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Feynman V3 OCR

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Scanned and OCR'd copy of The Feynman Lectures on Physics, Vol. III (Feynman, Leighton, Sands, Caltech lectures of 1961-64). It opens with Feynman's preface and Matthew Sands's foreword, then the contents. Chapters cover quantum behavior, probability amplitudes, identical particles, spin, the Hamiltonian matrix, the ammonia maser, two-state systems, crystal lattices, semiconductors, and the Schrodinger equation, ending with a superconductivity seminar. This is a published textbook in Phil's downloads folder, not his own work.

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Feynman ’sPreface These arethelectures inphysics thatlgave lastyear andtheyear before tothe freshman and sophomore classes atCaltech. The lectures are, ofcourse, not verbatim—they have been edited, sometimes extensively andsometimes lessso. Thelectures Form only part ofthecomplete course. Thewhole group of180 students gathered inabiglecture room twice aweek tohear these lectures ’and then they broke upinto small groups of15to20students inrecitation sections under theguidance ofateaching assistant. lnaddition, there wasalaboratory session once aweek. Thespecial problem wetried togetatwith these lectures wastomaintain the interest ol‘thevery enthusiastic andrather smart students coming outofthehigh schools andintoCaltech. They have heard alotabout howinteresting andexcit- ingphysics is--~the theory ofrelativity, quantum mechanics, and other modern ideas. Bytheendoftwoyears ofourprevious course, many would bevery dis- couraged because there were really very fewgrand, new, modern ideas presented tothem. They were made tostudy inclined planes, electrostatics, and soforth, andalter twoyears itwasquite stultifying. The problem waswhether ornotwe could make acourse which would save themore advanced andexcited student by maintaining hisenthusiasm. The lectures here arenotinanyway meant tobeasurvey course, butarevery serious. lthought toaddress them tothemost intelligent intheclass andtomake sure, ifpossible, thateven themost intelligent student wasunable tocompletely encompass everything thatwasinthelecturesﬁby putting insuggestions ofappli- cations oftheideas andconcepts invarious directions outside themain lineof attack. Forthis reason, though, ltried very hard tomake allthestatements as accurate aspossible, topoint outinevery casewhere theequations andideas ﬁtted intothebody ofphysics, andhow-—when they learned more things would be modiﬁed. lalso feltthat forsuch students itisimportant toindicate what itis that they should—il‘ they aresufﬁciently clever-be able tounderstand bydeduc- tion from what hasbeen said before, and what isbeing putinassomething new. When new ideas came in,lwould tryeither todeduce them ifthey were deducible, ortoexplain that itwasanew idea which hadn’t anybasis interms ofthings they hadalready learned and which was notsupposed tobeprovable—but wasjust added in. Atthestart oi‘these lectures, lassumed that thestudents knew something when theycame outofhighschool—such things asgeometrical optics, simple chemistry ideas, andsoon. lalso didn’t seethat there wasanyreason tomake thelectures 3 Foreword Agreat triumph oftwentieth-century physics, thetheory ofquantum mechanics, isnow nearly 40years old,yetwehave generally been giving ourstudents their introductory course inphysics (formany students, their last) with hardly more than acasual allusion tothiscentral partofourknowledge ofthephysical world. Weshould dobetter bythem. These lectures areanattempt topresent them with thebasic andessential ideas ofthequantum mechanics inaway that would, hopefully, becomprehensible. Theapproach youwillﬁndhereisnovel, particu- larly atthelevel ofasophomore course, andwasconsidered verymuch anexperi- ment. After seeing howeasily some ofthestudents take toit,however, Ibelieve thattheexperiment wasasuccess. There is,ofcourse, room forimprovement, anditwillcome with more experience intheclassroom. What youwillﬁndhere isarecord ofthatﬁrstexperiment. Inthetwo-year sequence oftheFeynman Lectures onPhysics which were given from September I961 through May 1963 fortheintroductory physics course at Caltech, theconcepts ofquantum physics were brought inwhenever they were necessary foranunderstanding ofthephenomena being described. Inaddition, thelasttwelve lectures ofthesecond year were given over toamore coherent introduction tosome oftheconcepts ofquantum mechanics. Itbecame clear as thelectures drew toaclose, however, that notenough time had been leftforthe quantum mechanics. Asthematerial wasprepared, itwascontinually discovered that other important and interesting topics could betreated with theelementary tools that hadbeen developed. There wasalso afearthat thetoobrief treatment oftheSchrodinger wave function which had been included inthetwelfth lecture would notprovide asufficient bridge tothemore conventional treatments ofmany books thestudents might hope toread. Itwastherefore decided toextend the series with seven additional lectures; they were given tothesophomore class in May ofI964. These lectures rounded outand extended somewhat thematerial developed intheearlier lectures. lnthis volume wehave puttogether thelectures from both years with some adjustment ofthe sequence. Inaddition, twolectures originally given tothefresh- man class asanintroduction toquantum physics have been lifted bodily from Volume I(where theywere Chapters 37and38)andplaced astheﬁrsttwochapters here~to make this volume aself-contained unit, relatively independent ofthe ﬁrsttwo. Afewideas about thequantization ofangular momentum (including a discussion oftheStern-Gerlach experiment) hadbeen introduced inChapters 34 and35ofVolume II,andfamiliarity with them isassumed; fortheconvenience ofthose whowillnothave thatvolume athand, those twochapters arereproduced here asanAppendix. This setoflectures tries toelucidate from thebeginning those features ofthe quantum mechanics which aremost basic andmost general. Theﬁrstlectures tackle head ontheideas ofaprobability amplitude, theinterference ofamplitudes, theabstract notion ofastate, andthesuperposition andresolution ofstates—and theDirac notation isused from thestart. Ineach instance theideas areintroduced together with adetailed discussion ofsome speciﬁc examples—to trytomake the physical ideas asrealaspossible. Thetime dependence ofstates including states ofdeﬁnite energy comes next, andtheideas areapplied atonce tothestudy of two-state systems. Adetailed discussion ofthe ammonia maser provides theframe- 7 work fortheintroduction toradiation absorption andinduced transitions. The lectures then goontoconsider more complex systems, leading toadiscussion ol thepropagation ofelectrons inacrystal, andtoarather complete treatment ofthe quantum mechanics ofangular momentum. Ourintroduction toquantum me- chanics ends inChapter 20with adiscussion oftheSchrodinger wave function, itsdifferential equation, andthesolution forthehydrogen atom. Thelastchapter ofthisvolume isnotintended tobeapartofthe“course. Itisa“seminar” onsuperconductivity andwasgiven inthespirit ofsome ofthe entertainment lectures oftheﬁrsttwovolumes, with theintent ofopening tothe students abroader view oftherelation ofwhat they were learning tothegeneral culture ofphysics. Feynman’s “epilogue” serves astheperiod tothethree- volume series. Asexplained intheForeword toVolume l,these lectures were butoneaspect ofaprogram forthedevelopment ofanewintroductory course carried outatthe California Institute ofTechnology under thesupervision ofthePhysics Course Revision Committee (Robert Leighton, Victor Neher, andMatthew Sands). The program was made possible byagrant from theFord Foundation. Many people helped with thetechnical details ofthepreparation ofthisvolume: Marylou Clayton, Julie Curcio, James Hartle, Tom Harvey, Martin Israel, Patricia Preuss. Fanny Warren, and Barbara Zimmerman. Professors Gerry Neugebauer and Charles Wilts contributed greatly totheaccuracy and clarity ofthematerial by reviewing carefully much ofthemanuscript. Butthestory ofquantum mechanics you willﬁnd here isRichard Feynman"s. Our labors willhave been well spent ifWehave been able tobring toothers even some oftheintellectual excitement weexperienced aswesawtheideas unfold in hisreal-life Lectures onPhysics. December, 1964 M/xrruew SANDS 8 Contents CHAPTER 1.QUANTUM BEHAvIoR >—>—>—~»->--A»---I-AOC\lO'\lJl-{>14-lI\-2"‘—Atomic mechanics 1-1 — Anexperiment with bullets 1-1 - Anexperiment with waves 1-3 -Anexperiment withelectrons 1-4 - The interference ofelectron waves 1-5 — Watching theelectrons 1-6 -First principles ofquantum mechanics 1-9 -Theuncertainty principle 1-11 CHAPTER 2.THE RELATION orWAVE AND PARTICLE VIEwPoINTs I\JI\)I\JI\JI\)l\)U\U|-l>'a~JI\J*-‘— Probability wave amplitudes 2-1 -Measurement ofposition andmomentum 2-2 -Crystal diffraction 2-4 -Thesizeofanatom 2-5 -Energy levels 2-7 -Philosophical implications 2-8 CHAPTER 3.PROBABILITY AMPLITUDES b~)L:JbJL¢-3 -[>14-l|\)r—K— The laws ofcombining amplitudes 3-1 -Thetwo-slit interference pattern 3-5 -Scattering from acrystal 3-7 -Identical particles 3-9 CHAPTER 4.IDENTICAL PARTIcI.Es -{>-l>-J>J>J>J>J>\lO’\\IlJ>laJl\J>—'- Bose particles and Fermi particles 4-1 -States with twoBose particles 4-3 -States with nBose particles 4-6 -Emission andabsorption ofphotons 4-7 -Theblackbody spectrum 4-8 - Liquid helium 4-12 -Theexclusion principle 4-12 CHAPTER 5.SPIN ONE LI1£I|LI|L!1£J|LII(J1lJ\O0\lO\U1->bJl\J'—‘-Filtering atoms with aStern-Gerlach apparatus 5-1 — Experiments with ﬁltered atoms 5-5 -Stern-Gerlach ﬁlters inseries 5-6 -Base states 5-8 —Interfering amplitudes 5-10 -Themachinery ofquantum mechanics 5-12 Transforming toadilferent base 5-15 Other situations 5-16CHAPTER 6.SPIN ONE-HALF O\O\O\O‘\O\O'\O\LJ1-l>bJI\J>-*- Transforming amplitudes 6-1 - Transforming toarotated coordinate system 6-3 - Rotations about thez-axis 6-6 —Rotations of180°and90°about y6-9 -Rotations about x6-11 -Arbitrary rotations 6-12 CHAPTER 7.THE DEPENDENCE OFAMPLITUDES oNTIME \l\I\l\I\lU1-l>UJI\J'—‘-Atoms atrest; stationary states 7-1 -Uniform motion 7-3 -Potential energy; energy conservation 7-6 — Forces; theclassical limit 7-9 -The“precession” ofaspinone-half particle 7-10 CHAPTER 8.THE HAMILTONIAN MATRIX OOOOOOOOOOOOO\l.!|.bLoJI\)'—*-Amplitudes andvectors 8-1 , -Resolving state vectors 8-3 -What arethebasestates oftheworld? 8-5 -I-Iow states change with time 8-7 -TheHamiltonian matrix 8-10 — The ammonia molecule 8-11 CHAPTER 9.THE AMMQNIA MAsER \O\O\O\O\D\OO\lJ1-{Rb-\I\J*-I-Thestates ofanammonia molecule 9-1 -Themolecule inastatic electric ﬁeld9-5 -Transitions inatime-dependent ﬁeld9-9 -Transitions atresonance 9-11 —Transitions offresonance 9-13 -Theabsorption oflight 9-14 CHAPTER 10. OTHER Two-STATE SYsTEMs 10-1 Thehydrogen molecular ion10-1 10-2 Nuclear forces 10-6 10-3 Thehydrogen molecule 10-8 10-4 Thebenzene molecule 10-10 10-5 Dyes 10-12 10-6 The Hamiltonian ofaspin one-half particle in magnetic ﬁeld 10-12 10-7 Thespinning electron inamagnetic ﬁeld10-15 IHAPTER 11.MoRE Two-STATE SYsTEMs CHAPTER 17. SYMMETRY AND CoNsERvATIoN LAws 11-1 ThePauli spinmatrices 11-1 17-1 11-2 Thespinmatrices asoperators 11-5 17-2 11-3 Thesolution ofthetwo-state equations 11-8 17-3 11-4 Thepolarization states ofthephoton 11-9 17-4 11-5 Theneutral K-meson 11-12 17-5 11-6 Generalization toN-state systems 11-20 17-6 IHAPTER 12.THE HYPERPINE SPLITTING INHYDRQGEN 12-1 12-2 12-3 12-4 12-5 12-6Base states forasystem withtwospinone-half particles 12-1 TheHamiltonian fortheground state ofhydrogen 12-3 Theenergy levels 12-7 TheZeeman splitting 12-9 Thestates inamagnetic ﬁeld 12-12 Theprojection matrix forspinone12-14 ZHAPTER 13.PRoPAoATIoN INACRYSTAL LATTIcE 13-1 13-2 13-3 13-4 13-5 13-6 13-7 13-8States foranelectron inaone-dimensional lattice 13-1 States ofdeﬁnite energy 13-3 Time-dependent states 13-6 Anelectron inathree-dimensional lattice 13-7 Other states inalattice 13-8 Scattering byimperfections inthelattice 13-10 Trapping byalattice imperfection 13-12 Scattering amplitudes andbound states 13-13 CHAPTER 14. SEMIcoNDucToRs 14-1 14-2 14-3 14-4 14-5 14-6Electrons andholes insemiconductors 14-1 Impure semiconductors 14-4 TheHalleffect 14-7 Semiconductor junctions 14-8 Rectiﬁcation atasemiconductor junction 14-10 Thetransistor 14-11 CHAPTER 15.THE INDEPENDENT PARTICLE APPROxIMATIoN 15-2 Two spinwaves 15-4Symmetry 17-1 Symmetry andconservation 17-3 Theconservation laws 17-7 Polarized light 17-9 Thedistintegration oftheA°17-11 Summary oftherotation matrices 17-15 CHAPTER 18. ANGULAR MoMENTuM 18-1 18-2 18-3 18-4 18-5 18-6Electric dipole radiation 18-1 Light scattering 18-3 Theannihilation ofpositronium 18-5 Rotation matrix foranyspin18-9 Measuring anuclear spin18-13 Composition ofangular momentum 18-14 Added Note 1:Derivation oftherotation matrix 18-19 Added Note 2:Conservation ofparity inphoton emission 18-22 CHAPTER 19.THE HYDROGEN ATOM AND THE 19-1 19-2 19-3 19-4 19-5 19-6PERIoDIc TABLE Schrodinger‘s equation forthehydrogen atom 19-1 Spherically symmetric solutions 19-2 States with anangular dependence 19-6 Thegeneral solution forhydrogen 19-10 Thehydrogen wave functions 19-12 Theperiodic table 19-13 CHAPTER 20. OPERAToRs 20-1 20-2 20-3 20-4 20-5 20-6 20-7Operations andoperators 20-1 Average energies 20-3 Theaverage energy ofanatom 20-6 Theposition operator 20-8 Themomentum operator 20-9 Angular momentum 20-14 Thechange ofaverages withtime 20-15 CHAPTER 21. THE ScHRoDINGER EQUATION INACLAssIcAL 154 SP1"WP“/65 154 CoNTExT: ASEMINAR 0NSuPERcoNDucTIvITY 15-3 Independent particles 15-6 21-1 Schrodinger’s equation inamagnetic ﬁeld21-1 15-4 Thebenzeng molecule 15-7 21-2 Theequation ofcontinuity forprobabilities 21-3 21-3 Two kinds ofmomentum 21-4 21-4 Themeaning ofthewave function 21-615-5 15-6More organic chemistry 15-10 Other usesoftheapproximation 15-12 CHAPTER 16.THE DEPENDENcE orAMPLITuDEs 16-1 16-2 16-3 16-4 16-5 16-6 100NPosITIoN Amplitudes onaline16-1 Thewave function 16-5 States ofdeﬁnite momentum 16-7 Normalization ofthestates inx16-9 TheSchrodinger equation 16-11 Quantized energy levels 16-1421-5 Superconductivity 21-7 21-6 TheMeissner effect 21-8 21-7 Flux quantization 21-10 21-8 Thedynamics ofsuperconductivity 21-12 21-9 TheJosephson junction 21-14 FEYNMAN’s EPILoouE APPENDIX INDEX I Quantum Behavior 1-1Atomic mechanics “Quantum mechanics” isthedescription ofthebehavior ofmatter andlight inallitsdetails and, inparticular, ofthehappenings onanatomic scale. Things onavery small scale behave like nothing that you have anydirect experience about. They donotbehave likewaves, they donotbehave likeparticles, they do notbehave likeclouds, orbilliard balls, orweights onsprings, orlikeanything thatyouhave everseen. Newton thought thatlight wasmade upofparticles, butthen itwasdiscovered thatitbehaves likeawave. Later, however (inthebeginning ofthetwentieth century), itwas found that light didindeed sometimes behave like aparticle. Historically. theelectron, forexample, wasthought tobehave likeaparticle, and then itwasfound thatinmany respects itbehaved likeawave. Soitreally behaves likeneither. Now wehave given up. Wesay: “Itislikeneither.” There isonelucky break, however—electrons behave just like light. The quantum behavior ofatomic objects (electrons, protons, neutrons, photons, and soon)isthesame forall,they areall“particle waves,” orwhatever youwant to callthem. Sowhat welearn about theproperties ofelectrons (which weshall use forourexamples) willapply also toall“particles,” including photons oflight. The gradual accumulation ofinformation about atomic andsmall-scale be- havior during theﬁrstquarter ofthiscentury, which gave some indications about how small things dobehave, produced anincreasing confusion which wasﬁnally resolved in1926 and 1927 bySchrodinger, Heisenberg, and Born. They ﬁnally obtained aconsistent description ofthebehavior ofmatter onasmall scale. We take upthemain features ofthat description inthischapter. Because atomic behavior issounlike ordinary experience, itisvery difﬁcult togetused to,anditappears peculiar andmysterious toeveryone—both tothe novice andtotheexperienced physicist. Even theexperts donotunderstand it theway they would liketo,anditisperfectly reasonable that they should not, because allofdirect, human experience andofhuman intuition applies tolarge objects. Weknow how large objects willact,butthings onasmall scale justdo notactthat way. Sowehave tolearn about them inasortofabstract orimagi- native fashion andnotbyconnection with ourdirect experience. Inthischapter weshall tackle immediately thebasic element ofthemysterious behavior initsmost strange form. Wechoose toexamine aphenomenon which is impossible, absolutely impossible, toexplain inanyclassical way, andwhich has inittheheart ofquantum mechanics. Inreality, itcontains theonly mystery. Wecannot make themystery goaway by“explaining” how itworks. Wewilljust tellyouhowitworks. Intelling youhowitworks wewillhave toldyouabout the basic peculiarities ofallquantum mechanics. 1-2 Anexperiment with bullets Totrytounderstand thequantum behavior ofelectrons, weshall compare and contrast their behavior, inaparticular experimental setup, with themore familiar behavior ofparticles likebullets, and with thebehavior ofwaves like water waves. Weconsider ﬁrst thebehavior ofbullets intheexperimental setup shown diagrammatically inFig.l-l.Wehave amachine gunthatshoots astream ofbullets. Itisnotavery good gun, inthatitsprays thebullets (randomly) over a fairly large angular spread, asindicated intheﬁgure. Infront ofthegunwehave 1-11-1Atomic mechanics 1-2Anexperiment with bullets 1-3Anexperiment with waves 1-4Anexperiment with electrons 1-5Theinterference ofelectron waves 1-6Watching theelectrons 1-7First principles ofquantum mechanics 1-8Theuncertainty principle Note: This chapter isalmost exactly thesame asChapter 37ofVolume I Fig. l—'l. Interference experiment with bullets.awall(made ofarmor plate) thathasinittwoholes justabout bigenough toleta bullet through. Beyond thewallisabackstop (sayathick wallofwood) which will “absorb” thebullets when they hitit.Infront ofthewallwehave anobject which weshall calla“detector” ofbullets. Itmight beaboxcontaining sand. Any bullet thatenters thedetector willbestopped andaccumulated. When wewish. wecan empty thebox and count thenumber ofbullets that have been caught. The detector canbemoved back andforth (inwhat wewillcallthex-direction). With thisapparatus, wecanﬁndoutexperimentally theanswer tothequestion: “What istheprobability that abullet which passes through theholes inthewall will arrive atthebackstop atthedistance xfrom thecenter?“ First, you should realize that weshould talk about probability, because wecannot saydeﬁnitely where anyparticular bullet willgo.Abullet which happens tohitoneoftheholes may bounce offtheedges ofthehole, andmay endupanywhere atall.By“prob- ability” wemean thechance thatthebullet willarrive atthedetector. which wecan measure bycounting thenumber which arrive atthedetector inacertain time and then taking theratio ofthisnumber tothetotal number thathitthebackstop during thattime. Or,ifweassume thatthegunalways shoots atthesame rateduring the measurements, theprobability wewant isjust proportional tothenumber that reach thedetector insome standard time interval. MOVA osregibn PI '5':l x / "r;/\\\‘ E}1ér~I-~- ~—~-I -cum i/\/ //iyw ~K§§ + :PN1:watt. aocxsrop r;2=F; (0) (bl (cl Forourpresent purposes wewould liketoimagine asomewhat idealized experiment inwhich thebullets arenotrealbullets. butare1'ndesiruciil>/c bullets~ they cannot break inhalf. lnourexperiment weﬁndthatbullets always arrive in lumps, andwhen weﬁndsomething inthedetector, itisalways onewhole bullet. Iftherateatwhich themachine gunﬁresismade verylow,weﬁndthatatanygiven moment either nothing arrives. oroneandonly one—exactly one—bullet arrives atthebackstop. Also, thesizeofthelump certainly does notdepend ontherate ofﬁring ofthe gun. Weshall say:"Bullets always arrive inidentical lumps." What wemeasure with ourdetector istheprobability ofarrival otalump. And wemeas- uretheprobability asafunction ofx.Theresult ofsuch measurements with this apparatus (wehave notyetdone theexperiment. sowearereally imagining the result) areplotted inthegraph drawn inpart(c)ofFig. l—l. Inthegraph weplot theprobability totheright andxvertically, sothattheat-scale ﬁtsthediagram of theapparatus. Wecalltheprobability P,-2because thebullets may have come either through hole lorthrough hole 2.You willnotbesurprised that P12is large earthemiddle ofthegraph butgets small ifxisvery large. You may wonddlg however, whyP12hasitsmaximum value atxIO.Wecanunderstand thisfactifwedoourexperiment again after covering uphole 2.andonce more while covering uphole l.When hole 2iscovered. bullets canpass only through hole l,andwegetthecurve marked P,inpart (b)oftheﬁgure. Asyouwould expect, themaximum ot"P, occurs atthevalue ofxwhich isonastraight linewith thegunandhole l.When hole lisclosed, wegetthesymmetric curve P2drawn intheﬁgure. P2istheprobability distribution forbullets thatpass through hole 2.Comparing parts (b)and(c)ofFig. l—l,weﬁndtheimportant result that P1g:P1+P2. I-2 Theprobabilities justaddtogether. Theeﬁect withboth holes open isthesumof theeffects with each holeopen alone. Weshall callthisresult anobservation of “nointerference," forareason that youwillseelater. Somuch forbullets. They come inlumps, andtheir probability ofarrival shows nointerference. 1-3Anexperiment with waves Now wewish toconsider anexperiment with water waves. Theapparatus is shown diagrammatically inFig. 1-2. Wehave ashallow trough ofwater. Asmall object labeled the“wave source” isjiggled upanddown byamotor andmakes circular waves. Totheright ofthesource wehave again awall with twoholes, andbeyond that isasecond wall, which, tokeep things simple, isan“absorber,” sothat there isnoreﬂection ofthewaves that arrive there. This canbedone by building agradual sand “beach.” Infront ofthebeach weplace adetector which canbemoved back andforth inthex-direction, asbefore. Thedetector isnow a device which measures the“intensity” ofthewave motion. You canimagine a gadget which measures theheight ofthewave motion, butwhose scale iscalibrated inproportion tothesquare oftheactual height, sothatthereading isproportional totheintensity ofthewave. Ourdetector reads, then, inproportion totheenergy being carried bythewave—or rather, therate atwhich energy iscarried tothe detector. 3 ._gX \\\ ¢é\ \ ,3) I 1%//l/.l1,/1//// . Fig. I-2. Interference experime WALL AB$QR3ER 1|=1y,.|2 1|2=|h|~hZ|2 with water waves. I2=l“2i2 (0) (bl (cl With ourwave apparatus, theﬁrst thing tonotice isthat theintensity can have anysize. Ifthesource justmoves avery small amount, then there isjust a little bitofwave motion atthedetector. When there ismore motion atthesource, there ismore intensity atthedetector. The intensity ofthewave canhave any value atall.Wewould notsaythatthere wasany“lumpiness” inthewave intensity. Now letusmeasure thewave intensity forvarious values ofx(keeping the wave source operating always inthesame way). Wegettheinteresting-looking curve marked [12inpart (c)oftheﬁgure. Wehave already worked outhow such patterns cancome about when we studied theinterference ofelectric waves inVolume I.Inthiscase wewould observe that theoriginal wave isdiffracted attheholes, andnew circular waves spread outfrom each hole. Ifwecover onehole atatime andmeasure theintensity distribution attheabsorber weﬁndtherather simple intensity curves shown inpart (b)oftheﬁgure. I1istheintensity ofthewave from hole 1(which weﬁnd by measuring when hole 2isblocked off)andI2istheintensity ofthewave from hole 2(seen when hole lisblocked). Theintensity 1,2observed when both holes areopen iscertainly notthesum ofI1andI2.Wesaythatthere is“interference” ofthetwowaves. Atsome places (where thecurve 112hasitsmaxima) thewaves are“inphase” andthewave peaks addtogether togivealarge amplitude and,therefore, alarge intensity. We saythat thetwowaves are“interfering constructively” atsuch places. There will besuch constructive interference wherever thedistance from thedetector toone holeisawhole number ofwavelengths larger (orshorter) than thedistance from thedetector totheother hole. 1-3 Fig. l~3. Interference experiment with electrons.Atthose places where thetwowaves arrive atthedetector withaphase differ- enceof1r(where theyare“outofphase”) theresulting wave motion atthedetector willbethedifference ofthetwoamplitudes. TheWaves “interfere dcstructively,“ andwegetalowvalue forthewave intensity. Weexpect such lowvalues wherever thedistance between hole 1andthedetector isdifferent from thedistance between hole 2andthedetector byanoddnumber ofhalf-wavelengths. Thelowvalues of I12inFig. 1-2correspond totheplaces where thetwowaves interfere destructively. You willremember that thequantitative relationship between I1.lg,andI12 canbeexpressed inthefollowing way; Theinstantaneous height ofthewater wave atthedetector forthewave from hole lcanbewritten as(therealpart of)/i,c"“”, where the“amplitude” /11is,ingeneral, acomplex number. The intensity is proportional tothemean squared height or,when weusethecomplex numbers, totheabsolute value squared lh‘l2- Similarly. forhole 2theheight is/i._»c"‘°' andthe intensity isproportional tolh2l2- When both holes areopen, thewave heights addtogive theheight(/11 +/1-;)c"“" andtheintensity l/I1—l—/lglg. ()mitting the constant ofproportionality forour present purposes, theproper relations for interfering waves are 11: :l/1,12. 1,,Il/1,+/Iglg. (1.2) You willnotice thattheresult isquite different from thatobtained with bullets (Eq. l—l). lfwe expand I/11+/l2l2 weseethat |h,+hzlz=|h1|2+l/121*+Zl/illl/!2lCOS 5. (1.3) where 6isthephase difference between /11and/12.Interms oftheintensities. we could write 1,2=1,+/2+2\/ﬁljeos 5. (1.4) The lastterm in(1.4) isthe“interference term.” Somuch forwater waves. The intensity canhave anyvalue, anditshows interference. 1-4Anexperiment withelectrons Now weimagine asimilar experiment with electrons. Itisshown diagram- matically inFig. 1-3. Wemake anelectron gunwhich consists ofatungsten wire heated byanelectric current andsurrounded byametal boxwith ahole init.If thewire isatanegative voltage with respect tothebox, electrons emitted bythe wire willbeaccelerated toward thewalls and some willpass through thehole. Alltheelectrons which come outofthegunwillhave (nearly) thesame energy. lnfront ofthegunisagain awall (just athin metal plate) with twoholes init. Beyond thewall isanother plate which willserve asa“backstop.” lnfront ofthe backstop weplace amovable detector. Thedetector might beageigcr counter or. perhaps better, anelectron multiplier, which isconnected toaloudspeaker. Weshould sayright away that youshould nottrytosetupthisexperiment (asyoucould have done with thetwowehave already described). This experiment DETECTOR *11.32-iI 1-35 : ELECTRTW/ ’4int\Ill N_all _'U>< __Itan-an,1,.1,:3’>""/\f{:13 watt BACKSTOP la,+az|‘ ~'U (ﬂl (bl (Cl l-4 hasnever been done injustthisway. Thetrouble isthattheapparatus would have tobemade onanimpossibly small scale toshow theeffects weareinterested in. Wearedoing a“thought experiment,” which wehave chosen because itiseasy to think about. Weknow theresults that would beobtained because there aremany experiments that have been done, inwhich thescale and theproportions have been chosen toshow theeﬁects weshall describe. The ﬁrst thing wenotice with ourelectron experiment isthat wehear sharp “clicks” from thedetector (that is,from theloudspeaker). And all“clicks" are thesame. There areno“half-clicks.” Wewould alsonotice thatthe“clicks” come very erratically. Something like: click .....click-click. ..click ........click ....click-click ......click ..., etc., just asyou have, nodoubt, heard ageiger counter operating. lfwecount theclicks which arrive inasufﬁciently long time——say formany minutesfand then count again foranother equal period, weﬁndthat thetwonumbers arevery nearly thesame. Sowecanspeak oftheaverage rateatwhich theclicks areheard (so-and-so-many clicks perminute ontheaverage). Aswemove thedetector around, therateatwhich theclicks appear isfaster orslower, butthesize(loudness) ofeach click isalways thesame. Ifwelower the temperature ofthewire inthegun, therate ofclicking slows down, butstilleach click sounds thesame. Wewould notice alsothat ifweputtwoseparate detectors atthebackstop, oneortheother would click, butnever both atonce. (Except that once inawhile, ifthere were twoclicks very close together intime, ourearmight notsense theseparation.) Weconclude, therefore, that whatever arrives atthe backstop arrives in“lumps.” Allthe“lumps” arethesame size: only whole “lumps” arrive, and they arrive oneatatime atthebackstop. Weshall say: “Electrons always arrive inidentical lumps.” Just asforourexperiment with bullets, wecannow proceed toﬁnd experi- mentally theanswer tothequestion: “What istherelative probability that an electron ‘lump’ willarrive atthebackstop atvarious distances xfrom thecenter?” Asbefore, weobtain therelative probability byobserving therateofclicks, holding theoperation ofthegunconstant. The probability that lumps willarrive ata particular xisproportional totheaverage rateofclicks atthat x. The result ‘bfourexperiment istheinteresting curve marked P12 inpart (c) ofFig.I-3. Yes! That isthewayelectrons go. 1-5Theinterference ofelectron waves Now letustrytoanalyze thecurve ofFig. l—3toseewhether wecanunder- stand thebehavior oftheelectrons. Theﬁrstthing wewould sayisthat since they come inlumps, each lump, which Wemay aswellcallanelectron, hascome either through hole 1orthrough hole 2.Letuswrite thisintheform ofa“Proposition”: Proposition A:Each electron either goes through hole loritgoes through hole 2. Assuming Propositon A,allelectrons that arrive atthebackstop canbedi- vided intotwoclasses: (I)those thatcome through hole l,and(2)those thatcome through hole 2.Soourobserved curve must bethesum oftheeffects oftheelec- trons which come through hole landtheelectrons which come through hole 2. Letuscheck thisideabyexperiment. First, wewillmake ameasurement forthose electrons that come through hole l.Weblock oilhole 2andmake ourcounts of theclicks from thedetector. From theclicking rate, wegetP1. Theresult ofthe measurement isshown bythecurve marked P1inpart (b)ofFig. 1-3. Theresult seems quite reasonable. lnasimilar way, wemeasure P2,theprobability distribu- tion fortheelectrons that come through hole 2.The result ofthismeasurement isalsodrawn intheﬁgure. Theresult P12obtained with both holes open isclearly notthesum ofP1and P2,theprobabilities foreach hole alone. Inanalogy with ourwater-wave experi- 1-5 ment, wesay: “There isinterference.” Forelectrons: P12 ¢P1+P2. (1.5) How cansuch aninterference come about‘? Perhaps weshould say: “Well, that means, presumably, that itisnottruethat thelumps goeither through hole 1orhole 2,because ifthey did,theprobabilities should add. Perhaps they goina more complicated way. They split inhalfand..”Butno! They cannot, they always arrive inlumps ...“Well, perhaps some ofthem gothrough l,andthen they goaround through 2,andthen around afewmore times, orbysortie other complicated path .then byclosing hole 2,wechanged thechance that anelec- tron that starred outthrough hole lwould ﬁnally gettothebackstop "Bttt notice! There aresome points atwhich very fewelectrons arrive when both holes areopen, butwhich receive many electrons ifweclose onehole, soclosing one hole fnereased thenumber from theother. Notice, however, that atthecenter ofthe pattern, P12ismore than twice aslarge asP1+P2.Itisasthough closing onehole decreased thenumber ofelectrons which come through theother hole. ltseems hard toexplain bot/1 effects byproposing that theelectrons travel in complicated paths. Itisallquite mysterious. And themore youlook atitthemore mysterious itseems. Many ideas have been concocted totrytoexplain thecurve forP12in terms ofindividual electrons going around incomplicated ways through theholes. None ofthem hassucceeded. None ofthem cangettheright cttrve forP12in terms ofP1andP2. Yet, surprisingly enough, thenmt/zemuties forrelating P1andP210 P12is extremely simple. For P12isjust likethecurve I12ofFig. 1-2, and t/tut was simple. What isgoing onatthebackstop canbedescribed bytwocomplex numbers thatwecancall¢1and¢2(they arefunctions ofx,ofcourse). Theabsolute square of¢1 gives thectTect with only hole lopen. That is,P1=l¢1I2 Theetlect with only hole 2open isgiven by¢>2inthesame way. That is,P2:l¢2l2 And the combined effect ofthetwo holes isjust P12 =|¢>1+(1)2 The /nu!/re/m1lt't'.s' isthesatne asthat wehadforthewater waves! (ltishard toseehow onecould getsuch asimple result from acomplicated game ofelectrons going back andforth through theplate onsome strange trajectory.) Weconclude thefollowing: Theelectrons arrive inlumps. likeparticles, and theprobability ofarrival ofthese lumps isdistributed likethedistribution of intensity ofawave. Itisinthissense that anelectron behaves “sometimes likea particle andsometimes likeawave." Incidentally, when wewere dealing with classical waves wedeﬁned thein- tensity asthemean over time ofthesquare ofthewave amplitude, andweused complex numbers asamathematical trick tosimplify theanalysis. Butinquantum mechanics itturns outthat theamplitudes ml/.\'{ berepresented bycomplex num- bers. Therealparts alone willnotdo.That isatechnical point, forthemoment, because theformulas lookjust thesame. Since theprobability ofarrival through both holes isgiven sosimply, although itisnotequal to(P1+P2),that isreally allthere istosay. Butthere area large number ofsubtleties involved inthefactthat nature does work thisway. We would liketoillustrate some ofthese subtleties foryounow. First, since thenttm— berthatarrives ataparticular point isnotequal tothenumber thatarrives through lplus thenumber that arrives through 2,aswewould have concluded from Proposition A,undoubtedly weshould conclude that Pro/Josirioiz Ais_fZt/.\"e. ltis nottruethat theelectrons goeit/zer through hole 1orhole 2.Butthatconclusion canbetested byanother experiment. l-6Watching theelectrons Weshall now trythefollowing experiment. Toourelectron apparatus we addavery strong light source, placed behind thewall andbetween thetwoholes, asshown inFig. 1-4. Weknow that electric charges scatter light. Sowhen an 1-6 X X I / Pt l Fl: souncs /- 3 r __l___J \-_\\ stzcrnou ‘curtlLIGHT § \\\\\\_ﬂ////// %=W+% (O) (bl (cl electron passes, however itdoes pass, onitswaytothedetector, itwillscatter some light tooureye,andwecanseewhere theelectron goes. If,forinstance, anelectron were totake thepath viahole 2that issketched inFig. 1-4,weshould seeaﬂash oflight coming from thevicinity oftheplace marked Aintheﬁgure. lfanelectron passes through hole l,wewould expect toseeaﬁash from thevicinity oftheupper hole. Ifitshould happen that wegetlight from both places atthesame time, because theelectron divides inhalf. ..Letusjustdotheexperiment! Here iswhat wesee:every time that wehear a“click” from ourelectron de- tector (atthebackstop), wealsoseeaﬁash oflight either near hole lornear hole 2.butnever both atonce! And weobserve thesame result nomatter where weput thedetector. From thisobservation weconclude thatwhen welook attheelectrons weﬁndthattheelectrons goeither through onehole ortheother. Experimentally, Proposition Aisnecessarily true. What. then, iswrong with ourargument against Proposition A? Why isn’t P12just equal toP1+P2? Back toexperiment! Letuskeep track ofthe electrons andﬁndoutwhat they aredoing. Foreach position (x-location) ofthedetector wewillcount theelectrons thatarrive andalsokeep track ofwhich holetheywent through, bywatching fortheﬂashes. Wecankeep track ofthings this way: Whenever weheara“click” wewillputacount inColumn 1ifweseetheﬂash near hole l,andifweseetheflash near hole 2,wewillrecord acount inColumn 2. Every electron which arrives isrecorded inoneoftwoclasses: those which come through 1andthose which come through 2.From thenumber recorded inColumn 1wegettheprobability P1that anelectron willarrive atthedetector viahole l; andfrom thenumber recorded inColumn 2wegetP2,theprobability that an electron willarrive atthedetector viahole 2.Ifwenow repeat such ameasurement formany values ofx,wegetthecurves forP1andP2shown inpart (b)ofFig. 1-4. Well, that isnottoosurprising! WegetforP1something quite similar to what wegotbefore forP1byblocking offhole 2;andP2issimilar towhat wegot byblocking hole l.Sothere isnotanycomplicated business likegoing through both holes. When wewatch them, theelectrons come through justaswewould expect them tocome through. Whether theholes areclosed oropen, those which weseecome through hole 1aredistributed inthesame waywhether hole 2isopen orclosed. Butwait! What dowehave nowforthetotal probability, theprobability that anelectron willarrive atthedetector byanyroute? Wealready have thatinforma- tion. Wejust pretend that wenever looked atthelight flashes, andwelump to- gether thedetector clicks which wehave separated into thetwocolumns. We must justaddthenumbers. Fortheprobability that anelectron willarrive atthe backstop bypassing through either hole, wedoﬁnd P12 =P1—l—P2. That is, although wesucceeded inwatching which hole ourelectrons come through, we nolonger gettheoldinterference curve P12, butanew one, P12, showing no interference! Ifweturnoutthelight P12isrestored. Wemust conclude that when welook attheelectrons thedistribution ofthem onthescreen isdifferent than when wedonotlook. Perhaps itisturning onour light source that disturbs things? ltmust bethat theelectrons arevery delicate, andthelight, when itscatters offtheelectrons, gives them ajoltthatchanges their 1-7Fig. l~4. Adifferent electron periment. motion. Weknow that theelectric ﬁeld ofthelight acting onacharge willexert aforce onit.Soperhaps weshould expect themotion tobechanged. Anyway, thelight exerts abiginﬂuence ontheelectrons. Bytrying to“watch” theelectrons wehave changed their motions. That is,thejoltgiven totheelectron when the photon isscattered byitissuch astochange theelectron’s motion enough sothat ifitmight have gone towhere P12wasatamaximum itwillinstead landwhere P12wasaminimum; that iswhy wenolonger seethewavy interference effects. You may bethinking: “Don’t usesuch abright source! Turn thebrightness down! Thelight waves willthen beweaker andwillnotdisturb theelectrons so much. Surely, bymaking thelight dimmer and dimmer, eventually thewave willbeweak enough that itwillhave anegligible effect.“ O.K. Let’s tryit.The ﬁrst thing weobserve isthat theﬂashes oflight scattered from theelectrons as they pass bydoes notgetweaker. Itisalways thesame-sizedﬂash. Theonly thing that happens asthelight ismade dimmer isthat sometimes wehear a“click“ frotn thedetector butseenoflash atall.Theelectron hasgone bywithout being “seen.” What weareobserving isthat light alsoactslikeelectrons, weknew that itwas“wavy,” butnow weﬁndthat itisalso “lumpy.” ltalways arrives——or is scattered~in lumps that wecall"photons." Asweturn down theintensity of thelight source wedonotchange thesizeofthephotons, only therateatwhich they areemitted. That explains why, when oursource isdim, some electrons get bywithout being seen. There didnothappen tobeaphoton around atthetime theelectron went through. This isallalittle discouraging. lfitistruethatwhenever we“see” theelectron weseethesame-sized ﬂash. then those electrons weseearealways thedisturbed ones. Letustrytheexperiment with adimlight anyway. Now whenever wehear aclick inthedetector wewillkeep acount inthree columns: inColumn (l)those electrons seen byhole l,inColumn (2)those electrons seen byhole 2.and in Column (3)those electrons notseen atall.When wework upourdata (computing theprobabilities) weﬁndthese results: Those “seen byhole 1”have adistribution likeP1;those “seen byhole 2”have adistribution likeP1(sothat those “seen by either hole lor2”have adistribution likeP12): andthose “notseenatall“havea “wavy” distribution just likeP12ofFig. 1-3! lftheelectrons arenotseen, we have interference! That isunderstandable. When wedonotseetheelectron, nophoton disturbs it,andwhen wedoseeit,aphoton hasdisturbed it.There isalways thesame amount ofdisturbance because thelight photons allproduce thesame-sized effects and theeffect ofthephotons being scattered isenough tosmear outanyinter- ference effect. Isthere notsome way wecanseetheelectrons without disturbing them? Welearned inanearlier chapter that themomentum carried bya“photon” isinversely proportional toitswavelength (pIh/A). Certainly thejoltgiven totheelectron when thephoton isscattered toward ottreyedepends onthe momentum that photon carries. Aha! lfwewant todisturb theelectrons only slightly weshould nothave lowered theintensity ofthelight, weshould have lowered itsfrequency (the same asincreasing itswavelength). Letususelight of aredder color. Wecould even useinfrared light, orradiowaves (like radar), and “see” where theelectron went with thehelp ofsome equipment that can“sec” light ofthese longer wavelengths. Ifweuse“gentler” light perhaps wecanavoid disturbing theelectrons somuch. Letustrytheexperiment with longer waves. Weshall keep repeating ourex- periment, each time with light ofalonger wavelength. Atiirst, nothing seems to change. Theresults arethesame. Then aterrible thing happens. You remember thatwhen wediscussed themicroscope wepointed outthat, duetothewave nature ofthelight, there isalimitation onhow close twospots canbeandstillbeseen astwoseparate spots. This distance isoftheorder ofthewavelength oflight. So now, when wemake thewavelength longer than thedistance between ourholes, weseeabigfuzzy ﬂash when thelight isscattered bytheelectrons. Wecanno longer tellwhich hole theelectron went through! Wejust know itwent somewhere! And itisjust with light ofthiscolor thatweﬁndthatthejolts given totheelectron 1-8 aresmall enough sothatP12begins tolook likeP12—that webegin togetsome interference effect. Anditisonlyforwavelengths much longer than theseparation ofthetwoholes (when wehave nochance atalloftelling where theelectron went) that thedisturbance duetothelight gets suﬂiciently small that weagain getthe curve P12shown inFig.I-3. Inourexperiment weﬁnd that itisimpossible toarrange thelight insuch a waythatonecantellwhich holetheelectron went through, andatthesatne time notdisturb thepattern. Itwassuggested byHeisenberg that thethen new laws of nature could only beconsistent ifthere were sotne basic limitation onourexperi- mental capabilities notpreviously recognized. Heproposed, asageneral principle, hisuncertainty principle, which wecanstate interms ofourexperiment asfollows: “Itisimpossible todesign anapparatus todetermine which hole theelectron passes through, thatwillnotatthesame timedisturb theelectrons enough todestroy the interference pattern.” Ifanapparatus iscapable ofdetermining which hole the electron goesthrough, itcannot besodelicate thatitdoes notdisturb thepattern in anessential way. Noonehasever found (oreven thought of)away around the uncertainty principle. Sowemust assume thatitdescribes abasic characteristic ofnature. Thecomplete theory ofquantum mechanics which wenow usetodescribe atoms and, infact, allmatter, depends onthecorrectness oftheuncertainty prin- ciple. Since quantum mechanics issuch asuccessful theory, ourbelief inthe uncertainty principle isreinforced. Butifawayto“beat” theuncertainty principle were ever discovered, quantum mechanics would give inconsistent results and would have tobediscarded asavalid theory ofnature. “Well,” yousay,“what about Proposition A? Isittrue, orisitnottrue, thattheelectron either goes through hole Ioritgoes through hole 2?” The only answer thatcanbegiven isthatwehave found from experiment thatthere isa certain special way that wehave tothink inorder that wedonotgetinto incon- sistencies. What wemust say(toavoid making wrong predictions) isthefollowing. Ifonelooks attheholes or,more accurately, ifonehasapiece ofapparatus which iscapable ofdetermining whether theelectrons gothrough hole Iorhole2,then onecansaythat itgoes either through hole lorhole 2.But, when onedoes not trytotellwhich waytheelectron goes, when there isnothing intheexperiment to disturb theelectrons, then onemay notsaythat anelectron goes either through hole 1orhole2.Ifonedoes saythat, andstarts tomake anydeductions from the statement, hewill make errors intheanalysis. This isthelogical tightrope on which wemust walk ifwewish todescribe nature successfully. Ifthemotion ofallmatter—as wellaselectrons—must bedescribed interms ofwaves, what about thebullets inourﬁrst experiment? Why didn’t weseean interference pattern there? Itturns outthatforthebullets thewavelengths were so tiny that theinterference patterns became very ﬁne. Soﬁne, infact, that with any detector ofﬁnite sizeonecould notdistinguish theseparate maxima andminima. What wesawwasonly akind ofaverage, which istheclassical curve. InFig. I-5 wehave tried toindicate schematically what happens with large-scale objects. Part (a)oftheﬁgure shows theprobability distribution onemight predict for bullets, using quantum mechanics. Therapid wiggles aresupposed torepresent theinterference pattern onegetsforwaves ofvery short wavelength. Any physical detector, however, straddles several wiggles oftheprobability curve, sothatthe measurements show thesmooth curve drawn inpart (b)oftheﬁgure. 1-'7First principles ofquantum mechanics Wewill‘now write asummary ofthemain conclusions ofourexperiments. Wewill, however, puttheresults inaform which makes them trueforageneral class ofsuch experiments. Wecanwrite oursummary more simply ifweﬁrst deﬁne an“ideal experiment” asoneinwhich there arenouncertain external inﬂuences, i.e.,nojiggling orother things going onthat wecannot take into ac- 1-9x E If? F|’2(smoothed) to) (bl Fig. l—5. Interference pctttern with bullets: (0) uctuol (schematic), lb)ob- served. count. Wewould bequite precise ifwesaid: “An ideal experiment isoneinwhich alloftheinitial andﬁnal conditions oftheexperiment arecompletely speciﬁed." What wewillcall“anevent” is.ingeneral. justaspeciﬁc setofinitial andﬁnal conditions. (For example: “anelectron leaves thegun, arrives atthedetector, and nothing elsehappens.”) Now foroursummary. SUMMARY (1)The probability ofanevent inanideal experiment isgiven bythesquare of theabsolute value ofacomplex number 4;which iscalled theprobability amplitude: P=probability, ¢=probability amplitude, (1.6) PW (2)When anevent canoccur inseveral alternative ways, theprobability ampli- tude fortheevent isthesum oftheprobability amplitudes foreach way considered separately. There isinterference: 05=¢1+ ¢2, PIl¢1'l' ¢2l2 (1-7) (3)Ifan experiment isperformed which iscapable ofdetermining whether oneor another alternative isactually taken, theprobability oftheevent isthesum oftheprobabilities foreach alternative. Theinterference islost: P=P1—l—P2. (l.8) Onemight stillliketoask: “How does itwork? What isthemachinery behind thelaw?” Noonehasfound anymachinery behind thelaw. Noonecan“explain” anymore than wehave just“explained.” Noonewillgiveyouanydeeper repre- sentation ofthesituation. Wehave noideas about amore basic mechanism from which these results canbededuced. Wewould liketoemphasize avery important difference between ela.s'sica/ and quantum mechanics. Wehave been talking about theprobability that anelectron willarrive inagiven circumstance. Wehave implied that inourexperimental arrangement (oreven inthebest possible one) itwould beimpossible topredict exactly what would happen. Wecanonly predict theodds! This would mean. if itwere true, that physics hasgiven upontheproblem oftrying topredict exactly what willhappen inadeﬁnite circumstance. Yes! physics hasgiven up. Wedo notknow howtopredict what would happen inagiven circumstance. andwebelieve now that itisimpossible—that theonly thing that canbepredicted istheprob- ability ofdiﬁerent events. ltmust berecognized that thisisaretrenchment inour earlier ideal ofunderstanding nature. Itmay beabackward step, butnoone hasseen awaytoavoid it. Wemake now afewremarks onasuggestion that hassometimes been made totrytoavoid thedescription wehave given: “Perhaps theelectron hassome kind ofinternal works~some inner variables—that wedonotyetknow about. Perhaps that iswhy wecannot predict what willhappen. Ifwecould look more closely at theelectron, wecould beable totellwhere itwould endup.“ Sofarasweknow, thatisimpossible. Wewould stillbeindifficulty. Suppose wewere toassume that inside theelectron there issome kind ofmachinery that determines where itis going toendup. That machine must also determine which hole itisgoing togo through onitsway. Butwemust notforget thatwhat isinside theelectron should notbedependent onwhat wedo,andinparticular upon whether weopen orclose oneoftheholes. Soifanelectron, before itstarts, hasalready made upitsmind (a)which hole itisgoing touse,and(b)where itisgoing toland, weshould lll1(l P1forthose electrons that have chosen/hole l,P2forthose thathave chosen hole 2,andnecessarily thesum P1+P2forthose that arrive through thetwoholes. There seems tobenowayaround this. Butwehave veriﬁed experimentally that that isnotthecase. And noonehasﬁgured away outofthispuzzle. Soatthe I-10 present time wemust limit ourselves tocomputing probabilities. Wesay“atthe present time,” butwesuspect very strongly thatitissomething thatwillbewith usforever—that itisimpossible tobeat thatpuzzle—that thisisthewaynature really is. 1-8Theuncertainty principle This isthewayHeisenberg stated theuncertainty principle originally: lfyou make themeasurement onanyobject, andyoucandetermine thex-component of itsmomentum with anuncertainty Ap,youcannot, atthesame time, know its x-position more accurately than Ax=h/Ap, where hisadeﬁnite ﬁxed number given bynature. Itiscalled “Planck’s constant,” andisapproximately 6.63 X IOT34 joule-seconds. The uncertainties intheposition and momentum ofa particle atanyinstant must have their product greater than Planck’s constant. This isaspecial case oftheuncertainty principle that was stated above more generally. Themore general statement wasthatonecannot design equipment in anyway todetermine which oftwo alternatives istaken, without, atthesame time, destroying thepattern ofinterference. Letusshow foroneparticular case that thekind ofrelation given byHeisen- bergmust betrueinorder tokeep from getting intotrouble. Weimagine amodiﬁ- cation oftheexperiment ofFig. 1-3,inwhich thewall with theholes consists ofa plate mounted onrollers sothatitcanmove freely upanddown (inthex-direction), asshown inFig. 1-6. Bywatching themotion oftheplate carefully wecantryto tellwhich holeanelectron goesthrough. Imagine what happens when thedetector isplaced atx=0.Wewould expect that anelectron which passes through hole 1 must bedeﬂected downward bytheplate toreach thedetector. Since thevertical component oftheelectron momentum ischanged, theplate must recoil with an equal momentum intheopposite direction. Theplate willgetanupward kick. Iftheelectron goes through thelower hole, theplate should feeladownward kick. Itisclear thatforevery position ofthedetector, themomentum received bythe plate willhave adifferent value foratraversal viahole lthan foratraversal via hole2.SolWithout disturbing theelectrons atall,butjustbywatching theplate, wecantellwhich path theelectron used. Now inorder todothisitisnecessary toknow what themomentum ofthe screen is,before theelectron goes through. Sowhen wemeasure themomentum after theelectron goesby,wecanﬁgure outhowmuch theplate’s momentum has changed. Butremember, according totheuncertainty principle wecannot atthe same time know theposition oftheplate withanarbitrary accuracy. Butifwedo notknow exactly where theplate is,wecannot sayprecisely where thetwoholes are. They willbeinadifferent place forevery electron thatgoesthrough. This means that thecenter ofourinterference pattern willhave adiﬂ'erent location foreach electron. Thewiggles oftheinterference pattern willbesmeared out.Weshall show quantitatively inthenext chapter that ifwedetermine themomentum oftheplate sufficiently accurately todetermine from therecoil measurement which hole was used, then theuncertainty inthex-position oftheplate will, according totheun- certainty principle, beenough toshift thepattern observed atthedetector upand down inthex-direction about thedistance from amaximum toitsnearest minimum. Such arandom shiftisjustenough tosmear outthepattern sothatnointerference isobserved. Theuncertainty principle “protects” quantum mechanics. Heisenberg recog- nized thatifitwere possible tomeasure themomentum andtheposition simultane- ously with agreater accuracy, thequantum mechanics would collapse. Sohe proposed that itmust beimpossible. Then people satdown andtried toﬁgure out ways ofdoing it,andnobody could ﬁgure outawaytomeasure theposition and themomentum ofanything——a screen, anelectron, abilliard ball, anything—with anygreater accuracy. Quantum mechanics maintains itsperilous butstillcorrect existence. 1-11ROLLERS P he kl /1 I ll: K ,j/'// \\\ IITECTOR-f:-/ I \~\ .__________ : 6 \\\\\\ T Z Z Z ELECTRON \—‘sunll[Al v"Aa'°5NDTION FREEl RCLLERS WALL BACKSTOP Fig. l—6. An experiment inwhich therecoil ofthewall ismeasured. 2 The Relation ofWave and Particle Viewpoints 2-1Probability wave amplitudes Inthis chapter weshall discuss therelationship ofthewave and particle viewpoints. Wealready know, from thelastchapter, thatneither thewave view- point northeparticle viewpoint iscorrect. Wewould always liketopresent things accurately, oratleast precisely enough thattheywillnothave tobechanged when welearn more—it may beextended, butitwillnotbechanged! Butwhen wetry totalkabout thewave picture ortheparticle picture, both areapproximate, and both willchange. Therefore what welearn inthischapter willnotbeaccurate ina certain sense; wewilldealwith some half-intuitive arguments which willbemade more precise later. Butcertain things willbechanged alittle bitwhen weinterpret them correctly inquantum mechanics. Wearedoing thissothatyoucanhave some qualitative feeling forsome quantum phenomena before wegetinto the mathematical details ofquantum mechanics. Furthermore, allourexperiences arewith waves and with particles, andsoitisrather handy tousethewave and particle ideas togetsome understanding ofwhat happens ingiven circumstances before weknow thecomplete mathematics ofthequantum-mechanical amplitudes. Weshall trytoindicate theweakest places aswegoalong. butmost ofitisvery nearly correct-it isjustamatter ofinterpretation. First ofall,weknow thatthenewwayofrepresenting theworld inquantum mechanics thenew framework——is togive anamplitude forevery event that can occur, andiftheevent involves thereception ofoneparticle, then wecangivethe amplitude toﬁnd that oneparticle atdifferent places andatdifferent times. The probability ofﬁnding theparticle isthen proportional totheabsolute square of theamplitude. Ingeneral, theamplitude toﬁnd aparticle indifferent places at different times varies with position andtime. Insome special case itcanbethat theamplitude varies sinusoidally inspace andtime likee“‘°'T""), where risthevector position from some origin. (Do not forget that these amplitudes arecomplex numbers, notreal numbers.) Such an amplitude varies according toadeﬁnite frequency wandwave number k.Then itturns outthat thiscorresponds toaclassical limiting situation where wewould have believed thatwehave aparticle whose energy Ewasknown andisrelated to thefrequency by E=hw, (2.1) andwhose momentum pisalsoknown andisrelated tothewave number by p=hk. (2.2) (The symbol hrepresents thenumber hdivided by21r;h=h/21r.) This means that theidea ofaparticle islimited. The idea ofaparticle—its location, itsmomentum, etc.—which weusesomuch, isincertain ways unsatis- factory. Forinstance, ifanamplitude toﬁnd aparticle atdifferent places isgiven bye“‘“"""'), whose absolute square isaconstant, thatwould mean thattheprob- ability ofﬁnding aparticle isthesame atallpoints. That means wedonotknow where itis-it canbeanywhere—there isagreat uncertainty initslocation. Ontheother hand, iftheposition ofaparticle ismore orlesswell known and wecanpredict itfairly accurately, then theprobability ofﬁnding itindifferent places must beconﬁnpd toacertain region, whose length wecallAx. Outside this region, theprobability iszero. Now thisprobability istheabsolute square ofan amplitude, and iftheabsolute square iszero, theamplitude isalso zero, sothat 2-12-1 Probability wave amplitudes 2-2 Measurement ofposition and momentum 2-3 Crystal diffraction 2-4 The sizeofanatom 2-5Energy levels 2-6Philosophical implications Note: This chapter isalmost exactly thesame asChapter 38ofVolume 1. <——--— AX - Fig. 2-1. Awove pocket oflength Ax. C __> 1’,/.l§9:____._.IB‘ Fig. 2-2. Diffraction of pcirticles passing through 0slit.wehave awave train whose length isAx(Fig. 2-1), andthewavelength (the distance between nodes ofthewaves inthetrain) ofthatwave train iswhat corre- sponds totheparticle momentum. _ Here weencounter astrange thing about waves; avery simple thing which has nothing todowith quantum mechanics strictly. Itissomething that anybody who works with waves, even ifheknows noquantum mechanics, knows: namely, wecannot deﬁne aunique wavelength forashort wave train. Such awave train does nothave adeﬁnite wavelength; there isanindeﬁniteness inthewave number that isrelated totheﬁnite length ofthetrain, and thus there isanindeﬁniteness in themomentum. 2-2 Measurement ofposition andmomentum Letusconsider twoexamples ofthisidea—to seethereason that there isan uncertainty intheposition and/or themomentum, ifquantum mechanics isright. Wehave alsoseen before that ifthere were notsuch athing—if itwere possible to measure theposition andthemomentum ofanything simultaneously—we would have aparadox; itisfortunate that wedonothave such aparadox, andthefact that such anuncertainty comes naturally from thewave picture shows thatevery- thing ismutually consistent. Here isoneexample which shows therelationship between theposition and themomentum inacircumstance that iseasy tounderstand. Suppose wehave a single slit,andparticles arecoming from very faraway with acertain energy—so that they areallcoming essentially horizontally (Fig. 2-2). Wearegoing to concentrate onthevertical components ofmomentum. Allofthese particles have acertain horizontal momentum p0,say, inaclassical sense. So,intheclassical sense, thevertical momentum pg,before theparticle goes through thehole, is deﬁnitely known. Theparticle ismoving neither upnordown. because itcame from asource that isfaraway—and sothevertical momentum isofcourse zero. But now letussuppose that itgoes through ahole whose width isB.Then after ithas come outthrough thehole, weknow theposition vertically—the y-position—with considerable accuracy—namely iB.T That is,theuncertainty inposition, Ay,is oforder B.Now wemight also want tosay, since weknown themomentum is absolutely horizontal, that Apyiszero; butthat iswrong. Weonce knew themo- mentum washorizontal, butwedonotknow itanymore. Before theparticles passed through thehole, wedidnotknow their vertical positions. Now that we have found thevertical position byhaving theparticle come through thehole, we have lostourinformation onthevertical momentum! Why? According tothe wave theory, there isaspreading out, ordiffraction, ofthewaves after they go through theslit,just asforlight. Therefore there isacertain probability that particles coming outoftheslitarenotcomingl exactly straight. The pattern is spread outbythediffraction effect, andtheangle ofspread, which wecandeﬁne astheangle oftheﬁrstminimum, isameasure oftheuncertainty intheﬁnal angle. How does thepattern become spread? Tosayitisspread means that there is some chance fortheparticle tobemoving upordown, thatis,tohave acomponent ofmomentum upordown. Wesaychance andparticle because wecandetect this diffraction pattern with aparticle counter, and when thecounter receives the particle, sayatCinFig. 2-2, itreceives theentire particle, sothat, inaclassical sense, theparticle hasavertical momentum, inorder togetfrom theslituptoC. Togetarough idea ofthespread ofthemomentum, thevertical momentum pyhasaspread which isequal topt)A0,where p(,isthehorizontal momentum. And how bigisA6inthespread-out pattern‘? Weknow that theﬁrst minimum occurs atanangle A0such that thewaves from oneedge oftheslithave totravel onewavelength farther than thewaves from theother side—we worked thatout before (Chapter 30ofVol. I).Therefore A6isA/B, andsoA/2,,inthisexperiment ispox/B. Note thatifwemake Bsmaller andmake amore accurate measurement '1'More precisely, theerror inourknowledge ofyis=B/2. Butwearenow only in- terested inthegeneral idea, sowewon’t worry about factors of2. 2-2 oftheposition oftheparticle, thediffraction pattern getswider. Sothenarrower wemake theslit,thewider thepattern gets,andthemore isthelikelihood thatwe would ﬁndthattheparticle hassidewise momentum. Thus theuncertainty inthe vertical momentum isinversely proportional totheuncertainty ofy.Infact, we seethattheproduct ofthetwoisequal topox. But>\isthewavelength and[)0is themomentum, andinaccordance withquantum mechanics, thewavelength times themomentum isPlanck’s constant h.Soweobtain therulethattheuncertainties inthevertical momentum andinthevertical position have aproduct oftheorder h: AyAp,~h. (2.3) Wecannot prepare asystem inwhich weknow thevertical position ofaparticle andcanpredict how itwillmove vertically with greater certainty than given by (2.3). That is,theuncertainty inthevertical momentum must exceed h/Ay, where Ayistheuncertainty inourknowledge oftheposition. Sometimes people sayquantum mechanics isallwrong. When theparticle arrived from theleft,itsvertical momentum waszero. And now that ithasgone through theslit,itsposition isknown. Both position andmomentum seem to beknown with arbitrary accuracy. Itisquite true that wecanreceive aparticle, and onreception determine what itsposition isand what itsmomentum would have hadtohave been tohave gotten there. That istrue, butthatisnotwhat the uncertainty relation (2.3) refers to. Equation (2.3) refers tothepredictability ofasituation, notremarks about thepast. Itdoes nogood tosay“Iknew what themomentum wasbefore itwent through theslit,andnow Iknow theposition," because now themomentum knowledge islost. Thefactthat itwent through the slitnolonger permits ustopredict thevertical momentum. Wearetalking about apredictive theory, notjustmeasurements after thefact. Sowemust talkabout what wecanpredict. Now letustake thething theother wayaround. Letustake another example ofthesame phenomenon, alittle more quantitatively. Intheprevious example wemeasured themomentum byaclassical method. Namely, weconsidered the direction andthevelocity andtheangles, etc., sowegotthemomentum byclassical analysis. Butsince momentum isrelated towave number, there exists innature stillanother way tomeasure themomentum ofaparticle—photon orotherwise- which hasnoclassical analog, because itusesEq.(2.2). Wemeasure thewave- lengths ofthewaves. Letustrytomeasure momentum inthisway. Suppose wehave agrating with alarge number oflines (Fig. 2-3), andsend abeam ofparticles atthegrating. Wehave often discussed thisproblem: ifthe particles have adeﬁnite momentum, then wegetavery sharp pattern inacertain direction, because oftheinterference. And wehave also talked about how accu- rately wecandetermine that momentum, that istosay,what theresolving power ofsuch agrating is.Rather than derive itagain, werefer toChapter 30ofVolume I,where wefound that therelative uncertainty inthewavelength that canbe measured with agiven grating is1/Nm, where Nisthenumber oflines onthegrat- ingandmistheorder ofthediffraction pattern. That is, AA/)\ =1/Nm. (2.4) Now formula (2.4) canberewritten as xx/x2 =1/Nmx =1/L, (2.5) where Listhedistance shown inFig. 2-3. This distance isthedifference between thetotal distance that theparticle orwave orwhatever itishastotravel ifitis reﬂected from thebottom ofthegrating. andthedistance that ithastotravel if itisreﬂected from thetopofthegrating. That is,thewaves which form thediffrac- tionpattern arewaves which come from different parts ofthegrating. Theﬁrst onesthatarrive come from thebottom endofthegrating, from thebeginning of thewave train, andtherestofthem come from laterparts ofthewave train, coming fromdifferent parts ofthegrating, untilthelastoneﬁnally arrives, andthatinvolves apoint inthewave train adistance Lbehind theﬁrstpoint. Soinorder thatwe 2-3‘\NrnX=L <-- Q / / / 4? l / /\_ - /\. / \ / \..\ \ Fig. 2-3. Determination ofmomen tumbyusing odiffraction grciting. \ / Q: /Z// is“\.®\‘\w /dE»..ds|n8 \..\ \ //1// Fig. 2-4. Scattering ofwaves by crystal planes.shall have asharp lineinourspectrum corresponding toadeﬁnite momentum, withanuncertainty given by(2.4), wehave tohave awave train ofatleast length L.Ifthewave train istooshort, wearenotusing theentire grating. The waves which form thespectrum arebeing reﬂected from only avery short sector ofthe grating ifthewave train istooshort, andthegrating willnotwork right—we will ﬁndabigangular spread. Inorder togetanarrower one, weneed tousethewhole grating, sothat atleast atsome moment thewhole wave train isscattering simul- taneously from allparts ofthegrating. Thus thewave train must beoflength L inorder tohave anuncertainty inthewavelength lessthan that given by(2.5). Incidentally, Ax/x2 =A(1/)\) =Ak/21r. (2.6) Therefore Ak=21:-/L, (2.7) where Listhelength ofthewave train. This means that ifwehave awave train whose length islessthan L,theun- certainty inthewave number must exceed 21r/L. Ortheuncertainty inawave number times thelength ofthewave train—-we willcallthat foramoment Ax- exceeds 21r. WecallitAxbecause that istheuncertainty inthelocation ofthe particle. Ifthewave train exists only inaﬁnite length, then thatiswhere wecould ﬁndtheparticle, within anuncertainty Ax. Now thisproperty ofwaves, that the length ofthewave train times theuncertainty ofthewave number associated with itisatleast 21r,isaproperty that isknown toeveryone who studies them. Ithas nothing todowith quantum mechanics. Itissimply that ifwehave aﬁnite train, wecannot count thewaves initvery precisely. Letustryanother way toseethereason forthat. Suppose that wehave a ﬁnite train oflength L;thenbecause ofthewayithastodecrease attheends, as inFig.2-1,thenumber ofwaves inthelength Lisuncertain bysomething like=h1. Butthenumber ofwaves inLiskL/21r. Thus kisuncertain, andweagain getthe result (2.7), aproperty merely ofwaves. Thesame thing works whether thewaves areinspace andkisthenumber ofradians percentimeter andListhelength of thetrain, orthewaves areintime andtoisthenumber ofoscillations persecond andTisthe“length” intime thatthewave train comes in.That is,ifwehave awave train lasting onlyforacertain ﬁnite timeT,thentheuncertainty inthefre- quency isgiven by Aw=21r/T. (2.8) Wehave tried toemphasize thatthese areproperties ofwaves alone, andtheyare well known, forexample, inthetheory ofsound. Thepoint isthatinquantum mechanics weinterpret thewave number as being ameasure ofthemomentum ofaparticle, with therule thatp=hk,so thatrelation (2.7) tellsusthatApzh/Ax. This, then, isalimitation oftheclassi- calidea ofmomentum. (Naturally, ithastobelimited insome ways ifweare going torepresent particles bywaves!) lt~is nice that wehave found arule that gives ussome ideaofwhen there isafailure ofclassical ideas. 2-3Crystal diffraction Next letusconsider thereﬂection ofparticle waves from acrystal. Acrystal isathick thing which hasawhole lotofsimilar atoms—we willinclude some com- plications later—in anicearray. Thequestion ishowtosetthearray sothatwe getastrong reﬂected maximum inagiven direction foragiven beam of,say,light (x-rays), electrons, neutrons, oranything else. Inorder toobtain astrong reﬂection, thescattering from alloftheatoms must beinphase. There cannot beequal num- bersinphase andoutofphase, orthewaves willcancel out. Thewaytoarrange things istoﬁndtheregions ofconstant phase, aswehave already explained; they areplanes which make equal angles with theinitial andﬁnal directions (Fig. 2-4). Ifweconsider twoparallel planes, asinFig.2-4,thewaves scattered from the twoplanes willbeinphase, provided thedifference indistance traveled byawave 2-4 front isanintegral number ofwavelengths. This difference canbeseen tobe 2dsin0,where distheperpendicular distance between theplanes. Thus the condition forcoherent reﬂection is 2dsin 6=n>\ (n=1,2,. ..). (2.9) If,forexample, thecrystal issuchthattheatoms happen tolieonplanes obey- ingcondition (2.9) with n=1,then there willbeastrong reﬂection. If,onthe other hand, there areother atoms ofthesame nature (equal indensity) halfway between, then theintermediate planes willalsoscatter equally strongly andwill interfere with theothers andproduce noeﬂect. Sodin(2.9) must refer toad- jacent planes; wecannot take aplane ﬁvelayers farther back andusethisformula! Asamatter ofinterest, actual crystals arenotusually assimple asasingle kind ofatom repeated inacertain way. Instead, ifwemake atwo-dimensional analog, theyaremuch likewallpaper, inwhich there issome kind ofﬁgure which repeats allover thewallpaper. By“ﬁgure” wemean, inthecaseofatoms, some arrangement—calcium andacarbon andthree oxygens, etc.,forcalcium carbonate, andsoon—which may involve arelatively large number ofatoms. Butwhatever itis,theﬁgure isrepeated inapattern. This basic ﬁgure iscalled aunitcell. Thebasic pattern ofrepetition deﬁnes what wecallthelattice type; thelattice typecanbeimmediately determined bylooking atthereﬂections andseeing what their symmetry is.Inother words, where weﬁndanyreﬂections atalldetermines thelattice type, butinorder todetermine what isineach oftheelements ofthe lattice onemust take intoaccount theintensity ofthescattering atthevarious directions. Which directions scatter depends onthetypeoflattice, buthowstrongly each scatters isdetermined bywhat isinside each unitcell,andinthatwaythe structure ofcrystals isworked out. Two photographs ofx-ray diffraction patterns areshown inFigs. 2-5and 2-6; theyillustrate scattering from rock saltandmyoglobin, respectively. Incidentally, aninteresting thing happens ifthespacings ofthenearest planes arelessthan A/2. Inthiscase (2.9) hasnosolution forn.Thus ifAisbigger than twice thedistance between adjacent planes, then there isnosidediffraction pattern, andthelight—or whatever itis—will goright through thematerial with- outbouncing offorgetting lost. Sointhecaseoflight, where Aismuch bigger than thespacing, ofcourse itdoes gothrough andthere isnopattern ofreﬂection from theplanes ofthecrystal. This factalsohasaninteresting consequence inthecaseofpiles which make neutrons (these areobviously particles, foranybody’s moneyl). Ifwetake these neutrons andletthem into along block ofgraphite, theneutrons diffuse and work their wayalong (Fig. 2-7). They diffuse because they arebounced bythe atoms, butstrictly, inthewave theory, they arebounced bytheatoms because ofdiffraction from thecrystal planes. Itturns outthatifwetakeaverylong piece ofgraphite, theneutrons thatcome outthefarendarealloflong wavelength! Infact,ifoneplots theintensity asafunction ofwavelength, wegetnothing except forwavelengths longer than acertain minimum (Fig. 2-8). Inother words, we cangetvery slow neutrons thatway. Only theslowest neutrons come through; theyarenotdiffracted orscattered bythecrystal planes ofthegraphite, butkeep going right through likelight through glass, andarenotscattered outthesides. There aremany other demonstrations ofthereality ofneutron waves andwaves ofother particles. 2-4Thesizeofanatom Wenow consider another application oftheuncertainty relation, Eq.(2.3). Itmust notbetaken tooseriously; theideaisright buttheanalysis isnotvery accurate. Theideahastodowith thedetermination ofthesizeofatoms, andthe factthat, classically, theelectrons would radiate light andspiral inuntil theysettle down right ontopofthenucleus. Butthatcannot beright quantum-mechanically because then wewould know where each electron wasandhowfastitwasmoving. 2-5Fig.2-5. The pattern produced by thediffraction ofabeam ofx-rays ina crystal ofsodium chloride. 1\- Fig.2-6. Thex-ray diffraction pat- ternofmyoglobin. SHORT-X NEUTRONS //—' —-tone-xus: GRAPHITE __NEUTRONS \\snonr-x ncumous Fig. 2-7. Diffusion ofpile neutrons through graphite block. .- 7/..xmlnntensy Fig.2-8. Intensity ofneutrons outof graphite rodasfunction ofwavelength. spectral frequencies wasnoted before quantum mechanics wasdiscovered, anditis called theRitzcombination principle. This isagain amystery from thepoint of view ofclassical mechanics. Letusnotbelabor thepoint thatclassical mechanics isafailure intheatomic domain; weseem tohave demonstrated that pretty well. Wehave already talked about quantum mechanics asbeing represented by amplitudes which behave likewaves, with certain frequencies andwave numbers. Letusobserve howitcomes about from thepoint ofview ofamplitudes thatthe atom hasdeﬁnite energy states. This issomething wecannot understand from what hasbeen saidsofar.butweareallfamiliar with thefactthatconﬁned waves have deﬁnite frequencies. Forinstance, ifsound isconﬁned toanorgan pipe, orany- thing likethat, then there ismore than onewaythatthesound canvibrate, but foreach such way there isadeﬁnite frequency. Thus anobject inwhich thewaves areconﬁned hascertain resonance frequencies. ltistherefore aproperty ofwaves inaconﬁned space—a subject which wewilldiscuss indetail with formulas later on—that they exist only atdeﬁnite frequencies. And since thegeneral relation exists between frequencies oftheamplitude and energy, wearenotsurprised to ﬁnddeﬁnite energies associated with electrons bound inatoms. 2-6Philosophical implications Letusconsider brieﬂy some philosophical implications ofquantum mechanics. Asalways, there aretwoaspects oftheproblem: oneisthephilosophical implica- tion forphysics, and theother istheextrapolation ofphilosophical matters to other ﬁelds. When philosophical ideas associated with science aredragged into another ﬁeld, they areusually completely distorted. Therefore weshall conﬁne ourremarks asmuch aspossible tophysics itself. First ofall,themost interesting aspect istheidea ofthe uncertainty principle; making anobservation aﬂects thephenomenon. Ithasalways been known that making observations affects aphenomenon, butthepoint isthat theeffect cannot bedisregarded orminimized ordecreased arbitrarily byrearranging theapparatus. When welook foracertain phenomenon wecannot help butdisturb itinacertain minimum way, andthedisturbance isnecessary fortheconsistency oftheviewpoint. The observer was sometimes important inprequantum physics, butonly ina trivial sense. Theproblem hasbeen raised: ifatreefalls inaforest andthere isnobody there tohear it,does itmake anoise? Arealtreefalling inarealforest makes asound, ofcourse, even ifnobody isthere. Even ifnooneispresent tohear it,there areother traces left. The sound willshake some leaves, andifwewere careful enough wemight ﬁndsomewhere thatsome thorn hadrubbed against a leafand made atiny scratch that could notbeexplained unless weassumed the leafwere vibrating. S0inacertain sense wewould have toadmit thatthere isa sound made. Wemight ask: wasthere asensation ofsound? No,sensations have todo,presumably, with consciousness. And whether ants areconscious and whether there were ants intheforest, orwhether thetreewasconscious, wedonot know. Letusleave theproblem inthatform. Another thing that people have emphasized since quantum mechanics was developed istheideathatweshould notspeak about those things which wecannot measure. (Actually relativity theory also said this.) Unless athing canbedeﬁned bymeasurement, ithasnoplace inatheory. And since anaccurate value ofthe momentum ofalocalized particle cannot bedeﬁned bymeasurement ittherefore hasnoplace inthetheory. Theideathatthisiswhat wasthematter withclassical theory isufalse position. Itisacareless analysis ofthesituation. Just because we cannot measure position andmomentum precisely does notapriori mean thatwe cannot talkabout them. Itonly means thatweneed nottalkabout them. The situation inthesciences isthis: Aconcept oranideawhich cannot bemeasured orcannot bereferred directly toexperiment may ormay notbeuseful. Itneed notexist inatheory. Inother words, suppose wecompare theclassical theory of theworld with thequantum theory oftheworld, and suppose that itistrue ex- perimentally thatwecanmeasure position andmomentum onlyimprecisely. The question iswhether theideas oftheexact position ofaparticle and theexact 2-8 momentum ofaparticle arevalid ornot. The classical theory admits theideas; thequantum theory does not. This does notinitself mean that classical physics iswrong. When thenewquantum mechanics wasdiscovered, theclassical people- which included everybody except Heisenberg, Schrodinger, and Born—said: “Look, your theory isnotanygood because youcannot answer certain questions like: what istheexact position ofaparticle?, which hole does itgothrough?, andsome others.” Heisenberg’s answer was: “ldonotneed toanswer such ques- tions because youcannot asksuch aquestion experimentally.” Itisthat wedo nothave to.Consider twotheories (a)and(b);(a)contains anidea thatcannot be checked directly butwhich isused intheanalysis, and theother, (b),does not contain theidea. Ifthey disagree intheir predictions, onecould notclaim that (b)isfalse because itcannot explain thisideathatisin(a),because thatideais oneofthethings thatcannot bechecked directly. Itisalways good toknow which ideas cannot bechecked directly, butitisnotnecessary toremove them all. Itis nottrue that wecanpursue science completely byusing only those concepts which aredirectly subject toexperiment. Inquantum mechanics itself there isaprobability amplitude, there isa potential, andthere aremany constructs thatwecannot measure directly. Thebasis ofascience isitsability topredict. Topredict means totellwhat willhappen inan experiment thathasnever been done. How canwedothat? Byassuming thatwe know what isthere, independent oftheexperiment. Wemust extrapolate the experiments toaregion where theyhave notbeen done. Wemust takeourcon- cepts andextend them toplaces where theyhave notyetbeen checked. Ifwedo notdothat, wehave noprediction. Soitwasperfectly sensible fortheclassical physicists togohappily along andsuppose thattheposition——which obviously means something forabaseball—meant something alsoforanelectron. Itwas notstupidity. Itwasasensible procedure. Today wesaythatthelawofrelativity issupposed tobetrueatallenergies, butsomeday somebody maycome along and sayhowstupid wewere. Wedonotknow where weare“stupid” until we“stick ourneck out,” andsothewhole ideaistoputourneck out. Andtheonlywayto ﬁndoutthatwearewrong istoﬁndoutwhat ourpredictions are. Itisabsolutely necessary tomake constructs. Wehave already made afewremarks about theindeterminacy ofquantum mechanics. That is,thatweareunable nowtopredict what willhappen inphysics inagiven physical circumstance which isarranged ascarefully aspossible. If wehave anatom thatisinanexcited state andsoisgoing toemit aphoton, we cannot saywhen itwillemit thephoton. Ithasacertain amplitude toemit the photon atanytime, andwecanpredict onlyaprobability foremission; wecannot predict thefuture exactly. Thishasgiven risetoallkinds ofnonsense andquestions onthemeaning offreedom ofwill,andoftheideathattheworld isuncertain. Ofcourse wemust emphasize thatclassical physics isalsoindeterminate, ina sense. Itisusually thought thatthisindeterminacy, thatwecannot predict the future, isanimportant quantum-mechanical thing, andthisissaidtoexplain the behavior ofthemind, feelings offreewill,etc.Butiftheworld were classical——if thelaws ofmechanics were classical—it isnotquite obvious thatthemind would notfeelmore orlessthesame. Itistrueclassically thatifweknew theposition and thevelocity ofevery particle intheworld, orinaboxofgas,wecould predict ex- actly what would happen. And therefore theclassical world isdeterministic. Suppose, however, that wehave aﬁnite accuracy anddonotknow exactly where justoneatom is,saytoonepartinabillion. Then asitgoesalong ithitsanother atom, andbecause wedidnotknow theposition better than toonepartinabillion, weﬁndaneven larger error intheposition after thecollision. And thatisampliﬁed, ofcourse, inthenext collision, sothat ifwestart with only atinyerror itrapidly magniﬁes toaverygreat uncertainty. Togiveanexample: ifwater fallsoveradam, itsplashes. Ifwestand nearby, every nowandthenadrop willland onournose. This appears tobecompletely random, yetsuch abehavior would bepredicted bypurely classical laws. The exact position ofallthedrops depends upon the precise wigglings ofthewater before itgoes over thedam. How? Thetiniest irregularities aremagniﬁed infalling, sothat wegetcomplete randomness. Ob- 2-9 viously, wecannot really predict theposition ofthedrops unless weknow the motion ofthewater absolutely exactly. Speaking more precisely. given anarbitrary accuracy. nomatter how precise, onecanﬁnd atime long enough that wecannot make predictions valid forthat long atime. Now thepoint isthatthislength oftime isnotvery large. ltisnot that thetime ismillions ofyears iftheaccuracy isonepart inabillion. The time goes. infact, only logarithmically with theerror, anditturns outthatinonly a very, very tinytime weloseallourinformation. lftheaccuracy istaken tobeone partinbillions andbillions andbillions—no matter how many billions wewish, provided wedostop somewhere—then wecanﬁnd atime lessthan thetime it took tostate theaccuracy—after which wecannolonger predict what isgoing tohappen! Itistherefore notfairtosaythat from theapparent freedom and indeterminacy ofthehuman mind, weshould have realized thatclassical “deter- ministic” physics could notever hope tounderstand it,andtowelcome quantum mechanics asarelease from a“completely mechanistic” universe. Foralready in classical mechanics there wasindeterminability from apractical point ofview. 2-10 3 Probability Amplitudes 3-1Thelaws forcombining amplitudes When Schrodinger ﬁrstdiscovered thecorrect laws ofquantum mechanics, hewrote anequation which described theamplitude toﬁndaparticle invarious places. This equation wasverysimilar totheequations thatwere already known toclassical physicists—equations thatthey hadused indescribing themotion of airinasound wave, thetransmission oflight, andsoon.Somost ofthetime at thebeginning ofquantum mechanics wasspent insolving thisequation. Butatthe same timeanunderstanding wasbeing developed. particularly byBorn andDirac, ofthebasically new physical ideas behind quantum mechanics. Asquantum mechanics developed further, itturned outthatthere were alarge number ofthings which were notdirectly encompassed intheSchrodinger equation—such asthe spin ofthe electron, andvarious relativistic phenomena. Traditionally, allcourses inquantum mechanics have begun inthesame way. retracing thepath followed in thehistorical development ofthesubject. One ﬁrstlearns agreat deal about clas- sical mechanics sothathewillbeabletounderstand howtosolve theSchrodinger equation. Then hespends along time working outvarious solutions. Only after adetailed study ofthisequation does hegettothe“advanced” subject ofthe electron’s spin. Wehadalsooriginally considered thattheright waytoconclude these lectures onphysics wastoshow how tosolve theequations ofclassical physics incompli- cated situations——such asthedescription ofsound waves inenclosed regions, modes ofelectromagnetic radiation incylindrical cavities, andsoon.That wastheoriginal planforthiscourse. However, wehave decided toabandon thatplanandtogive instead anintroduction tothequantum mechanics. Wehave come tothecon- clusion thatwhat areusually called theadvanced parts ofquantum mechanics are, infact, quite simple. The mathematics that isinvolved isparticularly simple, involving simple algebraic operations andnodifferential equations oratmost only very simple ones. The only problem isthat wemust jump thegapofno longer being abletodescribe thebehavior indetail ofparticles inspace. Sothis iswhat wearegoing totrytodo:totellyouabout what conventionally would be called the“advanced” parts ofquantum mechanics. Buttheyare,weassure you, byallodds thesimplest parts—in adeep sense oftheword—as well asthemost basic parts. This isfrankly apedagogical experiment; ithasnever been done before, asfarasweknow. Inthissubject wehave, ofcourse, thedifficulty thatthequantum mechanical behavior ofthings isquite strange. Nobody hasaneveryday experience tolean ontogetarough, intuitive ideaofwhat willhappen. Sothere aretwoways of presenting thesubject: Wecould either describe what can happen inarather rough physical way, telling youmore orlesswhat happens without giving the precise laws ofeverything; orwecould, ontheother hand, give theprecise laws intheir abstract form. But,then because oftheabstractions, youwouldn’t know what they were allabout, physically. The latter method isunsatisfactory because itiscompletely abstract, andtheﬁrstwayleaves anuncomfortable feeling because onedoesn’t know exactly what istrue andwhat isfalse. Wearenotsure how to overcome thisdifﬁculty. You willnotice, infact, thatChapters land2showed thisproblem. Theﬁrstchapter wasrelatively precise; butthesecond chapter was arough description ofthecharacteristics ofdifferent phenomena. Here, wewill trytoﬁndahappy medium between thetwoextremes. 3-13 CQUJUJ-1Thelaws forcombining BUJNamplitudes Thetwo-slit interference pattern Scattering from acrystal Identical particles \\\_ M<'\\\\__'U "U/ DETECTOR I L‘ X IZ /i\t\"’//\\\//\\\I aw ELECTRON__:____/_ ___ __ K sun \ 72 \\\\\\\\NY7 WALL BACKSTOP t0) tbl tel 3el. lnterterence experiment withelectrons. Wewillbegin inthischapter bydealing with some general quantum me- chanical ideas. Some ofthestatements willbequite precise. others onlypartially precise. Itwillbehard totellyouaswegoalong which iswhich, butbythetime youhave ﬁnished therestofthebook, youwillunderstand inlooking back which parts hold upandwhich parts were onlyexplained roughly. Thechapters which follow thisonewillnotbesoimprecise. Infact,oneofthereasons wehave tried carefully tobeprecise inthesucceeding chapters issothatwecanshow youoneof themost beautiful things about quantum mechanics—how much canbededuced from solittle. Webegin bydiscussing again thesuperposition ofprobability amplitudes. Asanexample wewillrefer totheexperiment described inChapter l,andshown again hereinFig.3-l. There isasource sofparticles, sayelectrons; then there isawallwith twoslitsinit;after thewall. there isadetector located atsome position x.Weaskfortheprobability thataparticle willbefound atx.Ourﬁrst general principle inquantum mechanics isthattheprobability thataparticle will arrive atx,when letoutatthesource s,canberepresented quantitatively bythe absolute square ofacomplex number called aprobability ampIitude—in thiscase, the“amplitude thataparticle from swillarrive atx.”Wewillusesuchamplitudes sofrequently that wewilluseashorthand notation——invented byDirac and generally used inquantum mechanics—to represent thisidea. Wewrite theproba- bility amplitude thisway: (Particle arrives atxIparticle leaves s). (3.1) Inother words, thetwobrackets ()areasign equivalent to"the amplitude that"; theexpression attheright ofthevertical linealways gives thestarting condition, andtheoneattheleft,theﬁnal condition. Sometimes itwillalso beconvenient to abbreviate stillmore anddescribe theinitial andﬁnal conditions bysingle letters. Forexample, wemay onoccasion write theamplitude (3.1) as (Xls). (3.2) Wewant toemphasize thatsuch anamplitude is,ofcourse, justasingle number— acomplex number. . Wehave already seeninthediscussion ofChapter Ithatwhen there aretwo ways fortheparticle toreach thedetector. theresulting probability isnotthe sumofthetwoprobabilities, butmust bewritten astheabsolute square ofthe sumoftwoamplitudes. Wehadthattheprobability thatanelectron arrives atthe detector when both paths areopen is Ptz =l¢1-l-¢>2l2- (3-3) 3-2 /ﬁt‘//\\// I \\3|l\\O-\\IQ\\\\\ _>121 \\\\\\\\\\\r /|/ _ F13 :4---e_’/ \\\_/ 3 =;__"*/ _"__“—$_"'_'_'__“-5 R‘ "' \ "2 ‘ Wewish now toputthisresult interms ofournew notation. First, however, we want tostate oursecond general principle ofquantum mechanics: When aparticle canreach agiven state bytwopossible routes, thetotal amplitude fortheprocess isthesum oftheamplitudes forthetworoutes considered separately. Inournew notation wewrite that <xlS>bot,h holes open =<xlSlthrough l+ <-XlS>thFO\1gl1 2' Incidentally, wearegoing tosuppose thattheholes Iand2aresmall enough that when wesayanelectron goesthrough thehole, wedon’t have todiscuss which part ofthehole. Wecould, ofcourse, split each hole intopieces with acertain amplitude thattheelectron goestothetopoftheholeandthebottom oftheholeandsoon. Wewillsuppose thattheholeissmall enough sothatwedon’t have toworry about thisdetail. That ispartoftheroughness involved; thematter canbemade more precise, butwedon’t want todosoatthisstage. Now wewant towrite outinmore detail what wecansayabout theamplitude fortheprocess inwhich theelectron reaches thedetector atxbywayofhole l. Wecandothatbyusing ourthirdgeneral principle: When aparticle goesbysome particular route theamplitude forthatroute canbewritten astheproduct ofthe amplitude togopart way with theamplitude togotherestoftheway. Forthe setup ofFig.3-1theamplitude togofrom stoxbywayofhole lisequal tothe amplitude togofrom stol,multiplied bytheamplitude togofrom ltox. (XlS>via. 1:(xi Again thisresult isnotcompletely precise. Weshould also include afactor forthe amplitude that theelectron willgetthrough thehole at1;butinthepresent case itisasimple hole, andwewilltake thisfactor tobeunity. You willnote that Eq.(3.5) appears tobewritten inreverse order. Itisto bereadfrom right toleft:Theelectron goes from stolandthen from Itox. Insummary, ifevents occur insuccession-—that is,ifyoucananalyze oneofthe routes oftheparticle bysaying itdocs this, then itdoes this, then itdoes that—the resultant amplitude forthatroute iscalculated bymultiplying insuccession the amplitude foreach ofthesuccessive events. Using thislawwecanrewrite Eq. (3.4)as (x|s)b0,h =(x|l)(l ls)+(x|2)(2 ls). Now wewish toshow thatjustusing these principles wecancalculate amuch more complicated problem liketheoneshown inFig.3-2. Here wehave two walls, onewith twoholes, Iand2,andanother which hasthree holes, a,b,andc. Behind thesecond wall there isadetector atx,andwewant toknow theamplitude foraparticle toarrive there. Well, onewayyoucanﬁndthisisbycalculating the superposition, orinterference, ofthewaves thatgothrough; butyoucanalsodo itbysaying thatthere aresixpossible routes andsuperposing anamplitude for each. Theelectron cangothrough hole l,then through‘ hole a,andthen tox;or itcould gothrough hole 1,thenthrough holeb,andthentox;andsoon.Accord- ingtooursecond principle, theamplitudes foralternative routes add,soweshould 3-3Fig. 3-2 Amore complicated inter- ference experiment beable towrite theamplitude from stoxasasum ofsixseparate amplitudes. Ontheother hand, using thethird principle, each ofthese separate amplitudes canbewritten asaproduct ofthree amplitudes. Forexample, oneofthem isthe amplitude forsto1,times theamplitude forItoa,times theamplitude foratox. Using ourshorthand notation, wecanwrite thecomplete amplitude togofrom st0xas (X18) =<X|¢1><a| 1>(1lS> +<X|b><l>l l><lls)+ +<Xl¢)(¢l2>(2lS)- Wecansave writing byusing thesummation notation <XlS>= Z(Xl<X>(<1li><ils)- (3-6) ..f==..fi?. Inorder tomake anycalculations using these methods, itis,naturally, neces- sarytoknow theamplitude togetfrom oneplace toanother. Wewillgivearough idea ofatypical amplitude. Itleaves outcertain things likethepolarization of light orthespinoftheelectron, butaside from such features itisquite accurate. Wegiveitsothatyoucansolve problems involving various combinations ofslits. Suppose aparticle with adeﬁnite energy isgoing inempty space from alocation r1toalocation r2.Inother words, itisafreeparticle with noforces onit.Except foranumerical factor infront, theamplitude togofrom r1tor2is e1_P"'12/5 ('2l'1)=*_i’ (3-7)V12 where P12—r2—r1,andpisthemomentum which isrelated totheenergy E bytherelativistic equation 22_ 2__ 22pC "_E (W106 )9 orthenonrelativistic equation Ei—Kinetic ener2m_ gy' Equation (3.7) saysineffect thattheparticle haswavelike properties, theamplitude propagating asawave with awave number equal tothemomentum divided byh. Inthemost general case, theamplitude andthecorresponding probability willalsoinvolve thetime. Formost ofthese initial discussions wewillsuppose thatthesource always emits theparticles with agiven energy sowewillnotneed to worry about thetime. Butwecould, inthegeneral case, beinterested insome other questions. Suppose thataparticle isliberated atacertain place Patacertain time, andyouwould liketoknow theamplitude forittoarrive atsome location, sayr,atsome later time. This could berepresented symbolically astheamplitude (r,t=t1lP,t=0).Clearly, thiswilldepend upon both randt.You willget different results ifyou putthedetector indifferent places andmeasure atdifferent times. This function ofrandt,ingeneral, satisﬁes adifferential equation which is awave equation. Forexample, inanonrelativistic case itistheSchrodinger equa- tion. Onehasthen awave equation analogous totheequation forelectromagnetic waves orwaves ofsound inagas. However, itmust beemphasized that thewave function that satisﬁes theequation isnotlikearealwave inspace; onecannot picture anykind ofreality tothiswave asonedoes forasound wave. Although onemay betempted tothink interms of“particle waves” when dealing with oneparticle, itisnotagood idea, forifthere are,say,twoparticles, theamplitude toﬁndoneatr1andtheother atr2isnotasimple wave inthree- dimensional space, butdepends onthesixspace variables r1andr2.Ifweare, forexample, dealing with two(ormore) particles, wewillneed thefollowing additional principle: Provided thatthetwoparticles donotinteract, theamplitude thatoneparticle willdoonething andtheother onesomething elseistheproduct ofthetwoamplitudes thatthetwoparticles would dothetwothings separately. Forexample, if(aI$1)istheamplitude forparticle Itogofrom s1toa,and(b|s2) 3-4 istheamplitude forparticle 2togofrom s2tob,theamplitude thatboththings willhappen together is (QlS1><b|S2)- There isonemore point toemphasize. Suppose that wedidn’t know where theparticles inFig.3-2come from before arriving atholes land2oftheﬁrst wall. Wecanstillmake aprediction ofwhat willhappen beyond thewall (for example, theamplitude toarrive atx)provided thatwearegiven twonumbers: theamplitude tohave arrived atIandtheamplitude tohave arrived at2.Inother words, because ofthefactthattheamplitude forsuccessive events multiplies, as shown inEq.(3.6), allyouneed toknow tocontinue theanalysis istwonumbers- inthisparticular case (1|s)and(2Is).These twocomplex numbers areenough topredict allthefuture. That iswhat really makes quantum mechanics easy. It turns outthatinlaterchapters wearegoing todojustsuchathing when wespecify astarting condition interms oftwo(orafew)numbers. Ofcourse, these numbers depend upon where thesource islocated andpossibly other details about the apparatus, butgiven thetwonumbers. wedonotneed toknow anymore about suchdetails. \\\\\\\\\\\\\‘Q\\\\\\‘$9\\\\ 'LIGHTi——i souncs ; ii-\'— —E Tlq/1; ;—/—/ ____ ELECTRON ‘r’sun 2 Q02 3-2The two-slit interference pattern Now wewould liketoconsider amatter which wasdiscussed insome detail inChapter l.This time wewilldoitwith thefullglory oftheamplitude idea toshow you how itworks out. Wetake thesame experiment shown inFig. 3—l, butnow with theaddition ofalight source behind thetwoholes, asshown inFig. 3-3. InChapter l,wediscovered thefollowing interesting result. If welooked behind slit1andsawaphoton scattered from there, then thedistribu- tionobtained fortheelectrons atxincoincidence withthese photons wasthesame asthough slit2were closed. The total distribution forelectrons that hadbeen “seen” ateither slitlorslit2wasthesumoftheseparate distributions andwas completely different from thedistribution with thelight turned off. This wastrue atleastifweused light ofshort enough wavelength. Ifthewavelength wasmade longer sowecould notbesure atwhich hole thescattering hadoccurred, the distribution became more liketheonewith thelight turned off. Let’s examine what ishappening byusing ournew notation andtheprinciples ofcombining amplitudes. Tosimplify thewriting, wecanagain let¢1stand for theamplitude that theelectron willarrive atxbyway ofhole l,that is, ¢>1= <Xl1><l ls)- Similarly, we’ll let4>2stand fortheamplitude thattheelectron getstothedetector bywayofhole2: ¢2=<Xl2><2 ls)- These aretheamplitudes togothrough thetwoholes andarrive atxifthere isno light. Now ifthere islight, weaskourselves thequestion: What istheamplitude fortheprocess inwhich theelectron starts atsandaphoton isliberated bythe 3-5Fig. 3-3 An experiment todeter mine which holetheelectron goes through it I x1 w?(0) lb) (Cl Fig. 3e11, The probability ofcount- ingonelectron catxincoincidence with Cl photon olDinthe experiment ofFig. 33;(Q)forb"<O;(blforbIo;lcl forO<b~§o.light source L,ending with theelectron atxandaphoton seen behind slitl? Suppose thatweobserve thephoton behind slit1bymeans ofadetector D1,as shown inFig.3-3, anduseasimilar detector D2tocount photons scattered behind hole2.There willbeanamplitude foraphoton toarrive atD1andan electron atx,andalsoanamplitude foraphoton toarrive atD2andanelectron atx.Let’s trytocalculate them. Although wedon’t have thecorrect mathematical formula forallthefactors that gointothiscalculation, youwillseethespirit ofitinthefollowing discussion. First, there istheamplitude (lls)thatanelectron goes from thesource tohole l. Then wecansuppose that there isacertain amplitude that while theelectron isat hole 1itscatters aphoton intothedetector D1. Letusrepresent thisamplitude by a.Then there istheamplitude (xI1}thattheelectron goes from slitltotheelec- tron detector atx.Theamplitude that theelectron goes from stoxviaslitland scatters aphoton intoD1isthen (xi l)u(l Or,inourprevious notation, itisjust11¢1. There isalsosome amplitude thatanelectron going through slit2willscatter aphoton intocounter D1. Yousay,“That's impossible; howcanitscatter into counter D1ifitisonly looking athole l?" Ifthewavelength islong enough, there arediffraction effects, anditiscertainly possible. Iftheapparatus isbuilt welland ifweusephotons ofshort wavelength, then theamplitude that aphoton willbe scattered into detector l.from anelectron at2isvery small. Buttokeep the discussion general wewant totake intoaccount thatthere isalways some such amplitude, which wewillcallb.Then theamplitude thatanelectron goes via slit2andscatters aphoton intoD1is (Xl2>l><2|S>=bd>2- The amplitude toﬁnd theelectron atxandthephoton inD1isthesum of twoterms, oneforeach possible path fortheelectron. Each term isinturn made upoftwofactors: ﬁrst, thattheelectron went through ahole, andsecond, thatthe photon isscattered bysuchanelectron intodetector l;wehave electron atxelectron from s\_ <photon atD1 photon from L/Tad)‘+bdw‘ 6'8) Wecangetasimilar expression when thephoton isfound intheother detector D2. lfweassume forsimplicity that thesystem issymmetrical, then uisalso the amplitude foraphoton inD2when anelectron passes through hole2,andbis theamplitude foraphoton inD2when theelectron passes through hole l.The corresponding total amplitude foraphoton atD2andanelectron atxis /electron atxlelectron from s \photon atD» >Zum+b¢1' (39) photon from L Now weareﬁnished. Wecaneasily calculate theprobability forvarious situations. Suppose thatwewant toknow with what probability wegetacount inD1and anelectron atx.That willbetheabsolute square oftheamplitude given inEq.(3.8), namely, just ll1¢>1 +b4>2l2. Let“s look more carefully atthis expression. First ofall,if[2iszero—which isthewaywewould liketodesign the apparatus——then theanswer issimply l¢>1l2 diminished intotal amplitude bythe factor |a|2. This istheprobability distribution that youwould getifthere were only onehole—as shown inthegraph ofFig.3—4(a). Ontheother hand, ifthe wavelength isvery long, thescattering behind hole2intoD1maybejustabout thesame asforhole l.Although there maybesome phases involved inaandb. wecanaskabout asimple ease inwhich thetwophases areequal. lfaispractically equal tob,then thetotal probability becomes |¢1+¢>2l2 multiplied bylulz, since thecommon factor acanbetaken out. This, however, isjusttheprobability 3-6 it it x1 w?(0) lb) (Cl Fig. 3\4. The probability ofcount- ingonelectron catxincoincidence with Cl photon olDinthe experiment ofFig. 33;(Q)forb"<O;(blforbIo;lcl forO<b~§o.light source L,ending with theelectron atxandaphoton seen behind slitl? Suppose thatweobserve thephoton behind slit1bymeans ofadetector D1,as shown inFig.3-3, anduseasimilar detector D2tocount photons scattered behind hole2.There willbeanamplitude foraphoton toarrive atD1andan electron atx,andalsoanamplitude foraphoton toarrive atD2andanelectron atx.Let’s trytocalculate them. Although wedon’t have thecorrect mathematical formula forallthefactors that gointothiscalculation, youwillseethespirit ofitinthefollowing discussion. First, there istheamplitude (lls)thatanelectron goes from thesource tohole l. Then wecansuppose that there isacertain amplitude that while theelectron isat hole 1itscatters aphoton intothedetector D1. Letusrepresent thisamplitude by a.Then there istheamplitude (xI1}thattheelectron goes from slitltotheelec- tron detector atx.Theamplitude that theelectron goes from stoxviaslitland scatters aphoton intoD1isthen (xi l)u(l Or,inourprevious notation, itisjust11¢1. There isalsosome amplitude thatanelectron going through slit2willscatter aphoton intocounter D1. Yousay,“That's impossible; howcanitscatter into counter D1ifitisonly looking athole l?" Ifthewavelength islong enough, there arediffraction eﬁ“ects, anditiscertainly possible. Iftheapparatus isbuilt welland ifweusephotons ofshort wavelength, then theamplitude that aphoton willbe scattered into detector l.from anelectron at2isvery small. Buttokeep the discussion general wewant totake intoaccount thatthere isalways some such amplitude, which wewillcallb.Then theamplitude thatanelectron goes via slit2andscatters aphoton intoD1is (Xl2>l><2|S>=bd>2- The amplitude toﬁnd theelectron atxandthephoton inD1isthesum of twoterms, oneforeach possible path fortheelectron. Each term isinturn made upoftwofactors: ﬁrst, thattheelectron went through ahole, andsecond, thatthe photon isscattered bysuchanelectron intodetector l;wehave electron atxelectron from s\_ <photon atD1 photon from L/Tad)‘+him‘ 6'8) Wecangetasimilar expression when thephoton isfound intheother detector D2. lfweassume forsimplicity that thesystem issymmetrical, then uisalso the amplitude foraphoton inD2when anelectron passes through hole2,andbis theamplitude foraphoton inD2when theelectron passes through hole l.The corresponding total amplitude foraphoton atD2andanelectron atxis /electron atxlelectron from s \photon atD» >Zum+b¢1' (39) photon from L Now weareﬁnished. Wecaneasily calculate theprobability forvarious situations. Suppose thatwewant toknow with what probability wegetacount inD1and anelectron atx.That willbetheabsolute square oftheamplitude given inEq.(3.8), namely, just ll1¢>1 +b4>2l2. Let“s look more carefully atthis expression. First ofall,ifbiszero—which isthewaywewould liketodesign the apparatus——then theanswer issimply l¢>1l2 diminished intotal amplitude bythe factor |a|2. This istheprobability distribution that youwould getifthere were only onehole—as shown inthegraph ofFig.3—4(a). Ontheother hand, ifthe wavelength isvery long, thescattering behind hole2intoD1maybejustabout thesame asforhole l.Although there maybesome phases involved inaandb. wecanaskabout asimple ease inwhich thetwophases areequal. lfaispractically equal tob,then thetotal probability becomes |¢1+¢>2l2 multiplied bylulz, since thecommon factor acanbetaken out. This, however, isjusttheprobability 3-6 distribution wewould have gotten without thephotons atall.Therefore, inthe casethatthewavelength isvery long——and thephoton detection ineﬁ‘ective—you return totheoriginal distribution curve which shows interference effects, asshown inFig.3—4(b). Inthecasethatthedetection ispartially effective, there isaninter- ference between alotof¢1andalittle of¢2,andyouwillgetanintermediate distribution such asissketched inFig.3—4(c). Needless tosay,ifwelook for coincidence counts ofphotons atD2andelectrons atx,wewillgetthesame kinds ofresults. Ifyouremember thediscussion inChapter l,youwillseethatthese results giveaquantitative description ofwhat wasdescribed there. Now wewould liketoemphasize animportant point sothatyouwillavoid acommon error. Suppose thatyouonly want theamplitude thattheelectron ar- rives atx,regardless ofwhether thephoton wascounted atD1orD2.Should you addtheamplitudes given inEqs. (3.8) and(3.9)? No! You must never add amplitudes fordifferent anddistinct ﬁnal states. Once thephoton isaccepted by oneofthephoton counters, wecanalways determine which alternative occurred ifwewant, without anyfurther disturbance tothesystem. Each alternative hasa probability completely independent oftheother. Torepeat, donotaddamplitudes fordifferent ﬁnal conditions, where by“ﬁnal” wemean atthat moment the probability isdesired—that is,when theexperiment is“ﬁnished.” Youdoaddthe amplitudes forthedifferent indistinguishable alternatives inside theexperiment, before thecomplete process isﬁnished. Attheendoftheprocess youmaysaythat you“don’t want tolook atthephoton.” That’s your business, butyoustilldonot addtheamplitudes. Nature does notknow what youarelooking at,andshe behaves thewaysheisgoing tobehave whether youbother totakedown thedata ornot. Soherewemust notaddtheamplitudes. Weﬁrstsquare theamplitudes forallpossible different ﬁnal events and then sum. The correct result foran electron atxandaphoton ateither D1orD2is /eatx efroms \2+ /eatx efroms \ \phatD1 phfrom L/ \phatD2 phfrom L/ =h¢1+-b¢a2-+la¢2+-b¢A”- (310) 3-3Scattering from acrystal Ournext example isaphenomenon inwhich wehave toanalyze theinter- ference ofprobability amplitudes somewhat carefully. Welook attheprocess of thescattering ofneutrons from acrystal. Suppose wehave acrystal which hasa lotofatoms withnuclei attheir centers, arranged inaperiodic array, andaneutron beam thatcomes from faraway. Wecanlabel thevarious nuclei inthecrystal by anindex i,where iruns over theintegers l,2,3,...N,with Nequal tothetotal number ofatoms. Theproblem istocalculate theprobability ofgetting aneutron intoacounter with thearrangement shown inFig.3-5. Foranyparticular atom 1',theamplitude thattheneutron arrives atthecounter Cistheamplitude thatthe neutron getsfrom thesource Stonucleus i,multiplied bytheamplitude athatit getsscattered there, multiplied bytheamplitude thatitgetsfrom itothecounter C.Let’s write thatdown: (neutron atCIneutron from S)1,1,, 1=(CIi)a(iIS). (3.1l) lnwriting thisequation wehave assumed thatthescattering amplitude aisthe same forallatoms. Wehave here alarge number ofapparently indistinguishable routes. They areindistinguishable because alow-energy neutron isscattered from anucleus without knocking theatom outofitsplace inthecrystal—no “record” isleftofthescattering. According totheearlier discussion, thetotal amplitude foraneutron atCinvolves asumofEq.(3.11) over alltheatoms: N (neutron atCIneutron from S)=2(CIi)a(iIS). (3.12) -i=1 3-7NEUTRON SOURCE CRYSTAL [jg ___a_____ C NEUTRON COUNTER Fig. 3-5. Measuring the scattering ofneutrons byucrystal. counrms“RATE <0) $PmFLIP llPROBABILITY (bl ll COUNTING RATE 9to Fig. 3-6. Theneutron counting rote ctscafunction ofangle: lo)forspin zero nuclei; (b)the probability ofscattering with spin ﬂip; (c)theobserved counting rote with ospin one-half nucleus.Because weareadding amplitudes ofscattering from atoms with different space positions, theamplitudes willhave different phases giving thecharacteristic inter- ference pattern thatwehave already analyzed inthecaseofthescattering oflight from agrating. Theneutron intensity asafunction ofangle insuch anexperiment isindeed often found toshow tremendous variations, with very sharp interference peaks andalmost nothing inbetween—as shown inFig.3—6(a). However, forcertain kinds ofcrystals itdoes notwork thisway, andthere is—along with theinterference peaks discussed above—a general background ofscattering inalldirections. We must trytounderstand theapparently mysterious reasons forthis. Well, wehave notconsidered oneimportant property oftheneutron. Ithasaspinofone-half, andsothere aretwoconditions inwhich itcanbe:either spin “up” (say perpendicu- lartothepage inFig.3-5)orspin“down.” lfthenuclei ofthecrystal have no spin, theneutron spin doesn't have anyeffect. Butwhen thenuclei ofthecrystal alsohaveaspin, sayaspinofone-half, youwillobserve thebackground ofsmeared- outscattering described above. The explanation isasfollows. Iftheneutron hasonedirection ofspinandtheatomic nucleus hasthesame spin, then nochange ofspin canoccur inthescattering process. lfthe neutron and atomic nucleus have opposite spin, then scattering canoccur bytwoprocesses, oneinwhich thespins areunchanged andanother inwhich thespin directions are exchanged. Thisrulefornonetchange ofthesumofthespins isanalogous toour classical lawofconservation ofangular momentum. Wecanbegin tounderstand thephenomenon ifweassume thatallthescattering nuclei aresetupwith spins in onedirection. Aneutron with thesame spin willscatter with theexpected sharp interference distribution. What about onewithopposite spin‘? lfitscatters without spin ﬂip, then nothing ischanged from theabove; butifthetwospins ﬂipover in thescattering, wecould, inprinciple, ﬁndoutwhich nucleus haddone thescatter- ing,since itwould betheonlyonewithspinturned over. Well, ifwecantellwhich atom didthescattering, what have theother atoms gottodowith it?Nothing, of course. Thescattering isexactly thesame asthatfrom asingle atom. Toinclude thiseffect, themathematical formulation ofEq.(3.12) must be modiﬁed since wehaven’t described thestates completely inthatanalysis. Let’s start with allneutrons from thesource having spin upandallthenuclei ofthe crystal having spindown. First, wewould liketheamplitude thatatthecounter thespin oftheneutron isupandallspins ofthecrystal arestilldown. This is notdifferent from ourprevious discussion. Wewillletubetheamplitude to scatter with noﬂiporspin. Theamplitude forscattering from theithatom is,of course, (Cup, crystal alldown IS1,,,,crystal alldown) =(CIi)(I(iIS). Since alltheatomic spins arestillclown, thevarious alternatives (different values ofi)cannot bedistinguished. There isclearly noway totellwhich atom didthe scattering. Forthisprocess, alltheamplitudes interfere. Wehave another case, however, where thespin ofthedetected neutron is down although itstarted from Swithspinup.lnthecrystal, oneofthespins must bechanged totheupdirection—let‘s saythat ofthekthatom. Wewillassume that there isthesame scattering amplitude with spin ﬁipforevery atom, namely h. (Inarealcrystal there isthedisagreeable possibility that thereversed spin moves tosome other atom, butlet‘stakethecaseofacrystal forwhich thisprobability isvery low.) The scattering amplitude isthen (C,1.,w11, nucleus kupISup,crystal alldown) :(CIk)b(k‘IS). (3.13) Ifweaskfortheprobability ofﬁnding theneutron spindown andthekthnucleus spinup,itisequal totheabsolute square ofthisamplitude, which issimply IbI2 times I(CIk)(l< IS)|2. Thesecond factor isalmost independent oflocation inthe crystal, andallphases have disappeared intaking theabsolute square. The 3-8 probability ofscattering from anynucleus inthecrystal with spin ﬂipisnow lblgEli|<Cl/<></<l5>l2,k=1 which willshow asmooth distribution asinFig.3—6(b). You may argue, “Idon’t carewhich atom isup.” Perhaps youdon’t, but nature knows; andtheprobability is,infact.what wegave above-there isn’tany interference. Ontheother hand, ifweaskfortheprobability thatthespinisupat thedetector andalltheatoms stillhave spindown, thenwemust taketheabsolute square ofN Z<<r|i>a<i|s>. Since theterms inthissum have phases, they dointerfere, and wegetasharp interference pattern. Ifwedoanexperiment inwhich wedon’t observe thespin ofthedetected neutron, then both kinds ofevents canoccur; and theseparate probabilities add. The total probability (orcounting rate) asafunction ofangle then looks likethegraph inFig.3—6(c). Let’s review thephysics ofthisexperiment. lfyoucould, inprinciple, distin- guish thealternative/inul states (even though youdonotbother todoso),thetotal, ﬁnal probability isobtained bycalculating theprobability foreach state (not the amplitude) and then adding them together. lfyou cannot distinguish theﬁnal states even inprinciple, then theprobability amplitudes must besummed before taking theabsolute square toﬁnd theactual probability. The thing you should notice particularly isthat ifyou were totrytorepresent theneutron byawave alone, youwould getthesame kind ofdistribution forthescattering ofadown- spinning neutron asforanup-spinning neutron. You would have tosaythat the “wave” would come from allthedifferent atoms andinterfere justasfortheup- spinning onewith thesame wavelength. Butweknow thatisnotthewayitworks. Soaswestated earlier, wemust becareful nottoattribute toomuch reality tothe waves inspace. They areuseful forcertain problems butnotforall. 3-4Identical particles Thenextexperiment wewilldescribe isonewhich shows oneofthebeautiful consequences ofquantum mechanics. Itagain involves aphysical situation in which athing canhappen intwoindisringzris/rublc ways, sothatthere isaninter- ference ofamplitudes asisalways true insuch circumstances. Wearegoing to discuss thescattering, atrelatively lowenergy, ofnuclei onother nuclei. We start bythinking ofoz-p£il‘tiClCS (which, asyouknow, arehelium nuclei) bombarding, say,oxygen. Tomake iteasier forustoanalyze thereaction. wewilllook atitin thecenter-of-mass system, inwhich theoxygen nucleus and thea-particle have their velocities inopposite directions before thecollision andagain inexactly opposite directions after thecollision. SeeFig.3—7(a). (The magnitudes ofthe velocities are,ofcourse, diﬁerent, since themasses arediﬂicrent.) Wewillalso suppose that there isconservation ofenergy andthat thecollision energy islow enough thatneither particle isbroken uporleftinanexcited state. Thereason that thetwoparticles deﬂect each other is,ofcourse. thateach particle carries apositive charge and, classically speaking. there isanelectrical repulsion asthey goby. The scattering willhappen atdifferent angles with different probabilities. andwe would liketodiscuss something about theangle dependence ofsuch scatterings. (ltispossible, ofcourse. tocalculate thisthing classically, anditisoneofthe most remarkable accidents ofquantum mechanics that theanswer tothis problem comes outthesame asitdoesclassically. Thisisacurious point because ithappens fornoother force except theinverse square law—so itisindeed anaccident.) Theprobability ofscattering indifferent directions canbemeasured byan experiment asshown inFig.3~7(a). Thecounter atposition lcould bedesigned todetect only a-particles; thecounter atposition 2could bedesigned todetect 3-9 D‘ Dt G O Q 9 9aPARTICLE OXYGEN G-PARTICLE OXYGEN Q F40 O—-t :1-Q 1r-9 Q G D2 (0) D2 (b) Fig.3-7. Thescattering ofor-particles from oxygen nuclei, asseen inthecenter-of-muss system. only oxygen-—just asacheck. (Inthelaboratory system thedetectors would not beopposite; butintheCMsystem theyare.) Ourexperiment consists inmeasuring theprobability ofscattering invarious directions. Let’s callf(0)theamplitude to scatter intothecounters when they areattheangle 6;then |f(6)l2 willbeour experimentally determined probability. Now wecould setupanother experiment inwhich ourcounters would respond toeither theor-particle ortheoxygen nucleus. Then wehave towork outwhat happens when wedonotbother todistinguish which particles arecounted. Of course, ifwearetogetanoxygen intheposition 9,there must beanat-pﬂl'llClC on theopposite sideattheangle (rr—6),asshown inFig.3—7(b). Soiff(9)isthe amplitude foror-scattering through theangle 0,thenf(1r —6)istheamplitude foroxygen scattering through theangle 6.1‘Thus, theprobability forhaving some particle inthedetector atposition lis: Probability ofsome particle inD1=[f(0)|2 -l-lf(1r —6)l2. (3.14) Note that thetwostates aredistinguishable inprinciple. Even though inthis experiment wedonordistinguish them, wecould. According totheearlier dis- cussion, then, wemust addtheprobabilities, nottheamplitudes. Theresult given above iscorrect foravariety oftarget nuclei—for a-particles onoxygen, oncarbon, onberyllium, onhydrogen. Butitiswrong fora-particles on or-p8I'tiCl€S. Fortheonecase inwhich both particles areexactly thesame, the experimental data disagree with theprediction of(3.14). Forexample, the scattering probability at90°isexactly twice what theabove theory predicts and hasnothing todowith theparticles being “helium” nuclei. Ifthetarget isHe“, buttheprojectiles area-particles (He“), then there isagreement. Only when the target isI-le4—so itsnuclei areidentical with theincoming 01-p3rIiClC—(lO6S the scattering vary inapeculiar waywith angle. Perhaps youcanalready seetheexplanation. There aretwoways togetan a-particle intothecounter: byscattering thebombarding 04-pal‘tlClC atanangle 6. orbyscattering itatanangle of(vr—0).How canwetellwhether thebombarding particle orthetarget particle entered thecounter? Theanswer isthatwecannot. Inthecaseofat-particles withor-particles there aretwoalternatives thatcannot be distinguished. Here, wemust lettheprobability amplitudes interfere byaddition, 1'Ingeneral, ascattering direction should, ofcourse. bedescribed bytwoangles, the polar angle ¢,aswellastheazimuthal angle 0.Wewould thensaythatanoxygen nucleus at(6,¢)means thattheor-particle isat(‘Ir-0,¢+1r).However, forCoulomb scattering (andformany other cases), thescattering amplitude isindependent of4>.Then theampli- tude togetanoxygen at0isthesame astheamplitude togetthea-particle at(vr—6). 3-10 DI SPINUP 9ELECTRON ELECTRON ELECTRON Q—r 1—Q I r8Dr SPINUP ELECTRON SPIN SPIN SPIN UP UP UP SPIN st-mUP UP O() D D2 Z (blI"-1O SPIN UP Fig. 3-8. The scattering ofelectrons onelectrons. Iftheincoming electrons have pcirollel spins, the processes (oi)and(b)areindistinguishable. andtheprobability ofﬁnding anor-p3.l'tlClC inthecounter isthesquare oftheir sum: Probability ofanot-p3l‘llClC atD1=|f(6) +f(1r -—0)l2. (3.l5) This isquite adifferent result than that inEq.(3.14). Wecantake anangle of1r/2asanexample, because itiseasytoﬁgure out. For0=1r/2,weobviously have f(6) =f(rr ——6),sotheprobability inEq. (3.15) becomes lf('rr/2) —l— f(1r/2)|2 =4|f(rr/2)l2' Ontheother hand, ifthey didnotinterfere, theresult ofEq. (3.14) gives only 2lf(1r/2)l2. Sothere istwice asmuch scattering at90°aswemight have expected. Ofcourse, atother angles theresults willalso bedifferent. And soyou have theunusual result thatwhen particles areidentical, acertain newthing hap- pens that doesn’t happen when particles canbedistinguished. lnthemathematical description youmust addtheamplitudes foralternative process inwhich thetwo particles simply exchange roles andthere isaninterference. Aneven more perplexing thing happens when wedothesame kind ofexperi- ment byscattering electrons onelectrons. orprotons onprotons. Neither ofthe above results isthencorrect! Forthese particles, wemust invoke stillanewrule. amost peculiar rule, which isthefollowing: When youhave asituation inwhich theidentity oftheelectron which isarriving atapoint isexchanged with another one, thenew amplitude interferes with theoldonewith anopposite phase. ltis interference allright, butwith aminus sign. Inthecaseofor-pHI‘tiCl€S, when you exchange theor-particle entering thedetector, theinterfering amplitudes interfere withthepositive sign. Inthecaseofelectrons, theinterfering amplitudes forexchange interfere with anegative sign. Except foranother detail tobediscussed below, the proper equation forelectrons inanexperiment liketheoneshown inFig.3-8is Probability ofeatD1=|f(6) —f(1r —6)l2. (3.16) Theabove statement must bequaliﬁed, because wehave notconsidered the spinoftheelectron (oi-particles have nospin). Theelectron spinmaybeconsidered tobeeither “up” or“down” with respect totheplane ofthescattering. Ifthe energy oftheexperiment islowenough, themagnetic forces duetothecurrents willbesmall andthespinwillnotbeallected. Wewillassume thatthisisthecase forthepresent analysis, sothatthere isnochance thatthespins arechanged during thecollision. Whatever spintheelectron has,itcarries along with it.Now you seethere aremany possibilities. Thebombarding andtarget particles canhave both spins up,both down, oropposite spins. Ifboth spins areup,asinFig.3-8 (orifboth spins aredown), thesame willbetrueoftherecoil particles andthe amplitude fortheprocess isthedrﬂerence oftheamplitudes forthetwopossibilities 3-11 0 ELECTRON ELECTRON ELECTRONQ m SPIN UP SPIN DOWN ELECTRON .5*»: SPl1N ' '1—5p|N UP UP1 I SPIN DOWN DOWN 1r—9 5"" spmur> DOWN D2 (O) D2 Fig. 3-9. Thescattering ofelectrons withontipurollel spins. shown inFig. 3—8(a) and (b). Theprobability ofdetecting anelectron inD1is then given byEq.(3.16). Suppose, however, the“bombarding” spin isupandthe“target” spin isdown. Theelectron entering counter 1canhave spin uporspin down, andbymeasuring thisspinwecantellwhether itcame from thebombarding beam orfrom thetarget. The twopossibilities areshown inFig. 3—9(a) and(b);they aredistinguishable in principle, andhence there willbenointerference-—merely anaddition ofthetwo probabilities. Thesame argument holds ifboth oftheoriginal spins arereversed—— thatis,iftheleft-hand spinisdown andtheright-hand spinisup. Now ifwetakeourelectrons atrandom—as from atungsten ﬁlament inwhich theelectrons arecompletely unpolarized—then theodds areﬁfty-ﬁfty thatany particular electron comes outwith spinuporspindown. Ifwedon‘t bother to measure thespinoftheelectrons atanypoint intheexperiment, wehave what we callanunpolarized experiment. Theresults forthisexperiment arebestcalculated bylisting allofthevarious possibilities aswehave done inTable 3-1. Aseparate probability iscomputed foreach distinguishable alternative. Thetotal probability isthen thesumofalltheseparate probabilities. Note thatforunpolarized beams theresult for6=1r/2isone-half thatoftheclassical result with independent particles. Thebehavior ofidentical particles hasmany interesting consequences; wewilldiscuss them ingreater detail inthenextchapter. Table 3-1 Scattering ofunpolarized spinone-half particles Fraction Spin of Spin of Spin at Spin at ofcases particle 1 particle 2 D1 D2 Probability Inw-/e—eP l.rw>~for—wit I/<@>|‘*’ ‘ I/(tr-ml? lf(1r-9)l2 I/(wigup up up up 4:»-\in-— down down down down up down % up down down up up down down up1- down up Total probability =%lf(9) —f(1r~9)l2+%|f(9)l2 +%lf(1r —9)l2 3-12 4 Identical Particles 4-1Bose particles andFermi particles Inthelastchapter webegan toconsider thespecial rules fortheinterference that occurs inprocesses with two identical particles. Byidentical particles we mean things likeelectrons which caninnowaybedistinguished onefrom another. Ifaprocess involves two particles that areidentical, reversing which onearrives atacounter isanalternative which cannot bedistinguished and—like allcases of alternatives which cannot bedistinguished interferes with theoriginal, un- exchanged case. Theamplitude foranevent isthen thesumofthetwointerfering amplitudes; but,interestingly enough, theinterference isinsome cases with the same phase and, inothers, with theopposite phase. Suppose wehave acollision oftwoparticles aandbinwhich particle ascatters inthedirection 1andparticle bscatters inthedirection 2,assketched inFig. 4—l(a). Let’s callf(6)theamplitude forthisprocess; then theprobability P1of observing such anevent isproportional to|f(6)12Ofcourse, itcould alsohappen thatparticle bscattered intocounter landparticle awent intocounter 2,asshown inFig. 4-l(b). Assuming that there arenospecial directions deﬁned byspins orsuch, theprobability P2forthisprocess isjust|f(1r -—6)|2,because itisjust equivalent totheﬁrst process with counter lmoved over totheangle 1r—0. You might alsothink thattheamplitude forthesecond process isjustf(‘If—0). Butthat isnotnecessarily so,because there could beanarbitrary phase factor. That is,theamplitude could be Q“f(1r-0). Such anamplitude stillgives aprobability P2equal to|f(1r —0)|2 Now let’sseewhat happens ifaandbareidentical particles. Then thetwo different processes shown inthetwodiagrams ofFig.4-1cannot bedistinguished. There isanamplitude thateither aorbgoes intocounter 1,while theother goes intocounter 2.This amplitude isthesumoftheamplitudes forthetwoprocesses shown inFig.4-1. Ifwecalltheﬁrst onef(6), then thesecond oneise’5‘5f(1r —6), where nowthephase factor isveryimportant because wearegoing tobeadding twoamplitudes. Suppose wehave tomultiply theamplitude byacertain phase factor when weexchange theroles ofthetwoparticles. Ifweexchange them again weshould getthesame factor again. Butwearethen back totheﬁrstprocess. I C)» e*OO4-1 Bose particles andFermi particles 4-2States withtwoBose particles 4-3 States with nBose particles 4-4 Emission andabsorption of photons 4-5 The blackbody spectrum 4-6 Liquid helium 4~7Theexclusion principle Review: Blackbody radiation in: Chapter 41,Vol. I,TheBrown ianMovement Chapter 42,Vol. I,Applica lions ofKinetic Theory \e 6 , 1Q 0 b ° b 2<0) 21r-9 Fig.4—l. Inthescattering oftwo identical particles, theprocesses (ct)and (b) areindistinguishable. 4—l PROTONNEUTRON a-particle_l (0/a " >/ l (b) Fig.4—2. Thescattering oftwo a-porticles. ln(clthetwo particles retain their identity; inlb)oneutron isexchanged during thecollision. Thephase factor taken twice must bring usback where westarted—its square must beequal tol.There areonly twopossibilities: eiaisequal to+1,orisequal to—l.Either theexchanged casecontributes with thesame sign, oritcontributes with theopposite sign. Both cases exist innature, each foradifferent class ofpar- ticles. Particles which interfere with apositive signarecalled Bose particles and those which interfere with anegative sign arecalled Fermi particles. TheBose particles arethephoton, themesons, andthegraviton. TheFermi particles are theelectron, themuon, theneutrinos, thenucleons, andthebaryons. Wehave, then, thattheamplitude forthescattering ofidentical particles is: Bose particles: (Amplitude direct) —l—(Amplitude exchanged). (4.1) Fermi particles.‘ (Amplitude direct) ~(Amplitude exchanged). (4.2) Forparticles with spin—like electrons—there isanadditional complication. Wemust specify notonlythelocation ofthe particles butthedirection oftheir spins. ltisonlyforidentical particles withidentical spinstates thattheamplitudes interfere when theparticles areexchanged. lfyouthink ofthescattering ofunpolarized beams—which areamixture ofditlerent spin states—there issome extra arithmetic. Now aninteresting problem arises when there aretwoormore particles bound tightly together. Forexample, ana-particle hasfour particles init——two neutrons andtwoprotons. When twoat-particles scatter, there areseveral possibilities. Itmay bethat during thescattering there isacertain amplitude that oneofthe neutrons willleapacross from oneoi-particle totheother, while aneutron from the other or-p3l‘llCl6 leaps theother waysothatthetwoalphas which come outofthe scattering arenottheoriginal ones—there hasbeen anexchange ofapair of neutrons. SeeFig.4-2. Theamplitude forscattering with anexchange ofapair ofneutrons willinterfere with theamplitude forscattering with nosuch exchange, andtheinterference must bewith aminus signbecause there hasbeen anexchange ofonepairofFermi particles. Ontheother hand, iftherelative energy ofthetwo at-p3I‘liCl6S issolowthat they stay fairly farapart—say, duetotheCoulomb repulsion-—and there isnever anyappreciable probability ofexchanging anyof theinternal particles, wecanconsider thea-particle asasimple object, andwedo notneed toworry about itsinternal details. lnsuch circumstances, there areonly twocontributions tothescattering amplitude. Either there isnoexchange, orall four ofthenucleons areexchanged inthescattering. Since theprotons andthe 4-2 neutrons intheat-particle areallFermi particles, anexchange ofanypairreverses thesignofthescattering amplitude. Solong asthere arenointernal changes in thea-particles. interchanging thetwoat-particles isthesame asinterchanging four pairs ofFermi particles. There isachange insignforeach pair, sothenetresult isthattheamplitudes combine withapositive sign. Theat-particle behaves likea Bose particle. Sotheruleisthatcomposite objects, incircumstances inwhich thecomposite object canbeconsidered asasingle object, behave likeFermi particles orBose particles, depending onwhether theycontain anoddnumber oraneven number ofFermi particles. Alltheelementary Fermi particles wehave mentioned—such astheelectron, theproton, theneutron, andsoon—have aspinj =1/2. Ifseveral such Fermi particles areputtogether toform acomposite object, theresulting spinmaybe either integral orhalf-integral. Forexample, thecommon isotope ofhelium, He4, which hastwoneutrons andtwoprotons, hasaspin ofzero, whereas Li? which hasthree protons andfourneutrons, hasaspinof3/2.Wewilllearn laterthe rules forcompounding angular momentum, andwilljustmention nowthatevery composite object which hasahalf-integral spinimitates aFermi particle, whereas every composite object with anintegral spinimitates aBoseparticle. Thisbrings upaninteresting question: Why isitthatparticles withhalf-integral spin areFermi particles whose amplitudes add with theminus sign, whereas particles withintegral spinareBose particles whose amplitudes addwiththeposi- tivesign? Weapologize forthefactthat wecannot give youanelementary ex- planation. Anexplanation hasbeen worked outbyPauli from complicated argu- ments ofquantum ﬁeld theory andrelativity. Hehasshown that thetwomust necessarily gotogether, butwehave notbeen abletoﬁndawayofreproducing his arguments onanelementary level. Itappears tobeoneofthefewplaces inphysics where there isarule which canbestated very simply, butforwhich noone hasfound asimple andeasy explanation. Theexplanation isdeep down inrela- tivistic quantum mechanics. This probably means thatwedonothave acomplete understanding ofthefundamental principle involved. Forthemoment, youwill justhave totakeitasoneoftherules oftheworld. 4-2 States with twoBose particles Now wewould liketodiscuss aninteresting consequence oftheaddition rule forBose particles. Ithastodowith their behavior when there areseveral particles present. Webegin byconsidering asituation inwhich twoBose particles arescat- tered from twodifferent scatterers. Wewon’t worry about thedetails ofthescatter- ingmechanism. Weareinterested only inwhat happens tothescattered particles. Suppose wehave thesituation shown inFig. 4-3. The particle aisscattered into thestate 1.Byastate wemean agiven direction andenergy, orsome other given condition. Theparticle bisscattered into thestate 2.Wewant toassume that the twostates land2arenearly thesame. (What wereally want toﬁndouteventually istheamplitude that thetwo particles arescattered into identical directions, or states; butitisbest ifwethink ﬁrst about what happens ifthestates arealmost thesame andthen work outwhat happens when they become identical.) Suppose that wehadonly particle a;then itwould have acertain amplitude forscattering indirection l,say(1|a).And particle balone would have theampli- tude (2lb)forlanding indirection 2.Ifthetwoparticles arenotidentical, the amplitude forthetwoscatterings tooccur atthesame timeisjusttheproduct <1la><2Ib>- Theprobability forsuchaneventisthen l<1l@>(2Ib>l2, l<1la>|2l(2 lb>|2-which isalsoequal to 4-3/tiFig. 4-3. Adouble scattering into nearby ﬁnol states. Tosavewriting forthepresent arguments, wewillsometimes set <1Ia>=a1i <2Ibi=b2- Then theprobability ofthedouble scattering is Ia1I2Ib2I2- Itcould alsohappen that particle bisscattered intodirection l,while particle agoes into direction 2.The amplitude forthisprocess is <2Ia)(lIb). andtheprobability ofsuch anevent is I<2Ir1><1Ibllz=I@2I2Ib1I2- lmagine now thatwehave apairoftiny counters thatpick upthetwoscattered particles. The probability P2that they willpick uptwoparticles together isjust thesum P2=I"1I2Ib2I2 +la2l2Ib1I2~ (4-3) Now let’s suppose that thedirections 1and 2arevery close together. We expect that ashould vary smoothly with direction, soa1anda2must approach each other asland2getclose together. lfthey areclose enough, theamplitudes a1 anda2willbeequal. Wecanseta1=asandcallthem both justa;similarly, we setbl=bg=b.Then wegetthat P2=2IaI2IbI2. (4.4) Now suppose, however, that aandbareidentical Bose particles. Then the process ofagoing into landbgoing into 2cannot bedistinguished from theex- changed process inwhich agoes into2andbgoes into l.lnthiscase theamplitudes forthetwo diﬁerent processes can interfere. The total amplitude toobtain a particle ineach ofthetwocounters is <1I¢1><2 I11>+<2Ia>(l|b>- (4-5) And theprobability that wegetapair istheabsolute square ofthisamplitude, P2=Ia1b2 -I"a2b1I2 I4IaI2IbI2' (4-6) Wehave theresult that itistwice aslikely toﬁnd two identical Bose particles scattered into thesame state asyou would calculate assuming t/1eparticles were different. Although wehave been considering that thetwo particles areobserved in separate counters, thisisnotessential—as wecanseeinthefollowing way. Let’s imagine that both thedirections land 2would bring theparticles into asingle small counter which issome distance away. Wewillletthedirection lbedeﬁned bysaying thatitheads toward theelement ofarea dS1ofthecounter. Direction 2 heads toward thesurface element (LS2ofthecounter. (We imagine thatthecounter presents asurface atright angles tothelinefrom thescatterings.) Now wecannot give aprobability that aparticle willgointo aprecise direction ortoaparticular point inspace. Such athing isimpossible—the chance foranyexact direction is zero. When wewant tobesospeciﬁc, weshall have todeﬁne ouramplitudes so that they give theprobability ofarriving perunitarea ofacounter. Suppose that wehadonly particle a;itwould have acertain amplitude forscattering indirection l.Let’s deﬁne (1|a) a1tobetheamplitude that awillscatter intoaunitarea ofthecounter inthedirection l.lnother words, thescale ofa1ischosen—we sayitis“normalized” sothat theprobability that itwillscatter intoanelement ofarea (IS,is lCl>I2dS1 =l(11I2dS1. 4-4 Ifourcounter hasthetotal areaAS,andweletdS1range overthisarea, thetotal probability thattheparticle awillbescattered intothecounter is /AS|a1|2dS1. (4.8) Asbefore, wewant toassume thatthecounter issuﬂiciently small sothatthe amplitude a,doesn’t vary signiﬁcantly over thesurface ofthecounter; a1isthen a constant amplitude which wecancalla.Then theprobability thatparticle ais scattered somewhere into thecounter is p,=|a|2AS. (4.9) Inthesame way, wewillhave thattheprobability thatparticle b—-when itis a1one—scatters intosome element ofarea, saya'S2, is |b2|2dS2. (We usedS2instead ofdS1because wewilllater want aandbtogointo diﬂ“erent directions.) Again wesetb2equal totheconstant amplitude b;thentheprobability that particle biscounted inthedetector is pi,=1b|2AS. (4.10) Now when both particles arepresent, theprobability that aisscattered into dS1andbisscattered intodS2is |a1b2|2dS1dS'2 : |al21b|2dS1 Ifwewant theprobability thatbothaandbgetintothecounter, weintegrate both dS1anddS2over ASandﬁnd that P2=|a|2|b|2 (AS)2. (4.12) Wenotice, incidentally, thatthisisjustequal topa-pb,justasyouwould suppose assuming thattheparticles aandbactindependently ofeach other. When thetwoparticles areidentical, however, there aretwoindistinguishable possibilities foreach pair ofsurface elements dS1 anddS2. Particle agoing into dS2andparticle bgoing intodS1isindistinguishable from aintodS1andbinto dS2, sotheamplitudes forthese processes will interfere. (When wehad two dzﬂerenl particles above—although wedidnotinfact care which particle went where inthecounter—~we could, inprinciple, have found out; sothere was no interference. Foridentical particles wecannot tell,even inprinciple.) Wemust write, then, that theprobability that thetwo particles arrive atdS1 and dS2 is |a1b2 + G2b1|2dS1 Now, however, when weintegrate over thearea ofthecounter, wemust becareful. IfweletdS1anddS2range overthewhole areaAS,wewould count each part of thearea twice since (4.13) contains everything that canhappen with anypair of surface elements dS1anddS2T Wecanstilldotheintegral thatway, ifwecorrect forthedouble counting bydividing theresult by2.Wegetthen thatP2foridentical Bose particles is P2(Bose) =%{4|a\2|b\2(AS)2} =2\a|21b|2(As)2. (4.14) Again, thisisjusttwice what wegotinEq.(4.12) fordistinguishable particles. Ifweimagine foramoment that weknew that thebchannel hadalready sent itsparticle intotheparticular direction, wecansaythattheprobability thata second particle willgointo thesame direction istwice asgreat aswewould have TIn(4.11) interchanging dS1anddSggives adifferent event, soboth surface elements should range over thewhole area ofthecounter. In(4.13) wearetreating dS1anddS2 asapairandincluding everything thatcanhappen. Iftheintegrals include again what happens when dS1anddS2arereversed, everything iscounted twice. 4-5 I 2 Fig. 4-4. The scattering ofnpct cles into nearby ﬁnal states.expected ifwehadcalculated itasanindependent event. Itisaproperty ofBose particles that ifthere isalready oneparticle inacondition ofsome kind, the probability ofgetting asecond oneinthesame condition istwice asgreat asit would beiftheﬁrstonewere notalready there. This factisoften stated inthe following way: Ifthere isalready oneBose particle inagiven state, theamplitude forputting anidentical oneontopofitis\/2greater than ifitweren’t there. (This isnotaproper wayofstating theresult from thephysical point ofview we have taken, butifitisusedconsistently asarule, itwill,ofcourse, givethecorrect result.) 4-3States withnBose particles Let’s extend ourresult toasituation inwhich there arenparticles present. Weimagine thecircumstance shown inFig.4-4. Wehave nparticles a,b,c,..., which arescattered andendupinthedirections l,2,3,..,n.Allndirections areheaded toward asmall counter along distance away. Asinthelastsection, wechoose tonormalize alltheamplitudes sothat theprobability thateach particle acting alone would gointoanelement ofsurface dSofthecounter is |<>|2d$- First, let’sassume thattheparticles arealldistinguishable; thentheprobability thatnparticles willbecounted together inndifferent surface elements is |a1b2c3.. l2dS1 Again wetake thattheamplitudes don’t depend onwhere dSislocated inthe counter (assumed small) andcallthem simply a,b,c,...The probability (4.15) becomes [al2lb|2lc|2...dS1dS2 dS3.. (4.16) Integrating each dSoverthesurface ASofthecounter, wehave thatP,(different), theprobability ofcounting ndifferent particles atonce, is P,(different) =|al2|b]2|c[2 ..(AS)" (4.17) This isjusttheproduct oftheprobabilities foreach particle toenter thecounter separately. They allactindependently—the probability foronetoenter does not depend onhowmany others arealsoentering. Now suppose thatalltheparticles areidentical Bose particles. Foreach set ofdirections 1,2,3, ...there aremany indistinguishable possibilities. Ifthere were, forinstance, justthree particles, wewould have thefollowing possibilities: a—>1 a—~>l a—>2 b—>2 b—>3 b—>l c—»3 c—>2 c—>3 a—>2 a~>3 a—>3 b~>3 b—>l b—>2 c—+l c—>2 c->1 There aresixdifferent combinations. With nparticles, there aren!different, but indistinguishable, possibilities forwhich wemust addamplitudes. Theprobability thatnparticles willbecounted innsurface elements isthen |(l1b2C3.. + a1b3C2. + a2b1c3.. + l12b3C1.. +CtC. +CtC.|2dS1dS2dS3...dSn. Once more weassume thatallthedirections aresoclose thatwecanseta,= a2= =a=an,andsimilarly forb,c,.;theprobability of(4.18) becomes Inlabc. ..l2dS1dS2...dS,,. (4.19) 4—~6 When weintegrate each a'Sover theareaASofthecounter, each possible product ofsurface elements iscounted n!times; wecorrect forthisbydividing byn!andget P,,(Bose) = |n!abc ...12(AS)" OI‘ P,,(Bose) =nl|abc...12(AS)". (4.20) Comparing thisresult with Eq.(4.17), Weseethat theprobability ofcounting n Bose particles together isn!greater than wewould calculate assuming that the particles were alldistinguishable. Wecansummarize ourresult thisway: P,,(Bose) =n!P,,(different). (4.21) Thus, theprobability intheBose case islarger byn!than you would calculate assuming that theparticles acted independently. Wecanseebetter what thismeans ifweaskthefollowing question: What is theprobability that aBose particle willgointo aparticular state when there are already nothers present? Let’s callthenewly added particle w.Ifwehave (n+1) particles, including w,Eq.(4.20) becomes P,,+1(Bose) =(n+1)!|abc...wl2(AS)”'+‘. (4.22) Wecanwrite thisas P,,+,(Bose) ={(n-1-1)|w|2 AS}nl labc...12AS" t\A or ~*”"r P,,+1(Bose) =(n+1)lw|2_AS_P,,(Bose).__ (4.23) Wecanlook atthisresult inthefollowing way: The number |w|2ASisthe probability forgetting particle wintothedetector ifnoother particles were present; P,,(Bose) isthechance that there arealready nother Bose particles present. So Eq.(4.23) says that when there arenother identical Bose particles present, the probability thatonemore particle willenter thesame state isenhanced bythefactor (n-1-1).Theprobability ofgetting aboson, where there arealready n,is(n+1) times stronger than itwould beifthere were none before. Thepresence oftheother particles increases theprobability ofgetting onemore. 4-4Emission andabsorption ofphotons Throughout ourdiscussion wehave talked about aprocess likethescattering ofoi-particles. Butthatisnotessential; wecould have been speaking ofthecreation ofparticles, asforinstance theemission oflight. When thelight isemitted, a photon is“created.” Insuch acase, wedon’t need theincoming lines inFig. 4~4; wecanconsider merely that there arenatoms a,b,c,...emitting light, asin Fig.4-5. Soourresult canalso bestated: Theprobability thatanatom willemit aphoton intoaparticular ﬁnal state isincreased bythefactor (n+1)ifthere are already nphotons inthatstate. People liketosummarize thisresult bysaying thattheamplitude toemit a photon isincreased bythefactor \/n+1when there arealready nphotons present. Itis,ofcourse, another wayofsaying thesame thing ifitisunderstood to mean that thisamplitude isjusttobesquared togettheprobability. Itisgenerally true inquantum mechanics that theamplitude togetfrom any condition ¢toanyother condition Xisthecomplex conjugate oftheamplitude to getfrom Xto¢: <><I¢>=<¢I><>*. (4-24) Wewilllearn about thislawalittle later, butforthemoment wewilljustassume itistrue. Wecanuseittoﬁndouthow photons arescattered orabsorbed outofa given state. Wehave thattheamplitude thataphoton willbeadded tosome state, sayi,when there arealready nphotons present is,say, (n+ lln) =\/n+ la, (4.25) 4-7atFig. 4-5. The creation ofnphotons innearby states. 8 1AE=T1w lenouno srarsq (<1) e 1AE=hw 1enouno sun: 9 (bl Fig. 4-6. Radiation and absorption ofaphoton with thefrequency at.where a=(iIa)istheamplitude when there arenoothers present. Using Eq. (4.24), theamplitude togotheother way—from (n+1)photons ton—is (n|n +1)=\/rt+la*. (4.26) Thisisn’tthewaypeople usually sayit;theydon’t liketothink ofgoing from (n+1)ton,butprefer always tostart with nphotons present. Then they say thattheamplitude toabsorb aphoton when there arenpresent—in other words, togofrom nto(n—l)—is (n—1In)=\/ha*. (4.27) which is,ofcourse, justthesame asEq.(4.26). Then theyhave trouble trying to remember when touse\/hor\/n+l.Here’s thewaytoremember: Thefactor isalways thesquare rootofthelargest number ofphotons present, whether itis before orafter thereaction. Equations (4.25) and(4.26) show thatthelawis really symmetric-it only appears unsymmetric ifyouwrite itasEq.(4.27). There aremany physical consequences ofthese newrules; wewant todescribe oneofthem having todowiththeemission oflight. Suppose weimagine asituation inwhich photons arecontained inabox—you canimagine aboxwithmirrors for walls. Now saythatintheboxwehave nphotons, allofthesame state—the same frequency, direction, andpo1arization—so they can’t bedistinguished, andthat alsothere isanatom inth_eboxthatcanemit another photon intothesame state. Then theprobability thatitwillemit aphoton is (H+1)|<1|2. (4-23) andtheprobability thatitwillabsorb aphoton is nlal2, (4.29) where \a|2istheprobability itwould emit ifnophotons were present. Wehave already discussed these rules inasomewhat different way inChapter 42ofVol. l. Equation (4.29) says that theprobability that anatom willabsorb aphoton and make atransition toahigher energy state isproportional totheintensity ofthe light shining onit.But, asEinstein ﬁrstpointed out,therateatwhich anatom will make atransition downward hastwo parts. There istheprobability that itwill make aspontaneous transition |al2,plustheprobability ofaninduced transition nla|2,which isproportional totheintensity ofthelight—that is,tothenumber of photons present. Furthermore, asEinstein said, thecoefficients ofabsorption and ofinduced emission areequal andarerelated totheprobability ofspontaneous emission. What welearn here isthat ifthelight intensity ismeasured interms of thenumber ofphotons present (instead ofastheenergy perunitarea, andpersec), thecoefficients ofabsorption ofinduced emission andofspontaneous emission are allequal. This isthecontent oftherelation between theEinstein coefficients AandBofChapter 42,Vol. I,Eq.(42.18). 4-5Theblackbody spectrum Wewould liketouseourrules forBose particles todiscuss once more the spectrum ofblackbody radiation (seeChapter 42,Vol.I).Wewilldoitbyﬁnding outhowmany photons there areinaboxiftheradiation isinthermal equilibrium with some atoms inthebox. Suppose thatforeach light frequency w,there area certain number Nofatoms which have twoenergy states separated bytheenergy AE=hw. SeeFig.4-6. We’ll callthelower-energy state the“ground” state andtheupper state the“excited” state. LetNaandN,betheaverage numbers of atoms intheground andexcited states; then inthermal equilibrium atthetem- perature T,wehave from statistical mechanics that %;=@"°E"” =@—-/". (4.30) 4~8 Each atom intheground state canabsorb aphoton andgointotheexcited state, andeach atom intheexcited state canemit aphoton andgototheground state. Inequilibrium, therates forthese twoprocesses must beequal. Therates areproportional totheprobability fortheevent and tothenumber ofatoms present. Let’s lethbetheaverage number ofphotons present inagiven state with thefrequency w.Then theabsorption ratefrom that state isNaﬁ|a| 2,andthe emission rateintothatstate isNe(ﬁ +1)|al2.Setting thetworates equal, wehave that Ngn=N,,(ﬁ +1). (4.31) Combining thiswith Eq.(4.30), wehave 71 =e_n.../tr E+1 ' Solving forE,wehave __ 1 which isthemean number ofphotons inanystate with frequency w,foracavity in thermal equilibrium. Since each photon hastheenergy hw,theenergy inthe photons ofagiven state isaha, or hw2575: - (4.33) Incidentally, weonce found asimilar equation inanother context [Chapter 41,Vol.I,Eq.(4l.l5)]. You remember thatforanyharmonic oscillator—such as aweight onaspring—the quantum mechanical energy levels areequally spaced with aseparation hw,asdrawn inFig.4-7. Ifwecalltheenergy ofthenthlevel nhw, weﬁndthat themean energy ofsuch anoscillator isalso given byEq.(4.33). Yetthisequation wasderived hereforphotons, bycounting particles, anditgives thesame results. That isoneofthemarvelous miracles ofquantum mechanics. Ifonebegins byconsidering akind ofstate orcondition forBose particles which donotinteract with each other (wehave assumed that thephotons donotinteract with each other), andthen considers thatintothisstate there canbeputeither zero, orone, ortwo, ...uptoanynumber nofparticles, oneﬁnds thatthissystem behaves forallquantum mechanical purposes exactly likeaharmonic oscillator. Bysuch anoscillator wemean adynamic system likeaweight onaspring ora standing wave inaresonant cavity. Andthatiswhyitispossible torepresent the electromagnetic ﬁeld byphoton particles. From onepoint ofview, wecananalyze theelectromagnetic ﬁeldinaboxorcavity interms ofalotofharmonic oscillators, treating each mode ofoscillation according toquantum mechanics asaharmonic oscillator. From adifferent point ofview, wecananalyze thesame physics in terms ofidentical Bose particles. And theresults ofboth ways ofworking are always inexact agreement. There isnowaytomake upyour mind whether the electromagnetic ﬁeldisreally tobedescribed asaquantized harmonic oscillator or bygiving howmany photons there areineach condition. Thetwoviews turnout tobemathematically identical. Sointhefuture wecanspeak either about the number ofphotons inaparticular state inaboxorthenumber oftheenergy level associated with aparticular mode ofoscillation oftheelectromagnetic ﬁeld. They aretwoways ofsaying thesame thing. Thesame istrueofphotons infreespace. They areequivalent tooscillations ofacavity whose walls have receded toinﬁnity. Wehave computed themean energy inanyparticular mode inaboxatthe temperature T;weneed only onemore thing togettheblackbody radiation law: Weneed toknow how many modes there areateach energy. (We assume that for every mode there aresome atoms inthebox—or inthewalls-—which have energy levels that canradiate into that mode, sothat each mode cangetinto thermal equilibrium.) Theblackbody radiation lawisusually stated bygiving theenergy perunitvolume carried bythelight inasmall frequency interval from cotow-1-Aw. Soweneed toknow howmany modes there areinaboxwith frequencies inthe 4-9E1 E GROUND STATE Fig. 4-7. The energy levels ofa harmonic oscillator.5hw 4ftw 3'50: Zﬁw ﬁat ._ 0 % O i/'\_/ ‘ lV\A/\A/M 1' t< L > Fig. 4-8. The standing wave modes onaline. 1---~-1 ___‘____ ___ ______ __ L, ‘F 1+4? ___.____Sl-.____-__,_,.\ __-'_v_:'_________ L- Fig. 4-9. Standing wave modes in twodimensions.interval Aw.Although thisquestion continually comes upinquantum mechanics, itispurely aclassical question about standing waves. Wewillgettheanswer onlyforarectangular box. ltcomes outthesame fora box ofanyshape, butit’svery complicated tocompute forthearbitrary case. Also, weareonly interested inaboxwhose dimensions areverylarge compared with awavelength ofthelight. Then there arebillions andbillions ofmodes; there willbemany inanysmall frequency interval Aw,sowecanspeak ofthe “average number” inanyAwatthefrequency w.Let's start byasking howmany modes there areinaone-dimensional case—as forwaves onastretched string. You know that each mode isasine wave that hastogotozero atboth ends; inother words, there must beanintegral number ofhalf-wavelengths inthelength oftheline, asshown inFig. 4-8. Weprefer tousethewave number k=21r/)\; calling k,-thewave number ofthejthmode, wehave that -1rk,= (4.34) where jisanyinteger. Theseparation 6kbetween successive modes is at=/<,+,-k,-= Wewant toassume thatkLissolarge thatinasmall interval Ak,there aremany modes. Calling Aittthenumber ofmodes intheinterval Ak,wehave Ak LA571—E—7_Ak. (4.35) Now theoretical physicists working inquantum mechanics usually prefer to saythatthere areone-half asmany modes; theywrite L Wewould liketoexplain why. They usually liketothink interms oftravelling waves—some going totheright (with apositive k)andsome going totheleft (with anegative k).Buta“mode” isastanding wave which isthesumoftwowaves, onegoing ineach direction. Inother words, theyconsider each standing wave ascontaining twodistinct photon “states.” SoifbyA91,oneprefers tomean the number ofphoton states ofagiven k(where nowkranges overpositive andnega- tivevalues), oneshould thentakeAETLhalfasbig.(Allintegrals must nowgofrom k==-—1-tok=-1-L, andthetotal number ofstates uptoanygiven absolute value ofkwillcome outO.K.) Ofcourse, wearenotthen describing standing waves verywell, butwearecounting modes inaconsistent way. Now wewant toextend theresults tothree dimensions. Astanding wave ina rectangular boxmust have anintegral number ofhalf-waves along eachaxis. The situation fortwoofthedimensions isshown inFig.4-9. Each wave direction andfrequency isdescribed byavector wave number k.whose x.y,andzcompo- nents must satisfy equations likeEq.(4.34). Sowehave that gjutk,_L1 k=lﬂ. H Lu /<=1” Thenumber ofmodes with k,inaninterval Ak,,is,asbefore, 5211'Akr’ andsimilarly forAk-yandAk,. IfwecallA%)‘L(k) thenumber ofmodes foravector 4-10 wave number kwhose x-component isbetween k,andk,+Ak,,whose y-com- ponent isbetween kgandk,,+Aky, andwhose z-component isbetween k,and k,-1-Ak,,then LzL1LzA&tt(k) =T2;-1)? Ak,,Ak,,Ak,. (4.37) Theproduct LIL/ZILZ isequal tothevolume Vofthebox. Sowehave theimportant result thatforhigh frequencies (wavelengths small compared with thedimensions), thenumber ofmodes inacavity isproportional tothevolume Voftheboxand tothe“volume ink-space” AkxAk,Ak,. This result comes upagain andagain in many problems andshould bememorized: 3kamt) =V1%),4 (4.38) Although wehave notproved it,theresult isindependent oftheshape ofthebox. Wewillnowapply thisresult toﬁndthenumber ofphoton modes forphotons with frequencies intherange Aw. Wearejust interested intheenergy invarious modes—but notinterested inthedirections ofthewaves. Wewould liketoknow thenumber ofmodes inagiven range offrequencies. Inavacuum themagnitude ofkisrelated tothefrequency by C0 1k1--6- (4.39) Soinafrequency interval Aw,these areallthemodes which correspond tok’s with amagnitude between kand k+Ak,independent ofthedirection. The “volume ink-space” between kandk-1-Akisaspherical shell ofvolume 41rk2 Ak. Thenumber ofmodes isthen 2 A3Z(w) =_"_“(’2L';)3“" . (4.40) However, since wearenowinterested infrequencies, weshould substitute k=w/c, soweget V41rw2 AwA9Z((0) — There isonemore complication. Ifwearetalking about modes ofanelectro- magnetic wave, foranygiven wave vector kthere canbeeither oftwopolarizations (atright angles toeach other). Since these modes areindependent, wemust-—for light-—double thenumber ofmodes. Sowehave V2a .A$)‘L(w) = (for11g1~11). (4.42) Wehave shown, Eq.(4.33), thateach mode (oreach “state”) hasonthe average theenergy _ hwnhw =-eftw/kT _1 Multiplying thisbythenumber ofmodes, wegettheenergy AEinthemodes that lieintheinterval Aw: hw Vw2Aw This isthelawforthefrequency spectrum ofblackbody radiation, which wehave already found inChapter 41ofVol. I.Thespectrum isplotted inFig. 4-10. You seenowthattheanswer depends onthefactthatphotons areBose particles, which 4-111.4- ‘£13_1.2- Wt U-I 1.0‘\_/ N 1 ‘§1> oe- oe- 0.4- oz- . . . 1 1 1 O 1 2 34 5 6 1e11 I5- ‘M;/kT Fig. 4-10. The frequency spectrum ofradiation inacavity inthermal equilib- rium, the"blackbody" spectrum. have atendency totrytogetallinto thesame state (because theamplitude for doing soislarge). You willremember, itwasPlanck's study oftheblackbody spectrum (which wasamystery toclassical physics), andhisdiscovery ofthefor- mula inEq.(4.43) thatstarted thewhole subject ofquantum mechanics. 4-6Liquid helium Liquid helium hasatlowtemperatures many oddproperties which wecannot unfortunately take thetime todescribe indetail right now, butmany ofthem arise from thefactthatahelium atom isaBose particle. One ofthethings isthatliquid helium ﬂows without anyviscous resistance. Itis,infact, theideal “dry” water wehave been talking about inone oftheearlier chapters-provided that the velocities arelowenough. Thereason isthefollowing. lnorder fora liquid tohave viscosity, there must beinternal energy losses; there must besome wayforonepart oftheliquid tohave amotion that isdifferent from that oftherestoftheliquid. This means thatitmust bepossible toknock some oftheatoms intostates that aredifferent from thestates occupied byother atoms. Butatsulliciently low temperatures, when thethermal motions getvery small, alltheatoms trytoget into thesame condition. So,ifsome ofthem aremoving along, then alltheatoms trytomove together inthesame state. There isakind ofrigidity tothemotion, and itishard tobreak themotion upinto irregular patterns ofturbulence. as would happen, forexample. with independent particles. Soinaliquid ofBose particles, there isastrong tendency foralltheatoms togointo thesame state which isrepresented bythe\/it‘?-Ml factor wefound earlier. (For 11bottle of liquid helium nis,ofcourse, 11very large number!) This cooperative motion does nothappen athigh temperatures, because then there issutllcient thermal energy toputthevarious atoms into various different higher states. Butat11 sufficiently lowtemperature there suddenly comes amoment inwhich allthehelium atoms trytogointothesame state. Thehelium becomes asuperﬂuid. Incidentally. thisphenomenon onlyappears withtheisotope ofhelium which hasatomic weight 4.For thehelium isotope ofatomic weight 3,theindividuul atoms areFermi particles, andtheliquid isanormal ﬂuid. Since superﬁuidity occurs only with He“, itisevidently aquantum mechanical elfect—due tot11cBose nature ofthe oi-particle. 4-7Theexclusion principle Fermi particles actinacompletely different way. Let’s seewhat happens ifwetrytoputtwoFermi particles intothesame state. Wewillgoback toour original example andaskfortheamplitude that twoidentical Fermi particles will bescattered intoalmost exactly thesame direction. Theamplitude thatparticle awillgoindirection 1andparticle bwillgoindirection 2is <1|¢1><2| 11>, whereas theamplitude thattheoutgoing directions willbeinterchanged is <21¢I><1 lb)- Since wehave Fermi particles, theamplitude fortheprocess isthedifference of these twoamplitudes: <11a>(21b> —<21(1)111b>< (4-44) Let’s saythat by“direction 1”wemean that theparticle hasnotonly acertain direction butalsoagiven direction ofitsspin, andthat“direction 2"isalmost exactly thesame asdirection 1andcorresponds tothesame spin direction. Then (11a)and (21a)arenearly equal. (This would notnecessarily betrue ifthe outgoing states 1and2didnothave thesame spin, because there might besome reason why theamplitude would depend onthespindirection.) Now ifwelet 4-12 onea. Two THREE">?<// //ELECTRON NUCLEUS ELECTRONS //' ELECTRONS ,@/ é??? / A1 /” (°l (bl7/ Fig. 4~ll.How otoms might look ifelectrons behaved likeBose particles. directions land2approach each other, thetotal amplitude inEq.(4.44) becomes zero. The result forFermi particles ismuch simpler than forBose particles. It justisn't possible atallfortwoFermi particles-—such astwoelectrons~to get intoexactly thesame state. Youwillnever ﬁndtwoelectrons inthesame position with their twospins inthesame direction. Itisnotpossible fortwoelectrons to have thesame momentum andthesame spin directions. Ifthey areatthesame location orwith thesame state ofmotion, theonly possibility isthat they must be spinning opposite toeach other. What aretheconsequences ofthis? There areanumber ofmost remarkable ellects which areaconsequence oftheFactthattwoFermi particles cannot getinto thesame state. lnfact, almost allthepeculiarities ofthematerial world hinge on thiswonderful fact. Thevariety thatisrepresented intheperiodic table isbasically aconsequence ofthisonerule. Oicourse. wecannot saywhat theworld would belikeifthisonerule were changed, because itisjust apartofthewhole structure ofquantum mechanics, andit isimpossible tosaywhat elsewould change iftheruleabout Fermi particles were dillerent. Anyway, let'sjust trytoseewhat would happen ifonly thisonerulewere changed. First, wecanshow that every atom would bemore orlessthesame. Let's start with thehydrogen atom. Itwould notbenoticeably atlected. The proton ofthenucleus would besurrounded byaspherically symmetric electron cloud, asshown inFig.4~ll(a). Aswehave described inChapter 2,theelectron isattracted tothecenter, buttheuncertainty principle requires that there be abalance between theconcentration inspace and inmomentum. The balance means thatthere must beacertain energy andacertain spread intheelectron distribution which determines thecharacteristic dimension ofthehydrogen atom. Now suppose thatwehave anucleus with twounits ofcharge, such asthe helium nucleus. This nucleus would attract twoelectrons, andiftheywere Bose particles, theywould—except fortheir electric repulsion—both crowd inasclose aspossible tothenucleus. Ahelium atom might look asshown inpart(b)ofthe ﬁgure. Similarly, alithium atom which hasatriply charged nucleus would have anelectron distribution likethatshown inpart (c)ofFig.4—ll.Every atom would look more orlessthesame~a little round ballwith alltheelectrons sitting near thenucleus, nothing directional andnothing complicated. Because electrons areFermi particles, however, theactual situation isquite diﬂerent. Forthehydrogen atom thesituation isessentially unchanged. Theonly dillerence isthattheelectron hasaspinwhich weindicate bythelittle arrow in Fig.4—l2(a). Inthecaseofahelium atom, however, wecannot puttwoelectrons ontopofeach other. Butwait, thatisonlytrueiftheir spins arethesame. Two electrons canoccupy thesame state iftheir spins areopposite. Sothehelium atom does notlook much diﬁerent either. Itwould appear asshown inpart (b)of Fig.4—l2. Forlithium, however, thesituation becomes quite different. Where canweputthethird electron‘? Thethird electron cannot goontopoftheother twobecause bothspindirections areoccupied. (You remember thatforanelectron oranyparticle with spin l/2there areonly twopossible directions forthespin.) Thethird electron can’t gonear theplace occupied bytheother two, soitmust take upaspecial condition inadiﬁerent kind ofstate farther away from the nucleus inpart(c)ofthe ﬁgure. (Wearespeaking onlyinarather rough wayhere, because inreality allthree electrons areidentical; since wecannot really distinguish which oneiswhich, ourpicture isonly anapproximate one.) 4~l3SPIN ONE_\‘7///// ELECTRON NUCLEUS / (<1) %/ Ettéatoig" //%(bl / /l;//////é W /// /A (<3l\\\\\\\\ 5\\%\\\\s\\\\ Fig. 4—l2. Atomic conﬁgurations for real, Fermi-type, spin one-half electrons. 4 / Fig. 4—l3. Thehydrogen molecule. ”//////%Fig. 4-14. Helium with one electron inahigher energy state. %%Fig.4-15. Thelikely mechanism ino ferromagnetic crystal; the conduction electron isantiparallel totheunpaired inner electrons.\\\\\\\ \Now wecanbegin toseewhy different atoms willhave diﬁerent chemical properties. Because thethird electron inlithium isfarther out,itisrelatively more loosely bound. Itismuch easier toremove anelectron from lithium than from helium. (Experimentally, ittakes 25volts toionize helium butonly 5volts to ionize lithium.) This accounts forthevalence ofthelithium atom. Thedirectional properties ofthevalence have todowith thepattern ofthewaves oftheouter electron, which wewillnotgointoatthemoment. Butwecanalready seetheim- portance oftheso-called exclusion princz'pIe——which states thatnotwoelectrons canbefound inexactly thesame state (including spin). The exclusion principle isalso responsible forthestability ofmatter ona large scale. Weexplained earlier thattheindividual atoms inmatter didnot collapse because oftheuncertainty principle; butthisdoes notexplain why itis thattwohydrogen atoms can’t besqueezed together asclose asyouwant—why itisthat alltheprotons don’t getclose together with onebigsmear ofelectrons around them. Theanswer is,ofcourse, thatsince nomore than twoelectrons— with opposite spins—can beinroughly thesame place, thehydrogen atoms must keep away from each other. Sothestability ofmatter onalarge scale isreally a consequence oftheFermi particle nature oftheelectrons. Ofcourse, iftheouter electrons ontwoatoms havespins inopposite directions, they cangetclose toeach other. This is,infact, just theway that thechemical bond comes about. Itturns outthattwoatoms together willgenerally have the lowest energy ifthere isanelectron between them. Itisakind ofanelectrical attraction forthetwopositive nuclei toward theelectron inthemiddle. Itis possible toputtwoelectrons more orlessbetween thetwonuclei solong astheir spins areopposite, andthestrongest chemical binding comes about thisway. There isnostronger binding, because theexclusion principle does notallow there tobemore than twoelectrons inthespace between theatoms. Weexpect the hydrogen molecule tolook more orlessasshown inFig. 4-13. Wewant tomention onemore consequence oftheexclusion principle. You remember that ifboth electrons inthehelium atom aretobeclose tothenucleus, their spins arenecessarily opposite. Now suppose thatwewould liketotryto arrange tohave both electrons with thesame spin aswemight consider doing by putting onafantastically strong magnetic ﬁeldthatwould trytolineupthespins inthesame direction. Butthen thetwoelectrons could notoccupy thesame state inspace. Oneofthem would have totakeonadifferent geometrical position, as indicated inFig. 4-14. Theelectron which islocated farther from thenucleus has lessbinding energy. Theenergy ofthewhole atom istherefore quite abithigher. Inother words, when thetwospins areopposite, there isamuch stronger total attraction. So,there isanapparent, enormous force trying tolineupspins opposite to each other when twoelectrons areclose together. Iftwoelectrons aretrying togo inthesame place, there isavery strong tendency forthespins tobecome lined opposite. Thisapparent force trying toorient thetwospins opposite toeach other ismuch more powerful than thetinyforce between thetwomagnetic moments of theelectrons. Youremember when wewere speaking offerromagnetism there was themystery ofwhy theelectrons indifferent atoms hadastrong tendency toline upparallel. Although there isstillnoquantitative explanation, itisbelieved that what happens isthat theelectrons around thecore ofoneatom interact through theexclusion principle withtheouter electrons which have become freetowander throughout thecrystal. This interaction causes thespins ofthefreeelectrons and theinner electrons totake onopposite directions. Butthefreeelectrons andthe inner atomic electrons canonly beopposite provided alltheinner electrons have thesame spindirection, asindicated inFig.4-15. Itseems probable thatitisthe effect oftheexclusion principle acting indirectly through thefreeelectrons that gives risetothestrong aligning forces responsible forferromagnetism. Wewillmention onefurther example ofthe inﬂuence oftheexclusion principle. Wehave saidearlier that thenuclear forces arethesame between theneutron and theproton, between theproton andtheproton, andbetween theproton andthe neutron. Why isitthen that aproton andaneutron canstick together tomake a 4—l4 deuterium nucleus, whereas there isnonucleus with justtwoprotons orwith just twoneutrons? Thedeuteron is,asamatter offact, bound byanenergy ofabout 2.2million volts, yet,there isnocorresponding binding between apair ofprotons tomake anisotope ofhelium with theatomic weight 2.Such nuclei donotexist. Thecombination oftwoprotons does notmake abound state. Theanswer isaresult oftwoeffects: ﬁrst, theexclusion principle; andsecond, thefactthatthenuclear forces aresomewhat sensitive tothedirection ofspin. The force between aneutron andaproton isattractive andsomewhat stronger when thespins areparallel than when they areopposite. Ithappens that these forces arejustdifferent enough thatadeuteron canonly bemade iftheneutron and proton have their spins parallel; when their spins areopposite, theattraction is notquite strong enough tobind them together. Since thespins oftheneutron and proton areeach one-half andareinthesame direction, thedeuteron hasaspin of one. Weknow, however, thattwoprotons arenotallowed tositontopofeach other iftheir spins areparallel. lfitwere notfortheexclusion principle, two protons would bebound, butsince theycannot exist atthesame place andwith thesame spin directions, theHe2 nucleus does notexist. Theprotons could come together with their spins opposite, butthen there isnotenough binding tomake astable nucleus, because thenuclear force foropposite spins istooweak to bind apairofnucleons. Theattractive force between neutrons andprotons of opposite spins canheseen byscattering experiments. Similar scattering experiments withtwoprotons withparallel spins show thatthere isthecorresponding attraction. Soitistheexclusion principle that helps explain why deuterium canexist when Hegcannot. 4-15 5 Spin 0ne 5-1Filtering atoms withaStern-Gerlach apparatus Inthischapter wereally begin thequantum mechanics proper—in thesense thatwearegoing todescribe aquantum mechanical phenomenon inacompletely quantum mechanical way. Wewillmake noapologies andnoattempt toﬁndcon- nections toclassical mechanics. Wewant totalkabout something newinanew language. Theparticular situation which wearegoing todescribe isthebehavior oftheso-called quantization oftheangular momentum, foraparticle ofspinone. Butwewon’t usewords like“angular momentum” orother concepts ofclassical mechanics until later. Wehave chosen thisparticular example because itisrela- tively simple, although notthesimplest possible example. Itissufﬁciently com- plicated thatitcanstand asaprototype which canbegeneralized forthedescription ofallquantum mechanical phenomena. Thus, although wearedealing with a particular example, allthelaws which wemention areimmediately generalizable, andwewillgivethegeneralizations sothatyouwillseethegeneral characteristics ofaquantum mechanical description. Webegin with thephenomenon ofthe splitting ofabeam ofatoms intothree separate beams inaStern-Gerlach experi- ment. You remember thatifwehave aninhomogeneous magnetic ﬁeld made bya magnet with apointed pole tipandwesend abeam through theapparatus, the beam ofparticles may besplit intoanumber ofbeams—the number depending ontheparticular kind ofatom anditsstate. Wearegoing totakethecaseofan atom which gives three beams, andwearegoing tocallthataparticle ofspinone. Youcandoforyourself thecaseofﬁvebeams, seven beams, twobeams, etc.-you justcopy everything down andwhere wehave three terms, youwillhave ﬁve terms, seven terms, andsoon. Imagine theapparatus drawn schematically inFig.5-1. Abeam ofatoms (orparticles ofanykind) iscollimated bysome slitsandpasses through anon- uniform ﬁeld. Let’s saythat thebeam moves inthey-direction andthat the magnetic ﬁeldanditsgradient areboth inthez-direction. Then, looking from the side,wewillseethebeam splitvertically intothree beams, asshown intheﬁgure. Now attheoutput endofthemagnet wecould putsmall counters which count therateofarrival ofparticles inanyoneofthethree beams. Orwecanblock offtwoofthebeams andletthethird onegoon. Suppose weblock offthelower twobeams andletthetop-most beam goon andenter asecond Stern-Gerlach apparatus ofthesame kind, asshown inFig. 5-2. What happens? There arenotthree beams inthesecond apparatus; there isonly thetopbeam.T This iswhat youwould expect ifyouthink ofthesecond apparatus assimply anextension oftheﬁrst. Those atoms which arebeing pushed upward continue tobepushed upward inthesecond magnet. a /l’°_,/;I_L I “Tva I5-1Filtering atoms witha Stern-Gerlach apparatus 5-2 Experiments with ﬁltered atoms 5-3Stern-Gerlach ﬁlters inseries 5-4Base states 5-5 Interfering amplitudes 5-6 The machinery ofquantum mechanics 5-7Transforming toadifferent base 5—8 Other situations Review: Chapter 35,Vol. II,Para- magnetism andMagnetic Res- onance. Foryour convenience thischapter isreproduced in theAppendix ofthisvolume. |SE2't'° IVB ,' I*|”YALl 21 Y Fig. 5—l. InaStern-Gerlach experi- ment, atoms ofspin one are split into three beams. Fig. 5-2. The atoms from one ofthe beams aresent into asecond identical apparatus TWeareassuming thatthedeflection angles areverysmall. 5-1 0) A I ‘:2 o L\ S N S -~———— —— —————->—----- _->- __>A < 5 N S N z Y + L r L 1 XL 5.‘! to) I _ Y Fig. 5-3. (a)Animagined modiﬁcation ofaStern-Gerloch apparatus. (b)The paths ofspin-one atoms. You canseethen that theﬁrst apparatus hasproduced abeam of"puriﬁed" objects—at0ms thatgetbent upward intheparticular inhomogeneous ﬁeld. The atoms, astheyenter theoriginal Stern-Gerlach apparatus, areofthree “varieties,” andthethree kinds takediﬁerent trajectories. Byﬁltering outallbutoneofthe varieties, wecanmake abeam whose future behavior inthesame kindofapparatus isdetermined andpredictable. Wewillcallthisaﬁltered beam, orapolarized beam, orabeam inwhich theatoms allareknown tobeinadeﬁnite stale. Fortherestofourdiscussion, itwillbemore convenient ifweconsider a somewhat modiﬁed apparatus oftheStern-Gerlach type. Theapparatus looks more complicated atﬁrst, butitwillmake allthearguments simpler. Anyway, since theyareonly “thought experiments,“ itdoesn’t costanything tocomplicate theequipment. (Incidentally, noonehasever done alloftheexperiments wewill describe injust thisway, butweknow what would happen from thelawsofquantum mechanics, which are,ofcourse, based onother similar experiments. These other experiments areharder tounderstand atthebeginning, sowewant todescribe some idealized—but possible—experiments.) Figure 5—3(a) shows adrawing ofthe“modiﬁed Stern-Gerlach apparatus” wewould liketouse. Itconsists ofasequence ofthree high-gradient magnets. Theﬁrstone(ontheleft)isjusttheusual Stern-Gerlach magnet andsplits the incoming beam ofspin-one particles into three separate beams. The second magnet hasthesame cross section astheﬁrst, butistwice aslong andthepolarity ofitsmagnetic ﬁeld isopposite theﬁeld inmagnet l.The second magnet pushes intheopposite direction ontheatomic magnets andbends their paths back toward theaxis, asshown inthetrajectories drawn inthelower part oftheﬁgure. The third magnet isjustliketheﬁrst, andbrings thethree beams back together again, sothat leaves theexithole along theaxis. Finally, wewould liketoimagine that infront oftheholeatAthere issome mechanism which cangettheatoms started from restandthat after theexithole atBthere isadecelerating mechanism that brings theatoms back torestatB.That isnotessential, butitwillmean thatin S-2 ouranalysis wewon’t have toworry about including anyeﬁects ofthemotion as theatoms come out,andcanconcentrate onthose matters having onlytodowith thespin. Thewhole purpose ofthe“improved” apparatus isjusttobring allthe particles tothesame place, andwith zerospeed. Now ifwewant todoanexperiment liketheoneinFig.5-2,wecanﬁrst make aﬁltered beam byputting aplate inthemiddle oftheapparatus thatblocks twoofthebeams, asshown inFig.5-4. Ifwenowputthepolarized atoms through asecond identical apparatus, alltheatoms willtake theupper path, ascanbe veriﬁed byputting similar plates inthewayofthevarious beams ofthesecond Sﬁlter andseeing whether particles getthrough. [_i__ _______ ~ \\ I? \ / *__/Z+ ______I""11IIII +IIIIII ______I r__ \<IIIIIIIIIIl___ [__IIIII mlIIIII|__ S Fig.5-4. The"improved" Stern-Gerlach apparatus Suppose wecalltheﬁrstapparatus bythename S.(Wearegoing toconsider allsorts ofcombinations, andwewillneed labels tokeep things straight.) Wewill saythattheatoms which takethetoppath inSareinthe“plus state with respect toS”;theones which takethemiddle path areinthe“zero state with respect to S”;andtheones which takethelowest path areinthe“minus state with respect toS.” (Inthemore usual language wewould saythatthez-component ofthe angular momentum was+lh,0,and—lh,butwearenotusing thatlanguage now.) Now inFig.5-4thesecond apparatus isoriented justliketheﬁrst, sotheﬁltered atoms willallgoontheupper path. Orifwehadblocked oiltheupper andlower beams intheﬁrstapparatus andletonly thezero state through, alltheﬁltered atoms would gothrough themiddle path ofthesecond apparatus. And ifwe hadblocked oﬁ“allbutthelowest beam intheﬁrst, there would beonly alow beam inthesecond. Wecansaythat ineach case ourﬁrst apparatus has produced aﬁltered beam inapure state with respect toS(+,O,or—),andwe cantestwhich state ispresent byputting theatoms through asecond, identical apparatus. Wecanmake oursecond apparatus sothat ittransmits only atoms ofa particular state—by putting masks inside itaswedidfortheﬁrstone~and then Wecantestthestate oftheincoming beam justbyseeing whether anything comes outthefarend. Forinstance, ifweblock offthetwolower paths inthesecond apparatus, 100percent oftheatoms willstillcome through; butifweblock oﬁthe upper path, nothing willgetthrough. Tomake thiskind ofdiscussion easier, wearegoing toinvent ashorthand symbol torepresent oneofourimproved Stern-Gerlach apparatuses. Wewilllet thesymbol + 0 (5.1) S stand foronecomplete apparatus. (This isnotasymbol youwilleverﬁndused in quantum mechanics; we’ve justinvented itforthischapter. Itissimply meant to beashorthand picture oftheapparatus ofFig.5-3.) Since wearegoing towant touseseveral apparatuses atonce, andwith various orientations, wewillidentify each with aletter underneath. Sothesymbol in(5.1) stands fortheapparatus S. When weblock oﬁoneormore ofthebeams inside, wewillshow thatbysome 5-3asaﬁlter. _‘" (0){ti _’ + I "I=(b) {ti} I ~\O -“Z -| ,1’ (cl +' I \ o = ——\—k_ (dl Fig. 5-5. Special shorthand symbols forStern-Gerlach type ﬁlters.vertical barsindicating which beam isblocked, likethis: + (5.2) S Thevarious possible combinations wewillbeusing areshown inFig.5—5. Ifwehave twoﬁlters insuccession (asinFig.5-4), wewillputthetwosym- bolsnexttoeach other, likethis: + + 0| 0- (5.3) S S Forthissetup, everything thatcomes through theﬁrstalsogetsthrough thesecond. Infact, even ifweblock oﬁthe“zero” and“minus” channels ofthesecond apparatus, sothatwehave + + 0I 0It (5.4) s s westillgetI00percent transmission through thesecond apparatus. Ontheother hand, ifwehave + + ol 0t (5.5) s s nothing atallcomes outofthefarend. Similarly, I?'IIIIs s would givenothing out. Ontheother hand. + +{on {on (5.7) I I + 0 s byitself. Now wewant todescribe these experiments quantum mechanically. Wewill saythatanatom isinthe(+S) state ifithasgone through theapparatus ofFig. 5—5(b), that itisina(OS)state ifithasgone through (c),andina(~S) state if ithasgone through (d).T Then welet(bIa)betheamplitude thatanatom which isinstate awillgetthrough anapparatus intothebstate. Wecansay: (b|a)is theamplitude foranatom inthestate atogelinto thestate b.The experiment (5.4) gives usthatwould bejustequivalent to (+5 I+5) =1, I‘Read: (+S) =“plus-S”; (OS) =“zero-S”; (—S) =“minus-S.” 5-4 . Lywhereas (5.5) gives us (—S I—I—S) =0. Similarly, theresult of(5.6) is <+5I -5)=0, andof(5.7) is (—-S| —S) =1. Aslongaswedealonlywith“pure” states—that is,wehave onlyonechannel open—there arenine such amplitudes, andwecanwrite them inatable: from +S OS—S 00 (5.8) F?o <3-FC/JV) oo- @\—~ --o -s This array ofninenumbers—called amatrix——summarizes thephenomena we’ve been describing. 5-2Experiments withﬁltered atoms Now comes thebigquestion: What happens ifthesecond apparatus istipped toadifferent angle, sothat itsﬁeld axis isnolonger parallel totheﬁrst? It could benotonly tipped, butalso pointed inadifferent direction~for instance, itcould takethebeam offat90°with respect totheoriginal direction. Totakeit easy atﬁrst, let’s ﬁrst think about anarrangement inwhich thesecond Stern- Gerlach experiment istilted bysome angle aabout they-axis, asshown inFig. 5-6. We’ll callthesecond apparatus T.Suppose thatwenowsetupthefollowing experiment: + + 0| 0|» Z I ortheexperiment: + +|0| 0t -—Is T What comes outatthefarendinthese cases? Theanswer isthis: Iftheatoms areinadeﬁnite state with respect toS,they arenotinthesame state with respect toT—a (+S)state isnotalsoa(+T)state. There is,however, acertain amplitude toﬁndtheatom ina(+T) state—or a(OT) state ora(—T) state. Inother words, ascareful aswehave been tomake sure that wehave the atoms inadeﬁnite condition, thefactofthematter isthat ifitgoes through an apparatus which istilted atadifferent angle ithas, sotospeak, to“reorient” Q Fig. 5-6 Two Stern Gerlach type 5-55 T ﬁlters insertes, thesecond Istilted atthe angle ctwtth respect totheﬁrst itself—which itdoes, don’t forget, byluck. Wecanputonlyoneparticle through atatime, andthen wecanonlyaskthequestion: What istheprobability thatit getsthrough? Some oftheatoms thathave gone through Swillendina(+T) state, some ofthem willendina(0T),andsome ina(—T) state—all withdifferent odds. These odds canbecalculated bytheabsolute squares ofcomplex amplitudes; what wewant issome mathematical method, orquantum mechanical description, forthese amplitudes. What weneed toknow arevarious quantities like bywhich wemean theamplitude thatanatom initially inthe(+S)state canget intothe(—T) condition (which isnotzero unless TandSarelined upparallel toeachother). There areother amplitudes like (+T I0S), or (0TI—S), etc. There are,infact,ninesuchamplitudes—another matrix——that atheory ofparticles should tellushowtocalculate. JustasF=matellsushowtocalculate what hap- pens toaclassical particle inanycircumstance, thelaws ofquantum mechanics permit ustodetermine theamplitude thataparticle willgetthrough aparticular apparatus. Thecentral problem, then, istobeabletocalculate—for anygiven tiltangle oz,orinfactforanyorientation whatever—the nineamplitudes: @rTI+$X VFTIOSX @+TI*5% Wecanalready ﬁgure outsome relations among these amplitudes. First, according toourdeﬁnitions, theabsolute square K+TI+sn* istheprobability thatanatom ina(+S) state willenter a(+D state. Wewilloften ﬁnditmore convenient towrite such squares intheequivalent form (+TI+5X+TI+$W- Inthesame notation thenumber (OTI +S)(0T| +S)"‘ istheprobability thataparticle inthe(+S) state willenter the(0T)state, and <—TI+SX—TI+SY istheprobability thatitwillenter the(—T) state. Butthewayourapparatuses aremade, every atom which enters theTapparatus must befound insome oneof thethree states oftheTapparatus—there’s nowhere elseforagiven kind ofatom togo.Sothesumofthethree probabilities we’ve justwritten must beequal to 100percent. Wehave therelation <+TI+SX+TI+Sh-+<OTI+SX0T|+SY+(—TI +S)(—TI +S)* =l. (5.10) There are,ofcourse, twoother such equations thatwegetifwestartwith a(0S) ora(—S) state. Buttheyareallwecaneasily get,soWe’ll goontosome other general questions. 5-3Stern-Gerlach ﬁlters inseries Here isaninteresting question: Suppose wehadatoms ﬁltered intothe(+S) state, then weputthem through asecond ﬁlter, sayintoa(0T)state, andthen through another +Sﬁlter. (We’ll callthelastﬁlter S’justsowecandistinguish 5-6 itfrom theﬁrstS-ﬁter.) Dotheatoms remember thattheywere once ina(+S) state? Inother words, wehave thefollowing experiment: Iill(‘IIIZIIIS T S’ Wewant toknow whether allthose thatgetthrough Talsogetthrough S’.They donot. Once theyhave been ﬁltered byT,theydonotremember inanywaythat theywere ina(+S)state when theyentered T.Note thatthesecond Sapparatus in(5.11) isoriented exactly thesame astheﬁrst, soitisstillanS-type ﬁlter. Thestates ﬁltered byS’are,ofcourse, still(+S), (0S),and(—S). Theimportant point isthis: IftheTﬁlter passes onlyonebeam, thefraction thatgetsthrough thesecond Sﬁlter depends onlyonthesetup oftheTﬁlter, and iscompletely independent ofwhat precedes it.Thefactthatthesame atoms were oncesorted byanSﬁlter hasnoinﬂuence whatever onwhat theywilldoonce they have been sorted again intoapure beam byaTapparatus. From then on,the probability forgetting intodifferent states isthesame nomatter what happened before theygotintotheTapparatus. Asanexample, let’scompare theexperiment of(5.ll) with thefollowing experiment: +I +I +0 0 0I (5.12)_I _| _ S T S’ inwhich onlytheﬁrstSischanged. Let’s saythattheangle a(between SandT) issuch thatinexperiment (5.11) one-third oftheatoms thatgetthrough Talso getthrough S’.Inexperiment (5.12), although there will,ingeneral, beadifferent number ofatoms coming through T,thesame fraction ofthese—one-third—will alsogetthrough S’. Wecan,infact,show from what youhave learned earlier thatthefraction of theatoms thatcome outofTandgetthrough anyparticular S’depends only on TandS’,notonanything that happened earlier. Let’s compare experiment (5.12)with 0 0 0- (5.13) —I —I —IS T S’ Theamplitude thatanatom thatcomes outofSwillalsogetthrough both Tand S’is,fortheexperiments of(5.12), <+SIOT)(oTIOs). Thecorresponding probability is I<+SI0T>(0TI 0$>I2 =I(+$I0T>I” I(°TI°5>l2- Theprobability forexperiment (5.13) is I<05l0T><0TI0$>l2 =l<0$I0T>I’ I<0TI0$>|"- Theratio is I<0$I0T>I2 I<+$I0T>I‘-’ anddepends only onTandS’,andnotatallonwhich beam (+S),(OS),or(——S) isselected byS.(The absolute numbers maygoupanddown together depending onhowmuch getsthrough T.)Wewould, ofcourse, ﬁndthesame result ifwe compared theprobabilities thattheatoms would gointotheplusortheminus 5-7 states with respect toS’,ortheratio oftheprobabilities togointothezero or minus states. Infact, since these ratios depend only onwhich beam isallowed topass through T,andnotontheselection made bytheﬁrstSﬁlter, itisclear thatwe would getthesame result even ifthelastapparatus were notanSﬁlter. Ifweuse forthethird apparatus—which wewillnowcallR—one rotated bysome arbitrary angle withrespect toT,wewould ﬁndthataratio suchasI(0RI0T)I2/I(+R I0T)I2 wasindependent ofwhich beam waspassed bytheﬁrstﬁlter S. 5-4Base states These results illustrate oneofthebasic principles ofquantum mechanics: Any atomic system canbeseparated byaﬁltering process intoacertain setof what wewillcallbasestates, andthefuture behavior oftheatoms inanysingle given basestate depends onlyonthenature ofthebasestate—it isindependent of anyprevious history.T Thebase states depend, ofcourse, ontheﬁlter used; for instance, thethree states (+T),(0T),and(—T)areonesetofbasestates; thethree states (+S), (0S),and(—S) areanother. There areanynumber ofpossibilities each asgood asanyother. Weshould becareful tosaythatweareconsidering good ﬁlters which do indeed produce “pure” beams. If,forinstance, ourStern-Gerlach apparatus didn't produce agood separation ofthethree beams sothatwecould notseparate them cleanly byourmasks, then wecould notmake acomplete separation intobase states. Wecantellifwehave pure basestates byseeing whether ornotthebeams canbesplitagain inanother ﬁlter ofthesame kind. Ifwehave apure (+T) state, forinstance, alltheatoms willgothrough .+. .Q., K I 1' andnone willgothrough!+ N <0 rs t I T orthrough +.0. K t T Ourstatement about basestates means thatitispossible toﬁlter tosome purestate, sothatnofurther ﬁltering byanidentical apparatus ispossible. Wemust also point outthat what wearesaying isexactly true only inrather idealized situations. InanyrealStern-Gerlach apparatus, wewould have toworry about diffraction bytheslitsthatcould cause some atoms togointostates corre- sponding todifferent angles, orabout whether thebeams might contain atoms with different excitations oftheir internal states, andsoon. Wehave idealized the situation sothatwearetalking only about thestates thataresplitinamagnetic ﬁeld; weareignoring things having todowith position, momentum, internal excitations, andthelike. Ingeneral, onewould need toconsider alsobase states which aresorted outwith respect tosuch things al§o. Buttokeep theconcepts simple, weareconsidering only oursetofthree states, which issufficient forthe exact treatment oftheidealized situation inwhich theatoms don’t gettornupin IWedonotintend theword “base state” toimply anything more than what issaid here. They arenottobethought ofas“basic” inanysense. Weareusing theword base with thethought ofabasis foradescription, somewhat inthesense thatonespeaks of “numbers tothebaseten.” 5-8 going through theapparatus, orotherwise badly treated, andcome torestwhen theyleave theapparatus. You willnote thatwealways begin ourthought experiments bytaking a ﬁlter with only onechannel open, sothatwestart with some deﬁnite base state. Wedothisbecause atoms come outofafurnace invarious states determined at random bytheaccidental happenings inside thefurnace. (Itgives what iscalled an“unpolarized” beam.) Thisrandomness involves probabilities ofthe“classical” kind—as incoin tossing——which aredifferent from thequantum mechanical probabilities weareworrying about now.) Dealing with anunpolarized beam would getusintoadditional complications thatarebetter toavoid until after we understand thebehavior ofpolarized beams. Sodon’t trytoconsider atthispoint what happens iftheﬁrstapparatus letsmore than onebeam through. (Wewill tellyouhowyoucanhandle such cases attheendofthechapter.) Let’s nowgoback andseewhat happens when wegofrom abase state for oneﬁlter toabasestate foradifferent ﬁlter. Suppose westart again with lllll'10I 0I- s T Theatoms which come outofTareinthebasestate (0T)andhave nomemory thattheywere once inthestate (+S).Some people would saythatintheﬁltering byTwehave “lost theinformation” about theprevious state (+S) because we have “disturbed” theatoms when weseparated them into three beams inthe apparatus T.Butthatisnottrue. Thepastinformation isnotlostbytheseparation into three beams, butbytheblocking masks that areputin—as wecanseebythe following setofexperiments. Westart with a+Sﬁlter andwillcallNthenumber ofatoms thatcome through it.Ifwefollow thisby?a0Tﬁlter, thenumber ofatoms thatcome outis some fraction oftheoriginal number, sayaN. Ifwethen putanother +Sﬁlter, onlysome fraction )8ofthese atoms willgettothefarend. Wecanindicate this inthefollowing way: + +| +oll.0.2-Q.0Bin (5.14)_ _| _ S T S’ Ifourthird apparatus S’selected adifferent state, saythe(0S)state, adifferent fraction, say7,would getthrough.1' Wewould have {+ +| +|0|L0:1.0| (5.15)_ _| _ S T S’ Now suppose werepeat these twoexperiments butremove allthemasks from T. Wewould thenﬁndtheremarkable results asfollows: 0*_N_,0L,0‘L» (5.16) E I Z {+ + +|0| 0L.0|_1>_.- (5.17) S T S’ Tlnterms ofourearlier notation a=[(0T| +S)l2, 5= +S|0T)|2, andY= l(0S|0T)|2. 5-9 Alltheatoms getthrough S’intheﬁrstcase, butnone inthesecond case! This is oneofthegreat laws ofquantum mechanics. That nature works thiswayisnot self-evident, buttheresults wehave given correspond forouridealized situation tothequantum mechanical behavior observed ininnumerable experiments. 5-5Interfering amplitudes How canitbethatingoing from (5.15) to(5.l7)——by opening more channels —we letfewer atoms through ?Thisistheold,deep mystery ofquantum mechanics —the interference ofamplitudes. lt’sthesame kind ofthing weﬁrstsawinthe two-slit interference experiment with electrons. Wesawthatwecould getfewer electrons atsome places with both slitsopen than wegotwith oneslitopen. It works quantitatively thisway. Wecanwrite theamplitude thatanatom willget through TandS’intheapparatus of(5.17) asthesumofthree amplitudes. one foreach ofthethree beams inT;thesum isequal tozero: <05‘!+T><+T| +5)+(0Sl0T)(0Tl +S>+<05]—T)(—T| +5)=0. (5.18) None ofthethree individual amplitudes iszer0—for example, theabsolute square ofthesecond amplitude is'Ya,see(5.l5)—but thesumiszero. Wewould have alsothesame answer ifS’were settoselect the(—S) state. However, inthesetup of(5.16), theanswer isdifferent. Ifwecallatheamplitude togetthrough Tand S’,inthiscasewehave'[' 11=(+51 +T>(+T| +3)+(+$|0T><0T| +5) +(+51 —T><"-Tl +3)=1- (5-19) Intheexperiment (5.16) thebeam hasbeen split and recombined. Humpty Dumpty hasbeen putback together again. Theinformation about theoriginal (+S) state isretained—it isjustasthough theTapparatus were notthere atall. This istruewhatever isputafter the“wide-open” Tapparatus. Wecould follow itwith anRﬁlter——a ﬁlter atsome oddangle—or anything wewant. The answer willalways bethesame asiftheatoms were taken directly from theﬁrstSﬁlter. Sothisistheimportant principle: ATﬁlter—or anyﬁ1ter—with wide-open masks produces nochange atall.Weshould make oneadditional condition. The wide-open ﬁlter must notonlytransmit allthree beams, butitmust alsonotproduce unequal disturbances onthethree beams. Forinstance, itshould nothave astrong electric ﬁeld near onebeam andnottheothers. Thereason isthat even ifthis extra disturbance would stillletalltheatoms through theﬁlter, itcould change the phases ofsome oftheamplitudes. Then theinterference would bechanged, and theamplitudes inEqs. (5.18) and(5.19) would bedifferent. Wewillalways assume thatthere arenosuch extra disturbances. Let’s rewrite Eqs. (5.18) and(5.19) inanimproved notation. Wewilllet istand foranyoneofthethree states (+T), (0T),or(—T); thentheequations can bewritten: Z(os|i)(i|+s>=0 (5.20)all1' and Z)<+S|i><i|+s>=1. (5-21>alli Similarly, foranexperiment where S’isreplaced byacompletely arbitrary ﬁlter R,wehave + + + 111{O11°11S T R TWereally cannot conclude from theexperiment thata=1,butonly that[@112=1, soamight beeff,butitcanbeshown thatthechoice 6=0represents noreallossof generality. 5-10 Theresults willalways bethesame asiftheTapparatus wereleftoutandwehad llf|lE111Or,expressed mathematically, Z<+R|i><i1+S> =<+R1+s>. (5-23)alli Thisisourfundamental law,anditisgenerally truesolongasistands forthethree basestates ofanyﬁlter. Youwillnotice thatintheexperiment (5.22) there isnospecial relation of SandRtoT.Furthermore, thearguments would bethesame nomatter what states theyselected. Towrite theequation inageneral way,without having to refertothespeciﬁc states selected bySandR,let’scall¢(“phi”) thestate prepared bytheﬁrstﬁlter (inourspecial example, +S) andX(“khi”) thestate tested by theﬁnalﬁlter (inourexample, +R). Then wecanstate ourfundamental lawof Eq.(5.23) intheform <><1¢>=Z<><1»'><»"1¢>. (5.24)all1' where iistorange overthethree base states ofsome particular ﬁlter. Wewant toemphasize again what wemean bybase states. They arelikethe three states which canbeselected byoneofourStern-Gerlach apparatuses. One condition isthatifyouhave abasestate, thenthefuture isindependent ofthepast. Another condition isthatifyouhave acomplete setofbase states, Eq.(5.24) is trueforanysetofbeginning andending states ¢andX.There is,however, no unique setofbase states. Webegan byconsidering base states withrespect toa particular apparatus T.Wecould equally wellconsider adiﬁferent setofbase states withrespect toanapparatus S,orwithrespect toR,etc.'l' Weusually speak ofthebasestates “inacertain representation.” Another condition onasetofbase states inanyparticular representation is thatthey areallcompletely diﬂ"erent. Bythatwemean thatifwehave a(+T) state, there isnoamplitude forittogointoa(0T)ora(—T) state. Ifweletiand jstand foranytwobase‘states ofaparticular set,thegeneral rules discussed in connection with (5.8) arethat (JI1')=0 foralliandjthatarenotequal. Ofcourse, weknow that (ili)=1. These twoequations areusually written as (I|i)=51¢, (5-25) where 6,-,-(the“Kronecker delta”) isasymbol thatisdeﬁned tobezerofori¢j, andtobeonefori=j. Equation (5.25) isnotindependent oftheother laws wehave mentioned. Ithappens thatwearenotparticularly interested inthemathematical problem of ﬁnding theminimum setofindependent axioms thatwillgiveallthelawsasconse- quences.I Wearesatisﬁed ifwehave asetthatiscomplete andnotapparently inconsistent. Wecan, however, show that Eqs. (5.25) and(5.24) arenotinde- pendent. Suppose welet45inEq.(5.24) represent oneofthebase states ofthe 1'Infact, foratomic systems with three ormore base states, there exist other kinds of ﬁlters—quite different from aStern-Gerlach apparatus—which canbeusedtogetmore choices forthesetofbasestates (each setwiththesame number ofstates). IRedundant truth doesn’t bother us! 5-11 same setas1',saythejthstate; thenwehave <wn=Zamwn ButEq.(5.25) saysthat(ilj)iszerounless i=j,sothesumbecomes just(X|j) andwehave anidentity, which shows thatthetwolawsarenotindependent. Wecanseethat there must beanother relation among theamplitudes ifboth Eqs. (5.10) and(5.24) aretrue. Equation (5.10) is (-1-Tl +5)(+T| +$)* +(0Tl +5)(0Tl +5)* +(—T| +5)(—Tl +$)* =1- lfwewrite Eq.(5.24), letting both 45andXbethestate (+S), theleft-hand side is(-1-SI +S), which isclearly =1; sowegetonce more Eq.(5.19), (+Sl +T)(+Tl +S)+(+Sl0T)(9Tl +5)+(+S| —T)(—Tl +S) =1. These twoequations areconsistent (forallrelative orientations oftheTandS apparatuses) only if (+31 +T)=(+T! +S)‘, (+$l QT)=(°T| +5)‘, (+51 —T)=("Tl +$)*- Anditfollows thatforanystates ¢andX, (¢lX)=(XI¢)*- (5-26) Ifthiswere nottrue, probability wouldn’t be“conserved,” andparticles would get“lost.” Before going on,wewant tosummarize thethree important general lawsabout amplitudes. They areEqs. (5.24), (5.25), and(5.26): I =5113 uam=Zamwt ownll'1' m@m=WW Inthese equations theiandjrefer toallthebasestates ofsome onerepresentation, while ¢andXrepresent anypossible states oftheatom. Itisimportant tonotethat IIisvalid only ifthesumiscarried outover allthebase states ofthesystem (in ourcase, three: +T, 0T,—T). These laws saynothing about what weshould choose forabase foroursetofbase states. Webegan byusing aTapparatus, which isaStern-Gerlach experiment withsome arbitrary orientation; butanyother orientation, sayW,would bejustasgood. Wewould have adifferent setofstates touseforZandj, butallthelawswould stillbegood—there isnounique set.One ofthegreat games ofquantum mechanics istomake useofthefactthat things canbecalculated inmore than oneway. 5-6Themachinery ofquantum mechanics Wewant toshow youwhythese lawsareuseful. Suppose wehave anatom in agiven condition (bywhich wemean thatitwasprepared inacertain way), and wewant toknow what willhappen toitinsome experiment. Inother words, we startwithouratom inthestate ¢~andwant toknow what aretheoddsthatitwillgo through some apparatus which accepts atoms only inthecondition X.Thelaws saythatwecandescribe theapparatus completely interms ofthree complex num- bers (XIi),theamplitudes foreach base state tobeinthecondition X;andthat wecantellwhat willhappen ifanatom isputintotheapparatus ifwedescribe the state oftheatom bygiving three numbers (iI¢),theamplitudes fortheatom inits original condition tobefound ineach ofthethree basestates. Thisisanimportant idea. 5-12 Let’s consider another illustration. Think ofthefollowing problem: Westart withanSapparatus; thenwehave acomplicated mess ofjunk, which wecancall A,andthen anRapparatus—~like this: ti:ii1;:ByAwemean anycomplicated arrangement ofStern-Gerlach apparatuses with masks orhalf-masks, oriented atpeculiar angles, with oddelectric andmagnetic ﬁelds ...almost anything youwant toput. (It’s nicetodothought experiments— youdon’t have togotoallthetrouble ofactually building theapparatus!) The problem then is:With what amplitude does aparticle that enters thesection A ina(+S) state come outofAinthe(OR)state, sothatitwillgetthrough thelast Rﬁlter? There isaregular notation forsuch anamplitude; itis (0RIAI+S). Asusual, itistoberead from right toleft(like Hebrew): (ﬁnish Ithrough Istart). Ifbychance Adoesn’t doanything—but isjustanopen channel—then wewrite (0R|1|+$)= (°Rl+$); (519) thetwosymbols areequivalent. Foramore general problem, wemight replace (+S)byageneral starting state ¢and(OR)byageneral ﬁnishing state X,andwe would want toknow theamplitude (><|Al¢)- Acomplete analysis oftheapparatus Awould have togivetheamplitude (XIAI¢) forevery possible pair ofstates 45and X—an inﬁnite number ofcombinations! How then canwegive aconcise description ofthebehavior oftheapparatus A? Wecandoitinthefollowing way. Imagine thattheapparatus of(5.28) ismodiﬁed ti:ti1:ti1;:This isreally nomodiﬁcation atallsince thewide-open Tapparatuses don’t do anything. Buttheydosuggest howwecananalyze theproblem. There isacertain setofamplitudes (iI7|-S)thattheatoms from Swill getintotheistate ofT.Then there isanother setofamplitudes thatanistate (with respect toT)entering A willcome outasajstate (with respect toT).And ﬁnally there isanamplitude thateachjstate willgetthrough thelastﬁlter asa(0R)state. Foreach possible alternative path, there isanamplitude oftheform (0Rl1')(J'lAli)(i| +5), andthetotal amplitude isthesumoftheterms wecangetwithallpossible combi- nations ofiandj.The amplitude wewant is Z<oR|i><i|A|i><i1+s>_ (5.31) If(0R)and(+S) arereplaced bygeneral states Xand¢,wewould have thesame kind ofexpression; sowehave thegeneral result <><IA1¢>=Z<><l1'>(jl AIi><i1¢>. (5.32)ii 5-13 Now notice thattheright-hand sideofEq.(5.32) isreally “simpler” than the left-hand side. Theapparatus Aiscompletely described bytheninenumbers (jIAI1')which telltheresponse ofAwith respect tothethree base states ofthe apparatus T.Once weknow these ninenumbers, wecanhandle anytwoincoming andoutgoing states ¢andXifwedeﬁne each interms ofthethree amplitudes for going into, orfrom, each ofthethree basestates. Theresult ofanexperiment is predicted using Eq.(5.32). This then isthemachinery ofquantum mechanics foraspin-one particle. Every state isdescribed bythree numbers which aretheamplitudes tobeineach ofsome selected setofbasestates. Every apparatus isdescribed byninenumbers which aretheamplitudes togofrom onebase state toanother intheapparatus. From these numbers anything canbecalculated. Thenine amplitudes which describe theapparatus areoften written asa square matrix—called thematrix (jIAIi): from + 0 — to + (+1/11+) (+1/110) <+lAl——) 0 (01/1|+) (01/110) (01/11-) (5-33) - <—lA|+> <—|/110) <—l/11-) Themathematics ofquantum mechanics isjustanextension ofthisidea. We willgiveyouasimple illustration. Suppose wehaveanapparatus Cthatwewishto analyze-that is,wewant tocalculate thevarious (jICI1').Forinstance, wemight want toknow what happens inanexperiment like 11)}IsllljilButthenwenotice thatCisjustbuiltoftwopieces ofapparatus AandBinseries- theparticles gothrough Aandthen through B—so wecanwrite symbolically 1:-1:18:WecancalltheCapparatus the“product” ofAandB.Suppose alsothatwe already know howtoanalyze thetwoparts; sowecangetthematrices (with respect toT)ofAandB.Ourproblem isthensolved. Wecaneasily ﬁnd <><ICl¢> foranyinput andoutput states. First wewrite that <><|C:¢> =Z<><:B:/<></<:A|¢>-k Doyouseewhy? (Hint: Imagine putting aTapparatus between AandB.)Then ifweconsider thespecial caseinwhich ¢>andXarealsobase states (ofT),sayi andj,wehave <i|c1i>=Z)<1:B:/<><I<1A1i>. (5.36)le This equation gives thematrix forthe“product” apparatus Cinterms ofthetwo matrices oftheapparatuses AandB.Mathematicians callthenewmatrix (jICIi) ——formed from twomatrices (jIBIi)and(jIAIi)according tothesumspeciﬁed inEq.(5.36)—the “product” matrix BAofthetwomatrices BandA.(Note thattheorder isimportant, AB;éBA.) Thus, wecansaythatthematrix fora succession oftwopieces ofapparatus isthematrix product ofthematrices forthe twoapparatuses (putting theﬁrstapparatus ontheright intheproduct). Anyone whoknows matrix algebra thenunderstands thatwemean justEq.(5.36). 5-14 5-7Transforming toadifferent base Wewant tomake oneﬁnalpoint about thebasestates usedinthecalculations. Suppose wehave chosen towork withsome particular base-—say theSbase—and another fellow decides todothesame calculations with adifferent base—say the Tbase. Tokeep things straight let’scallourbase states the(iS)states, where i=+,O,—.Similarly, wecancallhisbase states (jT). How canwecompare ourwork with his? Theﬁnal answers fortheresult ofanymeasurement should come outthesame, butinthecalculations thevarious amplitudes andmatrices usedwillbedifferent. How aretheyrelated? Forinstance, ifweboth start with thesame ¢,wewilldescribe itinterms ofthethree amplitudes (iSI¢)that45 goesintoourbasestates intheSrepresentation, whereas hewilldescribe itbythe amplitudes (jTI¢)thatthestate ¢goesintothebasestates ishisTrepresentation. How canwecheck thatwearereally both describing thesame state ¢?Wecando itwiththegeneral ruleIIin(5.27). Replacing XbyanyoneofhisstatesjT, wehave (JTI¢>=Z‘,(jTI1S><i$:¢>. (5.31)J Torelate thetworepresentations, weneed onlygivetheninecomplex numbers of thematrix (jTIiS).This matrix canthen beused toconvert allofhisequations toourform. Ittellsushowtotransform from onesetofbase states toanother. (For thisreason (jTIiS)issometimes called “the transformation matrix from representation Storepresentation T.”Bigwords!) Forthecaseofspin-one particles forwhich wehave only three base states (forhigher spins, there aremore) themathematical situation isanalogous towhat wehave seen invector algebra. Every vector canberepresented bygiving three numbers—the components along theaxes x,y,andz.That is,every vector can beresolved into three “base” vectors which arevectors along thethree axes. But suppose someone elsechooses touseadifferent setofaxes—x’, y’,andz’.Hewill beusing different numbers torepresent anyparticular vector. Hiscalculations will lookdifferent, buttheﬁnalresults willbethesame. Wehaveconsidered thisbefore andknow therules fortransforming vectors from onesetofaxes toanother. Youmaywant toseehowthequantum mechanical transformations work by trying some out;sowewillgivehere, without proof, thetransformation matrices forconverting thespin-one amplitudes inonerepresentation Stoanother repre- sentation T,forvarious special relative orientations oftheSandTﬁlters. (We willshow youinalater chapter howtoderive these same results.) First case: TheTapparatus hasthesame y-axis (along which theparticles move) astheSapparatus, butisrotated about thecommon y-axis bytheangle at(asinFig. 5-6). (Tobespeciﬁc, asetofcoordinates x’,y’,z’isﬁxed intheT apparatus, related tothex,y,zcoordinates oftheSapparatus by:2’=zcosa+ xsinoz,x’=xcosa—zsinoz,y’=y.)Then thetransformation amplitudes are: (+T +S) =%(l+cosOZ), (or+s>=-»\‘7is1n .1, (—T +S) =%(1—cosoz), (+Tes)=+5/Esina, (OT 0S)=cos(X, (5.38) (—T OS) =—-1—sina, W (+T -S) =—§(l—cosoz), (OT —S) =—I—%Slnot, (-TI -S) =%(1+cosoz). 5-15 Second Case: TheTapparatus hasthesame z-axis asS,butisrotated around thez-axis bytheangle ti.(The coordinate transformation isz’=2,x’= xcosﬁ +ysin/3,y’=ycost? —xsin6.)Then thetransformation amplitudes are: <+r|+S)=W”. (OT IOS) =1, _- 5.39 tﬁ, ( ) allothers =0. Note thatanyrotations ofTwhatever canbemade upofthetworotations described. Ifastate ¢isdeﬁned bythethree numbers andthesame state isdescribed from thepoint ofview ofTbythethree numbers C;=<+Tl~:>>.C6=<<>T1¢>. C’.=<—Tl¢>. (5-41) then thecoeﬁicients (jTIiS)of(5.38) or(5.39) givethetransformation connect- ingCIandCf.Inother words, theC,-arevery much likethecomponents ofa vector thatappear different from thepoint ofview ofSandT. Foraspin-one particle only—because itrequires three amplitudes—the cor- respondence with avector isvery close. Ineach case, there arethree numbers that must transform with coordinate changes inacertain deﬁnite way. lnfact, there isasetofbase states which transform just likethethree components ofa vector. Thethree combinations l 1C,,=————C —C_, C=———C—I—C_, Cz=C 5.42 \/i(+ ) 1/ X/i(+ ) n( ) transform toCL,Cj,and C;just theway that x,y,ztransform tox’,y’,2’.[You cancheck thatthisissobyusing thetransformation laws (5.38) and(5.39).] Now youseewhyaspin-one particle isoften called a“vector particle." 5-8Other situations Webegan bypointing outthatourdiscussion ofspin-one particles would be aprototype foranyquantum mechanical problem. Thegeneralization hasonly todowith thenumbers ofstates. Instead ofonly three base states, anyparticular situation may involve nbase states.I Ourbasic laws inEq.(5.27) have exactly thesame form—with theunderstanding thatiandjmust range over allnbase states. Any phenomenon canbeanalyzed bygiving theamplitudes thatitstarts ineach oneofthebasestates andends inanyother oneofthebasestates, andthen summing over thecomplete setofbase states. Anyproper setofbase states can beused, andifsomeone wishes touseadifferent set,itisjustasgood; thetwocan beconnected byusing annbyrztransformation matrix. Wewillhave more to saylater about such transformations. Finally, wepromised toremark onwhat todoifatoms come directly from a furnace, gothrough some apparatus, sayA,andarethenanalyzed byaﬁlter which selects thestate X.Youdonotknow what thestate ¢>isthattheystart outin.lt isperhaps bestifyoudon’t worry about thisproblem justyet,butinstead concen- trate onproblems thatalways start outwith pure states. Butifyouinsist, hereis howtheproblem canbehandled. First, youhave tobeabletomake some reasonable guess about thewaythe states aredistributed intheatoms thatcome from thefurnace. Forexample, if 1'Thenumber ofbase states nmay be,andgenerally is,inﬁnite. 5-16 there were nothing “special” about thefurnace, youmight reasonably guess thatatoms would leave thefurnace with random “orientations.” Quantum me- chanically, that corresponds tosaying that you don’t know anything about the states, butthatone-third areinthe(+S) state, one-third areinthe(0S)state, andone-third areinthe(—S) state. Forthose thatareinthe(+S) state the amplitude togetthrough is(XIAI+S) andtheprobability isI(XIAI—I—S)I2, andsimilarly fortheothers. Theoverall probability isthen %I(XIAI+5)I2 +%I(><I/1 I0$)I2 +%I(><IAI—5)I2- Why didweuseSrather than, say,T?Theanswer is,surprisingly, thesame no matter what wechoose forourinitial resolution—so long aswearedealing with completely random orientations. Itcomes about inthesame waythat Z1<><1iS>:2 =Z:<><|tT>|21 .7 foranyX.(Weleave itforyoutoprove.) Note that itisnotcorrect tosaythattheinput state hastheamplitudes \/W3’ tobein(+S), \/U3 tobein(0S),and\/T tobein(—S); thatwould imply that certain interferences might bepossible. Itissimply thatyoudonotknow what theinitial state is;youhave tothink interms oftheprobability thatthesystem starts outinthevarious possible initial states, andthenyouhave totakeaweighted average over thevarious possibilities. 5-17 6 Spin One-Ilalfi 6-1Transforming amplitudes Inthelastchapter, using asystem ofspin oneasanexample, weoutlined thegeneral principles ofquantum mechanics: Any state \//canbedescribed interms ofasetofbase states bygiving theamplitudes tobeineach ofthebase states. Theamplitude togofrom anystate toanother can, ingeneral, bewritten asasumofproducts, each product being theamplitude togointoone ofthebase states times theamplitude togofrom thatbase state tothe ﬁnal condition, with thesumincluding aterm foreach basestate: <><|¢>=Z<><|i><i|¢>- (6-1) Thebase states areorthogonal—the amplitude tobeinoneifyouare intheother iszero: (iI1')=511- (6-2) Theamplitude togetfrom onestate toanother directly isthecomplex conjugate ofthereverse: (XI1P)*=(1PIX)- (6-3) Wealsodiscussed alittle bitabout thefactthatthere canbemore than one base forthestates andthatwecanuseEq.(6.1) toconvert from onebase to another. Suppose, forexample, thatwehave theamplitudes (iSI1//)toﬁndthe state itinevery oneofthebase states iofabase system S,butthatwethen decide thatwewould prefer todescribe thestate interms ofanother setofbase states, saythestates jbelonging tothebase T.Inthegeneral formula, Eq.(6.1), we could substitute jTforXandobtain thisformula: <1TIt>=Z<tT|is><iS|¢>- (6.4) Theamplitudes forthestate ((1/)tobeinthebase states (iT)arerelated tothe amplitudes tobeinthebase states (iS)bythesetofcoeﬂicients (jTIiS).Ifthere areNbase states, there areN2such coefﬁcients. Such asetofcoefﬁcients isoften called the“transformation matrix togofrom theS-representation totheT-represen- tation.” This looks rather formidable mathematically, butwith alittle renaming wecanseethatitisreally notsobad. IfwecallC,theamplitude thatthestate up isinthebase state iS—that is,C,=(iSI1,!/)—and call thecorresponding amplitudes forthebase system T—that is,Cf=(jTI((1),then Eq.(6.4) canbe written as C:=ZR.-.-c.~. (6.5: where R,»,~means thesame thing as(jTIiS).Each amplitude C,’isequal toasum I"This chapter isarather long andabstract sidetour, anditdoes notintroduce any idea which wewillnotalso come tobyya dilferent route inlater chapters. You can, therefore. skipover it,andcome back later ifyouareinterested. 6-16-1 Transforming amplitudes 6-2Transforming toarotated coordinate system 6-3Rotations about thez-axis 6-4Rotations of180° and90° about y 6-5Rotations about x 6-6Arbitrary rotations over alliofoneofthecoeﬁicients R),times each amplitude C,-.Ithasthesame form asthetransformation ofavector from onecoordinate system toanother. Inorder toavoid being tooabstract fortoolong, wehave given yousome examples ofthese coefficients forthespin-one case, soyoucanseehow touse them inpractice. Ontheother hand, there isavery beautiful thing inquantum mechanics—that from thesheer factthat there arethree states andfrom the symmetry properties ofspace under rotations, these coeﬁicients canbefound purely byabstract reasoning. Showing yousuch arguments atthisearly stage has adisadvantage inthat you areimmersed inanother setofabstractions before we get“down toearth.” However, thething issobeautiful thatwearegoing todo itanyway. Wewillshow youinthischapter howthetransformation coefficients canbe derived forspinone-half particles. Wepickthiscase, rather thanspinone,because itissomewhat easier. Our problem istodetermine thecoefficients R,-,fora particle——an atomic system-which issplit into twobeams inaStern-Gerlach apparatus. Wearegoing toderive allthecoeﬁicients forthetransformation from onerepresentation toanother bypure reasoning—plus afewassumptions. Some assumptions arealways necessary inorder touse“pure” reasoning! Although thearguments willbeabstract andsomewhat involved, theresult wegetwillbe relatively simple tostate andeasy tounderstand—-and theresult isthemost important thing. Youmay, ifyouwish, consider thisasasortofcultural excursion. Wehave, infact, arranged that alltheessential results derived here arealso derived insome other waywhen theyareneeded inlater chapters. Soyouneed have nofearoflosing thethread ofourstudy ofquantum mechanics ifyouomit thischapter entirely, orstudy itatsome later time. Theexcursion is“cultural” inthesense thatitisintended toshow thattheprinciples ofquantum mechanics arenotonlyinteresting, butaresodeep thatbyadding onlyafewextra hypotheses about thestructure ofspace, wecandeduce agreat many properties ofphysical systems. Also, itisimportant thatweknow where thedifferent consequences of quantum mechanics come from, because solong asourlaws ofphysics arein- complete—as weknow they are—it isinteresting toﬁndoutwhether theplaces where ourtheories failtoagree with experiment iswhere ourlogic isthebestor where ourlogic istheworst. Until now, itappears thatwhere ourlogic isthemost abstract italways gives correct results—it agrees with experiment. Only when we trytomake speciﬁc models oftheinternal machinery ofthefundamental particles andtheir interactions areweunable toﬁnd atheory that agrees with experiment. Thetheory then thatweareabout todescribe agrees with experiment wherever ithasbeen tested—for thestrange particles aswell asforelectrons, protons, andsoon. Oneremark onanannoying, butinteresting, point before weproceed: Itis notpossible todetermine thecoefficients R),uniquely, because there isalways some arbitrariness intheprobability amplitudes. Ifyouhave asetofamplitudes ofanykind, saytheamplitudes toarrive atsome place byawhole lotofdifferent routes, and ifyou multiply every single amplitude bythesame phase factor——~ saybyef"—you have another setthatisjustasgood. So.itisalways possible to make anarbitrary change inphase ofalltheamplitudes inanygiven problem if youwant to. Suppose youcalculate some probability bywriting asumofseveral amplitudes, say(A—I—B—I—C—I—---)andtaking theabsolute square. Then somebody else calculates thesame thing byusing thesumoftheamplitudes (A’—I—B’+C’+ ---)andtaking theabsolute square. IfalltheA’,B’,C’,etc.,areequal tothe A,B,C,etc., except forafactor eff,allprobabilities obtained bytaking theabsolute squares willbeexactly thesame, since (A’+B’+C’+---)isthen equal to ei‘(A +B+C+--~). Orsuppose, forinstance, that Wewere computing something with Eq.(6.1), butthen wesuddenly change allofthephases ofa certain base system. Every oneoftheamplitudes (iI1//)would bemultiplied by thesame factor eff. Similarly, theamplitudes (iIX)would alsobechanged by eff,buttheamplitudes (XIi)arethecomplex conjugates oftheamplitudes (iIX); therefore, theformer getschanged bythefactor e‘“. Theplus andminus i5’s 6-2 intheexponents cancel out, andwewould have thesame expression wehad before. Soitisageneral rule that ifwechange alltheamplitudes with respect toagiven basesystem bythesame phase—or even ifwejustchange alltheampli- tudes inanyproblem bythesame phase—it makes nodifference. There is,there- fore,some freedom tochoose thephases inourtransformation matrix. Every now andthen wewillmake such anarbitrary choice—usually following theconventions thatareingeneral use. 6-2Transforming toarotated coordinate system Weconsider again the“improved” Stern-Gerlach apparatus described inthe lastchapter. Abeam ofspin one-half particles, entering attheleft, would, in general. besplit into twobeams, asshown schematically inFig.6-1. (There were three beams forspinone.) Asbefore, thebeams areputback together again unless oneortheother ofthem isblocked offbya“stop” which intercepts the beam atitshalf-way point. Intheﬁgure weshow anarrow which points inthe direction ofthe increase ofthemagnitude ofthe ﬁeld—say toward themagnet pole withthesharp edges. Thisarrow wetaketorepresent the“up” axisofanyparticular apparatus. Itisﬁxed relative totheapparatus andwillallow ustoindicate the relative orientations when weuseseveral apparatuses together. Wealsoassume thatthedirection ofthemagnetic ﬁeld ineach magnet isalways thesame with respect tothearrow. Wewillsaythatthose atoms which gointhe“upper” beam areinthe(+) state withrespect tothatapparatus andthatthose inthe“lower” beam areinthe (—)state. (There isno“zero” state forspinone-half particles.) Now suppose weputtwoofourmodiﬁed Stern-Gerlach apparatuses in sequence. asshown inFig.6—2(a). Theﬁrstone,which wecallS,canbeused to prepare apure (+S) orapure (—S) state byblocking onebeam ortheother. [Asshown itprepares apure (+S) state.] Foreach condition, there issome amplitude foraparticle thatcomes outofStobeineither the(+T) orthe(-T) beam ofthesecond apparatus. There are,infact,justfouramplitudes: theampli- tude togofrom (+S) to(+T), from (+S) to(~T), from (—S) to(+T), from (—S)to(—T). These amplitudes arejustthefourcoefﬁcients ofthe transformation matrix R,-,<togofrom theS-representation totheT-representation. Wecancon- sider that theﬁrst apparatus “prepares” aparticular state inonerepresentation andthatthesecond apparatus “analyzes” thatstate interms ofthesecond repre- sentation. Thekind ofquestion wewant toanswer, then, isthis: Ifanatom has been prepared inagiven condition—say the(+S)state—by blocking oneofthe beams intheapparatus S,what isthechance that itwillgetthrough thesecond apparatus Tifthisissetfor,say,the(—T) state. Theresult willdepend, ofcourse, ontheangles between thetwosystems SandT. Weshould explain whyitisthatwecould have anyhope ofﬁnding theco- eﬁicients R,-Ibydeduction. You know that itisalmost impossible tobelieve that ifaparticle hasitsspinlined upinthe+2-direction, thatthere issome chance of ﬁnding thesame particle with itsspin pointing inthe—I-x-direction—-or inany other direction atall.Infact,itisalmost impossible, butnotquite. ltissonearly impossible thatthere isonlyonewayitcanbedone, andthatisthereason wecan ﬁndoutwhat thatunique wayis. Theﬁrst kind ofargument wecanmake isthis. Suppose wehave asetup like theoneinFig.6—2(a). inwhich wehave thetwoapparatuses SandT.with T cocked attheangle atwith respect toS,andweletonly the(—I—) beam through S andthe(—) beam through T.Wewould observe acertain number forthe probability that theparticles coming outofSgetthrough T.Now suppose we make another measurement with theapparatus ofFig. 6—2(b). The relative orientation ofSandTisthesame, butthewhole system sitsatadifferent angle in space. Wewant toassume thatboth ofthese experiments givethesame number forthechance that aparticle inapure state with respect toSwillgetinto some particular state withrespect toT.Weareassuming, inother words, thattheresult ofanyexperiment ofthistypeisthesame—that thephysics isthesame—no matter 6-3SIDE VIEW I *______~_I | I | I ll : II y : 1g________ ____| \F|ELD /GRADIENT TOP VIEW / /_'__T__ —_\\ / \/ \ I 1/ /._/ /'< \\\____ \\ X / Fig. 6-l. Top ond side views ofon "improved" Stern-Gerlach opporotus with beams ofc1spin one-half particle. /\/ \ / \/ \ / / / / / / .____---< /~1 I / I/ v<1 | ,/ I |____ ____l_ S (<1) /\\ \\ \ //’\ / \ ,’ \\\ /I \ / \ )\/ \ \\ \\ /\ A s\ < ,-\ / \ ,’ \/._ \/ ‘T \.= lb) Fig. 6-2. Two equivalent experi- ments. 0 , / \F“"_*___|\\ / F_''__~7 \ \ I l \/ ' l//\\ \ / \ 2) ,’ , (b) / / / // & Fig. 6-,»’\/’/’ \(z \ \ I) \z/z 0 /‘lZ/ \z// //// I\,/6 I_____.' |_s s 3.IfTis“wide open," (b)isequivalent to(cu). howthewhole apparatus isoriented inspace. (You say,“That’s obvious.” But itisanassumption, anditis“right” only ifitisactually what happens.) That means that thecoefficients R,-idepend only ontherelation inspace ofSandT, andnotontheabsolute situation ofSandT.Tosaythisinanother way, R,-1‘ depends only ontherotation which carries StoT,forevidently what isthesame in Fig.6—2(a) andFig.6—2(b) isthethree-dimensional rotation which would carry apparatus Sintotheorientation ofapparatus T.When thetransformation matrix R,-.-depends onlyonarotation, asitdoes here, itiscalled arotation matrix. Forournextstepwewillneed onemore piece ofinformation. Suppose we addathird apparatus which wecancallU,which follows Tatsome arbitrary angle, asinFig. 6—3(a). (lt’s beginning tolook horrible, butthat's thefunof abstract thinking——you canmake themost weird experiments justbydrawing lines!) Now what istheS——>T—>Utransformation? What wereally want to askforistheamplitude togofrom some state with respect toStosome other state withrespect toU,when weknow thetransformation from StoTandfrom T toU.Wearethen asking about anexperiment inwhich both channels ofTare open. Wecangettheanswer byapplying Eq. (6.5) twice insuccession. For going from theS-representation totheT-representation, wehave cg=ZR,T,-Sc.-, (6.6)i where Weputthesuperscripts TSontheR,sothat wecandistinguish itfrom the coeﬂicients RUT wewillhave forgoing from TtoU. Assuming theamplitudes tobeinthebase states oftheU-representation Cl’,wecanrelate them totheT-amplitudes byusing Eq.(6.5) once more; weget cg=ZR,‘,’,Tc;-. (6.7) 1' Now wecancombine Eqs. (6.6) and(6.7) togetthetransformation toUdirectly from S.Substituting from Eq.(6.6) inEq.(6.7), wehave c,';=ZR,‘.’,TZR,T,-Sc.-. (6.8)] '1- Or,since idoes notappear inRZT, wecanputthei-summation alsoinfront, and write c,'.*=ZZR,@TR,T.-Sc. (6.9)‘I J This istheformula foradouble transformation. Notice, however, thatsolong asallthebeams inTareunblocked, thestate coming outofTisthesame astheonethatwent in.Wecould justaswellhave made atransformation from theS-representation directly totheU-representa- tion. Itshould bethesame asputting theUapparatus right after S,asinFig. 6-4 6—3(b). Inthatcase, wewould have written cg=ZRt’,-Sc,-, (6.10)i with thecoefficients Rﬁfs belonging tothistransformation. Now, clearly, Eqs. (6.9) and(6.10) should givethesame amplitudes C,’,',andthisshould betrueno matter what theoriginal state ¢waswhich gave ustheamplitudes C,-.Soitmust bethat R5,’,-S=2R,‘,’,-T12,“-",-*. (6.11) I Inother words, foranyrotation S—>Uofareference base, which isviewed asa compounding oftwosuccessive rotations S—>TandT—>U,therotation matrix 12,14“canbeobtained from thematrices ofthetwopartial rotations byEq.(6.11). lfyouwish, youcanﬁndEq.(6.11) directly from Eq.(6.1), foritisonlyadifferent notation for(kU| iS)=Z,(kU|jT)(jT| iS). Tobethorough, weshould addthefollowing parenthetical remarks. They arenot terribly important, however, soyoucanskiptothenext section ifyouwant. What we have saidisnotquite right. Wecannot really saythat Eq.(6.9) andEq.(6.10) must giveexactly thesame amplitudes. Only thephysics should bethesame; alltheamplitudes could bedifferent bysome common phase factor likee“without changing theresult of anycalculation about therealworld. So,instead ofEq.(6.11), allwecansay,really, is that ewlzgs =2R%TR,-Tﬁ, (6.12)J where 6issome realconstant. What thisextra factor ofe“means, ofcourse, isthatthe amplitudes wegetifweusethematrix RUSmight alldiffer bythesame phase (e—i‘) from theamplitude wewould getusing thetworotations RUTandR”. Weknow thatitdoesn’t matter ifallamplitudes arechanged bythesame phase, sowecould justignore thisphase factor ifwewanted to.Itturns out,however, thatifwedeﬁne allofourrotation matrices inaparticular way, thisextra phase factor willnever appear—the 6inEq.(6.12) will always bezero. Although itisnotimportant fortherestofourarguments, wecangivea quick proof byusing amathematical theorem about determinants. [Ifyoudon't yetknow much about determinants, don’t worry about theproof andjustskiptothedeﬁnition of Eq.(6.15).] First, weshould saythatEq.(6.11) isthemathematical deﬁnition ofa“product” oftwomatrices. (Itisjustconvenient tobeabletosay:“RUS istheproduct ofRUTand RTS3’) Second, there isatheorem ofmathematics—which youcaneasily prove forthe two-by-two matrices wehave here—which says thatthedeterminant ofa“product” of twomatrices istheproduct oftheir determinants. Applying thistheorem toEq.(6.12), weget em(Det RUS) =(Det RUT)- (Det R”). (6.13) (Weleave offthesubscripts, because they don‘t tellusanything useful.) Yes, the26is right. Remember thatwearedealing with two-by-two matrices; every term inthematrix Riffismultiplied bye"’,soeach product inthedeterminant—which hastwofactors—gets multiplied byem. Now let’stake thesquare root ofEq.(6.13) anddivide itinto Eq. (6.12); weget ___RZS__ =Z__1ilL ii. (614) \/Det RUS ,-\/Det R‘/T\/DetRTS ' Theextra phase factor hasdisappeared. Now itturns outthatifwewant allofouramplitudes inanygiven representation tobenormalized (which means, youremember, thatZ,»(¢|i)(il¢) =1),therotation matrices willallhave determinants thatarepure imaginary exponentials, likee"°‘. (We won’t prove it;youwillseethatitalways comes outthatway.) Sowecan,ifwewish, choose tomake allourrotation matrices Rhave aunique phase bymaking DetR=1. Itisdone likethis. Suppose weﬁndarotation matrix Rinsome arbitrary way. Wemake itaruleto“convert” itto“standard form” bydeﬁning RRanM=1- (6_15) atdd \/DetR 6-5 (0) j\X i~<FIELD GRADIENT M/‘Wecandothisbecause wearejustmultiplying each term ofRbythesame phase factor, togetthephases wewant. Inwhat follows, wewillalways assume thatourmatrices have been putinthe“standard form”; then wecanuseEq.(6.11) without having anyextra phase factors. 6-3Rotations about thez-axis Wearenowready toﬁndthetransformation matrix R,~,~between twodifferent representations. With ourrule forcompounding rotations and ourassumption that space hasnopreferred direction, wehave thekeys weneed forﬁnding the matrix ofanyarbitrary rotation. There isonly onesolution. Webegin with the transformation which corresponds toarotation about thez-axis. Suppose we have twoapparatuses SandTplaced inseries along astraight linewith their axes parallel andpointing outofthepage, asshown inFig.6—4(a). Wetake our“z-axis” inthisdirection. Surely, ifthebeam goes “up” (toward +2) intheSapparatus, itwilldothesame intheTapparatus. Similarly, ifitgoes down inS,itwillgo down inT.Suppose, however, that theTapparatus were placed atsome other angle, butstillwith itsaxisparallel totheaxisofS,asinFig.6—4(b). lntuitively, youwould saythata(-1-)beam inSwould stillgowith a(+) beam inT,because theﬁelds and ﬁeld gradients arestillinthesame physical direction. And that would bequite right. Also, a(—) beam inSwould stillgointo a(—) beam inT. Thesame result would apply foranyorientation ofTinthexy-plane ofS.What does thistellusabout therelation between C’+=(+T) tp),CL=(-TI t//)and C+=(+S |t//),C_=(—-S150)? You might conclude that anyrotation about thez-axis ofthe“frame ofreference” forbase states leaves theamplitudes C+to be“up” and“down,” thesame asbefore. Wecould write C:._=C+andCL=C_ —but thatiswrong. Allwecanconclude isthatforsuch rotations theprobabilities tobeinthe“up” beam arethesame fortheSandTapparatuses. That is, ICQLI=lC+| and lC'-|=|C-|- Wecannot saythatthephases oftheamplitudes referred totheTapparatus may notbedifferent forthetwodifferent orientations in(a)and(b)ofFig.6—4. lb) /"\ >1l*< i<____7 \\_______,’-1 ~<\ \ i——,—_ /_i‘__._\ \;_ \ K \ g’ "1\\P / \ / - '1“ 1. PQJ6”Q/ 1, 'v___S_,_/ L___.T_.._._/ ——g—- Fig. 6—4. Rotating 90° about thez-axis. The twoapparatuses in(a)and (b)ofFig. 6-4are,infact, different. aswe canseeinthefollowing way. Suppose thatweputanapparatus infront ofSwhich produces apure (+x) state. (The x-axis points toward thebottom oftheﬁgure.) Such particles would besplit into (+2) and(-2) beams inS.butthetwobeams would berecombined togive a(+x) state again atP1—the exitofS.The same thing happens again inT.Ifwefollow Tbyathird apparatus U,whose axisisin the(+x) direction and, asshown inFig. 6—5(a). alltheparticles would gointo the(+) beam ofU.Now imagine what happens ifTand Uareswung around together by90°tothepositions shown inFig. 6-5(b). Again, theTapparatus puts outjustwhat ittakes in.sotheparticles thatenter Uareina(+x) state with respect toS.ButUnow analyzes forthe(+y) state with respect toS,which is different. (Bysymmetry, wewould now expect only one-half oftheparticles to getthrough.) 6-6 (bl (0) y y Xi Xi (+x) 1/-“*“\\. //--—-\\ Jr/-,:;'\-\-1 (-11) (4-;() I/——— —\ # \\____/ Pl \\____/I l___l____l ¢ \____ S T U S Fig. 6—5. Particle in0i+X) state behaves differently in(0)and What could have changed‘? Theapparatuses TandUarestillinthesame physical relationship toeach other. Can thephysics bechanged justbecause T andUareinadifferent orientation? Ouroriginal assumption isthat itshould not. Itmust bethattheamplitudes withrespect toTare different inthetwocases shown inFig.6—5—and, therefore, also inFig.6—4. There must besome way fora particle toknow thatithasturned thecorner atP1.How could ittell? Well, all wehave decided isthat themagnitudes ofC1andcgarethesame inthetwocases, buttheycould—in fact, must—have different phases. Weconclude thatCftand C+must berelated by cg.=e“c.,, andthatCLandC_must berelated by C’_=e”‘C_, where Aand)1.arerealnumbers which must berelated insome waytotheangle between SandT. Theonlything wecansayatthemoment about >\and/.tisthattheymust not beequal [except forthespecial caseshown inFig.6—5(a), when Tisinthesame orientation asS].Wehave seenthatequal phase changes inallamplitudes have nophysical consequence. Forthesame reason, wecanalways addthesame arbitrary amount toboth )\and,LLwithout changing anything. Sowearepermitted tochoose tomake Aand/.tequal toplusandminus thesame number. That is,we canalways take >\r:)\_()\";l-f), 'ur:'u_(>\'gl~“)_ Then ,_§_g__,)‘_2 2‘ “' Soweadopt theconventioni" that/1.=—)\. Wehave then thegeneral rulethat forarotation ofthereference apparatus bysome angle about thez-axis, thetrans- formation is ca=@+“c+, cg=e-“c_. (616) Theabsolute values arethesame, only thephases aredifferent. These phase factors areresponsible forthedifferent results inthetwoexperiments ofFig.6-5. Now wewould liketoknow thelawthat relates Xtotheangle between S andT.Wealready know theanswer foronecase. Iftheangle iszero, Aiszero. Now wewill assume that thephase shift Aisacontinuous function ofangle qb between SandT(seeFig.6-4) as¢goes tozero——as only seems reasonable. In TLooking atitanother way, wearejustputting thetransformation inthe“standard form” described inSection 6-2byusing Eq.(6.15). 6-7(7) 1--7 __c_ _J[__ 1' '\/ \ __.;1_ /’\_// Lx I f |\ /| Pl therwords, ifwerotate Tfrom thestraight linethrough Sbythesmall angle 6,the Aisalsoasmall quantity, sayme,where missome number. Wewrite itthisway because wecanshow that>\must beproportional to6.Suppose wewere toput after Tanother apparatus T’which makes theangle ewith T,and, therefore, the angle 26withS.Then, withrespect toT,wehave cs,=@“c+, andwith respect toT’,wehave cg;=ﬂag.=@”*c+. Butweknow thatweshould getthesame result ifweputT’right after S.Thus, when theangle isdoubled, thephase isdoubled. Wecanevidently extend the argument andbuild upanyrotation atallbyasequence ofinﬁnitesimal rotations. Weconclude thatforanyangle qt,Aisproportional totheangle. Wecan,therefore, write >\=m¢. Thegeneral result weget,then, isthatforTrotated about thez-axis bythe angle ¢withrespect toS cg.=@*'"¢c+, cg=r1""¢c_. (6.17) Fortheangle ¢>,andforallrotations wespeak ofinthefuture, weadopt thestand- ardconvention thatapositive rotation isariglzt-handed rotation about thepositive direction ofthereference axis. Apositive ¢hasthesense ofrotation ofaright- handed screw advancing inthepositive z-direction. Now wehave toﬁndwhat mmust be.First, wemight trythisargument: Suppose Tisrotated by360°; then, clearly, itisright back atzerodegrees, andwe should have C:*_=C+andC'_=C_,or,what isthesame thing, e’”‘2" =l. Wegetm=l.Thisargument iswrong! Toseethatitis,consider thatTisrotated by180°. Ifmwere equal to1,wewould have Cir=e’”'C+ =—C+ andC’_= e*i"C_ =——C_. However, thisisjusttheoriginal state allover again. Both amplitudes arejustmultiplied by—lwhich gives back theoriginal physical system. (Itisagain acaseofacommon phase change.) Thismeans thatiftheangle between Tand SinFig.6—5(b) isincreased to180°, thesystem (with respect toT)would be indistinguishable from thezero-degree situation, andtheparticles would again gothrough the(+)state oftheUapparatus. At180°, though, the(+)state of theUapparatus isthe(—x) state oftheoriginal Sapparatus. Soa(+x) state would become a(—x) state. Butwehave done nothing tochange theoriginal state; theanswer iswrong. Wecannot have rn=1. Wemust have thesituation thatarotation by360° andnosmaller angle reproduces thesame physical state. This willhappen ifm=5Then, andonly then, willtheﬁrst angle that reproduces thesame physical state be¢>=360°.T Itgives C5,=—C+ 360°6666:z-axis. (6.1s) C’_=—C_ Itisverycurious tosaythatifyouturntheapparatus 360°yougetnewamplitudes. They aren’t really new, though, because thecommon change ofsigndoesn’t give anydifferent physics. Ifsomeone elsehaddecided tochange allthesigns ofthe amplitudes because hethought hehad turned 360°, that’s allright; hegetsthe same physics}: Soourﬁnalanswer isthatifweknow theamplitudes C+andC_for spinone-half particles with respect toareference frame S,andwethenuseabase TItappears thatm=—%would alsowork. However, weseein(6.17) thatthechange insignmerely redeﬁnes thenotation foraspin-up particle. IAlso, ifsomething hasbeen rotated byasequence ofsmall rotations whose netre- sultistoreturn ittotheoriginal orientation, itispossible todeﬁne theideathatithas been rotated 360°—as distinct from zero netrotation—if youhave kept track ofthe whole history. (Interestingly enough, thisisnottrueforanetrotation of720°.) 6-8 system referred toTwhich isobtained from Sbyarotation of¢around thez-axis, thenewamplitudes aregiven interms oftheoldby cg.=at/20+ ¢about z. (6.19) c'_=e-’¢’2c_ 6-4Rotations of180° and90°about y Next, wewilltrytoguess thetransformation forarotation ofTwith respect toSof180° around anaxisperpendicular tothez-axis—say, about they-axis. (Wehave deﬁned thecoordinate axesinFig.6-1.) Inother words, westart with twoidentical Stern-Gerlach equipments, with thesecond one, T,turned “upside down” withrespect totheﬁrstone,S,asinFig.6-6. Now ifwethink ofourpar- ticles aslittle magnetic dipoles, aparticle thatisthe(+S) state—so thatitgoeson the“upper” path intheﬁrstapparatus—will alsotake the“upper” path inthe second, sothatitwillbeintheminus state with respect toT.(Intheinverted Tapparatus, both thegradients andtheﬁeld direction arereversed; foraparticle withitsmagnetic moment inagiven direction, theforce isunchanged.) Anyway, what is“up” withrespect toSwillbe“down” withrespect toT.Forthese relative positions ofSandT,then, weknow thatthetransformation must give ICQLI=IC-I, IC’-l=lC+l- Asbefore, wecannot ruleoutsome additional phase factors; wecould have (for 180°about they-axis) cg.=e"’c_ and cg=wear, (6.20) where BandYarestilltobedetermined. What about arotation of360° about they-axis? Well, wealready know the answer forarotation of360° about thez-axis—the amplitude tobeinanystate changes sign. Arotation of360° around anyaxisalways brings usback tothe original position. Itmust bethatforany360° rotation, theresult isthesame as a360°rotation about thez-axis—all amplitudes simply change sign. Now suppose weimagine twosuccessive rotations of180°about y—using Eq.(6.20)—we should gettheresult ofEq.(6.18). Inother words, ca;=ei’3C'_=@"’e*'*c,. =-c. and (6.21) Cl=eilCQ_ =eilel-BC_ =—C_. Thismeans that ewe" =-1 or ei”=—e““’. Sothetransformation forarotation of180°about they-axis canbewritten C’+=e"BC_, C’_=—e"”3C+. (622) Thearguments wehavejustusedwould apply equally welltoarotation of180° about anyaxisinthexy-plane, although different axes can, ofcourse, givedifferent numbers for[-3.However, thatistheonlywaytheycandiffer. Now there isacer- tainamount ofarbitrariness inthenumber 6,butonce itisspeciﬁed foroneaxis ofrotation inthexy-plane itisdetermined foranyother axis. Itisconventional tochoose tosetB=0fora180°rotation about they-axis. Toshow thatwehave thischoice, suppose weimagine thatBwasnotequal tozeroforarotation about they-axis; then wecanshow thatthere issome other axisinthexy-plane, forwhich thecorresponding phase factor willbezero. Let’s ﬁndthephase factor BAforanaxisAthatmakes theangle Oiwith they-axis, as shown inFig.6—7(a). (For clarity, theﬁgire isdrawn with ctequal toanegative number, butthatdoesn’t matter.) Now ifwetakeaTapparatus which isinitially lined upwith theSapparatus andisthen rotated 180°about theaxisA,itsaxes- which wewillcallx”,y",andz"—will beasshown inFig6—7(a). Theamplitudes 6-9IF_—_T"-7 F——_—“'71: : | 1: ; I i | | | |________I L_______1 S T Z‘ Y Fig.6-6. Arototion of180° about they-axis. Z 180° /x“, (01 1/T \ a Y ‘a \¢\A* \I yll 2.4 Z 7I8 ,2 (bl "—--—* y 6\1'1 /,7 (Cl //1 >< N- -;__.___7/ / \<_/7’mQ~< Fig. 6-7. Al80° rotation about the axis Aisequivalent to<1rotation ofl80° about y,followed byorotation about z’.with respect toTwillthenbe cs;=e""11c_, ca=-6-”’Ac+. (6.23) Wecannow think ofgetting tothesame orientation bythetwosuccessive rotations shown in(b)and(c)oftheﬁgure. First, weimagine anapparatus U which isrotated withrespect toSby180°about they-axis. Theaxesx’,y’,andz’ ofUwillbeasshown inFig.6—7(b), andtheamplitudes withrespect toUare given by(6.22). Now notice thatwecangofrom UtoTbyarotation about the“z-axis” ofU,namely about z’,asshown inFig.6—7(c). From theﬁgure youcanseethat theangle required istwotimes theangle abutintheopposite direction (with respect toz’).Using thetransformation of(6.19) with 4)=-201, weget cg=e-“cgt, c1=e+"“c'_. (6.24) Combining Eqs. (6.24) and(6.22), wegetthat cg;=@*'“‘*“>c_, ca=-e-“"’—“>c+. (6.25) These amplitudes must, ofcourse, bethesame aswegotin6.23). SoBAmust berelated toozand/3by 6,1=[3—oz. (6.26) Thismeans thatiftheangle abetween theA-axis andthey-axis (ofS’)isequal to B,thetransformation forarotation of180°about Awillhave BA=0. Now solong assome axisperpendicular tothez-axis isgoing tohave 5=0, wemay aswell take ittobethey-axis. Itispurely amatter ofconvention, andwe adopt theoneingeneral use. Ourresult: Forarotation of180°about they-axis, wehave CQL=C_ 180°abouty. (6.27) C’_ = —C_|_ While wearethinking about they-axis, let’snext askforthetransformation matrix forarotation of90°about y.Wecanﬁnditbecause weknow thattwo successive 90°rotations about thesame axismust equal one180°rotation. We start bywriting thetransformation for90°inthemost general form: Cir=aC+ +bC_, C’_=cC+ +a’C_. (6.28) Asecond rotation of90°about thesame axiswould have thesame coefﬁcients: C1=aC5, +bC’_, CZ=cCﬂF +a'C’_. (6.29) Combining Eqs. (6.28) and(6.29), wehave C1 =a(aC+ + +b(CC+ +dC_), (6.30) Cl=c(aC+ +bC_) +d(cC+ +dC_). However, from (6.27) weknow that CQL=C_, CZ=—C+, sothatwemust have that ab-1-bd=1, a2+be=0, ac—I—cd=-1, bc+dz=0.(6.31) These four equations areenough todetermine allourunknowns: a,b,c,andd. 6-10 Itisnothard todo. Look atthesecond andfourth equations. Deduce that a2=d2,which means thata=dorelsethata=—d. Buta=—a'isout, because thentheﬁrstequation wouldn’t beright. Soa’=a.Using this,wehave immediately thatb=l/2a andthatc=-1/2a. Now wehave everything in terms ofa.Putting, say,thesecond equation allinterms ofa,wehave a2—4L‘ﬂ=O or a4=%. Thisequation hasfourdifferent solutions, butonlytwoofthem givethestandard value forthedeterminant. Wemight aswell take a=l/\/2; then'l a=1/\/2, b=1/\/2, 6=-1/\/2, d=1/V2. lnother words, fortwoapparatuses SandT,with Trotated with respect to Sby90°about they-axis, thetransformation is 1C’=—(C +C__) 90°about y. (6.32) ca={)2<~C++c_> Wecan, ofcourse, solve these equations forC+and C_,which willgive us thetransformation forarotation ofminus 90°about y.Changing theprimes around, wewould conclude that C5.=X»;-5(C.—Co _9o°abouty. (6.33) 1 0-=72(C+ '1‘C-) 6-5Rotations about x You may bethinking: “This isgetting ridiculous. What arethey going to donext, 47°around y,then 33°about x,andsoon,forever?” No,wearealmost ﬁnished. With justtwoofthetransformations wehave—90° about y,andanarbi- trary angle about z(which wedidﬁrst ifyou remember)—we cangenerate any rotation atall. Asanillustration, suppose thatwewant theangle aaround x.Weknow how todealwiththeangle aaround z,butnowwewant itaround x.How doweget it?First, weturn theaxiszdown onto x—whieh isarotation of+90° about y, asshown inFig. 6-8. Then Weturn through theangle ozaround z’.Then we rotate —90° about y”.Thenetresult ofthethree rotations isthesame asturning around xbytheangle a.Itisaproperty ofspace. (These facts ofthecombinations ofrotations, andwhat theyproduce, arehard tograsp intuitively. Itisrather strange, because weliveinthree dimensions, but itishard forustoappreciate what happens ifweturnthiswayandthen thatway. Perhaps, ifwewere ﬁshorbirds andhadarealappreciation ofwhat happens when weturnsomersaults inspace, wecould more easily appreciate such things.) Anyway, let’swork outthetransformation forarotation byozaround the x-axis byusing what weknow. From theﬁrstrotation by—l-90° around ythe amplitudes goaccording toEq.(6.32). Calling therotated axes x’,y’,andz’,the ‘(The other solution changes allsigns ofa,b,c,anddandcorresponds toa-270” rotation. 6-11Z (0) 90°yl -Q-_____>1\‘< z// X Z (blyll /7 ,/a \ Y Z" \ °\X‘X1, \zIII z (Cl \ \ ylll — \ /IZ // Y / / X X/11 Fig. 6-8. Arotation by ozabout thex-axis isequivalent to:la)arotation by+900 about y,followed bylb)a rotation byaabout 2',followed by(cla rotation of—90° about y”. nextrotation bytheangle ozaround z’takes ustoaframe x”,y",z”,for cg;=e""”c5,, cz=8-“/’c'_. Thelastrotation of—90° about y”takes ustox”’,y”’,z”’;by(6.33), cg’=é(cg;-ca), cw=é(cg;+ca). Combining these lasttwotransformations, weget §_cap=—(e+‘“/205, -e-i“/2c'_), 1 . . ca’=—(e+“*/20;. +e-1"/2c'_).\/2 Using Eqs. (6.32) forC;andC’_,wegetthecomplete transformation: C1’=a{e+“*”<c.. +c_>-e"‘“’2<—¢+ +c_>}. C1’=a{e+"“’”<C+ +C.)+e"i“”(— 0++Co}- Wecanputthese formulas inasimpler form byremembering that ei’+e_“ =2c0s 0, and e"—e“=2isin0. Weget (1 . -I1 C1! ___(COS +l(S111 b t aaoux. Cl’=i<sin -g)C+ +(cos %)C_ Here isourtransformation forarotation about thex-axis byanyangle oz. onlyalittle more complicated than theothers. 6—6Arbitrary rotations Now wecanseehow todoanyangle atall.First, notice thatanyre orientation oftwocoordinate frames canbedescribed interms ofthree ang shown inFig.6-9. Ifwehave asetofaxesx’,y’,andz’oriented inanyway withrespect tox,y,andz,wecandescribe therelationship between thetwofr bymeans ofthethree Euler angles a,B,andV,which deﬁne three successi tations thatwillbring thex,y,zframe intothex’,y’,z’frame. Starting atx werotate ourframe through theangle )6about thez-axis, bringing thex-a thelinex1.Then, werotate byaabout thistemporary x-axis, tobring zdo z’.Finally, arotation about thenewz-axis (that is,z’)bytheangle 'Ywill thex-axis intox’andthey-axis intoy'.1' Weknow thetransformations for ofthethree rotations—they aregiven in(6.19) and(6.34). Combining th theproper order, weget (1 ' .~ (Z _' _ C;=cos5e"'3+“”/2C+ +ism5emg7’/2C_, (I__--Ei<ﬂ—1)/2 5—i<ﬂ+~/>/2C__—lS1l’12€ C++cos2e C_. Sojuststarting from some assumptions about theproperties ofspace, we derived theamplitude transformation foranyrotation atall.That means t TWith alittle work youcanshow thattheframe x,y,zcanalsobebrought in frame x’,y’,z’bythefollowing three rotations about theoriginal axes: (1)rotate angle ‘Yaround theoriginal z-axis; (2)rotate bytheangle aaround theoriginal x (3)rotate bytheangle 6around theoriginal z-axis. 6-12which (6.34) Itis lative les,as atall EIITICS vero- Qy! Z! xlsto wnto bring each emin 6.35) have hatif tothe bythe -axis; Z 1 Z1 6 ..,\ . .<5 \ ..Fig. 6-9. The orientation of any Fig. 6—lO An OXIS Adeﬁned by coordinate frame x’,y’,z’relative to thepolar angles 0and¢>.N V Qgge another frame x,y,2canbedeﬁned in terms ofEuler's angles oz,B,‘Y. weknow theamplitudes foranystate ofaspinone-half particle togointothetwo beams ofaStern-Gerlach apparatus S,whose axes arex,y,andz,wecancalculate what fraction would gointoeither beam ofanapparatus Twith theaxes x’,y’, andz’.Inother words, ifwehave astate 4/ofaspin one-half particle, whose amplitudes areC+=(+ll//>andC_=(—lip)tobe“up” and“down” with respect tothez-axis ofthex,y,zframe, wealsoknow theamplitudes CﬁrandC’_ tobe“up” and“down” with respect tothe2’-axis ofanyother frame x’,y’,z’. Thefour coefficients inEqs. (6.35) aretheterms ofthe“transformation matrix” with which wecanproject theamplitudes ofaspin one-half particle into any other coordinate system. Wewillnowwork outafewexamples toshow youhow itallworks. Let’s takethefollowing simple question. Weputaspinone-half atom through aStern- Gerlach apparatus thattransmits onlythe(+2) state. What istheamplitude that itwillbeinthe(+x) state? The+xaxisisthesame asthe+2’axisofasystem rotated 90°about they-axis. Forthisproblem, then, itissimplest touseEqs. (6.32)—-although you could, ofcourse, usethecomplete equations of(6.35). Since C+=1andC_=0,wegetCQ.=1/\/2. Theprobabilities aretheabso- lutesquare ofthese amplitudes; there isa50percent chance that theparticle will gothrough anapparatus thatselects the(+x) state. lfwehadasked about Fie (—x) state theamplitude would have been —1/\/2, which alsogives aprobabi ity l/2-—as youwould expect from thesymmetry ofspace. Soifaparticle isinthe (+2) state, itisequally likely tobein(+x) or(—x), butwith opposite phase. There’s noprejudice inyeither. Aparticle inthe(+2) state hasa50-5O chance ofbeing in(-l—y) orin(—y). However, forthese (using theformula for rotating —90° about x),theamplitudes are1/\ﬁ and-i/\/2. Inthiscase, the twoamplitudes have aphase difference of90°instead of180°, astheydidforthe (+x) and(—x). Infact,that’s howthedistinction between xandyshows up. Asourﬁnalexample, suppose thatweknow thataspinone-half particle isin astate itsuch thatitispolarized “up” along some axisA,deﬁned bytheangles 0and4»inFig.6-10. Wewant toknow theamplitude (C+l11/)thattheparticle is“up” along zandtheamplitude (C_|1//)thatitis“down” along z.Wecanﬁnd these amplitudes byimagining that Aisthez-axis ofasystem whose x-axis liesin some arbitrary direction—say intheplane formed byAandz.Wecanthen bring theframe ofAintox,y,zbythree rotations. First, wemake arotation by—1r/2 about theaxisA,which brings thex-axis intothelineBintheﬁgure. Then we rotate by0about lineB(thenewx-axis offrame A)tobring Atothez-axis. Finally, werotate bytheangle (1r/2——¢)about x.Remembering thatwehave only a(+) 6-13 state with respect toA,weget 9-£45/2 -9+i¢/2C+ =COS 5e , C_. =Sln 2e . Wewould like,ﬁnally, tosummarize theresults ofthischapter inaform that willbeuseful forourlater work. First, weremind youthatourprimary result in Eqs. (6.35) canbewritten inanother notation. Note that Eqs. (6.35) mean justthesame thing asEq.(6.4). That is,inEqs. (6.35) thecoefficients ofC+= (+S |4/)andC_=(—S |1/1)arejusttheamplitudes (jT] iS)ofEq.(6.4)—the amplitudes thataparticle inthei-state with respect toSwillbeinthej-state with respect toT(when theorientation ofTwith respect toSisiven interms ofthe angles 04,6,andV).Wealsocalled them R,-7,15inEq.(6.6). (Vachave aplethora of notations!) Forexample, R153, =(—T| +S)isthecoefficient ofC+intheformula forC’_,namely, isin(a/2) e“"_"” 2.Wecan,therefore, make asummary ofour results intheform ofatable, aswehave done inTable 6-1. Itwilloccasionally behandy tohave these amplitudes already worked out forsome simple special cases. Let’s letR,(¢) stand forarotation bytheangle ¢> about thez-axis. Wecanalsoletitstand forthecorresponding rotation matrix (omitting thesubscripts iandj,which aretobeimplicitly understood). Inthe same spirit R,,(¢) andR,,(¢) willstand forrotations bytheangle ¢about the x-axis orthey-axis. WegiveinTable 6-2thematrices—the tables ofamplitudes (jTliS)—which project theamplitudes from theS-frame intotheT-frame, where Tisobtained from Sbytherotation speciﬁed. Table 6-2 Theamplitudes (jTliS)forarotation R(¢) bytheangle ¢ about thez-axis, x-axis, ory-axis Table6-1 RM) Theamplitudes (jT|iS)forarotation deﬁned bythe </.Tl"S> +5 -5 Euler angles oz,B,'YofFig.6-9 T“ +7. ens/2 O ___ A_ _ -T 0 e"'¢/2 (jTliS) l +5 -s —l—TOi. _.oi . ~ . -Cos5ei<ﬁ+~n/2 [Sm5e-.<ﬂ—~/>/2R..(¢) (jTliS) +S —S —' lisin(5e“¢‘—‘/l/2 cos9e-*<ﬁ+v>/2 +T C95¢/2 551"¢/22 2ll —T isin¢/2 cos¢/2 R..() (jT|iS) +s -s +T cos¢/2 sin¢/2 —T —sin ¢/2 cos¢/2 6-14 7 The Dependence ofAmplitudes onTime 7-1Atoms atrest; stationary states Wewant now totalkalittle bitabout thebehavior ofprobability amplitudes intime. Wesaya“little bit,” because theactual behavior intime necessarily involves thebehavior inspace aswell. Thus, wegetimmediately intothemost complicated possible situation ifwearetodoitcorrectly andindetail. Weare always inthedifficulty that wecaneither treat something inalogically rigorous butquite abstract way, orwecandosomething which isnotatallrigorous but which gives ussome ideaofarealsituation—postponing untillateramore careful treatment. With regard toenergy dependence, wearegoing totakethesecond course. Wewillmake anumber ofstatements. Wewillnottrytoberigorous—but willjustbetelling youthings thathave been found out,togiveyousome feeling forthebehavior ofamplitudes asafunction oftime. Aswegoalong, theprecision ofthedescription willincrease, sodon’t getnervous thatweseem tobepicking things outoftheair.Itis,ofcourse, alloutoftheair—the airofexperiment and oftheimagination ofpeople. Butitwould takeustoolongtogooverthehistorical development, sowehave toplunge insomewhere. Wecould plunge intotheab- stract anddeduce everything-which youwould notunderstand—or wecould gothrough alarge number ofexperiments tojustify each statement. Wechoose todosomething inbetween. Anelectron alone inempty space can,under certain circumstances, have a certain deﬁnite energy. Forexample, ifitisstanding still(soithasnotranslational motion, nomomentum, orkinetic energy), ithasitsrestenergy. Amore compli- cated object likeanatom canalsohave adeﬁnite energy when standing still,but itcould alsobeinternally excited toanother energy level. (Wewilldescribe later themachinery ofthis.) Wecanoften think ofanatom inanexcited state ashaving adeﬁnite energy, butthisisreally only approximately true. Anatom doesn’t stayexcited forever because itmanages todischarge itsenergy byitsinteraction with theelectromagnetic ﬁeld. Sothere issome amplitude thatanewstate is generated-—with theatom inalower state, andtheelectromagnetic ﬁeldinahigher state, ofexcitation. Thetotal energy ofthesystem isthesame before andafter, buttheenergy oftheatom isreduced. Soitisnotprecise tosayanexcited atom hasadeﬁnite energy; butitwilloften beconvenient andnottoowrong tosaythat itdoes. [Incidentally, whydoesitgoonewayinstead oftheother way’? Why doesan atom radiate light? Theanswer hastodowithentropy. When theenergy isinthe electromagnetic ﬁeld, there aresomany different ways itcanbe—so many different places where itcanwander—that ifwelook fortheequilibrium condition, we ﬁndthatinthemost probable situation theﬁeldisexcited withaphoton, andthe atom isde-excited. Ittakes averylongtimeforthephoton tocome back andﬁnd thatitcanknock theatom back upagain. lt’squite analogous totheclassical problem: Why doesanaccelerating charge radiate? Itisn’tthatit“wants” tolose energy, because, infact,when itradiates, theenergy oftheworld isthesame asit wasbefore. Radiation orabsorption goesinthedirection ofincreasing entr0py.] Nuclei canalsoexistindifferent energy levels, andinanapproximation which disregards theelectromagnetic effects, wecansaythatanucleus inanexcited state stays there. Although weknow thatitdoesn’t staythere forever, itisoften useful tostart outwith anapproximation which issomewhat idealized andeasier to think about. Also itisoften alegitimate approximation under certain circum- stances. (When weﬁrstintroduced theclassical lawsofafalling body, wedidnot include friction, butthere isalmost never acaseinwhich there isn’tsome friction.) 7-17-1Atoms atrest; stationary states 7-2Uniform motion 7-3Potential energy; energy conservation 7-4 Forces; theclassical limit 7-5The “precession” ofaspin one-half particle Review: Chapter 17,Vol.1, Space-Time Chapter 48,Vol. l,Beats Then there arethesubnuclear “strange particles,” which havevarious masses. Buttheheavier onesdisintegrate intoother lightparticles, soagain itisnotcorrect tosaythattheyhave aprecisely deﬁnite energy. That would betrueonlyifthey lasted forever. Sowhen wemake theapproximation thatthey have adeﬁnite energy, weareforgetting thefactthattheymust blow up.Forthemoment, then, wewillintentionally forget about suchprocesses andlearn later howtotakethem intoaccount. Suppose wehave anatom-—or anelectron, oranyparticle—which atrest would have adeﬁnite energy E0.Bytheenergy E0wemean themass ofthewhole thing times 02.Thismass includes anyinternal energy; soanexcited atom hasa mass which isdifferent from themass ofthesame atom intheground state. (The ground statemeans thestate oflowest energy.) WewillcallE0the“energy atrest.” Foranatom atrest,thequantum mechanical amplitude toﬁndanatom ata place isthesame everywhere; itdoesnotdepend onposition. Thismeans, ofcourse, thattheprobability ofﬁnding theatom anywhere isthesame. Butitmeans even more. Theprobability could beindependent ofposition, andstillthephase ofthe amplitude could varyfrom point topoint. Butforaparticle atrest,thecomplete amplitude isidentical everywhere. Itdoes, however, depend onthetime. Fora particle inastateofdeﬁnite energy E0,theamplitude toﬁndtheparticle at(x,y,z) atthetimetis ae-’<”°”‘", (7.1) where aissome constant. Theamplitude tobeatanypoint inspace isthesame forallpoints, butdepends ontime according to(7.1). Weshall simply assume thisruletobetrue. Ofcourse, wecould alsowrite (7.1) as ae-M, (7.2)with ho.)=E0=Mcz, where Mistherestmass oftheatomic state, orparticle. There arethree different ways ofspecifying theenergy: bythefrequency ofanamplitude, bytheenergy in theclassical sense, orbytheinertia. They areallequivalent; theyarejustdifferent ways ofsaying thesame thing. You may bethinking thatitisstrange tothink ofa“particle” which has equal amplitudes tobefound throughout allspace. After all,weusually imagine a“particle” asasmall object located “somewhere.” Butdon’t forget theuncer- tainty principle. Ifaparticle hasadeﬁnite energy, ithasalsoadeﬁnite momentum. Iftheuncertainty inmomentum iszero, theuncertainty relation, ApAx=ii, tellsusthattheuncertainty intheposition must beinﬁnite, andthatisjustwhat wearesaying when wesaythatthere isthesame amplitude toﬁndtheparticle atallpoints inspace. Iftheinternal parts ofanatom areinadifferent state with adifferent total energy, then thevariation oftheamplitude with time isdifferent. Ifyoudon’t know inwhich state itis,there willbeacertain amplitude tobeinonestate anda certain amplitude tobeinanother—and each ofthese amplitudes willhave adif- ferent frequency. There willbeaninterference between these different components —like abeat-note—which canshow upasavarying probability. Something will be“going on”inside oftheatom~even though itis“atrest” inthesense thatits center ofmass isnotdrifting. However, iftheatom hasonedeﬁnite energy, the amplitude isgiven by(7.1), andtheabsolute square ofthisamplitude does not depend ontime. Yousee,then, thatifathing hasadeﬁnite energy andifyouask anyprobability question about it,theanswer isindependent oftime. Although theamplitudes vary with time, iftheenergy isdeﬁnite theyvaryasanimaginary exponential, andtheabsolute value doesn’t change. That’s whyweoften saythatanatom inadeﬁnite energy levelisinastationary state. Ifyoumake anymeasurements ofthethings inside, you’ll ﬁndthatnothing (inprobability) willchange intime. Inorder tohave theprobabilities change in 7-2 time, wehavetohavetheinterference oftwoamplitudes attwodifferent frequencies, andthatmeans thatwecannot know what theenergy is.Theobject willhave one amplitude tobeinastate ofoneenergy andanother amplitude tobeinastate of another energy. That’s thequantum mechanical description ofsomething when itsbehavior depends ontime. Ifwehave a“condition” which isamixture oftwodifferent states withdiffer- entenergies, thentheamplitude foreachofthetwostates varies withtimeaccording toEq.(7.2), forinstance, as e-"<E"">‘ and e-‘(EM)’. (7.3) Andifwehave some combination ofthetwo,wewillhave aninterference. But notice thatifweadded aconstant tobothenergies, itwouldn’t make anydifference. Ifsomebody elsewere touseadifferent scale ofenergy inwhich alltheenergies wereincreased (ordecreased) byaconstant amount—say, bytheamount A-then theamplitudes inthetwostates would, from hispoint ofview, be e—-i(E1+A)!/71 and 6,-—i(E2+-4)!/7i_ Allofhisamplitudes would bemultiplied bythesame factor e“"(“/'9‘, andall linear combinations, orinterferences, would have thesame factor. When wetake theabsolute squares toﬁndtheprobabilities, alltheanswers would bethesame. Thechoice ofanorigin forourenergy scale makes nodifference; wecanmeasure energy from anyzerowewant. Forrelativistic purposes itisnicetomeasure the energy sothattherestmass isincluded, butformany purposes thataren’t rela- tivistic itisoften nicetosubtract some standard amount from allenergies that appear. Forinstance, inthecaseofanatom, itisusually convenient tosubtract theenergy M_.,c2, where M,isthemass ofalltheseparate pieces—the nucleus and theelectrons—which is,ofcourse, different from themass oftheatom. Forother problems itmaybeuseful tosubtract from allenergies theamount M,,c2, where Mgisthemass ofthewhole atom intheground state; thentheenergy thatappears isjusttheexcitation energy oftheatom. So,sometimes wemayshiftourzeroof energy bysome verylarge constant, butitdoesn’t make anydifference, provided weshiftalltheenergies inaparticular calculation bythesame constant. Somuch foraparticle standing still. 7-2Uniform motion Ifwesuppose thattherelativity theory isright, aparticle atrestinoneinertial system canbeinuniform motion inanother inertial system. Intherestframe of theparticle, theprobability amplitude isthesame forallx,y,andzbutvaries with t.Themagnitude oftheamplitude isthesame forall1,butthephase depends ont. Wecangetakindofapicture ofthebehavior oftheamplitude ifweplotlines of equal phase-—say, lines ofzerophase—as afunction ofxandt.Foraparticle at rest,these equal-phase lines areparallel tothex-axis andareequally spaced in thet-coordinate, asshown bythedashed linesinFig.7—l. Inadifferent frame—x’, y’,2’,t’—that ismoving withrespect totheparticle in,say,thex-direction, thex’andt’coordinates ofanyparticular point inspace arerelated toacandzbytheLorentz transformation. Thistransformation canbe represented graphically bydrawing x’andt’axes, asisdone inFig.7-1. (See Chapter 17,Vol.I,Fig.17-2.) Youcanseethatinthex’-t’system, points ofequal phase'l' have adifferent spacing along thet’-axis, sothefrequency ofthetime variation isdifferent. Also there isavariation ofthephase withx’,sotheprob- ability amplitude must beafunction ofx’. 1'Weareassuming thatthephase should have thesame value atcorresponding points inthetwosystems. This isasubtle point, however, since thephase ofaquantum me- chanical amplitude is,toalarge extent, arbitrary. Acomplete justiﬁcation ofthisassump- tionrequires amore detailed discussion involving interferences oftwoormore amplitudes. 7-3A‘ 1' L’ Fig. 7—l. Relativistic transformation oftheamplitude ofuparticle atrest m thex-tsystems.> X Under aLorentz transformation forthevelocity v,sayalong thenegative x-direction, thetime 1isrelated tothetime t’by t’-x’v/c2z=-\/l-v2/c2 soouramplitude now varies as e—(i/ﬁ)E0t _e—(t'/ﬁ)(E9t’/\/1—v2/e2—EQ1>x'/c2\/I-712/1:2) Intheprime system itvaries inspace aswellasintime. Ifwewrite theamplitude as —<'/ﬁ>(E' t'—'’> e I P PI ’ weseethat E1’,=E0/\/lé—‘zﬁ/T istheenergy computed classically fora particle ofrestenergy E0travelling atthevelocity v,andp’=E{,v/c2 isthe corresponding particle momentum. Youknow thatx,,=(t,x,y,z)andp,,=(E,p,,,pg,P2)arefour-vectors, and thatp,,x,, =El—p-xisascalar invariant. Intherestframe oftheparticle, p,,x,,isjustEl;soifwetransform toanother frame, Etwillbereplaced by E/tr __pl_xi- Thus, theprobability amplitude ofaparticle which hasthemomentum pwillbe proportional to e-—(1/7l)(l'7pf—P'”) , where E,,istheenergy oftheparticle whose momentum isp,thatis, E.=\/(pot+E5. (7.6) where E0is,asbefore, therestenergy. Fornonrelativistic problems, wecanwrite E71=Mscz +W21, (7-7) where W,istheenergy over andabove therestenergy M_,c2 oftheparts ofthe atom. Ingeneral, W,,would include both thekinetic energy oftheatom aswell asitsbinding orexcitation energy, which wecancallthe“internal” energy. We would write 2 W,=W,,,,+{T4-» (7.8) andtheamplitudes would be e-<‘/"><W»‘"""‘>. (7.9) Because wewillgenerally bedoing nonrelativistic calculations, wewillusethis form fortheprobability amplitudes. Note that ourrelativistic transformation hasgiven usthevariation ofthe amplitude ofanatom which moves inspace without anyadditional assumptions. Thewave number ofthespace variations is,from (7.9), _12- k-h, (7.10) sothewavelength is 271' hA--E- (7.11) This isthesame wavelength wehave used before forparticles with themomentum p.This formula wasﬁrstarrived atbydeBroglie injustthisway. Foramoving particle, thefrequency oftheamplitude variations isstillgiven by hat=W,,. (7.12) 7-4 Theabsolute square of(7.9) isjustl,soforaparticle inmotion with a deﬁnite energy, theprobability ofﬁnding itisthesame everywhere anddoes not change withtime. (Itisimportant tonotice thattheamplitude isacomplex wave. Ifweused arealsinewave, thesquare would vary from point topoint, which would notberight.) Weknow, ofcourse, thatthere aresituations inwhich particles move from place toplace sothattheprobability depends onposition andchanges withtime. How dowedescribe such situations? Wecandothatbyconsidering amplitudes which areasuperposition oftwoormore amplitudes forstates ofdeﬁnite energy. Wehave already discussed thissituation inChapter 48ofVol.I—even forprob- ability amplitudes! Wefound thatthesumoftwoamplitudes withdifferent wave numbers k(that is,momenta) andfrequencies to(that is,energies) gives inter- ference humps, orbeats, sothat thesquare oftheamplitude varies with space andtime. Wealsofound thatthese beats move withtheso-called “group velocity” given by ,._E,""Ak where AkandAwarethedifferences between thewave numbers andfrequencies forthetwowaves. Formore complicated waves—made upofthesumofmany amplitudes allnearthesame frequency——the group velocity is a0,,=i- (7.13) Taking w=E1,/it andk=p/h,weseethat dEv,=T; (7.14) Using Eq.(7.6), wehave £12_21. 715dp CE, (') ButE,,=Mei’, so dE,, _pZ;_M (7.16) which isjusttheclassical velocity oftheparticle. Alternatively, ifweusethenon- relativistic expressions, wehave W w=713 and k=i, and 2dw=a'W: d(_p>= ps (717) dk dp dp 2M M which isagain theclassical velocity. Our result, then, isthat ifwehave several amplitudes forpure energy states ofnearly thesame energy, their interference gives “lumps” intheprobability that move through space with avelocity equal tothevelocity ofaclassical particle ofthatenergy. Weshould remark, however, that when wesaywecanaddtwo amplitudes ofdifferent wave number together togetabeat-note that willcorre- spond toamoving particle, wehave introduced something new—something that wecannot deduce from thetheory ofrelativity. Wesaid what theamplitude did foraparticle standing stillandthendeduced what itwould doiftheparticle were moving. Butwecannot deduce from these arguments what would happen when there aretwowaves moving with different speeds. Ifwestop one,wecannot stop theother. Sowehave added tacitly theextra hypothesis that notonly is(7.9) a possible solution, butthatthere canalsobesolutions withallkinds ofp’sforthe same system, andthatthedifferent terms willinterfere. 7-5 _._..__---lI|III|1IIIIIIIIII1 nnxxcunwnwmamrtunxixzu-—-I-w l y’ ¢ M + \\\\\\ Fig. 7-2. Aparticle ofmass Mand momentum pinaregion ofconstant potential. 4'1: if: -1--1% ReAmp -1*1r —+2+~[FOR ¢2<4,’) Fig. 7-3. The amplitude forapar- ticle intransit from one potential to another.7-3Potential energy; energy conservation Now wewould liketodiscuss what happens when theenergy ofaparticle canchange. Webegin bythinking ofaparticle which moves inaforce ﬁeldde- scribed byapotential. Wediscuss ﬁrsttheeffect ofaconstant potential. Suppose thatwehave alarge metal canwhich wehave raised tosome electrostatic potential (1),asinFig.7-2. Ifthere arecharged objects inside thecan, their potential energy willbeqq>,which wewillcallV,andwillbeabsolutely independent ofposition. Then there canbenochange inthephysics inside, because theconstant potential doesn’t make anydifference sofarasanything going oninside thecanisconcerned. Now there isnowaywecandeduce/ what theanswer should be,sowemust make aguess. Theguess which works ismore orlesswhat youmight expect: Forthe energy, wemust usethesum ofthepotential energy Vandtheenergy E7,-which isitself thesumoftheinternal andkinetic energies. Theamplitude isproportional to e--(1./h)[(Ep+V)5_P‘xl. Thegeneral principle isthatthecoefﬁcient oft,which wemaycallw,isalways given bythetotalenergy ofthesystem: internal (or“mass”) energy, pluskinetic energy, pluspotential energy: ha:=E,+V. (7.19) Or,fornonrelativistic situations, ‘7 ha=W...++V. (7-20) Now what about physical phenomena inside thebox? Ifthere areseveral different energy states, what willweget? Theamplitude foreach state hasthe same additional factor e_(t/mvt over what itwould have with V=0.That isjust likeachange inthezero ofour energy scale. Itproduces anequal phase change inallamplitudes, butaswehave seen before, thisdoesn’t change anyoftheprobabilities. Allthephysical phenomena arethesame. (We have assumed thatwearetalking about different states ofthe same charged object, sothatq¢isthesame forall.Ifanobject could change its charge ingoing from onestate toanother, wewould have quite another result, butconservation ofcharge prevents this.) Sofar,ourassumption agrees with what wewould expect forachange of energy reference level. Butifitisreally right, itshould hold forapotential energy that isnotjust aconstant. Ingeneral, Vcould vary inanyarbitrary way with both time andspace, andthecomplete result fortheamplitude must begiven in terms ofadifferential equation. Wedon’t want togetconcerned with thegeneral case right now, butonly want togetsome idea about how some things happen, sowewillthink only ofapotential thatisconstant intime andvaries very slowly inspace. Then wecanmake acomparison between theclassical andquantum ideas. Suppose wethink ofthesituation inFig. 7-3, which hastwoboxes held at theconstant potentials ¢1and¢2andaregion inbetween where wewillassume that thepotential varies smoothly from onetotheother. Weimagine thatsome particle hasanamplitude tobefound inanyoneoftheregions. Wealsoassume thatthemomentum islarge enough sothatinanysmall region inwhich there are many wavelengths, thepotential isnearly constant. Wewould then think thatin anypartofthespace theamplitude ought tolook like(7.18) with theappropriate Vforthatpart ofthespace. Let’s think ofaspecial case inwhich 451=O,sothat thepotential energy there iszero, butinwhich q¢2isnegative, sothat classically theparticle would have more energy inthesecond box. Classically, itwould begoing faster inthe second box—it would have more energy and, therefore, more momentum. Let’s/ seehow thatmight come outofquantum mechanics. 7-6 With ourassumption, theamplitude intheﬁrstboxwould beproportional to e—(i/ﬂ)l(Wint+Pf/v2M+Vi)l—P1‘*1 ’ andtheamplitude inthesecond boxwould beproportional to e-—('i/ﬁ)l(Wint+Pg/2M+Vz)l—P2"l D (Let’s saythattheinternal energy isnotbeing changed, butremains thesame in both regions.) Thequestion is:How dothese twoamplitudes match together through theregion between theboxes? Wearegoing tosuppose thatthepotentials areallconstant intime—so that nothing intheconditions varies. Wewillthensuppose thatthevariations ofthe amplitude (that is,itsphase) have thesame frequency everywhere—because, so tospeak, there isnothing inthe“medium” thatdepends ontime. Ifnothing in thespace ischanging, wecanconsider thatthewave inoneregion “generates” subsidiary waves allover space which willalloscillate atthesame frequency——— justaslight waves going through materials atrestdonotchange their frequency. Ifthefrequencies in(7.21) and(7.22) arethesame, wemust have that .Pf _.11?. Wint +517 ‘l" V1 —Wint + + V2- Both sides arejusttheclassical total energies, soEq.(7.23) isastatement ofthe conservation ofenergy. Inother words, theclassical statement oftheconservation ofenergy isequivalent tothequantum mechanical statement thatthefrequencies foraparticle areeverywhere thesame iftheconditions arenotchanging withtime. Itallﬁtswiththeideathathm=E. Inthespecial example that V1=0and V2isnegative, Eq.(7.23) gives that p2isgreater thanpl,sothewavelength ofthewaves isshorter inregion 2.The surfaces ofequal phase areshown bythedashed lines inFig.7-3. Wehave also drawn agraph oftherealpart oftheamplitude, which shows again how the wavelength decreases ingoing from region 1toregion 2.Thegroup velocity of thewaves, which isp/M, alsoincreases inthewayonewould expect from the classical energy conservation, since itisjustthesame asEq.(7.23). There isaninteresting special casewhere V2getssolarge thatV2—V1is greater thanpf/2M. Then pg,which isgiven by 2 pi=2ME-,1!-V2+V1]- (7.24) isnegative. That means thatp2isanimaginary number, say,ip’.Classically, we would saythattheparticle never getsintoregion 2—it doesn’t have enough energy toclimb thepotential hill. Quantum mechanically, however, theamplitude isstill given byEq.(7.22); itsspace variation stillgoesas en‘/MP2-==_ Butifp2isimaginary, thespace dependence becomes arealexponential. Saythat theparticle wasinitially going inthe+x-direction; then theamplitude would varyas e-P"/". (7.25) Theamplitude decreases rapidly withincreasing x. Imagine thatthetworegions atdifferent potentials were veryclose together, sothatthepotential energy changed suddenly from V1toV2,asshown inFig. 7—4(a). Ifweplottherealpartoftheprobability amplitude, wegetthedependence shown inpart(b)oftheﬁgure. Thewave intheﬁrstregion corresponds toa particle trying togetintothesecond region, buttheamplitude there falls off 7-7 <1 ///// \\\\\ \ / 2/2m>Ol (n"’/2m<o ///// Vs\\§\ ..\\\/(\\\ lb) mp)Bu-Ra ~' ReA __1L?__ Fffl "*1 Re(Amp.)I/\\//\d _ I Fig. 7-4. Theamplitude foraparticle approaching Fig. 7-5. The penetration oftheamplitude through astrongly repulsive potential. apotential barrier. vtrtl (0) E VI 1 r, r / 1. (bl r~ReAmpl I’I Fig. 7-6. (a)The potential function forana-particle inauranium nucleus. lb)Thequalitative form oftheprobability amplitude.rapidly. There issome chance thatitwillbeobserved inthesecond region—where itcould never getclassically-but theamplitude isvery small except right near theboundary. Thesituation isverymuch likewhat wefound forthetotal internal reﬂection oflight. Thelight doesn’t normally getout,butwecanobserve itifwe putsomething within awavelength ortwoofthesurface. Youwillremember thatifweputasecond surface close totheboundary where lightwastotally reﬂected, wecould getsome lighttransmitted intothesecond piece ofmaterial. Thecorresponding thing happens toparticles inquantum mechanics. Ifthere isanarrow region with apotential V,sogreat thattheclassical kinetic energy would benegative, theparticle would classically never getpast. Butquan- tummechanically, theexponentially decaying amplitude canreach across the region andgiveasmall probability thattheparticle willbefound ontheother side where thekinetic energy isagain positive. Thesituation isillustrated inFig.7-5. Thiseffect iscalled thequantum mechanical “penetration ofabarrier.” Thebarrier penetration byaquantum mechanical amplitude gives theex- p1anation—or description—of thea-particle decay ofauranium nucleus. The potential energy ofana-particle, asafunction ofthedistance from thecenter, is shown inFig.7-6(a). Ifonetried toshoot an<1-particle with theenergy Einto thenucleus, itwould feelanelectrostatic repulsion from thenuclear charge zand would, classically, getnocloser thanthedistance r1where itstotal energy isequal tothepotential energy V.Closer in,however, thepotential energy ismuch lower because ofthestrong attraction oftheshort-range nuclear forces. How isitthen thatinradioactive decay weﬁnda-particles which started outinside thenucleus coming outwith theenergy E?Because theystart outwith theenergy Einside thenucleus and“leak” through thepotential barrier. Theprobability amplitude isroughly assketched inpart(b)ofFig.7-6,although actually theexponential decay ismuch larger than shown. Itis,infact,quite remarkable thatthemean lifeofana-particle intheuranium nucleus isaslongas4%billion years, when the natural oscillations inside thenucleus aresoextremely rapid—-about 1022persec! How canonegetanumber like109years from l0‘22 sec‘? Theanswer isthatthe exponential gives thetremendously small factor ofabout eT“5—which gives the very small, though deﬁnite, probability ofleakage. Once theoz-p21I'ilCl6 isinthe nucleus. there isalmost noamplitude atallforﬁnding itoutside; however, ifyou take many nuclei andwait long enough, youmay belucky andﬁndonethathas come out. 7-8 7-KYT 11.owv/ °y 1 X L// " It|F=-AV/Byl 89 -~p D —>- 89pl II P YO r / 3 J | /l ‘ 1, , ‘T; 88 j b |l AX ,1-1101-1 v,'F71 W____| WAVE NODE // Fig. 7-7. The deﬂection ofaparticle by a Fig. 7-8. The probability amplitude inaregionH l-1-—w—>1OI< transverse potential gradient. with atransverse potential gradient. 7-4Forces; theclassical limit Suppose thatwehave aparticle moving along andpassing through aregion where there isapotential thatvaries atright angles tothemotion. Classically, we would describe thesituation assketched inFig.7-7. Iftheparticle ismoving along thex-direction andenters aregion where there isapotential thatvaries withy,theparticle willgetatransverse acceleration from theforce F=—6V/6y. Iftheforce ispresent onlyinalimited region ofwidth w,theforce willactonlyfor thetimew/17. Theparticle willbegiven thetransverse momentum P11=F':¥' Theangle ofdeﬂection 60isthen 50 —— £2 -—- Q1) 9 P PU where pistheinitial momentum. Using —6V/6y forF,weget as=-g%;- (7.26) Itisnowuptoustoseeifourideathatthewaves goas(7.20) willexplain thesame result. Welookatthesame thing quantum mechanically, assuming that everything isonavery large scale compared with awavelength ofourprobability amplitudes. Inanysmall region wecansaythattheamplitude varies as e-(i/h)1<W+p’/2M+V) 2-P-~1_ (727) Canweseethatthiswillalsogiverisetoadeﬂection oftheparticle when Vhas atransverse gradient? Wehave sketched inFig. 7-8what thewaves ofprob- ability amplitude willlooklike. Wehave drawn asetof“wave nodes” which you canthink ofassurfaces where thephase oftheamplitude iszero. Inevery small region, thewavelength——the distance between successive nodes—is )\= -{ls P where pisrelated toVthrough P2 W + W + V= COHSIL. Intheregion where Vislarger, pissmaller, andthewavelength islonger. Sothe angle ofthewave nodes getschanged asshown intheﬁgure. Toﬁnd thechange inangle ofthewave nodes wenotice that forthetwo paths aandbinFig. 7-8there isadifference ofpotential AV=(6V/6y)D, so there isadifference Apinthemomentum along thetwo tracks which canbe 7-9 obtained from (7.28): AL2=lap =—AV (729)2M M ' ' Thewave number p/his,therefore, different along thetwopaths, which means thatthephase isadvancing atadifferent rate. Thedifference intherateofincrease ofphase isAk=Ap/h, sotheaccumulated phase difference inthetotal distance wis A(phase) =Ak-W=9" =-ii;AV-W. (7.30) This istheamount bywhich thephase onpath bis“ahead” ofthephase onpath aasthewave leaves thestrip. Butoutside thestrip, aphase advance ofthisamount corresponds tothewave node being ahead bytheamount Ax=7);;A(phase) =%A(phase) or Ax=—-2%AV- w. (7.31) Referring toFig. 7-8, weseethat thenew wavefronts willbeattheangle 60 given by Ax=D60; (7.32) sowehave nae=-1%AV-w. (7.33) This isidentical toEq.(7.26) ifwereplace p/m byvandAV/D by6V/6y. Theresult wehave justgotiscorrect only ifthepotential variations areslow andsmooth-—in what wecalltheclassical limit. Wehave shown thatunder these conditions wewillgetthesame particle motions wegetfrom F=ma,provided weassume thatapotential contributes aphase totheprobability amplitude equal toVt/h. Intheclassical limit, thequantum mechanics willagree with Newtonian mechanics. 7-5The“precession” ofaspinone-half particle Notice thatWehave notassumed anything special about thepotential energy-— itisjustthatenergy whose derivative gives aforce. Forinstance, intheStern- Gerlach experiment wehadtheenergy U=—[.|.-B,which gives aforce ifBhasa spatial variation. Ifwewanted togiveaquantum mechanical description, we would have saidthattheparticles inonebeam hadanenergy thatvaried oneway andthat those intheother beam hadanopposite energy variation. (We could putthemagnetic energy Uinto thepotential energy Vorinto the“internal” energy W;itdoesn’t matter.) Because oftheenergy variation, thewaves are refracted, andthebeams arebent upordown. (We seenow that quantum me- chanics would giveusthesame bending aswewould compute from theclassical mechanics.) From thedependence oftheamplitude onpotential energy wewould also expect that ifaparticle sitsinauniform magnetic ﬁeld along thez-direction, its probability amplitude must bechanging withtimeaccording to e-(i/M-1i,B>t_ (We canconsider that thisis,ineffect, adeﬁnition of11,.) Inother words, ifwe place aparticle inauniform ﬁeld Bforatime T,itsprobability amplitude willbe multiplied by e-lHﬁM—#zBW 7-10 overwhat itwould beinnoﬁeld. Since foraspinone-half particle, [J2canbe either plus orminus some number, say,u,thetwopossible states inauniform ﬁeldwould have their phases changing atthesame ratebutinopposite direc- tions. Thetwoamplitudes getmultiplied by e*""’“"”’. (7.34) This result hassome interesting consequences. Suppose wehave aspin one- halfparticle insome state thatisnotpurely spinuporspindown. Wecandescribe itscondition interms oftheamplitudes tobeinthepure upandpure down states. Butinamagnetic ﬁeld, these twostates willhave phases changing atadifferent rate. Soifweasksome question about theamplitudes, theanswer willdepend onhowlongithasbeen intheﬁeld. Asanexample, weconsider thedisintegration ofthemuon inamagnetic ﬁeld. When muons areproduced asdisintegration products of7r-mesons, they are polarized (inother words, they have apreferred spin direction). Themuons, in turn, disintegrate—in about 2.2microseconds ontheaverage-—emitting anelectron andtwoneutrinos: it->e+1/+1/. Inthisdisintegration itturns outthat(foratleastthehighest energies) theelectrons areemitted preferentially inthedirection opposite tothespindirection ofthemuon. Suppose then thatweconsider theexperimental arrangement shown inFig. 7-9. Ifpolarized muons enter from theleftandarebrought torestinablock of material atA,theywill, alittle while later, disintegrate. Theelectrons emitted will,ingeneral, gooffinallpossible directions. Suppose, however, thatthemuons allenter thestopping block atAwith their spins inthex-direction. Without a magnetic ﬁeldthere would besome angular distribution ofdecay directions; we would liketoknow howthisdistribution ischanged bythepresence ofthemag- neticﬁeld. Weexpect thatitmayvaryinsome waywithtime. Wecanﬁndout what happens byasking, foranymoment, what theamplitude isthat themuon willbefound inthe(+x) state. Wecanstate theproblem inthefollowing way: Amuon isknown tohave itsspininthe+x-direction att=0;what istheamplitude thatitwillbeinthe same state atthetime"r?Now wedonothave anyruleforthebehavior ofaspin one-half particle inamagnetic ﬁeld atright angles tothespin, butwedoknow what happens tothespin upandspindown states with respect totheﬁeld—their ampli- tudcs getmultiplied bythefactor (7.34). Ourprocedure then istochoose the representation inwhich thebase states arespin upandspin down with respect tothez-direction (theﬁeld direction). Any question canthen beexpressed with reference totheamplitudes forthese states. Let’s saythattl/(t)represents themuon state. When itenters theblock A,its stateis11/(0), andwewant toknow (I/(T)atthelatertime1.Ifwerepresent thetwo base states by(+2) and(—-z) weknow thetwoamplitudes (+2I1p(0)) and (-2|¢(0))-we know these amplitudes because weknow that1//(0)represents a statewiththespininthe(+x) state. From theresults ofthelastchapter, these amplitudes aref (+z|+x) =c,=I;-5 and (7.35) <-Zl+><>= C-=-‘5—5- They happen tobeequal. Since these amplitudes refer tothecondition atI=0, let’scallthem C+(0) andC_(0). TIfyouskipped Chapter 6,youcanjusttake (7.35) asanunderived rulefornow. Wewillgivelater (inChapter 10)amore complete discussion ofspinprecession, including aderivation ofthese amplitudes. 7-11B Z L--X SPIN }I- C 1- n: A Et Fig.7-9. Amuon-decay experi- ment. Fig. 7-IO. Time dependence ofthe probability that aspin one-half particle willbeina(+) state with respect tothe x-axis.Now weknow what happens tothese two amplitudes with time. Using (7.34), wehave C+(t) :C+(O)e—(1T/7i)uBI and (7.36) C_(t) :C(_(0)e-I-(i/ft)/-tlil Butifweknow C+(t) andC_(t), wehave allthere istoknow about thecondition att.The only trouble isthat what wewant toknow istheprobability that att thespin willbeinthe+x-direction. Ourgeneral rules can, however, take care of thisproblem. Wewrite thattheamplitude tobeinthe(+x) state attime t,which wemay callA+(t), is A+(l) =(+Xl1//(l)> =<+Xl+Z>(+Z I11/(f)> +if-Xl —Z)<—Zl1l/(1)) O1‘ -4+0) =<-l-XI+Z>C+(l) +<+Xl —Z>C-(l)- (7-37) Again using theresults ofthelastchapter—or better theequality (4)lX)= (X|¢)*from Chapter 5—we know that <+x1+z> =é <+><1-z> =J5 Soweknow allthequantities inEq.(7.37). Weget /1+0) =%e(i/7l)}lBi +_%e—*(7-/ﬁ)|l,Bt. OI‘ BA+(t) =cosif,-1. Aparticularly simple result! Notice that theanswer agrees with what weexpect fort=0.WegetA+(0) =1,which isright, because weassumed thatthemuon wasinthe(+x) state att=0. The probability P+that themuon will befound inthe(+x) state attis (A+)2 or 2BtP+= cos The probability oscillates between zero and one. asshown inFig. 7-10. Note that theprobability returns tooneforpBt/h =7r(not 27r). Because wehave squared thecosine function, theprobability repeats itself with thefrequency Z/.1B/h. PROB.TOHASPNN+xDR. q--__._VE 21r >‘u,B -FT Thus, weﬁndthat thechance ofcatching thedecay electron intheelectron counter ofFig.7-9varies periodically with thelength oftime themuon hasbeen sitting inthemagnetic ﬁeld. Thefrequency depends onthemagnetic moment ii. The magnetic moment ofthemuon has, infact, been measured injustthisway. Wecan, ofcourse, usethesame method toanswer anyother questions about themuon decay. Forexample, how does thechance ofdetecting adecay electron 7-12 inthey-direction at90°tothex-direction butstillatright angles totheﬁelddepend on1?Ifyou work itout, theamplitude tobeinthe(+y) state varies as cos2 {(uBr/ii) —1r/4}, which oscillates with thesame period butreaches itsmax- imum one-quarter cycle later, when /.tBt/h =1r/4. Infact, what ishappening is thatastimegoeson,themuon goesthrough asuccession ofstates which correspond tocomplete polarization inadirection thatiscontinually rotating about thez-axis. Wecandescribe thisbysaying thatthespinisprecessing atthefrequency Zpl? = 7.38) You canbegin toseetheform that ourquantum mechanical description willtake when wearedescribing how things behave intime. 7-13 8 The Hamiltonian Matrix 8-1Amplitudes andvectors Before webegin themain topic ofthischapter, wewould liketodescribe a number ofmathematical ideas that areused alotintheliterature ofquantum mechanics. Knowing them willmake iteasier foryoutoread other books or papers onthesubject. Theﬁrstideaistheclose mathematical resemblance between theequations ofquantum mechanics andthose ofthescalar product oftwovectors. You remember that ifXandqtaretwostates, theamplitude tostart in48andend upinXcanbewritten asasumoveracomplete setofbasestates oftheamplitude togofrom ¢intooneofthebasestates andthenfrom thatbasestate outagain intoX: <nw=Zammo on1ali Weexplained thisinterms ofaStern-Gerlach apparatus, butweremind youthat there isnoneed tohave theapparatus. Equation (8.1) isamathematical lawthat isjustastruewhether weputtheﬁltering equipment inornot—it isnotalways necessary toimagine thattheapparatus isthere. Wecanthink ofitsimply asa formula fortheamplitude (X|¢). Wewould liketocompare Eq.(8.1) totheformula forthedotproduct of twovectors BandA.It"BandAareordinary vectors inthree dimensions, wecan write thedotproduct thisway: §j<B~axa-Al (analli withtheunderstanding thatthesymbol e,-stands forthethree unitvectors inthe x,y,andz-directions. Then B-eliswhat weordinarily callBx;B-e2iswhat we ordinarily callBy;andsoon.SoEq.(8.2) isequivalent to Ba;/42 +B1/Au +B2/42> which isthedotproduct B-A. Comparing Eqs. (8.1) and (8.2), wecan seethefollowing analogy: The states Xand¢correspond tothetwovectors AandB.Thebase states icorrespond tothespecial vectors e,-towhich werefer allother vectors. Any vector canbe represented asalinear combination ofthethree “base vectors” e,-.Furthermore, ifyouknow thecoefficients ofeach “base vector" inthiscombination—that is, itsthree c0mp0nents—you know everything about avector. Inasimilar way, anyquantum mechanical state can bedescribed completely bytheamplitude (i1¢)togointo thebase states; andifyou know these coefﬁcients, you know everything there istoknow about thestate. Because ofthisclose analogy, what wehave called a“state” isoften alsocalled a“state vector." Since thebase vectors e,»areallatright angles, wehave therelation Bi'83'=51']: Thiscorresponds totherelations (5.25) among thebasestates i, (iIf)=5u- (3-4) You seenow why onesays thatthebase states iareall“orthogonal.” 8-18-1Amplitudes andvectors 8-2 Resolving state vectors 8-3 What arethebase states ofthe world? 8-4How states change with time 8-5TheHamiltonian matrix 8-6Theammonia molecule Review: Chapter 49,Vol. I,Modes There isoneminor difference between Eq.(8.1) andthedotproduct. We have that (¢lX) =<><l¢>*- (8-5)Butinvector algebra, A-B=B-A. With thecomplex numbers ofquantum mechanics wehave tokeep straight the order oftheterms, whereas inthedotproduct, theorder doesn’t matter. Now consider thefollowing vector equation: A=Ze,~(e,~-A). (8.6) It’salittle unusual, butcorrect. Itmeans thesame thing as A=ZA,-e,=A,e_,+Ave,+Azez. (8.7) Notice, though, thatEq.(8.6) involves aquantity which isdiﬂerent from adot product. Adotproduct isjustanumber, whereas Eq.(8.6) isavector equation. Oneofthegreat tricks ofvector analysis wastoabstract away from theequations theideaofavector itself. Onemight besimilarly inclined toabstract athing that istheanalog ofa“vector” from thequantum mechanical formula Eq.(8.l)—and onecanindeed. Weremove the(XIfrom both sides Eq.(8.1) and write the following equation (don’t getfrightened——it’s justanotation andinafewminutes youwillﬁndoutwhat thesymbols mean): I¢>=Z1i><iI¢>. (8.8)1 Onethinks ofthebracket (XI4»)asbeing divided intotwopieces. Thesecond piece I¢)isoften called aket,andtheﬁrstpiece (XIiscalled abra(puttogether, theymake a“bra-ket”—a notation proposed byDirac); thehalf-symbols (XIand I¢)arealsocalled state vectors. lnanycase, theyarenotnumbers, and,ingeneral, wewanttheresults ofourcalculations tocome outasnumbers; sosuch“unﬁnished” quantities areonlypart-way steps inourcalculations. Ithappens thatuntil nowwehave written allourresults interms ofnumbers. How havewemanaged toavoid vectors? Itisamusing tonotethateveninordinary vector algebra wecould make allequations involve onlynumbers. Forinstance, instead ofavector equation like F=ma, wecould always have written C-F=C~(ma). Wehave then anequation between dotproducts that istrue foranyvector C. ButifitistrueforanyC,ithardly makes sense atalltokeep writing theCl Now look atEq.(8.1). Itisanequation that istrue foranyX.Sotosave writing, weshould justleave outtheXandwrite Eq.(8.8) instead. Ithasthesame information provided weunderstand thatitshould always be“ﬁnished” by“multi- plying ontheleftby”-—which simply means reinserting——some (XIonboth sides. SoEq.(8.8) means exactly thesame thing asEq.(8.l)—no more, noless. When youwant numbers, youputinthe(XIyouwant. Maybe youhave already wondered about the4)inEq.(8.8). Since theequa- tionistrue forany¢,why dowekeep it?Indeed, Dirac suggests thatthe¢also canjustaswellbeabstracted away, sothatwehave only I=Z|i><i|- (8.9) Andthisisthegreat lawofquantum mechanics! (There isnoanalog invector analysis.) Itsaysthatifyouputinanytwostates Xand¢ontheleftandright of both sides, yougetback Eq.(8.1). Itisnotreally very useful, butit’sanice reminder thattheequation istrueforanytwostates. 8-2 8-2Resolving state vectors Let’slookatEq.(8.8) again; wecanthink ofitinthefollowing way. Any statevector I¢)canberepresented asalinear combination withsuitable coefficients ofasetofbase “vectors”—or, ifyouprefer, asasuperposition of“unit vectors” insuitable proportions. Toemphasize thatthecoeﬂicients (iI¢)arejustordinary (complex) numbers, suppose wewrite <iI45>=Ci- ThenEq.(8.8)isthesame as |¢>=Z|i>c.». (8-10)t Wecanwrite asimilar equation foranyother state vector, sayIX),with, ofcourse, different coefﬁcients——say D,-.Then wehave |><)=Ii)D,-. (8.11) TheD,arejusttheamplitudes (iIX). Suppose wehad started byabstracting the¢from Eq.(8.1). Wewould havehad <><I= (X|1)<tI. (8.12) Remembering that(XIi)=(iIX)*,wecanwrite thisas (XI=2D?(iI. (8.13) Now theinteresting thing isthat wecanjustmultiply Eq.(8.13) andEq.(8.10) togetback (XI¢).When wedothat, wehave tobecareful ofthesummation indices, because theyarequite distinct inthetwoequations. Let’s ﬁrstrewrite Eq.(8.13) as <><l=ZD3‘<11.1' which changes nothing. Then putting ittogether withEq.(8.10), wehave <><1¢>=Z)1>;“<1'|i>c.~. (8-14)if Remember, though, that(jIi)=6,-,',sothatinthesumwehave leftonly the terms withj =i.Weget <><1¢>=ZD?ct. <8-15> where, ofcourse, D,-*=(iIX)*=(XIi),andCi=(iI¢).Again weseethe closeanalogy withthedotproduct A-B= 2,4,-B,-. Theonly difference isthecomplex conjugate onD). SoEq.(8.15) says that if thestate vectors (XIandI¢)areexpanded interms ofthebase vectors (iIorIi), theamplitude togofrom ¢toXisgiven bythekind ofdotproduct inEq.(8.15). Thisequation is,ofcourse, justEq.(8.1) written with different symbols. Sowe havejustgone inacircle togetused tothenewsymbols. Weshould perhaps emphasize again thatwhile space vectors inthree dimen- sionsaredescribed interms ofthree orthogonal unitvectors, thebase vectors Ii) ofthequantum mechanical states must range overthecomplete setapplicable to anyparticular problem. Depending onthesituation, two,orthree, orﬁve,oran inﬁnite number ofbasestates maybeinvolved. Wehave also talked about what happens when particles gothrough an apparatus. Ifwestarttheparticles outinacertain state¢,thensendthem through 8-3 anapparatus, andafterward make ameasurement toseeiftheyareinstate X,the result isdescribed bytheamplitude (XIAI¢)- (8-16) Such asymbol doesn’t have aclose analog invector algebra. (Itiscloser totensor algebra, buttheanalogy isnotparticularly useful.) WesawinChapter 5,Eq. (5.32), thatwecould write (8.16) as <nMo=ZommAmmo. omii Thisisjustanexample ofthefundamental ruleEq.(8.9), used twice. Wealsofound thatifanother apparatus Bwasadded inseries withA,thenwe could write <nMm=ZomWMmMMMm- weilk Again, thiscomes directly from Dirac’s method ofwriting Eq.(8.9)—remember thatwecanalways place abar(I),which isjustlikethefactor 1,between BandA. Incidentally, wecanthink ofEq.(8.17) inanother way. Suppose wethink oftheparticle entering apparatus Ainthestate 4»andcoming outofAinthestate 1,0(“psi”). Inother words, wecould askourselves thisquestion: Canweﬁnda1/1 suchthattheamplitude togetfrom ¢toXisalways identically andeverywhere the same astheamplitude (XIAI¢)?Theanswer isyes. Wewant Eq.(8.17) tobe replaced by <no=Zammo om Wecanclearly dothisif wo=Z@Mmmo=mMo omJ which determines 11/.“But itdoesn’t determine 1/1,”yousay;“itonlydetermines (iI1/1).” However, (iI¢)doesdetermine up,because ifyouhave allthecoefﬁcients thatrelate 1,0tothebasestates i,then1/1isuniquely deﬁned. Infact,wecanplay withournotation andwrite thelastterm ofEq.(8.20) as Mo=ZummMo am1 Then, since thisequation istrueforalli,wecanwrite simply 1o=ZwwMw) om1 Then wecansay:“The state1/1iswhat wegetifwestartwith¢andgothrough the apparatus A.” One ﬁnal example ofthetricks ofthetrade. Westart again with Eq.(8.17). Since itistrueforanyXand¢,wecandrop them both! WethengetI' A=ZHamAtmn- ow)if What doesitmean? Itmeans nomore, noless,thanwhat yougetifyouputback the¢andX.Asitstands, itisan“open” equation andincomplete. lfwemultiply it“ontheleft” byI¢),itbecomes Mo=ZwWAmmo omif TYou might think weshould write IAIinstead ofjust A.Butthen itwould look like thesymbol for“absolute value ofA,”sothebarsareusually dropped. Ingeneral, the bar(I)behaves much likethefactor one. 8-4 which isjustEq.(8.22) allover again. Infact, wecould have justdropped the j’sfromthatequation andwritten I11/)=AI¢)- (3-25) Thesymbol Aisneither anamplitude, noravector; itisanewkindofthing called anoperator. Itissomething which “operates on”astate toproduce anew state—Eq. (8.25) says that Iip)iswhat results ifAoperates onI¢).Again, itis stillanopen equation until itiscompleted withsome bralike(XItogive (XI10)=(XIAI¢)- (3-26) Theoperator Ais,ofcourse, described completely ifwegivethematrix ofampli- tudes (iIAIj)Aalso written A),-—in terms ofanysetofbasevectors. Wehave really added nothing newwith allofthisnewmathematical notation. Onereason forbringing itallupwastoshow youtheway ofwriting pieces of equations, because inmany books you will ﬁnd theequations written inthe incomplete forms, andthere's noreason foryoutobeparalyzed when youcome across them. Ifyou prefer, youcanalways addthemissing pieces tomake an equation between numbers thatwilllook likesomething more familiar. Also, asyouwillsee,the“bra” and“ket” notation isavery convenient one. Foronething, wecanfrom now onidentify astate bygiving itsstate vector. When wewant torefer toastate ofdeﬁnite momentum pwecansay:“thestate Ip)”. Orwemay speak ofsome arbitrary state Itp). Forconsistency wewill always usetheket,writing I11/),toidentify astate. (Itis,ofcourse anarbitrary choice; wecould equally wellhave chosen tousethebra,(I0I.) 8-3What arethebase states oftheworld? Wehave discovered thatanystate intheworld canberepresented asasuper- positionea linear combination with suitable coeff1cients—of base states. You mayask.ﬁrstofall,what base states? Well, there aremany different possibilities. Youcan,forinstance, project aspin inthez-direction orinsome other direction. There aremany, many different representations, which aretheanalogs ofthediffer- entcoordinate systems onecanusetorepresent ordinary vectors. Next, what coefficients? Well, thatdepends onthephysical circumstances. Different setsof coefficients correspond todifferent physical conditions. The important thing to know about isthe“space” inwhich youareworking—in other words, what the basestates mean physically. Sotheﬁrstthing youhave toknow about, ingen- eral,iswhat thebase states arelike. Then youcanunderstand how todescribe a situation interms ofthese base states. Wewould liketolook ahead alittle andspeak abitabout what thegeneral quantum mechanical description ofnature isgoing tobe—in terms ofthenow current ideas ofphysics, anyway. First, onedecides onaparticular representation forthebase states~different representations arealways possible. Forexample, foraspin one-half particle wecanusetheplus andminus states with respect tothe z-axis. Butthere’s nothing special about thez-axis—you cantakeanyother axis youlike. Forconsistency we’ll always pickthez-axis, however. Suppose webegin withasituation with oneelectron. Inaddition tothetwopossibilities forthespin (“up” and“down” along thez-direction), there isalsothemomentum oftheelectron. Wepick asetofbase states, each corresponding toonevalue ofthemomentum. What iftheelectron doesn’t have adeﬁnite momentum? That’s allright; we’re just saying what thebase states are. Iftheelectron hasn't gotadeﬁnite momentum, ithassome amplitude tohave onemomentum andanother amplitude tohave another momentum, and soon. And ifitisnotnecessarily spinning up,ithassome amplitude tobespinning upgoing atthismomentum, andsome amplitude tobespinning down going atthat momentum, and soon. The complete description ofanelectron, sofarasweknow, requires only that the basestates bedescribed bythemomentum andthespin. Sooneacceptable setof base states Ii)forasingle electron refer todifferent values ofthemomentum and 8-5 whether thespinisupordown. Different mixtures ofamplitudes—that is,differ- entcombinations oftheC’sdescribe different circumstances. What anyparticular electron isdoing isdescribed bytelling with what amplitude ithasanup-spin ora down-spin andonemomentum oranother—for allpossible momenta. Soyou canseewhat isinvolved inacomplete quantum mechanical description ofa single electron. What about systems with more than oneelectron? Then thebase states get more complicated. Let’s suppose thatwehave twoelectrons. Wehave, ﬁrstofall, four possible states with respect tospin: both electrons spinning up,theﬁrstone down andthesecond oneup,theﬁrstoneupandthesecond onedown, orboth down. Also wehave tospecify thattheﬁrstelectron hasthemomentum pl,and thesecond electron, themomentum pg.Thebase states fortwoelectrons require thespeciﬁcation oftwomomenta andtwospincharacters. With seven electrons, wehave tospecify seven ofeach. Ifwehave aproton andanelectron, wehave tospecify thespindirection ofthe proton anditsmomentum, andthespindirection oftheelectron anditsmomen- tum. Atleast that’s approximately true. Wedonotreally know what thecorrect representation isfortheworld. Itisallvery welltostart outbysupposing thatif youspecify thespinintheelectron anditsmomentum, andlikewise foraproton, youwillhave thebase states; butwhat about the“guts” oftheproton? Let’s look atitthisway. Inahydrogen atom which hasoneproton andoneelectron. wehave many different base states todescribe——up anddown spins oftheproton andelectron andthevarious possible momenta oftheproton andelectron. Then there aredifferent combinations ofamplitudes C,:which together describe the character ofthehydrogen atom indifferent states. Butsuppose welook atthe whole hydrogen atom asa“particle.” Ifwedidn’t know thatthehydrogen atom wasmade outofaproton andanelectron, wemight have started outandsaid: “Oh, Iknow what thebase states are—they correspond toaparticular momentum ofthehydrogen atom.” No, because thehydrogen atom hasinternal parts. Itmay, therefore, have various states ofdifferent internal energy, anddescribing therealnature requires more detail. Thequestion is:Does aproton have internal parts? Dowehave todescribe aproton bygiving allpossible states ofprotons, andmesons, andstrange particles? Wedon’t know. And even though wesuppose thattheelectron issimple, sothat allwehave totellabout itisitsmomentum anditsspin, maybe tomorrow wewill discover thattheelectron alsohasinner gears andwheels. ltwould mean thatour representation isincomplete, orwrong, orapproximate——in thesame waythata representation ofthehydrogen atom which describes only itsmomentum would be incomplete, because itdisregarded thefactthat thehydrogen atom could have become excited inside. Ifanelectron could become excited inside andturn into something elselike, forinstance, amuon, then itwould bedescribed notjustby giving thestates ofthenewparticle, butpresumably interms ofsome more com- plicated internal wheels. Themain problem inthestudy oft/iefundamentul particles today istodiscover what arethecorrect representations forthedescription of nature. Atthepresent time, weguess thatfortheelectron itisenough tospecify itsmomentum andspin. Wealsoguess thatthere isanidealized proton which has its1r-mesons, andk-mesons, andsoon,thatallhave tobespeciﬁed. Several dozen particles—that’s crazy! Thequestion ofwhat isafundamental particle andwhat isnotafundamental particle—a subject youhear somuch about these days—is thequestion ofwhat istheﬁnal representation going tolook likeintheultimate quantum mechanical description oftheworld. Will theelectron’s momentum stillbetheright thing with which todescribe nature? Oreven, should thewhole question beputthiswayatall!This question must always come upinanyscientiﬁc investigation. Atanyrate, weseeaproblem—how toﬁndarepresentation. We don’t know theanswer. Wedon’t even know whether wehave the“right” problem, butifwedo,wemust ﬁrstattempt toﬁndoutwhether anyparticular particle is “fundamental” ornot. Inthenonrelativistic quantum mechanics—if theenergies arenottoohigh, sothatyoudon’t disturb theinner workings ofthestrange particles andsoforth- 8-6 youcandoapretty good jobwithout worrying about these details. You canjust decide tospecify themomenta andspins oftheelectrons andofthenuclei; then everything willbeallright. Inmost chemical reactions andother low-energy happenings, nothing goes oninthenuclei; they don’t getexcited. Furthermore, ifahydrogen atom ismoving slowly andbumping quietly against other hydrogen atoms—never getting excited inside, orradiating, oranything complicated like that, butstaying always intheground state ofenergy forinternal motion—you canuseanapproximation inwhich youtalkabout thehydrogen atom asone object, orparticle, andnotworry about thefactthatitcandosomething inside. Thiswillbeagood approximation aslong asthekinetic energy inanycollision iswellbelow 10electron volts—the energy required toexcite thehydrogen atom to adifferent internal state. Wewilloften bemaking anapproximation inwhich wedonotinclude thepossibility ofinner motion, thereby decreasing thenumber ofdetails thatwehave toputintoourbase states. Ofcourse, wethen omit some phenomena which would appear (usually) atsome higher energy, butbymaking suchapproximations wecansimplify verymuch theanalysis ofphysical problems. Forexample, wecandiscuss thecollision oftwohydrogen atoms atlowenergy—or anychemical process—without worrying about thefactthat theatomic nuclei could beexcited. Tosummarize, then, when wecanneglect theeffects ofany internal excited states ofaparticle wecanchoose abasesetwhich arethestates of deﬁnite momentum andz-component ofangular momentum. Oneproblem thenindescribing nature istoﬁndasuitable representation for thebasestates. Butthat’s onlythebeginning. Westillwant tobeabletosaywhat “happens.” Ifweknow the“condition” oftheworld atonemoment, wewould like toknow thecondition atalater moment. Sowealsohave toﬁndthelaws that determine howthings change with time. Wenow address ourselves tothissecond partoftheframework ofquantum mechanics——how states change with time. 8-4How states change withtime Wehave already talked about howwecanrepresent asituation inwhich we putsomething through anapparatus. Now oneconvenient, delightful “apparatus” toconsider ismerely await ofafewminutes; thatis,youprepare astate ¢,and thenbefore youanalyze it,youjustletitsit.Perhaps youletitsitinsome particular electric ormagnetic ﬁeld—it depends onthephysical circumstances intheworld. Atanyrate, whatever theconditions are,youlettheobject sitfrom time t1to timet2.Suppose thatitisletoutofyour ﬁrstapparatus inthecondition ¢att1. Andthen itgoes through an“apparatus,” butthe“apparatus” consists ofjust delay until I2.During thedelay, various things could begoing on—external forces applied orother shenanigans—so thatsomething ishappening. Attheendofthe delay, theamplitude toﬁndthething insome state Xisnolonger exactly thesame asitwould have been without thedelay. Since “waiting” isjustaspecial caseof an“apparatus,” wecandescribe what happens bygiving anamplitude with the same form asEq.(8.17). Because theoperation of“waiting” isespecially impor- tant, we’ll callitUinstead ofA,andtospecify thestarting andﬁnishing times t1 andt2,we’ll write U(t2, t1).Theamplitude wewant is <8I1102.1.) I¢>. <8-27> Like anyother such amplitude, itcanberepresented insome base system orother bywriting it Z<><Ii><-"Iveg.toIj><JI¢>- (8.28)ij Then Uiscompletely described bygiving thewhole setofamplitudes—the matrix (iIU02,ti)If>- (3-29) Wecanpoint out,incidentally, thatthematrix (iIU(t2, t1)Ij)gives much more detail than may beneeded. Thehigh-class theoretical physicist working in 8-7 high-energy physics considers problems ofthefollowing general nature (because it’stheway experiments areusually done). Hestarts with acouple ofparticles, likeaproton andaproton, coming together from inﬁnity. (Inthelab,usually one particle isstanding still, andtheother comes from anaccelerator that ispractically atinﬁnity onatomic level.) Thethings gocrash andoutcome, say,twok-mesons, sixrr-mesons, and twoneutrons incertain directions with certain momenta. What’s theamplitude forthistohappen? The mathematics looks likethis: The qs-state speciﬁes thespins and momenta oftheincoming particles. The X would bethequestion about what comes out. Forinstance, with what amplitude doyougetthesixmesons going insuch-and-such directions, andthetwoneutrons going offinthese directions, with their spins so-and-so. Inother words, Xwould bespeciﬁed bygiving allthemomenta, andspins, andsoonoftheﬁnal products. Then thejobofthetheorist istocalculate theamplitude (8.27). However, heis really only interested inthespecial casethatt1is—ooand:2is+oo.(There is noexperimental evidence onthedetails oftheprocess, only onwhat comes in andwhat goes out.) Thelimiting case ofU(t2, t1)ast1—>——m and12—>+00 iscalled S,andwhat hewants is <><ISI¢>. Or,using theform (8.28), hewould calculate thematrix (iI511'), which iscalled theS-matrix. Soifyouseea,theoretical physicist pacing theﬂoor andsaying, “All Ihave todoiscalculate theS-matrix,” youwillknow what he isworried about. How toanalyze—l1ow tospecify thelaws for——the S-matrix isaninteresting question. Inrelativistic quantum mechanics forhigh energies, itisdone oneway, butinnonrelativistic quantum mechanics itcanbedone another way, which is veryconvenient. (This other waycanalsobedone intherelativistic case, butthen itisnotsoconvenient.) Itistowork outtheU-matrix forasmall interval oftime- inother words fort2andt1close together. Ifwecanﬁndasequence ofsuch U's forsuccessive intervals oftime wecanwatch howthings goasafunction oftime. Youcanappreciate immediately thatthiswayisnotsogood forrelativity, because youdon’t want tohave tospecify howeverything looks “simultaneously” every- where. Butwewon’t worry about that—-we’re justgoing toworry about non- relativistic mechanics. Suppose wethink ofthematrix Uforadelay from I1until t3which isgreater than t2.Inother words. let’stakethree successive times: t1lessthan 12lessthan I3. Then weclaim thatthematrix thatgoes between t1andt3istheproduct insuc- cession ofwhat happens when youdelay from I1until t2andthenfrom t2until 13. It’sjustlikethesituation when wehadtwoapparatuses BandAinseries. Wecan then write, following thenotation ofSection 5—6. U(la.-11) =I/((3, 12)‘ U(12,l1)- (8-30) Inother words, wecananalyze anytime interval ifwecananalyze asequence of short timeintervals inbetween. Wejustmultiply together allthepieces; tl1at’s the waythatquantum mechanics isanalyzed nonrelativistically. Ourproblem. then, istounderstand thematrix U(t2, t1)foraninﬁnitesimal time interval—for t2=11+At.Weaskourselves this: Ifwehave astate 4) now, what does thestate look likeaninﬁnitesimal timeAtlater'.’Let’s seehowwe write thatout. Callthestate atthetime t¢(t)) (weshow thetime dependence oflltobeperfectly clear thatwemean thecondition atthetime t).Now weask thequestion: What isthecondition after thesmall interval oftimeAtlater? The answer is III/(t +At)) =U(t—I—At,t)I¢(t)). (8.31) This means thesame aswemeant by(8.25), namely, that theamplitude to 8-8 ﬁndXatthetime t+At,is (XIto+A1)>=(XIU0+At,t)I=t(t)>- (8-32) Since we’re notyettoogood atthese abstract things, let’sproject ourampli- tudes into adeﬁnite representation. Ifwemultiply both sides ofEq.(8.31) by(iI,weget <1‘Ito+A0)=<iI110+Ar.t)Iv(1)>- (8-33) Wecanalsoresolve theI1//(t)) intobase states andwrite <iI¢(1+ A0)=Z<iIvo+At.t)I)"><)'II/(1)) (8.34)J Wecanunderstand Eq.(8.34) inthefollowing way. IfweletC,-(t)= (iI1//(t)) stand fortheamplitude tobeinthebase state iatthetime t,then wecanthink ofthisamplitude (just anumber, remember!) varying with time. Each C,becomes afunction oft.And wealso have some information onhow theamplitudes C;varywith time. Each amplitude at(t+At)isproportional toalloftheother amplitudes attmultiplied byasetofcoefficients. Let’s calltheU-matrix U,~,~,by which wemean U.)=(iIU11)- Then wecanwrite Eq.(8.34) as c,-(1+At)=ZU,-,-(t+At,t)c,-(1). (8.35).7 This, then, ishow thedynamics ofquantum mechanics isgoing tolook. Wedon’t know much about theU,-,-yet,except foronething. Weknow that ifAtgoes tozero, nothing canhappen—we should getjusttheoriginal state. So, U,-I—>1andU,-,2—>0,ifi;éj.Inother words, U,-,-—>6,)forAt—>0.Also, we cansuppose thatforsmall At,each ofthecoefficients U,-,<should differ from 6,-,~ byamounts proportional toAt;sowecanwrite U5)" =dij —I—Ki; AI. However, itisusual totake thefactor (—i/h)I outofthecoefficients K,-,~, for historical andother reasons; weprefer towrite U,-,-(t+At,t)=8,,-£11,-,-(1) At. (8.37) Itis,ofcourse, thesame asEq.(8.36) and, ifyouwish, justdeﬁnes thecoefficients H,-,-(t). Theterms H,-,~arejustthederivatives with respect tot2ofthecoefficients U,-,-(t2, t1),evaluated att2=t1=t. Using thisform forUinEq.(8.35), wehave C,;(l —I—Al) = Ila-[j — H-[j(t) AII Cj(l). .7 Taking thesumover the6,-,~term, wegetjustC,(t), which wecanputontheother sideoftheequation. Then dividing byAt,wehave what werecognize asaderivative §ﬂl =_ H1.j(;)Cj(;) Or ihLL39=EHrj(t)C,-(t). (8.39)J" I‘Weareinabitgf trouble here with notation. Inthefactor (—i/ft), theimeans the imaginary unit\/—l, andnottheindex ithatrefers totheithbase state! Wehope that youwon’t ﬁndittooconfusing. 8-9 You remember that C,-(t) istheamplitude (iI1/)toﬁndthestate /inoneof thebase states i(atthetime t).SoEq.(8.39) tellsushow each ofthecoefficients (iI1/)varies with time. Butthatisthesame assaying thatEq.(8.39) tellsushow thestate 1/varies with time, since wearedescribing 1/interms oftheamplitudes (iI1/).Thevariation of1/intimeisdescribed interms ofthematrix H,,~.which has toinclude, ofcourse, thethings wearedoing tothesystem tocause ittochange. Ifweknow theH,-,-——which contains thephysics ofthesituation andcan,ingeneral, depend onthetime—we have acomplete description ofthebehavior intime ofthe system. Equation (8.39) isthen thequantum mechanical lawforthedynamics oftheworld. (We should saythatwewillalways take asetofbase states which areﬁxed anddonotvary with time. There arepeople who usebase states thatalsovary. However, that’s likeusing arotating coordinate system inmechanics, andwe don’t want togetinvolved insuch complications.) 8-5TheHamiltonian matrix Theidea, then, isthattodescribe thequantum mechanical world weneed to pick asetofbase states iandtowrite thephysical laws bygiving thematrix of coefficients H,-,~. Then wehave everything-—we cananswer anyquestion about what willhappen. Sowehave tolearn what therules areforﬁnding theH’stogo with anyphysical situation—what corresponds toamagnetic ﬁeld, oranelectric ﬁeld, andsoon.And that’s thehardest part. Forinstance, forthenewstrange particles, wehave noidea what Hi,-’s touse. Inother words, nooneknows the complete H,-jforthewhole world. (Part ofthedifficulty isthatonecanhardly hope todiscover theH,-,~when nooneeven knows what thebase states are!) Wedohave excellent approximations fornonrelativistic phenomena andforsome other special cases. Inparticular, wehave theforms thatareneeded forthemotions ofelectrons inatoms-—to describe chemistry. Butwedon’t know thefulltrue Hforthe whole universe. Thecoefficients H,-jarecalled theHamiltonian matrix or,forshort, justthe Hamiltonian. (How Hamilton, who worked inthel830’s, gothisname ona quantum mechanical matrix isataleofhistory.) Itwould bemuch better called theenergy matrix, forreasons thatwillbecome apparent aswework with it.So theproblem is:Know your Hamiltonian! TheHamiltonian hasoneproperty thatcanbededuced right away, namely, that H2“,=H,-,-. (8.40) This follows from thecondition that thetotal probability that thesystem isin some state does notchange. Ifyoustart with aparticle anobject ortheworld- thenyou’ve stillgotitastimegoeson.Thetotal probability ofﬁnding itsomewhere is ZICt(F)I2. which must notvary with time. Ifthisistobetrueforanystarting condition 4), then Eq.(8.40) must alsobetrue. Asourﬁrstexample, wetake asituation inwhich thephysical circumstances arenotchanging with time; wemean theexternal physical conditions, sothatH isindependent oftime. Nobody isturning magnets onandoff. Wealsopick a system forwhich only onebase state isrequired forthedescription; itisanap- proximation wecould make forahydrogen atom atrest, orsomething similar. Equation (8.39) thensays .dClhT‘=Hncl. (8.41) Only oneequation—that’s all! And if11isconstant, thisdifferential equation iseasily solved togive cl=(<><>nst)e—"/Wu‘ (8.42) 8-10 This isthetime dependence ofastate with adeﬁnite energy E=H11. You see whyH1)ought tobecalled theenergy matrix. Itisthegeneralization oftheenergy formore complex situations. Next, tounderstand alittle more about what theequations mean, welook atasystem which hastwobase states. Then Eq.(8.39) reads 171% =Hllcl -1"H1262, (8.43) .dClhﬁ =H21C1 -I"Hggcg. IftheH’sareagain independent oftime, you caneasily solve these equations. Weleave youtotryforfun,andwe’ll come back anddothem later. Yes, youcan solve thequantum mechanics without knowing theH’s, solong asthey arein- dependent oftime. 8-6Theammonia molecule Wewant now toshow youhow thedynamical equation ofquantum mechanics canbeused todescribe aparticular physical circumstance. Wehave picked an interesting butsimple example inwhich, bymaking some reasonable guesses about theHamiltonian, wecanwork outsome important—and even practical—results. Wearegoing totake asituation describable bytwostates: theammonia molecule. The ammonia molecule hasonenitrogen atom and three hydrogen atoms located inaplane below thenitrogen sothatthemolecule hastheform ofapyramid, asdrawn inFig.8—l(a). Now thismolecule, likeanyother, hasaninﬁnite number ofstates. Itcanspin around anypossible axis; itcanbemoving inanydirection: itcanbevibrating inside, andsoon,and soon.Itis,therefore, notatwo-state system atall.Butwewant tomake anapproximation thatallother states remain ﬁxed, because they don’t enter into what weareconcerned with atthemoment. Wewillconsider only that themolecule isspinning around itsaxis ofsymmetry (asshown intheﬁgure), that ithaszero translational momentum, andthat itis vibrating aslittle aspossible. That speciﬁes allconditions except one: there arestill thetwopossible positions forthenitrogen atom——the nitrogen may beononeside oftheplane ofhydrogen atoms orontheother, asshown inFig. 8—l(a) and(b). Sowewilldiscuss themolecule asthough itwere atwo-state system. Wemean thatthere areonly twostates wearegoing toreally worry about, allother things being assumed tostay put. You see,even ifweknow that itisspinning with a certain angular momentum around theaxis andthat itismoving with acertain momentum andvibrating inadeﬁnite way, there arestilltwopossible states. We willsaythat themolecule isinthestate I1)when thenitrogen is“up,” asin Fig.8-1(a),andisinthestate I2)when thenitrogen is“down,” asin(b).Thestates II)andI2)willbetaken asthesetofbase states forouranalysis ofthebehavior oftheammonia molecule. Atanymoment, theactual state I1/)ofthemolecule canberepresented bygiving C1=(II/),theamplitude tobeinstate I1),and C2=(2I1/),theamplitude tobeinstate I2). Then, using Eq.(8.8) wecan write thestate vector I1/)as |1I»)= |1>(1|¢)+ I2>(2I¢>01' I11/)=I1>C1 +I2)C2- (3-44) Now theinteresting thing isthatifthemolecule isknown tobeinsome state atsome instant, itwillnotbeinthesame state alittle while later. The two C-coefficients willbechanging with time according totheequations (8.43)——which hold foranytwo-state system. Suppose, forexample, that you hadmade some observation—or had made some selection ofthemolecules—so that you know thatthemolecule isinitially inthestate I1).Atsome later time, there issome chance thatitwillbefound instate I2).Toﬁndoutwhat thischance is,wehave tosolve thedifferential equation which tellsushowtheamplitudes change with time. 8-ll$45.Fig. 8—-1. Two equivalent geometric arrangements ofthecimmonio molecule.I> 12> Theonly trouble isthatwedon’t know what touseforthecoefﬁcients Hijin Eq.(8.43). There aresome things wecansay,however. Suppose that once the molecule was inthestate II)there was nochance that itcould ever getinto |2), and vice versa. Then H12 and H21 would both bezero, and Eq. (8.43) would read .dC .dClh-?l =H11C1, lhiﬁ =H22C2. Wecaneasily solve these twoequations; weget C1=(const)e"(” “H11‘, C2=(c0nst)e““/'9” 21‘. (8.45) These arejust theamplitudes forstationary states with theenergies E1=H11 andE2=H22. Wenote, however, thatfortheammonia molecule thetwostates |I)and |2)have adeﬁnite symmetry. Ifnature isatallreasonable, thematrix elements H11 andH22 must beequal. We’ll callthem both E0,because they correspond totheenergy thestates would have ifH12andH21were zero. But Eqs. (8.45) donottelluswhat ammonia really does. Itturns outthatitispossible Forthenitrogen topush itswaythrough thethree hydrogens andﬁiptotheother side. Itisquite diﬁicult; togethalf-way through requires alotofenergy. How canitgetthrough ifithasn’t gotenough energy? There issome amplitude thatit willpenetrate theenergy barrier. Itispossible inquantum mechanics tosneak quickly across aregion which isillegal energetically. There is,therefore, some small amplitude that amolecule which starts in|I)willgettothestate |2).The coefficients H12 andH21 arenotreally zero. Again, bysymmetry, they should both bethesame—at least inmagnitude. Infact, wealready know that. ingeneral, H,-1must beequal tothecomplex conjugate ofH_,-,~, sothey candiﬁer only bya phase. Itturns out, asyouwillsee,that there isnolossofgenerality ifwetake them equal toeach other. Forlater convenience wesetthem equal toanegative number; wetake H12 -H21 =—A. Wethen have thefollowing pair of equations: ih%1_=EQC1-AC2, (8.46) ifi =EOC2 —-AC1. (8.47) These equations aresimple enough andcanbesolved inanynumber ofways. One convenient wayisthefollowing. Taking thesum ofthetwo, weget it2.‘;(C1+C2)=(E0—Am+C2), whose solution isC1 + C2 =ae—(t/ﬁ)(Ii‘n—>A)t. Then, taking thedifference of(8.46) and(8.47), weﬁndthat it(C1-cg)=(E0+Axe.-cg). which gives C1—C2=be““”“‘E°+"”. (8.49) Wehave called thetwointegration constants aandb;they are,ofcourse, tobe chosen togive theappropriate starting condition forany particular physical problem. Now, byadding andsubtracting (8.48) and(8.49), wegetC1and C2: C10) :%e—('i/77)(E'0—-4)! +ge——(i/ﬁ>(11'0+A)l, C20) =ge-<i/n1<1='t»-A>r _ge-<»"m><1:0+A>¢_ (851) They arethesame except forthesign ofthesecond term. 8-12 Wehavethesolutions; nowwhat dotheymean? (The trouble withquantum mechanics isnotonly insolving theequations butinunderstanding what the solutions mean!) First, notice thatifb=0,both terms have thesame frequency w=(E0~A)/li. Ifeverything changes atonefrequency, itmeans thatthesystem isinastate ofdeﬁnite energy—here, theenergy (E0—A).Sothere isastationary state ofthisenergy inwhich thetwoamplitudes C1andC2areequal. Wegetthe result that theammonia molecule hasadeﬁnite energy (E11—A)ifthere areequal amplitudes forthenitrogen atom tobe“up” andtobe“down.” There isanother stationary statepossible ifa=0;bothamplitudes thenhave thefrequency (E1,+A)/ft. Sothere isanother state with thedeﬁnite energy (E1,+A)ifthetwoamplitudes areequal butwith theopposite sign; C2=—C1. These aretheonly twostates ofdeﬁnite energy. Wewilldiscuss thestates ofthe ammonia molecule inmore detail inthenext chapter; wewillmention here only a couple ofthings. Weconclude thatbecause there issome chance thatthenitrogen atom can ﬂipfrom oneposition totheother, theenergy ofthemolecule isnotjustE11,aswe would have expected, butthatthere aretwoenergy levels (E11+A)and(E0—A). Every oneofthepossible states ofthemolecule, whatever energy ithas,is“split” intotwolevels. Wesayevery oneofthestates because, youremember, wepicked outoneparticular state ofrotation, andinternal energy, andsoon. Foreach possible condition ofthatkind there isadoublet ofenergy levels because ofthe ﬂip-ﬂop ofthemolecule. Let’s now askthefollowing question about anammonia molecule. Suppose thatatt=O,weknow thatamolecule isinthestate |I)or,inother words, that C1(0) =1andC2(0) =O.What istheprobability thatthemolecule willbefound inthestate I2)atthetime t,orwillstillbefound instate II)atthetime t?Our starting condition tells uswhat aandbareinEqs. (8.50) and (8.51). Letting t=0,wehave that b —b 01(0)=5‘»’2’—=1,62(0)=12—=0. Clearly, a=b=l.Putting these values into theformulas forC1(t) and C2(1) andrearranging some terms, wehave 1 (vi/ii)At _<z/mat C1(t) =e'”/ME“! , (i/MA! —(t'/it)/it_'5E 8 —6 ' Z 6 (ll )Ot ' Wecanrewrite these as c1(t)=e—(i/mE"tcos ail, (8.52) c2(¢)=ze-<”’“E~‘sin all’ (8.53) Thetwoamplitudes have amagnitude thatvaries harmonically with time. The probability that themolecule isfound instate l2)atthetime tisthe absolute square ofC2(t): |C2(r)|2 =sinz%5- (8.54) Theprobability starts atzero (asitshould), rises toone,andthen oscillates back and forth between zero andone, asshown inthecurve marked P2ofFig. 8-2. The probability ofbeing intheI1)state does not, ofcourse, stayatone. It“dumps” intothesecond state until theprobability ofﬁnding themolecule intheﬁrststate iszero, asshown bythecurve P1ofFig. 8-2. Theprobability sloshes back and forth between thetwo. Along time agowesawwhat happens when Wehave twoequal pendulums withaslight coupling. (See Chapter 49,Vol. I.)When weliftoneback andletgo, 8-13 P |.O -\ Pl /’ \\ /I’/ / \ / O5 I \ / i / P/ \ /2/ \ // / \ . .. \ /Fig. 8—2. The probability P1that QI l 1 >4-I 1 anammonia molecule instate ll)at I Z 177' 77 5477 f iIOwillbefound instate ll)atf.The ___ probability P2that itwill befound in t fi stcite‘21>. unl S0 A itswings, butthen gradually theother onestarts toswing. Pretty soon thesecond pendulum haspicked upalltheenergy. Then, theprocess reverses, andpendulum number onepicks uptheenergy. Itisexactly thesame kind ofathing. Thespeed atwhich theenergy isswapped back andforth depends onthecoupling between thetwo pendulums—the rate atwhich the“oscillation” isable toleak across. Also, youremember, with thetwopendulums there aretwospecial motions——each with adeﬁnite frequency—which wecallthefundamental modes. lfwepullboth pendulums outtogether, they swing together atonefrequency. Ontheother hand, ifwepulloneoutoneway andtheother outtheother way, there isanother sta- tionary mode alsoatadeﬁnite frequency. Well, here wehave asimilar situation—the ammonia molecule ismathe- matically likethepairofpendulums. These arethetwofrequencies (E11+A)/h and(E11——A)/h-—for when they areoscillating together, oroscillating opposite. Thependulum analogy isnotmuch deeper than theprinciple that thesame equations have thesame solutions. Thelinear equations fortheamplitudes (8.39) arevery much likethelinear equations ofharmonic oscillators. (Infact, thisis thereason behind thesuccess ofourclassical theory oftheindex ofrefraction, in which wereplaced thequantum mechanical atom byaharmonic oscillator, even though, classically, thisisnotareasonable view ofelectrons circulating about a nucleus.) Ifyou pull thenitrogen tooneside. then you getasuperposition of these twofrequencies, andyougetakind ofbeat note, because thesystem isnot inoneortheother states ofdeﬁnite frequency. Thesplitting oftheenergy levels oftheammonia molecule is,however, strictly aquantum mechanical eﬁect. The splitting oftheenergy levels oftheammonia molecule hasimportant practical applications which wewilldescribe inthenext chapter. Atlong lastwe have anexample ofapractical physical problem thatyoucanunderstand with the quantum mechanics! 8-14 9 Tho Ammonia Maser 9-1Thestates ofanammonia molecule Inthischapter wearegoing todiscuss theapplication ofquantum mechanics toapractical device. theammonia maser. You may wonder why westop our formal development ofquantum mechanics todoaspecial problem, butyouwill ﬁndthat many ofthefeatures ofthisspecial problem arequite common inthe general theory ofquantum mechanics, andyouwilllearn agreat dealbyconsidering thisoneproblem indetail. Theammonia maser isadevice forgenerating electro- magnetic waves, whose operation isbased ontheproperties oftheammonia molecule which wediscussed brieﬂy inthelastchapter. Webegin bysummarizing what wefound there. Theammonia molecule hasmany states. butweareconsidering itasatwo- state system, thinking now only about what happens when themolecule isinany speciﬁc state ofrotation ortranslation. Aphysical model forthetwostates can bevisualized asfollows. Iftheammonia molecule isconsidered toberotating about anaxis passing through thenitrogen atom andperpendicular totheplane ofthehydrogen atoms, asshown inFig.9-1, there arestilltwopossible conditions —the nitrogen may beononesideoftheplane ofhydrogen atoms orontheother. Wecallthese twostates I1)andI2).They aretaken asasetofbase states forour analysis ofthebehavior oftheammonia molecule. lnasystem with two base states, anystate It//)ofthesystem canalways bedescribed asalinear combination ofthetwo base states; that is,there isa certain amplitude C1tobeinonebase state andanamplitude C2tobeinthe other. Wecanwrite itsstate vector as lil/>=l1>C1 +l2>C2, (9-1) where C1=<1l\l’> and C2=<2l¢>- These twoamplitudes change with time according totheHamiltonian equa- tions. Eq.(8.43). Making useofthesymmetry ofthetwostates oftheammonia molecule, wesetH11: H22 :E11, and H12 =H21: —A, and getthe 9-19-1Thestates ofanammonia molecule 9-2Themolecule inastatic electric ﬁeld 9-3Transitions inatime-dependent ﬁeld 9-4Transitions atresonance 9-5Transitions offresonance 9-6Theabsorption oflight MASER :Microwave Amplification byStimulated Emission ofRadiation 6 Dipole Q Moment 9 QS *‘ »-(Q ‘Q CenIer 0 o M Q w ass Q Fig. 9-1. Aphysical model oftwo base states forthe ammonia molecule. These states have the electric dipole I|> I2> moments ,u solution [seeEqs.(8.50) and(8.51)] C1:ge-<1‘/ii)<E11-A)i _I_2e—(i/h)(E11+A)t’ (92) __i __ l)_- C2 =ge (1/71)(E'o AH _Ee(1/ﬁ>(Eo+A)i_ (9.3) Wewant nowtotakeacloser look atthese general solutions. Suppose that themolecule wasinitially putintoastate It//1;) forwhich thecoeﬁicient bwasequal tozero. Then att=0theamplitudes tobeinthestates II)andI2)areidentical, andtheystay thatwayforalltime. Their phases both vary with time inthesame way-—with thefrequency (E11-A)/h. Similarly, ifwewere toputthemolecule intoastate Ii//1)forwhich a=0,theamplitude C2isthenegative ofC1,andthis relationship would stay thatway forever. Both amplitudes would now vary with timewiththefrequency (E11+A)/ii. These aretheonlytwopossibilities ofstates forwhich therelation between C1andC2isindependent oftime. Wehave found twospecial solutions inwhich thetwoamplitudes donotvary inmagnitude and, furthermore, have phases which vary atthesame frequencies. These arestationary slates aswedeﬁned them inSection 7-l, which means that they arestates ofdeﬁnite energy. Thestate I$11) hastheenergy E” =E11—A, andthestate I¢1)hastheenergy E1=E11—I—A.They aretheonly twostationary states thatexist, soweﬁndthatthemolecule hastwoenergy levels, with theenergy difference 2A. (We mean, ofcourse, twoenergy levels fortheassumed state of rotation andvibration which wereferred toinourinitial assumptions.)'I Ifwehadn’t allowed forthepossibility ofthenitrogen flipping back andforth, wewould have taken Aequal tozero andthetwoenergy levels would beontopof each other atenergy E11. Theactual levels arenotthisway; their average energy isE11,butthey aresplit apart byiA,giving aseparation of2Abetween theenergies ofthetwostates. Since Ais,infact, very small, thedifference inenergy isalso verysmall. Inorder toexcite anelectron inside anatom, theenergies involved arerela- tively veryhigh—requiring photons intheoptical orultraviolet range. Toexcite thevibrations ofthemolecules involves photons intheinfrared. Ifyoutalkabout exciting rotations, theenergy differences ofthestates correspond tophotons in thefarinfrared. Buttheenergy difference 2Aislower than anyofthose andis,in fact, below theinfrared andwell into themicrowave region. Experimentally, it hasbeen found that there isapair ofenergy levels with aseparation of10“ electron volt—corresponding toafrequency 24,000 megacycles. Evidently this means that2A=hf,withf=24,000 megacycles (corresponding toawavelength of1%cm). Sohere wehave amolecule thathasatransition which does notemit light intheordinary sense, butemits microwaves. Forthework that follows weneed todescribe these twostates ofdeﬁnite energy alittle bitbetter. Suppose wewere toconstruct anamplitude C11bytaking thesum ofthetwonumbers C1andC2: CII= C1"l"C2= <1l‘i’>-l-<2l‘I”l- (9-4) What would thatmean? Well, thisisjusttheamplitude toﬁndthestate III>)ina newstate III)inwhich theamplitudes oftheoriginal base states areequal. That is,writing C”=(III<P),wecanabstract the14>)away from Eq.(‘l.4)—because itistrue forany—and get <11l=<1l+<-9|, which means thesame as III) =I1)+I2). (9.5) I‘Inwhat follows itishelpful—in reading toyourself orintalking tosomeone else~t0 have ahandy way ofdistinguishing between theArabic land2andtheRoman IandII. Weﬁnditconvenient toreserve thenames “one” and“two’” fortheArabic numbers, and tocallIandIIbythenames “eins"" and“zwei" (although “unus” and “duo” might be more logicall). 9-2 Theamplitude forthestate II1)tobeinthestate I1)is (1|11>= <1|1)+(1l2), which is,ofcourse, just 1,since II)and I2)arebase states. The amplitude for thestate III)tobeinthestate I2)isalsol,sothestate III)isonewhich hasequal amplitudes tobeinthetwobase states II)andI2). Weare,however, inabitoftrouble. The state III)hasatotal probability greater than oneofbeing insome base state orother. That simply means, however, thatthestate vector isnotproperly “normalized.” Wecantake care ofthat by remembering that weshould have (III1])=1,which must besoforanystate. Using thegeneral relation that <><l>=Z<><|i><il<1>>, letting both andXbethestate II,andtaking thesum over thebase states II) andI2),wegetthat <11]11>=<11|1>(1|11> +(11|2)<2|11>. Thiswillbeequal tooneasitshould ifwechange ourdeﬁnition ofC”—in Eq. (9.4)—to read 1C =—_[C+C]. 11 X/2 1 2 Inthesame waywecanconstruct anamplitude cu=$2[C1—C21, Of CI=é[<1l<1>>—<2l<1>>1- <9-6) Thisamplitude istheprojection ofthestate IQ)intoanewstate II)which has opposite amplitudes tobeinthestates I1)andI2).Namely, Eq.(9.6) means thesame as <11=inn—<2|1, Or 11>=in1>-12>], <91) from which itfollows that <1|1>=ti;=—<2|1>- Now thereason wehave done allthisisthatthestates II)andII1)canbe taken asanewsetofbase slates which areespecially convenient fordescribing the stationary states oftheammonia molecule. You remember that therequirement forasetofbase states isthat <5Ij>=5w‘- Wehave already ﬁxed things sothat (III) =(IIIII) =1. You caneasily show from Eqs. (9.5) and(9.7) that (IIII) =<11|1> =0. Theamplitudes C1=(II<b)andC1; =(III)foranystate tobeinour new base states II)and III)must also satisfy aHamiltonian equation with the 9-3 form ofEq.(8.39). Infact, ifwejustsubtract thetwoequations (9.2) and(9.3) anddifferentiate withrespect toz,weseethat ih52% =(E0—I—A)C1 =EICI. (9.8) And taking thesum ofEqs. (9.2) and(9.3), weseethat atiﬁfl=(E0~A)c,, =E,,c,,. (9.9) Using II)and III)forbase states, theHamiltonian matrix hasthesimple form H1,1 =E1, H1,11 =0, H1,11 =0, H1111 =E11- Note that each oftheEqs. (9.8) and(9.9) look justlikewhat wehadinSection 8—6fortheequation ofaone-state system. They have asimple exponential time dependence corresponding toasingle energy. Astimegoeson,theamplitudes to beineach state actindependently. The two stationary states IILI) and Iyb”) wefound above are, ofcourse, solutions ofEqs. (9.8) and (9.9). The state I1//I) (for which C,I—C2) has CI=e—(-i/ﬁ)(E0+A)t, CH=0' (9_10) Andthestate Ii//1;)(forwhich C1=C2)has c,=0,C11=e-“/""‘E<>-A“. (9.11) Remember that theamplitudes inEq.(9.10) are C1=<1I\//1), and C11=<”I¢1>; soEq.(9.10) means thesame thing as WI) =Ine—(i/ﬁ)(E0+A)t_ That is,thestate vector ofthestationary state I(1/1)isthesame asthestate vector ofthebase state II)except fortheexponential factor appropriate totheenergy of thestate. Infactatt=0 W1) =I1); thestate II)hasthesame physical conﬁguration asthestationary state ofenergy E0+A.Inthesame way, wehave forthesecond stationary state that III/HI :I”)e-(i/ﬁ)(E0-—A)t' Thestate II1)isjustthestationary state ofenergy E0—Aatt=O.Thus our twonewbasestates I1)andIII)have physically theform ofthestates ofdeﬁnite energy, with theexponential time factor taken outsothat they canbetime- independent base states. (Inwhat follows wewillﬁnditconvenient nottohave todistinguish always between thestationary states I|//1)andIt//11) andtheir base states II)andIII),since they differ only bytheobvious time factors.) Insummary, thestate vectors I1)andIII)areapairofbasevectors which areappropriate fordescribing thedeﬁnite energy states oftheammonia molecule. They arerelated toouroriginal basevectors by I 1 I1)=—[|1>-|3>l, I11)=—[|1>+I3>]~ (9-12)\/_2 \/5 Theamplitudes tobeinI1)andII1)arerelated toC1andC2by CI=$161 -C21, C11=$16. +ca. (9.13) 9-4 Anystate atallcanberepresented byalinear combination ofI1)andI2)—with thecoefﬁcients C1andC2—or byalinear combination ofthedeﬁnite energy base states II)andIII)—with thecoeﬂicients C1andC11. Thus, I4‘)=|1)C1 -I"|3>Cz of I> =II)C1 —I—III>C1']. Thesecond form gives ustheamplitudes forﬁnding thestate I)inastate with theenergy E1=EU+Aorinastate withtheenergy E11=E0—A. 9-2Themolecule inastatic electric ﬁeld Iftheammonia molecule isineither ofthetwostates ofdeﬁnite energy andwe disturb itatafrequency wsuch thatitw=E1—E11=2A,thesystem may make atransition from onestate totheother. Or,ifitisintheupper state, itmay change tothelower state andemit aphoton. Butinorder toinduce such transitions you must have aphysical connection tothestates—some wayofdisturbing thesystem. There must besome external machinery foraffecting thestates, such asmagnetic orelectric ﬁelds. Inthisparticular case, these states aresensitive toanelectric ﬁeld. Wewill, therefore, look next attheproblem ofthebehavior oftheammonia molecule inanexternal electric ﬁeld. Todiscuss thebehavior inanelectric ﬁeld, wewillgoback totheoriginal base system II)andI2),rather than using II)andIII). Suppose thatthere isan electric ﬁeld inadirection perpendicular totheplane ofthehydrogen atoms. Disregarding forthemoment thepossibility ofﬂipping back andforth, would itbe truethattheenergy ofthis molecule isthesame forthetwopositions ofthenitrogen atom? Generally, no.Theelectrons tend toliecloser tothenitrogen than tothe hydrogen nuclei, sothehydrogens areslightly positive. The actual amount depends onthedetails ofelectron distribution. Itisacomplicated problem to ﬁgure outexactly what thisdistribution is,butinanycasethenetresult isthatthe ammonia molecule hasanelectric dipole moment, asindicated inFig. 9-1. We cancontinue ouranalysis without knowing indetail thedirection oramount of displacement ofthecharge. However, tobeconsistent with thenotation ofothers, let’s suppose that theelectric dipole moment is11,with itsdirection point from thenitrogen atom andperpendicular totheplane ofthehydrogen atoms. Now, when thenitrogen ﬂipsfrom onesidetotheother, thecenter ofmass willnotmove, buttheelectric dipole moment willﬁipover. Asaresult ofthis moment, theenergy inanelectric ﬁeld 8willdepend onthemolecular orientation.'I With theassumption made above, thepotential energy willbehigher ifthenitrogen atom points inthedirection oftheﬁeld, andlower ifitisintheopposite direction; theseparation inthetwoenergies willbe2,1/.8. lnthediscussion uptothispoint, wehave assumed values ofE0andAwithout knowing how tocalculate them. According tothecorrect physical theory, it should bepossible tocalculate these constants interms ofthepositions and motions ofallthenuclei andelectrons. Butnobody hasever done it.Such a system involves tenelectrons and four nuclei and that’s just toocomplicated a problem. AsamatterIof fact, there isnoonewho knows much more about this molecule than wedo. Allanyone cansayisthat when there isanelectric ﬁeld, theenergy ofthetwostates isdifferent, thedifference being proportional tothe electric ﬁeld. Wehave called thecoefficient ofproportionality 2,11,butitsvalue must bedetermined experimentally. Wecanalso saythat themolecule hasthe amplitude Atoﬂipover, butthiswillhave tobemeasured experimentally. Nobody cangive usaccurate theoretical values of,uandA,because thecalculations are toocomplicated todoindetail. TWearesorry thatwehave tointroduce anewnotation. Since wehave been using pandEformomentum andenergy, wedon’t want tousethem again fordipole moment andelectric ﬁeld. Remember, inthissection ,u.istheelectric dipole moment. 9-5 For theammonia molecule inanelectric ﬁeld, ourdescription must be changed. Ifweignored theamplitude forthemolecule toﬂipfrom oneconﬁgura- tiontotheother, wewould expect theenergies ofthetwostates I1)andI2)tobe (E0i118). Following theprocedure ofthelastchapter, wetake H11 = E0 "I" [18, H22 = E0 _' Also wewillassume thatfortheelectric ﬁelds ofinterest theﬁelddoes notaffect appreciably thegeometry ofthemolecule and, therefore, does notaffect the amplitude that thenitrogen willjump from oneposition totheother. Wecan thentakethatH12andH21arenotchanged; so H12 =H21 =—A. Wemust now solve theHamiltonian equations, Eq.(8.43), with these new values ofH1,-. Wecould solve them justaswedidbefore, butsince wearegoing tohave several occasions towant thesolutions fortwo-state systems, let’ssolve theequa- tions once andforallinthegeneral case ofarbitrary H,-,'——assuming only thatthey donotchange with time. Wewant thegeneral solution ofthepair ofHamiltonian equations in%=H1101+H1202. <9-16> @552 = HZIC1 + H22C2. Since these arelinear differential equations with constant coeﬁicients, wecanalways ﬁndsolutions which areexponential functions ofthedependent variable 1.We willﬁrstlook forasolution inwhich C1andC2both have thesame time depen- dence; wecanusethetrialfunctions C1 = a1€_i°’i, C2 = a2e_i“". Since such asolution corresponds toastate ofenergy E=hm,wemay aswellwrite right away (WM C1=919-1 J”, (913) C2 =a2e—(i/ii)Et’ (9_19) where Eisasyetunknown andtobedetermined sothatthedifferential equations (9.16) and(9.17) aresatisﬁed. When wesubstitute C1and C2from (9.18) and (9.19) inthedifferential equations (9.16) and (9.17), thederivatives give usjust —1'E/h times C1orC1, sotheleftsides become just EC1 andEC2. Cancelling thecommon exponential factors, weget E91: H1191+ H1292» E92 =H2191+ H2292- Or,rearranging theterms, wehave (E_H11)91 "H1292 =or (9-20) —H2191 +(E—H22)92 =0- (921) With such asetofhomogeneous algebraic equations, there willbenonzero solu- tions fora1and112only ifthedeterminant ofthecoefficients ofa1and111iszero, thatis,if E--H -11Det 11 12=0. (9.22)—H21 E—H22 9-6 However, when there areonly twoequations andtwounknowns, wedon’t need such asophisticated idea. The twoequations (9.20) and (9.21) each give aratio forthetwocoefﬁcients a1and112,andthese tworatios must beequal. From (9.20) wehave that 91 H10 ~=—A, 9.2392 E- H11 ( ) andfrom (9.21) that G1 E—H22 ~=————~ 9.2492 H21 ( ) Equating these tworatios, wegetthatEmust satisfy (E—H11)(E _H22) "H12H21= O- This isthesame result wewould getbysolving Eq.(9.22). Either way, wehave aquadratic equation forEwhich hastwosolutions: E:.H11 -2522 =1;\/§H11 -4H22)2 —I—H12H21. (9.25) There aretwo possible values fortheenergy E.Note that both solutions give realnulnlw/'.s' fortheenergy. because H11andH22arereal,andH12H21 isequal toH12H’I‘2 =IH12|2, which isboth realandpositive. Using thesame convention wetook before, wewillcalltheupper energy E1andthelower energy E11. Wehave H H2 7-'1-H...2Ti“TE1=41%’ -1-J%i)— -1-H12H21, (9-26) El! : H11 T; H22 _ {£1111 ;_H22)2 _I_ H12H21. Using each ofthese twoenergies separately inEqs. (9.18) and (9.19), wehave theamplitudes forthetwostationary states (thestates ofdeﬁnite energy). Ifthere arenoexternal disturbances, asystem initially inoneofthese states willstaythat wayforever—only itsphase changes. Wecancheck ourresults fortwospecial cases. IfH12 =H21 =O,wehave that E1=H11 and E11 =H22. This iscertainly correct. because then Eqs. (9.16) and(9.17) areuncoupled, andeach represents astate ofenergy H11 and H22. Next. ifwesetH11 =H22 =E11and H21 =H12 =—A, wegetthe solution wefound before: EIIIEQ-I-A and E][=EO—A. Forthegeneral case, thetwosolutions E1andE11refer totwostates—which wecanagain callthestates II”) :II>e—(i/ii)EIz and Ill/11> :I”>e—(i/1‘1)E”1_ These states willhave C1and C2asgiven inEqs. (9.18) and (9.19), where a1 and112arestill tobedetermined. Their ratio isgiven byeither Eq.(9.23) or Eq.(9.24). They must alsosatisfy onemore condition. lfthesystem isknown to beinoneofthestationary states, thesum oftheprobabilities thatitwillbefound inI1)orI2)must equal one. Wemust have that lC1|2+[C212=1. (9-28)or,equivalently, I(l1I2 ‘I’Ia2I2 =1. These conditions donotuniquely specify a1anda2;they arestillundetermined 9-7 Fig. 9-2. Energy levels ofthe om monio molecule incmelectric ﬁeld.EA /2 22 /Eo+ A+11¢\ // \t\ /‘4\Eo‘l'l*8\ / I / E1,+A.-/ /,//. 1 1 : E<>< \0.5 1.0 |l5 2.0/:8 E°—A \\ I \‘\ \ \\- E0“/-L8Eo_ /A2_#2€2 </\ - \ byanarbitrary phase—in other words, byafactor likeei“. Although general solutions forthea’scanbewritten downff itisusually more convenient towork them outforeach special case. Let’s goback now toourparticular example oftheammonia molecule inan electric ﬁeld. Using thevalues forH11, H22, andH12 given in(9.14) and(9.15), wegetfortheenergies ofthetwostationary states E1=E0+\/A2 —I—11282, E11 =E11—\/A2 +1126?. (9.30) These twoenergies areplotted asafunction oftheelectric ﬁeld strength £3inFig. 9-2. When theelectric ﬁeld iszero, thetwoenergies are,ofcourse, justE0=hA. When anelectric ﬁeld isapplied, thesplitting between thetwo levels increases. Thesplitting increases atﬁrstslowly with 8,buteventually becomes proportional to8.(The curve isahyperbola.) Forenormously strong ﬁelds, theenergies arejust E1=E0+113=H11, E11=E0—I-13IH22 (9-31) Thefact thatthere isanamplitude forthenitrogen toﬂiphack ant/forth haslittle effect when thetwopositions have verydiﬂerent energies. This isaninteresting point which wewillcome back toagain later. Weareatlastready tounderstand theoperation oftheammonia maser. The idea isthefollowing. First. weﬁnd away ofseparating molecules inthe state II)from those inthestate III).I1 Then themolecules inthehigher energy state II)arepassed through acavity which hasaresonant frequency of24,000 mega- cycles. The molecules candeliver energy tothecavity—in away wewilldiscuss later—and leave thecavity inthestate Ill). Each molecule that makes such a transition willdeliver theenergy E=E1—E11tothecavity. The energy from themolecules willappear aselectrical energy inthecavity. How canweseparate thetwomolecular states? One method isasfollows. The ammonia gasisletoutofalittle jetandpassed through apair ofslits to give anarrow beam, asshown inFig. 9—3. The beam isthen setthrough a TForexample, thefollowing setisoneacceptable solution, asyoucaneasily verify: ,,1 ,,,=_i—__e_~__.l(E—H102 +H12H21]1(2 [(5-H102 +Ht2H21ll/2 1;From now onwewillwrite II)andIII)instead ofI11/1)andI\p11). You must remember that theactual states It//1)and It/qr) aretheenergy base states multiplied bytheappro- priate exponential factor. 9-8 region inwhich there isalarge transverse electric ﬁeld. Theelectrodes toproduce theﬁeld areshaped sothattheelectric ﬁeld varies rapidly across thebeam. Then thesquare oftheelectric ﬁeld 8~8willhave alarge gradient perpendicular tothe beam. Now amolecule instate II)hasanenergy which increases with 82,and therefore thispart ofthebeam willbedeﬂected toward theregion oflower 82. Amolecule instate II1)will, ontheother hand, bedeﬂected toward theregion oflarger 82,since itsenergy decreases as82increases. Incidentally, with theelectric ﬁelds which canbegenerated inthelaboratory, theenergy 118isalways much smaller than A.Insuch cases, thesquare root in Eqs.(9.30) canbeapproximated by 1 22 A<1+5 (9.32) Sotheenergy levels are,forallpractical purposes, H282 and 22 E1] =E0 -"A— ' And theenergies vary approximately linearly with 82.Theforce onthemolecules isthen 2 _L 2F-2AV8. (9.35) Many molecules have anenergy inanelectric ﬁeld which isproportional to82. Thecoefficient isthepolarizability ofthemolecule. Ammonia hasanunusually high polarizability because ofthesmall value ofAinthedenominator. Thus, ammonia molecules areunusually sensitive toanelectric ﬁeld. (What would you expect forthedielectric coeﬂicient ofNH3 gas?) I wmsea cavnvFREQUENCY w/ / \,’ \ -_...._...__/ \///\ //\/,/ \T‘\\/\//\//1-1 -Vt"""'"""'*I -upI- . D> .\\“\\\‘\\\\\\I_TT—TTH l NH3 I 1 l |II |______ INCREASING cI SLITS Fig. 9-3. The ammonia beam may beseparated byonelectric ﬁeld in which 82hosctgradient perpendicular to thebeum. Q \/ electrilc fieldé Fig. 9-4. Schematic diagram ofthe I VT -Z>I ammonia moser. 9-3Transitions inatime-dependent ﬁeld Intheammonia maser, thebeam with molecules inthestate II)andwith the energy E1issentthrough aresonant cavity, asshown inFig.9-4. Theother beam isdiscarded. Inside thecavity, there willbeatime-varying electric ﬁeld, sothe next problem wemust discuss isthebehavior ofamolecule inanelectric ﬁeld that varies with time. Wehave acompletely dilferent kind ofaproblem—one with a time-varying Hamiltonian. Since H,-,~depends upon 8,theH,-1vary with time, and wemust determine thebehavior ofthesystem inthiscircumstance. Tobegin with, wewrite down theequations tobesolved: .dC1117‘=(E0+,1a)c1 -AC2, (9.36) .dClh T2 = '_AC1 —I— (E0 ”" 9-9 Tobedeﬁnite, let’ssuppose thattheelectric ﬁeldvaries sinusoidally; thenwecan write _ I 8=280coswt=8(1(e“"’ —I—e_‘°°'). (9.37) Inactual operation thefrequency 0.1willbeverynearly equal totheresonant fre- quency ofthemolecular transition we=2A/ii, butforthetime being wewant tokeep things general, sowe’ll letithave anyvalue atall.Thebestwaytosolve ourequations istoform linear combinations ofC1andC2aswedidbefore. So weaddthetwoequations, divide bythesquare root of2,andusethedeﬁnitions ofC1andC11thatwehadinEq.(9.13). Weget 1% =(E0 —A)C]] —I—,U.8C1. You’ll note thatthisisthesame asEq.(9.9) with anextra term duetotheelectric ﬁeld. Similarly, ifwesubtract thetwoequations (9.36), weget ih%=(E11+A)C1+111.01,. (9.39) Now thequestion is,howtosolve these equations? They aremore ditﬁcult than ourearlier set,because 8depends ont;and, infact, forageneral 8(1)the solution isnotexpressible inelementary functions. However, wecangetagood approximation solongastheelectric ﬁeldissmall. First wewillwrite CI=-I,Ie—i(E0+A)t/ti =-yIe—i(E1)t/ii, (9.40)C” =-YIIe—'£(E0—A)t/it :VH8-i<E,,>1/rt Ifthere were noelectric ﬁeld, these solutions would becorrect with ‘Y1andV11 justchosen astwocomplex constants. Infact, since theprobability ofbeing in state II)istheabsolute square ofC1andtheprobability ofbeing instate III)isthe absolute square ofC11, theprobability ofbeing instate II)orinstate III)is justIY1I2 orIY11I2. Forinstance, ifthesystem were tostart originally instate I11) sothatV1waszero andIY11I2Wasone, thiscondition would goonforever. There would benochance, ifthemolecule were originally instate III), ever toget intostate II). Now theidea ofwriting ourequations intheform ofEq.(9.40) isthat if 118issmall incomparison with A,thesolutions canstillbewritten inthisway, but then Y1and ‘V11become slowly varying functions oftime—where by“slowly varying” wemean slowly incomparison with theexponential functions. That is thetrick. Weusethefact that Y1and V11vary slowly togetanapproximate solution. Wewant now tosubstitute C1from (9.40) inthedifferential equation (9.39), butwemust remember that71isalsoafunction oft.Wehave dc! —'F1/r d'Y1 -'1?1/r71-=E7 "I ‘ 'h— 1' ‘. 1dt 11e —I—1dte Thedifferential equation becomes <E1'r1 +ih%t’>2-‘”"”EI‘ =E1111»-<‘/WI’ +151111»-<"/“'1'. (9.41) Similarly, theequation ina'C11/dt becomes <E11‘Y11 +in‘%)2-"'/M11’ =12111112-<"””"’II’ +,m1@r<"’””". (9.42) Now youwillnotice thatwehave equal terms onboth sides ofeach equation. We cancel these terms, andwealso multiply theﬁrst equation bye+‘”I”" and the 9-10 second bye"""EII‘/". Remembering that (E1—E11) =2A=hwo, wehave ﬁnally, ‘ll511%}=1»8(t)e“’°‘m,(9.43) ih%=,1a(t)e'“"°‘v1. Now wehave anapparently simple pairofequations—and theyarestillexact, ofcourse. Thederivative ofonevariable isafunction oftime ;18(t)e“"9‘, multiplied bythesecond variable; thederivative ofthesecond isasimilar time function, multiplied bytheﬁrst. Although these simple equations cannot besolved ingeneral, wewillsolve them forsome special cases. Weare,forthemoment atleast, interested only inthecase ofanoscillating electric ﬁeld. Taking 8(t)asgiven inEq.(9.37), weﬁnd that theequations for 'Y1andV11become .dv - _~_ lh_?I :#g0[e't(t->-I-wQ)t_I_ e1.(w w0)t],yII’ (9.44) ih% =“80[e1'(1.1-211,): _I_e—i(w-I-w0)t].YI. Now if80issutﬁciently small, therates ofchange of'r1and‘V11arealsosmall. ThetwoV’swillnotvary much with t,especially incomparison with therapid variations duetotheexponential terms. These exponential terms have realand imaginary parts thatoscillate atthefrequency to+tooor11>—wo.Theterms with to—I—0.111oscillate very rapidly about anaverage value ofzero and, therefore, donot contribute very much ontheaverage totherateofchange of'Y.Sowecanmake a reasonably good approximation byreplacing these terms bytheir average value, namely, zero. Wewilljustleave them out,andtake asourapproximation: % =[.LSQ€_i(w_w0)t9’I[, (9.45) ih12% =/180e"(“’_“’°)t'Y1. Even theremaining terms, with exponents proportional to(w—we),willalso vary rapidly unless avisnear 0.10.Only then willtheright-hand sidevary slowly enough that anyappreciable amount willaccumulate when weintegrate the equations with respect tot.Inother words, with aweak electric ﬁeldtheonly signiﬁcant frequencies arethose near(.00. With theapproximation made ingetting Eq.(9.45), theequations canbe solved exactly, butthework isalittle elaborate, sowewon’t dothatuntil later when wetake upanother problem ofthesame type. Now we’ll justsolve them ap- proximately—or rather, we’ll ﬁndanexact solution forthecaseofperfect reso- nance, L0=wu,andanapproximate solution forfrequencies near resonance. 9-4Transitions atresonance Let’s takethecaseofperfect resonance ﬁrst. Ifwetakew=wo,theexpo- nentials areequal tooneinboth equations of(9.45), andwehavejust (1771 i_H.180 (I711 __Z‘/J.g0 W —- T 7]], ifit — ‘T 71. Ifweeliminate ﬁrst“t1andthenV11from these equations, weﬁndthateachsatisﬁes thedifferential equation ofsimple harmonic motion: dz) s2W=- 1. (9.47) Thegeneral solutions forthese equations canbemade upofsines andcosines. 9-ll Asyoucaneasily verify, thefollowing equations areasolution: 'Y1=acos t+bsin t, 711=ibcos<%>t —iasin t, where aandbareconstants tobedetermined toﬁtanyparticular physical situation. Forinstance, suppose that att=0ourmolecular system wasintheupper energy state |I),which would require——from Eq.(9.40)—that ‘Y1=landV11=0 att=0.Forthissituation wewould need a=1andb=O.The probability thatthemolecule isinthestate lI>atsome later tistheabsolute square of‘Y1,or(9.48) P1=|v1|2=@0521. (9.49) Similarly, theprobability thatthemolecule willbeinthestate |II)isgiven bythe absolute square ofV11, P,,=vi,=sinz 1. (9.50) Solong as8issmall andweareonresonance, theprobabilities aregiven bysimple oscillating functions. Theprobability tobeinstate lI)falls from onetozero and back again, while theprobability tobeinthestate lII)rises from zero tooneand back. Thetime variation ofthetwoprobabilities isshown inFig.9—5. Needless tosay, thesum ofthetwoprobabilities isalways equal toone; themolecule is always insome state! P I /\ / P ’ ‘i /I / \ / / \ / / \ / / \ / T \\g\ Z / \\\ / \ / Fig 9—5 Probabilities for the two 1’ \,| / states ofthe ammonia molecule ina I 2 t sinusoidal electric ﬁeld . A ,h tinunits of11'/2;.t€o Let’s suppose thatittakes themolecule thetime Ttogothrough thecavity. Ifwemake thecavity justlong enough sothat;.i80T/h =1r/2, then amolecule which enters instate [1)willcertainly leave itinstate lII). Ifitenters thecavity intheupper state, itwillleave thecavity inthelower state. Inother words, its energy isdecreased, andtheloss ofenergy can’t goanywhere elsebutinto the machinery which generates theﬁeld. Thedetails bywhich youcanseehow the energy ofthemolecule isfedinto theoscillations ofthecavity arenotsimple; however, wedon’t need tostudy these details, because wecanusetheprinciple ofconservation ofenergy. (Wecould study them ifwehadto,butthen wewould have todeal with thequantum mechanics oftheﬁeld inthecavity inaddition to thequantum mechanics oftheatom.) Insummary: themolecule enters thecavity, thecavity ﬁeld—oscillating at exactly theright frequency——induces transitions from theupper tothelower state, andtheenergy released isfedinto theoscillating ﬁeld. Inanoperating maser themolecules deliver enough energy tomaintain thecavity oscillations—not only providing enough power tomake upforthecavity losses buteven providing small amounts ofexcess power thatcanbedrawn from thecavity. Thus, themolecular energy isconverted intotheenergy ofanexternal electromagnetic ﬁeld. 9-12 Remember thatbefore thebeam enters thecavity, wehave touseaﬁlter which separates thebeam sothatonlytheupper state enters. Itiseasytodemon- strate thatifyouwere tostartwithmolecules inthelower state, theprocess willgo theother wayandtakeenergy outofthecavity. Ifyouputtheunﬁltered beam in, asmany molecules aretaking energy outasareputting energy in,sonothing much would happen. Inactual operation itisn’t necessary, ofcourse, tomake (/.i8OT/h) exactly 1r/2. Foranyother value (except anexact integral multiple of7r),there is some probability fortransitions from state |1)tostate III). Forother values, however, thedevice isn’t 100percent eﬁicient; many ofthemolecules which leave thecavity could have delivered some energy tothecavity butdidn’t. Inactual use,thevelocity ofallthemolecules isnotthesame; they have some kind ofMaxwell distribution. This means that theideal periods oftime for different molecules willbedifferent, anditisimpossible toget100percent efficiency forallthemolecules atonce. Inaddition, there isanother complication which is easytotakeintoaccount, butwedon’t want tobother withitatthisstage. You remember thattheelectric ﬁeld inacavity usually varies from place toplace across thecavity. Thus, asthemolecules driftacross thecavity, theelectric ﬁeldatthe molecule varies inaway that ismore complicated than thesimple sinusoidal oscillation intimethatwehave assumed. Clearly, onewould have touseamore complicated integration todotheproblem exactly, butthegeneral ideaisstillthe same. There areother ways ofmaking masers. Instead ofseparating theatoms in state II)from those instate |II)byaStern-Gerlach apparatus, onecanhave the atoms already inthecavity (asagasorasolid) andshift atoms from state III) tostate |I)bysome means. Onewayisoneusedintheso-called three-state maser. Forit,atomic systems areused which have three energy levels, asshown inFig. 9-6, with thefollowing special properties. The system will absorb radiation (say, light) offrequency hwlandgofrom thelowest energy level E11tosome high-energy level E’,andthenwillquickly emitphotons offrequency hwgandgo tothestate [I)withenergy E1.Thestate I1)hasalonglifetime soitspopulation canberaised, andtheconditions arethenappropriate formaser operation between states II)andIII).Although such adevice iscalled a“three-state” maser, the maser operation really works justasatwo-state system such aswearedescribing. Alaser (Light Ampliﬁcation byStimulated Emission ofRadiation) isjusta maser working atoptical frequencies. The“cavity” foralaser usually consists of justtwoplane mirrors between which standing waves aregenerated. 9-5Transitions offresonance Finally, wewould liketoﬁndouthow thestates vary inthecircumstance that thecavity frequency isnearly, butnotexactly, equal tomo. Wecould solve this problem exactly, butinstead oftrying todothat, we’ll take theimportant case thattheelectric ﬁeld issmall andalsotheperiod oftime Tissmall, sothat;.t80T/ft ismuch lessthan one. Then, even inthecase ofperfect resonance which wehave justworked out,theprobability ofmaking atransition issmall. Suppose thatwe start again with 71=land‘Y1;=0.During thetime Twewould expect ‘/1to remain nearly equal toone, andY1;toremain very small compared with unity. Then theproblem isvery easy. Wecancalculate Y”from thesecond equation in (9.45), taking 71equal tooneandintegrating from t=0tot=T.Weget _M80 1:ei(w—wn)7] ’y]I — “T wo This‘Y11,usedwithEq.(9.40), gives theamplitude tohavemade atransition from thestate |I)tothestate III)during thetime interval T.Theprobability P(I—>II) tomake thetransition isI“/Hlz, or T2-2 _ 1/ P(I—> 11)=l71Il2 = §] (9.52) 9-136‘ E, Fig. 9-6. The energy levels ofa "three-state" maser.‘hm, hwz E1 Mo E11 AIj i’ i;,,,<<»)/e,,,(<»<,i5/7// O __ /Itisinteresting toplotthisprobability foraﬁxed length oftime asafunction ofthefrequency ofthecavity inorder toseehow sensitive itistofrequencies near theresonant frequency wo.Weshow such aplotofP(1—+II)inFig.9-7. (The vertical scale hasbeen adjusted tobelatthepeak bydividing bythevalue ofthe probability when w=0:0.) Wehave seen acurve likethisinthediffraction theory, soyoushould already befamiliar with it.Thecurve falls rather abruptly tozero for(w—we)=21r/T andnever regains signiﬁcant sizeforlarge frequency devia- tions. lnfact, byfarthegreatest part ofthearea under thecurve lieswithin the range iTl"/T. Itispossible toshow1' thatthearea under thecurve isjust211'/T and isequal tothearea oftheshaded rectangle drawn intheﬁgure. Let’s examine theimplication ofourresults forarealmaser. Suppose that theammonia molecule isinthecavity forareasonable length oftime, sayforone millisecond. Then forf0=24,000 megacycles, wecancalculate that theprob- ability foratransition falls tozero forafrequency deviation of(f—fa)/f0 = l/f,,T, which isﬁveparts inI08. Evidently thefrequency must bevery close totoo togetasigniﬁcant transition probability. Such aneffect isthebasis ofthegreat precision that canbeobtained with “atomic” clocks. which work onthemaser principle. \A.9(w) , 'i.‘l(w°) li l\ I I'Jlw) ,‘\ I \ I l 1r/T l li/ I | ‘6/ l‘5?: '<—2'rr/T I, \\\ I /| \/’ L», wo <0 \I we T)-\ ___ / \ ___ , _' \/ _-— \.--~. Fig. 9—7. Transition probability fortheammonia Fig 9-8 Thes ectral'nt 'tSl) . . p iensiywcanbeapprox- molecule asaFunction offrequency. imated byitsvalue at0:0. 9-6Theabsorption oflight Our treatment above applies toamore general situation than theammonia maser. Wehave treated thebehavior ofamolecule under theinﬂuence ofan electric ﬁeld, whether thatﬁeld wasconﬁned inacavity ornot. Sowecould be simply shining abeam of“light“—at microwave frequencies—at themolecule andaskfortheprobability ofemission orabsorption. Ourequations apply equally well tothiscase, butlet's rewrite them interms oftheintensity oftheradiation rather than theelectric ﬁeld. Ifwedeﬁne theintensity 9tobetheaverage energy ﬂow perunit area persecond, then from Chapter 27ofVolume II,wecanwrite 9=eqc2]8 ><B1,“,:%e0c2(8 ><3),,“ =zencsi-’,. (The maximum value of8is280.) Thetransition probability now becomes: 2 -2 _ P(1-»11)=21r[Z?_—:;T1ﬁ] .<IT2 - (9.53) l‘Using theformula ffw(sinz x/x2) dx=1r. 9-14 Ordinarily thelight shining onsuch asystem isnotexactly monochromatic. Itis,therefore, interesting tosolve onemore problem—that is,tocalculate the transition probability when thelight hasintensity 5(w) perunitfrequency interval, covering abroad range which includes wo.Then, theprobability ofgoing from |I)to[II)willbecome anintegral: 2 °° -2 P(I->11)=211"?/Q) 9(w) dw. (9.54) Ingeneral, 9(0))willvarymuch more slowly withatthanthesharp resonance term. Thetwofunctions might appear asshown inFig. 9-8. Insuch cases, wecanre- place 9(0))byitsvalue §(w0) atthecenter ofthesharp resonance curve andtake itoutside oftheintegral. What remains isjusttheintegral under thecurve of Fig.9-7,which is,aswehave seen, justequal to21r/T. Wegettheresult that 2 P(I->11)=41%’ sl(w0)T. (9.55) Thisisanimportant result, because itisthegeneral theory oftheabsorption oflight byanymolecular oratomic system. Although webegan byconsidering a caseinwhich state lI)hadahigher energy thanstate lII),none ofourarguments depended onthatfact. Equation (9.55) stillholds ifthestate ll)hasalower energy thanthestate |II);thenP(I—->II)represents theprobability foratransition with theabsorption ofenergy from theincident electromagnetic wave. The absorption oflight byanyatomic system always involves theamplitude fora transition inanoscillating electric ﬁeld between twostates separated byan energy E=hwo. Foranyparticular case, itisalways worked outinjustthe waywehave done hereandgives anexpression likeEq.(9.55). We,therefore, emphasize thefollowing features ofthisresult. First, theprobability ispro- portional toT.Inother words, there isaconstant probability perunit time thattransitions willoccur. Second, thisprobability isproportional totheintensity ofthelight incident onthesystem. Finally, thetransition probability ispropor- tional to#2,where, youremember, p8deﬁned theshift inenergy duetothe electric ﬁeld8.Because ofthis,p8alsoappeared inEqs.(9.38) and(9.39) asthe coupling termthatisresponsible forthetransition between theotherwise stationary states |I)andlll). Inother words, forthesmall 8wehave been considering, p8istheso-called “perturbation term” intheHamiltonian matrix element which connects thestates |I)andIII). Inthegeneral case, wewould have thatits getsreplaced bythematrix element (II|H|I) (seeSection 5-6). InVolume I(Section 42-5) wetalked about therelations among light absorp- tion, induced emission, andspontaneous emission interms oftheEinstein A-and B-coeﬁicients. Here, wehave atlastthequantum mechanical procedure for computing these coeﬁicients. What wehave called P(1—>II)forourtwo-state ammonia molecule corresponds precisely totheabsorption coeﬁicient BM,ofthe Einstein radiation theory. Forthecomplicated ammonia molecule—which istoo difficult foranyone tocalculate—we have taken thematrix element (II|H|I)as /.i8,saying that/.Listobegotten from experiment. Forsimpler atomic systems, the um,which belongs toanyparticular transition canbecalculated from thedeﬁnition 1.i,,,,,8 =(m|H|n) =Hm, (9.56) where H,,,,, isthematrix element oftheHamiltonian which includes theeffects of aweak electric ﬁeld. Thepmcalculated inthiswayiscalled theelectric dipole matrix element. Thequantum mechanical theory oftheabsorption andemission oflightis,therefore, reduced toacalculation ofthese matrix elements forparticular atomic systems. Ourstudy ofasimple two-state system hasthusledustoanunderstanding ofthegeneral problem oftheabsorption andemission oflight. 9-15 I0 Other Two-State Systems 10-1 Thehydrogen molecular ion Inthelastchapter wediscussed some aspects oftheammonia molecule under theapproximation thatitcanbeconsidered asatwo-state system. Itis,ofcourse, notreally atwo-state system-there aremany states ofrotation, vibration, transla- tion, andsoon—but each ofthese states ofmotion must beanalyzed interms of twointernal states because oftheﬂip-flop ofthenitrogen atom. Here wearegoing toconsider other examples ofsystems which, tosome approximation orother, canbeconsidered astwo-state systems. Lots ofthings willbeapproximate because there arealways many other states, andinamore accurate analysis they would have tobetaken into account. Butineach ofourexamples wewillbeable to understand agreat deal byjustthinking about twostates. Since wewillonly bedealing with two-state systems, theHamiltonian we need willlookjust liketheoneweused inthelastchapter. When theHamiltonian isindependent oftime, weknow thatthere aretwostationary states with deﬁnite- andusually different energies. Generally, however, westart ouranalysis with a setofbase states which arenotthese stationary states, butstates which may, perhaps, have some other simple physical meaning. Then, thestationary states ofthesystem willberepresented byalinear combination ofthese base states. Forconvenience, wewillsummarize theimportant equations from Chapter 9.Lettheoriginal choice ofbase states beII)and I2).Then anystate Ii//)is represented bythelinear combination li//>=l1><1l‘l’>+ l2><2l1l/> =lI>C1 +l3>C2- (101) Theamplitudes C,’(bywhich wemean either C1orC2)satisfy thetwolinear differ- ential equations .dC,ih-5ZZ11,,-c,-, (10.2) 1' where both 1'andj take onthevalues 1and2. When theterms oftheHamiltonian H,»,-donotdepend ont,thetwostates of deﬁnite energy (thestationary states), which wecall Ii/I> :II>e—(i/MEI! and I‘!/”> :I1I>e—(i/ME!!!’ have theenergies “tTTi2i_T'_H11 +H22 IH11 H22E1~‘ff —I— (mwmciz )—I—Hi2H2i 5,,:!?'..1i_*2;f_2z _I(!?.'1.%_Hi2)2 +HIZHZI. ThetwoCsforeach ofthese states have thesame time dependence. The state vectors II)andIII)which gowith thestationary states arerelated toouroriginal base states I1)andI2)by(10.3) :l1>aI + l2>a29 l11>=l1)<1’1+ |3>¢1é- 10-110-1 10-2 10-3 10-4 10-5 10-6 10-7Thehydrogen molecular ion Nuclear forces Thehydrogen molecule Thebenzene molecule Dyes TheHamiltonian ofaspinone- halfparticle inamagnetic ﬁeld Thespinning electron ina magnetic ﬁeld , / l2>0 Fig. lO—l. Asetofbase states for two protons and anelectron.Thea’sarecomplex constants, which satisfy lallg +la2l2 =1’ Q=l , (105) 02 E1*"H11 lull”+lash=1, ﬁ=_i. 106 ab EIr—H11 (') IfH11andH22areequal—say both areequal toE0—and H12 =H21 _—A, thenE1=E0+A,E”=E0—A,andthestates II)andIII)areparticularly simple: =_1_ _ =L|I>ViI|1>|2>I. |11>X/5II1>+12>I_ (10.1) Now wewillusethese results todiscuss anumber ofinteresting examples taken from theﬁelds ofchemistry andphysics. Theﬁrstexample isthehydrogen molecular ion. Apositively ionized hydrogen molecule consists oftwoprotons withoneelectron worming itswayaround them. Ifthetwoprotons areveryfar apart, what states would weexpect forthissystem? The answer ispretty clear: Theelectron willstayclose tooneproton andform ahydrogen atom initslowest state, andtheother proton willremain alone asapositive ion. So,ifthetwo protons arefarapart, wecanvisualize onephysical state inwhich theelectron is “attached” tooneoftheprotons. There is,clearly, another state symmetric to thatoneinwhich theelectron isneartheother proton, andtheﬁrstproton isthe onethatisanion. Wewilltakethese twoasourbasestates, andwe’ll callthem I1)andI2).They aresketched inFig.10-1. Ofcourse, there arereally many states ofanelectron nearaproton, because thecombination canexistasanyone oftheexcited states ofthehydrogen atom. Wearenotinterested inthatvariety ofstates now; wewillconsider onlythesituation inwhich thehydrogen atom isin thelowest state—its ground state—and wewill,forthemoment, disregard spin oftheelectron. Wecanjustsuppose thatforallourstates theelectron hasits spin“up” along thez-axis.'I Now toremove anelectron from ahydrogen atom requires 13.6electron volts ofenergy. Solongasthetwoprotons ofthehydrogen molecular ionar'efarapart, itstillrequires about thismuch energy—which isforourpresent considerations a great dealofenergy—to gettheelectron somewhere nearthemidpoint between the protons. Soitisimpossible, classically, fortheelectron tojump from oneproton totheother. However, inquantum mechanics itispossible—though notvery likely. There issome small amplitide fortheelectron tomove from oneproton totheother. Asaﬁrstapproximation, then, each ofourbase states II)andI2) willhave theenergy E0,which isjusttheenergy ofonehydrogen atom plus one proton. Wecantake that theHamiltonian matrix elements H11 andH22 are both approximately equal toE0.Theother matrix elements H12 andH21, which aretheamplitudes fortheelectron togoback andforth, wewillagain write as—A. You seethat thisisthesame game weplayed inthelasttwochapters. Ifwe disregard thefactthat theelectron canﬂipback andforth, wehave twostates of exactly thesame energy. This energy will, however, besplit intotwoenergy levels bythepossibility oftheelectron going back andforth—the greater theprobability ofthetransition, thegreater thesplit. Sothetwoenergy levels ofthesystem are E0+AandE0—A,andthestates which have these deﬁnite energies aregiven byEqs.(10.7). ‘I’Thisissatisfactory solongasthere arenoimportant magnetic ﬁelds. Wewilldiscuss theeffects ofmagnetic ﬁelds ontheelectron later inthischapter, andthevery small effects ofspininthehydrogen atom inChapter 12. 10-2 From oursolution weseethat ifaproton andahydrogen ionareputany- where near together, theelectron willnotstay ononeoftheprotons butwillflip back andforth between thetwoprotons. Ifitstarts ononeoftheprotons, itwill oscillate back andforth between thestates I1)and I2)-giving atime-varying solution. Inorder tohave thelowest energy solution (which does notvary with time), itisnecessary tostart thesystem with equal amplitudes fortheelectron to bearound each proton. Remember, there arenottwoelectrons—we arenotsaying thatthere isanelectron around each proton. There isonly oneelectron, andit hasthesame amplitude—l/\/T inmagnitude—to beineither position. Now theamplitude Aforanelectron which isnear oneproton togettothe other onedepends ontheseparation between theprotons. Thecloser theprotons aretogether, thelarger theamplitude. You remember that wetalked inChapter 7about theamplitude foranelectron to“penetrate abarrier,” which itcould not doclassically. Wehave thesame situation here. The amplitude foranelectron togetacross decreases roughly exponentially with thedistance—for large distances. Since thetransition probability, andtherefore A,getslarger when theprotons are closer together, theseparation oftheenergy levels willalsogetlarger. Ifthesystem isinthestate II),theenergy E0+Aincreases with decreasing distance, sothese quantum mechanical effects make arepulsive force tending tokeep theprotons apart. Ontheother hand, ifthesystem isinthestate III),thetotal energy decreases iftheprotons arebrought closer together; there isanattractive force pulling the protons together. Thevariation ofthetwoenergies with thedistance between the twoprotons should beroughly asshown inFig. 10-2. Wehave, then, aquantum- mechanical explanation ofthebinding force thatholds theH;iontogether. E‘ asEH 0.3-l \ = + I E E A I I 0 0.2— \ \ o.i—\ \ \\‘E0 D’. O- DISTANCE BETWEEN -0.1- PROTONS "0.2-— EI=Eo-A Fig. lO—2. The energies ofthe two stationary Fig. lO—3. Theenergy levels oftheH;ionasa states oftheHQ‘ionasafunction ofthedistance function oftheinterproton distance D(Eh=I36ev between thetwo protons. Wehave, however, forgotten onething. Inaddition totheforce wehave just described, there isalso anelectrostatic repulsive force between thetwoprotons. When thetwoprotons arefarapart—as inFig. l0—l—the "bare" proton seesonly aneutral atom, sothere isanegligible electrostatic force. Atvery close distances, however, the“bare” proton begins toget“inside” theelectron distribution—that is,itiscloser totheproton ontheaverage than totheelectron. Sothere begins tobesome extra electrostatic energy which is,ofcourse, positive. This energy- which also varies with theseparation—should beincluded inE0. SoforE0we should take something likethebroken-line curve inFig. 10-2 which rises rapidly fordistances lessthan theradius ofahydrogen atom.We should addandsubtract theflip-flop energy Afrom thisE0.When wedothat, theenergies E1andE11will vary with theinterproton distance Dasshown inFig. 10-3. [Inthisﬁgure, we have plotted theresults ofamore detailed calculation. Theinterproton distance 10-3i I | l I 2 3 4O D(A isgiven inunits of1A(l0‘8 cm),andtheexcess energy overaproton plusahydro- genatom isgiven inunits ofthebinding energy ofthehydrogen atom—the so- called “Rydberg” energy, 13.6ev.]Weseethatthestate III)hasaminimum-en- ergy point. This willbetheequilibrium conﬁgiiration—the lowest energy condition —for theH?ion.Theenergy atthispoint islower thantheenergy ofaseparated proton andhydrogen ion,sothesystem isbound. Asingle electron actstohold thetwoprotons together. Achemist would callita“one-electron bond.” This kind ofchemical binding isalso often called “quantum mechanical resonance” (byanalogy with thetwo coupled pendulums wehave described before). Butthatreally sounds more mysterious thanitis,it’sonlya“resonance” ifyoustart outbymaking apoor choice foryour base states—as wedidalso! Ifyoupicked thestate II1),youwould have thelowest energy state—that’s all. Wecanseeinanother waywhy such astate should have alower energy than aproton andahydrogen atom. Let’s think about anelectron near twoprotons with some ﬁxed, butnottoolarge, separation. You remember thatwith asingle proton theelectron is“spread out” because oftheuncertainty principle. Itseeks abalance between having alowcoulomb potential energy andnotgetting con- ﬁned into toosmall aspace, which would make ahigh kinetic energy (because of theuncertainty relation ApAxzii).Now ifthere aretwoprotons, there ismore space where theelectron canhave alowpotential energy. Itcanspread out— lowering itskinetic energy—without increasing itspotential energy. The net result isalower energy than ahydrogen atom. Then whydoes theother state II) have ahigher energy? Notice thatthisstate isthedifference ofthestates II)and I2). Because ofthesymmetry ofI1)and I2), thedifference must have zero amplitude toﬁndtheelectron half-way between thetwoprotons. This means that theelectron issomewhat more conﬁned, which leads toalarger energy. Weshould saythat ourapproximate treatment oftheH2+ionasatwo-state system breaks down pretty badly once theprotons getasclose together asthey areattheminimum inthecurve ofFig. 10-3, andso,willnotgiveagood value fortheactual binding energy. Forsmall separations, theenergies ofthetwo “states” weimagined inFig.6-1arenotreally equal toE0;amore reﬁned quan- tummechanical treatment isneeded. Suppose weasknow what would happen ifinstead oftwoprotons, wehad twodifferent objects—as, forexample, oneproton andonelithium positive ion (both particles stillwith asingle positive charge). Insuch acase, thetwoterms H11 andH22 oftheHamiltonian would nolonger beequal; they would, infact, bequite different. Ifitshould happen that thedifference (H11 —H22) is,in absolute value, much greater than A=—H12, theattractive force getsvery weak, aswecanseeinthefollowing way. IfweputH12H21 =A2intoEqs. (10.3) weget __H11‘l' H22 H11'_ H22 4/12 _ E"“am **;>.%\l‘ +***(11-"i~1—)211 22 When H11—H22 ismuch greater than A2,thesquare root isvery nearly equal to 2/12l ii. -T(H11—Hm Thetwoenergies arethen A2 E=H Li » I 11 + (H11 — H22) A2 E=11.2-A-—---H T (H11 —H22) They arenow very nearly just theenergies H11 andH22 oftheisolated atoms, pushed apart only slightly bytheflip-ﬂop amplitude A. Theenergy difference E1—E11is 2142(H11 *H22) “l' 10-4 Theadditional separation from theﬂip-ﬂop oftheelectron isnolonger equal to 2A;itissmaller bythefactor A/(H11 ——H22),which wearenowtaking tobe much lessthan one. Also, thedependence ofE1—E11ontheseparation ofthe twonuclei ismuch smaller than fortheHQ"ion—it isalsoreduced bythefactor A/(H11 —H22). Wecannow seewhythebinding ofunsymmetric diatomic molecules isgenerally veryweak. Inourtheory oftheH3"ionwehave discovered anexplanation forthe mechanism bywhich anelectron shared bytwoprotons provides, ineffect, an attractive force between thetwoprotons which canbepresent even when the protons areatlarge distances. Theattractive force comes from thereduced energy ofthesystem duetothepossibility oftheelectron jumping from oneproton to theother. Insuch ajump thesystem changes from theconﬁguration (hydrogen atom, proton) totheconﬁguration (proton, hydrogen atom), orswitches back. Wecanwrite theprocess symbolically as (H,P)i(P,H)- Theenergy shift duetothisprocess isproportional totheamplitude Athatan electron whose energy is—WH(itsbinding energy inthehydrogen atom) can getfrom oneproton totheother. Forlarge distances Rbetween thetwoprotons, theelectrostatic potential energy oftheelectron isnearly zeroovermost ofthespace itmust gowhen it makes itsjump. Inthisspace, then, theelectron moves nearly likeafreeparticle inempty space—but with anegative energy! Wehave seen inChapter 3[Eq. (3.7)] that theamplitude foraparticle ofdeﬁnite energy togetfrom oneplace toanother adistance raway isproportional to eti/ﬁmr **-— s I‘ where pisthemomentum corresponding tothedeﬁnite energy. Inthepresent case(using thenonrelativistic formula), pisgiven by L2——W (l09)2m_ H' ' This means thatpisanimaginary number, p=ix/2mWH (theother signfortheradical gives nonsense here). Weshould expect, then, thattheamplitude AfortheH;ionwillvary as e—(\/2mWH/7i)R A~___R___ (10.10) forlarge separations Rbetween thetwoprotons. Theenergy shift duetothe electron binding isproportional toA,sothere isaforce pulling thetwoprotons together which isproportional~for large R—to thederivative of(10.10) with respect toR. Finally, tobecomplete, weshould remark thatinthetwo-proton, one-electron system there isstilloneother effect which gives adependence oftheenergy onR. Wehave neglected ituntil now because itisusually rather unimportant——the exception isjustforthose verylarge distances where theenergy oftheexchange term Ahasdecreased exponentially tovery small values. Thenew effect weare thinking ofistheelectrostatic attraction oftheproton forthehydrogen atom, which comes about inthesame wayanycharged object attracts aneutral object. Thebareproton makes anelectric ﬁeld8(varying as1/R2)attheneutral hydrogen atom. Theatom becomes polarized, taking onaninduced dipole moment [.4 proportional toS.Theenergy ofthedipole is148,which isproportional to82——or tol/R4. Sothere isaterm intheenergy ofthesystem which decreases withthe fourth power ofthedistance. (Itisacorrection toE0.) Thisenergy fallsoffwith l0—5 distance more slowly than theshift Agiven by(l0.l0); atsome large separation Ritbecomes theonlyremaining important term giving avariation ofenergy with R—and, therefore, theonlyremaining force. Note thattheelectrostatic term has thesame signforboth ofthebasestates (theforce isattractive, sotheenergy is negative) andsoalsoforthetwostationary states, whereas theelectron exchange term Agives opposite signs forthetwostationary states. 10-2 Nuclear forces Wehave seenthatthesystem ofahydrogen atom andaproton hasanenergy ofinteraction duetotheexchange ofthesingle electron which varies atlarge separations Ras e-aR ——R— , (10.11) withoz=\/2mWH/h. (One usually saysthatthere isanexchange ofa“virtual” electron when—as here—the electron hastojump across aspace where itwould have anegative energy. More speciﬁcally, a“virtual exchange" means thatthe phenomenon involves aquantum mechanical interference between anexchanged state andanonexchanged state.) Now wemight askthefollowing question: Could itbethatforces between other kinds ofparticles have ananalogous origin? What about, forexample, the nuclear force between aneutron andaproton, orbetween twoprotons‘? Inan attempt toexplain thenature ofnuclear forces, Yukawa proposed thattheforce between twonucleons isduetoasimilar exchange eﬂ‘ect—only, inthiscase, due tothevirtual exchange, notofanelectron, butofanewparticle, which hecalled a“meson.” Today, wewould identify Yukawa’s meson with the1r-meson (or “pion”) produced inhigh-energy collisions ofprotons orother particles. Let’s see,asanexample, what kind ofaforce wewould expect from theex- change ofapositive pion (1r+) ofmass mwbetween aproton andaneutron. Just asahydrogen atom H0cangointoaproton p+bygiving upanelectron e_ H°—>p++e‘, (10.12) aproton p+cangointoaneutron n°bygiving upa1r+meson: p+_>n°+wt. (10.13) Soifwehave aproton ataandaneutron atbseparated bythedistance R,the proton canbecome aneutron byemitting a1r+which isthen absorbed bythe neutron atb,turning itintoaproton. There isanenergy ofinteraction ofthe two-nucleon (plus pion) system which depends ontheamplitude Aforthepion exchange—just aswefound fortheelectron exchange intheHQ"ion. Intheprocess (10.12), theenergy oftheH0atom islessthanthatoftheproton byW11 (calculating nonrelativistically, and omitting therestenergy mcz ofthe electron), sotheelectron hasanegative kinetic energy—or imaginary momentum— asinEq.(10.9). Inthenuclear process (l0.l3), theproton andneutron have almost equal masses, sothe11-+willhave zerototalenergy. Therelation between thetotal energy Eandthemomentum pforapionofmass Wlvris 2 22 24E=pc +m,,c Since Eiszero (oratleast negligible incomparison withm,,), themomentum is again imaginary: p=imwc. Using thesame arguments wegave fortheamplitude thatabound electron would penetrate thebarrier inthespace between twoprotons, wegetforthenuclear caseanexchange amplitude Awhich should—for large R~go as e-——(m.,,c/ﬁ)11’ -T- - (10.14) 10-6 Theinteraction energy isproportional toA,andsovaries inthesame way. We getanenergy variation intheform oftheso-called Yukawa potential between twonucleons. Incidentally, weobtained thissame formula earlier directly from thedifferential equation forthemotion ofapion infreespace [seeChapter 28, Vol. ll.Eq.(28.l8)]. Wecan, following thesame lineofargument, discuss theinteraction between two protons (orbetween two neutrons) which results from theexchange ofa neutral pion (1r°). Thebasic process isnow p+—>p+-l—1r°. (10.15) Aproton canemitavirtual 1r°,butthenitremains stillaproton. Ifwehave two protons, proton No. 1canemit avirtual 1r°which isabsorbed byproton No.2. Attheend, westillhave twoprotons. This issomewhat different from theHQion. There theH0went into adiﬁ'erent condition—the proton—after emitting the electron. Now weareassuming thataproton canemit a1r°without changing its character. Such processes are,infact, observed inhigh-energy collisions. The process isanalogous tothewaythatanelectron emits aphoton andends upstill anelectron: e——>e+photon. (10.16) Wedonot“see” thephotons inside theelectrons before theyareemitted orafter they areabsorbed, andtheir emission does notchange the“nature” oftheelectron. Going back tothetwoprotons, there isaninteraction energy which arises from theamplitude Athat oneproton emits aneutral pion which travels across (with imaginary momentum) totheother proton and isabsorbed there. This amplitude isagain proportional to(10.14), with m,,themass oftheneutral pion. Allthesame arguments give anequal interaction energy fortwoneutrons. Since thenuclear forces (disregarding electrical eﬂects) between neutron and proton, between proton andproton, between neutron andneutron arethesame, wecon- clude thatthemasses ofthecharged andneutral pions should bethesame. Experi- mentally, themasses areindeed very nearly equal, andthesmall difference isabout what one would expect from electric self-energy corrections (see Chapter 28, Vol. ll). There areother kinds ofparticles—like K-mesons—which canbeexchanged between twonucleons. Itisalso possible fortwopions tobeexchanged atthe same time. Butallofthese other exchanged “objects” have arestmass m,higher than thepion mass mn,andlead toterms intheexchange amplitude which vary as e—(mIe/ﬁ)R i . These terms dieoutfaster with increasing Rthan theone-meson term. Noone knows, today, how tocalculate these higher-mass terms, butforlarge enough values ofRonly theone-pion term survives. And, indeed, those experiments which involve nuclear interactions only atlarge distances doshow that theinteraction energy isaspredicted from theone-pion exchange theory. Intheclassical theory ofelectricity andmagnetism, thecoulomb electrostatic interaction andtheradiation oflightbyanaccelerating charge areclosely related- bothcome outoftheMaxwell equations. Wehaveseeninthequantum theory that lightcanberepresented asthequantum excitations oftheharmonic oscillations of theclassical electromagnetic ﬁelds inabox. Alternatively, thequantum theory canbesetupbydescribing light interms ofparticles——photons—which obey Bose statistics. Weemphasized inSection 4-5that thetwoalternative points ofview always giveidentical predictions. Canthesecond point ofview becarried through completely toinclude allelectromagnetic effects? Inparticular, ifwewant to describe theelectromagnetic ﬁeld purely interms ofBose particles—that is,in terms ofphotons—what isthecoulomb force dueto? From the“particle” point ofview thecoulomb interaction between two electrons comes from theexchange ofavirtual photon. Oneelectron emits aphoton —asinreaction (l0.l6)—which goes over tothesecond electron, where itis absorbed inthereverse ofthesame reaction. Theinteraction energy isagain given 10-7 byaformula like(10.14), butnowwithm,,replaced bytherestmass ofthephoton —which iszero. Sothevirtual exchange ofaphoton between twoelectrons gives aninteraction energy thatvaries simply inversely asR,thedistance between the twoelectrons—just thenormal coulomb potential energy! Inthe“particle” theory ofelectromagnetism, theprocess ofavirtual photon exchange gives risetoallthe phenomena ofelectrostatics. yo ELECTRONS% b I'>/ PROTONS \ .a.1 ll 0.4m i >—<En 02* O“W _,_at a a _o_2_. —o.4___m__.l__ L__L..l____.L_.__.__O l 2 3 D(l) Fig. l0—5. The energy levels ofthe H2molecule for different interproton distances D.(Eh=13.6 ev.)10-3 Thehydrogen molecule Asournext two-state system wewilllook attheneutral hydrogen molecule H2. Itis,naturally, more complicated tounderstand because ithastwoelectrons. Again, westart bythinking ofwhat happens when thetwo protons arewell separated. Only nowwehave twoelectrons toadd. Tokeep track ofthem, we’ll calloneofthem “electron a”andtheother “electron b.”Wecanagain imagine twopossible states. Onepossibility isthat“electron a”isaround theﬁrstproton and“electron b”isaround thesecond, asshown inFig.l0—4(a). Wehave simply twohydrogen atoms. Wewillcallthisstate I1).There isalsoanother possibility: that “electron b”isaround theﬁrst proton andthat “electron a”isaround the second. Wecallthisstate I2).From thesymmetry ofthesituation, those two possibilities should beenergetically equivalent, but,aswewillsee,theenergy of thesystem isnotjusttheenergy oftwohydrogen atoms. Weshould mention that there aremany other possibilities. Forinstance, “electron a”might benear the ﬁrstproton and“electron b”might beinanother state around thesame proton. We’ll disregard such acase, since itwillcertainly have higher energy (because of thelarge coulomb repulsion between thetwoelectrons). Forgreater accuracy, we would have toinclude such states, butwecangettheessentials ofthemolecular binding byconsidering justthetwostates ofFig. 10.4. Tothisapproximation we candescribe anystate bygiving theamplitude (II¢)tobeinthestate II)andan amplitude (2I¢)tobeinstate I2).Inother words, thestate vector I¢)canbe written asthelinear combination l¢>=Ii><il¢>. Toproceed, weassume—as usual—that there issome amplitude Athat the electrons canmove through theintervening space and exchange places. This possibility ofexchange means thattheenergy ofthesystem issplit, aswehave seen forother two-state systems. Asforthehydrogen molecular ion,thesplitting is verysmall when thedistance between theprotons islarge. Astheprotons approach each other, theamplitude fortheelectrons togoback andforth increases, sothe splitting increases. Thedecrease ofthelower energy state means that there isan attractive force which pulls theatoms together. Again theenergy levels risewhen theprotons getvery close together because ofthecoulomb repulsion. The net ﬁnal result isthatthetwostationary states have energies which vary with thesep- aration asshown inFig.10-5. Ataseparation ofabout 0.74A,thelower energy 10-8 levelreaches aminimum; thisistheproton-proton distance ofthetruehydrogen molecule. Now youhave probably been thinking ofanobjection. What about thefact thatthetwoelectrons areidentical particles? Wehavebeencalling them “electron a”and“electron b,”butthere really isnowaytotellwhich iswhich. Andwehave saidinChapter 4that forelectrons—which areFermi particles—if there aretwo ways something canhappen byexchanging theelectrons, thetwoamplitudes will interfere withanegative sign. Thismeans thatifweswitch which electron iswhich, thesignoftheamplitude must reverse. Wehave justconcluded, however, that thebound state ofthehydrogen molecule would be(att=0) |11>=\if2<|1>+|2>>. However, according toourrules ofChapter 4,thisstate isnotallowed. Ifwe reverse which electron iswhich, wegetthestate l $03)-1- lll). andwegetthesame signinstead oftheopposite one. These arguments arecorrect ifbothelectrons havethesame spin. Itistruethat ifboth electrons have spinup(orboth have spin down), theonly state thatisper- mitted is 11>-\§i<|1>-12>). Forthisstate, aninterchange ofthetwoelectrons gives (I2)—l1)), which is-II),asrequired. Soifwebring twohydrogen atoms near toeach other with their electrons spinning inthesame direction, they cangointo the state II)andnotstate III). Butnotice thatstate II)istheupper energy state. Itscurve ofenergy versus separation hasnominimum. The two hydrogens will always repel andwillnotform amolecule. Soweconclude that thehydrogen molecule cannot existwithparallel electron spins. Andthatisright. Ontheother hand, ourstate III)isperfectly symmetric forthetwoelectrons. Infact, ifweinterchange which electron wecallaandwhich wecallbwegetback exactly thesame state. WesawinSection 4-7that iftwoFermi particles arein thesame state, they must have opposite spins. So,thebound hydrogen molecule must have oneelectron with spin upandonewith spin down. Thewhole story ofthehydrogen molecule isreally somewhat more compli- cated ifwewant toinclude theproton spins. Itisthen nolonger right tothink of themolecule asatwo-state system. Itshould really belooked atasaneight-state system—there arefourpossible spinarrangements foreach ofourstates II)and I2)—so wewere cutting things alittle short byneglecting thespins. Ourﬁnal conclusions are,however, correct. Weﬁndthatthelowest energy state—the onlybound state—of theH2mole- culehasthetwoelectrons withspins opposite. Thetotal spinangular momentum oftheelectrons iszero. Ontheother hand, twonearby hydrogen atoms with spins parallel—and sowith atotal angular momentum h—must beinahigher (unbound) energy state; theatoms repel each other. There isaninteresting correlation be- tween thespins andtheenergies. Itgives another illustration ofsomething we mentioned before, which isthatthere appears tobean“interaction” energy be- tween twospins because thecaseofparallel spins hasahigher energy than the opposite case. Inacertain sense youcould saythatthespins trytoreach an antiparallel condition and,indoing so,have thepotential toliberate energy—n0t because there isalarge magnetic force, butbecause oftheexclusion principle. 10-9 Fig. lO—6. The benzene molecule, c¢H6.\ /I-I C ll c cH’ s/‘HcWesawinSection 10-1thatthebinding oftwodiﬂerent ionsbyasingle elec- tron islikely tobequite weak. This isnottrueforbinding bytwoelectrons. Sup- posethetwoprotons inFig.10-4werereplaced byanytwoions(with closed inner electron shells and asingle ionic charge), and that thebinding energies ofan electron atthetwoions aredifferent. Theenergies ofstates II)andI2)would stillbeequal because ineach ofthese states wehave oneelectron bound toeach ion. Therefore, wealways have thesplitting proportional toA.Two-electron binding isubiquitous—it isthemost common valence bond. Chemical binding usually involves thisﬁip-ﬁop game played bytwoelectrons. Although twoatoms canbebound together byonly oneelectron, itisrelatively rare—because itre- quires justtheright conditions. Finally, wewant tomention thatiftheenergy ofattraction foranelectron to onenucleus ismuch greater than totheother, then what wehave saidearlier about ignoring other possible states isnolonger right. Suppose nucleus a(oritmaybe apositive ion)hasamuch stronger attraction foranelectron than does nucleus b. Itmay then happen thatthetotal energy isstillfairly loweven when both electrons areatnucleus a,andnoelectron isatnucleus b.Thestrong attraction may more than compensate forthemutual repulsion ofthetwoelectrons. Ifitdoes, the lowest energy state may have alarge amplitude toﬁndboth electrons ata(making anegative ion)andasmall amplitude toﬁndanyelectron atb.Thestate looks like anegative ionwith apositive ion. This is,infact, what happens inan“ionic” molecule likeNaCl. You canseethatallthegradations between covalent binding andionic binding arepossible. You cannow begin toseehow itisthat many ofthefacts ofchemistry can bemost clearly understood interms ofaquantum mechanical description. 10-4 Thebenzene molecule Chemists have invented nice diagrams torepresent complicated organic molecules. Now wearegoing todiscuss oneofthemost interesting ofthem—the benzene molecule shown inFig. 10-6. Itcontains sixcarbon andsixhydrogen atoms inasymmetrical array. Each barofthediagram represents apair ofelec- trons, with spins opposite, doing thecovalent bond dance. Each hydrogen atom contributes one electron and each carbon atom contributes four electrons to make upthetotal of30electrons involved. (There aretwomore electrons close to thenucleus ofthecarbon which form theﬁrst, orK,shell. These arenotshown since they aresotightly bound thatthey arenotappreciably involved inthecova- lentbinding.) Soeach barintheﬁgure represents abond, orpair ofelectrons, andthedouble bonds mean thatthere aretwopairs ofelectrons between alternate pairs ofcarbon atoms. There isamystery about thisbenzene molecule. Wecancalculate what energy should berequired toform thischemical compound, because thechemists have measured theenergies ofvarious compounds which involve pieces ofthering—for instance, they know theenergy ofadouble bond bystudying ethylene, andsoon. Wecan, therefore, calculate thetotal energy weshould expect forthebenzene H HI | C CH\ 4 \ /Br H\ / § /Br C C c c<01 | II (b) II I C C c cH’ QC/ \Br |-|/ \ y \Br C l I H H Fig. 10-7. Two possibilities oforthodibromobenzene. Thetwo bromines could beseparated byasingle bond orbycdouble bond. 10-10 molecule. Theactual energy ofthebenzene ring, however, ismuch lower thanwe getbysuch acalculation; itismore tightly bound than wewould expect from what iscalled an“unsaturated double bond system.” Usually adouble bond system which isnotinsuch aringiseasily attacked chemically because ithasarelatively high energy—the double bonds canbeeasily broken bytheaddition ofother hydrogens. Butinbenzene theringisquite permanent andhard tobreak up. Inother words, benzene hasamuch lower energy thanyouwould calculate from thebond picture. Then there isanother mystery. Suppose wereplace twoadjacent hydrogens bybromine atoms tomake ortho-dibromobenzene. There aretwoways todothis, asshown inFig. 10-7. Thebromines could beontheopposite ends ofadouble bond asshown inpart(a)oftheﬁgure, orcould beontheopposite ends ofasingle bond asin(b). One would think that ortho-dibromobenzene should have two different forms, butitdoesn’t. There isonly onesuch chemical.'I' Now wewant toresolve these mysteries——and perhaps you have already guessed how: bynoticing, ofcourse, that the“ground state” ofthebenzene ring isreally atwo-state system. Wecould imagine that thebonds inbenzene could beineither ofthetwoarrangements shown inFig. 10-8. You say,“But they are really thesame; they should have thesame energy.” Indeed, they should. And for thatreason they must beanalyzed asatwo-state system. Each state represents a different conﬁguration ofthewhole setofelectrons, andthere issome amplitude Athatthewhole bunch canswitch from onearrangement totheother—there isa chance thattheelectrons canﬂipfrom onedance totheother. Aswehave seen, thischance ofﬂipping makes amixed state whose energy is lower than youwould calculate bylooking separately ateither ofthetwopictures inFig.10-8. Instead, there aretwostationary states—one withanenergy above andonewith anenergy below theexpected value. Soactually, thetrue normal state (lowest energy) ofbenzene isneither ofthepossibilities shown inFig.10-8, butithastheamplitude 1/\/i tobeineach ofthestates shown. Itistheonly state that isinvolved inthechemistry ofbenzene atnormal temperatures. In- cidentally, theupper state alsoexists; wecantellitisthere because benzene hasa strong absorption forultraviolet light atthefrequency w=(E1—E11)/h. You willremember thatinammonia, where theobject ﬂipping back andforth wasthree protons, theenergy separation wasinthemicrowave region. Inbenzene, the objects areelectrons, andbecause theyaremuch lighter, theyﬁnditeasier toﬂip back and forth, which makes thecoeﬂicient Avery much larger. The result is that theenergy difference ismuch larger—about 1.5ev,which istheenergy of anultraviolet photon.I What happens ifwesubstitute bromine? Again thetwo“possibilities” (a) and(b)inFig. 10-7 represent thetwodifferent electron conﬁgurations. Theonly difference isthatthetwobase states westart with would have slightly different energies. Thelowest energy stationary state willstillinvolve alinear combination ofthetwostates, butwith unequal amplitudes. Theamplitude forstate II)might haveavalue something like\/Z73‘, say,whereas stateI2)would havethemagnitude I‘Weareoversimplifying alittle. Originally, thechemists thought thatthere should befour forms ofdibromobenzene: twoforms withthebromines onadjacent carbon atoms (ortho-dibromobenzene), athird form withthebromines onnext-nearest carbons (meta- dibromobenzene), andafourth form with thebromines opposite toeach other (para- dibromobenzene). However, they found only three forms—there isonly oneform of theortho-molecule. IWhat wehave saidisalittle misleading. Absorption ofultraviolet light would be veryweak inthetwo-state system wehave taken forbenzene, because thedipole moment matrix element between thetwostates iszero. [Thetwostates areelectrically symmetric, soinourformula Eq.(9.55) fortheprobability ofatransition, thedipole moment a iszeroandnolight isabsorbed.] Ifthese were theonlystates, theexistence oftheupper state would have tobeshown inother ways. Amore complete theory ofbenzene, how- ever, which begins with more basestates (such asthose having adjacent double bonds) shows thatthetruestationary states ofbenzene areslightly distorted from theones we have found. Theresulting dipole moments permit thetransition wementioned inthetext tooccur bytheabsorption ofultraviolet light. 10-11HI cH\C 4\C/H l1> l ll C /\Hc/%H cIH H|cH\ / %/H c c |2> ll I c cH/ \%\H<31H Fig. 10-8. Asetofbase states for thebenzene molecule. H2N C :§H2 |I> 11> HEN—®: c H2 |2> 12> Fig. 10-9. Two base states forthe molecule ofthedye mcigento.\/1/3. Wecan’t sayforsurewithout more information, butonce thetwoenergies H11andH22arenolonger equal, thentheamplitudes C1andC2nolonger have equal magnitudes. Thismeans, ofcourse, thatoneofthetwopossibilities inthe ﬁgure ismore likely than theother, buttheelectrons aremobile enough sothat there issome amplitude forboth. Theother state hasdifferent amplitudes (like \/% and—\/iﬁ) butliesatahigher energy. There isonlyonelowest state, nottwoasthenaive theory ofﬁxed chemical bonds would suggest. 10-5 Dyes Wewillgiveyouonemore chemical example ofthetwo-state phenomenon- thistime onalarger molecular scale. Ithastodowith thetheory ofdyes. Many dyes—in fact, most artiﬁcial dyes—have aninteresting characteristic; they have a kind ofsymmetry. Figure 10-9 shows anionofaparticular dyecalled magenta, which hasapurplish redcolor. The molecule hasthree ring structures-two of which arebenzene rings. The third isnotexactly thesame asabenzene ring because ithasonlytwodouble bonds inside thering. Theﬁgure shows twoequally satisfactory pictures, andwewould guess that they should have equal energies. Butthere isacertain amplitude thatalltheelectrons canﬂipfrom onecondition totheother, shifting theposition ofthe“unﬁlled” position totheopposite end. With somany electrons involved, theﬂipping amplitude issomewhat lower than it isinthecase ofbenzene, andthedifference inenergy between thetwostationary states issmaller. There are,nevertheless, theusual twostationary states II)andIII) which arethesum anddifference combinations ofthetwobase states shown inthe ﬁgure. Theenergy separation ofII)andIII)comes outtobeequal totheenergy ofaphoton intheoptical region. Ifoneshines light onthemolecule, there isa very strong absorption atonefrequency, anditappears tobebrightly colored. That’s why it’sadye! Another interesting feature ofsuch adyemolecule isthat inthetwo base states shown, thecenter ofelectric charge islocated atdifferent places. Asaresult, themolecule should bestrongly affected byanexternal electric ﬁeld. Wehada similar eﬁ'ect intheammonia molecule. Evidently wecananalyze itbyusing exactly thesame mathematics, provided weknow thenumbers E0andA.Gener- ally,these areobtained bygathering experimental data. Ifonemakes measure- ments withmany dyes, itisoften possible toguess what willhappen with some related dyemolecule. Because ofthelarge shift intheposition ofthecenter of electric charge thevalue ofItinformula (9.55) islarge andthematerial hasahigh probability forabsorbing light ofthecharacteristic frequency 2A/ft. Therefore, itisnotonlycolored butverystrongly so—a small amount ofsubstance absorbs alotoflight. Therateofﬂipping—and, therefore, A—is verysensitive tothecomplete struc- tureofthemolecule. Bychanging A,theenergy splitting, andwith itthecolor of thedye,canbechanged. Also, themolecules donothave tobeperfectly sym- metrical. Wehave seen thatthesame basic phenomenon exists with slight modiﬁca- tions, even ifthere issome small asymmetry present. So,onecangetsome modi- ﬁcation ofthecolors byintroducing slight asymmetries inthemolecules. For example, another important dye,malachite green, isverysimilar tomagenta, but hastwoofthehydrogens replaced byCH3. It’sadifferent color because theAis shifted andtheﬂip-ﬂop rateischanged. 10-6 TheHamiltonian ofaspinone-half particle inamagnetic ﬁeld Now wewould liketodiscuss atwo-state system involving anobject ofspin one-half. Some ofwhat wewillsayhasbeencovered inearlier chapters, butdoing itagain may help tomake some ofthepuzzling points alittle clearer. Wecan think ofanelectron atrestasatwo-state system. Although wewillbetalking in thissection about “anelectron,” what weﬁndoutwillbetrueforanyspin one-half particle. Suppose wechoose forourbasestates I1)andI2)thestates inwhich the z-component oftheelectron spinis+h/2 and—h/2. 10-12 These states are,ofcourse, thesame ones wehave called (—I—)and(—)in earlier chapters. Tokeep thenotation ofthischapter consistent, though, wecall the“plus” spinstate I1)andthe“minus” spinstate I2)—where “plus” and“minus” refer totheangular momentum inthez-direction. Any possible state titfortheelectron canbedescribed asinEq.(10.1) by giving theamplitude C1thattheelectron isinstate II),andtheamplitude C2 thatitisinstate I2).Totreat thisproblem, wewillneed toknow theHamiltonian forthistwo-state system—that is,foranelectron inamagnetic ﬁeld. Webegin withthespecial caseofamagnetic ﬁeldinthez-direction. Suppose thatthevector Bhasonlyaz-component B2.From thedeﬁnition ofthetwobase states (that is,spins parallel andantiparallel toB)weknow that they arealready stationary states with adeﬁnite energy inthemagnetic ﬁeld. State II)corresponds toanenergy'I‘ equal to—uB, andstate I2)to-I-I.¢B,. The Hamiltonian must bevery simple inthiscase since C1,theamplitude tobeinstate I1),isnotaffected byC2,andviceversa: =E1C1 =-'MBzC1, (10.17) ihifi =E2C =—l—,uB2C2. Forthisspecial case, theHamiltonian is H11 =-I431, H12 =0, H21 =0, H22 ='l'#B2~ (10-18) Soweknow what theHamiltonian isforthemagnetic ﬁeldinthez-direction, and weknow theenergies ofthestationary states. Now suppose theﬁeld isnotinthez-direction. What istheHamiltonian? How arethematrix elements changed iftheﬁeld isnotinthez-direction? We aregoing tomake anassumption that there isakind ofsuperposition principle fortheterms oftheHamiltonian. More speciﬁcally, wewant toassume that if twomagnetic ﬁelds aresuperposed, theterms intheHamiltonian simply add-if weknow theH,-jforapure B,andweknow theH,»,~forapure BI,then theH,-j forboth B,andB,together issimply thesum. This iscertainly trueifweconsider only ﬁelds inthez-direction—if wedouble B2,then alltheH,-,~aredoubled. So let's assume that Hislinear intheﬁeld B.That’s allweneed tobeable toﬁnd theH,-Iforanymagnetic ﬁeld. Suppose wehave aconstant ﬁeld B.Wecould have chosen ourz-axis inits direction, andwewould have found twostationary states with theenergies =FItB. Just choosing ouraxes inadifferent direction won’t change thephysics. Our description ofthestationary states willbedifferent, buttheir energies willstillbe ¥=;.tB—-that is, E1=-vvzsi +B5+B3 and (10.19) E11=+/A/125+ B§+B3. Therestofthegame iseasy. Wehave heretheformulas fortheenergies. Wewant aHamiltonian which islinear inBr,By,andB2,andwhich willgivethese energies when used inourgeneral formula ofEq.(10.3). Theproblem: ﬁnd the Hamiltonian. First, notice thattheenergy splitting issymmetric, with anaverage value ofzero. Looking atEq.(10.3), wecanseedirectly thatthatrequires H22=_H1 1- (Note thatthischecks with what wealready know when B,andB1,areboth zero; ‘IWearetaking therestenergy mot-2 asour“zero” ofenergy andtreating themagnetic moment Itoftheelectron asanegative number, since itpoints opposite tothespin. I0-13 inthatcaseH11 =—/.tB, andH22 =uB,.) Now ifweequate theenergies of Eq.(10.3) withwhat weknow from Eq.(10.19), wehave (5§l%5Hmm=ma+%+£> mm (Wehave alsomade useofthefactthatH21 =H’f2, sothatH12H21 canalso bewritten asIH12I2.) Again forthespecial caseofaﬁeldinthez-direction, this gives 11233-1‘ lH12l2 =I123?- Clearly, IH12I must bezero inthisspecial case, which means that H12 cannot have anyterms inB2.(Remember, wehave saidthatallterms must belinear in B2,By’andB,.) Sofar,then, wehave discovered that H11 andH22 have terms inB2,while H12andH21donot. Wecanmake asimple guess thatwillsatisfy Eq.(10.20) if wesaythat H11=_MBz, H22 =l'l'Bz> (10-21) and lH12l2 =I~¢2(B§ "l'Bl’)- Anditturns outthatthat’s theonlywayitcanbedone! “Wait”—you say—“H12 isnotlinear inB;Eq.(10.21) gives H12 = /.t\/B2 +Bf.” Notnecessarily. There isanother possibility which islinear, namely, H12 = —I— There are,infact,several such possibilities—most generally, wecould write H12 =I-‘(Br =*i311)‘-“.6, where 6issome arbitrary phase. Which sign andphase should weuse? Itturns outthat youcanchoose either sign, andanyphase youwant, andthephysical results willalways bethesame. Sothechoice isamatter ofconvention. People ahead ofushave chosen tousetheminus signandtotakeell"=-1. Wemight aswellfollow suitandwrite H12 =“P-(Ba T‘iBy)» H21 =*l"(Br ‘l'i811)- (lncidentally, these conventions arerelated to,andconsistent with, some ofthe arbitrary choices wemade inChapter 6.) Thecomplete Hamiltonian foranelectron inanarbitrary magnetic ﬁeldis, then H11 =-1432, H12 =-/~l(B.r —lBu),(10.22) H21 =_l~l'(Ba: —I—iBy), H22 =+1131- Andtheequations fortheamplitudes C1andC2are .dC .n7¥-mc+@-mm(10.23) .ac .lh-5%=-,l[(B, +1B,)c, -12,02]. Sowehave discovered the“equations ofmotion forthespinstates“ ofan electron inamagnetic ﬁeld. Weguessed atthem bymaking some physical argu- ment, butthereal testofanyHamiltonian isthat itshould give predictions in agreement with experiment. According toanytests that have been made, these equations areright. Infact,although wemade ourarguments onlyforconstant ﬁelds, theHamiltonian wehave written isalso right formagnetic ﬁelds which vary with time. Sowecannow useEq.(10.23) tolook atallkinds ofinteresting problems. 10-14 10-7 Thespinning electron inamagnetic ﬁeld Example number one:Westartwithaconstant ﬁeldinthez-direction. There arejustthetwostationary states with energies =F;.tB,. Suppose weaddasmall ﬁeldinthex-direction. Then theequations look likeouroldtwo-state problem. Wegettheﬂip-ﬂop business once more, andtheenergy levels aresplit alittle farther apart. Now let’sletthex-component oftheﬁeldvarywithtime—say, as coswt.Theequations arethen thesame aswehadwhen weputanoscillating electric ﬁeld ontheammonia molecule inChapter 9.You canwork outthede- tailsinthesame way. You willgettheresult thattheoscillating ﬁeld causes transitions from the+2-state tothe—z-state—and viceversa—when thehori- zontal ﬁeld oscillates near theresonant frequency 0:0=2aB,/h. Thisgives the quantum mechanical theory ofthemagnetic resonance phenomena wedescribed inChapter 35ofVolume II(seeAppendix). Itisalsopossible tomake amaser which usesaspinone-half system. A Stern-Gerlach apparatus isused toproduce abeam ofparticles polarized in,say, the+2-direction, which aresent into acavity inaconstant magnetic ﬁeld. The oscillating ﬁelds inthecavity cancouple with themagnetic moment andinduce transitions which giveenergy tothecavity. Now let’slook atthefollowing question. Suppose wehave amagnetic ﬁeld Bwhich points inthedirection whose polar angle is0andazimuthal angle is ¢,asinFig. 10-10. Suppose, additionally. thatthere isanelectron which hasbeen prepared with itsspin pointing along thisﬁeld. What aretheamplitudes C1and C2forsuch anelectron‘? Inother words, calling thestate oftheelectron It/1). wewant towrite lit)=l1>C1 +l3>C2,where C1andC2are C1 = C2 : where byII)andI2)wemean thesame thing weused tocallI+)andI—> (referred toourchosen z-axis). Theanswer tothisquestion isalsoinourgeneral equations fortwo-state systems. First, weknow thatsince thee1ectron’s spinisparallel toBitisina stationary state with energy E1=—uB. Therefore, both C1andC2must vary asc_lEI””', asin(9.18); andtheir coefﬁcients a1anda2aregiven by(10.5), namely. at H12—=i- 10.24172 Er-"H11 ( ) Anadditional condition isthata1anda2should benormalized sothatIa1I2 —I- Ia2I2 =1.WecantakeH11andH12from (10.22) using B,=Bcos 0, B,=Bsin0cos¢, By=Bsinﬂsin ¢. Sowehave H11 = —[.LBCOS0, _ (10.25)H12 =—/.tBsin 0(cos¢> —ism¢). Thelastfactor inthesecond equation is,incidentally, e_“, soitissimpler towrite H12=-,tBsin6e““’. (10.26) Using these matrix elements inEq.(l0.l6)—and canceling —;.1B from numer- atoranddenominator—we ﬁnd a1 sin0e_l“_=i-_- .7a2 1—cos0 (102 ) With thisratio and thenormalization condition, wecanﬁnd both a1and a2. That’s nothard, butwecanmake ashort cutwith alittle trick. Notice that 10-15X 4»Z 9 Y Fig. 10-10. The direction ofBis deﬁned bythe polar ongle 0ond the dzimuthol angle dz. l—cos9=2sin2 (6/2), and that sin6=2sin (6/2) cos(0/2). Then Eq. (10.27) isequivalent to a cos; e“l“’1 _=i-—- 10.28 (12 Sm 6 ( ) Z Soonepossible answer is 0. .9a1=cos5e_"", a2:S1115, (l0.29) since itﬁtswith (10.28) andalsomakes |“1|2+|a2|2=1- Asyou know, multiplying both a1anda2byanarbitrary phase factor doesn’t change anything. People generally prefer tomake Eqs. (10.29) more symmetric bymultiplying both bye“"/2. Sotheform usually used is 19 - .9 -. a1=cos; e‘“’/2, a2=S1115 e+"“"‘”, (10.30) andthisistheanswer toourquestion. Thenumbers a1anda2aretheamplitudes toﬁndanelectron with itsspin upordown along thez-axis when weknow that itsspin isalong theaxisat9and¢.(The amplitudes C1andC2arejusta,and a2times eTiEI””.) Now wenotice aninteresting thing. The strength Bofthemagnetic ﬁeld does notappear anywhere in(l0.30). Theresult isclearly thesame inthelimit that Bgoes tozero. This means thatwehave answered ingeneral thequestion ofhow torepresent aparticle whose spin isalong anarbitrary axis. The amplitudes of (10.30) aretheprojection amplitudes forspin one-half particles corresponding to theprojection amplitudes wegave inChapter 5[Eqs. (5.38)] forspin-one par- ticles. Wecannow ﬁndtheamplitudes forﬁltered beams ofspinone-half particles togothrough anyparticular Stern-Gerlach ﬁlter. Let|+2) represent astate with spin upalong thez-axis, and|—z> represent thespindown state. if1+2’) represents astate with spinupalong az’-axis which makes thepolar angles 0and¢with thez-axis, then inthenotation ofChapter 5,wehave (+z| +2’) =cosge_’i¢'/2, (—z |+z') =sin%e+‘4“’/2. (lO.3l) These results areequivalent towhat wefound inChapter 6,Eq.(6.36), bypurely geometrical arguments. (Soifyoudecided toskip Chapter 6,younow have the essential results anyway.) Asourﬁnal example letslook again atonewhich we’ve already mentioned a number oftimes. Suppose that weconsider thefollowing problem. Westart with anelectron whose spin isinsome given direction, then turn onamagnetic ﬁeld inthez-direction for25minutes, andthen turn itoff.What istheﬁnal state? Again let’s represent thestate bythelinear combination Itl/)=I1)C1 +I2)C2. Forthisproblem, however, thestates ofdeﬁnite energy arealso ourbase states |1)andI2).SoC1andC2only vary inphase. Weknow that 61(1)=C1(0)@"“"”” =C1(0)e+"“""",and ~ ‘ C2(t)=c2(0)@-"W/" =c2(0)@-“M”. Now initially wesaid theelectron spin wassetinagiven direction. That means that initially C1and C2aretwonumbers given byEqs. (10.30). After wewait foraperiod oftime T,thenew C1andC2arethesame twonumbers multiplied respectively bye"’“B1T/” and e_"*‘B=T/l. What state isthat? That’s easy. lt’s exactly thesame asiftheangle ¢hadbeen changed bythesubtraction of2/.tB,T/h andtheangle 6hadbeen leftunchanged. That means that attheendofthetime l0-16 T,thestate I1/1)represents anelectron lined upinadirection which differs from theoriginal direction onlybyarotation about thez-axis through theangle A¢= 2/.iB,T/h. Since thisangle isproportional toT,wecanalsosaythedirection ofthe spinprecesses attheangular velocity 2#Bz/h around thez-axis. This result we discussed several times previously inalesscomplete andrigorous manner. Now wehave obtained acomplete and accurate quantum mechanical description of theprecession ofatomic magnets. Itisinteresting thatthemathematical ideas wehave justgone over forthe spinning electron inamagnetic ﬁeld canbeapplied toanytwo-state system. That means that bymaking amathematical analogy tothespinning electron, anyproblem about two-state systems canbesolved bypure geometry. Itworks likethis. First youshift thezero ofenergy sothat(H11 +H22) isequal to zero sothatH11 =—H22. Then anytwo-state problem isformally thesame astheelectron inamagnetic ﬁeld. Allyouhave todoisidentify —;uB2 with H11 and—a(B,, —iB,,)with H12. Nomatter what thephysics isoriginally—an ammonia molecule, orwhatever—you cantranslate itinto acorresponding electron problem. Soifwecansolve theelectron problem ingeneral, wehave solved alltwo-state problems. Andwehave thegeneral solution fortheelectron! Suppose youhave some state tostartwiththathasspin“up” insome direction, andyouhave amagnetic ﬁeldBthatpoints insome other direction. Youjustrotate thespindirection around theaxisofBwith thevector angular velocity w(t)equal toaconstant times the vector B(namely to:2aB/h). AsBvaries with time, youkeep moving theaxis oftherotation tokeep itparallel with B,andkeep changing thespeed ofrotation sothatitisalways proportional tothestrength ofB.SeeFig.10-1l.Ifyoukeep doing this,youwillendupwithacertain ﬁnalorientation ofthespinaxis,andthe amplitudes C1andC2arejustgiven bytheprojections—using (l0.30)—into your coordinate frame. You see,it’sjust ageometric problem tokeep track ofwhere you endupafter alltherotating. Although it’seasytoseewhat’s involved, thisgeo- metric problem (ofﬁnding thenetresult ofarotation with avarying angular velocity vector) isnoteasy tosolve explicitly inthegeneral case. Anyway, wesee, inprinciple, thegeneral solution toanytwo-state problem. Inthenext chapter wewilllook some more into themathematical techniques forhandling theim- portant caseofaspinone-half particle——and, therefore, forhandling two-state systems ingeneral. 10-17z w(U : /i X/ / / é//1~<i\\\ \ \ \,\L Fig. lO-l l.The spin direction ofon electron inctvorying magnetic ﬁeld Bl!) precesses citthefrequency coll) about on axis porullel toB. II More Two-State Systems 11-1 ThePauli spinmatrices Wecontinue ourdiscussion oftwo-state systems. Attheendofthelast chapter wewere talking about aspinone-half particle inamagnetic ﬁeld. We described thespinstate bygiving theamplitude C1thatthez-component ofspin angular momentum is+h/2 andtheamplitude C2thatitis—h/2. Inearlier chapters wehave called these base states I—I—)andI—). Wewillnowgoback tothatnotation, although wemayoccasionally ﬁnditconvenient touseI+)or II),andI—)orI2),interchangeably. Wesawinthelastchapter thatwhen aspinone-half particle withamagnetic moment Itisinamagnetic ﬁeld B=(B,,By’B,),theamplitudes C+(=C1) andC_(= C2)areconnected bythefollowing differential equations: .a’ .at-€i=-1t[B,c+ +(B,-1B,,)c_],(11.1) ih1%:=-y.[(B;1; +iB,,)C+ -B,C_]. Inother words, theHamiltonian matrix H,-1is H11 =—P-B2, H12 =“MB: "iBu)>(11.2) H21 =—l"(B:a: -In131;), H22 =+1-132- AndEqs.(11.1) are,ofcourse, thesame as ih%=ZH,-,-c,-, (11.3)J where iandjtakeonthevalues +and—(or1and2). Thetwo-state system oftheelectron spin issoimportant thatitisvery useful tohave aneater way ofwriting things. Wewillnow make alittle mathematical digression toshow you how people usually write theequations ofatwo-state system. Itisdone thisway: First, note thateach term intheHamiltonian is proportional toitandtosome component ofB;wecanthen-—purely formally—— write that Hi;=—;.t[0"§,~B,, —I—0'71,-B1, —I—0'f,~Bz]. (11.4) There isnonewphysics here; thisequation justmeans thatthecoefﬁcients of), 0'11,-, andof-J-—there are4X3=12ofthem—can beﬁgured outsothat (11.4) isidentical with (11.2). Let’s seewhat theyhave tobe.Westart with B,.Since B,appears onlyin H11 andH22, everything willbeO.K. if Fit=1, Viz=0, or§1=0, JZ2=—l. Weoften write thematrix H,-,-asalittletable likethis: ,-_. H__=iI<H11 H12)_U H21 H22 ll-111-1 ThePauli spin matrices 11-2 Thespinmatrices asoperators 11-3 Thesolution ofthetwo-state equations 11-4 Thepolarization states ofthe photon 11-5 Theneutral K-mesoni 11-6 Generalization toN-state systems Review: Chapter 35,Vol. I,Polariza tion ‘IThis section should beomitted onthe ﬁrstreading ofthisbook. Itismore ad- vanced thanisappropriate inaﬁrstcourse Table 11-1 ThePauli spinmatrices a,= o',,= a,,= 1: /—-\/’“‘\./’“‘\/"“\Q.--O--QOr-— I—'@@v~<.@v-—-’—‘@\~_¢/\\_,/\__,/\~_»/FortheHamiltonian ofaspinone-half particle inthemagnetic ﬁeld B2,thisis thesame as jio t .H” = _/-‘B2 "l“"(Bz _1B1/)> _ _l*¢(B:c + 'l"'HBz Inthesame way, wecanwrite thecoefﬁcients of)asthematrix ]'——v of;=il(1 °)- (11.5) o-1 Working with thecoefﬁcients ofBI,wegetthattheterms of<1,have tobe ‘Tit=O, Viz=1, F31=1, “$2=0- 11:,=(0‘)- (11.6) 10 Finally, looking atBy’wegetOr,inshorthand, all=0. viz=—i, u_- u_.1721-1, 0'22—0, O1‘ .0 _ 0?;=( '>- (11.7) 10 With these three sigma matrices, Eqs. (11.2) and (11.4) areidentical. Toleave room forthesubscripts iandj,wehave shown which 0'goes with which component ofBbyputting x,y,andzassuperscripts. Usually, however, the1'andjareomitted —it’s easy toimagine they arethere—and thex,y,zarewritten assubscripts. Then Eq.(11.4) iswritten H=_1“'l:aa:Bz: +all/BU "l“¢7zBz]~ Because thesigma matrices aresoimportant—they areused allthetime bythe professionals—we have gathered them together inTable 11-1. (Anyone whois going towork inquantum physics really hastomemorize them.) They arealso called thePauli spinmatrices after thephysicist who invented them. Inthetable wehave included onemore two-by-two matrix which isneeded if wewant tobeabletotakecareofasystem which hastwospinstates ofthesame energy, orifwewant tochoose adifferent zero energy. Forsuch situations wemust addEOC+ totheﬁrstequation in(11.1) andE0C_ tothesecond equation. Wecaninclude thisinthenewnotation ifwedeﬁne theunitmatrix “l”as6,-,-, 1=5,,=<10)’ (11.9) 01 H=E05,,-,1(<1,B, +11,12,+0,3,). (11.10)andrewrite Eq.(11.8) as Usually, itisunderstood thatanyconstant likeE0isautomatically tobemultiplied bytheunitmatrix; thenonewrites simply H=E11_1i(¢,B, +11,12,+11.12,). (11.11) Onereason thespinmatrices areuseful isthatanytwo-by-two matrix atall canbewritten interms ofthem. Anymatrix youcanwrite hasfour numbers init,say, M=<11 b>_ cd ll—2 Itcanalways bewritten asalinear combination offourmatrices. Forexample, M=a(1o)+b<01>+c<o 0>+d<o o>_ oo oo to o1 There aremany ways ofdoing it,butonespecial wayistosaythatMisacertain amount of11,,plusacertain amount of111,,andsoon,likethis: M=111+1311,,+we,+511,, where the“amounts” oz,18,7,and6may, ingeneral, becomplex numbers. Since anytwo-by-two matrix canberepresented interms oftheunitmatrix andthesigma matrices, wehave allthatweeverneed foranytwo-state system. Nomatter what thetwo-state system—the ammonia molecule, themagenta dye, anything—the Hamiltonian equation canbewritten interms ofthesigmas. Although thesigmas seem tohave ageometrical signiﬁcance inthephysical situation ofanelectron inamagnetic ﬁeld, theycanalsobethought ofasjust useful matrices, which canbeused foranytwo-state problem. Forinstance, inonewayoflooking atthings aproton andaneutron canbe thought ofasthesame particle ineither oftwostates. Wesaythenucleon (proton orneutron) isatwo-state system—in thiscase, twostates withrespect toitscharge. When looked atthatway, theI1)state canrepresent theproton andtheI2) state canrepresent theneutron. People saythatthenucleon hastwo“isotopic- spin” states. Since wewillbeusing thesigma matrices asthe“arithmetic” ofthequantum mechanics oftwo-state systems, let’sreview quickly theconventions ofmatrix algebra. Bythe“sum” ofanytwoormore matrices wemean justwhat wasobvious inEq.(11.4). Ingeneral, ifwe“add” twomatrices AandB,the“sum” Cmeans thateach term C,»,-isgiven by Cij=A,-j—I—Bi,-. Each term ofCisthesumoftheterms inthesame slots ofAandB. InSection 5-6wehave already encountered theideaofamatrix “product.” Thesame ideawillbeuseful indealing withthesigma matrices. Ingeneral, the “product” oftwomatrices AandB(inthatorder) isdeﬁned tobeamatrix C whose elements are c,~,~=ZA,-1,B;,,-. (11.12)lc Itisthesumofproducts ofterms taken inpairs from theithrowofAandthekth column ofB.Ifthematrices arewritten outintabular form asinFig.ll-1, there isagood “system” forgetting theterms oftheproduct matrix. Suppose youare calculating C23. Yourunyour leftindex ﬁnger along thesecond rowofAandyour right index ﬁnger down thethird column ofB,multiplying each pairandadding asyougo.Wehave tried toindicate howtodoitintheﬁgure. A11 A12 A12 A111 31.1 \ \ \\ \'\\ \\ \\\ \\ , A31 A52 Ass A511 B51 A111 A112 “us Amt B111 Exuple‘ C25'A21B15*A22B25*A25B351'A211BusB12 B22 B112\s \na23\ \}\\52 ‘.93\ C =ZAik J311+ C11 B211 C21 B311 C51 5\§§§3+5\But C111 13Bjk Fig. ll—l. Multiplying two matrices. 11-3C13 \\ \ ea\ \ C33 “usC111 C211 ‘B11 cut Table 11-2 Products ofthespinmatrices of=1 03=1 0f=1 0,0,, =-—0,,0, =1'0, 01,0, =—0,0,, =i0, 0,0, =—0,0, =i01,Itis,ofcourse, particularly simple fortwo-by-two matrices. Forinstance, ifwemultiply 0,,times 01,,weget 2 <01)<01)<10)ﬁx = 0'1 ~ax = ~ 2 7 10 l0 0l which isjusttheunitmatrix l.Or,foranother example, let’s work out0,01,: -<2III?-11-<1.11>)Referring toTable ll-1, youseethat theproduct isjust itimes thematrix 0,. (Remember thatanumber times amatrix justmultiplies each term ofthematrix.) Since theproducts ofthesigmas taken twoatatime areimportant—as well as rather amusing—we have listed them allinTable ll-2. You canwork them outas wehave done for0fand0,01,. There’s another veryimportant andinteresting point about these 0matrices. Wecanimagine, ifwewish, thatthethree matrices 0,,01,,and0,areanalogous to thethree components ofavector—it issometimes called the“sigma vector” and iswritten 0.Itisreally a“matrix vector” ora“vector matrix.” Itisthree different matrices-one matrix associated with each axis, x,y,andz.With it,wecanwrite theHamiltonian ofthesystem inanice form which works inanycoordinate system: H=-110-B. (11.13) Although wehave written ourthree matrices intherepresentation inwhich “up” and“down” areinthez-direction—so that0,hasaparticular simplicity- wecould ﬁgure outwhat thematrices would look likeinsome other representation. Although ittakes alotofalgebra, youcanshow thattheychange among themselves likethecomponents ofavector. (Wewon’t, however, worry about proving it right now. Youcancheck itifyouwant.) Youcanuse0indifferent coordinate systems asthough itisavector. Youremember thattheHisrelated toenergy inquantum mechanics. Itis, infact,justequal totheenergy inthesimple situation where there isonly onestate. Even fortwo-state systems oftheelectron spin, when wewrite theHamiltonian asinEq.(11.13), itlooks very much liketheclassical formula fortheenergy ofa littlemagnet withmagnetic momenta inamagnetic ﬁeld. BClassically, wewould say U=—pi-B, (11.14) where ptistheproperty oftheobject andBisanexternal ﬁeld. Wecanimagine thatEq.(11.14) canbeconverted to(11.13) ifwereplace theclassical energy by theHamiltonian andtheclassical p.bythematrix ,u.0. Then, after thispurely formal substitution, weinterpret theresult asamatrix equation. Itissometimes saidthattoeach quantity inclassical physics there corresponds amatrix inquantum mechanics. Itisreally more correct tosaythat theHamiltonian matrix corre- sponds totheenergy, andanyquantity thatcanbedeﬁned viaenergy hasacorre- sponding matrix. Forexample, themagnetic moment canbedeﬁned viaenergy bysaying that theenergy inanexternal ﬁeld Bis—]L-B.This deﬁnes themagnetic moment vector a.Then welook attheformula fortheHamiltonian ofareal(quantum) object inamagnetic ﬁeld andtrytoidentify whatever thematrices arethatcorre- spond tothevarious quantities intheclassical formula. That’s thetrick bywhich sometimes classical quantities have their quantum counterparts. Youmaytry,ifyouwant, tounderstand howaclassical vector isequal toa matrix 1.10,andmaybe youwilldiscover something—-but don’t break your head onit.That’s nottheidea—-they arenotequal. Quantum mechanics isadifferent kind ofatheory torepresent theworld. Itjusthappens thatthere arecertain correspondences which arehardly more than mnemonic devices—things tore- member with. That is,youremember Eq.(11.14) when youlearn classical physics; 11-4 thenifyouremember thecorrespondence p.—>/10',youhave ahandle forre- membering Eq.(11.13). Ofcourse, nature knows thequantum mechanics, and theclassical mechanics isonly anapproximation; sothere isnomystery inthe factthatinclassical mechanics there issome shadow ofquantum mechanical laws— which aretruly theonesunderneath. Toreconstruct theoriginal object from the shadow isnotpossible inanydirect way, buttheshadow does helpyoutore- member what theobject looks like. Equation (11.13) isthetruth, andEq.(11.14) istheshadow. Because welearn classical mechanics ﬁrst, wewould liketobe abletogetthequantum formula from it,butthere isnosure-ﬁre scheme for doing that. Wemust always goback totherealworld anddiscover thecorrect quantum mechanical equations. When theycome outlooking likesomething in classical physics, weareinluck. lfthewarnings above seem repetitious andappear toyoutobebelaboring self-evident truths about therelation ofclassical physics toquantum physics, please excuse theconditioned reﬂexes ofaprofessor who hasusually taught quantum mechanics tostudents whohadn’t heard about Pauli spinmatrices until theywereingraduate school. Then theyalways seemed tobehoping that,somehow, quantum mechanics could beseentofollow asalogical consequence ofclassical mechanics which they hadlearned thoroughly years before. (Perhaps they wanted toavoid having tolearn something new.) Youhave learned theclassical formula, Eq.(11.14), only afewmonths ago—and then with warnings thatitwasinade- quateiso maybe youwillnotbesounwilling totake thequantum formula, Eq.(11.13), asthebasic truth. ll—2 Thespinmatrices asoperators While weareonthesubject ofmathematical notation, wewould liketode- scribe stillanother wayofwriting things——a waywhich isused veryoften because itissocompact. Itfollows directly from thenotation introduced inChapter 8. lfwehave asystem inastate It//(1)), which varies with time, wecan—as we didinEq.(8.3l)—write theamplitude thatthesystem would beinthestate Ii) att—I—Aras (1110+A1»=Z<i|U(1.r+ Ar)I1><1'1¢<1>>_1' The matrix element (iIU(t,t +At)Ij)istheamplitude that thebase state Ij) willbeconverted intothebasestate Ii)inthetimeinterval At.Wethendeﬁned H,~,-bywriting <zIz/(1,;+At)|,->=5,-,~-filer,-,-(1) At, andweshowed thattheamplitudes C,-(1) =(iI¢(t)) were related bythediffer- ential equations .dc,-lhW=ZH,-,-c,-. (11.15) i Ifwewrite outtheamplitudes C,explicitly, thesame equation appears as .d. .#1E<11¢>=H1,-<1l¢>. (11-16> Now thematrix elements Hfjarealsoamplitudes which wecanwrite as(iIHIj); ourdillerential equation looks likethis: ih%<i11> =Z<i1H11></"1¢>- (11.11) Weseethat -1’/h (iIHIj)istheamplitude that—under thephysical conditions described byH-—a state Ij)will, during thetime dt,“generate” thestate Ii). (Allofthisisimplicit inthediscussion ofSection 8-4.) 11-5 Now following theideas ofSection 8-2,wecandrop outthecommon term (iIinEq.(1l.l7)—since itistrueforanystateIi)—and write thatequation simply as 115’;11>=ZH11"><111>. (11.18) Or,going onestepfurther, wecanalsoremove thejandwrite ihg;I1//) =HI¢). (11.19) InChapter 8wepointed outthat when things arewritten this way, theHin HIj)orHI1//)iscalled anoperator. From now onwewillputthelittle hat (A)overanoperator toremind youthatitisanoperator andnotjustanumber. Wewillwrite HI1!/). Although thetwoequations (11.18) and(11.19) mean exactly thesame thing asEq.(11.17) orEq.(11.15), wecanthink about them ina different way. Forinstance, wewould describe Eq.(11.18) inthisway: “The timederivative ofthestate vector I1,1)isequal towhat yougetbyoperating with theHamiltonian operator Honeach base state, multiplying bytheamplitude (jI1//)that11/isinthestatej, andsumming overa11j.” OrEq.(11.19) isdescribed thisway. “The time derivative (times ih)ofastate Iil/)isequal towhat youget ifyouoperate withtheHamiltonian Honthestate vector I1//).” It’sjustashort- hand wayofsaying what isinEq.(11.17), but,asyouwillsee,itcanbeagreat convenience. Ifwewish, wecancarry the“abstraction” ideaonemore step. Equation (11.19) istrueforanystate I1!/).Alsotheleft-hand side,ihd/dt, isalsoanoperator —it’s theoperation “differentiate bytandmultiply byih.”Therefore, Eq.(11.19) canalsobethought ofasanequation between operators——the operator equation .dlhJt—ﬁ. TheHamiltonian operator (within aconstant) produces thesame result asdoes d/dt when acting onanystate. Remember that thisequation aswell asEq. (11.19)—is notastatement thattheHoperator isjusttheidentical operation as d/dt. Theequations arethedynamical lawofnature—the lawofm0tion—for a quantum system. Just togetsome practice with these ideas, wewillshow youanother way we could gettoEq.(11.18). You know that wecanwrite anystate I11/)interms of itsprojections intosome base set[seeEq.(8.8)], 11>=Z11><1|1>_ <11-20> How does I1,//)change withtime? Well, justtakeitsderivative: $1-1>= 11><111>. (11.21) Now thebasestates Ii)donotchange withtime(atleastwearealways taking them asdeﬁnite ﬁxed states), buttheamplitudes (iI1/»)arenumbers which may vary. SoEq.(11.21) becomes %|1/>=Zij|1>%<i11>_ <11-22> Since weknow d(iI41)/dt from Eq.(11.16), weget ,%111>= -f;1i>;H.~.~<1|1> =-%Z11><11H11><111> =—gZH11><111>. ThisisEq.(11.18) alloveragain. 11-6 Sowehave many ways oflooking attheHamiltonian. Wecanthink ofthe setofcoefﬁcients H,-jasjustla bunch ofnumbers, orwecanthink ofthe“ampli- tudes” (iIHIj), orwecanthink ofthe“matrix” H,-J-, orwecanthink ofthe operator” H.Itallmeans thesame thing. Now let’s goback toourtwo-state systems. Ifwewrite theHamiltonian in terms ofthesigma matrices (with suitable numerical coefﬁcients like B,,etc.), wecanclearly also think ofof,asanamplitude (iIa,Ij)or,forshort, asthe operator 6,.Ifweusetheoperator idea, wecanwrite theequation ofmotion ofa state I1//)inamagnetic ﬁeldas 11$1,11>=-111.11. +B16.+B.<1.>11>. <11-21> When wewant to“use” such anequation wewillnormally have toexpress I1/») interms ofbase vectors (just aswehave toﬁndthecomponents ofspace vectors when wewant speciﬁc numbers). Sowewillusually want toputEq.(11.23) in thesomewhat expanded form: 1'1§,11>=-1;(11.11.+Rm+11.1.)11><111>_ (11.24) Now youwillseewhytheoperator ideaissoneat. TouseEq.(11.24) we needtoknow what happens when the1?operators work oneachofthebasestates. Let’s ﬁndout. Suppose wehave 62I—I—);itissome vector I‘2).butwhat? Well, let'smultiply itontheleftby(+I;wehave (-I-I5'zI'I'> =<T11 =1 (using Table ll-1). Soweknow that (—I—I7)=l. (11.25) Now let’s multiply 6'2I+)ontheleftby(—I.Weget <-1azl+> =1121- 0;SO <-|?>=0. (11.26) There isonly onestate vector thatsatisﬁes both (11.25) and(11.26); itisI+). Wediscover then that 6'zI_+->1 I—I—). (11.27) Bythiskind ofargument youcaneasily show thatalloftheproperties ofthesigma matrices canbedescribed intheoperator notation bythesetofrules given in Table 11-3. Ifwehave products ofsigma matrices, they goover intoproducts ofoperators. When twooperators appear together asaproduct, youcarry outﬁrsttheoperation with theoperator which isfarthest totheright. Forinstance, bymay I+)we aretounderstand 6-1;(6-H I—l—)). From Table 11-3, weget6,,I+)=iI—),so 173511I-1-)=<T1(iI —))- (11-23) Now anynumber—like i—-just moves through anoperator (operators work only onstate vectors); soEq.(11.28) isthesame as 5'19:/I‘l*> =i511I—> =iI'l'>- Ifyoudothesame thing for6,6,, I—),youwillﬁndthat <?ﬁ1l—> =-1‘!—)- Looking atTable ll-3, you seethat 6,6” operating onI—I—)orI—)gives just what yougetifyouoperate with 6‘,andmultiply by—i. Wecan, therefore, say 11-7Properties ofthe6-operatorTable 11-3 <tzI+> 0'2I_> a1|+) 0'1I_> 111,]-1-) 0'11I_>I+> —I—) |—> |+) I'I—> —i|+ thattheoperation 6,63,isidentical withtheoperation 1a,,andwrite thisstatement asanoperator equation: 6,6,, =121,. (11.29) Notice thatthisequation isidentical with oneofourmatrix equations ofTable 11-2. Soagain weseethecorrespondence between thematrix andoperator points ofview. Each oftheequations inTable ll-2 can,therefore, alsobeconsidered asequations about thesigma operators. You cancheck thatthey doindeed follow from Table 11-3. Itisbest, when working with these things, nottokeep track ofwhether aquantity like0orHisanoperator oramatrix. Alltheequations arethesame either way,soTable 11-2isforsigma operators, orforsigma matrices, asyouwish. ll-3 Thesolution ofthetwo-state equations Wecannow write ourtwo-state equation invarious forms, forexample, either as .dC-lhTil =Z H,-,-C; or ’ (11.30) ih‘I?=H11). They both mean thesame thing. Foraspinone-half particle inamagnetic ﬁeld, theHamiltonian Hisgiven byEq.(11.8) orbyEq.(11.13). Iftheﬁeldisinthez-direction, then—as wehave seenseveral times bynow- thesolution isthatthestate I1//),whatever itis,precesses around thez-axis (just asifyouwere totakethephysical object androtate itbodily around thez-axis) atanangular velocity equal totwice themagnetic ﬁeldtimes 1.1/h. Thesame is true, ofcourse, foramagnetic ﬁeldalong anyother direction, because thephysics isindependent ofthecoordinate system. Ifwehave asituation where themagnetic ﬁeldvaries from timetotimeinacomplicated way, thenwecananalyze thesitua- tion inthefollowing way. Suppose youstart with thespin inthe+2-direction andyouhave anx-magnetic ﬁeld. Thespin starts toturn. Then ifthex—ﬁeld is turned off,thespinstops turning. Now ifaz-ﬁeld isturned on,thespinprecesses about z,andsoon. Sodepending onhow theﬁelds vary intime, youcanﬁgure outwhat theﬁnal state is—along what axisitwillpoint. Then youcanrefer that state back totheoriginal I-I-)andI—)with respect tozbyusing theprojection formulas wehad inChapter 10(orChapter 6). lfthestate ends upwith its spininthedirection (1-9,4»), itwillhave anup-amplitude cos(0/2)e_"°’/2 anda down-amplitude sin(9/2)e+‘°‘/ 2.That solves anyproblem. Itisaword description ofthesolution ofthedifferential equations. Thesolution justdescribed issufficiently general totake care ofanytwo-state system. Let’s take ourexample oftheammonia molecule—including theeffects of anelectric ﬁeld. Ifwedescribe thesystem interms ofthestates II)andIII),the equations look likethis: 1'7115% =+/1C1 -I"115C111 (11.31) ifi‘-1% =—/{C11 —I—,U.SC1. Yousay,“No, Iremember there wasanE0inthere.” Well, wehave shifted the origin ofenergy tomake theE0zero. (You canalways dothatbychanging both amplitudes bythesame factor—e‘E°T/ﬁ—and getridofanyconstant energy.) Now ifcorresponding equations always have thesame solutions, then wereally don’t have todoittwice. Ifwelook atthese equations andlook atEq.(11.1), thenwecanmake thefollowing identiﬁcation. Let’s callII)thestate I+)and III)thestate I—).That does notmean thatwearelining-up theammonia inspace, orthatI+)andI—)hasanything todowith thez-axis. Itispurely artiﬁcial. 11-8 Wehave anartiﬁcial space thatwemight “cal1 theammonia molecule repre- sentative space,” orsomething—a three-dimensional “diagram” inwhich being “up” corresponds tohaving themolecule inthestate II)and being “down” along thisfalse z-axis represents having amolecule inthestate III). Then, the equations willbeidentiﬁed asfollows. First ofall,youseethattheHamiltonian canbewritten interms ofthesigma matrices as H=+Ao', +1.1817,. (11.32) Or,putting itanother way, ;.1B,inEq.(11.1) corresponds to—AinEq.(11.32), and;.tB,corresponds to—;.18. Inour“model” space, then, wehave aconstant B ﬁeld along thez-direction. Ifwehave anelectric ﬁeld 8which ischanging with time, then wehave aBﬁeld along thex-direction which varies inproportion. S0thebehavior ofanelectron inamagnetic ﬁeld with aconstant component inthe z-direction andanoscillating component inthex-direction ismathematically analo- gous andcorresponds exactly tothebehavior ofanammonia molecule inanoscillating electric ﬁeld. Unfortunately, wedonothave thetime togoanyfurther into the details ofthiscorrespondence, ortowork outanyofthetechnical details. We onlywished tomake thepoint thatallsystems oftwostates canbemade analogous toaspin one-half object precessing inamagnetic ﬁeld. 11-4 Thepolarization states ofthephoton There areanumber ofother two-state systems which areinteresting tostudy, andtheﬁrstnewonewewould liketotalkabout isthephoton. Todescribe a photon wemust ﬁrstgiveitsvector momentum. Forafreephoton, thefrequency isdetermined bythemomentum, sowedon’t have tosayalsowhat thefrequency is.After that, though, westillhave aproperty called thepolarization. Imagine thatthere isaphoton coming atyouwith adeﬁnite monochromatic frequency (which willbekeptthesame throughout allthisdiscussion sothatwedon’t have avariety ofmomentum states). Then there aretwodirections ofpolarization. Intheclassical theory, light canbedescribed ashaving anelectric ﬁeld which oscillates horizontally oranelectric ﬁeldwhich oscillates vertically (forinstance); these twokinds oflightarecalled x-polarized andy-polarized light. Thelight can alsobepolarized insome other direction, which canbemade upfrom thesuper- position ofaﬁeldinthex-direction andoneinthey-direction. Orifyoutake thex-andthey-components outofphase by90°,yougetanelectric ﬁeld that rotates—the light iselliptically polarized. (This isjustaquick reminder ofthe classical theory ofpolarized light thatwestudied inChapter 35,Vol.I.) Now, however, suppose wehaveasingle photon—just one. There isnoelectric ﬁeldthatwecandiscuss inthesame way. Allwehaveisonephoton. Butaphoton hastohave theanalog oftheclassical phenomena ofpolarization. There must be atleast twodifferent kinds ofphotons. Atﬁrst,youmight think there should be aninﬁnite variety-—after all,theelectric vector canpoint inallsorts ofdirections. Wecan,however, describe thepolarization ofaphoton asatwo-state system. Aphoton canbeinthestate Ix)orinthestate Iy).ByIx)wemean thepolariza- tionstate ofeach oneofthephotons inabeam oflight which classically isx-polar- izedlight. Ontheother hand, byIy)wemean thepolarization state ofeachofthe photons inay-polarized beam. AndwecantakeIx)andIy)asourbasestates ofaphoton ofgiven momentum pointing atyou—in hatwewillcallthez-direc- tion. Sothere aretwobasestates Ix)andIy),andHueyareallthatareneeded todescribe anyphoton atall. Forexample, ifwehave apiece ofpolaroid setwithitsaxistopasslightpolar- izedinwhat wecallthex-direction, andwesendinaphoton which weknow isin thestate Iy),itwillbeabsorbed bythepolaroid. Ifwesendinaphoton which we know isinthestate Ix),itwillcome right through asIx).Ifwetakeapiece of calcite which takes abeam ofpolarized lightandsplits itintoanIx)beam anda Iy)beam, thatpiece ofcalcite isthecomplete analog ofaStern-Gerlach apparatus which splits abeam ofsilver atoms intothetwostates I+)andI—). Soevery- 11-9 Fig.ll—2. Coordinates at right angles tothe momentum vector ofthe photon.Y’ Y Fig. l1-3. Two sheets ofpolaroid with angle 0between planes ofpolariza- tion.thing wedidbefore with particles andStern-Gerlach apparatuses, wecando again with light andpieces ofcalcite. And what about light ﬁltered through a piece ofpolaroid setatanangle 9?Isthatanother state? Yes,indeed, itisanother state. Let’s calltheaxisofthepolaroid x’todistinguish itfrom theaxesofour base states. SeeFig.11-2. Aphoton thatcomes outwillbeinthestate Ix’). However, anystatecanberepresented asalinear ‘combination ofbasestates, and theformula forthecombination is,here, Ix’)=cos6IIx) +sin0Iy). (11.33) That is,ifaphoton comes through apiece ofpolaroid setattheangle 0(with respect tox),itcanstillberesolved into Ix)andIy)beams—by apiece ofcalcite, forexample. Oryoucan, ifyouwish, justanalyze itintox-andy-components in your imagination. Either way, youwillﬁndtheamplitude cos6tobeintheIx) state andtheamplitude sinBtobeintheIy)state. Now weaskthisquestion: Suppose aphoton ispolarized inthex’-direction byapiece ofpolaroid setattheangle Bandarrives atapolaroid attheangle zero— asinFig.11-3; what willhappen? With what probability willitgetthrough? Theanswer isthefollowing. After itgetsthrough theﬁrstpolaroid, itisdeﬁnitely inthestate Ix’).Thesecond polaroid willletthephoton through ifitisinthe state Ix)(butabsorb itifitisthestate Iy)).Soweareasking withwhat probability does thephoton appear tobeinthestate Ix)?Wegetthatp(obabi1ity from the absolute square ofamplitude (xIx’)thataphoton inthestate Ix’)isalsoin thestate Ix).What is(xIx’)? Justmultiply Eq.(11.33) by(xItoget (xIx’) =cos6(xIx)—l—sin0(xIy). Now (XIY)=0,from thephysics—as they must beifIx)andIy)arebase states —and (xIx)=1.Soweget (xIx’)=cos0, andtheprobability iscosz0.Forexample, iftheﬁrstpolaroid issetat30°,a photon willgetthrough 3/4ofthetime, and1/4ofthetime itwillheatthepolaroid bybeing absorbed therein. Y AXIS OFPOLARIZER I L|GHT //{X I /1/ix Z STATE Ix’) Now letusseewhat happens classically inthesame situation. Wewould have abeam oflight with anelectric ﬁeld which isvarying insome wayoranother—say “unpolarized.” After itgetsthrough theﬁrstpolaroid, theelectric ﬁeldisoscillat- inginthex’-direction with asize8;wewould draw theﬁeld asanoscillating vector with apeak value 8,,inadiagram likeFig. 11-4. Now when thelight arrives atthesecond polaroid, only thex-component, 80cos0,oftheelectric ﬁeld gets through. The intensity isproportional tothesquare oftheﬁeld and, therefore, toS3cos2 B.Sotheenergy coming through iscosz 0weaker than the energy which wasentering thelastpolaroid. 11-10 Theclassical picture andthequantum picture givesimilar results. Ifyou were tothrow 10billion photons atthesecond polaroid, andtheaverage prob- ability ofeach onegoing through is,say,3/4,youwould expect 3/4of10billion would getthrough. Likewise, theenergy thatthey would carry would be3/4 oftheenergy thatyouattempted toputthrough. Theclassical theory saysnothing about thestatistics ofthething—it simply saysthattheenergy thatcomes through willbeprecisely 3/4oftheenergy which youwere sending in.That is,ofcourse, impossible ifthere isonlyonephoton. There isnosuchthing as3/4ofaphoton. Itiseither allthere, oritisn’tthere atall.Quantum mechanics tellsusitisall there 3/4ofthetime. Therelation ofthetwotheories isclear. What about theother kinds ofpolarization? Forexample, right-hand circular polarization? Intheclassical theory, right-hand circular polarization hasequal components inxandywhich are90°outofphase. Inthequantum theory, aright-hand circularly polarized (RHC) photon hasequal amplitudes to bepolarized Ix)orIy),andtheamplitudes are90°outofphase. Calling aRHC photon astate IR)andaLHC photon astate IL),wecanwrite (seeVol.I,Section 33-1) 1o=iﬂm+um.‘/2 (11.34) 11>--é<1x>-111>>. —the 1/\/2 isputintogetnormalized states. With these states youcancalculate anyﬁltering orinterference effects youwant, using thelaws ofquantum theory. Ifyouwant, youcanalsochoose IR)andIL)asbasestates andrepresent every- thing interms ofthem. Youonlyneed toshow ﬁrstthat(RIL)=O-—which you candobytaking theconjugate form oftheﬁrstequation above [seeEq.(8.l3)] and multiplying itbytheother. Youcanresolve light intox-andy-polarizations, or intox’-andy’-polarizations, orintoright andleftpolarizations asabasis. Justasanexample, let’strytoturnourformulas around. Canwerepresent thestate Ix)asalinear combination ofright andleft? Yes,hereitis: 11>=#<1R>+ 11>).\/5‘ (11.35) l Iy>=~ (IR)—lL))- Proof: Addandsubtract thetwoequations in(11.34). Itiseasytogofrom onebasetotheother. Onecurious point hastobemade, though. Ifaphoton isright circularly polarized, itshouldn’t have anything todowith thex-andy-axes. Ifwewere tolook atthesame thing from acoordinate system turned atsome angle about thedirection ofﬂight, thelightwould stillberight circularly polarized—and simi- larlyforleft.Theright andleftcircularly polarized lightarethesame foranysuch rotation; thedeﬁnition isindependent ofanychoice ofthex-direction (except thatthephoton direction isgiven). Isn’t thatnice—it doesn’t takeanyaxesto deﬁne it.Much better than xandy.Ontheother hand, isn’titrather amiracle thatwhen youaddtheright andlefttogether youcanﬁndoutwhich direction x was? If“right” and“1eft” donotdepend onxinanyway, howisitthatwecan putthem back together again andgetx?Wecananswer thatquestion inpart bywriting outthestate IR’),which represents aphoton RHC polarized inthe frame x’,y’.Inthatframe, youwould write 1R'>=\%<1><'>+11y'>>. ll-llY 3*so 6°Cos6+] x Fig. ll~4. The clossicol picture of theelectric vector 8. How does suchastate lookintheframe x,y?Justsubstitute x’from Eq.(l1. 33) andthecorresponding Iy’)—we didn’t write itdown, butitis(—sin 0)Ix)—I- (cos0)Iy).Then IR’) =%[cos0Ix)+sin19Iy)— isin0Ix)+icos0Iy)] =-/l—i[(cos0— isin0)Ix)+ i(cos6—isin0)Iy)] =\%(Ix)+1|y))(¢<>se -181111). Theﬁrstterm isjustIR),andthesecond ise+“; ourresult isthat IR’)=11-"|R). (11.36) Thestates IR’)andIR)arethesame except forthephase factor e"". Ifyouwork outthesame thing forIL’),yougetthatI' IL’)=er“IL). (11.37) Now youseewhat happens. IfweaddIR)andIL),wegetsomething different from what wegetwhen weaddIR’)andIL’). Forinstance, anx-polarized photon is[Eq. (11.35)] thesum ofIR)andIL),butay-polarized photon isthesum with thephase ofoneshifted 90°backward andtheother 90°forward. That isjust what wewould getfrom thesum ofIR’)andIL’)forthespecial angle 0=90°, andthat’s right. Anx-polarization intheprime frame isthesame asay-polariza- tion intheoriginal frame. Soitisnotexactly true that acircularly polarized photon looks thesame foranysetofaxes. Itsphase (thephase relation ofthe right andleftcircularly polarized states) keeps track ofthex-direction. 11-5 Theneutral K-mesoni Wewillnowdescribe atwo-state system intheworld ofthestrange particles- asystem forwhich quantum mechanics gives amost remarkable prediction. To describe itcompletely would involve usinalotofstuff about strange particles, sowewill,unfortunately, have tocutsome corners. Wecanonlygiveanoutline ofhowacertain discovery wasmade-—to show youthekindofreasoning thatwas involved. Itbegins withthediscovery byGell-Mann andNishijima oftheconcept ofstrangeness andofanewlawofconservation ofstrangeness. Itwaswhen Gell- Mann andPaiswere analyzing theconsequences ofthese newideas thattheycame across theprediction ofamost remarkable phenomenon wearegoing todescribe. First, though, wehave totellyoualittle about “strangeness.” Wemust begin withwhat arecalled thestrong interactions ofnuclear particles. These aretheinteractions which areresponsible forthestrong nuclear forces- asdistinct, forinstance, from therelatively weaker electromagnetic interactions. The interactions are“strong” inthesense that iftwoparticles getclose enough tointeract atall,theyinteract inabigwayandproduce other particles veryeasily. IIt’s similar towhat wefound (inChapter 6)foraspinone-half particle when we rotated thecoordinates about thez-axis—then wegotthephase factors eilf’/2. Itis,in fact,exactly what wewrote down inSection 5-7fortheI—I—)andI—)states ofaspin-one particle—which isnocoincidence. Thephoton isaspin-one particle which has,however, no“zero” state. IWenowfeelthatthematerial ofthissection islonger andharder than isappropriate atthispoint inourdevelopment. Wesuggest thatyouskipitandcontinue withSection 11-6. Ifyouareambitious andhave time youmaywish tocome back toitlater. We leave ithere, because itisabeautiful example-taken from recent work inhigh-energy physics-—of what canbedone with ourformulation ofthequantum mechanics oftwo- state systems. 11-12 Thenuclear particles have alsowhat iscalled a“weak interaction” bywhich cer- tainthings canhappen, such asbetadecay, butalways veryslowly onanuclear time scale—the weak interactions aremany, many orders ofmagnitude weaker than thestrong interactions andeven much weaker than electromagnetic inter- actions. When thestrong interactions were being studied with thebigaccelerators, people were surprised toﬁndthatcertain things that“should” happen—that were expected tohappen-—did notoccur. Forinstance, insome interactions aparticle ofacertain typedidnotappear when itwasexpected. Gell-Mann andNishijima noticed thatmany ofthese peculiar happenings could beexplained atonce by inventing anewconservation law:theconservation ofstrangeness. They proposed thatthere wasanewkind ofattribute associated witheach particle—which they called its“strangeness” number—and thatinanystrong interaction the“quantity ofstrangeness” isconserved. Suppose, forinstance, thatahigh-energy negative K-meson—with, say,an energy ofmany Bev—co11ides with aproton. Outoftheinteraction maycome many other particles: 1r-mesons, K-mesons, lambda particles, sigma particles- anyofthemesons orbaryons listed inTable 2-2ofVol.I.Itisobserved, however, thatonlycertain combinations appear, andnever others. Now certain conservation laws were already known toapply. First, energy andmomentum arealways conserved. Thetotal energy andmomentum after anevent must bethesame as before theevent. Second, there istheconservation ofelectric charge which says thatthetotal charge oftheoutgoing particles must beequal tothetotal charge carried bytheoriginal particles. Inourexample ofaK-meson andaproton coming together, thefollowing reactions dooccur: K_+p—>p-I-K_+1r++ 1r_+1r° Of (11.38) K_—I-p—+Z_‘ —I—7l'+. Wewould never get: K“+p—>p+K_+1r+ or K_—I-p->A0—l—1r+. (11.39) because oftheconservation ofcharge. Itwasalsoknown thatthenumber of baryons isconserved. Thenumber ofbaryons outmust beequal tothenumber ofbaryons in.Forthislaw,anantiparticle ofabaryon iscounted asminus one baryon. Thismeans thatwecan-and do—see K_+p—>A°+1r° m nmm K‘+p-p+K‘+p+5 (where pistheantiproton, which carries anegative charge). Butwenever see K_+p—>K_+1r+—l-11'° or (11.41) K_+P—’P+K_+11 (even when there isplenty ofenergy), because baryons would notbeconserved. These laws, however, donotexplain thestrange factthatthefollowing re- actions—which donotimmediately appear tobeespecially different from some of those in(11.38) or(11.40)—are alsonever observed: K‘+P-p+K“+K°OI‘ K_+p-p+T_ umnO1‘ K-+p-AW+W. Theexplanation istheconservation ofstrangeness. With each particle goes a number—its strangeness S—and there isalawthatinanystrong interaction, the 11-13 Table 11-4 Thestrangeness numbers ofthestrongly interacting particles 1 S I -2 -1 0 +1 Baryons I Z+ p I I 410, 2 0 n _ 2.. Mesons 1r+ K+IIIIIIO 75 Hc78o I 1 K‘1'“Note: The-tr“istheantiparticle ofthe1r+(orviceverso). total strangeness outmust equal thetotal strangeness thatwent in.Theproton and antiproton (p,5),theneutron andantineutron (n,ii),andthe1r-mesons (1r+, 1r°, 1r_)allhave thestrangeness number zero;theK+andK0mesons have strangeness +1; theK‘andK0(the anti-K°),'I theA0andtheZ-particles (+, 0,~—)have strangeness -1. There isalso aparticle with strangeness —2—~the E-particle (capital “ksi”)—and perhaps others asyetunknown. Wehavemade alistofthese strangenesses inTable 11-4. Let’s seehow thestrangeness conservation works insome ofthereactions we havewritten down. IfwestartwithaK“andaproton, wehaveatotalstrangeness of(-1 +0)=—1.Theconservation ofstrangeness saysthatthestrangeness ofproducts after thereaction must alsoaddupto—l.Youseethatthatissofor thereactions of(11.38) and(11.40). Butinthereactions of(11.42) thestrangeness oftheright-hand sideiszeroineachcase. Such reactions donotconserve strange- ness, anddonotoccur. Why? Nobody knows. Nobody knows anymore than what wehave justtoldyouabout this. Nature justworks thatway. Now let’s look atthefollowing reaction: atr‘hitsaproton. You might, forinstance, getaA0particle plus aneutral K-particle-—two neutral particles. Now which neutral Kdoyouget? Since theA-particle hasastrangeness —land the1randp+haveastrangeness zero, andsince thisisafastproduction reaction, thestrangeness must notchange. The K-particle must have strangeness +1-it must therefore betheK“.Thereaction is 1-+p->11°+K“.with S=0+0=—1—I—-I-1 (conserved). IftheK0were there instead oftheK0,thestrangeness ontheright would be-2 —which nature does notpermit, since thestrangeness ontheleftside iszero. Ontheother hand, aK0canbeproduced inother reactions, such as n—l—n->n+p+Iz°-I-K1’, S=O—l—O=O+O—I—+l+-1 or K“—I—p->n +K0, S: -1+0=0+—1. You may bethinking, “That’s allalotofstuff, because how doyouknow whether itisaK0oraK0? They lookexactly thesame. They areantiparticles of each other, sothey have exactly thesame mass, andboth have zero electric charge. TRead as:“K-naught-bar,” or“K-zero-bar.” 11-14 \.,,5\ 1'- —- \ \ \ 1r- ~_\u,/ _+INTERACTION /\A°-decoy "70 _% L A0 I *> * E .......r... ~.KO '. NUCLEAR INTERACTION ,<<>_“co, /I '- LIQUID HYDROGEN1r*\;_"~.__----- 1r— NUCLEAR LIQUID HYDROGEN (CI) lb) Fig. ll—5. High-energy events asseen inohydrogen bubble chamber. (a)A7r—meson interacts with ahydrogen nucleus (proton) producing ciA0particle and aK0meson. Both particles decay in thechamber. (b)AR0meson interacts with aproton producing a1r+meson and aA0particle which then decays. (The neutral particles leave notracks. Their inferred traiectories areindicated here bylight dashed lines.) d 1)” 1 0How 0youdistinguish them. Bythereactions they produce. For examp e, aK0caninteract with matter toproduce aA-particle, likethis: K°+p—->A°+1r+, butaKcannot. There isnowayaK0canproduce aA-particle when itinteracts with ordinary matter (protons andneutrons).'l' Sotheexperimental distinction between theK0andtheK0would bethat oneofthem willandoneofthem will notproduce A’s. One ofthepredictions ofthestrangeness theory isthen this—-if, inanexperi- ment with high-energy pions, aA-particle isproduced with aneutral K-meson. then thatneutral K-meson going intoother pieces ofmatter willnever produce aA.Th . . .. . 'I _eexperiment might runsomething likethis. You send abeam of1r-mesons into alarge hydrogen bubble chamber. A'rr*track disappears, butsomewhere elseapair oftracks appear (aproton anda1r_)indicating that aA-particle has disintegratedI——see Fig. 11-5. Then youknow thatthere isaK0somewhere which youcannot see. You can, however, ﬁgure outwhere itisgoing byusing theconservation 0ofmomentum andenergy. [Itcould reveal itself later bydisintegrating into two charged particles, asshown inFig. ll—5(a).] AstheK0goes ﬂying along, itmay interact with oneofthehydrogen nuclei (protons), producing perhaps some other particles. The prediction ofthestrangeness theory isthatitwillnever produce a A-particle inasimple reaction like. say, K°+p——>A°+1r°, although aKcandojust that. That is,inabubble chamber aK0might produce theevent sketched inFig.ll—5(b)——in which theA°isseenbecause itdecays—but aK0willnot. That’s theﬁrstpartofourstory. That's theconservation ofstrange- ness. Theconservation ofstrangeness is,however, notperfect. There arevery slow disintegrations ofthestrange particles—decays taking alongil time like lO'1° second inwhich thestrangeness isnotconserved. These arecalled the“weak” Odecays. Forexample, theKdisintegrates into apair of1r—mesons (—l-and —) TExcept, ofcourse, ifitalsoproduces twoK+’s orother particles withatotal strange- nessof+2.Wecanthink hereofreactions inwhich there isinsufﬁcient energy toproduce these additional strange particles. ;tThe freeA-particle decays slowly viaaweak interaction (sostrangeness need notbe conserved). Thedecay products areeither apanda1r_,orannanda1r°.Thelifetime is2.2X10"” sec. TiAtypical time forstrong interactions ismore like10-23 sec. 11-15 with alifetime ofl0‘1° second. That was, infact, thewayK-particles were ﬁrstseen. Notice thatthedecay reaction K°—>1r"'+1r' does notconserve strangeness, soitcannot go“fast” bythestrong interaction; itcanonlygothrough theweak decay process. Now theK”alsodisintegrates inthesame way—into a1r"‘anda1r“— and alsowiththesame lifetime K°—>1r"+1r+. Again wehave aweak decay because itdoesnotconserve strangeness. There isa principle thatforanyreaction there isthecorresponding reaction with “matter” replaced by“antimatter” andviceversa. Since theK‘)istheantiparticle ofthe K°,itshould decay intotheantiparticles ofthe1r"'andtr“,buttheantiparticle ofa1r+isthe1r‘.(Or,ifyouprefer, viceversa. Itturns outthatforthe1r-mesons itdoesn’t matter which oneyoucall“matter.”) Soasaconsequence oftheweak decays, theK°andK‘)cangointothesame ﬁnalproducts. When “seen” through their decays—as inabubble chamber—they look likethesame particle. Only their strong interactions aredifferent. Atlastweareready todescribe thework ofGell-Mann andPais. They ﬁrstnoticed thatsince theK°andtheK‘)canboth turnintostates oftwo1r-mesons there must besome amplitude thataK°canturnintoaK0,andalsothataK‘) canturnintoaK0.Writing thereactions asonedoesinchemistry, wewould have K0ﬁr»1r_+1r+‘:>K‘). (11.43) These reactions imply thatthere issome amplitude perunittime, say—i/h times (K°|W|K0), thataK0willturn intoaK0through theweak interaction re- sponsible forthedecay intotwo1r-mesons. And there isthecorresponding amplitude (K0lWlK°)forthereverse process. Because matter andantimatter behave inexactly thesame way, these twoamplitudes arenumerically equal; we’ll callthem both A: <K°|w|K°) =(K°|w|i{°) =A. (11.44) Now—said Gell-Mann andPais—here isaninteresting situation. What people have been calling twodistinct states oftheworld-—the K°andthel{°— should really beconsidered asonetwo-state system, because there isanamplitude togofrom onestatetotheother. Foracomplete treatment, onewould, ofcourse, have todealwithmore than twostates, because there arealsothestates of2-tr’s, andsoon;butsince theywere mainly interested intherelation ofK°andK‘), theydidnothave tocomplicate things andcould make theapproximation ofa two-state system. Theother states weretaken intoaccount totheextent thattheir elfects appeared implicitly intheamplitudes ofEq.(11.44). Accordingly, Gell-Mann andPais analyzed theneutral particle asatwo- state system. They began bychoosing astheir twobasestates thestates |K0)and lK0). (From hereon,thestory goesverymuch asitdidfortheammonia mole- cule.) Any state I11/)oftheneutral K-particle could thenbedescribed bygiving theamplitudes thatitwasineither basestate. We’ll callthese amplitudes 6+=<K°lv>.0-=<K°I=t>- (11.45) Thenext stepwastowrite theHamiltonian equations forthistwo-state system. Ifthere were nocoupling between theK”andtheK0,theequations would besimply .dClh7+=1500+, (11.46) ihtg =ECdt °—' ll-16 Butsince there istheamplitude (K0IWIK°)fortheK0toturnintoaK0there should betheadditional term <K°IwIi<°>c_ =AC_ added totheright-hand side oftheﬁrstequation. And similarly, theterm AC+ should beinserted intheequation fortherateofchange ofC_. Butthat’s notall.When thetwo-pion effect istaken intoaccount there isan additional amplitude fortheK0toturn into itseb‘ through theprocess K0-—>1r_—l—1r+—>K°. Theadditional amplitude, which wewould write (K0IWIK0), isjust equal to theamplitude (K0IWIK0), since theamplitudes togotoandfrom apair of 1r-mesons areidentical fortheK0andtheK0. Ifyouwish, theargument canbe written outindetail likethis. First write'I' <K°|W|K°>= <_I{°lWl21r><21rlWlK°> and (K0IWIKO)=(K0IWI2-tr)(21r IWIK0). Because ofthesymmetry ofmatter andantimatter <21rlWlK°>=<21FlW|K°), andalso _ (K°IWI21r> =(K°IWI21r). Itthenfollows that(K0IWIK0)=(K0IWIK0),andalsothat(K0IWIK0)= (K0IWIK0), aswesaid earlier. Anyway, there arethetwoadditional ampli- tudes (K°IWIK0)and(K0IWIKO), both equal toA,which should beincluded intheHamiltonian equations. Theﬁrstgives aterm AC+ ontheright-hand side oftheequation fordC+/dt, andthesecond gives anewterm AC_ intheequation fordC_/dt. Reasoning thisway, Gell-Mann andPaisconcluded thattheHamil- tonian equations fortheK°K0system should be itfigfﬂ =1500++AC_+AC+, (11.47) ih%:=E(,C_+AC++AC_. Wemust now correct something wehave said inearlier chapters: that two amplitudes like (K0IWIK0) and (K0IWIK0) which arethereverse ofeach other, arealways complex conjugates. That wastruewhen wewere talking about particles that didnotdecay. But ifparticles candecay—and can, therefore, become “lost”—-the twoamplitudes arenotnecessarily complex conjugates. So theequality of(1l.44) does notmean that theamplitudes arerealnumbers; they areinfactcomplex numbers. The coefficient Ais,therefore, complex; andWe can’t justincorporate itintotheenergy E0. Having played often with electron spins andsuch, ourheroes knew that the Hamiltonian equations of(11.47) meant that there wasanother pair ofbase states which could alsobeused torepresent theK-particle system andwhich would have especially simple behaviors. They said, “Let’s take thesum anddifference ofthese twoequations. Also, let’s measure allourenergies from E0,anduseunits for IWearemaking asimpliﬁcation here. The2-/r-system canhave many states corre- sponding tovarious momenta ofthe1r-mesons, andweshould make theright-hand side ofthisequation intoasumoverthevarious basestates ofthe1r’s.Thecomplete treatment stillleads tothesame conclusions. ll-17 energy andtime thatmake it=1.”(That’s what modern theoretical physicists always do. Itdoesn’t change thephysics butmakes theequations take ona simple form.) Their result: 1%(c++c_)=2,4(c., +C_), 15%(c+-c_)=0. (11.48) Itisapparent that thecombinations ofamplitudes (C_I_—I—C_) and (C+——C_)actindependently from each other (corresponding, ofcourse, to thestationary states wehave been studying earlier). Sotheyconcluded thatit would bemore convenient touseadifferent representation fortheK-particle. They deﬁned thetwostates |1<.>=$<1K°>+1r<°>>. |K2>=\%(|K°>—lK°>)- (11.49) They saidthatinstead ofthinking oftheK°andK°mesons, wecanequally well think interms ofthetwo“partic1es” (that is,“states”) K1andK2.(These corre- spond, ofcourse, tothestates wehave usually called II)andIII). Wearenot using ouroldnotation because wewant nowtofollow thenotation oftheoriginal authors—and theoneyouwillseeinphysics seminars.) Now Gell-Mann andPaisdidn’t doallthisjusttogetdifferent names for thepartic1es——there isalsosome strange newphysics init.Suppose thatC1and C2aretheamplitudes thatsome state I11)willbeeither aK1oraK2meson: C1=(K1I‘!/>, C2=(K2I\//>- From theequations of(11.49), c.=é<c++C_), C2=X}?(ct—¢_>- (11.50) Then theEqs.(11.48) become .a'C1 _ _1?_2,4c., 1dt_0. (11.51) Thesolutions are _ C1(t) =Ci(0)@_‘“‘, C2(t)=C2(0), (11-52) where, ofcourse, C1(O) andC2(0) aretheamplitudes att=0. These equations saythatifaneutral K-particle starts outinthestate IK1) att=0[then C1(0) =1andC2(0) =0],theamplitudes atthetimetare C1(t) =9-H“, C2(t) =0- Remembering thatAisacomplex number, itisconvenient totake A= o1—iI6.(Since theimaginary partof2Aturns outtobenegative, wewrite itas minus i/3.)With thissubstitution, C1(t) reads C1(t)=C1(0)€_'s‘€_iM. (11.53) Theprobability ofﬁnding aK1particle attistheabsolute square ofthisampli- tude, which ise‘2”‘.And, from Eqs.(11.52), theprobability ofﬁnding theK2state atanytime iszero. That means thatifyoumake aK-particle inthestate IK1), theprobability ofﬁnding itinthesame state decreases exponentially withtime—- butyouwillnever ﬁnditinstate IK2). Where doesitgo?Itdisintegrates intotwo 1r-mesons with themean lifeT=1/26which is,experimentally, 10”“) sec. We made provisions forthatwhen wesaidthatAwascomplex. Ontheother hand, Eq.(11.52) saysthatifwemake aK-particle completely intheK2state, itstays thatwayforever. Well, that’s notreally true. Itisobserved experimentally todisintegrate intothree 1r-mesons, but600times slower than the 11-18 two-pion decay wehave described. Sothere aresome other small terms we have leftoutinourapproximation. Butsolong asweareconsidering onlythe two-pion decay, theK2lasts“forever.” Now toﬁnish thestory ofGell-Mann andPais. They wentontoconsider what happens when aK-particle isproduced withaA0particle inastrong interaction. Since itmust thenhave astrangeness of+1,itmust beproduced intheK°state. Soatt=0itisneither aK1noraK2butamixture. Theinitial conditions are C+(0) =1, C_(0) =0. Butthatmeans—from Eq.(ll.50)—that 1 1CO=—, C0)=—» i() ‘/5 2( \/5 and—from Eq.(11.51)—that C1(t)=X-I-2e-‘"6-‘"‘, c2(t)= (11.54) Now remember thatK1andK2areeach linear combinations ofK0andK0. InEqs. (11.54) theamplitudes have been chosen sothatatt=0theK0parts cancel each other outbyinterference, leaving onlyaK0state. ButtheIK1)state changes withtime, andtheIK2)state doesnot. After t=Otheinterference of C1andC2willgiveﬁnite amplitudes forboth K0andK°. What does allthismean? Let’s goback andthink oftheexperiment we sketched inFig.11-5. A1r"meson hasproduced aA0particle andaK0meson which istooting along through thehydrogen inthechamber. Asitgoes along, there issome small butuniform chance thatitwillcollide withahydrogen nucleus. Atﬁrst, wethought thatstrangeness conservation would prevent theK-particle from making aA°insuch aninteraction. Now, however, weseethatthatisnot right. Foralthough ourK-particle starts outasaK°—which cannot make a A°——it does notstaythisway. After awhile, there issome amplitude thatitwill have ﬂipped totheK0state. Wecan,therefore, sometimes expect toseeaA° produced along theK-particle track. Thechance ofthishappening isgiven by theamplitude C_,which wecan[byusing Eq.(11.50) backwards] relate toC1 andC2.Therelation is 1 - —'ia c_=72(c,-c2)=%(e“e‘-1). (11.55) AsourK-particle goesalong, theprobability thatitwill“actlike” aK°isequal toIC_I2, which is Ic_|2=in+e'25’-2e'5‘coscal). (11.56) Acomplicated andstrange result! This, then, istheremarkable prediction ofGell-Mann andPais: when aK° isproduced, thechance thatitwillturnintoaK°—as itcandemonstrate bybeing abletoproduce aA°—varies withtime according toEq.(11.56). Thisprediction came from using only sheer logic andthebasic principles ofthequantum me- chanics——with noknowledge atalloftheinner workings oftheK-particle. Since nobody knows anything about theinner machinery, thatisasfarasGell-Mann andPaiscould go.They could notgiveanytheoretical values foratandB.And nobody hasbeen abletodosotothisdate. They were abletogiveavalue ofB obtained from theexperimentally observed rateofdecay intotwo1r’s(28= 101°sec),buttheycould saynothing about oz. Wehave plotted thefunction ofEq.(11.56) fortwovalues ofainFig.11-6. Youcanseethattheform depends verymuch ontheratio ofatoI6.There isno K0probability atﬁrst; thenitbuilds up.Ifozislarge, theprobability would have ll-19 2A|c_1 (O) |.O"" q=41yB 0.75- O.50>— O.25'- —"‘-——___"--" "—23=|o'°seclarge oscillations. Ifasissmall, there willbelittle ornoosci1lation—the prob- ability willjustrisesmoothly to1/4. Now, typically, theK-particle willbetravelling ataconstant speed near the speed oflight. Thecurves ofFig. 11-6 then also represent theprobability along thetrack ofobserving aK°—-with typical distances ofseveral centimeters. You canseewhy thisprediction issoremarkably peculiar. You produce asingle particle andinstead ofjust disintegrating, itdoes something else. Sometimes it disintegrates, andother times itturns into adifferent kind ofaparticle. Itschar- acteristic probability ofproducing aneffect varies inastrange way asitgoes along. There isnothing elsequite likeitinnature. And thismost remarkable prediction wasmade solely byarguments about theinterference ofamplitudes. (bl a=1rB 2lC_l 102=100.15 B 5” 0.50 Q_g5_____._ ___ ____.i_.. O t(|0"°sec) Fig.11-6. The0 I 1 1 1 1 I I 1,0 0.25 0.50 0.15 1.0 0 I 2 3 4 t(lO"°secl function ofEq.(11-56): la)foroz=rrti,ib)foroz=41rd (with 26=101°sec). Ifthere isanyplace where wehave achance totestthemain principles of quantum mechanics inthepurest way—does thesuperposition ofamplitudes work ordoesn’t it?—this isit.Inspite ofthefactthatthiseffect hasbeen pre- dicted nowforseveral years, there isnoexperimental determination thatisvery clear. There aresome rough results which indicate thattheatisnotzero, andthat theeffect really occurs—they indicate that(Xisbetween 25and46.That’s allthere is,experimentally. Itwould bevery beautiful tocheck outthecurve exactly tosee iftheprinciple ofsuperposition really stillworks insuch amysterious world as thatofthestrange partic1es—with unknown reasons forthedecays, andunknown reasons forthestrangeness. Theanalysis wehavejustdescribed isverycharacteristic ofthewayquantum mechanics isbeing used today inthesearch foranunderstanding ofthestrange particles. Allthecomplicated theories thatyoumayhearabout arenomore and nolessthan thiskind ofelementary hocus-pocus using theprinciples ofsuper- position andother principles ofquantum mechanics ofthatlevel. Some people claim thattheyhave theories bywhich itispossible tocalculate thetiandoz,or atleast theozgiven theB,butthese theories arecompletely useless. Forinstance, thetheory thatpredicts thevalue ofct,given theB,tellsusthatthevalue ofa should beinﬁnite. Thesetofequations with which theyoriginally start involves two11'-mesons andthengoesfrom thetwo1r’sback toaK°,andsoon.When it’s allworked out,itdoes indeed produce apairofequations liketheones wehave here; butbecause there areaninﬁnite number ofstates oftwo1r’s,depending on their momenta, integrating over allthepossibilities gives anL!which isinﬁnite. Butnature’s atisnotinﬁnite. Sothedynamical theories arewrong. Itisreally quite remarkable that thephenomena which canbepredicted atallintheworld ofthestrange particles come from theprinciples ofquantum mechanics atthe level atwhich youarelearning them now. 11~20 11-6 Generalization toN-state systems Wehave ﬁnished with allthetwo-state systems wewanted totalkabout. Inthefollowing chapters wewillgoontostudy systems withmore states. The extension toN-state systems oftheideas wehave worked outfortwostates is pretty straightforward. Itgoeslikethis. Ifasystem hasNdistinct states, wecanrepresent anystate I¢(t)) asalinear combination ofanysetofbasestates |i),where i=1,2,3,...,N; I1/»(r)>=Z|i>c.-<1). (11.51)all1' Thecoefﬁcients C,~(t) aretheamplitudes (iI(l»(l)). Thebehavior oftheamplitudes C,withtimeisgoverned bytheequations .dC,~zlhTo =ZH,-,~c,-, (11.58) i where theenergy matrix H,-,»describes thephysics oftheproblem. Itlooks the same asfortwostates. Only now, both iandjmust range overallNbasestates, andtheenergy matrix H,,-—or, ifyouprefer, theHamiltonian—is anNbyN matrix withN2numbers. Asbefore, Hf}=H,-,—so longasparticles areconserved —and thediagonal elements H,-,-arerealnumbers. Wehave found ageneral solution fortheC’sofatwo-state system when the energy matrix isconstant (doesn’t depend ont).Itisalsonotdiﬁicult tosolve Eq.(1l.58) foranN-state system when Hisnottime dependent. Again, webegin bylooking forapossible solution inwhich theamplitudes allhave thesame time dependence. Wetry c,-=a,-e-<1/"W. (11.59) When these C/saresubstituted into(11.58), thederivatives dC,-(t)/dt become just (—i/h)EC,~. Canceling thecommon exponential factor from allterms, weget Ea,=2H,~,-a,. (11.60) J This isasetofNlinear algebraic equations fortheNunknowns a1,a2,...,a,,, andthere isasolution onlyifyouarelucky—only ifthedeterminant oftheeo- efﬁcients ofallthea’siszero. Butit’snotnecessary tobethatsophisticated; you canjuststarttosolve theequations anywayyouwant, andyouwillﬁndthatthey canbesolved onlyforcertain values ofE.(Remember thatEistheonlyadjustable thing wehave intheequations.) Ifyouwant tobeformal, however, youcanwrite Eq.(11.60) as Z(H,,--a,,~E)a,- =0. (11.61) 1 Then youcanusetherule—if youknow it—that these equations willhave asolu- tiononlyforthose values ofEforwhich DC! (Hij —5,']'E) = Each term ofthedeterminant isjustH,-_,-,except thatEissubtracted from every diagonal element. That is,(11.62) means just H11_E H12 H13 H21 H22 -E H23 D =O. 11.63ct H31 E32 H33 —E ( ) 11-21 This is,ofcourse, justaspecial way ofwriting analgebraic equation forEwhich isthesum ofabunch ofproducts ofalltheterms taken acertain way. These products willgiveallthepowers ofEuptoEN. Sowehave anNthorder polynomial equal tozero, andthere are,ingeneral, Nroots. (We must remember, however, thatsome ofthem may bemultiple roots—meaning thattwoormore roots areequal.) Let’s calltheNroots E1,E11, E1”,...,E,,,...,EN. (11.64) (Wewillusentorepresent thenthRoman numeral, sothatntakes onthevalues I,II,...,N.)Itmay bethatsome ofthese energies areequal—say E”=E;;;— butwewillstillchoose tocallthem bydifferent names. The equations (1l.60)— or(l1.61)—have onesolution foreach value ofE.If youputanyoneoftheE’s—say E,,—into (11.60) andsolve forthea,-,yougeta setwhich belongs totheenergy En.Wewillcallthisseta,-(n). Using these a,(n) inEq.(11.59), wehave theamplitudes C,-(n) thatthedeﬁnite energy states areinthebasestate Ii).Letting In)stand forthestate vector ofthe deﬁnite energy state att=0,wecanwrite =<1-In>e(i/MED!’ with (iIn)=a,~(n). (11.65) Thecomplete deﬁnite energy state I¢,,(t)) canthenbewritten as 1~//1-(1))=Z1i>a.~(n>e"“”‘”‘""', OI‘ i |¢,,(z)) =In)e"‘/“E->'. (11.66) Thestate vectors In)describe theconﬁguration ofthedeﬁnite energy states, but have thetime dependence factored out. Then theyareconstant vectors which canbeused asanewbasesetifwewish. Each ofthestates In)hastheproperty——as youcaneasily show—that when operated onbytheHamiltonian operator Hitgives justE,,times thesame state: H111) =E..In). (11.67) Theenergy Enis,then, anumber which isacharacteristic oftheHamiltonian operator H.Aswehave seen, aHamiltonian will, ingeneral, have several char- acteristic energies. Inthemathematician’s world these would becalled the“char- acteristic values” ofthematrix H,~,~. Physicists usually callthem the“eigenvalues” ofH.(“Eigen” istheGerman word for“characteristic” or“proper.”) With each eigenvalue ofIil—in other words, foreach energy—there isthestate of deﬁnite energy, which wehave called the“stationary state.” Physicists usually callthestates In)“theeigenstates ofH.” Each eigenstate corresponds toapar- ticular eigenvalue En. Now, generally, thestates In)—of which there areN-—can alsobeused asa baseset.Forthistobetrue, allofthestates must beorthogonal, meaning that foranytwoofthem, sayIn)andIm), (n|m)=0. (11.68) This willbetrueautomatically ifalltheenergies aredifferent. Also, wecan multiply allthea,~(n) byasuitable factor sothatallthestates arenormalized—by which wemean that (nIn) =l (11.69) foralln. When ithappens thatEq.(11.63) accidentally hastwo(ormore) roots with thesame energy, there aresome minor complications. First, there arestilltwo different setsofa,~’swhich gowiththetwoequal energies, butthestates theygive 11-22 maynotbeorthogonal. Suppose yougothrough thenormal procedure andﬁnd twostationary states with equal energies—let’s callthem I,u)andI1/).Then it willnotnecessarily besothattheyareorthogonal—if youareunlucky, (u11/>#0- Itis,however, always truethatyoucancook uptwonewstates, which wewill callIa’)andI1/’),thathave thesame energies andarealsoorthogonal, sothat (;.t'I1/’)=0. (11.70) You candothisbymaking In’) and I1/’)asuitable linear combination ofIp.) andI11),with thecoefﬁcients chosen tomake itcome outsothatEq.(11.70) is true. Itisalways convenient todothis. Wewillgenerally assume thatthishas been done sothatwecanalways assume thatourproper energy states In)are allorthogonal. Wewould like,forfun,toprove thatwhen twoofthestationary states have different energies theyareindeed orthogonal. Forthestate In)with theenergy En,wehave that mil) =E,,In). (11.71) This operator equation really means thatthere isanequation between numbers. Filling themissing parts, itmeans thesame as Z)<ilﬁ|1><11n> =E..<i1n>- (11.72) Ifwetakethecomplex conjugate ofthisequation, weget Z<i|1?|j>*<1\n>* =E£<i1n>*- (11.73)1' Remember nowthatthecomplex conjugate ofanamplitude isthereverse ampli- tude, so(11.73) canberewritten as Z<-=\1><1'1H1i>= E.f<n1i>- (11.74)1 Since thisequation isvalid foranyi,its“short form” is (nIH=Ej(n|, (11.75) which iscalled theatﬁoint toEq.(11.71). _ Now wecaneasily prove thatEnisarealnumber. Wemultiply Eq.(11.71) by(nItoget (n[HIn)=En, (11.76) since (nIn)=1.Then wemultiply Eq.(11.75) ontheleftbyIn)toget (nIHIn)=E:. (11.77) Comparing (11.76) with (11.77) itisclear that E,,=5:, (11.78) which means thatEnisreal. Wecanerase thestaronEninEq.(11.75). Finally weareready toshow thatthedifferent energy states areorthogonal. LetIn)andIm)beanytwoofthedeﬁnite energy basestates. Using Eq.(11.75) forthestate m,andmultiplying itbyIn),wegetthat (111W111) =Emfm |11>- 11-23 Butifwemultiply (11.71) by(mI,weget (mIHIn)=E,,(m In). Since theleftsides ofthese twoequations areequal, theright sides are,also: E,,,(m In)=E,,(m In). (11.79) IfEn,=Entheequation doesnottellusanything. Butiftheenergies ofthetwo states Im)andIn)arediﬂerent (Em95En),Eq.(11.79) saysthat(mIn)must bezero, aswewanted toprove. Thetwostates arenecessarily orthogonal solong asEnandEmarenumerically different. ll—24 I2 The Hyperfine Splitting inHydrogen 12-1 Base states forasystem withtwospinone-half particles Inthischapter wetake upthe“hyperﬁne splitting” ofhydrogen, because itisaphysically interesting example ofwhat wecanalready dowith quantum mechanics. It’sanexample with more than twostates, anditwillbeillustrative of themethods ofquantum mechanics asapplied toslightly more complicated prob- lems. Itisenough more complicated that once youseehow thisoneishandled youcangetimmediately thegeneralization toallkinds ofproblems. Asyouknow, thehydrogen atom consists ofanelectron sitting intheneigh- borhood oftheproton, where itcanexist inanyoneofanumber ofdiscrete energy states ineachoneofwhich thepattern ofmotion oftheelectron isdiﬁerent. Theﬁrst excited state, forexample, lies3/4ofaRydberg, orabout 10electron volts, above theground state. Buteven theso-called ground state ofhydrogen isnotreally asingle, deﬁnite-energy state, because ofthespins oftheelectron and theproton. These spins areresponsible forthe“hyperﬁne structure” intheenergy levels, which splits alltheenergy levels intoseveral nearly equal levels. The electron canhave itsspin either “up” or“down” and, theproton can alsohave itsspineither “up” or“down.” There are,therefore,f0ur possible spin states forevery dynamical condition oftheatom. That is,when people say“the ground state” ofhydrogen, they really mean the“four ground states,” andnot justthevery lowest state. Thefour spin states donotallhave exactly thesame energy; there areslight shifts from theenergies wewould expect with nospins. The shifts are,however, much, much smaller than the10volts orsofrom the ground state tothenextstate above. Asaconsequence, each dynamical state has itsenergy splitintoasetofveryclose energy levels—the so-called hyperﬁne splitting. Theenergy differences among thefour spinstates iswhat wewant tocalculate inthischapter. Thehyperﬁne splitting isduetotheinteraction ofthemagnetic moments oftheelectron and proton, which gives aslightly different magnetic energy foreach spin state. These energy shifts areonly about ten-millionths ofanelectron volt—really very small compared with 10volts! Itisbecause of thislarge gapthat wecanthink about theground state ofhydrogen asa“four- state” system, without worrying about thefactthatthere arereally many more states athigher energies. Wearegoing tolimit ourselves here toastudy ofthe hyperﬁne structure oftheground state ofthehydrogen atom. Forourpurposes wearenotinterested inanyofthedetails about thepositions oftheelectron andproton because thathasallbeenworked outbytheatom soto speak—it hasworked itself outbygetting into theground state. Weneed know only thatwehave anelectron andproton intheneighborhood ofeach other with some deﬁnite spatial relationship. Inaddition, they canhave various different relative orientations oftheir spins. Itisonly theeffect ofthespins thatwewant to look into. Theﬁrstquestion wehavetoanswer is:What arethebasestates forthesystem? Now thequestion hasbeen putincorrectly. There isnosuch thing as“the” base states, because, ofcourse, thesetofbase states youmaychoose isnotunique. New setscanalways bemade outoflinear combinations oftheold. There are always many choices forthebase states, andamong them, anychoice isequally legitimate. Sothequestion isnotwhat isthebaseset,butwhat could abase set be?Wecanchoose anyonewewishforourownconvenience. Itisusually best tostart with abase setwhich isphysically theclearest. Itmay notbethesolution 12-112-1 Base states forasystem with twospinone-half particles 12-2 TheHamiltonian fortheground state ofhydrogen 12-3 Theenergy levels 12-4 TheZeeman splitting 12-5 Thestates inamagnetic ﬁeld 12-6 Theprojection matrix forspin one LECTRON PROTON / /Fig. 12-1. Asetofbase states for theground ﬂute ofthe hydrogen atom.toanyproblem, ormay nothave anydirect importance, butitwillgenerally make iteasier tounderstand what isgoing on. Wechoose thefollowing fourbasestates: State I:Theelectron andproton areboth spin“up.” State 2:Theelectron is“up” andtheproton is“down.” State 3:Theelectron is“down” andtheproton is“up.” State 4:Theelectron andproton areboth “down.” Weneed ahandy notation forthese fourstates, sowe’ll represent them thisway: State 1:I++);electron up,proton up. State 2:I+—);electron up,proton down. State 3:I—+);electron down, proton up. (121) State 4:I——);electron down, proton down. Youwillhave toremember thattheﬁrstplusorminus signrefers totheelectron andthesecond, totheproton. Forhandy reference, we’ve alsosummarized the notation inFig.12-1. Sometimes itwillalsobeconvenient tocallthese states I1),I2),I3>,=1nd I4)-You maysay,“But theparticles interact, andmaybe these aren’t theright basestates. Itsounds asthough youareconsidering thetwoparticles indepen- dently.” Yes,indeed! Theinteraction raises theproblem: what istheHamiltonian forthesystem, buttheinteraction isnotinvolved inthequestion ofhowtodescribe thesystem. What wechoose forthebase states hasnothing todowith what happens next. Itmaybethattheatom cannot everstayinoneofthese basestates, even ifitisstarted thatway. That’s another question. That’s thequestion: How dotheamplitudes change withtimeinaparticular (ﬁxed) base? Inchoosing thebasestates, wearejustchoosing the“unit vectors” forourdescription. While we’re onthesubject, let’slook atthegeneral problem ofﬁnding aset ofbasestates when there ismore than oneparticle. Youknow thebasestates for asingle particle. Anelectron, forexample, iscompletely described inreallife—not inoursimpliﬁed cases, butinreallife—by giving theamplitudes tobeineach of thefollowing states: Ielectron “up” withmomentum p) or Ielectron “down” withmomentum p). There arereally twoinﬁnite setsofstates, onestate foreach value ofp.That is tosaythatanelectron state I1/»)iscompletely described ifyouknow alltheampli- tudes <+1p I and <_1p I‘pl: where the+and—represent thecomponents ofangular momentum along some axis—usually thez-axis—and pisthevector momentum. There must, therefore, betwoamplitudes forevery possible momentum (amulti-inﬁnite setofbase states). That isallthere istodescribing asingle particle. When there ismore than oneparticle, thebase states canbewritten ina similar way. Forinstance, ifthere were anelectron andaproton inamore com- plicated situation thanweareconsidering, thebasestates could beofthefollowing kind: Ianelectron withspin“up,” moving 'with momentum pland aproton withspin“down,” moving withmomentum pg). And soonforother spincombinations. Ifthere aremore than twopartic1es— same idea. Soyouseethattowrite down thepossible basestates isreally veryeasy. Theonlyproblem is,what istheHamiltonian? Forourstudy oftheground state ofhydrogen wedon’t need tousethefull setsofbase states forthevarious momenta. Wearespecifying particular mo- 12-2 mentum states fortheproton andelectron when wesay“theground state.” The details oftheconﬁguration—the amplitudes forallthemomentum basestates canbecalculated, butthatisanother problem. Now weareconcerned onlywith theeffects ofthespin, sowecantake only thefourbase states of(12.1). Our nextproblem is:What istheHamiltonian forthissetofstates? 12-2 TheHamiltonian fortheground state ofhydrogen We’ll tellyouinamoment what itis.Butﬁrst,weshould remind youofone thing: anystate canalways bewritten asalinear combination ofthebasestates. Foranystate I(I/)wecanwrite I~t>=|++><++I=t>+l+ -><+—|t>+|— +><—+|t> +I——>(- —I¢>- (12-2) Remember thatthecomplete brackets arejustcomplex numbers, sowecanalso write them intheusual fashion asC,-,where i=1,2,3,or4,andwrite Eq.(12.2) as I‘!/>=I++>C1 —I—I-1"—>C2 -I"I-+>C3 *1-I_ "‘>C4- (12-3) Bygiving thefouramplitudes C,wecompletely describe thespinstate IIL).If these fouramplitudes change withtime, astheywill,therateofchange intimeis given bytheoperator H.Theproblem istoﬁndtheH. There isnogeneral ruleforwriting down theHamiltonian ofanatomic system, andﬁnding theright formula ismuch more ofanartthanﬁnding asetof basestates. Wewere abletotellyouageneral ruleforwriting asetofbasestates foranyproblem ofaproton andanelectron, buttodescribe thegeneral Hamilton- ianofsuch acombination istoohard atthislevel. Instead, wewillleadyoutoa Hamiltonian bysome heuristic argument—and youwillhave toaccept itasthe correct onebecause theresults willagree withthetestofexperimental observation. You willremember thatinthelastchapter wewere abletodescribe the Hamiltonian ofasingle, spinone-half particle byusing thesigma matrices——or the exactly equivalent sigma operators. Theproperties oftheoperators aresum- marized inTable 12—1. These operators—which arejustaconvenient, shorthand way ofkeeping track ofthematrix elements ofthetype (—I—I0,I—I—)——were useful fordescribing thebehavior ofasingle particle ofspinone-half. Thequestion is:Canweﬁndananalogous device todescribe asystem with twospins? The answer isyes,verysimply, asfollows. Weinvent athing which wewillcall“sigma electron,” which werepresent bythevector operator er”,andwhich hasthe x-,y-,andz-components, 0;,0;,0'2.Wenow make theconvention thatwhen one ofthese things operates onanyoneofourfourbasestates ofthehydrogen atom, itactsonlyontheelectron spin, andinexactly thesame wayasiftheelectron were allbyitself. Example: What is0;I——I—)? Since 0,,onanelectron “down” is—itimes thecorresponding state withtheelectron “up”, ¢I§I~ +)= —iI++>- (When 0;actsonthecombined state itﬂipsovertheelectron, butdoesnothing to theproton andmultiplies theresult by—i.) Operating ontheother states, of, would give <1Z|++>=i|—+>, 0-;I+ ___) = _>! -21-->=—t1+->. Justremember thattheoperators 11°work onlyontheﬁrstspinsymbol—that is, ontheelectron spin. Next wedeﬁne thecorresponding operator “sigma proton” fortheproton spin. Itsthree components oi’,(III,ofactinthesame wayas11°,only onthe 12-3Table 12-1 <1.I+>=+I+> <T=I~>=-I—) ‘7rI+>=‘I"I'“> U==I_>:‘I'I‘I'> at/I+>= ‘I'iI_> <1tI—>=—i1+> proton spin. Forexample, ifwehaveaﬁ,’acting oneach ofthefourbasestates, we get—always using Table 12-1- <1§I++>=I+—), <1§|+—>=I++>. <r§I—+)=I-->, <r§I-—)=I—+>- Asyoucansee,it’snotvery hard. Now inthemost general case wecould have more complex things. For instance, wecould have products ofthetwooperators likeojol’. When wehave such aproduct wedoﬁrstwhat theoperator ontheright says, andthendowhat theother onesays.1' Forexample, wewould have that tT§<T§I+—)=¢T§(<T§I+—>)=aZ(—I+-))=—<T§I+—>=—I--)- Note thatthese operators don’t doanything onpure numbers——we have used thisfactwhen wewrote o'f,(— 1)=(—1)o§. Wesaythattheoperators “commute” with pure numbers, orthat anumber “can bemoved through” theoperator. You canpractice byshowing thattheproduct ojoﬂ’ gives thefollowing results forthefourstates: vZaZI++>=+I—+>. a§<r‘§I+—)=—I-—), 6:d;|—+>=+|++>. <rZ<r‘Z|——>=—I+—>- Ifwetakeallthepossible operators, using each kind ofoperator onlyonce, there aresixteen possibilities. Yes, sixteen—provided weinclude alsothe“unit operator” l.First, there arethethree: oi,of},oi.Then thethree oﬁ,oI],o§—that makes six. Inaddition, there arethenine possible products oftheform ojoﬁ, which makes atotal of15.And there’s theunitoperator which justleaves any state unchanged. Sixteen inall. Now note thatforafour-state system, theHamiltonian matrix hastobe afour-by-four matrix ofcoefﬁcients—-—it willhave sixteen entries. Itiseasily demonstrated that anyfour-by-four matrix—and, therefore, theHamiltonian matrix inparticular—can bewritten asalinear combination ofthesixteen double- spinmatrices corresponding tothesetofoperators wehavejustmade up.There- fore, fortheinteraction between aproton andanelectron thatinvolves onlytheir spins, wecanexpect thattheHamiltonian operator canbewritten asalinear combination ofthesame 16operators. Theonlyquestion is,how? Well, ﬁrst, weknow thattheinteraction doesn’t depend onourchoice of axesforacoordinate system. Ifthere isnoexternal disturbance-—like amagnetic ﬁeld—-to determine aunique direction inspace, theHamiltonian can’t depend on ourchoice ofthedirection ofthex-,y-,andz-axes. That means that the Hamiltonian can’t have aterm likeojallbyitself. Itwould beridiculous, because thensomebody withadifferent coordinate system would getdifferent results. Theonlypossibilities areaterm withtheunitmatrix, sayaconstant a(times 1),andsome combination ofthesigmas thatdoesn’t depend onthecoordinates— some “invariant” combination. Theonly scalar invariant combination oftwo vectors isthedotproduct, which forouro’sis e P__ eP eP eD0'-0'-o,,o, +o",,o,, +a,o,. (12.4) This operator isinvariant with respect toanyrotation ofthecoordinate system. IForthese particular operators, youwillnotice itturns outthatthesequence ofthe operators doesn’t matter. 12-4 Sotheonlypossibility foraHamiltonian with theproper symmetry inspace isa constant times theunitmatrix plusaconstant times thisdotproduct, say, H=E,+Aas-Up. (12.5) That’s ourHamiltonian. It’stheonly thing thatitcanbe,bythesymmetry of space, solongasthere isnoexternal ﬁeld. Theconstant term doesn’t tellusmuch; itjustdepends onthelevel wechoose tomeasure energies from. Wemayjust aswelltakeE0=0.Thesecond term tellsusallweneed toknow toﬁndthe levelsplitting ofthehydrogen. Ifyouwant to,youcanthink oftheHamiltonian inadifferent way. Ifthere aretwomagnets near each other withmagnetic moments nuandpip,themutual energy willdepend onpg-;i,,—among other things. And, you‘remember, we found thattheclassical thing wecallneappears inquantum mechanics aspcﬂe. Similarly, what appears classically asppwillusually turnoutinquantum mechanics tobeapap (where ppisthemagnetic moment oftheproton, which isabout 1000 times smaller than he,andhastheopposite sign). SoEq.(12.5) saysthatthe interaction energy isliketheinteraction between twomagnets—only notquite, because theinteraction ofthetwomagnets depends ontheradial distance between them. ButEq.(12.5) could be—and, infact, is-—-some kind ofanaverage inter- action. Theelectron ismoving allaround inside theatom, andourHamiltonian gives onlytheaverage interaction energy. Allitsaysisthatforaprescribed ar- rangement inspace fortheelectron andproton there isanenergy proportional tothecosine oftheangle between thetwomagnetic moments, speaking classically. Such aclassical qualitative picture mayhelpyoutounderstand where itcomes from, buttheimportant thing isthatEq.(12.5) isthecorrect quantum mechanical formula. Theorder ofmagnitude oftheclassical interaction between twomagnets would betheproduct ofthetwomagnetic moments divided bythecube ofthe distance between them. Thedistance between theelectron andtheproton inthe hydrogen atom is,speaking roughly, onehalfanatomic radius, or0.5angstrom. Itis,therefore, possible tomake acrude estimate thattheconstant Ashould be about equal totheproduct ofthetwomagnetic moments It,andppdivided by thecube of1/2angstrom. Such anestimate gives anumber intheright ballpark. Itturns outthatAcanbecalculated accurately onceyouunderstand thecomplete quantum theory ofthehydrogen atom—which wesofardonot. Ithas,infact, been calculated toanaccuracy ofabout 30parts inonemillion. So,unlike the ﬂip-ﬂop constant Aoftheammonia molecule, which cou1dn’t becalculated at allwellbyatheory, ourconstant Aforthehydrogen canbecalculated from amore detailed theory. Butnever mind, wewillforourpresent purposes think oftheA asanumber which could bedetermined byexperiment, andanalyze thephysics ofthesituation. Taking theHamiltonian ofEq.(12.5), wecanuseitwiththeequation ihC,-=ZH,-,~c, (12.6)J toﬁndoutwhat thespininteractions dototheenergy levels. Todothat, weneed towork outthesixteen matrix elements H,-,»=(iIHIj)corresponding toeach pairofthefourbasestates in(12.1). Webegin byworking outwhat HIj)isforeach ofthefour base states. Forexample, HI++)=A6"¢"I++)=A{o§o§ +63¢;+63¢?)|++). (12.7) Using themethod wedescribed alittle ear1ier—it’s easyifyouhave memorized Table 12-1-—we ﬁndwhat each pairofa"sdoes onI++).Theanswer is <rZv§I++>=+|——), (12621++>=—I——>. (12.8) v:<1‘;|++>=+I++>-12-5 Spin operators forthehydrogen atomTable 12-2 3 p 1710': 9 I7 (7:52 9 I) ‘Tova: 0 P (7:72: e1> <71/7v er> 0'2/711 eP 0'1/0'11 9 p 0'1/711 via‘; <r§<r’l via’; 636‘;++>= +—>=_+)= ___): ++>= +—>=_.I_>= __)= ++)= +—)= __I.)= _._)=++++ + + + ++ + + + + ++) + +) +) +)So(12.7) becomes 1i|++>=A{I——>—I——>+|++>}=A|++). (12.9) Since ourfourbasestates areallorthogonal, thatgives usimmediately that (++|H|++)= <+-|HI++>= <-+Ir1|++>= <——|HI++>=A(++I-I-+)=A, -4<-I--I++)=0, A(—+I++)=0, A<——l++>=0-(12.10) Remembering that(jIHIi)=(iIHIj)*,wecanalready write down thediffer- ential equation fortheamplitudes C1: ihC1= H11C1-I“ H12C2 -I"H13C3 -1-H14C4 OI‘ ihcl = That’s all!Wegetonlytheoneterm.(12.11) Now togettherestoftheHamiltonian equations wehave tocrank through thesame procedure forHoperating ontheother states. First, wewillletyou practice bychecking outallofthesigma products wehave written down inTable 12-2. Then wecanusethem toget: til+-)=A{2 HI—+>=A{2|+ ->—I—+>}, <12-12> H|——>=AI— —>- Then, multiplying each oneinturnontheleftbyalltheother state vectors, we getthefollowing Hamiltonian matrix, H,-,-: 1'_-. ‘IA000 _> H”: 0—A2,4 0 +) 02.4—A 0 > 000- (12.15) A Itmeans, ofcourse, nothing more thanthatourdifferential equations forthefour amplitudes C,are ihC1=AC1, inc",=—AC2 +2AC3, (12.14) ihC3=2/1c,-AC3, ihC4=AC4. Before solving these equations wecan’t resist telling youabout aclever ruleduetoDirac—it willmake youfeelthatyouarereally advanced—a1though wedon’t need itforourwork. Wehave—from theequations (12.9) and(12.12)- that a°'o1pI++)= <1”-<="I+—)= <r°-¢‘°l—+>= <1‘-¢‘°I——>=12-6I++), 2I~ +)'"I+ —), (12-15) |——)- Look, saidDirac, Icanalsowrite theﬁrstandlastequations as a'°-a"I+-I-)=2I++)-|++), ¢°-<r"|——>=2|——)—|——>: thenthey areallquite similar. Now Iinvent anewoperator, which Iwillcall P,,p;,, ml,andwhich Ideﬁne tohave thefollowing properties:'I' Pepin exchI—I——I—)=I-I“-I-)1 Pepin cxchI+—)=I-+). Pspiii 651611I—-I-)=I—I—‘—), Pepin eXcl1I ——)=I'-—)- Alltheoperator does isinterchange thespindirections ofthetwoparticles. Then Icanwrite thewhole setofequations in(12.15) asasimple operator equation: as-6*’=2P,,,,,,,,.,,,-1. (12.16) That’s theformula ofDirac. His“spin exchange operator” gives ahandy ruleforﬁguring outa"'-op.(You see,youcandoeverything now. Thegates areopen.) 12-3 Theenergy levels Now weareready towork outtheenergy levels oftheground state ofhydro- genbysolving theHamiltonian equations (12.14). Wewant toﬁndtheenergies ofthestationary states. This means thatwewant toﬁndthose special states I¢)forwhich each amplitude C,»=(iII0)inthesetbelonging toI¢)hasthe same timedependence—namely, e_“"‘. Then thestatewillhavetheenergyE =hw. Sowewant asetforwhich c,-=12,-e‘-'~'/"W, (12.17) where thefourcoefficients a,-areindependent oftime. Toseewhether wecan getsuch amplitudes, wesubstitute (12.17) into Eq.(12.14) andseewhat happens. Each ihdC/dt inEq.(12.14) turns intoEC,and—after cancelling outthecommon exponential factor——each Cbecomes ana;weget Eal =Aal, E112 =—Aa2 —I—2Aa3, Ea3=2Aa2 —Aa3, Ea, =A04,(12.18) which wehave tosolve fora1,a2,a3,anda4.Isn’titnicethattheﬁrstequation is independent oftherest—that means wecanseeonesolution right away. Ifwe choose E=A, a1=1, a2=a3=a4=O, gives asolution. (Ofcourse, taking allthea’sequal tozero alsogives asolution, butthat’s nostate atall!) Let’s callourﬁrstsolution thestate II):1 I1)=I1)=I++). (12.19) Itsenergy is EI=A. "IThisoperator isnowcalled the“Pauli spinexchange operator.” IIThestate isreally II)e"(‘/ﬁwl‘; but,asusual wewillidentify thestates bythecon- stant vectors which areequal tothecomplete vectors att=0. 12-7 I,n,mEo+A so ---- - — AE=i'\w E0-3A N Fig. l2—2. Energy-level diagram for theground state ofatomic hydrogen.With thatclueyoucanimmediately seeanother solution from thelastequation in(12.18): a1=a2=a3=O, a4=l, E=A. We’ll callthatsolution state |I1): |11>=I4)=l-—), (12-20) Eff=A. Now itgetsalittle harder; thetwoequations leftin(12.18) aremixed up. Butwe’ve done itallbefore. Adding thetwo,weget E(a2 +a3)=A(a2 +a3). (12.21) Subtracting, wehave E(a2 —as)=—3A(a2 -—a3). (12.22) Byinspection—and remembering ammonia—we seethatthere aretwosolutions: £12 =G3, E=A and (12.23)(12 =-03, E= They aremixtures ofI2)and|3).Calling these states |III)andfIV),andputting inafactor l/\f2 tomake thestates properly normalized, wehave |I11>=i<|z>+ |3>)=~‘~<|+ »>+|—+>>.\/E ‘/2 (12.24) E111 =A and 111/>=i<|2>-|s>>=i<1+->-1-+>>,‘/5 \/Z (12.25) Ely =—3A. Wehave found four stationary states andtheir energies. Notice, incidentally, thatourfourstates areorthogonal, sothey alsocanbeused forbase states if desired. Ourproblem iscompletely solved. Three ofthestates have theenergy A,andthelasthastheenergy —-3A. Theaverage iszero-——which means thatwhen wetook E0=0inEq.(12.5), we were choosing tomeasure alltheenergies from theaverage energy. Wecandraw theenergy-level diagram fortheground state ofhydrogen asshown inFig.12-2. Now thedifference inenergy between state IIV) andanyoneoftheothers is4A.Anatom which happens tohave gotten intostate II)could fallfrom there tostate |IV)andemitlight. Notoptical light, because theenergy issotiny—it would emitamicrowave quantum. Or,ifweshine microwaves onhydrogen gas, wewillﬁndanabsorption ofenergy astheatoms instate IIV)pickupenergy and gointooneoftheupper states———but only atthefrequency w=4A/h. This frequency hasbeen measured experimentally; thebest result, obtained very recently,1' is f=w/21r =(l,420,405,75l.800 ¢0.028) cycles persecond. (12.26) Theerror isonlytwoparts inI00billion! Probably nobasic physical quantity is measured better thanthat—it’s oneofthemost remarkably accurate measurements inphysics. Thetheorists were veryhappy thattheycould compute theenergy to anaccuracy of3parts in105,butinthemeantime ithasbeenmeasured to2parts in 10‘‘—amillion times more accurate than thetheory. Sotheexperimenters are TCrampton, Kleppner, andRamsey; Physical Review Lelrers, Vol.11,page 338(1963). 12-8 wayahead ofthetheorists. Inthetheory oftheground stateofthehydrogen atom youareasgood asanybody. You, too,canjusttakeyour value ofAfrom experi- ment—that’s what everybody hastodointheend. Youhave probably heard before about the“21-centimeter line” ofhydrogen. That’s thewavelength ofthe1420 megacycle spectral linebetween thehyperﬁne states. Radiation ofthiswavelength isemitted orabsorbed bytheatomic hydrogen gasinthegalaxies. Sowith radio telescopes tuned into21-cm waves (or1420 megacycles approximately) wecanobserve thevelocities andthelocation ofcon- centrations ofatomic hydrogen gas. Bymeasuring theintensity, wecanestimate theamount ofhydrogen. Bymeasuring thefrequency shiftduetotheDoppler effect, wecanﬁndoutabout themotion ofthegasinthegalaxy. That isoneof thebigprograms ofradio astronomy. Sonow wearetalking about something that’s veryreal—it isnotanartiﬁcial problem. 12-4 TheZeeman splitting Although wehave ﬁnished theproblem ofﬁnding theenergy levels ofthe hydrogen ground state, wewould liketostudy thisinteresting system some more. Inorder tosayanything more about it—for instance, inorder tocalculate the rateatwhich thehydrogen atom absorbs oremits radio waves at21centimeters— wehave toknow what happens when theatom isdisturbed. Wehave todoaswe didfortheammonia molecule—after wefound theenergy levels wewent onand studied what happened when themolecule wasinanelectric ﬁeld. Wewere then abletoﬁgure outtheeffects from theelectric ﬁeld inaradio wave. Forthehydro- genatom, theelectric ﬁeld does nothing tothelevels, except tomove them allby some constant amount proportional tothesquare oftheﬁeld—which isnotof anyinterest because that won’t change theenergy dzﬂerences. Itisnow the magnetic ﬁeld which isimportant. Sothenext step istowrite theHamiltonian foramore complicated situation inwhich theatom sitsinanexternal magnetic ﬁeld. What, then, istheHamiltonian? We’ll justtellyoutheanswer, because we can’t giveyouany“proof” except tosaythatthisisthewaytheatom works. TheHamiltonian is H=Aw-G») -/.1808-B-l~‘pdp.B- (12.27) Itnow consists ofthree parts. The ﬁrst term Aa°-uprepresents themagnetic interaction between theelectron andtheproton—it isthesame onethat would bethere ifthere were nomagnetic ﬁeld. This istheterm wehave already had; andtheinﬂuence ofthemagnetic ﬁeld ontheconstant Aisnegligible. Theeffect oftheexternal magnetic ﬁeldshows upinthelasttwoterms. Thesecond term, -14.41" -B,istheenergy theelectron would have inthemagnetic ﬁeld ifitwere there alone.’[ Inthesame way, thelastterm —;tpaI’ -B,would have been the energy ofaproton alone. Classically, theenergy ofthetwoofthem together would bethesum ofthetwo, andthatworks alsoquantum mechanically. Inamagnetic ﬁeld, theenergy ofinteraction duetothemagnetic ﬁeldisjust thesumoftheenergy ofinteraction oftheelectron with theexternal ﬁeld, andoftheproton with the ﬁeld—both expressed interms ofthesigma operators. Inquantum mechanics these terms arenotreally theenergies, butthinking oftheclassical formulas for theenergy isaway ofremembering therules forwriting down theHamiltonian. Anyway, thecorrect Hamiltonian isEq.(12.27). Now wehave togoback tothebeginning anddotheproblem alloveragain. Much ofthework is,however, done—we need only toaddtheeffects ofthenew terms. Let’s takeaconstant magnetic ﬁeldBinthez-direction. Then wehave to ’tRemember thatclassically U=—pt-B,sotheenergy islowest when themoment isalong theﬁeld. Forpositive particles. themagnetic moment isparallel tothespinand fornegative particles itisopposite. SoinEq.(12.27), upisapositive number, but110is anegative number. 12-9 addtoourHamiltonian operator 1-7thetwonewpieces—which wecancallH’: HI =_'(/-Leo‘: +,up0'g)B- Using Table 12-1, wegetright away that 17'l++) =-(1-¢@+#1>)B|++), H'|+—>=-<1».—11,081+ —>. F?’I—+>=—<—~.+~..)Bl—+>. H'|——>=(a.+1»,.)BI— —>- How very convenient! TheI2’operating oneach state justgives anumber times thatstate. Thematrix (i|H’|j)has,therefore, onlydiagonal elements—we can justaddthecoefﬁcients in(12.28) tothecorresponding diagonal terms of(12.13), andtheHamiltonian equations of(12.14) become(12.28) iﬁdct/dl ={A-(Me+#193} C1, l'hdC2/dl‘ =*{/1 +(lie—I-¢p)B} C2+2/‘C3, ihdca/df =ZAC2 "{/4*(Me—M13153, lihdcrt/di ={A"1"(lie-I"#p)B}C4-(12.29) Theform oftheequations isnotdifferent—only thecoefﬁcients. Solong asBdoesn’t vary with time, wecancontinue aswedidbefore. Substituting C,-=a,»e““/‘)5’, weget—as amodiﬁcation of(12.18)— Ea, =A{— (,ue +ptp)B}a1, E92 =“(A +(He_I~*p)B}a2 "1"2/Ms, E413 =21402 —{A—(He_I1p)B}"s1 Eat={A+(Ue+#,»)B}¢14-(12.30) Fortunately, theﬁrstandfourth equations arestillindependent oftherest.sothe same technique works again. Onesolution isthestate |I)forwhich at=1,a2 =a3=a4=0,or II)=I1)=I-1-+), with (12.31) E1=A—(/it+Mp)B-Another is I11)=I4)=|——),with (12.32) E11 =A+(Me+/.tp)B. Alittle more work isinvolved fortheremaining twoequations, because the coefﬁcients ofa; anda3arenolonger equal. Butthey arejustlikethepairwehad fortheammonia molecule. Looking back atEq.(9.20), wecanmake thefollowing analogy (remembering that thelabels 1and2there correspond to2and3here): H11—>—/4 *(“'0-Mp)B, H12 ——>2A, H21—> 2A, H22 —’'-A+(Me—l1p)B»(12.33) Theenergies arethengiven by(9.25), which was E= It +Hl2H21_ (1234) l2-l0 Making thesubstitutions from (12.33), theenergy formula becomes E=—A1./(,1,_,.,,)2B2 +4,42. Although inChapter 9weused tocallthese energies E1andE”, andwearein thisproblem calling them E111 andE111, E111=A{~1 +2\/1+(Me—Mp)2B2/4/12}, (12.35)EIV=—A{1+2\/1+ta.-a,.)2B2/4A2}. Sowehave found theenergies ofthefour stationary states ofahydrogen atom inaconstant magnetic ﬁeld. Let’s check ourresults byletting Bgotozero andseeing whether wegetthesame energies wehadinthepreceding section. You seethat wedo. ForB=0,theenergies E1,E11, andE111 goto+A, andE111 goesto~3A. Even ourlabeling ofthestates agrees withwhat wecalled them be- fore. When weturn onthemagnetic ﬁeld though, alloftheenergies change ina different way. Let’s seehowtheygo. First, wehave toremember thatfortheelectron, asisnegative, andabout 1000 times larger than /.t,,—which ispositive. So[Le+ppandpg—/2,,areboth negative numbers, andnearly equal. Let’s callthem —;uand—/.t’: F‘=_(#e "l"I-‘p)> F‘)='_(P'e _P'p)' (Both itand)2’arepositive numbers. nearly equal tomagnitude of,u.,—which is about oneBohr magneton.) Then ourfourenergies are E1=A"l"I-‘B, E11=A—/13, E111 =A{—l -l-2 }, EIV=-Att+2~/7%} Theenergy E1starts atAandincreases linearly with B—with theslope /2.The L(12.37) LA l E 4- ,t>""e€ I 3- " / co ,/2i \\\’).\16)\ l / X1 l | t 1O l I l I 1 1 1 > // )u.B/A _|< é‘#§ \ 4\ \ #8 \ _ _4_/X /k’ /Q \S \ _5_ ‘Q’ X Fig. 12-3. Theenergy levels oftheground state Fig. 12-4. Transitions between the levels of ofhydrogen inamagnetic ﬁeld B. ground state energy levels ofhydrogen insome pctrticulor mognetic ﬁeld B 12-11 energy E1Ialsostarts atAbutdecreases linearly with increasing B~its slope is -12. These twolevels vary with Basshown inFig. l2—3. Weshow also inthe ﬁgure theenergies E”; andE1;/. They have adillerent B-dependence. Forsmall B.theydepend quadratically onB,sotheystart outwithhorizontal slopes. Then they begin tocurve, andforlarge Bthey approach straight lines with slopes in’,which arenearly thesame astheslopes ofE1andE;1. Theshiftoftheenergy levels ofanatom duetoamagnetic ﬁeldiscalled the Zeeman effect. Wesaythatthecurves inFig.12-3show theZeeman splitting of theground state ofhydrogen. When there isnomagnetic ﬁeld, wegetjustone spectral linefrom thehyperﬁne structure ofhydrogen. Thetransitions between state IIV)andanyoneoftheothers occurs with theabsorption oremission ofa photon whose frequency 1420 megacycles isl/htimes theenergy dillerence 4A. When theatom isinamagnetic ﬁeld B,however, there aremany more lines. There canbetransitions between anytwoofthefour states. Soifwehave atoms inallfo11r states, energy canbeabsorbed—or emitted—in anyone ofthesix transitions shown bythevertical arrows inFig. 12-4. Many ofthese transitions canbeobserved bytheRabi molecular beam technique wedescribed inVolume ll, Section 35-3 (seeAppendix). What makes thetransitions go?Thetransitions willoccur ifyouapply asmall disturbing magnetic ﬁeld that varies with time (inaddition tothesteady strong ﬁeld B).It’sjustaswesawforavarying electric ﬁeld ontheammonia molecule. Only here, itisthemagnetic ﬁeld which couples with themagnetic moments and does thetrick. Butthetheory follows through inthesame waythatweworked itoutfortheammonia. Thetheory isthesimplest ifyoutake aperturbing mag- netic ﬁeldthatrotates inthexy-plane—although anyhorizontal oscillating ﬁeld willdo.When youputinthisperturbing ﬁeld asanadditional term intheHam- iltonian, yougetsolutions inwhich theamplitudes vary with time—as wefound fortheammonia molecule. Soyoucancalculate easily andaccurately theprob- ability ofatransition from onestate toanother. And youﬁndthat itallagrees with experiment. 12-5 Thestates inIamagnetic ﬁeld Wewould likenow todiscuss theshapes ofthecurves inFig. 12-3. Inthe ﬁrstplace, theenergies forlarge ﬁelds areeasytounderstand, andrather interesting. ForBlarge enough (namely for;.tB/.4 >>1)wecanneglect thelintheformulas of(12.37). Thefour energies become EI=A+#Bs Ell:/4_'I"Ba (12.38) E111‘= -14-1-I/B, EIV =-A —14'3- These aretheequations ofthefourstraight linesinFig.12-3. Wecanunderstand these energies physically inthefollowing way. Thenature ofthestationary states inazero ﬁeld isdetermined completely bytheinteraction ofthetwomagnetic moments. Themixtures ofthebasestates I+—)andI—+)inthestationary states IIII)andIIV)areduetothisinteraction. Inlarge external ﬁelds, however, theproton andelectron willbeinﬂuenced hardly atallbytheﬁeld oftheother; each willactasifitwere alone intheexternal ﬁeld. Then—as wehave seen many times—~the electron spin will beeither parallel tooropposite totheexternal magnetic ﬁeld. Suppose theelectron spinis“up”—that is,along theﬁeld: itsenergy willbe —;ucB. The proton canstillbeeither way. Iftheproton spin isalso “up,” its energy is—;.tl,B. Thesum ofthetwois—(;i,, +,u,,)B =;.iB. That isjust what weﬁndforE1——which isﬁne, because wearedescribing thestate I—I—+)=II). There isstillthesmall additional term A(now /2B>>A)which represents the interaction energy oftheproton andelectron when their spins areparallel. (We originally took Aaspositive because thetheory wespoke ofsays itshould be. andexperimentally itisindeed so.) Ontheother hand, theproton canhave its spindown. Then itsenergy intheexternal ﬁeld goes to—/.iI,B, soitandtheelectron have theenergy —(;.t,, —;ip)B =,a’B. Andtheinteraction energy becomes —A. 12-12 Thesumisjusttheenergy E111 in(12.38). Sothestate IIII)must forlarge ﬁelds become thestate I+—). Suppose nowtheelectron spinis“down.” Itsenergy intheexternal ﬁeldis ueB. Iftheproton isalso“down,” thetwotogether havetheenergy (ue—I—up)B = /.iB,plustheinteraction energy A—since their spins areparallel. That makes just theenergy E11in(12.38) andcorresponds tothestate I——)=II1)—which is nice. Finally iftheelectron is“down” andtheproton is“up,” wegettheenergy (ue—up)B —A(minus Afortheinteraction because thespins areopposite) which isjustE1;/. And thestate corresponds toI—+). “But, wait amomentl”, you areprobably saying, “The states IIII) and IIV)arenotthestates I+—)andI—-+); they aremixtures ofthetwo.” Well, only slightly. They areindeed mixtures forB=O,butwehave notyetﬁgured outwhat they areforlarge B.When weused theanalogies of(12.33) inourformu- lasofChapter 9togettheenergies ofthestationary states, wecould also have taken theamplitudes that gowith then1. They come from Eq.(9.23), which is ﬂ_E—'H22. as H11 Theratio a2/as is,ofcourse, justC2/C3. Plugging intheanalogous quantities from (12.33), weget §g=E+A—(l»..-#p)B C3 2A O1‘ cEA'13é=i%_, (12.39) where forEwearetousetheappropriate energy—either EH1orEIv.Forinstance. forstate IIII)wehave C2 ~ll/B ~T “W Soforlarge Bthestate IIII)hasC2>>C3;thestate becomes almost completely thestate I2)=I+-—). Similarly, ifweputElyinto (12.39) weget(C2/C3)1v <<1;forhighﬁelds state IIV)becomes justthestate I3)=I—+).Youseethat thecoefﬁcients inthelinear combinations ofourbasestates which make upthe stationary states depend onB.Thestate wecallI111)isa50-50 mixture ofI+—) andI-—I—)atverylowﬁelds, butshifts completely overtoI+—)athighﬁelds. Similarly, thestate IIV),which atlowﬁelds isalsoa50-50 mixture (with opposite signs) ofI+-)andI—-+),goesoverintothestate I—+)when thespins are uncoupled byastrong external ﬁeld. Wewould alsoliketocallyour attention particularly towhat happens at verylowmagnetic ﬁelds. There isoneenergy—at -3A—-which doesnotchange when youturnonasmall magnetic ﬁeld. Andthere isanother energy—at +A- which splits intothree diﬁerent energy levels when youturnonasmall magnetic ﬁeld. Forweak ﬁelds theenergies vary withBasshown inFig.12-5. Suppose thatwehave somehow selected abunch ofhydrogen atoms which allhave the energy —3A. Ifweputthem through aStern-Gerlach experiment—with ﬁelds thatarenottoostrong-we would ﬁndthattheyjustgostraight through. (Since their energy doesn’t depend onB,there is—according totheprinciple ofvirtual work—no force onthem inamagnetic ﬁeldgradient.) Suppose, ontheother hand, wewere toselect abunch ofatoms with theenergy +A, andputthem through aStern-Gerlach apparatus, sayanSapparatus. (Again theﬁelds intheapparatus should notbesogreat thattheydisrupt theinsides oftheatom, bywhich wemean aﬁeldsmall enough thattheenergies varylinearly withB.)Wewould ﬁndthree beams. Thestates II)andIII)getopposite forces—their energies vary linearly withBwiththeslopes insotheforces arelikethose onadipole with/2,=$1.1; butthestate IIII) goes straight through. Soweareright back inChapter 5. Ahydrogen atom withtheenergy —|—Aisaspin-one particle. Thisenergy state isa “particle” forwhich j=1,anditcanbedescribed—with respect tosome setof 12-13El NH" -013+ Q >- _3A moo Fig. 12-5. Thestates ofthehydrogen otom forsmoll magnetic ﬁelds. Table 12-3 Zero ﬁeldstates ofthehydrogen atom State Ij,m) j m Ournotation l1.+1> I110) I1, aO>I0 Ob-1F-1i—1+1|I>-111)= 1I11)= 11/)|+S> l05> IS>axes inspace—in terms ofthebase states I+S), I0S),andI—S) weused inChap- ter5.Ontheother hand, when ahydrogen atom hastheenergy —3A, itisaspin- zero particle. (Remember, what wearesaying isonly strictly trueforinﬁnitesimal magnetic ﬁelds.) Sowecangroup thestates ofhydrogen inzero magnetic ﬁeld thisway: II)=I++> I-l-5) _|+—>+|—+> I111)_ WI spinlI0s) (12.41) l11>=l——> I—$> Irv)=Iilxg-ll-+l spin0. (12.42) Wehave saidinChapter 35ofVolume ll(Appendix) thatforanyparticle its component ofangular momentum along anyaxiscanhave only certain values always hapart. Thez-component ofangular momentum J,canbejh,(j—1)h, (j—2)h,...,(—j)h, wherej isthespinoftheparticle (which canbeaninteger or half-integer). Although weneglected tosaysoatthetime, people usually write J,=mh, (12.43) where mstands foroneofthenumbersj,j -l,j—2,...,-j. You will. there- fore, seepeople inbooks label thefour ground states ofhydrogen bytheso-called quantum numbers jandm[often called the“total angular momentum quantum number” (j),and“magnetic quantum number” (m)]. Then. instead ofourstate symbols II),III),andsoon,they willwrite astate asIj,m).Sothey would write ourlittle table ofstates forzero ﬁeld in(12.41) and(12.42) asshown inTable 12-3. It’snotnew physics, it’salljustamatter ofnotation. 12-6 Theprojection matrix forspinoneT Wewould likenow touseourknowledge ofthehydrogen atom todosome- thing special. Wediscussed inChapter 5that aparticle ofspinonewhich wasin oneofthebase states (+,O,or—)with respect toaStern-Gerlach apparatus ofa particular orientation—say anSapparatus—would have acertain amplitude to beineach ofthethree states with respect toaTapparatus with adifferent orienta- tion inspace. There arenine such amplitudes (jTI iS)which make upthepro- jection matrix. InSection 5-7wegave without proof theterms ofthismatrix forvarious orientations ofTwith respect toS.Now wewillshow youoneway theycanbederived. lnthehydrogen atom wehave found aspin-one system which ismade up oftwospin one-half particles. Wehave already worked outinChapter 6how totransform thespinone-half amplitudes. Wecanusethisinformation tocalculate thetransformation forspin one. This istheway itworks: Wehave asystem—a hydrogen atom with theenergy +A—which hasspin one. Suppose werunit through aStern-Gerlach ﬁlter S,sothat weknow itisinoneofthebase states with respect toS,sayI+S). What istheamplitude that itwillbeinoneofthe base states, sayI—|-T), with respect totheTapparatus? lfwecallthecoordinate system oftheSapparatus thex,y,zsystem, theI+S) state iswhat wehave been calling thestate I—I——I—). Butsuppose another guytook hisz-axis along theaxis ofT.Hewillbereferring hisstates towhat wewillcallthex’,y’,z’frame. His “up” and“down” states fortheelectron andproton would bedifferent from ours. His“plus-plus” state-which wecanwrite I—I—’—I—’). referring tothe“prime” frame—is theI—I—T) state ofthespin-one particle. What wewant is(-1-TI +S) which isjustanother way ofwriting theamplitude (+’—I—’I—l-+). TThose whochose tojump overChapter 6should skipthissection also. 12-14 Wecanﬁndtheamplitude (+’—I—’I—I——I—)inthefollowing way. Inourframe theelectron intheI+—I—}state hasitsspin “up”. That means that ithassome amplitude (—I-’I—I-)6 ofbeing “up” inhisframe, andsome amplitude (—’|+),, ofbeing “down” inthat frame. Similarly, theproton inthe|++>state has spin “up” inourframe andtheamplitudes (+’I-I-)1, and (—’I+)p ofhaving spin“up” orspin “down” inthe“prime” frame. Since wearetalking about two distinct particles, theamplitude that both particles will be“up” /ogerher inhis frame istheproduct ofthetwoamplitudes, <+’+’I++>=<+'I+)<>(+’ I—I->1» (12-44) Wehave putthesubscripts eandpontheamplitudes (+’|+)tomake itclear what wewere doing. Butthey areboth justthetransformation amplitudes fora spinone-half particle, sothey arereally identical numbers. They are.infact, just theamplitude wehave called (+TI +S) inChapter 6,andwhich welisted in thetables attheendofthatchapter. Now, however, weareabout togetinto trouble with notation. Wehave to beable todistinguish theamplitude (—I—TI +S) foraspin one-half particle from what wehave alsocalled (+T |+S) foraspin-one particle——yet theyarecompletely diﬁerent! Wehope itwon’t betooconfusing, butfor themoment atleast, wewill have tousesome different symbols forthespin one-half amplitudes. Tohelp youkeep things straight, wesummarize thenewnotation inTable 12-4. Wewill continue tousethenotation I+S), [0S),and|—S) forthestates ofaspin-one particle. With ournewnotation. Eq.(12.44) becomes simply <+'+'|+ +>=<12. andthisisjust thespin-one amplitude (+T| +S). Now, let’s suppose, forin- stance. thattheother guy’s coordinate frame—that is,theT.or“primed,” appara- tus—is justrotated with respect toourz-axis bytheangle ¢;then from Table 6-2, a=<+'I+>=ew- Sofrom (12.44) wehave thatthespin-one amplitude is (-I-TI -I-S) =(-I-’—I—’I+—I—)=(cw/2)2 =cw. (12.45) You canseehow itgoes. Now wewillwork through thegeneral case forallthestates. Iftheproton andelectron areboth “up” inourframe theS-frame—the amplitudes that it willbeinanyoneofthefour possible states intheother guy’s frame—the T-frame- are (+, +1 I+ :<+’ I+>e<+’ I+>11 =(I2: <+’—’|+ +>=<+’I+>e<_I I+>p=~11 (1246) <-'+'|++>=<—’I+>@<+’|+>,, =bu. ' <_, _/ I—I— :<_’ I+>e<_’ I+);: =b2- Wecan, then, write thestate |—I——I—)asthefollowing linear combination: I++>=H2I+’+’>+11b{I+’ —’>+I—’+’>}+bzl—’—’>- (12-47) Now wenotice that I+'+’)isthestate I—I—T), that {I—I—’—’)+ —’—I—’)} is just\/2limes thestate I0T>—see (12.4l)—and that |—’—’) =|—T). Inother words, Eq.(12.47) canberewritten as |+s>=(12|+T>+\/2abIor)+1,2|-T). (12.48) Inasimilar wayyoucaneasily show that |~s>=C2|+T>+\/§cd]0T) +d2|—T). (12.49) 12-15Table 12-4 Spin one-half amplitudes This chapter Chapter 6 ~=<+’I+> <+TI+s>1>=<—'I+> <-r»+s>c=<+'!—> <+TI—$>d=<—'|-> <—T|—s> ForI0S)it’salittlemore complicated, because 1I05)=-—{I+ —>+I— +>}-\/5 Butwecanexpress each ofthestates I+—)andI—-+)interms ofthe“prime” states andtakethesum. That is, I—I——)=acI—I—’+’)+adI+’—’)—I—bcI—’—I—’)+bdI—’—’) (12.50) and I—+)=acI+’+’)+bcI+’——')+odI—'-1-’)+bdl—‘—’). (12.51) Taking 1/\/2times thesum, weget 2 ad—I- bc 2 5=’ﬂ +'+’+*——~ +'—'+ —’-I-"+'~bd —‘——’. I0>\/EQCI > ﬂ {I )I >1 \/5 I > ltfollows that I0s)=\/2“ac I+7‘)+(ad+bc)I0T)+\/21111] -T). (12.52) Wehave now alloftheamplitudes wewanted. Thecoefﬁcients ofEqs. (12.48), (12.49), and(12.52) arethematrix elements (jTIiS).Let’s pullthem all together: LS4 1'TI a2 \/2ac c2 (jTIiS)= \/fab ad+bc \/2ca’ (12.53) b2 \/2bd d2 Wehave expressed thespin-one transformation interms ofthespin one-half amplitudes a,b,0,andd. Forinstance, iftheT-frame isrotated withrespect toSbytheangle orabout they-axis—as inFig. 5—6—the amplitudes inTable 12-4 arejust thematrix elements ofR,,(a) inTable 6-2. QRa= cos— b= -sing, (12.54) E.D Q9= _ ,1= 9-‘, C COS 2 Using these in(12.53), wegettheformulas of(5.38), which wegavethere without proof. What everhappened tothestate IIV)?! Well, itisaspin-zero system, soit hasonly onestate—it isthesame inallcoordinate systems. Wecancheck that everything works outbytaking thediﬂerence ofEq.(12.50) and(12.51); weget that 1+->—1-+>=(44—1>c>1|+' -'>-1-~ +'>}- But(ad-—bc)isthedeterminant ofthespin one-half matrix, andsoisequal tol. Wegetthat IIV’)=IIV) foranyrelative orientation ofthetwocoordinate frames. 12-16 I3 Propagation inaCrystal Lattice 13-1 States foranelectron inaone-dimensional lattice Youwould, atﬁrstsight, think thatalow-energy electron would have great difficulty passing through asolid crystal. Theatoms arepacked together with theircenters only afewangstroms apart, andtheetlective diameter oftheatom forelectron scattering isroughly anangstrom orso.That is,theatoms arelarge, relative totheir spacing, sothat you would expect themean free path between C0lllSlOl'1S tobeoftheorder ofafewangstroms—which ispractically nothing. Youwould expect theelectron tobump intooneatom oranother almost imme- diately. Nevertheless, itisaubiquitous phenomenon ofnature thatifthelattice isperfect, theelectrons areabletotravel through thecrystal smoothly andeasily- almost asifthey were inavacuum. This strange factiswhat letsmetals conduct electricity soeasily; ithasalso permitted thedevelopment ofmany practical devices. Itis,forinstance, what makes itpossible foratransistor toimitate the radio tube. Inaradio tube electrons move freely through avacuum, while inthe transistor they move freely through acrystal lattice. Themachinery behind the behavior ofatransistor willbedescribed inthischapter; thenextonewilldescribe theapplication ofthese principles invarious practical devices. Theconduction ofelectrons inacrystal isoneexample ofavery common phenomenon. Notonly canelectrons travel through crystals, butother “things” like atomic excitations canalsotravel inasimilar manner. Sothephenomenon which wewant todiscuss appears inmany ways inthestudy ofthephysics ofthesolid state. You willremember thatwehave discussed many examples oftwo-state sys- tems. Let‘s nowthink ofanelectron which canbeineither oneoftwopositions, ineach ofwhich itisinthesame kind ofenvironment. Let’s alsosuppose that there isacertain amplitude togofrom oneposition totheother, and, ofcourse, thesame amplitude togoback, justaswehave discussed forthehydrogen molec- ularioninSection 10—l. Thelawsofquantum mechanics then givethefollowing results. There aretwopossible states ofdeﬁnite energy fortheelectron. Each state canbedescribed bytheamplitude fortheelectron tobeineach ofthetwo basic positions. Ineither ofthedeﬁnite-energy states, themagnitudes ofthese twoamplitudes areconstant intime, andthephases vary intime with thesame frequency. Ontheother hand, ifwestart theelectron inoneposition, itwilllater have moved totheother, andstilllater willswing back again totheﬁrstposition. Theamplitude isanalogous tothemotions oftwocoupled pendulums. Now consider aperfect crystal lattice inwhich weimagine thatanelectron canbesituated inakind of“pit” atoneparticular atom andwith some particular energy. Suppose alsothattheelectron hassome amplitude tomove intoadifferent pitatoneofthenearby atoms. Itissomething likethetwo-state system—but with anadditional complication. When theelectron arrives attheneighboring atom, itcanafterward move ontostillanother position aswellasreturn toitsstarting point. Now wehave asituation analogous nottotwocoupled pendulums, butto aninﬁnite number ofpendulums allcoupled together. Itissomething likewhat youseeinoneofthose machines—made with along rowofbarsmounted ona torsion w1re~that isused inﬁrst-year physics todemonstrate wave propagation. Ifyouhave aharmonic oscillator which iscoupled toanother harmonic oscillator, andthatonetoanother, andsoon...,andifyoustart anirregularity inoneplace, theirregularity willpropagate asawave along theline. Thesame situation exists ifyouplace anelectron atoneatom ofalong chain ofatoms. l3—l13-1 l3-2 13-3 13-4 13-5 13—6 13-7 13-8States foranelectron ina one-dimensional lattice States ofdeﬁnite energy Time-dependent states Anelectron inathree- dimensional lattice Other states inalattice Scattering byimperfections inthelattice Trapping byalattice imperfection Scattering amplitudes and bound states O @—rU O_L O O O O/Atom (0) O O r\-3 n-2 n-I nnel n+2 n+3 ~ Electron \/ (b)OOO OOOOO In-1> (C) O O O O O O O O I"> Id) O O O O O O O O In-H) Fig. 13-1. The base states ofon electron incione-dimensional crystcil.Usually, thesimplest wayofanalyzing themechanical problem isnottothink interms ofwhat happens ifapulse isstarted atadeﬁnite place, butrather in terms ofsteady-wave solutions. There exist certain patterns ofdisplacements which propagate through thecrystal asawave ofasingle, ﬁxed frequency. Now thesamethinghappens withtheelectron--and forthesamereason, because it’s described inquantum mechanics bysimilar equations. You must appreciate onething, however; theamplitude fortheelectron to beataplace isanamplitude, notaprobability. Iftheelectron were simply leaking from oneplace toanother, likewater going through ahole, thebehavior would becompletely different. Forexample, ifwehadtwotanks ofwater connected byatube topermit some leakage from onetotheother, then thelevels would approach eachother exponentially. Butfortheelectron, what happens isamplitude leakage andnotjustaplain probability leakage. And it’sacharacteristic ofthe imaginary term—the iinthedifferential equations ofquantum mechanics—which changes theexponential solution toanoscillatory solution. What happens then isquite different from theleakage between interconnected tanks. Wewant now toanalyze quantitatively thequantum mechanical situation. lmagine aone-dimensional system made ofalong lineofatoms asshown in Fig. 13-1(a). (Acrystal is,ofcourse, three-dimensional butthephysics isvery much thesame; once youunderstand theone-dimensional caseyouwillbeable tounderstand what happens inthree dimensions.) Next, wewant toseewhat happens ifweputasingle electron onthislineofatoms. Ofcourse, inarealcrystal there arealready millions ofelectrons. Butmost ofthem (nearly allforanin- sulating crystal) takeuppositions insome pattern ofmotion each around itsown atom—and everything isquite stationary. However, wenowwant tothink about what happens ifweputanextra electron in.Wewillnotconsider what theother ones aredoing because wesuppose thattochange their motion involves alotof excitation energy. Wearegoing toaddanelectron asiftoproduce oneslightly bound negative ion. Inwatching what theoneextra electron'does wearemaking anapproximation which disregards themechanics oftheinside workings ofthe atoms. Ofcourse theelectron could then move toanother atom, transferring the negative iontoanother place. Wewillsuppose thatjustasinthecase ofan electron jumping between twoprotons, theelectron canjump from oneatom to theneighbor oneither sidewith acertain amplitude. Now how dowedescribe such asystem? What willbereasonable base states? Ifyouremember what wedidwhen wehadonly twopossible positions, youcan guess howitwillgo.Suppose thatinourlineofatoms thespacings areallequal; andthatwenumber theatoms insequence, asshown inFig.13-1(a). Oneofthe base states isthattheelectron isatatom number 6,another base state isthatthe electron isatatom number 7,oratatom number 8,andsoon.Wecandescribe thenthbase state bysaying thattheelectron isatatom number n.Let’s saythat thisisthebase state In).Figure 13-l shows what wemean bythethree base states In—l), In), and In+l). Using these base states, anystate I4»)ofourone-dimensional crystal canbede- scribed bygiving alltheamplitudes (nI¢)thatthestate I¢)isinoneofthe base states—which means theamplitude thatitislocated atoneparticular atom. Then wecanwrite thestate I¢)asasuperposition ofthebase states |4>>=ZIn><4I¢>. (13.1) Next, wearegoing tosuppose that when theelectron isatoneatom, there isa certain amplitude thatitwillleaktotheatom oneither side. And we’ll takethe simplest case forwhich itcanonly leaktothenearest neighbors—to gettothe next-nearest neighbor. ithastogointwosteps. We’ll takethattheamplitudes for theelectron jump from oneatom tothenextisiA/h (perunittime). 13-2 Forthemoment wewould liketowrite theamplitude (nI¢)tobeonthe nthatom asC".Then Eq.(13.1) willbewritten I¢>=EIn)C,,. (13.2) Ifweknew each oftheamplitudes C,,atagiven moment, wecould take their absolute squares andgettheprobability thatyouwould ﬁndtheelectron ifyou looked atatom natthattime. What willthesituation beatsome later time‘? Byanalogy with thetwo-state systems wehave studied, wewould propose thattheHamiltonian equations for thissystem should bemade upofequations likethis: ih% =E0C,,(t) -AC,,+1(t) -AC,,_1(t). (13.3) Theﬁrstcoefﬁcient ontheright, E0,is,physically, theenergy theelectron would have ifitcouldn’t leak away from oneoftheatoms. (Itdoesn’t matter what wecallE0;aswehave seenmany times. itrepresents really nothing butour choice ofthezero ofenergy.) Thenext term represents theamplitude perunit time thattheelectron isleaking intothenthpitfrom the(n+l)stpit;andthe lastterm istheamplitude forleakage from the(n—l)stpit. Asusual, we’ll assume thatAisaconstant (independent of1). Forafulldescription ofthebehavior ofanystate I<1>),wewould have one equation like(13.3) forevery oneoftheamplitudes C,,.Since wewant toconsider acrystal with avery large number ofatoms, we’ll assume thatthere areanin- deﬁnitely large number ofstates-that theatoms goonforever inboth directions. (Todotheﬁnite case, wewillhave topayspecial attention towhat happens atthe ends.) Ifthenumber Nofourbase states isindeﬁnitely large, then alsoourfull Hamiltonian equations areinﬁnite innumber! We’ll write down justasample: . -¢ . ih£6’;-1 =E.,c.,_, -AC,,_2 -AC", ih11%=E0C,,-AC,,_1 -AC,,+1, (13.4) ih% =E0C,,+1 -AC,,-AC,,+2,. I 13-2 States ofdeﬁnite energy Wecould study many things about anelectron inalattice, butﬁrstlet’stry toﬁndthestates ofdeﬁnite energy. Aswehave seeninearlier chapters thismeans thatwehave toﬁndasituation inwhich theamplitudes allchange atthesame frequency iftheychange with time atall.Welook forsolutions oftheform Cn=a,,e_'E””. (13.5) Thecomplex number antellusabout thenon-time-varying partoftheamplitude toﬁndtheelectron atthenthatom. Ifweputthistrialsolution intotheequations of(13.4) totestthem out,wegettheresult Ea" =Egan —Aa,,+1 —Aan_1. (13.6) Wehave aninﬁnite number ofsuch equations fortheinﬁnite number ofunknowns a,,—which israther petrifying. Allwehave todoistake thedeterminant. ..butwait! Determinants are ﬁnewhen there are2,3,or4equations. Butifthere arealarge number—or an inﬁnite number——of equations, thedeterminants arenotvery convenient. We’d better justtrytosolve theequations directly. First, let’slabel theatoms bytheir 13-3 Fig. l3—2. Variation ofthereolport ofC,,withx,,.posizions; we’ll saythattheatom nisatx,.andtheatom (n+1)isatx,,+1. If theatomic spacing isb—as inFig. l3—l—we willhave that xn+1 =xn—I—b. Bychoosing ourorigin atatom zero, wecaneven have itthatx,.=nb.Wecan rewrite Eq.(13.5) as c..=a(x,.)e—’E”“, (13.7) andEq.(13.6) would become Ea(x") =E0a(x,,+1) —Aa(x,,+1) —~Aa(x,,_1). (13.8) Or,using thefactthatx,.+1 =xn+b,wecould alsowrite Ea(x,.) =E0a(x") -Aa(x,. +b)-—Aa(x,. —-b). (13.9) This equation issomewhat similar toadifferential equation. Ittellsusthata quantity, a(x), atonepoint, (xn), isrelated tothesame physical quantity atsome neighboring points, (xn=bb).(Adifferential equation relates thevalue ofafunc- tion atapoint tothevalues atinﬁnitesimally nearby points.) Perhaps themethods weusually useforsolving diﬂerential equations willalsowork here, let’stry. Linear diﬂ‘erent1al equations with constant coefﬁcients canalways besolved interms ofexponential functions. Wecantrythesame thing here; let’stakeasa trialsolution a(x,,)=e""~. (13.10) Then Eq.(13.9) becomes Eezkxn =Eoetkzn _Ae1k(:z,,-1-b)‘ _Ae17c(:::,,—b)' Wecannowdivide outthecommon factor e"”»; weget E=E0-AW’-A@—"°”. (13.12) Thelasttwoterms arejustequal to(2Acoskb),so E=E0—2Acoskb. (13.13) Wehave found thatforanychoice atallfortheconstant kthere isasolution whose energy isgiven bythisequation. There arevarious possible energies depending onk,andeach kcorresponds toadifferent solution. There arean inﬁnite number ofsolut1ons—which 1snotsurprising, since westarted outwith aninﬁnite number ofbasestates. Let’s seewhat these solutions mean. Foreach k,thea’saregiven byEq. (13.10). Theamplitudes C,,arethen given by Cn =etka:,,e—(1/?i)Et, where youshould remember that theenergy Ealso depends onkasgiven inEq. (l3.l3). The space dependence oftheamplitudes ise"“". The amplitudes oscillate aswegoalong from oneatom tothenext. Wemean that, inspace, theamplitude goes asacomplex oscillation-—the magnitude isthesame atevery atom, butthephase atagiven timeadvances bythe amount (ikb) from oneatom tothenext. Wecanvisualize what isgoing onby plotting avertical linetoshow justtherealpartateach atom aswehave done in Fig. l3—2. Theenvelope ofthese vertical lines (asshown bythebroken-line curve) Re(C> bl\\Q\\i T/x<f/9’/b I: 13-4 is,ofcourse, acosine curve. Theimaginary partofC,,isalsoanoscillating function, butisshifted 90°inphase sothattheabsolute square (which isthesum ofthe squares oftherealandimaginary parts) isthesame foralltheC’s. Thus ifwepickak,wegetastationary state ofaparticular energy E.And foranysuch state, theelectron isequally likely tobefound atevery atom—there isnopreference foroneatom ortheother. Only thephase isdifferent fordifferent atoms. Also, astime goes onthephases vary. From Eq.(13.14) therealand imaginary parts propagate along thecrystal aswaves——namely astherealor imaginary parts of e“’°’"-‘E"‘>'1. (13.15) Thewave cantravel toward positive ornegative xdepending onthesignwehave picked fork. Notice thatwehave been assuming thatthenumber kthatweputinour trialsolution, Eq.(13.10), wasarealnumber. Wecanseenowwhythatmust be soifwehave aninﬁnite lineofatoms. Suppose thatkwere animaginary number, sayik'.Then theamplitudes anwould goase'°"‘», which means thattheamplitude would getlarger andlarger aswegotoward large x’s—or toward large negative x’sifk’isanegative number. This kind ofsolution would beO.K. ifwewere dealing with lineofatoms thatended, butcannot beaphysical solution foran inﬁnite chain ofatoms. Itwould giveinﬁnite amplitudes—and, therefore, inﬁnite probabilities——which can’t represent arealsituation. Later onwewillseeanex- ample inwhich animaginary kdoes make sense. Therelation between theenergy Eandthewave number kasgiven inEq. (13.13) isplotted inFig.13—3. Asyoucanseefrom theﬁgure, theenergy cango from (E0—2A)atk=0to(E0+2A)atk==*=1r/b. Thegraph isplotted forpositive A;ifAwere negative, thecurve would simply beinverted, butthe range would bethesame. Thesigniﬁcant result isthatanyenergy ispossible within acertain range or“band” ofenergies, butnoothers. According toour assumptions, ifanelectron inacrystal isinastationary state, itcanhave no energy other than values inthisband. According toEq.(13.10), thesmallest k’scorrespond tolow-energy states—- Ez(E0—2A). Askincreases inmagnitude (toward either positive ornegative values) theenergy atﬁrstincreases, butthen reaches amaximum atk==I=1r/b, asshown inFig.13-3. Fork’slarger than 1r/b, theenergy would start todecrease again. Butwedonotreally need toconsider such values ofk,because they do notgivenewstates—they justrepeat states wealready have forsmaller k.We canseethatinthefollowing way. Consider thelowest energy state forwhich k=0.Thecoefficient a(x,,) isthesame forallx,,.Now wewould getthesame energy fork=211'/b. Butthen, using Eq.(13.10), wehave that a(xn) =e'i(21rIb):e,, However, taking x0tobeattheorigin, wecansetx,,=nb;then a(x,,) becomes a(x,,) =ei2'" =1. Thestate described bythese a(x,.) isphysically thesame state wegotfork=0. Itdoes notrepresent adiﬁerent solution. Asanother example, suppose thatkwere 1r/4b. Therealpartofa(x,.) would vary asshown bycurve linFig.13-4. Ifkwere seven times larger (k=71r/4), therealpartofa(x,.) would varyasshown bycurve 2intheﬁgure. (The complete ReA(xn) 2 I \ Q,,/,1-\1”‘/1/1Tux 1 \ X \.'‘JM-/‘fUU l3—51: ' 1 \13/E0—2AH1O 1______cr O.Q\U-_______ QY Fig. 13-3. Theenergy ofthestation- ary states asafunction oftheparam- eterk. Fig. 13-4. Two values ofkwhich represent thesame physical situation; curve lisfork=1r/4, curve 2isfor k=71r/4. cosine curves don’t mean anything, ofcourse; allthatmatters istheir values at thepoints x,,.Thecurves arejusttohelpyouseehowthings aregoing.) You see thatbothvalues ofkgivethesame amplitudes atallofthex..’s. Theupshot isthatwehave allthepossible solutions ofourproblem ifwetake only k'sinacertain limited range. We’ll pick therange between —1r/b and +1r/b——the oneshown inFig. 13-3. lnthisrange, theenergy ofthestationary states increases uniformly with anincrease inthemagnitude ofk. Onesideremark about something youcanplaywith. Suppose thattheelec- troncannot onlyjump tothenearest neighbor with amplitude 1A/h, butalsohas thepossibility tojump inonedirect leaptothenext nearest neighbor with some other amplitude z'B/h. You willﬁndthatthesolution canagain bewritten inthe form an=e"'”"—this type ofsolution isuniversal You willalsoﬁndthatthe stationary states withwave number khave anenergy equal to(E0—2Acoskb— 2BcosZkb). Thisshows thattheshape ofthe curve ofEagainst kisnotuniversal, butdepends upon theparticular assumptions oftheproblem. lt1Snotalways a cosine wave—it’s noteven necessarily symmetrical about some horizontal line. ltistrue, however, thatthecurve always repeats itself outside oftheinterval from —1r/b to1r/b, soyounever need toworry about other values ofk. Let’s look alittle more closely atwhat happens forsmall k—that is,when thevariations oftheamplitudes from onex,.tothenextarequite slow. Suppose wechoose ourzero ofenergy bydeﬁning E0=2A;then theminimum ofthe curve inFig.13-3 isatthezero ofenergy. Forsmall enough k,wecanwrite that coskb=1-k2b2/2, andtheenergy ofEq.(13.13) becomes E=Ak2b2. (13.16) Wehave thattheenergy ofthestate isproportional tothesquare ofthewave number which describes thespatial variations oftheamplitudes C,.. 13-3 Time-dependent states Inthissection wewould liketodiscuss thebehavior ofstates intheone- dimensional lattice inmore detail. Iftheamplitude foranelectron tobeatx,, isC,,,theprobability ofﬁnding itthere is|C,,|2.Forthestationary states described byEq.(13.12), thisprobability isthesame forallxnanddoes notchange withtime. How canwerepresent asituation which wewould describe roughly bysaying an electron ofacertain energy islocalized inacertain region—so thatitismore likely tobefound atoneplace than atsome other place? Wecandothatbymaking asuperposition ofseveral solutions likeEq.(13.12) with slightly different values ofk—and, therefore, slightly different energies. Then att=0,atleast, theampli- tude C0willvary with position because oftheinterference between thevarious terms, justasonegetsbeats when there isamixture ofwaves ofdifferent wave- lengths (aswediscussed inChapter 48,Vol.I).Sowecanmake upa“wave packet” withapredominant wave number k0,butwithvarious other wave numbers neark0.1' Inoursuperposition ofstationary states, theamplitudes with diﬁerent k’s willrepresent states ofslightly different energies, and,therefore, ofslightly different frequencies; theinterference pattern ofthetotal C0will, therefore, alsovary with time—there willbeapattern of“beats.” Aswehave seeninChapter 48ofVolume I,thepeaks ofthebeats [theplace where |C(x,,)|2 islarge] willmove along inx astime goes on;they move with thespeed wehave called the“group velocity." Wefound thatthisgroup velocity wasrelated tothevariation ofkwithfrequency by d vl;I‘Oll]') :ﬁi 1'Provided wedonottrytomake thepacket toonarrow. 13-6 thesame derivation would apply equally wellhere. Anelectron state which isa “clump“—namely oneforwhich theC0vary inspace likethewave packet of Fig.l3—5—will move along ourone-dimensional “crystal” with thespeed vequal todco/clk, where w=E/h. Using (13.16) forE,wegetthat 2 1»=iiik. (13.18) lnother words, theelectrons move along with aspeed proportional tothetypical k.Equation (13.16) then saysthattheenergy ofsuch anelectron isproportional tothesquare ofitsvelocity—it acts likeaclassical particle. Solong aswelook atthings onascale gross enough thatwedon’t seetheﬁnestructure, ourquantum mechanical picture begins togiveresults likeclassical physics. Infact,ifwesolve Eq.(13.18) forkandsubstitute into(13.16), wecanwrite E=%m0;; 112, where met;isaconstant. Theextra “energy ofmotion” oftheelectron inapacket depends onthevelocity just asforaclassical particle. The constant merr—called the“effective mass”——is given by h2 mm = ' Also notice that wecanwrite meff I)= lfwechoose tocallme“vthe“momentum,” itisrelated tothewave number k inthewaywehave described earlier forafreeparllCl6. Don’t forget thatmet;hasnothing todowith therealmass ofanelectron. Itmaybequite difi"erent——although inrealcrystals itoften happens toturnouttobe thesame general order ofmagnitude, about 2to20times thefree-space mass of theelectron. Wehave now explained aremarkable mystery—how anelectron inacrystal (like anextra electron putintogermanium) canrideright through thecrystal and flow perfectly freely even though ithastohitalltheatoms. Itdoes sobyhaving itsamplitudes going pip-pip-pip from oneatom tothenext, working itswaythrough thecrystal. That ishowasolid canconduct electricity. 13-4 Anelectron inathree-dimensional lattice Let’s look foramoment athowwecould apply thesame ideas toseewhat happens toanelectron inthree dimensions. Theresults turnouttobeverysimilar. Suppose wehave arectangular lattice ofatoms with lattice spacings ofa,b,cin thethree directions (lfyou want acubic lattice, takethethree spacings allequal.) Also suppose thattheamplitude toleapinthex-direction toaneighbor is(IA,/h), toleap inthey-direction is(tA,,/h), andtoleap inthez-direction is(iA,/ii). Now how should wedescribe thebase states? Asintheone-dimensional case, one base state isthattheelectron isattheatom whose locations arex,y,z,where (x,y,z)isoneofthe lattice points. Choosing ourorigin atoneatom, these points areallat x=n.a, y=n,,b. and 2=nzc. where n,,,ny,nzareanythree integers. Instead ofusing subscripts toindicate such points, wewillnowjust usex,y,andz,understanding thattheytakeononly their values atthelattice points. Thus thebase state isrepresented bythesymbol Ielectron atx,y,2),andtheamplitude foranelectron insome state I¢>tobein thisbase state isC(x.y,z)=(electron atx,y,zlit/). l3—7ReCixnl —#—> > X 1 l Fig. 13-5. Thereal part ofC(x,,) as afunction ofxforasuperposition of several states ofsimilar energy. (The spacing bisvery small onthescale of xshown.) Asbefore, theamplitudes C(x,y,z)may vary with time. With ourassump- tions, theHamiltonian equations should belikethis: ———i =E0C(x, y,z)—A,,C(x +a,y,z)—A,C(x —a,y,2) —A,,C(x,y -l-b,z)—A,,C(x,y —b,z) —A,C(x,y,z +c)—A,C(x, y,z—c). (13.22),,,dang)».Z) Itlooks rather long, butyoucanseewhere each term comes from. Again wecantrytoﬁndastationary state inwhich alltheC'svarywithtime inthesame way. Again thesolution isanexponential: C(x,y,z)=e"*”""e‘<’“=‘+’"-”+"==>. (13.23) Ifyousubstitute thisinto(13.22) youseethatitworks, provided thattheenergy Eisrelated tok,,k,,,andk,inthefollowing way: E=E0—2A,,coskza—2A,,cosk,,b—2A,cosk,c. (13.24) Theenergy nowdepends onthethree wave numbers k,,k,,,k,,which, incidentally, arethecomponents ofathree-dimensional vector k.Infact, wecanwrite Eq. (13.23) invector notation as C(x,y,Z)=e_'E””e_""' (13.25) Theamplitude varies asactmplex plane wave inthree dimensions, moving inthe direction ofk,andwith thewave number k=(kg+kf+k§)1/2. Theenergy associated with these stationary states depends onthethree com- ponents ofkinthecomplicated waygiven inEq.(13.24). Thenature, ofthe variation ofEwithkdepends onrelative signs andmagnitudes ofA,,A,,,andA,. Ifthese three numbers areallpositive, andifweareinterested insmall values of k,thedependence isrelatively simple. Expanding thecosines aswedidbefore togetEq.(13.16), wecannowgetthat E=Em...+A,a2kf +Aybkf+A,ckf. (13.26) Forasimple cubic lattice with lattice spacing aweexpect thatA,andA, andA,would beequal—say allarejustA—-and wewould have just E=Emin Ti“Aa2(kZ + + Or E=Em,“+Aa2k2. (13.27) This isjustlikeEq.(13.16). Following thearguments used there, wewould con- clude thatanelectron packet inthree dimensions (made upbysuperposing many states with nearly equal energies) alsomoves likeaclassical particle with some effective mass. Inacrystal with alower symmetry than cubic (oreven inacubic crystal in which thestate oftheelectron ateachatom isnotsymmetrical) thethree coefﬁcients A,,A0,andA,aredifferent. Then the“effective mass” ofanelectron localized inasmall region depends onitsdirection ofmotion. Itcould, forinstance, have a different inertia formotion inthex-direction than formotion inthey-direction. (The details ofsuch asituation aresometimes described interms ofan“effective mass tensor.”) 13-S Other states inlllattice According toEq.(13.24) theelectron states wehave been talking about can have energies only inacertain “band” ofenergies which covers theenergy range from theminimum energy E0—2(A, -1-A,+A,) 13-8 tothemaximum energy E0 + + Ag + Other energies arepossible, butthey belong toadifferent class ofelectron states. Forthestates Wehave described, weimagined base states inwhich anelectron is placed onanatom ofthecrystal insome particular state, saythelowest energy state. Ifyouhave anatom inempty space, andaddanelectron tomake anion,the ioncanbeformed inmany ways. Theelectron cangooninsuch away astomake thestate oflowest energy, oritcangoontomake oneoranother ofmany possible “excited states” oftheioneach with adeﬁnite energy above thelowest energy. The same thing canhappen inacrystal. Let’s suppose that theenergy E0wepicked above corresponds tobase states which areions ofthelowest possible energy. Wecould alsoimagine anewsetofbase states inwhich theelectron sitsnear the nthatom inadifferent way—in oneoftheexcited states oftheion—so that the energy E0isnow quite abithigher. Asbefore there issome amplitude A(different from before) that theelectron willjump from itsexcited state atoneatom tothe same excited state ataneighboring atom. The whole analysis goes asbefore, we ﬁndaband ofpossible energies centered atahigher energy. There can, ingeneral, bemany such bands each corresponding toadifferent level ofexcitation. There arealso other possibilities. There may besome amplitude thatthe electron jumps from anexcited condition atoneatom toanunexcited condition atthenext atom. (This iscalled aninteraction between bands.) Themathematical theory gets more andmore complicated asyou take into account more andmore bands andaddmore andmore coefﬁcients forleakage between thepossible states. Nonew ideas areinvolved, however; theequations aresetupmuch aswehave done inoursimple example. Weshould remark alsothatthere isnotmuch more tobesaidabout thevari- ouscoefficients, such astheamplitude A,which appear inthetheory. Generally they arevery hard tocalculate, soinpractical cases very little isknown theoretically about these parameters and forany particular real situation wecan only take values determined experimentally. There areother situations where thephysics andmathematics arealmost exactly likewhat wehave found foranelectron moving inacrystal, butinwhich the“object” thatmoves isquite different. Forinstance, suppose thatouroriginal crystal—-or rather linear lattice—was alineofneutral atoms, each with aloosely bound outer electron. Then imagine thatwewere toremove oneelectron. Which atom haslostitselectron? LetC,,nowrepresent theamplitude thattheelectron ismissing from theatom atx,,.There will, ingeneral, besome amplitude iA/h thattheelectron ataneighboring atom—say the(n—l)statom—will jump to thenthleaving the(n—l)statom without itselectron. This isthesame assaying that there isanamplitude Aforthe“missing e1ectron” tojump from thenth atom tothe(n—l)statom. You canseethattheequations willbeexactly the same—of course, thevalue ofAneed notbethesame aswehadbefore. Again wewillgetthesame formulas fortheenergy levels, forthe“waves” ofprobability which move through thecrystal with thegroup velocity ofEq.(13.18), forthe effective mass, andsoon.Only nowthewaves describe thebehavior ofthemissing electr0n—-or “hole” asitiscalled. Soa“hole” actsjustlikeaparticle with a certain mass meg. You canseethatthisparticle willappear tohave apositive charge. We’ll have some more tosayabout such holes inthenextchapter. Asanother example, wecanthink ofalineofidentical neutral atoms oneof which hasbeen putinto anexcited state—that is,with more than itsnormal ground state energy. LetC"betheamplitude thatthenthatom hastheexcitation. Itcaninteract with aneighboring atom byhanding over toittheextra energy and returning totheground state. Call theamplitude forthisprocess iA/h. You canseethatit’sthesame mathematics alloveragain. Now theobject which moves iscalled anexciton. Itbehaves likeaneutral “particle” moving through thecrystal, carrying theexcitation energy. Such motion maybeinvolved incertain biological 13-9 processes suchasvision, orphotosynthesis. Ithasbeen guessed thattheabsorption oflight intheretina produces an“exciton” which moves through some periodic structure (such asthelayers intherods wedescribed inChapter 36,Vol. 1;see Fig.36-5) tobeaccumulated atsome special station where theenergy isused to induce achemical reaction. 13-6 Scattering from imperfections inthelattice Wewant now toconsider thecaseofasingle electron inacrystal which is notperfect. Ourearlier analysis saysthatperfect crystals have perfect conductivity —that electrons cangoslipping through thecrystal, asinavacuum, without friction. Oneofthemost important things thatcanstopanelectron from going onforever isanimperfection orirregularity inthecrystal. Asanexample, suppose that somewhere inthecrystal there isamissing atom; orsuppose thatsomeone put onewrong atom atoneoftheatomic sites sothatthings there aredifferent than attheother atomic sites. Saytheenergy, E0ortheamplitude Acould bedifferent. How would wedescribe what happens then? Tobespeciﬁc, wewillreturn totheone-dimensional caseandwewillassume thatatom number “zero” isan“impurity” atom andhasadifferent value ofE0 than anyoftheother atoms. Let’s callthisenergy (E0+F).What happens? When anelectron arrives atatom “zero” there issome probability thattheelectron isscattered backwards. Ifawave packet ismoving along anditreaches aplace where things arealittle bitdifferent, some ofitwillcontinue onward andsome of itwillbounce back. It’squite difficult toanalyze such asituation using awave packet, because everything varies intime. Itismuch easier towork with steady- state solutions. Sowewillwork with stationary states, which wewillﬁndcanbe made upofcontinuous waves which have transmitted andreﬂected parts. In three dimensions wewould callthereﬂected part thescattered wave, since it would spread outinvarious directions. Westart outwithasetofequations which arejustliketheones inEq.(13.6) except thattheequation forn=0isdifferent from alltherest. Theﬁveequations forn=-2,-1,0,+1,and+2look likethis: Ea_2 =E0a_2 —Aa_1 —Aa_3, Ea_1 =E0a1_ —Aa0 —Aa_2, Ea0=(E0+F)a0 —Aal—-Aa_1, (13.28) Ea, =E0a1 —A02 —Aa0, Eaz =E002 —Aa3 —Aal, There are,ofcourse, alltheother equations for|n]isgreater than 2.They will lookjustlikeEq.(13.16). Forthegeneral case, wereally ought touseadifferent Afortheamplitude thattheelectron jumps toorfrom atom “zero,” butthemain features ofwhat goes onwillcome outofasimpliﬁed example inwhich alltheA’sareequal. Equation (13.10) would stillwork asasolution foralloftheequations except theoneforatom “zero”—it isn’tright forthatoneequation. Weneed adifferent solution which wecancook upinthefollowing way. Equation (13.10) represents awave going inthepositive x-direction. Awave going inthenegative x-direction would have been anequally good solution. Itwould bewritten a(x,,) =e_"””". Themost general solution wecould have taken forEq.(13.6) would beacom- 13-10 bination ofaforward andabackward wave, namely ti,=a@“"‘~+13¢-‘W (13.29) Thissolution represents acomplex wave ofamplitude ozmoving inthe-1—x-direction andawave ofamplitude Bmoving inthe—x-direction. Now take alook atthesetofequations forournewproblem—the ones in (13.28) together with those foralltheother atoms. Theequations involving a,,’swith n31areallsatisﬁed byEq.(13.29), with thecondition thatkisrelated toEandthelattice spacing bby E=E0-2Acoskb. (13.30) The physical meaning isan“incident” wave ofamplitude aapproaching atom “zero” (the“scatterer”) from theleft,anda“scattered” or“reﬂected” wave of amplitude 6going back toward theleft. Wedonotloose anygenerality ifweset theamplitude atoftheincident wave equal to1.Then theamplitude Bis,in general, acomplex number. Wecansayallthesame things about thesolutions ofa,,fornZ1.The coefficients could bedifferent, sowewould have forthem 11,,='Ye'k’°"+5e_’k'", for n31. (13.31) Here, ’Yistheamplitude ofawave going totheright and5awave coming from theright. Wewant toconsider thephysical situation inwhich awave isoriginally started only from theleft, andthere isonly a“transmitted” wave that comes out beyond thescatterer—or impurity atom. Wewilltryforasolution inwhich 5=0.Wecan, certainly, satisfy alloftheequations forthea,,except forthe middle three inEq.(13.28) bythefollowing trialsolutions. 0,,(forn <0)=em" +6e_""", (13.32) an(forn >0)=“/e””'-. Thesituation wearetalking about isillustrated inFig.13-6. Byusing theformulas inEq.(13.32) fora_1anda+,, thethree middle equa- tions ofEq.(13.28) willallow ustosolve fora0andalsoforthetwocoefficients BandV.Sowehave found acomplete solution. Setting x,,=nb,wehave tosolve thethree equations (E_E0){etk(—b) +ﬁe-tk(-12)} =_A{a0 +etk(-2b) +ﬁe—tk(-212)}, (E-E0-F)a0=—A{Ve"°b +6”“-‘*1 +be-"‘<""’}, (13.33) (E-E0)v@“"' =-A{v@""<“’> +U0}. Remember thatEisgiven interms ofkbyEq.(13.30). Ifyousubstitute this value forEinto theequations, andremember that cosx=%(e”” +e_"), you getfrom theﬁrstequation that a0=1-1-5; (13.34) andfrom thethird equation that a0='Y. (13.35) These areconsistent only if V=1+B (13.36) This equation says that thetransmitted wave (7)isjusttheoriginal incident wave (1)with anadded wave (B)equal tothereﬂected wave. This isnotalways true, buthappens tobesoforascattering atoneatom only. lfthere were aclump of impurity atoms, theamount added totheforward wave would notnecessarily bethesame asthereﬂected wave. 13-llSCATTERED WAVE O I O I oI-----4 l\I OI aBA RANSMITTED WAVE INCIDENT WAVE F1—>-4 -3 -2 —l Fig. 13-6. Waves inaone-dimen- sional lattice with one "impurity" atom atn=O. PROBABILITY 2+2Kx 2-2KxC6 /\ C8\v/ \/ \ 8 § III]\\ EIZIZII1wI / / \p’, i F\\ " -4 T“m ex impurity Atom 0000I’/0 000 n--4-3-2-io1234 Fig. 13-7. The relative probabilities ofﬁnding atrapped electron atatomic sites near thetrapping impurity atom.Wecangettheamplitude Bofthereﬂected wave from themiddle equation ofEq.(13.33); weﬁndthat —F B- ' <13-37) Wehave thecomplete solution forthelattice with oneunusual atom. You may bewondering how thetransmitted wave canbe“more” than the incident wave asitappears inEq.(13.34). Remember, though, that5and3'are complex numbers andthatthenumber ofparticles (orrather, theprobability of ﬁnding aparticle) inawave isproportional totheabsolute square ofthe amplitude. Infact, there willbe“conservation ofe1ectrons” only if |t3|2+[312=1. (13.38) You canshow thatthisistrueforoursolution. 13-7 Trapping byalattice imperfection There isanother interesting situation thatcanarise ifFisanegative number. Iftheenergy oftheelectron islower attheimpurity atom (atn=O)than itis anywhere else,thentheelectron cangetcaught onthisatom. That is,if(E0+F) isbelow thebottom oftheband at(E0—2A),thentheelectron canget“trapped” inastate with E<E0—2A.Such asolution cannot come outofwhat wehave done sofar. Wecangetthissolution, however, ifwepermit thetrial solution we took inEq.(13.15) tohave animaginary number fork.Let’s setk=ixAgain, we canhave different solutions forn<0andforn>0.Apossible solution for n<Omight be an(forn <O)=ce+"I". (13.39) Wehave totake aplus sign intheexponent; otherwise theamplitude would get indeﬁnitely large forlarge negative values ofn.Similarly, apossible solution for n>0would be an(forn >O)=c’e“"”". (13.40) Ifweputthese trial solutions intoEq.(13.28) allbutthemiddle three are satisﬁed provided that E=E0-A(@*”+e—"b). (13.41) Since thesum ofthetwo exponential terms isalways greater than 2,thisenergy isbelow theregular band, andiswhat wearelooking for. Theremaining three equations inEq.(13.28) aresatisﬁed ifc=c’andifKischosen sothat A(e"b-e_"b)=—F. (13.42) Combining thisequation with Eq.(13.41) wecanﬁndtheenergy ofthetrapped electron; weget E=E0-\/4,12 +F2. (13.43) The trapped electron hasaunique energy—located somewhat below thecon- duction band. Notice thattheamplitudes wehave inEq.(13.39) and(13.40) donotsaythat thetrapped electron sitsright ontheimpurity atom. The probability ofﬁnding theelectron atnearby atoms isgiven bythesquare ofthese amplitudes. Forone particular choice oftheparameters itmight vary asshown inthebargraph of Fig. 13-7. Theprobability isgreatest forﬁnding theelectron ontheimpurity atom. Fornearby atoms theprobability drops oﬂexponentially with thedistance from theimpurity atom. This isanother example of“barrier penetration.” From thepoint-of-view ofclassical physics theelectron doesn’t have enough energy to getaway from theenergy “hole” atthetrapping center. Butquantum mechanically itcanleak outalittle way. 13-12 13-8 Scattering amplitudes andbound states Finally, ourexample canbeused toillustrate apoint which isvery useful these daysinthephysics ofhigh-energy particles. Ithastodowitharelationship between scattering amplitudes andbound states. Suppose wehave discovered- through experiment andtheoretical ana1ysis—the way that pions scatter from protons. Then anew particle isdiscovered andsomeone wonders whether maybe itisjustacombination ofapion andaproton heldtogether insome bound state (inananalogy tothewayanelectron isbound toaproton tomake ahydrogen atom). Byabound state wemean acombination which hasalower energy than thetwofree-particles. There isageneral theory which says that abound state willexist atthat energy atwhich thescattering amplitude becomes inﬁnite ifextrapolated alge- braically (themathematical term is“analytically continued") toenergy regions outside ofthepermitted band. Thephysical reason forthisisasfollows. Abound state isasituation in which there areonly waves tiedontoapoint andthere’s nowave coming intoget itstarted, itjustexists there byitself. Therelative proportion between theso-called “scattered” orcreated wave andthewave being “sent in”isinﬁnite. Wecantest thisideainourexample. Let’s write ourexpression Eq.(13.37) forthescattered amplitude directly interms oftheenergy Eoftheparticle being scattered (instead ofinterms ofk).Since Equation (13.30) canberewritten as 2Asinkb=\/4.42 —(E-—E0)?, thescattered amplitude is F__I-\ﬂi,4-1 __(E_E0)2 (13.44) From ourderivation, thisequation should beused onlyforrealstates—those with energies intheenergy band, E=E0=2A.Butsuppose weforget thatfactand extend theformula intothe“unphysical” energy regions where |E—E01>2A. Forthese unphysical regions wecanwritei \/4/12 -(E-E0)2=ix/(E-E0)?-4,12. Then the"scattering amplitude,” whatever itmaymean, is t3=T—%2 (13.45)F+\/(E E0)-4,42 Now weask:Isthere anyenergy Eforwhich Bbecomes inﬁnite (i.e.,forwhich the expression forBhasa“pole”)? Yes,solong asFisnegative, thedenominator of Eq(13.45) willbezerowhen (E-E0)2-4/12=F2, E=E01 \/4/11 +25. Theminus signgives justtheenergy wefound inEq.(13.43) forthetrapped energy. What about theplus sign? This gives anenergy above theallowed energy band. And indeed there isanother bound state there which wemissed when we solved theequations ofEq.(13.28). Weleave itasapuzzle foryoutoﬁndthe energy andamplitudes anforthisbound state. Therelation between scattering andbound states provides oneofthemost useful clues inthecurrent search foranunderstanding oftheexperimental ob- servations about thenewstrange particles.orwhen TThe signoftheroottobechosen hereisatechnical point related totheallowed signs ofKinEqs. (13.39) and(1340). Wewon’t gointoithere. 13-13 I4 Semiconductors 14-1 Electrons andholes insemiconductors One oftheremarkable anddramatic developments inrecent years hasbeen theapplication ofsolid state science totechnical developments inelectrical devices suchastransistors. Thestudy ofsemiconductors ledtothediscovery oftheir useful properties and toalarge number ofpractical applications. The ﬁeld is changing sorapidly thatwhat wetellyoutoday may beincorrect nextyear. lt willcertainly beincomplete. And itisperfectly clear that with thecontinuing study ofthese materials many newandmore wonderful things willbepossible astime goes on. You will notneed tounderstand thischapter forwhat comes laterinthisvolume, butyoumay ﬁnditinteresting toseethatatleast something ofwhat youarelearning hassome relation tothepractical world. There arelarge numbers ofsemiconductors known, butwe’ll concentrate onthose which now have thegreatest technical application. They arealso the onesthatarebestunderstood, andinunderstanding them weWlllobtain adegree ofunderstanding ofmany oftheothers. The semiconductor substances inmost common usetoday aresilicon andgermanium. These elements crystallize inthe diamond lattice, akind ofcubic structure inwhich theatoms have tetrahedral bonding withtheir fournearest neighbors. They areinsulators atverylowtempera- tures——near absolute zero—although they doconduct electricity somewhat at room temperature. They arenotmetals; theyarecalled semiconductors. lfwesomehow putanextra electron into acrystal ofsilicon orgermanium which isatalowtemperature, wewillhavejustthesituation wedescribed inthe lastchapter. The electron willbeable towander around inthecrystal jumping from oneatomic sitetothenext. Actually, wehave looked only atthebehavior ofelectrons inarectangular lattice, andtheequations would besomewhat different forthereallattice ofsilicon orgermanium. Allofthe essential points are,however. illustrated bytheresults fortherectangular lattice. AswesawinChapter 13.these electrons canhave energies only inacertain energy band—-called theconduction band. Within thisband theenergy isrelated tothewave-nuniber koftheprobability amplitude C(seeEq.13.24) by E=E0—2A,cosk.a—2A,,cosk,,h-2Acoskzc. (14.1) TheA’saretheamplitudes forjumping inthex-,y-,andz-directions, andu,h. andcarethelattice spacings inthese directions. Forenergies near thebottom oftheband, wecanapproximate Eq.(14.1) by E2 +/1,tﬁ/<3 +/1,112/<3 +,4.H1<;’ (14.2) (seeSection 13-4) lfwethink ofelectron motion insome particular direction, sothat thecom- ponents ofkarealways inthesame ratio, theenergy isaquadratic function of thewave number—and aswehave seen ofthemomentum oftheelectron. We canwrite E=E,,,,,, -1-otl<2, (14.3) where aissome constant. andwecanmake agraph ofEversus kasinFig. 14-1. We'll callsuch agraph an"energy diagram.” Anelectron inaparticular state of energy andmomentum canbeindicated byapoint such asSintheﬁgure I4-l14-1 Electrons andholes in semiconductors 14-2 Impure semiconductors 14-3 TheHall effect 14-4 Semiconductor junctions 14-5 Rectiﬁcation ata semiconductor junction 14-6 Thetransistor Reference‘ C.Kittcl. Introduction to So/id State Phys/cs", Chapters I3,14,and 18. ‘E s _—T——__ T—TTTT—Emin > lt Fig. 14-1. The energy diagram for anelectron inaninsulating crystal. Aswealsomentioned inChapter 13,wecanhave asimilar situation ifwe remove anelectron from aneutral insulator. Then, anelectron canjump over from anearby atom andﬁllthe“hole,” butleaving another “hole” attheatom it started from. Wecandescribe thisbehavior bywriting anamplitude toﬁndthe holeatanyparticular atom, andbysaying thattheholecanjump from oneatom to thenext. (Clearly, theamplitudes AthattheholejUn'|pS from atom atoatom b isjustthesame astheamplitude thatanelectron onatom bjumps intothehole atatom a.)Themathematics isjustthesame fortheholeasitwasfortheextra electron, andwegetagain thattheenergy oftheholeisrelated toitswave number byanequation justlikeEq.(14.1) or(14.2), except, ofcourse, with dillerent nu- merical values fortheamplitudes A,,A1,,andAZ.Theholehasanenergy related tothewave number ofitsprobability amplitudes. Itsenergy liesinarestricted band, andnear thebottom oftheband itsenergy varies quadratically with the wave number—or momentum——just asinFig.14—1. Following thearguments of Section 13-3, wewould ﬁndthat theholealsobehaves like<1Cll1S.SlC(1l particle with acertain elfective mass—except thatinnoncubic crystals themass depends onthedirection ofmotion. Sotheholebehaves likeap0S'lllV6’ particle moving through thecrystal. Thecharge ofthehole-particle ispositive, because itislocated atthesiteofamissing electron: andwhen itmoves inonedirection there areac- tually electrons moving intheopposite direction. Ifweputseveral electrons intoaneutral crystal, theywillmove around much liketheatoms ofalow-pressure gas. Ifthere arenottoomany, their interactions willnotbeveryimportant. Ifwethen putanelectric ﬁeldacross thecrystal, the electrons willstart tomove andanelectric current willﬂow. Eventually theywould allbedrawn tooneedge ofthecrystal, and, ifthere isametal electrode there, theywould becollected, leaving thecrystal neutral. Similarly wecould putmany holes intoacrystal. They would roam around atrandom unless there isanelectric ﬁeld. With aﬁeld they would ﬂow toward thenegative terminal, andwould be“col1ected"——what actually happens isthat theyareneutralized byelectrons fromthemetal terminal. Onecanalsohave both holes andelectrons together. Ifthere arenottoo many, they willallgotheir wayindependently. With anelectric ﬁeld, they will allcontribute tothecurrent. Forobvious reasons, electrons arecalled thenegative carriers andtheholes arecalled thepositive carriers. Wehave sofarconsidered thatelectrons areputintothecrystal from the outside, orareremoved tomake ahole. Itisalsopossible to“create” anelectron- holepairbytaking abound electron away from oneneutral atom andputting it some distance away inthesame crystal. Wethenhave afreeelectron andafree hole, andthetwocanmove about aswehave described. Theenergy required toputanelectron intoastate S—we sayto“create” thestate S—is theenergy ETshown inFig.14-2. Itissome energy above E;,,,. Theenergy required to“create” aholeinsome state S’istheenergy ETofFig. 14-3, which issome energy greater than E,tn. Now ifwecreate apairinthestates SandS’,theenergy required isjustET+ET. ‘E —————— ==—~E ‘——‘__ '—Eln|ﬂ I E. 1, I-E- It k Fig 14-2 Theenergy E“isrequired Fig. l4—3. Theenergy El’isrequired tocreate afree electron to“create” cihole inthestate S’. 14-2 Thecreation ofpairs isacommon process (aswewillseelater), somany people liketoputFig.14-2 andFig.14-3 together onthesame graph—with the holeenergy plotted downward, although itis,ofcourse apositive energy. Wehave combined ourtwographs inthiswayinFig. 14-4. Theadvantage ofsuch a graph isthat theenergy Emu, =ET-1-ETrequired tocreate apair with the electron inSandthehole inS’isjustthevertical distance between SandS’as shown inFig. 14-4. The minimum energy required tocreate apair iscalled the “gap” energy andisequal toE,§,,, +E,T,,,. Sometimes youwillseeasimpler diagram called anenergy leveldiagram which isdrawn when people arenotinterested inthekvariable. Such adiagram—shown inFig. l4—5—just shows thepossible energies fortheelectrons andholes.1' How canelectron-hole pairs becreated” There areseveral ways. Forex- ample, photons oflight (orx-rays) canbeabsorbed andcreate apairifthephoton energy isabove theenergy ofthegap. The rate atwhich pairs areproduced is proportional tothelight intensity. Iftwoelectrodes areplated onawafer ofthe crystal anda“bias” voltage isapplied, theelectrons andholes willbedrawn to theelectrodes. Thecircuit current willbeproportional totheintensity ofthelight. This mechanism isresponsible forthephenomenon ofphotoconductivity andthe operation ofphotoconductive cells. Electron hole pairs canalso beproduced byhigh-energy particles. When a fast-moving charged particle—-for instance, aproton orapion with anenergy of tensorhundreds ofMev—goes through acrystal, itselectric ﬁeldwillknock elec- trons outoftheir bound states creating electron-hole pairs. Such events occur hundreds ofthousands oftimes permillimeter oftrack. After thepassage ofthe particle, thecarriers canbecollected andindoing sowillgiveanelectrical pulse. This isthemechanism atplay inthesemiconductor counters recently puttouse forexperiments innuclear physics. Such counters donotrequire semiconductors; theycanalso bemade with crystalline insulators. Infact, theﬁrstofsuch counters wasmade using adiamond crystal which isaninsulator atroom temperature. Very pure crystals arerequired iftheholes and electrons aretobeable tomove freely totheelectrodes without being trapped. Thesemiconductors silicon and germanium areused because they canbeproduced with high purity inreasonable large sizes (centimeter dimensions). Sofarwehave been concerned with semiconductor crystals attemperatures nearabsolute zero. Atanyﬁnite temperature there isstillanother mechanism by which electron-hole pairs canbecreated. The pair energy canbeprovided from thethermal energy ofthecrystal. Thethermal vibrations ofthecrystal cantransfer their energy toapair-—giving riseto“spontaneous” creation. Theprobability perunittime thattheenergy aslarge asthegapenergy Em, willbeconcentrated atoneatomic siteisproportional toe_EK“P/“T, where Tisthe temperature andKisBoltzmann’s constant (seeChapter 40,Vol.I).Near absolute zero there isnoappreciable probability, butasthetemperature rises there is anincreasing probability ofproducing such pairs. Atanyﬁnite temperature the production should continue forever ataconstant rate giving more and more negative andpositive carriers. Ofcourse that does nothappen because after awhile theelectrons andholes accidentally ﬁndeach other—the electron drops into thehole andtheexcess energy isgiven tothelattice. Wesaythat theelectron andhole“annihilate.” There isacertain probability persecond thatahole meets anelectron andthetwothings annihilate each other. Ifthenumber ofelectrons perunitvolume isN"(nfornegative carriers) andthedensity ofpositive carriers isNp,thechance perunit time that anelectron andaholewillﬁndeach other andannihilate isproportional totheproduct N,,N,,. Inequilibrium thisratemust equal theratethatpairs arecreated. You seethatin 1'Inmany books thissame energy diagram isinterpreted inadifferent way. Theenergy scale refers only toelectrons. Instead ofthinking oftheenergy ofthehole, theythink of theenergy anelectron would have ifitﬁlled thehole. This energy islower than thefree- electron energy—in fact, justtheamount lower thatyouseeinFig. 14-5. With this interpretation oftheenergy scale, thegapenergy istheminimum energy which must be given toanelectron tomove itfrom itsbound state totheconduction band. 14-3A ELECTRONE(electron) S Emlﬂ EDOIF + —————— ———— —-— Emin HOLE VS. E+(holo)It (Positive energy downward) Fig. 14-4. Energy diagrams for an electron andahole drawn together. 1E(electron) ELECTRONCONDUCTIONBAND EQQP <——%‘§ % ////// HOLE CONDUCTION BAND Fig. l4-5. Energy level diagram for electrons and holes.E / E"(ho|e)'/2’/STATE s Emin + ﬂ'\ll1 TATE s’ equilibrium theproduct ofN,,andN,should begiven bysome constant times the Boltzmann factor: N,,N,, =const e_E“““""T. (l44) When wesayconstant, wemean nearly constant. Amore complete theory-which includes more details about how holes andelectrons “ﬁnd” each other~shows that the“constant” isslightly dependent upon temperature, butthemajor de- pendence ontemperature isintheexponential.- Let’s consider, asanexample, apure material which isoriginally neutral. Ataﬁnite temperature youwould expect thenumber ofpositive andnegative carriers tobeequal, N"=Np.Then each ofthem should vary with temperature ase_El=*P/ “T. Thevariation ofmany oftheproperties ofasuperconductor—the conductivity forexample—is mainly determined bytheexponential factor because alltheother factors vary much more slowly withtemperature. Thegapenergy for germanium isabout 0.72evandforsilicon l.lev. Atroom temperature KTisabout 1/40 ofanelectron volt. Atthese tempera- tures there areenough holes andelectrons togiveasigniﬁcant conductivity, while at,say,30°K~—one-tenth ofroom temperature—the conductivity isimperceptible. Thegapenergy ofdiamond is6or7evanddiamond isagood insulator atroom temperature. 14-2 Impure semiconductors Sofarwehave talked about twoways thatextra electrons canbeputintoan otherwise ideally perfect crystal lattice. Onewaywastoinject theelectron from anoutside source; theother way, wastoknock abound electron offaneutral atom creating simultaneously anelectron andahole. Itispossible toputelectrons intotheconduction band ofacrystal instillanother way. Suppose weimagine a crystal ofgermanium inwhich oneofthegermanium atoms isreplaced byan arsenic atom. Thegermanium atoms have avalence of4andthecrystal structure iscontrolled bythefour valence electrons. Arsenic, ontheother hand, hasa valence of5.Itturns outthatasingle arsenic atom cansitinthegermanium lattice (because ithasapproximately thecorrect size), butindoing soitmust actasa valence 4atom—-using fourofitsvalence electrons toform thecrystal bonds and having oneelectron leftover. This extra electron isvery loosely attached—the binding energy islessthan l/10ofavolt. Atroom temperature theelectron easily picks upthatmuch energy from thethermal energy ofthecrystal, andthen takes offonitsown—moving about inthelattice asafreeelectron. Animpurity atom such asthearsenic iscalled adonor sitebecause itcangiveupanegative carrier tothecrystal. Ifacrystal ofgermanium isgrown from amelttowhich averysmall amount ofarsenic hasbeen added, thearsenic donor sites willbedistributed throughout thecrystal andthecrystal willhave acertain density ofnegative carriers built in. Youmight think thatthese carriers would getswept away assoon asanysmall electric ﬁeldwasputacross thecrystal. This willnothappen, however, because thearsenic atoms inthebody ofthecrystal eachhave apositive charge. Ifthebody ofthecrystal istoremain neutral, theaverage density ofnegative carrier electrons must beequal tothedensity ofdonor sites. Ifyouputtwoelectrodes ontheedges ofsuch acrystal andconnect them toabattery, acurrent willﬂow; butasthe carrier electrons areswept outatoneend, new conduction electrons must be introduced from theelectrode ontheother endsothattheaverage density of conduction electrons isleftverynearly equal tothedensity ofdonor sites. Since thedonor sites arepositively charged, there willbesome tendency for them tocapture some oftheconduction electrons asthey diffuse around inside thecrystal. Adonor sitecan,therefore, actasatrapsuch asthose wediscussed inthelastsection. Butifthetrapping energy issutﬁciently small—as itisforarsenic -—-the number ofcarriers which aretrapped atanyonetime isasmall fraction ofthetotal. Foracomplete understanding ofthebehavior ofsemiconductors 14-4 onemust takeintoaccount thistrapping. Fortherestofourdiscussion, however, wewillassume thatthetrapping energy issufficiently lowandthetemperature is sufficiently high, thatallofthedonor siteshave given uptheir electrons. This is, ofcourse, justanapproximation. Itisalso possible tobuild into agermanium crystal some impurity atom whose valence is3,such asaluminum. Thealuminum atom tries toactasa valence 4object bystealing anextra electron. Itcansteal anelectron from some nearby germanium atom andendupasanegatively charged atom withaneffective valence of4.Ofcourse, when itsteals theelectron from agermanium atom, it leaves ahole there; andthishole canwander around inthecrystal asapositive carrier. Animpurity atom which canproduce ahole inthisway iscalled an acceptor because it“accepts” anelectron. Ifagermanium orasilicon crystal is grown from amelt towhich asmall amount ofaluminum impurity hasbeen added, thecrystal willhave built-in acertain density ofholes which canactas positive carriers. When adonor oranacceptor impurity isadded toasemiconductor, wesay thatthematerial hasbeen “doped.” When agermanium crystal with some built-in donor impurities isatroom temperature, some conduction electrons arecontributed bythethermally induced electron-hole paircreation aswellasbythedonor sites. Theelectrons from both sources are,naturally, equivalent, anditisthetotal number N"which comes into playinthestatistical processes thatleadtoequilibrium. Ifthetemperature isnot toolow,thenumber ofnegative carriers contributed bythedonor impurity atoms isroughly equal tothenumber ofimpurity atoms present. Inequilibrium Eq. (14.4) must stillbevalid; atagiven temperature theproduct NHNI, isdetermined. This means thatifweaddsome donor impurity which increases N,,,thenumber N,ofpositive carriers willhave todecrease bysuch anamount that N,N,, is unchanged. Iftheimpurity concentration ishigh enough, thenumber N,,ofnega- tivecarriers isdetermined bythenumber ofdonor sitesandisnearly independent oftemperature—a1l ofthevariation intheexponential factor issupplied byNp’ even though itismuch lessthan N,,.Anotherwise pure crystal with asmall con- centration ofdonor impurity willhave amajority ofnegative carriers; such a material iscalled an“n-type” semiconductor. Ifanacceptor-type impurity isadded tothecrystal lattice, some ofthenew holes willdrift around andannihilate some ofthefreeelectrons produced by thermal fluctuation. This process willgoonuntil Eq.(14.4) issatisﬁed. Under equilibrium conditions thenumber ofpositive carriers willbeincreased andthe number ofnegative carriers willbedecreased, leaving theproduct aconstant. A material with anexcess ofpositive carriers iscalled a“p-type” semiconductor. Ifweputtwoelectrodes onapiece ofsemiconductor crystal andconnect them toasource ofpotential difference, there Wlllbeanelectric ﬁeld inside the crystal. Theelectric ﬁeldwillcause thepositive andthenegative carriers tomove, andanelectric current willﬂow. Let’s consider ﬁrstwhat willhappen inan n-type material inwhich there isalarge majority ofnegative carriers. Forsuch material wecandisregard theholes, theywillcontribute verylittle tothecurrent because there aresofewofthem. Inanideal crystal thecarriers would move across without anyimpediment. Inarealcrystal ataﬁnite temperature, however,-— especially inacrystal with some impurities—the electrons donotmove completely freely. They arecontinually making collisions which knock them outoftheir original trajectories, thatis,changing their momentum. These collisions arejust exactly thescatterings wetalked about inthelastchapter andoccur atanyirregu- larity inthecrystal lattice. Inann-type material themain causes ofscattering are theverydonor sitesthatareproducing thecarriers. Since theconduction electrons have avery slightly different energy atthedonor sites, theprobability waves are scattered from thatpoint. Even inaperfectly pure crystal, however, there are (atanyﬁnite temperature) irregularities inthelattice duetothermal vibrations. From theclassical point ofview wecansaythattheatoms aren’t lined upexactly onaregular lattice, butare,atanyinstant, slightly outofplace duetotheir thermal 14-5 vibrations. Theenergy E0associated with each lattice point inthetheory we described inChapter 13varies alittle bitfrom place toplace sothatthewaves of probability amplitude arenottransmitted perfectly butarescattered inanirregular fashion. Atveryhigh temperatures orforverypure materials thisscattering may become important, butinmost doped materials used inpractical devices the impurity atoms contribute most ofthescattering. Wewould likenowtomake an estimate oftheelectrical conductivity ofsuch amaterial. When anelectric ﬁeld isapplied toann-type semiconductor, each negative carrier willbeaccelerated inthisﬁeld, picking upvelocity until itisscattered from oneofthedonor sites. This means thatthecarriers which areordinarily moving about inarandom fashion with their thermal energies willpick upanaverage drift velocity along thelines oftheelectric ﬁeldandgiverisetoacurrent through thecrystal. Thedriftvelocity isingeneral rather small compared with thetypical thermal velocities sothatwecanestimate thecurrent byassuming thattheaverage time thatthecarrier travels between scatterings isaconstant. Let's saythatthe negative carrier hasaneffective electric charge q...Inanelectric ﬁeld6,theforce onthecarrier willbeq,,8. InSection 43-3 ofVolume Iwecalculated theaverage drift velocity under such circumstances andfound that itisgiven byFr/m, where Fistheforce onthecharge, 'risthemean freetimebetween collisions, andmisthe mass. Weshould usetheeffective mass wecalculated inthelastchapter but since wewant tomake arough calculation wewillsuppose thatthiseffective mass isthesame inalldirections. Here wewillcallitm,,.With thisapproximation the average driftvelocity willbe vdrift = Knowing thedrift velocity wecanﬁndthecurrent. Electric current density jis justthenumber ofcarriers perunitvolume, N,,,multiplied bytheaverage drift velocity, andbythecharge oneach carrier. Thecurrent density istherefore 2 '=N,,v,i,,;,qn8 = ~ S. (14.6) J "1 Weseethatthecurrent density isproportional totheelectric ﬁeld; such asemi- conductor material obeys Ohm’s law. Thecoefficient ofproportionality between jand8,theconductivity 0",is 2 0-=ME . (14_7) mn Forann-type material theconductivity isrelatively independent oftemperature. First, thenumber ofmajority carriers N"isdetermined primarily bythedensity ofdonors inthecrystal (solong asthetemperature isnotsolowthattoomany ofthecarriers aretrapped). Second, themean timebetween collisions 1,,ismainly controlled bythedensity ofimpurity atoms, which is,ofcourse, independent of thetemperature. Wecanapply allthesame arguments toap-type material, changing only the values oftheparameters which appear inEq.(14.7). Ifthere arecomparable numbers ofboth negative andpositive carriers present atthesame time, wemust addthecontributions from each kind ofcarrier. Thetotal conductivity willbe given by N21' N21' __ nqn n q _ tr-T +-—"m'1:" (14.8) Forverypurematerials, NpandN,,willbenearly equal. They willbesmaller than inadoped material, sotheconductivity willbeless. Also they willvary rapidly with teinperature (like e_EK"P/“T, aswehave seen), sotheconductivity maychange extremely fastwith temperature. 14-6 14-3 TheHall effect Itiscertainly apeculiar thing thatinasubstance where theonly relatively freeobjects areelectrons, there should beanelectrical current carried byholes thatbehave likepositive particles. Wewould like,therefore, todescribe anexperi- ment thatshows inarather clear waythatthesignofthecarrier ofelectric current isquite deﬁnitely positive. Suppose wehave ablock made ofsemiconductor material-it could alsobeametal-—and weputanelectric ﬁeldonitsoastodraw a current insome direction, saythehorizontal direction asdrawn inFig. 14-6. Now suppose weputamagnetic ﬁeld ontheblock pointing ataright angle to thecurrent, sayintotheplane oftheﬁgure. Themoving carriers willfeelamag- netic force q(vXB).And since theaverage drift velocity iseither right orleft- depending onthesignofthecharge onthecarrier—the average magnetic force on thecarriers willbeeither upordown. No, that isnotright! Forthedirections wehave assumed forthecurrent andthemagnetic ﬁeldthemagnetic force onthe moving charges willalways beup.Positive charges moving inthedirection ofj (totheright) willfeelanupward force. Ifthe current iscarried bynegative charges. they willbemoving left(for thesame sign oftheconduction current) and they willalsofeelanupward force. Under steady conditions, however, there isno upward motion ofthecarriers because thecurrent canﬂow only from lefttoright. What happens isthatafewofthecharges initially ﬂowupward, producing asur- facecharge density along theupper surface ofsemiconductor—leaving anequal andopposite surface charge density along thebottom surface ofthecrystal. The charges pileuponthetopandbottom surfaces until theelectric forces theyproduce onthemoving charges justexactly cancel themagnetic force (ontheaverage) so thatthesteady current flows horizontally. Thecharges onthetopandbottom surfaces willproduce apotential difference vertically across thecrystal which can bemeasured with ahigh-resistance voltmeter, asshown inFig. 14-7. Thesign ofthepotential difference registered bythevoltmeter willdepend onthesign of thecarrier charges responsible forthecurrent. When such experiments were ﬁrstdone itwasexpected thatthesignofthe potential difference would benegative asonewould expect fornegative conduction electrons. People were, therefore, quite surprised toﬁndthatforsome materials thesignofthepotential difference wasintheopposite direction. Itappeared that thecurrent carrier wasaparticle with apositive charge. From ourdiscussion of doped semiconductors itisunderstandable thatann-type semiconductor should produce thesignofpotential difference appropriate tonegative carriers, andthat ap-type semiconductor should giveanopposite potential difference, since the current iscarried bythepositively charged holes. Theoriginal discovery oftheanomalous signofthepotential difference in theHall effect was made inametal rather than asemiconductor. Ithadbeen assumed thatinmetals theconduction wasalways byelectron; however, itwas found outthatforberylium thepotential difference hadthewrong sign. Itisnow understood thatinmetals aswellasinsemiconductors itispossible, incertain circumstances, thatthe“objects” responsible fortheconduction areholes. Al- though itisultimately theelectrons inthecrystal which dothemoving, neverthe- less,therelationship ofthemomentum andtheenergy, andtheresponse toexternal ﬁelds isexactly what onewould expect foranelectric current carried bypositive particles. Let’s seeifwecanmake aquantitative estimate ofthemagnitude ofthevolt- agedifference expected from theHalleffect. Ifthevoltmeter inFig14-7 draws a negligible current, then thecharges inside thesemiconductor must bemoving from lefttoright andthevertical magnetic force must beprecisely cancelled bya vertical electric ﬁeld which wewillcall8,,(the“tr”isfor“transverse”). Ifthis electric ﬁeld istocancel themagnetic forces, wemust have 8,. Z _U,|nfj XB. Using therelation between thedrift velocity andtheelectric current density given 14-7+ _ IL +(-) B® i —<+i&\\\\\\\\\\\\\\\\‘ § Fig. l4—6. TheHall effect comes from themagnetic forces onthecarriers, ELECTRONICVOLTMETER \_ O + *1- Fig. l4—7. Measuring theHall effect, _________L________U2Z<--////°/ .228 >1vto (bl iﬁjto1 || f\ I xi Fig. l4—9. Theelectric potential cind the ccirrier densities incin unbicised semiconductor junction.inEq.(14.6), weget l.gtr — Thepotential difference between thetopandthebottom ofthecrystal is,ofcourse, thiselectric ﬁeldstrength multiplied bytheheight ofthecrystal. Theelectric ﬁeld strength ﬁt,inthecrystal isproportional tothecurrent density and tothemag- netic ﬁeld strength. The constant ofproportionality I/qN iscalled theHall coefﬁcient andisusually represented bythesymbol R”. The Hall coefficient de- pends justonthedensity ofcarriers-—provided thatcarriers ofonesignareina large majority. Measurement oftheHall effect is,therefore. oneconvenient way ofdetermining experimentally thedensity ofcarriers inasemiconductor. 