...,rjsj,...,rNsN)doesn’tvaryunderexchangeoftwoidenticalparticles.Inotherwords,theHamiltonianisinvarianthere,nomatterhowmanyidenticalparticlesyouexchange.That’scalledexchangedegeneracy,andmathematically,itlookslikethis:H(r1s1,r2s2,...,risi,...,rjsj,...,rNsN)=H(r1s1,r2s2,...,rjsj,...,risi,...,rNsN)Thatmeans,incidentally,thattheexchangeoperator,Pij,isaninvariantofthemotionbecauseitcommuteswiththeHamiltonian:[H,Pij]=0Namethatcomposite:GroovingwiththesymmetrizationpostulateIntheearliersectiontitled“Symmetryandantisymmetry,”IshowthatthewavefunctionofasystemofNparticlesiseithersymmetricorantisymmetricundertheexchangeoftwoparticles:Symmetric:ψ(r1s1,r2s2,...,risi,...,rjsj,...,rNsN)=ψ(r1s1,r2s2,...,rjsj,...,risi,...,rNsN)Antisymmetric:ψ(r1s1,r2s2,...,risi,...,rjsj,...,rNsN)=–ψ(r1s1,r2s2,...,rjsj,...,risi,...,rNsN)Thisturnsouttobethebasisofthesymmetrizationpostulate,whichsaysthatinsystemsofNidenticalparticles,onlystatesthataresymmetricorantisymmetricexist—anditsaysthatstatesofmixedsymmetrydon’texist.Thesymmetrizationpostulatealsosays,asobservedfromnature,thatParticleswithhalf-odd-integralstates(1/2,3/2,5/2,...)arefermions,andtheyhaveantisymmetricstatesundertheinterchangeoftwoparticles.Particleswithintegralspin(0,1,2,...)arebosons,andtheyhavesymmetricstatesundertheinterchangeoftwoparticles.SothewavefunctionofNfermionsiscompletelyantisymmetric,andthewavefunctionofNbosonsiscompletelysymmetric.Determiningwhetheraparticleisafermionorabosonmayseemlikeaneasytask—justlookitup.Electrons,protons,andneutronsarefermions,forexample,withhalf-odd-integralspin.Andphotons,pimesons,andsoonarebosons,withintegralspins.Butwhatiftheparticleyou’restudyingisacompositeparticle?Whatif,forexample,youhaveanalphaparticle,whichismadeupoftwoprotonsandtwoneutrons?Isthatafermionoraboson?Infact,protonsandneutronsthemselvesaremadeupofthreequarks,andpimesonsaremadeupoftwo—andquarkshavespin1/2.Compositescanbeeitherfermionsorbosons—italldependsonwhetherthespinofthecompositeparticleendsupbeinghalf-odd-integralorintegral.Ifthecompositeparticle’sspinis1/2,3/2,5/2,andsoon,thenthecompositeparticleisafermion.Ifthecompositeparticle’sspinis0,1,2,andsoon,thenthecompositeparticleisaboson.Ingeneral,ifthecompositeparticleismadeupofanoddnumberoffermions,thenit’safermion.Otherwise,it’saboson.Soforexample,becausequarksarefermionsandbecausenucleonssuchasprotonsandneutronsaremadeupofthreequarks,thosenucleonsendupbeingfermions.Butbecausepimesonsaremadeupoftwoquarks,theyendupbeingbosons.Thealphaparticle,whichconsistsoftwoprotonsandtwoneutrons,isaboson.Youcanevenconsiderwholeatomstobecompositeparticles.Forexample,considerthehydrogenatom:Thatatomismadeupofaproton(afermion)andanelectron(anotherfermion),sothat’stwofermions.Andthatmakesthehydrogenatomaboson.BuildingSymmetricandAntisymmetricWaveFunctionsManyofthewavefunctionsthataresolutionstophysicalsetupslikethesquarewellaren’tinherentlysymmetricorantisymmetric;they’resimplyasymmetric.Inotherwords,theyhavenodefinitesymmetry.Sohowdoyouendupwithsymmetricorantisymmetricwavefunctions?Theansweristhatyouhavetocreatethemyourself,andyoudothatbyaddingtogetherasymmetricwavefunctions.Forexample,saythatyouhaveanasymmetricwavefunctionoftwoparticles,ψ(r1s1,r2s2).Tocreateasymmetricwavefunction,addtogetherψ(r1s1,r2s2)andtheversionwherethetwoparticlesareswapped,ψ(r2s2,r1s1).Assumingthatψ(r1s1,r2s2)andψ(r2s2,r1s1)arenormalized,youcancreateasymmetricwavefunctionusingthesetwowavefunctionsthisway—justbyaddingthewavefunctions:Youcanmakeanantisymmetricwavefunctionbysubtractingthetwowavefunctions:Thisprocessgetsrapidlymorecomplexthemoreparticlesyouadd,however,becauseyouhavetointerchangealltheparticles.Forexample,whatwouldasymmetricwavefunctionbasedontheasymmetricthree-particlewavefunctionψ(r1s1,r2s2,r3s3)looklike?Why,it’dlooklikethis:Andhowabouttheantisymmetricwavefunction?Thatlookslikethis:Andinthisway,atleasttheoretically,youcancreatesymmetricandantisymmetricwavefunctionsforanysystemofNparticles.WorkingwithIdenticalNoninteractingParticlesWorkingwithidenticalnoninteractingparticlesmakeslifeeasierbecauseyoucantreattheequationsindividuallyinsteadofcombiningthemintoonebigmess.SayyouhaveasystemofNidenticalparticles,eachofwhichexperiencesthesamepotential.YoucanseparatetheSchrödingerequationintoNidenticalsingle-particleequations:Andthetotalenergyisjustthesumoftheenergiesoftheindividualparticles:Butnowlookatthewavefunctionforthesystem.Earlierinthechapter(see“FloatingCars:TacklingSystemsofManyDistinguishableParticles”),youconsiderthewavefunctionofasystemofNdistinguishableparticlesandcomeupwiththeproductofalltheindividualwavefunctions:However,thatequationdoesn’tworkwithidenticalparticlesbecauseyoucan’tsaythatparticle1isinstateψ1(r1),particle2isinstateψ2(r2),andsoon—they’reidenticalparticleshere,notdistinguishableparticlesasbefore.Theotherreasonthisequationdoesn’tworkhereisthatithasnoinherentsymmetry—andsystemsofNidenticalparticlesmusthaveadefinitesymmetry.Soinsteadofsimplymultiplyingthewavefunctions,youhavetobealittlemorecareful.Wavefunctionsoftwo-particlesystemsHowdoyoucreatesymmetricandantisymmetricwavefunctionsforatwoparticlesystem?Startwiththesingle-particlewavefunctions(seetheearliersection“BuildingSymmetricandAntisymmetricWaveFunctions”):Byanalogy,here’sthesymmetricwavefunction,thistimemadeupoftwosingle-particlewavefunctions:Andhere’stheantisymmetricwavefunction,madeupofthetwosingle-particlewavefunctions:wherenistandsforallthequantumnumbersoftheithparticle.Noteinparticularthatψa(r1s1,r2s2)=0whenn1=n2;inotherwords,theantisymmetricwavefunctionvanisheswhenthetwoparticleshavethesamesetofquantumnumbers—thatis,whenthey’reinthesamequantumstate.Thatideahasimportantphysicalramifications.Youcanalsowriteψs(r1s1,r2s2)likethis,wherePisthepermutationoperator,whichtakesthepermutationofitsargument:Andalsonotethatyoucanwriteψa(r1s1,r2s2)likethis:wheretheterm(–1)Pis1forevenpermutations(whereyouexchangebothr1s1andr2s2andalson1andn2)and–1foroddpermutations(whereyouexchanger1s1andr2s2butnotn1andn2;oryouexchangen1andn2butnotr1s1andr2s2).Infact,peoplesometimeswriteψa(r1s1,r2s2)indeterminantformlikethis:Notethatthisdeterminantiszeroifn1=n2.Wavefunctionsofthree-or-more-particlesystemsNowyougettoputtogetherthewavefunctionofasystemofthreeparticlesfromsingle-particlewavefunctions.Thesymmetricwavefunctionlookslikethis:Andtheantisymmetricwavefunctionlookslikethis:Thisasymmetricwavefunctiongoestozeroifanytwosingleparticleshavethesamesetofquantumnumbers(ni=nj,i≠j).HowaboutgeneralizingthistosystemsofNparticles?IfyouhaveasystemofNparticles,thesymmetricwavefunctionlookslikethis:Andtheantisymmetricwavefunctionlookslikethis:ThebignewsisthattheantisymmetricwavefunctionforNparticlesgoestozeroifanytwoparticleshavethesamequantumnumbers(ni=nj,i≠j).Andthathasabigeffectinphysics,asyouseenext.It’sNotComeOne,ComeAll:ThePauliExclusionPrincipleTheantisymmetricwavefunctionvanishesifanytwoparticlesinanN-particlesystemhavethesamequantumnumbers.Becausefermionsarethetypeofparticlesthathaveantisymmetricwavefunctions,that’stheequivalentofsayingthatinasystemofNparticles,notwofermionscanhavethesamequantumnumbers—thatis,occupythesamestate.Thatidea,whichAustrianphysicistWolfgangPaulifirstformulatedin1925,iscalledthePauliexclusionprinciple.Thetopicofdiscussionatthattimewastheatom,andthePauliexclusionprincipleappliedtotheelectrons(atypeoffermion),whicharepresentinallatoms.ThePauliexclusionprinciplestatesthatnotwoelectronscanoccupythesamequantumstateinsideasingleatom.Andthatresultisimportantforthestructureofatoms.Insteadofjustpilingonwilly-nilly,electronshavetofillquantumstatesthataren’talreadytaken.Thesameisn’ttrueforbosons—forexample,ifyouhaveaheapofalphaparticles(bosons),theycanallbeinthesamequantumstate.Notsoforfermions.Therearevariousquantumnumbersthatelectronscantakeinanatom—n(theenergy),l(theangularmomentum),m(thezcomponentoftheangularmomentum),andms(thezcomponentofspin).Andusingthatinformation,youcanconstructtheelectronstructureofatoms.FiguringoutthePeriodicTableOneofthebiggestsuccessesoftheSchrödingerequation,togetherwiththePauliexclusionprinciple(seetheprecedingsection),isexplainingtheelectronstructureofatoms.Theelectronsinanatomhaveashellstructure,andtheyfillthatstructurebasedonthePauliexclusionprinciple,whichmaintainsthatnotwoelectronscanhavethesamestate:Themajorshellsarespecifiedbytheprincipalquantumnumber,n,correspondingtothedistanceoftheelectronfromthenucleus.Shells,inturn,havesubshellsbasedontheorbitalangularmomentumquantumnumber,l.Inturn,eachsubshellhassubshells—calledorbitals—whicharebasedonthezcomponentoftheangularmomentum,m.Soeachshellnhasn–1subshells,correspondingtol=0,1,2,...,n–1.Andinturn,eachsubshellhas2l+1orbitals,correspondingtom=–1,–l+1,...,l–1,l.Muchaswiththehydrogenatom,thevarioussubshells(l=0,1,2,3,4,andsoon)arecalledthes,p,d,f,g,h,andsoonstates.So,forexample,foragivenn,ansstatehasoneorbital(m=0),apstatehasthreeorbitals(m=–1,0,and1),adstatehasfiveorbitals(m=–2,–1,0,1,and2),andsoon.Inaddition,duetothezcomponentofthespin,ms,eachorbitalcancontaintwoelectrons—onewithspinup,andonewithspindown.Sohowdotheelectrons,asfermions,fillthestructureofanatom?Electronscan’tfillaquantumstatethat’salreadybeentaken.Foratomsinthegroundstate,electronsfilltheorbitalsinorderofincreasingenergy.Assoonasallofasubshell’sorbitalsarefilled,thenextelectrongoesontothenextsubshell;andwhenthesubshellisfilled,thenextelectrongoesontothenextshell,andsoon.Ofcourse,asyoufillthedifferentelectronshells,subshells,andorbitals,youendupwithadifferentelectronstructure.Andbecauseinteractionsbetweenelectronsformthebasisofchemistry,aselectronsfillthesuccessivequantumlevelsinvariousatoms,youendupwithdifferentchemicalpropertiesforthoseatoms—whichsetuptheperiod(row)andgroup(column)organizationoftheperiodictable.PartVGroupDynamics:IntroducingMultipleParticlesInthispart...Thispartintroducesyoutoworkingwithmultipleparticlesatthesametime.Now,alltheparticlesinthesystemcaninteractnotonlywithanoverallpotentialbutalsowitheachother.Youseehowtodealwithatoms(electronandnucleussystems)here,aswellassystemsofmanyatoms.Afterall,thewholeworldismadeupofmany-particlesystems.Goodthingquantumphysicsisuptothetask.Chapter11GivingSystemsaPush:PerturbationTheoryInThisChapterNondegenerateanddegenerateperturbationtheoryPerturbingharmonicoscillatorsTheStarkeffectandperturbinghydrogenatomsProblemsinquantumphysicscanbecomeprettytoughprettyfast—anotherwayofsayingthat,unfortunately,youjustcan’tfindexactsolutionstomanyquantumphysicsproblems.Thisisparticularlythecasewhenyoumergetwokindsofsystems.Forexample,youmayknowallabouthowsquarewellsworkandallabouthowelectronsinmagneticfieldswork,butwhatifyoucombinethetwo?Thewavefunctionsofeachsystem,whichyouknowexactly,arenolongerapplicable—youneedsomesortofmixinstead.Perturbationtheorytotherescue!Thistheoryletsyouhandlemixesofsituations,aslongastheinterferenceisn’ttoostrong.Inthischapter,youexploretime-independentperturbationtheoryanddegenerateandnondegenerateHamiltonians.Youalsolookatsomeexamplesthatplaceharmonicoscillatorsandhydrogenatomsinelectricfields.IntroducingTime-IndependentPerturbationTheoryTheideabehindtime-independentperturbationtheoryisthatyoustartwithaknownsystem—onewhosewavefunctionsyouknowandwhoseenergylevelsyouknow.Everythingisallsetuptothispoint.Thensomenewstimulus—aperturbation—comesalong,disturbingthestatusquo.Forexample,youmayapplyanelectrostaticormagneticfieldtoyourknownsystem,whichchangesthatsystemsomewhat.Perturbationtheoryletsyouhandlesituationslikethis—aslongastheperturbationisn’ttoostrong.Inotherwords,ifyouapplyaweakmagneticfieldtoyourknownsystem,theenergylevelswillbemostlyunchangedbutwithacorrection.(Note:That’swhyit’scalledperturbationtheoryandnotdrasticinterferencetheory.)Thechangeyoumaketothesetupisslightenoughsothatyoucancalculatetheresultingenergylevelsandwavefunctionsascorrectionstothefundamentalenergylevelsandwavefunctionsoftheunperturbedsystem.Sowhatdoesitmeantotalkofperturbationsinphysicsterms?SaythatyouhavethisHamiltonian:Here,H0isaknownHamiltonian,withknowneigenfunctionsandeigenvalues,andλWistheso-calledperturbationHamiltonian,whereλ<<1indicatesthattheperturbationHamiltonianissmall.FindingtheeigenstatesoftheHamiltonianinthisequationiswhatsolvingproblemslikethisisallabout—inotherwords,here’stheproblemyouwanttosolve:Thewayyousolvethisequationdependsonwhethertheexact,knownsolutionsofH0aredegenerate(thatis,severalstateshavethesameenergy)ornondegenerate.Thenextsectionsolvesthenondegeneratecase.WorkingwithPerturbationstoNondegenerateHamiltoniansStartwiththecaseinwhichtheunperturbedHamiltonian,H0,hasnondegeneratesolutions.Thatis,foreverystate|ϕn>,there’sexactlyoneenergy,En,thatisn’tthesameastheenergyforanyotherstate:(justasaone-to-onefunctionhasonlyonexvalueforanyy).YourefertothesenondegenerateenergylevelsoftheunperturbedHamiltonianasE(0)ntodistinguishthemfromthecorrectionsthattheperturbationintroduces,sotheequationbecomesFromhereon,IrefertotheenergylevelsoftheperturbedsystemasEn.Theideabehindperturbationtheoryisthatyoucanperformexpansionsbasedontheparameterλ(whichismuch,muchlessthan1)tofindthewavefunctionsandenergylevelsoftheperturbedsystem.Inthissection,yougouptotermsinλ2intheexpansions.Alittleexpansion:PerturbingtheequationsTofindtheenergyoftheperturbedsystem,En,startwiththeenergyoftheunperturbedsystem:En=E(0)n+...Addthefirst-ordercorrectiontotheenergy,λE(1)n:Andaddthesecond-ordercorrectiontotheenergy,λ2E(2)n,aswell:Nowwhataboutthewavefunctionoftheperturbedsystem,|ψn>?Startwiththewavefunctionoftheunperturbedsystem,|ϕn>:Addtoitthefirst-ordercorrection,λ|ψ(1)n>:Andthenaddtothatthesecond-ordercorrectiontothewavefunction,λ2|ψ(2)n>:Notethatwhenλ→0,energy:becomestheunperturbedEn=E(0)nAndbecomestheunperturbedwavefunction:SoyourtaskistocalculateE(1)nandE(2)n,aswellasψ(1)nandψ(2)n.Sohowdoyoudothatingeneral?Timetostartslingingsomemath.Youstartwiththreeperturbedequations:Hamiltonian:Energylevels:Wavefunctions:Combinethesethreeequationstogetthisjumboequation:MatchingthecoefficientsofλandsimplifyingYoucanhandlethejumboequationintheprecedingsectionbysettingthecoefficientsofλoneithersideoftheequalsignequaltoeachother.Equatingthezerothordertermsinλoneithersideofthisequation,here’swhatyouget:Nowforthefirst-ordertermsinλ;equatingthemoneithersideofthejumboequationgivesyouNowequatethecoefficientsofλ2inthejumboequation,givingyouOkay,that’stheequationyouderivefromthesecondorderinλ.NowyouhavetosolveforE(1)n,E(2)n,andsoonusingthezeroth-order,first-order,andsecond-orderequations.Startbynotingthattheunperturbedwavefunction,|ϕn>isn’tgoingtobeverydifferentfromtheperturbedwavefunction,|ψn>,becausetheperturbationissmall.Thatmeansthat.Infact,youcannormalize|ψn>sothat<ϕn|ψn>isexactlyequalto1:Giventhat,theequationbecomesAndbecausethecoefficientsofλmustbothvanish,yougetthefollowing:Thisequationisusefulforsimplifyingthemath.Findingthefirst-ordercorrectionsAftermatchingthecoefficientsofλandsimplifying(seetheprecedingsection),youwanttofindthefirst-ordercorrectionstotheenergylevelsandthewavefunctions.Findthefirst-ordercorrectiontotheenergy,E(1)n,bymultiplyingby<ϕn|:Thenthefirsttermcanbeneglectedandwecanuseoursimplificationabovetowritethefirstorderenergyperturbationas:Swell,that’stheexpressionyouuseforthefirst-ordercorrection,E(1)n.Nowlookintofindingthefirst-ordercorrectiontothewavefunction,|ψ(1)n>.Youcanmultiplythewave-functionequationbythisnextexpression,whichisequalto1:SoyouhaveNotethatthem=ntermiszerobecause<ϕn|ψ(1)n>=0.Sowhatis<ϕm|ψ(1)n>?Youcanfindoutbymultiplyingthefirst-ordercorrection,,by<ϕm|togiveyouAndsubstitutingthatintogivesyouOkay,that’syourtermforthefirst-ordercorrectiontothewavefunction,|ψ(1)n>.From,thewavefunctionlookslikethis,madeupofofzeroth-,first-,andsecond-ordercorrections:Ignoringthesecond-ordercorrectionforthemomentandsubstitutinginforthefirst-ordercorrectiongivesyouthisforthewavefunctionoftheperturbedsystem,tothefirstorder:That’sthewavefunctionoftheperturbedsystemintermsoftheperturbation.Butthat’sstillonlythefirst-ordercorrection.Howaboutthesecond?Readon.Findingthesecond-ordercorrectionsNowfindthesecond-ordercorrectionstotheenergylevelsandthewavefunctions(theprecedingsectioncoversfirst-ordercorrections).TofindE(2)n,multiplybothsidesofby<ϕn|:Thislookslikeatoughequationuntilyourealizethat<ϕn|ψ(1)n>isequaltozero,soyougetBecause<ϕn|ψ(2)n>isalsoequaltozero,andagainneglectingthefirstterm,yougetE(2)nisjustanumber,soyouhaveAndofcourse,because<ϕn|ϕn>=1,youhaveNotethatif|ψ(1)n>isaneigenstateofW,thesecond-ordercorrectionequalszero.Okay,soE(2)n=<ϕn|W|ψ(1)n>.Howcanyoumakethatsimpler?Well,fromusing.SubstitutingthatequationintogivesyouNowyouhaveand.Here’sthetotalenergywiththefirst-andsecond-ordercorrections:Sofromthisequation,youcansayThatgivesyouthefirst-andsecond-ordercorrectionstotheenergy,accordingtoperturbationtheory.Notethatforthisequationtoconverge,theterminthesummationmustbesmall.Andnoteinparticularwhathappenstotheexpansiontermiftheenergylevelsaredegenerate:Inthatcase,you’regoingtoendupwithanE(0)nthatequalsanE(0)m,whichmeansthattheenergy-correctionsequationblowsup,andthisapproachtoperturbationtheoryisnogood—whichistosaythatyouneedadifferentapproachtoperturbationtheory(cominguplaterin“WorkingwithPertubationstoDegenerateHamiltonians”)tohandlesystemswithdegenerateenergystates.Inthenextsection,IshowyouanexampletomaketheideaofperturbingnondegenerateHamiltoniansmorereal.PerturbationTheorytotheTest:HarmonicOscillatorsinElectricFieldsConsiderthecaseinwhichyouhaveasmallparticleoscillatinginaharmonicpotential,backandforth,asFigure11-1shows.Figure11-1:Aharmonicoscillator.Here’stheHamiltonianforthatparticle,wheretheparticle’smassism,itslocationisx,andtheangularfrequencyofthemotionisω:Nowassumethattheparticleischarged,withchargeq,andthatyouapplyaweakelectricfield,ε,asFigure11-2shows.Figure11-2:Applyinganelectricfieldtoaharmonicoscillator.Theforceduetotheelectricfieldinthiscaseistheperturbation,andtheHamiltonianbecomesInthissection,youfindtheenergyandwavefunctionsoftheperturbedsystemandcomparethemtotheexactsolutions.FindingexactsolutionsSowhataretheenergyeigenvaluesoftheprecedingHamiltonianfortheharmonicoscillatorinanelectricfield?Firstsolvefortheeigenvaluesexactly;thenuseperturbationtheory.Youcansolvefortheexactenergyeigenvaluesbymakingoneofthefollowingsubstitutions:SubstitutingtheequationsolvedforxintogivesyouThelasttermisaconstant,sotheequationisoftheformwhere.isjusttheHamiltonianofaharmonicoscillatorwithanaddedconstant,whichmeansthattheenergylevelsaresimplySubstitutinginforCgivesyoutheexactenergylevels:Great—that’stheexactsolution.ApplyingperturbationtheoryAssoonasyouhavetheexacteigenvaluesforyourchargedoscillator(seetheprecedingsection),youhavesomethingtocomparethesolutionfromperturbationtheoryto.Nowyoucanfindtheenergyandwavefunctionsoftheperturbedsystem.EnergyofthechargedoscillatorSowhatistheenergyofthechargedoscillator,asgivenbyperturbationtheory?YouknowthatthecorrectedenergyisgivenbywhereλWistheperturbationtermintheHamiltonian.Thatis,here,λW=qεx.NowtakealookatthecorrectedenergyequationusingqεxforλW.Thefirstordercorrectionis,which,usingλW=qεx,becomes<ϕn|qεx|ϕn>orqε<ϕn|x|ϕn>But<ϕn|x|ϕn>=0,becausethat’stheexpectationvalueofx,andharmonicoscillatorsspendasmuchtimeinnegativexterritoryasinpositivexterritory—thatis,theaveragevalueofxiszero.Sothefirst-ordercorrectiontotheenergy,asgivenbyperturbationtheory,iszero.Okay,what’sthesecond-ordercorrectiontotheenergy,asgivenbyperturbationtheory?Hereitis:AndbecauseλW=qεx,youhaveCastthisintermsofbrasandkets(seeChapter4),changing<ϕm|to<m|and|ϕn>to|n>,makingthesecond-orderenergycorrectionintothisexpression:Youcandecipherthisstepbystep.First,theenergyisThatmakesfiguringoutthesecond-orderenergyalittleeasier.Also,thefollowingexpressionsturnouttoholdforaharmonicoscillator:Withthesefourequations,you’rereadytotackle,thesecondordercorrectiontotheenergy.Omittinghigher-powerterms,thesummationinthisequationbecomesAndsubstitutingintheforE(0)n–E(0)n+1andE(0)n–E(0)n–1givesyouNow,substitutinginfor<n+1|x|n>and<n–1|x|n>givesyouorSothesecond-ordercorrectionisTherefore,accordingtoperturbationtheory,theenergyoftheharmonicoscillatorintheelectricfieldshouldbeComparethisresulttotheearlierequationfortheexactenergylevels,—they’rethesame!Inotherwords,perturbationtheoryhasgivenyouthesameresultastheexactanswer.How’sthatforagreement?Ofcourse,youcan’texpecttohitthesameanswereverytimeusingperturbationtheory,butthisresultisimpressive!WavefunctionsofthechargedoscillatorNowfigureoutwhatthechargedoscillator’swavefunctionlookslikeinthepresenceoftheelectricfield.Here’sthewavefunctionoftheperturbedsystem,tothefirstorder:Usingthe<n|and|n>brasandketsyou’reusedtoforharmonicoscillators,thisbecomesBecauseλW=qεx,thisbecomesEvidently,aswiththeenergy,onlytwotermscontribute,because<n|x|n>=0.Inparticular,thetwotermsthatcontributeareNotealsothatand.ThesefourequationsmeanthatNotewhatthisequationmeans:Addinganelectricfieldtoaquantumharmonicoscillatorspreadsthewavefunctionoftheharmonicoscillator.Originally,theharmonicoscillator’swavefunctionisjustthestandardharmonicoscillatorwavefunction,|ψn>=|n>.Applyinganelectricfieldspreadsthewavefunction,addingacomponentof|n–1>,whichisproportionaltotheelectricfield,ε,andthechargeoftheoscillator,q,likethis:Andthewavefunctionalsospreadstotheotheradjacentstate,|n+1>,likethis:Youendupmixingstates.Thatblendingbetweenstatesmeansthattheperturbationyouapplymustbesmallwithrespecttotheseparationbetweenunperturbedenergystates,oryouriskblurringthewholesystemtothepointthatyoucan’tmakeanypredictionsaboutwhat’sgoingtohappen.Inanycase,that’saniceresult—blendingthestatesinproportiontothestrengthoftheelectricfieldyouapply—andit’stypicaloftheresultyougetwithperturbationtheory.Okay,that’shownondegenerateperturbationtheoryworks.Asyoucansee,it’sstronglydependentonhavingtheenergystatesseparatesothatyoursolutioncanblendthem.Butwhathappenswhenyouhaveasystemwheretheenergiesaredegenerate?Youtakealookatthatinthenextsection.WorkingwithPerturbationstoDegenerateHamiltoniansThissectiontacklessystemsinwhichtheenergiesaredegenerate.TakealookatthisunperturbedHamiltonian:Inotherwords,severalstateshavethesameenergy.Saytheenergystatesaref-folddegenerate,likethis:Howdoesthisaffecttheperturbationpicture?ThecompleteHamiltonian,H,ismadeupoftheoriginal,unperturbedHamiltonian,H0,andtheperturbationHamiltonian,Hρ:Inzeroth-orderapproximation,youcanwritetheeigenfunction|ψn>asacombinationofthedegeneratestates|ϕnα>:Notethatinwhatfollows,youassumethat<ϕn|ϕn>=1and<ϕm|ϕn>=0ifmisnotequalton.Also,youassumethatthe|ψn>arenormalized—thatis,<ψn|ψn>=1.Pluggingthiszeroth-orderequationintothecompleteHamiltonianequation,yougetNowmultiplyingthatequationby<ϕnβ|givesyouUsingthefactthat<ϕn|ϕn>=1and<ϕm|ϕn>=0ifmisnotequaltongivesyouPhysicistsoftenwritethatequationaswhere.AndpeoplealsowritethatequationaswhereE(1)n=En–E(0)n.That’sasystemoflinearequations,andthesolutionexistsonlywhenthedeterminanttothisarrayisnonvanishing:ThedeterminantofthisarrayisanfthdegreeequationinE(1)n,andithasfdifferentroots,E(1)nα.Thosefdifferentrootsarethefirst-ordercorrectionstotheHamiltonian.Usually,thoserootsaredifferentbecauseoftheappliedperturbation.Inotherwords,theperturbationtypicallygetsridofthedegeneracy.Sohere’sthewayyoufindtheeigenvaluestothefirstorder—yousetupanfby-fmatrixoftheperturbationHamiltonian,Hρ,whereHραβ=<ϕnα|Hρ|ϕnβ>:ThendiagonalizethismatrixanddeterminethefeigenvaluesE(1)nαandthematchingeigenvectors:Thenyougettheenergyeigenvaluestofirstorderthisway:AndtheeigenvectorsareInthenextsection,youlookatanexampletoclarifythisidea.TestingDegeneratePerturbationTheory:HydrogeninElectricFieldsInthissection,youseewhetherdegenerateperturbationtheorycanhandlethehydrogenatom,whichhasenergystatesdegenerateindifferentangularmomentumquantumnumbers,whenyouremovethatdegeneracybyapplyinganelectricfield.ThissetupiscalledtheStarkeffect.Specifically,supposeyouapplyanelectricfield,ε,toahydrogenatominthen=2excitedstate.Thatstatehasfoureigenfunctionsthathavethesameenergy,wherethequantumnumbersare|nlm>(notethatyou’rerenamingtheseeigenfunctions|1>,|2>,andsoontomakethecalculationeasier):|1>=|200>|2>=|211>|3>=|210>|4>=|21–1>Alltheseunperturbedstateshavethesameenergy,E=–R/4,whereRistheRydbergconstant,13.6eV.Butatleastsomeofthesestateswillhavetheirenergieschangedwhenyouapplytheelectricfield.Whatdoestheelectricfield,ε,causetheperturbationHamiltonian,Hp,tobecome?Here’stheperturbationHamiltonian:Hp=eεzSoyouhavetoevaluatethisequationforthevariousstates.Forexample,whatisthefollowingexpressionequalto,where<1|=<200|and|3>=|210>?<1|Hp|3>YousolvefortheunperturbedhydrogenwavefunctionsinChapter9.Ingeneral,here’swhatthewavefunctionψnlm(r,θ,ϕ)lookslikeforhydrogen:whereLn–l–12l+1(2r/nr0)isageneralizedLaguerrepolynomial.Doingallthemathgivesyouthefollowingresult,wherea0istheBohrradiusoftheatom:The<1|Hp|3>isjustonetermyouhavetocompute,ofcourse.Here’sthefullmatrixfortheperturbationHamiltonianconnectingallstates,whereHpαβ=<α|Hp|β>:Doingthemathgivesyouthisremarkablysimpleresult:Diagonalizingthismatrixgivesyoutheseeigenvalues—thefirst-ordercorrectionstotheunperturbedenergies:E(1)1=–3eεa0E(1)2=0E(1)3=3eεa0E(1)4=0whereE(1)1isthefirst-ordercorrectiontotheenergyofthe|1>eigenfunction,E(1)2isthefirst-ordercorrectiontotheenergyofthe|2>eigenfunction,andsoon.Addingthesecorrectionstotheunperturbedenergyforthen=2stategivesyouthefinalenergylevels:whereRistheRydbergconstant.Notethisresult:TheStarkeffectremovestheenergydegeneracyin|200>and|210>(the|1>and|3>eigenfunctions),butthedegeneracyin|211>and|21–1>(the|2>and|4>eigenfunctions)remains.Chapter12Wham-Blam!ScatteringTheoryInThisChapterSwitchingbetweenlabandcenter-of-massframesSolvingtheSchrödingerequationFindingthewavefunctionPuttingtheBornapproximationtoworkYourNationalScienceFoundationgrantfinallycamethrough,andyoubuiltyournewsynchrotron—aparticleaccelerator.Electronsandanti-​electronsaccelerateatnearthespeedoflightalongagiantcirculartrackenclosedinavacuumchamberandcollide,lettingyouprobethestructureofthehigh-energyparticlesyoucreate.You’resittingattheconsoleofyourgiantnewexperiment,watchingthelightsflashingandthesignalsonthescreensapprovingly.Millionsofwattsofpowercoursethroughthethickcables,andtheradiationmonitorsarebeeping,indicatingthatthingsareworking.Cool.You’reacceleratingparticlesandsmashingthemagainsteachothertoobservehowtheyscatter.Butthisisslightlymorecomplexthanobservinghowpoolballscollide.Classically,youcanpredicttheexactangleatwhichcollidingobjectswillbounceoffeachotherifthecollisioniselastic(thatis,momentumandkineticenergyarebothconserved).Quantummechanically,however,youcanonlyassignprobabilitiestotheanglesatwhichthingsscatter.Physicistsuselargeparticleacceleratorstodiscovermoreaboutthestructureofmatter,andthatstudyiscentraltomodernphysics.Thischapterservesasanintroductiontothatfieldofstudy.Yougettotakealookatparticlescatteringonthesubatomiclevel.IntroducingParticleScatteringandCrossSectionsThinkofascatteringexperimentintermsofparticlesinandparticlesout.LookatFigure12-1,forexample.Inthefigure,particlesarebeingsentinastreamfromtheleftandinteractingwithatarget;mostofthemcontinueonunscattered,butsomeparticlesinteractwiththetargetandscatter.Figure12-1:Scatteringfromatarget.Thoseparticlesthatdoscatterdosoataparticularangleinthreedimensions—thatis,yougivethescatteringangleasasolidangle,dΩ,whichequalssinθdθdϕ,whereϕandθarethesphericalanglesIintroduceinChapter8.ThenumberofparticlesscatteredintoaspecificdΩperunittimeisproportionaltoaveryimportantquantityinscatteringtheory:thedifferentialcrosssection.Thedifferentialcrosssectionisgivenby,andit’sameasureofthenumberofparticlespersecondscatteredintodΩperincomingflux.Theincidentflux,J(alsocalledthecurrentdensity),isthenumberofincidentparticlesperunitareaperunittime.SoiswhereN(ϕ,θ)isthenumberofparticlesatanglesϕandθ.Thedifferentialcrosssectionhasthedimensionsofarea,socallingitacrosssectionisappropriate.Thecrosssectionissortoflikethesizeofthebull’seyewhenyou’reaimingtoscatterincidentparticlesthroughaspecificsolidangle.Thedifferentialcrosssectionisthecrosssectionforscatteringtoaspecificsolidangle.Thetotalcrosssection,σ,isthecrosssectionforscatteringofanykind,throughanyangle.Soifthedifferentialcrosssectionforscatteringtoaparticularsolidangleislikethebull’seye,thetotalcrosssectioncorrespondstothewholetarget.Youcanrelatethetotalcrosssectiontothedifferentialcrosssectionbyintegratingthefollowing:TranslatingbetweentheCenter-ofMassandLabFramesNowyoucanstartgettingintothedetailsofscattering,beginningwithadiscussionofthecenter-of-massframeversusthelabframe.Experimentstakeplaceinthelabframe,butyoudoscatteringcalculationsinthecenter-of-massframe,soyouhavetoknowhowtotranslatebetweenthetwoframes.Thissectionexplainshowtheframesdifferandshowsyouhowtorelatethescatteringanglesandcrosssectionswhenyouchangeframes.FramingthescatteringdiscussionLookatFigure12-2—that’sscatteringinthelabframe.Oneparticle,travelingatv1lab,isincidentonanotherparticlethat’satrest(v2lab=0)andhitsit.Afterthecollision,thefirstparticleisscatteredatangleθ1,travelingatv'1lab,andtheotherparticleisscatteredatangleθ2andvelocityv'2lab.Figure12-2:Scatteringinthelabframe.Nowinthecenter-of-massframe,thecenterofmassisstationaryandtheparticlesheadtowardeachotherwithvelocitiesv1candv2c,respectively.Aftertheycollide,theyheadawayfromeachotherwithvelocitiesv'1candv'2c,atanglesθandπ–θ.Youhavetomovebackandforthbetweenthesetwoframes—thelabframeandthecenter-of-massframe—soyouneedtorelatethevelocitiesandangles(inanonrelativisticway).RelatingthescatteringanglesbetweenframesTorelatetheanglesθ1andθ,youstartbynotingthatyoucanconnectv1labandv1cusingthevelocityofthecenterofmass,vcm,thisway:v1lab=v1c+vcmInaddition,here’swhatcansayaboutthevelocityofparticle1afteritcollideswithparticle2:v'1lab=v'1c+vcmNowyoucanfindthecomponentsofthesevelocities:v'1labcosθ1=v'1ccosθ+vcmv'1labsinθ1=v'1csinθDividingtheequationinthesecondbulletbytheoneinthefirstgivesyouButwouldn’titbeeasierifyoucouldrelateθ1andθbysomethingthatdidn’tinvolvethevelocities,onlythemasses,suchasthefollowing?Well,youcan.Toseethat,startwithAndyoucanshowthatYoucanalsousetheconservationofmomentumtosaywhathappensafterthecollision.Infact,becausethecenterofmassisstationaryinthecenter-of-massframe,thetotalmomentumbeforeandafterthecollisioniszerointhatframe,likethis:m1v1c+m2v2c=0ThereforeAndafterthecollision,m1v'1c+m2v'2c=0,whichmeansthatAlso,ifthecollisioniselastic(andyouassumeallcollisionsareelasticinthischapter),kineticenergyisconservedinadditiontomomentum,sothatmeansthefollowingistrue:Substitutingandintothisequationgivesyouv'1c=v1candv'2c=v2cGiventhesetwoequations,youcanredoasDividingthemagnitudeofeachsideofaboveequationgivesyouAndbecauseyousawearlierthatbythemagnitudeofthe,substitutingintothisequationgivesyouatlastOkay,thatrelatesθ1andθ,whichiswhatyouweretryingtodo.Usingtherelation,youcanrewriteasthefollowing:Youcanalsorelateθ2andθ.Youcanshowthattrig,meansthat,which,usingalittleOkay,nowyou’verelatedtheanglesbetweenthelabandcenter-of-massframes.Howaboutrelatingthecrosssectionsinthetwoframes?That’sinthenextsection.TranslatingcrosssectionsbetweentheframesTheprecedingsectionrelatesθ1andθandθ2—theanglesofthescatteredparticlesinthelabframeandthecenter-of-massframe.Nowhowaboutrelatingthedifferentialcrosssection—thebull’seyewhenyou’reaimingtoscattertheparticlesataparticularangle—betweenthelabandcenter-ofmassframes?Thedifferentialdσ(totalcrosssection)isinfinitesimalinsize,anditstaysthesamebetweenthetwoframes.ButtheanglesthatmakeupdΩ,thescatteringangle,varywhenyoutranslatebetweenframes.Yougettotakealookathowthatworksnow,relatingthelabdifferentialcrosssection:tothecenter-of-massdifferentialcrosssection:Inthelabframe,dΩ1=sinθ1dθ1dϕ1.Andinthecenter-of-massframe,dΩ=sinθdθdϕ.Becausedσlab=dσcm,thefollowingequationistrue:Puttingthatequationwiththeequationsforthelabframeandthecenter-ofmassframe,youhaveBecauseyouhavecylindricalsymmetryhere,ϕ=ϕ1,soYou’vealreadyseenthat,so.ThereforeYoucanalsoshowthatTryingalab-frameexamplewithparticlesofequalmassSayyouhavetwoparticlesofequalmasscollidinginthelabframe(whereoneparticlestartsatrest).Youwanttoshowthatthetwoparticlesenduptravelingatrightangleswithrespecttoeachotherinthelabframe.Notethatifm1=m2,thengivestan(θ1)=tan(θ/2),soθ1=θ/2.AndbecomesNotealsothattan(θ2)=cot(θ/2),ortan(θ2)=tan(π/2–θ/2).Youknowthatθ1=θ/2,andtan(θ2)=tan(π/2–θ/2)tellsyouthatthefollowingistrue:θ2=π/2–θ/2Sosubstitutingθ1=θ/2intotheprecedingequationgivesyouθ2=π/2–θ1θ2+θ1=π/2Therefore,θ2andθ1,theanglesoftheparticlesinthelabframeafterthecollision,adduptoπ/2—whichmeansθ2andθ1areatrightangleswithrespecttoeachother.Cool.Inthiscase,youcanusetherelationsyou’vealreadyderivedtogettheserelationsinthespecialcasewherem1=m2:TrackingtheScatteringAmplitudeofSpinlessParticlesIntheearliersection“TranslatingbetweentheCenter-of-MassandLabFrames,”youseehowtotranslatefromthelabframetothecenter-of-massframeandbackagain,andthosetranslationsworkclassicallyaswellasinquantumphysics(aslongasthespeedsinvolvedarenonrelativistic).Nowyoulookattheelasticscatteringoftwospinlessnonrelativisticparticlesfromthetime-independentquantumphysicspointofview.Assumethattheinteractionbetweentheparticlesdependsonlyontheirrelativedistance,|r1–r2|.Youcanreduceproblemsofthiskindtotwodecoupledproblems(seeChapter9fordetails).Thefirstdecoupledequationtreatsthecenterofmassofthetwoparticlesasafreeparticle,andthesecondequationisforaneffectiveparticleofmass.Thefirstdecoupledequation,thefree-particleequationofthecenterofmass,isofnointeresttoyouinscatteringdiscussions.Thesecondequationistheonetoconcentrateon,whereμ=:YoucanusetheprecedingequationtosolvefortheprobabilitythataparticleisscatteredintoasolidangledΩ—andyougivethisprobabilitybythedifferentialcrosssection, .Inquantumphysics,wavepacketsrepresentparticles.Intermsofscattering,thesewavepacketsmustbewideenoughsothatthespreadingthatoccursduringthescatteringprocessisnegligible(however,thewavepacketcan’tbesospreadthatitencompassesthewholelab,includingtheparticledetectors).Here’sthecrux:Afterthescattering,thewavefunctionbreaksupintotwoparts—anunscatteredpartandascatteredpart.That’showscatteringworksinthequantumphysicsworld.TheincidentwavefunctionAssumethatthescatteringpotentialV(r)hasaveryfiniterange,a.Outsidethatrange,thewavefunctionsinvolvedactlikefreeparticles.Sotheincidentparticle’swavefunction,outsidethelimitofV(r)—thatis,outsidetherangeafromtheotherparticle—isgivenbythisequation,becauseV(r)iszero:where.Theformistheequationforaplanewave,soϕinc(r)isϕinc(r)=·Aeik0 r,whereAisaconstantandk0·risthedotproductbetweentheincidentwave’swavevectorandr.Inotherwords,you’retreatingtheincidentparticleasaparticleofmomentum.ThescatteredwavefunctionAfterthescatteringofthespinlessparticles,thenonscatteredwavefunctionisn’tofmuchinteresttoyou,butthescatteredwavefunctionis.Althoughtheincidentwavefunctionhastheformϕinc(r)=Aeik0·r,thescatteredwavefunctionhasaslightlydifferentform:Thef(ϕ,θ)partiscalledthescatteringamplitude,andyourjobistofindit.Here,AisanormalizationfactorandwhereEistheenergyofthescatteredparticle.RelatingthescatteringamplitudeanddifferentialcrosssectionThescatteringamplitudeofspinlessparticlesturnsouttobecrucialtounderstandingscatteringfromthequantumphysicspointofview.Toseethat,takealookatthecurrentdensities,Jinc(thefluxdensityoftheincidentparticle)andJsc(thecurrentdensityforthescatteredparticle):Insertingyourexpressionsforϕincandϕscintotheseequationsgivesyouthefollowing,wheref(ϕ,θ)isthescatteringamplitude:Nowintermsofthecurrentdensity,thenumberofparticlesdN(ϕ,θ)scatteredintodΩandpassingthroughanareadA=r2dΩisdN(ϕ,θ)=Jscr2dΩPlugginginintotheprecedingequationgivesyouAlso,recallfromthebeginningofthechapterthat.YougetAndhere’sthetrick—forelasticscattering,k=k0,whichmeansthatthisisyourfinalresult:Theproblemofdeterminingthedifferentialcrosssectionbreaksdowntodeterminingthescatteringamplitude.FindingthescatteringamplitudeTofindthescatteringamplitude—andthereforethedifferentialcrosssection—ofspinlessparticles,youworkonsolvingtheSchrödingerequation:.YoucanalsowritethisasYoucanexpressthesolutiontothatdifferentialequationasthesumofahomogeneoussolutionandaparticularsolution:ψ(r)=ψh(r)+ψp(r)Thehomogeneoussolutionsatisfiesthisequation:Andthehomogeneoussolutionisaplanewave—thatis,itcorrespondstotheincidentplanewave:Totakealookatthescatteringthathappens,youhavetofindtheparticularsolution.YoucandothatintermsofGreen’sfunctions,sothesolutiontoiswhere.ThisintegralbreaksdowntoYoucansolvetheprecedingequationintermsofincomingand/oroutgoingwaves.Becausethescatteredparticleisanoutgoingwave,theGreen’sfunctiontakesthisform:YoualreadyknowthatSosubstitutingintotheprecedingequationgivesyouWow,that’sanintegralequationforψ(r),thewaveequation—howdoyougoaboutsolvingthiswhopper?Why,youusetheBornapproximation,ofcourse.TheBornApproximation:RescuingtheWaveEquationOkay,yourdilemmaistosolvethefollowingequationforψ(r),whereϕinc=Aeik0r:Youcandothatwithaseriesofsuccessiveapproximations,calledtheBornapproximation(thisisafamousresult).Tostart,thezerothorderBornapproximationisjustψ0(r)=ϕinc(r).Andsubstitutingthiszeroth-orderterm,ψ0(r),intothefirstequationinthissectiongivesyouthefirst-orderterm:which,usingψ0(r)=ϕinc(r)givesyouYougetthesecond-ordertermbysubstitutingthisequationinto:AndsubstitutingyouintotheprecedingequationgivesThepatterncontinuesforthehigherterms,whichyoucanfindbyplugginglower-ordertermsintohigherones.ExploringthefarlimitsofthewavefunctionNowthatyou’veusedtheBornapproximation(seetheprecedingsection),takealookatthecasewhererislarge—inscatteringexperiments,r>>r',whereristhedistancefromthetargettothedetectorandr'isthesizeofthedetector.Whathappensto,theexactintegralequationforthewavefunction,whenr>>r'?Here’stheanswer:Becauser>>r',youcansaythatk|r–r'|≈kr–k·r',wherek·r'isthedotproductofkandr'(kisthewavevectorofthescatteredparticle).AndUsingthelasttwoequationsingivesyouAndhereThedifferentialcrosssectionisgivenbybecomes,whichinthiscaseUsingthefirstBornapproximationIfthepotentialisweak,theincidentplanewaveisonlyalittledistortedandthescatteredwaveisalsoaplanewave.That’stheassumptionbehindthefirstBornapproximation,whichyoutakealookathere.Soifyoumaketheassumptionthatthepotentialisweak,youcandeterminefromtheequationthatOkay,sowhatisf(θ,ϕ)?WellAndthisequalsthefollowing,whereq=k0–k:Andbecause,youhaveWhenthescatteringiselastic,themagnitudeofkisequaltothemagnitudeofk0,andyouhaveq=|k0–k|=2ksin(θ/2)whereθistheanglebetweenk0andk.Inaddition,ifyousaythatV(r)issphericallysymmetric,andyoucanchoosethezaxisalongq,thenq.r'=qr'cosθ',soThatequalsYouknowthat,soYou’vecomefarinthischapter—fromtheSchrödingerequationallthewaythroughtheBornapproximation,andnowtotheprecedingequationforweak,sphericallysymmetricpotentials.Howaboutyouputthistoworkwithsomeconcretenumbers?PuttingtheBornapproximationtoworkInthissection,youfindthedifferentialcrosssectionfortwoelectricallychargedparticlesofchargeZ1eandZ2e.Here,thepotentiallookslikethis:Sohere’swhatthedifferentialcrosssectionlookslikeinthefirstBornapproximation:Andbecause,youknowthatAndbecauseq=2ksin(θ/2),thefollowingistrue:whereEisthekineticenergyoftheincomingparticle:.Nowgetmorespecific;saythatyou’resmashinganalphaparticle,Z1=4,againstagoldnucleus,Z2=79.Ifthescatteringangleinthelabframeis60°,whatisitinthecenter-of-massframe?Theratiooftheparticles'massinthiscase,m1/m2,is0.02,sothescatteringangleinthecenter-of-massframe,θ,isthefollowing,whereθlab=60°:Solvingthatequationforθgivesyouθ=61°.Sowhat’sthecrosssectionforthisscatteringangle?Takealook:Plugginginthenumbersiftheincidentalphaparticle’senergyis8MeVgivesyouthefollowing:That’sthesizeofthetarget—thecrosssection—youhavetohittocreatethescatteringangleseen.PartVIThePartofTensInthispart...Iletquantumphysicsofftheleashinthispart,anditgoeswild.Yougettoseethetenbestonlinetutorialshere,aswellastenmajortriumphsofquantumphysics.Researcherscreatedquantumphysicsbecauseoftheneedtohandleissuessuchasthewave-particleduality,theuncertaintyprinciple,andthephotoelectriceffect,andyourelivethosetriumphshere.Chapter13TenQuantumPhysicsTutorialsInThisChapterUnderstandingbasicconceptsandequationsViewingillustrationsandanimationsWhenscientistsstartmixingtalkofdice,billiardballs,andapossiblyundeadcat-in-a-box,youknowyou’redealingwithachallengingsubject.Luckily,youcanfindplentyofonlinetutorials,someofthemfeaturinganimation,tohelpyouwrapyourbrainaroundquantumphysics.Thischapterpresentsagoodstarterlist.AnIntroductiontoQuantumMechanicshttp://legacyweb.chemistry.ohio-state.edu/betha/qmWhatisawavefunction?Whatisanorbital?:AnIntroductiontoQuantumMechanicscomesfromNealMcDonald,MidoriKitagawa-DeLeon,AnnaTimasheva,HeathHanlin,ZilLilas,andSherwinJ.SingeratTheOhioStateUniversity.Thissiteincludestutorialsonprobability,particlesversuswaves,wavefunctions,andmore,includingShockwave-basedsound(thoughifyoudon’thaveShockwaveinstalled,that’snotaproblem).QuantumMechanicsTutorialwww.gilestv.com/tutorials/quantum.htmlThiscooltutorialisoneoftheFlash-animatedModernPhysicsTutorialsbyGilesHogben.Extensivelyillustrated,thistutorialprobesquestionssuchaswaveparticledualityandoffersagoodgeneralintroductiontoquantumphysics.GrainsofMystique:QuantumPhysicsfortheLaymanhttp://www.faqs.org/docs/qpThissiteprovidesgoodhistoricalandexperimentalbackgroundinfo—andthey’vedocumentedtheirsourcesandmadesomeattemptsatpeerreview.QuantumPhysicsOnlineVersion2.0www.quantum-physics.polytechnique.fr/index.htmlThisisacoolsetofprogramsthatruninyourbrowser,givingsimulationsofvariousquantumphysicsexperiments.It’sbyManuelJoffre,Jean-LouisBasdevant,andJeanDalibardoftheÉcolePolytechniqueinFrance.Lookforinformationonwavemechanics,quantization,quantumsuperposition,andspin1/2.ToddK.Timberlake’sTutorialfacultyweb.berry.edu/ttimberlake/qchaos/qm.htmlThistutorialisbyToddK.Timberlake,assistantprofessoroftheDepartmentofPhysics,Astronomy,&GeologyofBerryCollegeinGeorgia.It’safairlybriefbutwell-writtenintroductiontotheideasofquantummechanics.Physics24/7’sTutorialwww.physics247.com/physics-tutorial/quantum-physics-billiards.shtmlThisisatext-basedtutorialfromPhysics24/7.Itincludesmaterialonquanta,theuncertaintyprinciple,andquantumtunneling(aswellassomeads).StanZochowski’sPDFTutorialswww.cmmp.ucl.ac.uk/~swz/courses/SM355/SM355.htmlStanZochowski,fromthedepartmentofPhysics&AstronomyatUniversityCollegeLondon,puttogetherthesePDF-basedtutorialsonquantumphysics.ThesearetutorialhandoutsforaQuantumMechanicscourseattheUniversityCollege,andtheyserveasanexcellentintroductiontoquantumphysics.QuantumAtomTutorialwww.colorado.edu/physics/2000/quantumzone/index.htmlThisisafun,cartoon-centrictutorialonthequantumnatureoftheatomfromtheUniversityofColoradoPhysics2000project.CollegeofSt.Benedict’sTutorialwww.physics.csbsju.edu/QM/Index.htmlThisisacomprehensivequantumphysicstutorialfromtheCollegeofSt.Benedict.It’sagood,moreserious,textandequations-basedtutorialwithplentyofillustrations.AWeb-BasedQuantumMechanicsCourseelectron6.phys.utk.edu/qm1/Modules.htmThisone’sfromtheUniversityofTennessee,andit’sanextensiveonlinecourseinquantumphysics.Itincludesmodulesonsquarepotentials,harmonicoscillators,angularmomentum,spin,andsoon.Chapter14TenQuantumPhysicsTriumphsInThisChapterExplainingunexpectedresultsIdentifyingcharacteristicsofthequantumworldDevelopingnewmodelsQuantumphysicshasbeenverysuccessfulinexplainingmanyphysicalphenomena,suchaswave-particleduality.Infact,quantumphysicswascreatedtoexplainphysicalmeasurementsthatclassicalphysicscouldn’texplain.Thischapterisabouttentriumphsofquantumphysics,anditpointsyoutoresourcesontheWebthatexaminethosetriumphsforfurtherinformation.Wave-ParticleDualityIsthatparticleawave?Oristhatwaveaparticle?That’soneofthequestionsthatquantumphysicswascreatedtosolve,becauseparticlesexhibitedwavelikepropertiesinthelab,whereaswavesexhibitedparticle-likeproperties.TheseWebsitesoffermoreinsight:www.gilestv.com/tutorials/quantum.htmlwww.physics247.com/physics-tutorial/quantum-physics-billiards.shtmlThePhotoelectricEffectAnotherfoundingpillarofquantumphysicswasexplainingthephotoelectriceffect,inwhichexperimentersshonelightonametal.Nomatterhowstrongthelight,theenergyofejectedelectronsfromthemetaldidn’trise.Itturnsoutthattheenergyofelectronsgoesupwiththefrequencyofthelight,notitsintensity—whichgivessupporttothelightasastreamofdiscretephotonstheory.Formoreinfoonthephotoelectriceffect,checkoutwww.gilestv.com/tutorials/quantum.html.PostulatingSpinTheStern-Gerlachexperimentresultscouldn’tbeexplainedwithoutpostulatingspin,anothertriumphofquantumphysics.Thisexperimentsentelectronsthroughamagneticfield,andtheclassicalpredictionisthattheelectronstreamwouldcreateonespotofelectronsonascreen—butthereweretwo(correspondingtothetwospins,upanddown).ThisWebsitehasmoreinfo:electron6.phys.utk.edu/qm1/modules/m9/spin.htm.DifferencesbetweenNewton’sLawsandQuantumPhysicsInclassicalphysics,boundparticlescanhaveanyenergyorspeed,butthat’snottrueinquantumphysics.Andinclassicalphysics,youcandetermineboththepositionandmomentumofparticlesexactly,whichisn’ttrueinquantumphysics(thankstotheHeisenberguncertaintyprinciple).Andinquantumphysics,youcansuperimposestatesoneachother,andhaveparticlestunnelintoareasthatwouldbeclassicallyimpossible.Youcanfindanicediscussionofthedifferencesbetweenclassicalandquantumphysicsathttp://facultyweb.berry.edu/ttimberlake/qchaos/qm.html.HeisenbergUncertaintyPrincipleOneofthetriumphsofquantumphysicsistheHeisenberguncertaintyprinciple:Heisenbergtheorizedthatyoucan’tsimultaneouslymeasureaparticle’spositionandmomentumexactly.Thisisoneofthecentraltheoriesthathasdestroyedclassicalphysics.Here'swhereyoucanfindoneofthebestWebdiscussionsonthistopic:www.physics247.com/physics-tutorial/quantum-physics-billiards.shtml.QuantumTunnelingHowcanparticlesgowhere,classically,theydon’thaveenoughenergytogo?Forexample,howcananelectronwithenergyEgointoanelectrostaticfieldwhereyouneedtohavemorethanenergyEtopenetrate?Theanswerwaspostulatedwithquantumtunneling,andyoucanfindmoreinformationaboutthatatwww.physics247.com/physics-tutorial/quantum-physics-billiards.shtml.DiscreteSpectraofAtomsModelingthequantizednatureofatomsandorbitalsisanothertriumphofquantumphysics.Itturnsoutthatelectronscan’thaveanyoldenergyinanatom,butareonlyallowedparticularquantizedenergylevels—andthatwasoneofthefoundationsofquantumphysics.Foralotmoreonthetopic,visitwww.colorado.edu/physics/2000/quantumzone/index.html.HarmonicOscillatorQuantizingharmonicoscillatorsonthemicrolevelwasanothertriumphofquantumphysics.Classically,harmonicoscillatorscanhaveanyenergy—butnotquantummechanically.Andguesswhichonewasright?Readallaboutithere:www.physics.csbsju.edu/QM/Index.htmlelectron6.phys.utk.edu/qm1/modules/module8.htmSquareWellsLikeharmonicoscillators,quantizingparticlesboundinsquarewellsatthemicrolevelwasanothertriumphforquantumphysics.Classically,particlesinsquarewellscanhaveanyenergy,butquantumphysicssaysyoucanonlyhavecertainallowedenergies.There’splentyontheWebaboutit,includingthesetwogoodtreatments:www.physics.csbsju.edu/QM/Index.htmlelectron6.phys.utk.edu/qm1/modules/module2.htmSchrödinger’sCatSchrödinger’sCatisathoughtexperimentthatdetailssomeproblemsthatariseinthemacroworldfromthinkingofthespinofelectronsascompletelynon-determineduntilyoumeasurethem.Forexample,ifyouknowthespinofoneofapairofnewly-createdelectrons,youknowtheotherhastohavetheoppositespin.Soifyouseparatetwoelectronsbylightyearsandthenmeasurethespinofoneelectron,doestheotherelectron’sspinsuddenlysnaptotheoppositevalue—evenatadistancethatwouldtakeasignalfromthefirstelectronyearstocover?Trickystuff!Formore,takealookatwww.gilestv.com/tutorials/quantum.html.GlossaryHere’saglossaryofcommonquantumphysicsterms:amplitude:Themaximumamountofdisplacementofanoscillatingparticle.angularmomentum:Theproductofthedistanceaparticleisfromacertainpointanditsmomentummeasuredwithrespecttothepoint.annihilationoperator:Anoperatorthatlowerstheenergylevelofaneigenstatebyonelevel.anti-Hermitian:AnoperatorwhoseHermitianadjointisthesameastheoriginaloperator,butwithaminussign;inotherwords,anoperatorisantiHermitianifA†=–A.SeealsoHermitianoperator.blackbody:Abodythatabsorbsallradiationandradiatesitallaway.Bohrradius:Theaverageradiusofanelectron’sorbitinahydrogenatom,about10–10meters.boundstate:Astateinwhichaparticleisn’tfreetotraveltoinfinity.bosons:Particleswithintegerspins,includingphotons,pimesons,andsoon.bra-ketnotation:Abbreviatingthematrixformofastatevectorasaket,or|ψ>,andabbreviatingtheket’scomplexconjugate,orbra,as<ψ|.center-of-massframe:Inscatteringtheory,theframeinwhichthecenterofmassisstationaryandtheparticlesheadtowardeachotherandcollide.Seealsolabframe.centralpotential:Asphericallysymmetricalpotential.commute:Twooperatorscommutewitheachotheriftheircommutatorisequaltozero.ThecommutatorofoperatorsAandBis[A,B]=AB–BA.complexconjugate:Thenumberyougetbynegatingtheimaginarypartofacomplexnumber.The*symbolindicatesacomplexconjugate.Comptoneffect:Anincreaseofwavelength,dependingonthescatteringangle,thatoccursafterincidentlighthitsanelectronatrest.conservationofenergy:Thelawofphysicsthatsaystheenergyofaclosedsystemdoesn’tchangeunlessexternalinfluencesactonthesystem.creationoperator:Anoperatorthatraisestheenergylevelofaneigenstatebyonelevel.currentdensity:Seeincidentflux.electronvolts(eV):Theamountofenergyoneelectrongainsfallingthrough1volt.diagonalize:Writingamatrixsothattheonlynonzeroelementsappearalongthematrix’sdiagonal.differentialcrosssection:Inscatteringtheory,thecrosssectionforscatteringaparticletoaspecificsolidangle;it’slikeabull’s-eye.Diracnotation:Seebra-ketnotation.eigenvalue:Acomplexconstantthatrepresentsthechangeinmagnitudeofavectorwhenyouactonthatvectorwithanoperator.eigenvector:Avectorthatchangesinmagnitudebutnotdirectionafteryouapplyanoperator.elasticcollision:Acollisioninwhichkineticenergyisconserved.electricfield:TheforceonapositivetestchargeperCoulombduetootherelectricalcharges.electron:Anegativelychargedparticlewithhalf-integerspin.emissivity:Apropertyofasubstanceshowinghowwellitradiates.energy:Theabilityofasystemtodowork.energydegeneracy:Thenumberofstatesthathavethesameenergy.energywell:Seepotentialwell.expectationvalue:Theaveragevalueanoperatorwillreturn.fermions:Particleswithhalf-integerspin,includingelectrons,protons,neutrons,quarks,andsoon.frequency:Thenumberofcyclesofaperiodicoccurrencepersecond.Hamiltonian:Anoperatorforthetotalenergyofaparticle,bothkineticandpotential.Heisenberguncertaintyprinciple:Seeuncertaintyprinciple.Hermitianadjoint:Thecomplexconjugateofanumber,thebracorrespondingtoaketvectorortheketcorrespondingtoabravector,ortheconjugatetransposeA†ofanoperatorA.Hermitianoperator:OperatorsthatareequaltotheirHermitianadjoints;inotherwords,anoperatorisHermitianifA†=A.incidentflux:Thenumberofincidentparticlesperunitareaperunittime.inelasticcollision:Acollisioninwhichkineticenergyisn’tconserved.intensity(wave):Thetime-averagedrateofenergytransmittedbyawaveperunitofarea.Joule:TheMKSunitofenergy—oneNewton-meter.ket:Seebra-ketnotation.kineticenergy:Theenergyofanobjectduetoitsmotion.labframe:Inscatteringtheory,theframeinwhichoneparticleisincidentonaparticleatrestandhitsit.Seealsocenter-of-massframe.Laplacian:Anoperator,representedby∇2,thatyouusetofindtheHamiltonian.magneticfield:Theforceonamovingpositivetestcharge,perCoulomb,frommagnetsormovingcharges.magnitude:Thesizeorlengthassociatedwithavector(vectorsaremadeupofadirectionandamagnitude).mass:Thepropertythatmakesmatterresistbeingaccelerated.momentum:Theproductofmasstimesvelocity,avector.MKSsystem:Themeasurementsystemthatusesmeters,kilograms,and​‐seconds.Newton:TheMKSunitofforce—onekilogram-meterpersecond2.normalizedfunction:Afunctioninwhichtheprobabilityaddsupto1.orbitals:Differentangularmomentumstatesofanelectron,representedassubshellsinatomicstructure.orthogonal:Twokets,|ψ>and|ϕ>,forwhich<ψ|ϕ>=0.orthonormal:Twokets,|ψ>and|ϕ>,thatmeetthefollowingconditions:<ψ|ϕ>=0;<ψ|ψ>=1;and<ϕ|ϕ>=1.oscillate:Tomoveorswingsidetosideregularly.pairannihilation:Theconversionofanelectronandpositronintopurelight.pairproduction:Theconversionofahigh-poweredphotonintoanelectronandpositron.particle:Adiscretepieceofmatter.Pauliexclusionprinciple:Theideathatnotwoelectronscanoccupythesamestateinasingleatom.period:Thetimeittakesforonecompletecycleofarepeatingevent.perturbation:Astimulusmildenoughthatyoucancalculatetheresultingenergylevelsandwavefunctionsascorrectionstothefundamentalenergylevelsandwavefunctionsoftheunperturbedsystem.photoelectriceffect:Aresultinwhichthekineticenergyofelectronsemittedfromapieceofmetaldependsonlyonthefrequency—nottheintensity—oftheincidentlight.photon:Aquantumofelectromagneticradiation.Anelementaryparticlethatisitsownantiparticle.pimeson:Asubatomicparticlethathelpsholdthenucleusofanatomtogether.Planck’sconstant:Auniversalconstant,h,thatdescribestherelationshipbetweentheenergyandfrequencyofaphoton.Itequals6.626×10–34Jouleseconds.positron:Apositivelychargedanti-electron.potentialbarrier:Apotentialstepoflimitedextent;anelectronmaybeabletotunnelthroughthebarrierandcomeouttheotherside.potentialenergy:Anobject’senergybecauseofitspositionwhenaforceisactingonitoritsinternalconfiguration.potentialstep:Aregioninwhichtheenergypotentialformsastairshape;aparticlestrikingthestepmaybereflectedortransmitted.potentialwell:Aregioninwhichthere’sadipintheenergypotentialthreshold;particleswithoutenoughenergytoovercomethebarriercanbecometrappedinthewell,unabletoconvertthepotentialenergytokinetic.power:Therateofchangeinasystem’senergy.probabilityamplitude:Thesquarerootoftheprobabilitythataparticlewilloccupyacertainstate.probabilitydensity:Thelikelihoodthataparticlewilloccupyaparticularpositionorhaveaparticularmomentum.quantized:Comingindiscretevalues.quark:Particlesthatcombinewithantiquarkstoformprotons,neutrons,andsoon.radian:TheMKSunitofangle—2πradiansareinacircle.radiation:Aphysicalmechanismthattransportsheatandenergyaselectromagneticwaves.scalar:Asimplenumber(withoutadirection,whichavectorhas).Schrödingerequation:Anequationthattellsushowthewavefunction(whichdescribestheprobablelocationsofparticleslikeelectrons)changesovertime.simpleharmonicmotion:Repetitivemotionwheretherestoringforceisproportionaltothedisplacement.sphericalcoordinates:Coordinatesthatindicatelocationusingtwoanglesandthelengthofaradiusvector.spin:Theintrinsicangularmomentumofanelectron,classifiedasupordown.synchrotron:Atypeofcircularparticleaccelerator.statevector:Avectorthatgivestheprobabilityamplitudethatparticleswillbeintheirvariouspossiblestates.thresholdfrequency:Ifyoushinelightbelowthisfrequencyonmetal,noelectronsareemitted.totalcrosssection:Inscatteringtheory,thecrosssectionforanykindofparticlescattering,throughanyangle.tunneling:Thephenomenonwhereparticlescangetthroughregionsthatthey’reclassicallyforbiddentogo.ultravioletcatastrophe:ThefailureoftheRayleigh-JeansLawtoexplainblack-bodyradiationathighfrequencies.uncertaintyprinciple:Aprinciplethatsaysit’simpossibletoknowanobject’sexactmomentumandposition.vector:Amathematicalconstructthathasbothamagnitudeandadirection.velocity:Therateofchangeofanobject’sposition,expressedasavectorwhosemagnitudeisspeed.volt:TheMKSunitofelectrostaticpotential—oneJouleperCoulomb.wave:Atravelingenergydisturbance.wavelength:Thedistancebetweencrestsortroughsofawave.wave-particleduality:Theobservationthatlighthaspropertiesofbothwavesandparticles,dependingontheexperiment.wavepacket:Acollectionofwavefunctionssuchthatthewavefunctionsinterfereconstructivelyatonelocationandinterferedestructively(gotozero)atallotherlocations.work:Forcemultipliedbythedistanceoverwhichthatforceacts..