14-4 Semiconductor junctions Wewould liketodiscuss now what happens ifwetake twopieces ofgermanium orsilicon with different internal characteristics——say different kinds oramounts ofdoping—and putthem together tomake a“junction.” Let’s start outwithwhat iscalled ap-njunction inwhich wehave p-type germanium onone side ofthe boundary andn-type germanium ontheother sideoftheboundary—as sketched inFig. l4—8. Actually, itisnotpractical toputtogether two separate pieces of crystal andhave them inuniform contact onanatomic scale. Instead, junctions aremade outofasingle crystal which hasbeen modiﬁed inthetwo separate regions. One wayistoaddsome suitable doping impurity tothe“melt“ after only halfofthecrystal hasgrown. Another way istopaint alittle oftheimpurity element onthesurface andthen heatthecrystal causing some impurity atoms to diffuse intothebody ofthecrystal. Junctions made inthese ways donothavea sharp boundary, although theboundaries canbemade asthinas10-‘ centimeters orso.Forourdiscussions wewillimagine anideal situation inwhich these two regions ofthecrystal with different properties meeting atasharp boundary. Onthen-type sideofp-njunction there arefreeelectrons which canmove about, aswellastheﬁxed donor sites which balance theoverall electric charge. Onthep-type sidethere arefreeholes moving about andanequal number of negative acceptor sites keeping thecharge balanced. Actually, thatdescribes the situation before weputthetwo materials incontact. Once they areconnected together thesituation willchange near theboundary. When theelectrons in then-type material arrive attheboundary they willriotbereflected back asthey would atafreesurface, butareabletogoright onintothep-type material. Some oftheelectrons ofthen-type material will, therefore, tend todiffuse over intothe p-type material where there arefewer electrons. Thiscannot goonforever because asweloseelectrons from then-side thenetpositive charge there increases until ﬁnally anelectric voltage isbuilt upwhich retards thediffusion ofelectrons into thep-side. Inasimilar way. thepositive carriers ofthep-type material candiffuse across thejunction intothen-type material. When theydothistheyleave behind anexcess ofnegative charge. Under equilibrium conditions thenetdiffusion cur- rentmust bezero. This brought about bytheelectric ﬁelds which areestablished insuch away astodraw thepositive carriers back toward thep-type material. The two diffusion processes wehave been describing goonsimultaneously and, youwillnotice. both actinthedirection which willcharge upthen-type material inapositive sense andthep-type material inanegative sense. Because oftheﬁnite conductivity ofthesemiconductor material, thechange inpotential from thep-side tothen-side willoccur inarelatively narrow region near thebound- ary;themain body ofeach block ofmaterial willhave auniform potential. Let’s imagine anx-axis inadirection perpendicular totheboundary surface. Then the electric potential willvary with x,asshown inFig.l4—9(b). Wehave alsoshown inpart (c)oftheﬁgure theexpected variation ofthedensity N”ofn-carriers and thedensity N,ofp-carriers. Faraway from thejunction thecarrier densities N],andN"should bejusttheequilibrium density wewould expect forindividual blocks ofmaterials atthesame temperature. (We have drawn theﬁgure fora l4—8 junction inwhich thep-type material ismore heavily doped than then-type material.) Because ofthepotential gradient atthejunction, thepositive carriers have toclimb upapotential hilltogettothep-type side. This means thatunder equilibrium conditions there canbefewer positive carriers inthen-type material than there areinthep-type material Remembering thelaws ofstatistical me- chanics, weexpect thattheratio ofp-type carriers onthetwosides tobegiven by thefollowing equation: % =.»~’1»”/"1. (14.10) Theproduct q,,Vin thenumerator oftheexponential isjusttheenergy required to carry acharge ofqpthrough apotential difference V. Wehave aprecisely similar equation forthedensities ofthen-type carriers: Nn(n'side) __ —-qnv/KT Ifweknow theequilibrium densities ineach ofthetwomaterials, wecanuse either ofthetwoequations above todetermine thepotential difference across the junction. Notice that ifEqs. (14.10) and(14.11) aretogivethesame value forthe potential difference V,theproduct N,,N,, must bethesame forthep-side asfor then-side. (Remember thatq,,=——q,,.) Wehave seenearlier, however, thatthis product depends only onthetemperature andthegapenergy ofthecrystal. Provided both sides ofthecrystal areatthesame temperature, thetwoequations areconsistent with thesame value ofthepotential difference. Since there isapotential difference from onesideofthejunction totheother, itlooks something likeabattery. Perhaps ifweconnect awirefrom then-type side tothep-type sidewewillgetanelectrical current. That would benicebecause thenthecurrent would ﬂowforever without using upanymaterial andwewould have aninﬁnite source ofenergy inviolation ofthe second lawofthermodynamics! There is,however, nocurrent ifyouconnect awirefrom thep-side tothen-side. Andthereason iseasytosee.Suppose weimagine ﬁrstawiremade outofapiece ofundoped material. When weconnect thiswire tothen-type side, wehave a junction. There willbeapotential difference across thisjunction. Let’s saythat itisjustone-half thepotential difference from thep-type material tothen-type material. When weconnect ourundoped wire tothep-type sideofthejunction, there isalsoapotential difference atthisjunction—again, one-half thepotential drop across thep-njunction. Atallthejunctions thepotential differences adjust themselves sothatthere isnonetcurrent ﬂowinthecircuit. Whatever kind ofwire youusetoconnect together thetwosides ofthen-pjunction, youareproducing twonewjunctions, andsolong asallthejunctions areatthesame temperature, the potential jumps atthejunctions allcompensate each other andnocurrent will ﬂow inthecircuit. Itdoes turn out, however—if youwork outthedetails—that if some ofthejunctions areatadifferent temperature than theother junctions, currents willﬂow. Some ofthejunctions willbeheated andothers willbecooled bythiscurrent andthermal energy willbeconverted into electrical energy. This effect isresponsible fortheoperation ofthermocouples which areused formeasur- ingtemperatures, andofthermoelectric generators. Thesame effect isalsoused tomake small refrigerators. Ifwecannot measure thepotential difference between thetwosides ofan n-pjunction, how canwereally besure that thepotential gradient shown inFig. 14-9 really exists? One way istoshine light onthejunction. When thelight photons areabsorbed they can produce anelectron-hole pair. lnthestrong electric ﬁeldthatexists atthejunction (equal totheslope ofthepotential curve of Fig.l4—9) theholewillbedriven intothep—type region andtheelectron willbe driven intothen-type region. Ifthetwosides ofthejunction arenowconnected toanexternal circuit, these extra charges willprovide acurrent. Theenergy of thelight willbeconverted intoelectrical energy inthejunction. Thesolar cells which generate electrical power fortheoperation ofsome ofoursatellites operate onthisprinciple. l4—9 Inourdiscussion oftheoperation ofasemiconductor junction wehave been assuming thattheholes andtheelectrons actmore-or-less independently—except thattheysomehow getintoproper statistical equilibrium. When wewere describing thecurrent produced bylight shining onthejunction, wewere assuming thatan electron oraholeproduced inthejunction region would getintothemain body of thecrystal before being annihilated byacarrier oftheopposite polarity. Inthe immediate vicinity ofthejunction, where thedensity ofcarriers ofboth signs is approximately equal, theeffect ofelectron-hole annihilation (orasitisoften called, “recombination”) isanimportant effect, andinadetailed analysis ofasemi- conductor junction must beproperly taken intoaccount. Wehave been assuming thatahole oranelectron produced inajunction region hasagood chance of getting intothemain body ofthecrystal before recombining. Thetypical time foranelectron oraholetoﬁndanopposite partner andannihilate itisfortypical semiconductor materials intherange between l0'3 andl0_7 seconds. This time is,incidentally, much longer than themean freetime 'rbetween collisions with scattering sites inthecrystal which weused intheanalysis ofconductivity. In atypical n-pjunction, thetimeforanelectron orholeformed inthejunction region tobeswept away intothebody ofthecrystal isgenerally much shorter than the recombination time. Most ofthepairs will, therefore, contribute toanexternal current. 14-5 Rectiﬁcation atasemiconductor junction Wewould liketoshow nexthowitisthatap-njunction canactlikearectiﬁer. Ifweputavoltage across thejunction, alarge current willﬂowifthepolarity isin onedirection, butaverysmall current willﬁowifthesame voltage isapplied inthe opposite direction. Ifanalternating voltage isapplied across thejunction, anet current willflow inonedirection—the current is“rectiﬁed.” Let’s look again at what isgoing onintheequilibrium condition described bythegraphs ofFig. 14-9. Inthep-type material there isalarge concentration Npofpositive carriers. These carriers arediffusing around andacertain number ofthem each second approach thejunction. This current ofpositive carriers which approaches the junction isproportional toNp. Most ofthem, however, areturned back bythe high potential hillatthejunction andonly thefraction e"‘1V/“T gets through. There isalsoacurrent ofpositive carriers approaching thejunction from theother side. This current isalsoproportional tothedensity ofpositive carriers inthe n-type region, butthecarrier density hereismuch smaller than thedensity onthe p-type side. When thepositive carriers approach thejunction from then-type side, they ﬁndahillwith anegative slope andimmediately slide downhill tothe p-type sideofthejunction. Let’s callthiscurrent 10.Under equilibrium thecur- rents from thetwodirections areequal. Weexpect then thefollowing relation: 1.,~N,,(n-side) =N,,(p-side)eTqVf"T. (14.12) Youwillnotice thatthisequation isreally justthesame asEq.(14-10). Wehave justderived itinadifferent way. Suppose, however, thatwelower thevoltage onthen-side ofthejunction by anamount AV—which wecandobyapplying anexternal potential difference to thejunction. Now thedifference inpotential across thepotential hillisnolonger VbutV—AV. Thecurrent ofpositive carriers from thep-side tothen-side will now have thispotential difference initsexponential factor. Calling thiscurrent 11,wehave 1,~1v,(p-sid¢)e-@<V-“WT. This current islarger than I0byjustthefactor e"‘“’/“T. Sowehave thefollowing relation between I,andI.,: 1,=r,,@+‘1”'~". (14.13) Thecurrent from thep-side increases exponentially with theexternally applied voltage AV. The current ofpositive carriers from then-side, however, remains 14-10 constant solong asAVisnottoolarge. When they approach thebarrier, these carriers willstillﬁndadownhill potential andwillallfalldown tothep-side (IfAV islarger than thenatural potential difference V,thesituation would change, butwewillnotconsider what happens atsuch high voltages.) Thenetcurrent Iof positive carriers which ﬂows across thejunction isthen thedifference between the currents from thetwosides: 1=1.,(e+‘1“’/“T -1). (14.14) Thenetcurrent Iofholes ﬂows intothen-type region. There theholes diffuse intothebody ofthen-region, where theyareeventually annihilated bythemajority n-type carriers—the electrons. The electrons which arelostinthis annihilation willbemade upbyacurrent ofelectrons from theexternal terminal ofthen-type material. When AViszero. thenetcurrent inEq.(14.14) iszero Forpositive AVthe current increases rapidly with theapplied voltage. Fornegative AVthecurrent reverses insign, buttheexponential term soon becomes negligible andthenegative current never exceeds I,,——which under ourassumptions israther small. This backcurrent I0islimited bythesmall density oftheminority carriers onthen-side ofthejunction. Ifyougothrough exactly thesame analysis forthecurrent ofnegative carriers which ﬂows across thejunction, ﬁrst with nopotential difference andthen with a small externally applied potential difference AV,yougetagain anequation just like(14.14) forthenetelectron current. Since thetotal current isthesum ofthe currents contributed bythetwocarriers, Eq.(14.14) stillapplies forthetotal current provided weidentify 10asthemaximum current which canﬁow fora reversed voltage. Thevoltage-current characteristic ofEq.(14.14) isshown inFig. 14-10. It shows thetypical behavior ofsolid state di0des—such asthose used inmodern computers. Weshould remark thatEq.(14.14) istrueonly forsmall voltages. Forvoltages comparable toorlarger than thenatural internal voltage difference V,other effects come intoplayandthecurrent nolonger obeys thesimple equation. You may remember, incidentally, that wegotexactly thesame equation we havefound hereinEq.(14.14) when wediscussed the“mechanical rectiﬁer”—the ratchet andpawl—in Chapter 46ofVolume I.Wegetthesame equations inthe twosituations because thebasic physical processes arequite similar. 14-6 Thetransistor Perhaps themost important application ofsemiconductors isinthetransistor. The transistor consists oftwo semiconductor junctions very close together. Its operation isbased inpartonthesame principles thatwejustdescribed forthe semiconductor diode——the rectifying junction. Suppose wemake alittle barof germanium with three distinct regions, ap-type region, ann-type region, and another p-type region, asshown inFig. l4—ll(a). This combination iscalled a p-rt-p transistor. Each ofthetwojunctions inthetransistor willbehave much in thewaywehave described inthelastsection. Inparticular, there willbeapotential gradient ateachjunction having acertain potential drop from then-type region to eachp-type region. Ifthetwop-type regions have thesame internal properties, thevariation inpotential aswegoacross thecrystal willbeasshown inthegraph ofFig.l4—11(b). Now let'simagine thatweconnect each ofthethree regions toexternal voltage sources asshown inpart(a)ofFig.l4—l2Wewillrefer allvoltages totheterminal connected totheleft-hand p-region soitwillbe,bydeﬁnition, atzero potential. Wewillcallthisterminal theemitter. Then-type region iscalled thebaseanditis connected toaslightly negative potential. The right-hand p-type region iscalled thecollector, andisconnected toasomewhat larger negative potential. Under these circumstances thevariation ofpotential across thecrystal willbeasshown in thegraph ofFig.14—l2(b). Let’s ﬁrstseewhat happens tothepositive carriers, since itisprimarily their behavior which controls theoperation ofthep-n-p transistor. Since theemitter is 14-11tI/IO 6..- 5.- 4- 3.- 2.. |__ AV/KTQ > ii——-_——__§____' .. -2 ._ Fig. l4—lO. The current through u junction csofunction ofthevoltcige ocrossii. (0) 12%Fig. l4—ll. The potential distribu- tion incttransistor with noapplied voltages.I-§E‘,< **_'_——‘—_—_—§\\\\i\§\\\ v,=o vb<0 vc<<v (I ‘<-—|—40O’I-1 —>u_O _.>i-0 ‘°’ ________:___\>____l______| l l \__ ___ <n 1 Fig. l4—l2. The potential distribu- tioninonoperating transistor.b atarelatively more positive potential than thebase, acurrent ofpositive carriers willﬂowfrom theemitter region intothebase region. Arelatively large current ﬂows. since wehave ajunction operating witha“forward voltage”—corresponding totheright-hand halfofthegraph inFig.14-10. Witli these conditions, positive carriers orholes arebeing “emitted” from thep-type region intothen-type region. Youmight think thatthiscurrent would ﬂowoutofthen-type region through the base terminal b.Now, however, comes thesecret ofthetransistor. Then-type region ismade verythin—typically l0'3 cmorless,much narrower thanitstrans- verse dimensions. This means thatastheholes enter then-type region theyhave avery good chance ofdiffusing across totheother junction before theyareanni- hilated bytheelectrons inthen-type region. When they gettotheright-hand boundary ofthen-type region theyﬁndasteep downward potential hillandim- mediately fallintotheright-hand p-type region. This sideofthecrystal iscalled thecollector because it“collects” theholes after theyhavediffused across then-type region. Inatypical transistor, allbutafraction ofapercent oftheholecurrent which leaves theemitter andenters thebase iscollected inthecollector region, andonlythesmall remainder contributes tothenetbasecurrent. Thesumofthe base andcollector currents is,ofcourse, equal totheemitter current. Now imagine what happens ifwevary slightly thepotential V1,onthebase terminal. Since weareonarelatively steep part ofthecurve ofFig. l4—l0, a small variation ofthepotential V1,willcause arather large change intheemitter current 1,.Since thecollector voltage V,ismuch more negative than thebase voltage, these slight variations inpotential will noteffect appreciably thesteep potential hillbetween thebase andthecollector. Most ofthepositive carriers emitted intothen-region willstillbecaught bythecollector. Thus aswevary thepotential ofthebase electrode, there willbeacorresponding variation inthe collector current 1,.Theessential point, however, isthat thebase current Ii, always remains asmall fraction ofthecollector current. Thetransistor isan ampliﬁer; asmall current I1,introduced intothebaseelectrode gives alarge current -—100 orsotimes higher—at thecollector electrode. What about theelectrons—the negative carriers thatwehave been neglecting sofar? First, note thatwedonotexpect anysigniﬁcant electron current toflow between thebaseandthecollector. With alarge negative voltage onthecollector, theelectrons inthebasewould have toclimb averyhigh potential energy hilland theprobability ofdoing that isvery small. There isavery small current ofelec- trons tothecollector. Ontheother hand, theelectrons inthebase cangointotheemitter region. Infact,youmight expect theelectron current inthisdirection tobecomparable to thehole current from theemitter intothebase. Such anelectron current isn’t useful, and, onthecontrary, isbadbecause itincreases thetotal base current required foragiven current ofholes tothecollector. Thetransistor is,therefore, designed tominimize theelectron current totheemitter. Theelectron current is proportional toN,,(base), thedensity ofnegative carriers inthebase material while thehole current from theemitter depends onN,,(emitter), thedensity of positive carriers intheemitter region. Byusing relatively littledoping inthen-type material N,,(base) canbemade much smaller than N,,(emitter). (The very thin base region alsohelps agreat dealbecause thesweeping outoftheholes inthis region bythecollector increases signiﬁcantly theaverage hole current from the emitter intothebase, while leaving theelectron current unchanged.) Thenet result isthattheelectron current across theemitter-base junction canbemade much lessthan theholecurrent, sothattheelectrons donotplayanysigniﬁcant roleinoperation ofthep-n-p transistor. Thecurrents aredominated bymotion of theholes, andthetransistor performs asanampliﬁer aswehave described above. Itisalsopossible tomake atransistor byinterchanging thep-type andn-type materials inFig.14-11. Then wehave what iscalled anrt-p-n transistor. Inthe n—p—n transistor themain currents arecarried bytheelectrons which ﬂowfrom the emitter intothebaseandfrom there tothecollector. Obviously, allthearguments wehave made forthep-n-p transistor alsoapply tothen-p-n transistor ifthepo- tentials oftheelectrodes arechosen with theopposite signs. l4—l2 I5 The Independent Particle Approximation 15-1 Spin waves InChapter 13weworked outthetheory forthepropagation ofanelectron or ofsome other “particle,” such asanatomic excitation, through acrystal lattice. Inthelastchapter weapplied thetheory tosemiconductors. Butwhen wetalked about situations inwhich there aremany electrons wedisregarded anyinteractions between them. Todothisisofcourse only anapproximation. Inthischapter wewilldiscuss further theideathatyoucandisregard theinteraction between the electrons. Wewillalsousetheopportunity toshow yousome more applications ofthetheory ofthepropagation ofparticles. Since wewillgenerally continue to disregard theinteractions between particles, there isverylittle really newinthis chapter except forthenewapplications. Theﬁrstexample tobeconsidered is, however, oneinwhich itispossible towrite down quite exactly thecorrect equa- tions when there ismore than one“particle” present. From them wewillbeable toseehow theapproximation ofdisregarding theinteractions ismade. Wewill not,though, analyze theproblem very carefully. Asourﬁrstexample wewillconsider a“spin wave” inaferromagnetic crystal. Wehave discussed thetheory offerromagnetism inChapter 36ofVolume II. Atzerotemperature alltheelectron spins thatcontribute tothemagnetism inthe body ofaferromagnetic crystal areparallel. There isaninteraction energy between thespins, which islowest when allthespins aredown. Atanynonzero temperature, however, there issome chance thatsome ofthespins areturned over. Wecalculated theprobability inanapproximate manner inChapter 36.Thistimewewilldescribe thequantum mechanical theory—so youwillseewhat youwould have todoifyou wanted tosolve theproblem more exactly. (Wewillstillmake some idealizations byassuming thattheelectrons arelocalized attheatoms andthatthespins interact only with neighboring spins.) Weconsider amodel inwhich theelectrons ateach atom areallpaired except one,sothatallofthemagnetic effects come from onespin-% electron peratom. Further, weimagine thatthese electrons arelocalized attheatomic sites inthe lattice. The model corresponds roughly tometallic nickel. Wealsoassume thatthere 1Saninteraction between anytwoadjacent spinning electrons which gives aterm intheenergy ofthesystem E=—2Kd¢'G], (15.1)1,] where o"Srepresent thespins andthesummation isover alladjacent pairs of electrons. Wehave already discussed thiskind ofinteraction energy when we considered thehyperﬁne splitting ofhydrogen duetotheinteraction ofthemag- netic moments oftheelectron andproton inahydrogen atom. Weexpressed it thenasAir,-up.Now, foragiven pair, saytheelectrons atatom 4andiat atom 5, theHamiltonian would be—Ka4 -<15.Wehave aterm foreach such pair, and theHamiltonian is(asyouwould expect forclassical energies) thesumofthese terms foreach interacting pair. Theenergy iswritten with thefactor —Ksothat apositive Kwillcorrespond toferromagnetism—that is,thelowest energy results when adjacent spins areparallel. Inarealcrystal, there may beother terms which aretheinteractions ofnextnearest neighbors, andsoon,butwedon’t need tocon- sider such complications atthisstage. With theHamiltonian ofEq.(15.1) wehave acomplete description ofthe ferromagnet—within ourapproximation—and theproperties ofthemagnetization l5—l15-1 Spin waves 15-2 Two spinwaves 15-3 Independent particles 15-4 Thebenzene molecule 15-5 More organic chemistry 15-6 Other usesofthe approximation should come out. Weshould alsobeabletocalculate thethermodynamic proper- tiesduetothemagnetization. Ifwecanﬁndalltheenergy levels, theproperties ofthecrystal atatemperature Tcanbefound from theprinciple thattheprob- ability thatasystem willbefound inagiven state ofenergy Eisproportional to e_E'“T.This problem hasnever been completely solved. Wewillshow some oftheproblems bytaking asimple example inwhich all theatoms areinaline—a one-dimensional lattice. Youcaneasily extend theideas tothree dimensions. Ateach atomic location there isanelectron which hastwo possible states, either spinuporspindown, andthewhole system isdescribed by telling howallofthespins arearranged. WetaketheHamiltonian ofthesystem tobetheoperator oftheinteraction energy. Interpreting thespinvectors ofEq. (15.1)asthesigma-operators—or thesigma-matrices—we write forthelinear lattice A--PI=E--2-an-a,,+1. (15.2) Inthisequation wehave written theconstant asA/2forconvenience (sothatsome ofthelater equations willbeexactly thesame astheones inChapter 13). Now what isthelowest state ofthissystem? Thestate oflowest energy is theoneinwhich allthespins areparallel—let’s say, allup.'I' Wecanwrite this state as[---+—}—++---),or|gnd) forthe“ground,” orlowest, state. It’s easytoﬁgure outtheenergy forthisstate. Onewayistowrite outallthevector sigmas interms of6,,6'”,and6,,andwork through carefully what each term of theHamiltonian does totheground state, andthen addtheresults. Wecan, however, alsouseagood short cut. WesawinSection 12-2, that6,-6,could bewritten interms ofthePauli spinexchange operator likethis: a,~a,=(2PZ‘;"“”‘ _1), (15.3) where theoperator Pffi“"" interchanges thespins oftheithandjthelectrons. With thissubstitution theHamiltonian becomes H=—/1Z(P?§‘3i"i —%)- (15-4) Itisnoweasytowork outwhat happens todiflerent states. Forinstance ifiandj areboth up,then exchanging thespins leaves everything unchanged, soP”acting onthestatejustgives thesame state back, andisequivalent tomultiplying by+1. The expression (15,, —%)isjust equal toone-half. (From now onwewillleave offthedescriptive superscript ontheP.) Fortheground state allspins areup;soifyouexchange aparticular pair of spins, yougetback theoriginal state. Theground state isastationary state. If you operate onitwith theHamiltonian you getthesame state again multiplied byasumofterms, —(A/2) foreach pairofspins. That is,theenergy ofthesystem intheground state is—A/2 peratom Next wewould liketolook attheenergies ofsome oftheexcited states. It willbeconvenient tomeasure theenergies with respect totheground state—that is,tochoose theground state asourzero ofenergy. Wecandothat byadding the energy A/2toeach term intheHamiltonian. That Justchanges the“%”inEq. (15.4) to“l.” Our new Hamiltonian is H=—AZ (P,,_,+, -1). (15.5) With thisHamiltonian theenergy ofthelowest state iszero; thespinexchange operator isequivalent tomultiplying byunity (fortheground state) which is cancelled bythe“l"ineach term. ’[The ground state here isreally “degenerate”; there areother states with thesame energy-—for example, allspins down, orallinanyother direction. Theslightest external ﬁeldinthez-direction willgiveadifferent energy toallthese states. andtheonewehave chosen willbethetrueground state. 15-2 Fordescribing states other than theground state wewillneed asuitable set ofbasestates. Oneconvenient approach istogroup thestates according towhether oneelectron hasspindown, ortwoelectrons have spindown, andsoon.There are,ofcourse, many states withonespindown. Thedown spincould beatatom “4,”oratatom “S,”oratatom “6,”...Wecan,infact,choose justsuch states forourbase states. Wecould write them thisway: I4),I5),I6),...Itwill, however, bemore convenient later ifwelabel the“odd atom”—the onewith the down-spinning electron—by itscoordinate x.That is,we’ll deﬁne thestate Ix5) tobeonewith alltheelectrons spinning upexcept fortheoneontheatom atx5, which hasadown-spinning electron (seeFig.l5—l). Ingeneral, Ixn)isthestate withonedown spinthatislocated atthecoordinate x,,ofthenthatom. What istheaction oftheHamiltonian (15.5) onthestate Ix5)‘?Oneterm of theHamiltonian issay—/{(137, 8—1).Theoperator 157,8exchanges thetwospins oftheadjacent atoms 7,8.Butinthestate Ix5)these areboth up,andnothing happens; PH, isequivalent tomultiplying by1: 161,8 IX5)=|X5>- ltfollows that (pins -1)IXs)= 0- Thus alltheterms oftheHamiltonian givezero—except those involving atom 5, ofcourse. Onthestate I5),theoperation P4_5 exchanges thespinofatom 4(up) andatom 5(down). Theresult isthestate with allspins upexcept theatom at4. That is p4.5 IX5)=IX4>- Inthesame way 135,6 IX5)=IX6>~ Hence, theonly terms oftheHamiltonian which survive are—A(P4,5 —1) and—A(P5,6 —-1).Acting on|x5) they produce —AIx4)+A|x5) and -AIxﬁ)+AIx5),respectively. Theresult is 1t|><5>=—AZ(Pn.n+1—1>|><5>=—A{l><@> +|xi>-2|x5>}- (15.6) When theHamiltonian actsonstate Ix5)itgives risetosome amplitude tobe instates Ix4)andIx6). That justmeans thatthere isacertain amplitude tohave thedown spinjump overtothenextatom. Sobecause oftheinteraction between thespins, ifwebegin with onespindown, thenthere issome probability thatata latertimeanother onewillbedown instead. Operating onthegeneral state Ixn), theHamiltonian gives HIX»)=_A{Ixn+1>+I-xn-1) —2IXn>}- (15-7) Notice particularly thatifwetake acomplete setofstates with only onespin down, they willonly bemixed among themselves. TheHamiltonian willnever mixthese states with others thathave more spins down. Solong asyouonlyex- change spins younever change thetotal number ofdown spins. Itwillbeconvenient tousethematrix notation fortheHamiltonian, say H,,,,,, E(x,,IHIx,,,); Eq.(15.7) isequivalent to 1¥h,n ==14; Hn,n+1 =Hn,n—1 = H,,,,,,=0, for In—mI>l. Now what aretheenergy levels forstates with onespindown? Asusual we letC,,betheamplitude thatsome state I¢)isinthestate Ixn). IfI1//)istobea deﬁnite energy state, alltheC,,’smust varywith timeinthesame way, namely, C"=a,,e“'E'm. (l5.9) 15-3b|“‘| +4+t+t++.,+t--3-2-|O|234567 I———)T—-I Fig. 15-1. Thebase state Ix5) ofa linear array ofspins. Allthespins areup except theoneatx5,which isdown. 45:5:4iiI4~t~¢—<H'3‘2-lOl234567 Fig. l5—2. Astate with two down spins.Wecanputthistrial solution into ourusual Hamiltonian equation dC,lhTEFL :Z HIIIIIC/II1 using Eq(15.8) forthematrix elements. Ofcourse wegetaninﬁnite number of equations, buttheycanallbewritten as Ea" =2Au,, —/l(1n_1 _A£I,,+1 Wehave again exactly thesame problem weworked outinChapter 13,except that where wehadE0wenow have 2A. The solutions correspond toamplitudes C" (the down-spin amplitude) which propagate along thelattice with dpropagation constant kandanenergy E=2A(l —coskb), (15.12) where bisthelattice constant. The deﬁnite energy solutions correspond to“waves” ofdown spin—called “spin waves." And foreach wavelength there isacorresponding energy. For large wavelengths (small k)thisenergy varies as E=Ab2k2. (15.13) Just asbefore, wecanconsider alocalized wave packet (containing, however, only long wavelengths) which corresponds toaspin-down electron inonepart of thelattice. This down spin willbehave likea“particle.” Because itsenergy is related tokby(15.13) the“particle” willhave anellective mass: mo“ =~ 2Ab2 These “particles” aresometimes called “magnons.” 15-2 Twospinwaves Now wewould liketodiscuss what happens ifthere aretwodown spins. Again wepickasetofbase states. We’ll choose states inwhich there aredown spins attwoatomic locations, such asthestate shown inFig.15-2. Wecanlabel such astate bythex-coordinates ofthetwositeswithdown spins. Theoneshown canbecalled Ix2,x5). Ingeneral thebasestates areIx,,,x,,,)—a doubly inﬁnite set! Inthissystem ofdescription, thestate Ix.,,x9)andthestate Ixg,x4)are exactly thesame state, because each simply saysthatthere isadown spinat4and oneat9;there isnomeaning totheorder. Furthermore, thestate Ixi,x4)has nomeaning, there isn’tsuch athing. Wecandescribe anystate I1/1)bygiving the amplitudes tobeineach ofthebasestates. Thus Cm,” =(xm,x,,It//>nowmeans theamplitude forasystem inthestate Iti)tobeinastate inwhich both themth andnthatoms have adown spin. Thecomplications which now arise arenot complications ofideas—they aremerely complexities inbookkeeping. (One ofthe complexities ofquantum mechanics isjustthebookkeeping. With more and more down spins, thenotation becomes more andmore elaborate with lotsof indices andtheequations always look veryhorrifying, buttheideas arenotneces- sarily more complicated than inthesimplest case.) Theequations ofmotion ofthespinsystem arethedifferential equations for theC,,,,,,. They are dCﬂ,77l Ih—d—l~ =Z(H,,,,,_.,)c,,. (15.15) '11} Suppose wewant toﬁndthestationary states. Asusual, thederivatives withre- spect totimebecome Etimes theamplitudes andtheC,,,,,, canbereplaced bythe l5—4 coefﬁcients a.,,,.,. Next wehave towork outcarefully theeffect ofHonastate withspins mandndown. Itisnothard toﬁgure out. Suppose foramoment thatmandnarefarenough apart sothatwedon’t have toworry about theobvious trouble. Theoperation ofexchange atthelocation x,,willmove thedown spin either tothe(n—I—1)or(n—1)atom, and sothere’s anamplitude that the present state hascome from thestate Ix,,,,x,,+1) andalsoanamplitude thatit hascome from thestate Ixm, x,,_1). Oritmay have been theother spin that moved; sothere’s acertain amplitude that Cm," isfedfrom C,,,+1_" orfrom C,,,_1_,,. These effects should allbeequal. Theﬁnal result fortheHamiltonian equation onC,,,,,, is Eam,n = _'A(am-I-1,n +a"L—I,7|. +amnt-I-1 +am,n—1) ‘l’4Aam,n- Thisequation iscorrect except intwosituations. Ifm=nthere isnoequation atall,andifm=n=*=1,thentwooftheterms inEq.(15.16) should bemissing. Wearegoing todisregard these exceptions. Wesimply ignore thefactthatsome fewofthese equations areslightly altered. After all,thecrystal issupposed tobe inﬁnite, andwehave aninﬁnite number ofterms; neglecting afewmight not matter much. Soforaﬁrst rough approximation let’s forget about thealtered equations. Inother words, weassume that Eq.(15.16) istrue forallmand n,even formandnnexttoeach other. Thisistheessential part ofourapproxi- mation. Then thesolution isnothard toﬁnd. Wegetimmediately cm‘,=a,,,,e-1'”/t, (15.17) with am,=(const.) e*t1’me”‘2“'-, (15.18)where E=4A—2Acosklb—2Acoskzb. (15.19) Think foramoment what would happen ifwehadtwoindependent, single spinwaves (asintheprevious section) corresponding tok=klandk=k2; they would have energies, from Eq.(15.12), of e1=(2A—-2Acosklb) and e;=(2A—2Acoskgb). Notice thattheenergy EinEq.(15.19) isjusttheir sum, E=e(k1) +e(k2). (15.20) Inother words wecanthink ofoursolution inthisway. There aretwoparticles- thatis,twospinwaves. Oneofthem hasamomentum described bykl,theother byk2,andtheenergy ofthesystem isthesumoftheenergies ofthetwoobjects. Thetwoparticles actcompletely independently. That’s allthere istoit. Ofcourse wehave made some approximations, butwedonotwish todiscuss theprecision ofouranswer atthispoint. However, youmight guess thatina reasonable sizecrystal with billions ofatoms—and, therefore, billions ofterms in theHamiltonian—leaving outafewterms wouldn’t make much ofanerror. Ifwehadsomany down spins thatthere wasanappreciable density, thenwewould certainly have toworry about thecorrections. [Interestingly enough, anexact solution canbewritten down ifthere arejust thetwodown spins. The result isnotparticularly important. Butitisinteresting thattheequations canbesolved exactly forthiscase. Thesolution is: am”, =expl"°"("”'+’”"l] sinkIxm—x,,I, (15.21) with theenergy E=4A—2Acosklb——2Acosk2b, 15-5 andwiththewave numbers kcandkrelated toklandkgby kl=k,—k, k2=kc+k. (15.22) This solution includes the“interaction” ofthetwospins. Itdescribes thefact that when thespins come together there isacertain chance ofscattering. The spins actvery much likeparticles with aninteraction. Butthedetailed theory of their scattering goesbeyond what wewant totalkabout here.] 15-3 Independent particles Inthelastsection wewrote down aHamiltonian, Eq.(15.15), foratwo- particle system. Then using anapproximation which isequivalent toneglecting any“interaction” ofthetwoparticles, wefound thestationary states described byEqs. (15.17) and(15.18). This state isjust theproduct oftwosingle-particle states. Thesolution wehave given foram," inEq.(15.18) is,however, really not satisfactory. Wehave verycarefully pointed outearlier thatthestate Ixg,xl) isnotadilierent state from Ix.l,x9)—the order ofx,,,andx,,hasnosignificance. Ingeneral, thealgebraic expression fortheamplitude C,,,,,,must beunchanged if weinterchange thevalues ofx,,,andx,,,since thatdoesn’t change thestate. Either way, itshould represent theamplitude toﬁndadown spin atx,,,andadown spin atx,,. Butnotice that (15.18) isnotsymmetric inx,,,andx,,—since klandk2 caningeneral bediﬁerent. Thetrouble isthat wehave notforced oursolution ofEq.(15.15) tosatisfy thisadditional condition. Fortunately itiseasy toﬁxthings up.Notice ﬁrstthat asolution oftheHamiltonian equation justasgood as(15.18) is a,,_,=KJ'°"'~@"’“i=». (15.23) Itevenhasthesame energy wegotfor(15.18).Anylinear combination of(15.15) and(15.23) isalsoagood solution, andhasanenergy stillgiven byEq.(15.19). Thesolution weshould have chosen—because ofoursymmetry requirement—is justthesumof(15.15) and(15.23): am?” =K[eikl::meik2:c,l +e‘”C21me1..k1Zn]‘ Now, given anyklandk2theamplitude Cmmisindependent ofwhich Way we putxmandx,,—if weshould happen todeﬁne xmandxnreversed wegetthesame amplitude. Our interpretation ofEq.(15.24) interms of“magnons” must alsobe different. Wecannolonger saythattheequation represents oneparticle with wave number klandasecond particle with wave number k2. The amplitude (15.24) represents onestate with twoparticles (magnons). Thestate ischaracterized by thetwowave numbers klandk2.Oursolution looks likeacompound state of oneparticle with themomentum pl=h/kl andanother particle with themo- mentump2 =h/k2, butinourstate wecan’t saywhich particle iswhich. Bynow, thisdiscussion should remind you ofChapter 4andourstory of identical particles. Wehave justbeen showing thattheparticles ofthespinwaves- themagnons—behave likeidentical Bose particles. Allamplitudes must besym- metric inthecoordinates ofthetwoparticles—which isthesame assaying that ifwe“interchange thetwoparticles," wegetback thesame amplitude andwith thesame sign. But,youmaybethinking, whydidwechoose toaddthetwoterms inmaking Eq.(15.24). Why notsubtract? With aminus sign, interchanging x,,,andx,,would justchange thesign ofam,” which doesn't matter. Butinter- changing xmandx,,doesn’t change anything——all theelectrons ofthecrystal are exactly where they were before, sothere isnoreason foreven thesign ofthe amplitude tochange. Themagnons willbehave likeBose particles.I IIngeneral, thequasi particles ofthekind wearediscussing mayactlikeeither Bose particles orFermi particles, andasforfreeparticles, theparticles with integral spinare bosons andthose withhalf-integral spins arefermions. The“magnon” stands foraspin-up electron turned over. Thechange inspinisone. Themagnon hasanintegral spin, and isaboson. 15-6 The main points ofthisdiscussion have been twofold: First, toshow you something about spin waves, and, second, todemonstrate astate whose amplitude isaproduct oftwoamplitudes, andwhose energy isthesumoftheenergies corre- sponding tothetwoamplitudes. Forindependent particles theamplitude isthe product andtheenergy isthesum. You caneasily seewhy theenergy isthesum. Theenergy isthecoefficient oftinanimaginary exponential—it isproportional tothefrequency. Iftwoobjects aredoing something, oneofthem with theampli- tude e_"E1’/ll and theother with theamplitude e_”5¢‘/"L, and iftheamplitude forthetwothings tohappen together istheproduct oftheamplitudes foreach, then there isasingle frequency intheproduct which isthesum ofthetwofre- quencies. Theenergy corresponding totheamplitude product isthesumofthetwo energies. Wehave gone through arather long-winded argument totellyouasimple thing. When youdon’t take into account anyinteraction between particles, you canthink ofeach particle independently. They canindividually exist inthevarious diflerent states they would have alone, andthey willeach contribute theenergy theywould have hadiftheywere alone. However, youmust remember thatifthey areidentical particles, they may behave either asBose orasFermi particles de- pending upon theproblem. Two extra electrons added toacrystal, forinstance, would have tobehave likeFermi particles. When thepositions oftwoelectrons areinterchanged, theamplitude must reverse sign. Intheequation corresponding toEq.(15.24) there would have tobeaminus sign between thetwoterms onthe right. Asaconsequence, twoFermi particles cannot beinexactly thesame con- dition—with equal spins andequal k’s. Theamplitude forthisstate iszero. 15-4 Thebenzene molecule Although quantum mechanics provides thebasic laws that determine the structures ofmolecules, these laws canbeapplied exactly only tothemost simple compounds. The chemists have, therefore, worked outvarious approximate methods forcalculating some oftheproperties ofcomplicated molecules. We would now liketoshow youhow theindependent particle approximation isused bytheorganic chemists. Webegin with thebenzene molecule. Wediscussed thebenzene molecule from another point ofview inChapter 10. There wetook anapproximate picture ofthemolecule asatwo-state system, with thetwobase states shown inFig.l5—3. There isaring ofsixcarbons with a hydrogen bonded tothecarbon ateach location. With theconventional picture ofvalence bonds itisnecessary toassume double bonds between halfofthecarbon atoms, andinthelowest energy condition there arethetwopossibilities shown in theﬁgure. There arealsoother, higher-energy states. When wediscussed benzene inChapter 10,wejusttook thetwostates andforgot alltherest. Wefound that theground-state energy ofthemolecule wasnottheenergy ofoneofthestates in theﬁgure, butwaslower than that byanamount proportional totheamplitude toﬂipfrom oneofthese states totheother. _ Now we’re going tolook atthesame molecule from acompletely different point ofview—using adifferent kind ofapproximation. Thetwopoints ofview willgiveusdifferent answers, butifweimprove either approximation itshould leadtothetruth, avalid description ofbenzene. However, ifwedon’t bother to improve them, which isofcourse theusual situation, then you should notbe surprised ifthetwodescriptions donotagree exactly. Weshall atleast show that alsowith thenew point-of-view thelowest energy ofthebenzene molecule is lower than either ofthethree-bond structures ofFig. 15-3. Now wewant tousethefollowing picture. Suppose weimagine thesix carbon atoms ofabenzene molecule connected only bysingle bonds asinFig. 15-4. Wehave removed sixelectrons—since abond stands forapairofelectrons —so wehave asix-times ionized benzene molecule. Now wewillconsider what happens when weputback thesixelectrons oneatatime, imagining thateach onecanrunfreely around thering. Weassume also that allthebonds shown in Fig.15-4aresatisﬁed, anddon’t need tobeconsidered further. l5-7H H \ / C: H H H H \ /c—c// \\Ii> H—C C—H \<.//\ /C:c\|2> H—-C C—H\\ //cc/\H H Fig. 15-3. The two base states for thebenzene molecule used inChapter 10. H H\ /C—C / \ H—C 6+ C—H\C c/ / \H H Fig. 15-4. Abenzene ring with six electrons removed. H H \C—C//_\H H Fig. 15-5. Theethylene molecule. E1 i:,+A ~—~ 15,, ——————— —- Eo—A 1‘ Fig. 15-6. Thepossible energy levels forthe"extra" electrons intheethylene molecule. EA E°+A '-ii E0 _____ __ E,-A —$—§— Fig. 15-7. lntheextra bond ofthe ethylene molecule two electrons (one spin up,one spin down) can occupy the lowest energy level.What happens when weputoneelectron back into themolecular ion? It might, ofcourse, belocated inanyoneofthesixpositions around thering~ corresponding tosixbase states. Itwould alsohave acertain amplitude, sayA,to gofrom oneposition tothenext. Ifweanalyze thestationary states, there would becertain possible energy levels. That's only foroneelectron. Next putasecond electron in.And now wemake themost ridiculous ap- proximation thatyoucanthink of—that what oneelectron does isnotaffected by what theother isdoing. Ofcourse they really willinteract; they repel each other through theCoulomb force, andfurthermore when they areboth atthesame site, they must have considerably different energy than twice theenergy foronebeing there. Certainly theapproximation ofindependent particles isnotlegitimate when there areonly sixsites—particularly when wewant toputinsixelectrons. Nevertheless theorganic chemists have been able tolearn alotbymaking this kind ofanapproximation. Before wework outthebenzene molecule indetail, let’s consider asimpler example—the ethylene molecule which contains just twocarbon atoms with two hydrogen atoms oneither sideasshown inFig.l5—5. This molecule hasone“extra” bond involving twoelectrons between thetwocarbon atoms. Now remove one ofthese electrons; what dowehave? Wecanlook atitasatwo-state system—the remaining electron canbeatonecarbon ortheother. Wecananalyze itasatwo- state system. The possible energies forthesingle electron areeither (E0—A) or(El,+A),asshown inFig.15-6. Now addthesecond electron. Good, ifwehave two electrons, wecanput theﬁrstoneinthelower state andthesecond oneintheupper. Notquite; we forgot something. Each oneofthestates isreally double. When wesaythere’s apossible state with theenergy (Ell—A),there arereally two. Two electrons cangointo thesame state ifonehasitsspin upandtheother, itsspin down. (Nomore canbeputinbecause oftheexclusion principle.) Sothere really are twopossible states ofenergy (E0—A).Wecandraw adiagram, asinFig.15-7, which indicates both theenergy levels andtheir occupancy. Inthecondition of lowest energy both electrons willbeinthelowest state with their spins opposite. Theenergy oftheextra bond intheethylene molecule therefore is2(E0 —A)if weneglect theinteraction between thetwoelectrons. Let’s goback tothebenzene. Each ofthetwostates ofFig.15-3hasthree double bonds. Each ofthese isjust likethebond inethylene, andcontributes 2(E0 —A)totheenergy ifEllisnow theenergy toputanelectron onasitein benzene andAistheamplitude toﬂiptothenext site. Sotheenergy should beroughly 6(E0 —A). Butwhen westudied benzene before, wegotthat the energy waslower than theenergy ofthestructure with three extra bonds. Let’s see iftheenergy forbenzene comes outlower than three bonds from ournew point ofview. Westartwiththesix-times ionized benzene ringandaddoneelectron. Now wehave asix-state system. Wehaven’t solved such asystem yet,butweknow what todo.Wecanwrite sixequations inthesixamplitudes, andsoon.But let’s save some work—by noticing that we’ve already solved theproblem, when weworked outtheproblem ofanelectron onaninﬁnite lineofatoms. Ofcourse, thebenzene isnotaninﬁnite line, ithas6atomic sites inacircle. Butimagine that weopen outthecircle toaline, andnumber theatoms along thelinefrom 1to6. Inaninﬁnite linethenext location would be7,butifweinsist that thislocation beidentical with number landsoon,thesituation willbejustlikethebenzene ring. Inother words wecantake thesolution foraninﬁnite linewith anadded requirement thatthesolution must beperiodic with acycle sixatoms long. From Chapter 13theelectron onalinehasstates ofdeﬁnite energy when theamplitude ateach siteise”"i- =eikb". Foreach ktheenergy is E=E0—2Acoskb. (15.25) Wewant tousenow only those solutions which repeat every 6atoms. Let’s doﬁrstthegeneral case foraringofNatoms. Ifthesolution istohave aperiod 15-8 ofNatomic spacing, em” must beunity; orkbNmust beamultiple of21r.Taking storepresent anyinteger, ourcondition isthat kbN=21$. (15.26) Wehave seen before thatthere isnomeaning totaking k’soutside therange 11r/b. This means thatwegetallpossible states bytaking values ofsintherange ¢N/2. Weﬁndthen thatforanN-atom ringthere areNdeﬁnite energy statesl andtheyhave wave numbers k,given by 21rk,=W5s. (15.27) Each state hastheenergy (15.25). Wehave alinespectrum ofpossible energy levels. Thespectrum forbenzene (N=6)isshown inFig. l5—8(b). (The numbers inparentheses indicate thenumber ofdifferent states with thesame energy.) There’s anice way tovisualize thesixenergy levels, aswehave shown in part(a)oftheﬁgure. Imagine acircle centered onalevelwithEll,andwitharadius of2A.Ifwestart atthebottom andmark offsixequal arcs(atangles from the bottom point ofk,b=2rrs/N, or2rrs/6 forbenzene), thenthevertical heights of thepoints onthecircle arethesolutions ofEq.(15.25). Thesixpoints represent thesixpossible states. Thelowest energy level isat(E0—2A); there aretwo states withthesame energy (E0—A),andsoon.I These arepossible states for oneelectron. Ifwehave more thanoneelectron, two—with opposite spins—can gointoeach state. Forthebenzene molecule wehave toputinsixelectrons. Fortheground state theywillgointothelowest possible energy states—two ats=0,twoat s=+1,andtwoats=-1. According totheindependent particle approxima- tiontheenergy oftheground state is Eground =2(E0 —-2A)+4(E0 —A) =6E0 ——8A. (15.28) Theenergy isindeed lessthan thatofthree separate double bonds—~by theamount 2A. Bycomparing theenergy ofbenzene totheenergy ofethylene itispossible todetermine A.Itcomes outtobe0.8electron volt, or,intheunits thechemists like,18kilocalories permole. Wecanusethisdescription tocalculate orunderstand other properties of benzene. Forexample, using Fig. 15-8 wecandiscuss theexcitation ofbenzene bylight. What would happen ifwetried toexcite oneoftheelectrons? Itcould move uptooneoftheempty higher states. Thelowest energy ofexcitation would beatransition from thehighest ﬁlled level tothelowest empty level. That takes theenergy 2A. Benzene willabsorb light offrequency 1/when hll=2A. There willalsobeabsorption ofphotons withtheenergies 3Aand4A.Needless tosay, theabsorption spectrum ofbenzene hasbeenmeasured andthepattern ofspectral lines ismore orlesscorrect except that thelowest transition occurs intheultra- violet; andtoﬁtthedata onewould have tochoose avalue ofAbetween 1.4and 2.4electron volts. That is,thenumerical value ofAistwoorthree times larger thanispredicted from thechemical binding energy. What thechemist does insituations likethisistoanalyze many molecules ofasimilar kind and getsome empirical rules. Helearns, forexample: For calculating binding energy usesuch andsuch avalue ofA,butforgetting the absorption spectrum approximately right useanother value ofA.Youmayfeel ‘IYou might think thatforNaneven number there areN+1states. That isnot sobecause s=d=N/2 givethesame state. IWhen there aretwostates (which willhave different amplitude distributions) with thesame energy, wesaythatthetwostates are“degenerate.” Notice thatfourelectrons canhave theenergy E0—A. 15-9E 5:3_____ -_@°_+_2_A_ (I) 5=-g -=g_Eg+_A__ (2) s=-i€ _--——_-——— (2) E0‘A (ll= rr/6 ,VA~&‘W(<1) (bl Fig. 15-8. The energy levels ina ring with six electron locations (for example, abenzene ring). H H\ / c:c—c:c H/ \H Fig. 15-9. The valence bond repre- sentation ofthemolecule butadiene (1,3). -0~0go-0moE0__e.Q____.I.-, _ I Fig. 15-10. AlineofNmolecules. EA E0+|.6l8A E,+o.eis A W iiiifﬂl:21°-6:? 9‘ -1.68A Fig. 15-11. The energy levels of butadiene.that thissounds alittle absurd. Itisnotvery satisfactory from thepoint ofview ofaphysicist who istrying tounderstand nature from ﬁrst principles. Butthe problem ofthechemist isdifferent. Hemust trytoguess ahead oftime what isgoing tohappen with molecules that haven‘t been made yet,orwhich aren’t understood completely. What heneeds isaseries ofempirical rules; itdoesn’t make much difference where they come from. Soheuses thetheory inquite a different waythan thephysicist. Hetakes equations thathave some shadow ofthe truth inthem, butthen hemust alter theconstants inthem—making empirical corrections. Inthecase ofbenzene, theprincipal reason fortheinconsistency isour assumption that theelectrons areindependent—the theory westarted with is really notlegitimate. Nevertheless, ithassome shadow ofthetruth because its results seem tobegoing intheright direction. With such equations plus some empirical rules-including various exceptions—the organic chemist makes his waythrough themorass ofcomplicated things hechooses tostudy. (Don‘t forget that thereason aphysicist canreally calculate from ﬁrst principles isthat he chooses only simple problems. Henever solves aproblem with 42oreven 6 electrons init.Sofar,hehasbeen abletocalculate reasonably accurately only the hydrogen atom andthehelium atom.) 15-5 More organic chemistry Let's seehow thesame ideas canbeused tostudy other molecules. Consider amolecule likebutadiene (1,3)—it isdrawn inFig. 15-9 according totheusual valence bond picture. Wecanplay thesame game with theextra four electrons corresponding to thetwodouble bonds. Ifweremove four electrons, wehave four carbon atoms inaline. You already know how tosolve aline. You say,“Oh no,Ionly know how tosolve aninﬁnite line.” Butthesolutions fortheinﬁnite linealsoinclude theones foraﬁnite line. Watch. LetNbethenumber ofatoms onthelineand number them from 1toNas shown inFig. 15-10. Inwriting theequations forthe amplitude atposition 1you would nothave aterm feeding from position 0. Similarly, theequation forposition Nwould differ from theonethatwetised for aninﬁnite line because there would benothing feeding from position N+1. Butsuppose thatwecanobtain asolution fortheinﬁnite linewhich hasthefollow- ingproperty: theamplitude tobeatatom 0iszero andtheamplitude tobeat atom (N+1)isalso zero. Then thesetofequations forallthelocations from 1toNontheﬁnite linearealsosatisﬁed. You might think nosuch solution exists fortheinﬁnite linebecause oursolutions alllooked likee”°"'i which hasthesame absolute value oftheamplitude everywhere. Butyouwillremember that theen- ergy depends only ontheabsolute value ofk,sothat another solution, which is equally legitimate forthesame energy, would bee_””». And thesame istrueof anysuperposition ofthese twosolutions. Bysubtracting them wecangetthe solution sinkx,,, which satisﬁes therequirement that theamplitude bezero at x=0.Itstillcorresponds totheenergy (Ell—2Acoskb). Now byasuitable choice forthevalue ofkwecanalso make theamplitude zero atx,v+l. This requires that (N+l)kb beamultiple of1r,orthat kb=-_-L 15.29 (N+1)X, ( ) where sisaninteger from 1toN.(Wetake only positive k‘sbecause each solution contains -I—kand—k; changing thesignofkgives thesame state allover again.) Forthebutadiene molecule, N=4,sothere arefour states with kl): tr/5, 2rr/5, 31r/5, and 41r/5. (15.30) Wecanrepresent theenergy levels using acircle diagram similar totheone forbenzene. This time weuseasemicircle divided intoﬁveequal parts asshown inFig. 15-11. The point atthebottom corresponds tos=O,which gives no 15-10 state atall.The same istrue ofthepoint atthetop, which corresponds tos= N—I—1.The remaining 4points give usfour allowed energies. There arefour stationary states, which iswhat weexpect having started with four base states. Inthecircle diagram, theangular intervals are'rr/5or36degrees. The lowest energy comes out(El,—l.6l8A). (Ah, what wonders mathematics holds; the golden mean oftheGreeksi gives usthelowest energy state ofthebutadiene molecule according tothistheory!) Now wecancalculate theenergy ofthebutadiene molecule when weput infour electrons. With four electrons, weﬁllupthelowest twolevels, each with twoelectrons ofopposite spin. Thetotal energy is E=2(E,,-1.618/1) +2(E,,-0.6l8A) =4(E,,-A)-0.477/1. (15.31) This result seems reasonable. The energy isalittle lower than fortwosimple double bonds, butthebinding isnotsostrong asinbenzene. Anyway thisisthe waythechemist analyzes some organic molecules. Thechemist canusenotonly theenergies buttheprobability amplitudes as well. Knowing theamplitudes foreach state, andwhich states areoccupied, he cantelltheprobability ofﬁnding anelectron anywhere inthemolecule. Those places where theelectrons aremore likely tobeareapttobereactive inchemical substitutions which require that anelectron beshared with some other group of atoms. Theother sites aremore likely tobereactive inthose substitutions which have atendency toyield anextra electron tothesystem. The same ideas wehave been using cangive ussome understanding ofa molecule even ascomplicated aschlorophyll, oneversion ofwhich isshown in Fig. 15-12. Notice that thedouble andsingle bonds wehave drawn with heavy lines form along closed ring with twenty intervals. The extra electrons ofthe double bonds canrunaround thisring. Using theindependent particle method wecangetawhole setofenergy levels. There arestrong absorption lines from transitions between these levels which lieinthevisible part ofthespectrum, and give thismolecule itsstrong color. Similar complicated molecules such asthe xanthophylls, which make leaves turn red,canbestudied inthesame way. There isonemore idea which emerges from theapplication ofthis kind of theory inorganic chemistry. Itisprobably themost successful or,atleast ina certain sense, themost accurate. This idea hastodowith thequestion: Inwhat situations does onegetaparticularly strong chemical binding? Theanswer isvery interesting. Take theexample, ﬁrst, ofbenzene, andimagine thesequence ofevents thatoccurs aswestart with thesix-times ionized molecule andaddmore andmore electrons. Wewould then bethinking ofvarious benzene ions—negative or positive. Suppose weplottheenergy oftheion(orneutral molecule) asafunction ofthenumber ofelectrons. Ifwetake E0=0(since wedon’t know what itis), wegetthecurve shown inFig. 15-13. Fortheﬁrsttwoelectrons theslope ofthe function isastraight line. Foreach successive group theslope increases, and there isadiscontinuity inslope between thegroups ofelectrons. Theslope changes when onehasjustﬁnished ﬁlling asetoflevels which allhave thesame energy and must move uptothenext higher setoflevels forthenext electron. Theactual energy ofthebenzene ionisreally quite different from thecurve ofFig. 15-13 because oftheinteractions oftheelectrons andbecause ofelectro- static energies wehave been neglecting. These corrections will, however, vary with ninarather smooth way. Even ifwewere tomake allthese corrections, the resulting energy curve would stillhave kinks atthose values ofnwhich just ﬁll upaparticular energy level. Now consider avery smooth curve thatﬁtsthepoints ontheaverage likethe onedrawn inFig. 15-14. Wecansaythatthepoints above thiscurve have “higher- than-normal“ energies, andthepoints below thecurve have “lower-than-normal” ITheratio ofthesides ofarectangle which canbedivided intoasquare andasimilar rectangle. 15-11CH=CH2 H3 HBC / / CZHS // H30 \ / —c=o OCH3 c>_o_o0I1|\'Jl‘\)o_o—o oc2oHs9 Fig. 15-12. Achlorophyll molecule. ETOTAL tn5 ol'\7-AO’) I2 -8A Fig. 15-13. The sum ofalltheelec- tron energies when theIowest states in Fig. 15-8 are occupied bynelectrons ifwetake that E0=O. E -m ""6Ki =1o -re-4>-on Fig. 15-14. Thepoints ofFig.15-12 with asmooth curve. Molecules with n=2,6,1Oare more stable than the others. E _f:_°tAl__ (2) E0 Eo—2A —————————— —— (I) Fig. 15-15. Energy diagram for a ring ofthree. Fig. 15-16. The triphenyl cyclo- propanyl cation.energies. Wewould, ingeneral, expect thatthose conﬁgurations with alower-than- normal energy would have anabove average stability—chemically speaking. Notice thattheconﬁgurations farther below thecurve always occur attheendof oneofthestraight linesegments—namely when there areenough electrons toﬁll upan“energy shell," asitiscalled. This isthevery accurate prediction ofthe theory. Molecules—or ions—are particularly stable (incomparison with other similar conﬁgurations) when theavailable electrons justﬁllupanenergy shell. This theory hasexplained andpredicted some very peculiar chemical facts. Totake avery simple example, consider aring ofthree. lt’salmost unbelievable thatthechemist canmake aringofthree andhave itstable, butithasbeen done. Theenergy circle forthree electrons isshown inFig. 15-15. Now ifyouputtwo electrons inthelower state, youhave onlytwoofthethree electrons thatyoure- quire. Thethird electron must beputinatamuch higher level. Byourargument thismolecule should notbeparticularly stable, whereas thetwo-electron structure should bestable. Itdoes turn out, infact, that theneutral molecule oftriphenyl cyclopropenyl isvery hard tomake, butthatthepositive ionshown inFig. 15-16 is relatively easy tomake. The ring ofthree isnever really easy because there is always alarge stress when thebonds inanorganic molecule make anequilateral triangle. Tomake astable compound atall,thestructure must bestabilized in some way. Anyway ifyouaddthree benzene rings onthecorners, thepositive ioncanbemade. (The reason forthisrequirement ofadded benzene rings isnot really understood.) Inasimilar way theﬁve-sided ring canalso beanalyzed. Ifyoudraw the energy diagram, youcanseeinaqualitative waythatthesix-electron structure should beanespecially stable structure, sothatsuch amolecule should bemost stable asanegative ion. Now theﬁve-ring iswellknown andeasytomake and always actsasanegative ion. Similarly, youcaneasily verify thataringof4or8 isnotveryinteresting, butthataringof14orlO—like aringof6—should be especially stable asaneutral object. 15-6 Other usesoftheapproximation There aretwo other similar situations which wewilldescribe only brieﬂy. Inconsidering th/estructure ofanatom, wecanconsider that theelectrons ﬁll successive shells. TheSchrodinger theory ofelectron motion canbeworked out easily only forasingle electron moving ina“central” ﬁeld—one which varies only with thedistance from apoint. How canwethen understand what goes oninan atom which has22electrons?! Onewayistouseakind ofindependent particle approximation. First youcalculate what happens with oneelectron. You geta number ofenergy levels. Youputanelectron intothelowest energy state. You can,forarough model, continue toignore theelectron interactions andgoon ﬁlling successive shells, butthere isawaytogetbetter answers bytaking into account—in anapproximate way atleast—the effect oftheelectric charge carried bytheelectron. Each time youaddanelectron youcompute itsamplitude tobe atvarious places, andthen usethisamplitude toestimate akind ofspherically symmetric charge distribution. You usetheﬁeld ofthisdistribution—together with theﬁeld ofthepositive nucleus andalltheprevious electrons—to calculate thestates available forthenextelectron. Inthiswayyoucangetreasonably cor- rectestimates fortheenergies fortheneutral atom andforvarious ionized states. You ﬁndthatthere areenergy shells, justaswesawfortheelectrons inaring molecule. With apartially ﬁlled shell, theatom willshow apreference fortaking ononeormore extra electrons, orforlosing some electrons soastogetintothe most stable state ofaﬁlled shell. This theory explains themachinery behind thefundamental chemical properties which show upintheperiodic table oftheelements. Theinert gases are those elements inwhich ashell hasjustbeen completed, anditisespecially difficult tomake them react. (Some ofthem doreact ofcourse—with ﬂuorine andoxygen, forexample; butsuch compounds arevery weakly bound; theso-called inert gases arenearly inert.) Anatom which hasoneelectron more oroneelectron less 15-12 than aninert gaswilleasily loseorgain anelectron togetintotheespecially stable (low-energy) condition which comes from having acompletely ﬁlled shell—they arethevery active chemical elements ofvalence +1or-1. Theother situation isfound innuclear physics. Inatomic nuclei theprotons andneutrons interact with each other quite strongly. Even so,theindependent particle model canagain beused toanalyze nuclear structure. Itwasﬁrstdiscovered experimentally thatnuclei were especially stable ifthey contained certain particular numbers ofneutrons—namely 2,8,20,28,50,82. Nuclei containg protons inthese numbers arealsoespecially stable. Since there wasinitially noexplanation forthese numbers they were called the“magic numbers” ofnuclear physics. Itis well known that neutrons andprotons interact strongly with each other; people were, therefore, quite surprised when itwas discovered that anindependent particle model predicted ashell structure which came outwith theﬁrstfewmagic numbers. Themodel assumed thateach nucleon (proton orneutron) moved ina central potential which wascreated bytheaverage effects ofalltheother nucleons. This model failed, however, togivethecorrect values forthehigher magic numbers. Then itwas discovered byMaria Mayer, andindependently byJensen andhis collaborators, that bytaking theindependent particle model andadding only a correction forwhat iscalled the“spin-orbit interaction,” onecould make an improved model which gave allofthemagic numbers. (The spin-orbit interaction causes theenergy ofanucleon tobelower ifitsspinhasthesame direction asits orbital angular momentum from motion inthenucleus.) The theory gives even more—its picture oftheso-called “shell structure” ofthenuclei enables usto predict certain characteristics ofnuclei andofnuclear reactions. The independent particle approximation hasbeen found useful inawide range ofsubjects—from solid-state physics, tochemistry, tobiology, tonuclear physics. Itisoften only acrude approximation, butisabletogiveanunderstanding ofwhy there areespecially stable conditions—in shells. Since itomits allofthe complexity oftheinteractions between theindividual particles, weshould notbe surprised thatitoften fails completely togivecorrectly many important details. 15-13 I6 The Dependence ofAmplitudes onPosition 16-1 Amplitudes onaline Wearenow going todiscuss how theprobability amplitudes ofquantum mechanics vary inspace. Insome oftheearlier chapters you may have hada rather uncomfortable feeling that some things were being leftout. Forexample, when wewere talking about theammonia molecule, wechose todescribe itinterms oftwobase states. Foronebase state wepicked thesituation inwhich thenitrogen atom was“above” theplane ofthethree hydrogen atoms, andfortheother base state wepicked thecondition inwhich thenitrogen atom was“below” theplane ofthethree hydrogen atoms. Why didwepickjustthese twostates? Why isit notpossible thatthenitrogen atom could beat2angstroms above theplane ofthe three hydrogen atoms, orat3angstroms, orat4angstroms above theplane‘? Certainly, there aremany positions that thenitrogen atom could occupy. Again when wetalked about thehydrogen molecular ion,inwhich there isoneelectron shared bytwoprotons, weimagined twobase states: onefortheelectron inthe neighborhood ofproton number one, andtheother fortheelectron intheneigh- borhood ofproton number two. Clearly wewere leaving outmany details. The electron isnotexactly atproton number two butisonly intheneighborhood. Itcould besomewhere above theproton, somewhere below theproton, somewhere totheleftoftheproton, orsomewhere totheright oftheproton. Weintentionally avoided discussing these details. Wesaid that wewere interested inonly certain features oftheproblem, sowewere imagining thatwhen theelectron wasinthevicinity ofproton number one, itwould take upacertain rather deﬁnite condition. Inthat condition theprobability toﬁnd theelectron would have some rather deﬁnite distribution around theproton. butwewere not interested inthedetails. Wecanalso putitanother way. Inourdiscussion ofahydrogen molecular ionwechose anapproximate description when wedescribed thesituation interms oftwobasestates. Inreality there arelotsandlotsofthese states. Anelectron can take upacondition around aproton initslowest, orground, state, butthere are alsomany excited states. Foreach excited state thedistribution oftheelectron around theproton isdifferent. Weignored these excited states, saying that we were interested inonly theconditions oflowenergy. Butitisjust these other excited states which give thepossibility ofvarious distributions oftheelectron around theproton. Ifwewant todescribe indetail thehydrogen molecular ion, wehave totake into account also these other possible base states. Wecould do thisinseveral ways, andonewayistoconsider ingreater detail states inwhich the location oftheelectron inspace ismore carefully described. Wearenow ready toconsider amore elaborate procedure which willallow ustotalk indetail about theposition oftheelectron, bygiving aprobability amplitude toﬁndtheelectron anywhere andeverywhere inagiven situation. This more complete theory provides theunderpinning fortheapproximations wehave been making inourearlier discussions. Inasense, ourearly equations canbe derived asakind ofapproximation tothemore complete theory. You may bewondering why wedidnotbegin with themore complete theory andmake theapproximations aswewent along. Wehave feltthat itwould be much easier foryoutogain anunderstanding ofthebasic machinery ofquantum mechanics bybeginning with thetwo-state approximations andworking gradually uptothemore complete theory than toapproach thesubject theother wayaround. Forthisreason ourapproach tothesubject appears tobeinthereverse order to theoneyouwillﬁndinmany books. l6—l16-1 Amplitudes onaline 16-2 Thewave function 16-3 States ofdeﬁnite momentum 16-4 Normalization ofthestates inx 16-5 TheSchrodinger equation 16-6 Quantized energy levels Aswegointothesubject ofthischapter youwillnotice thatwearebreaking arulewehave always followed inthepast. Whenever wehave taken upany subject wehave always tried togiveamore orlesscomplete description ofthe physics—showing youasmuch aswecould about where theideas ledto.We have tried todescribe thegeneral consequences ofatheory aswellasdescribing some speciﬁc detail sothatyoucould seewhere thetheory would lead. Weare nowgoing tobreak thatrule; wearegoing todescribe howonecantalkabout probability amplitudes inspace andshow youthedifferential equations which theysatisfy. Wewillnothave time togoonanddiscuss many oftheobvious implications which come outofthetheory. Indeed wewillnoteven beabletoget farenough torelate thistheory tosome oftheapproximate formulations wehave usedearlier—for example, tothehydrogen molecule ortotheammonia molecule. Foronce, wemust leave ourbusiness unﬁnished andopen-ended. Weareapproach- ingtheendofourcourse, andwemust satisfy ourselves withtrying togiveyouan introduction tothegeneral ideas andwith indicating theconnections between what wehave been describing andsome oftheother ways ofapproaching thesubject ofquantum mechanics. Wehope togiveyouenough ofanideathatyoucango oﬁ"byyourself andbyreading books learn about many oftheimplications ofthe equations wearegoing todescribe. Wemust, after all,leave something forthe future. Let’s review once more what wehave found outabout howanelectron can move along alineofatoms. When anelectron hasanamplitude tojump from oneatom tothenext, there aredeﬁnite energy states inwhich theprobability ampli- tudeforﬁnding theelectron isdistributed along thelattice intheform ofatravel- ingwave. Forlong wavelengths—for small values ofthewave number k—the energy ofthestateisproportional tothesquare ofthewave number. Foracrystal lattice withthespacing b,inwhich theamplitude perunittimefortheelectron to jump from oneatom tothenextisiA/h, theenergy ofthestate isrelated tok (forsmall kb)by E=Akzbz (16.1) (seeSection 13-3). Wealsosawthatgroups ofsuch waves withsimilar energies would make upawave packet which would behave likeaclassical particle with a mass meffgiven by: h2 mt-rs =ﬂﬁ‘ (16-2) Since waves ofprobability amplitude inacrystal behave likeaparticle, one might wellexpect thatthegeneral quantum mechanical description ofaparticle would show thesame kindofwave behavior weobserved forthelattice. Suppose wewere tothink ofalattice onalineandimagine thatthelattice spacing bwere to bemade smaller andsmaller. Inthelimitwewould bethinking ofacaseinwhich theelectron could beanywhere along theline. Wewould have gone over toa continuous distribution ofprobability amplitudes. Wewould have theamplitude toﬁndanelectron anywhere along theline. Thiswould beonewaytodescribe themotion ofanelectron inavacuum. Inother words, ifweimagine thatspace can belabeled byaninﬁnity ofpoints allveryclose together andwecanwork outthe equations thatrelate theamplitudes atonepoint totheamplitudes atneighboring points, wewillhave thequantum mechanical laws ofmotion ofanelectron inspace. Let’s begin byrecalling some ofthegeneral principles ofquantum mechanics. Suppose wehave aparticle which canexist invarious conditions inaquantum mechanical system. Anyparticular condition anelectron canbefound in,wecall a“state,” which welabel withastatevector, sayI4>).Some other condition would belabeled withanother state vector, say|1/1).Wethenintroduce theideaofbase states. Wesaythatthere isasetofstates [1),|2),|3),|4),andsoon,which have thefollowing properties. First, allofthese states arequite distinct—we say theyareorthogonal. Bythiswemean thatforanytwoofthebasestates l1')and Ij\theamplitude (i|j)thatanelectron known tobeinthestate [1')isalsointhe 16-2 stateIj)isequal tozero—unless, ofcourse, Ii)andIj)stand forthesame state. Werepresent thissymbolically by (ill) =511- (16-3) Youwillremember that 6,;=0ifiandj aredifferent, and6,-j=1ifiandj are thesame number. Second, thebasestates Ii)must beacomplete set,sothatanystate atallcan bedescribed interms ofthem. That is,anystate Iqs)atallcanbedescribed com- pletely bygiving alloftheamplitudes (iI4:)that aparticle inthestate I¢)will alsobefound inthestate Ii).Infact, thestate vector I11>)isequal tothesum of thebase states each multiplied byacoefﬁcient which istheamplitude ofthe stateI¢)isalsointhestate Ii): l¢>=§jwxMo_ (ma) Finally, ifweconsider anytwostates Iqb)andI1/1),theamplitude thatthestate Itb)willalsobeinthestate I¢)canbefound byﬁrstprojecting thestate II/1)into thebase states andthen projecting from each base state into thestate I¢).We write thatinthefollowing way: Ww=ZMMW> mm Thesummation is,ofcourse, tobecarried outoverthewhole setofbasestate Ii). InChapter 13when wewere working outwhat happens with anelectron placed onalinear array ofatoms, wechose asetofbase states inwhich theelectron was localized atoneorother oftheatoms intheline. Thebase state In)represented thecondition inwhich theelectron waslocalized atatom number “n.” (There is, ofcourse, nosigniﬁcance tothefactthat wecalled ourbase states In)instead of Ii).)Alittle later, wefound itconvenient tolabel thebase states bythecoordinate xnoftheatom rather than bythenumber oftheatom inthearray. The state Ix,,)isjustanother wayofwriting thestate In).Then, following thegeneral rules, anystate atall,sayII0)isdescribed bygiving theamplitudes andthatanelectron inthestate I¢)isalsoinoneofthestates Ixn). Forconvenience wehave chosen toletthesymbol C,,stand forthese amplitudes, Cn='(xnI111). (16.6) Since thebasestates areassociated withalocation along theline,wecanthink oftheamplitude C,,asafunction ofthecoordinate xandwrite itasC(x,,). The amplitudes C(x,,) will, ingeneral, vary with time andare,therefore, alsofunctions of1.Wewillnotgenerally bother toshow explicitly thisdependence. InChapter 13wethenproposed thattheamplitudes C(x,,) should varywith time inaway described bytheHamiltonian equation (Eq. 13.3). Inournew notation thisequation is ih =E0C(x,,) -AC(x,, +b)-AC(x,, -b). (16.7) Thelasttwoterms ontheright-hand siderepresent theprocess inwhich anelectron atatom (n+I)oratatom (n—l)canfeedintoatom n. Wefound thatEq.(l6.7) hassolutions corresponding todeﬁnite energy states, which wewrote as C(x,,)=ME‘/”-@“"». (16.8) Forthelow-energy states thewavelengths arelarge (kissmall), andtheenergy is related tokby E=(E0-2A)+Akzbz, (16.9) or,choosing ourzeroofenergy sothat(E0—2A)=O,theenergy isgiven by Eq.(16.1). 16-3 Let’s seewhat might happen ifwewere toletthelattice spacing bgotozero, keeping thewave number kﬁxed. Ifthatisallthatwere tohappen thelastterm inEq.(16.9) would justgotozeroandthere would benophysics. Butsuppose Aandbarevaried together sothatasbgoes tozero theproduct Abziskept constantI'-—using Eq.(16.2) wewillwrite Abzastheconstant hz/2mm. Under these circumstances, Eq.(16.9) would beunchanged, butwhat would happen tothe differential equation (16.7)? First wewillrewrite Eq.(16.7) as #1 =(E0-2A>c<x..) +A12<r<><..> -co.+1»)~co.-or (16.10) Forourchoice ofE0,theﬁrsttermdrops out. Next, wecanthink ofacontinuous function C(x)thatgoessmoothly through theproper values C(x,,) ateachx,,.As thespacing bgoestozero, thepoints x,,getcloser andcloser together, and(ifwe keep thevariation ofC(x) fairly smooth) thequantity inthebrackets isjustpro- portional tothesecond derivative ofC(x). Wecanwrite—-as youcanseebymaking aTaylor expansion ofeach term——the equality 2C(x)_C(x+b)-C(x-6)z-629i%3‘l- (16.11) Inthelimit, then, asbgoestozero, keeping b2Aequal toK,Eq.(16.7) goesover into -§£<£)__“.2.6220). zhat - zmeﬂ ——ax2~— (16.12) Wehave anequation which saysthatthetimerateofchange ofC(x)—the ampli- tude toﬁndtheelectron atx-—depends ontheamplitude toﬁndtheelectron at nearby points inawaywhich isproportional tothesecond derivative ofthe amplitude withrespect toposition. Thecorrect quantum mechanical equation forthemotion ofanelectron in freespace wasﬁrstdiscovered bySchrodinger. Formotion along alineithas exactly theform ofEq.(16.12) ifwereplace meffbym,thefree-space mass ofthe electron. Formotion along alineinfreespace theSchrodinger equation is .6C(x) _ h’62C(x)_m_-a7- _-57"-5?; (16.13) Wedonotintend tohaveyouthink wehavederived theSchrodinger equation butonlywish toshow youonewayofthinking about it.When Schrodinger ﬁrst wrote itdown, hegaveakindofderivation based onsome heuristic arguments and some brilliant intuitive guesses. Some ofthearguments heusedwereevenfalse, but thatdoes notmatter; theonlyimportant thing isthattheultimate equation gives acorrect description ofnature. Thepurpose ofourdiscussion isthen simply to show youthatthecorrect fundamental quantum mechanical equation (16.13) hasthesame form yougetforthelimiting caseofanelectron moving along aline ofatoms. This means thatwecanthink ofthedifferential equation in(16.13) asdescribing thediffusion ofaprobability amplitude from onepoint tothenext along theline. That is,ifanelectron hasacertain amplitude tobeatonepoint, it will, alittle time later, have some amplitude tobeatneighboring points. Infact, theequation looks something likethediffusion equations which wehave used in Volume 1.Butthere isonemain difference: theimaginary coefficient infront of thetime derivative makes thebehavior completely different from theordinary diffusion suchasyouwould have foragasspreading outalong athintube. Ordi- nary diffusion gives risetorealexponential solutions, whereas thesolutions of Eq.(16.13) arecomplex waves. IYoucanimagine thatasthepoints x,,getcloser together, theamplitude Atojump from xnd:1tox,,willincrease. 16—4 16-2 Thewave function Now thatyouhave some ideaabout howthings aregoing tolook, wewant togoback tothebeginning andstudy theproblem ofdescribing themotion ofan electron along alinewithout having toconsider states connected with atoms ona lattice. Wewant togoback tothebeginning andseewhat ideas wehave touse ifwewant todescribe themotion ofafreeparticle inspace. Since weareinterested inthebehavior ofaparticle along acontinuum, wewillbedealing withaninﬁnite number ofpossible states and, asyouwillsee,theideas wehave developed for dealing with aﬁnite number ofstates willneed some technical modiﬁcations. Webegin byletting thestate vector Ix)stand forastate inwhich aparticle is located precisely atthecoordinate x.Forevery value xalong theline—for instance 1.73,or9.67, orl0.00—there isthecorresponding state. Wewilltakethese states Ix)asourbase states and, ifweinclude allthepoints ontheline, wewillhave acomplete setformotion inonedimension. Now suppose wehave adifferent kindofastate, sayItp),inwhich anelectron isdistributed insome wayalong the line. Onewayofdescribing thisstate istogivealltheamplitudes thattheelectron willbealsofound ineach ofthebase states Ix).Wemust give aninﬁnite setof amplitudes, oneforeach value ofx.Wewillwrite these amplitudes as(xI50). Each ofthese amplitudes isacomplex number andsince there isonesuch complex number foreach value ofx,theamplitude (xI1//)isindeed justafunction ofx, Wewillalsowrite itasC(x), C(x) E(xIIL). (16.14) Wehave already considered suchamplitudes which varyinacontinuous way with thecoordinates when wetalked about thevariations ofamplitude with time inChapter 7.Weshowed there, forexample, that aparticle with adeﬁnite mo- mentum should beexpected tohave aparticular variation ofitsamplitude in space. Ifaparticle hasadeﬁnite momentum pandacorresponding deﬁnite energy E,theamplitude tobefound atanyposition xwould look like (xI1l/)=cot)<=<6+6"/". (16.15) Thisequation expresses animportant general principle ofquantum mechanics which connects thebasestates corresponding todifferent positions inspace toanother system ofbasestates—all thestates ofdeﬁnite momentum. Thedeﬁnite momentum states areoften more convenient than thestates inxforcertain kinds ofproblems. Either setofbase states is,ofcourse, equally acceptable foradescription ofa quantum mechanical situation. Wewillcome back later tothematter ofthe connection between them. Forthemoment wewant tostick toourdiscussion of adescription interms ofthestates Ix). Before proceeding, wewant tomake onesmall change innotation which we hope willnotbetooconfusing. The function C(x), deﬁned inEq.(16.14), will ofcourse have aform which depends ontheparticular state I1,0)under considera- tion. Weshould indicate thatinsome way. Wecould, forexample, specify which function C(x)wearetalking about byasubscript say,C¢(x). Although thiswould beaperfectly satisfactory notation, itisalittle bitcumbersome andisnottheone youwillﬁndinmost books. Most people simply omit theletter Candusethe symbol I0todeﬁne thefunction 11/(X)EC¢(X) =(XI~l/>- (16-16) Since thisisthenotation used byeverybody elseintheworld, youmight aswell getusedtoitsothatyouwillnotbefrightened when youcome across itsomewhere else. Remember though, thatwewillnowbeusing upintwodifferent ways. In Eq.(16.14),#1stands foralabel wehave given toaparticular physical state ofthe electron. Ontheleft-hand sideofEq.(16.16), ontheother hand, thesymbol 1/1 isused todeﬁne amathematical function ofxwhich isequal totheamplitude to beassociated witheachpoint xalong theline.Wehope itwillnotbetooconfusing 16-5 once yougetaccustomed totheidea. Incidentally, thefunction 1/1(x) isusually called “the wave function”——because itmore often than nothastheform ofacom- plex wave initsvariables. Since wehave deﬁned ¢(x)tobetheamplitude thatanelectron inthestate1/ willbefound atthelocation x,wewould liketointerpret theabsolute square of 1/1tobetheprobability ofﬁnding anelectron attheposition x.Unfortunately, the probability ofﬁnding aparticle exactly atanyparticular point iszero. Theelectron will, ingeneral, besmeared outinacertain region oftheline, andsince, inany small piece oftheline, there areaninﬁnite number ofpoints, theprobability that itwillbeatanyoneofthem cannot beaﬁnite number. Wecanonly describe the probability ofﬁnding anelectron interms ofaprobability (/i.s'!ributi0n'I' which gives there/alive probability ofﬁnding theelectron atvarious approximate locations along theline. Let's letprob (x,Ax)stand forthechance ofﬁnding theelectron inasmall interval Axlocated near x.Ifwegotoasmall enough scale inany physical situation, theprobability willbevarying smoothly from place toplace, andtheprobability ofﬁnding theelectron inanysmall ﬁnite linesegment Axwill beproportional toAx.Wecanmodify ourdeﬁnitions totakethisintoaccount. Wecanthink oftheamplitude (xI1b)asrepresenting akind of“amplitude density" forallthebase states Ix)inasmall region. Since theprobability of ﬁnding anelectron inasmall interval Axatxshould beproportional totheinterval Ax,wechoose ourdeﬁnition of(xI1/1)sothat thefollowing relation holds: prob (x,Ax)=I(xI1l1)I2Ax. Theamplitude (xI1/)istherefore proportional totheamplitude thatanelectron inthestate 1/1willbefound inthebase state xandtheconstant ofproportionality ischosen sothattheabsolute square oftheamplitude (xI1/)gives theprobability density ofﬁnding anelectron inanysmall region. Wecanwrite, equivalently, prob (x,Ax)=I1/(x)I2 Ax. (16.17) Wewillnow have tomodify some ofourearlier equations tomake them compatible withthisnewdeﬁnition ofaprobability amplitude. Suppose wehave anelectron inthestate I10)andwewant toknow theamplitude forﬁnding itina different state I11>)which may correspond toadifferent spread-out condition oftheelectron. When wewere talking about aﬁnite setofdiscrete states, wewould have used Eq.(16.5). Before modifying ourdeﬁnition oftheamplitudes wewould have written <¢-11>=Z)<¢1><><><|11>. (16.18)all2: Now ifboth ofthese amplitudes arenormalized inthesame wayaswehave de- scribed above, thenasumofallthestates inasmall region ofxwould beequivalent tomultiplying byAx,andthesum over allvalues ofxsimply becomes anintegral. With ourmodiﬁed deﬁnitions, thecorrect form becomes <¢111>-/U<¢1><><x1i>d»<. 116.191 Theamplitude (xI1/)iswhat wearenowcalling 1/1(x)and,inasimilar way, wewillchoose tolettheamplitude (xI1/1)berepresented by¢(x). Remembering that(41Ix)isthecomplex conjugate of(xI41),wecanwrite Eq.(16.18) as <11I11>=/1»*<><>i<x> dx. <16-20> With ournewdeﬁnitions everything follows with thesame formulas asbefore if youalways replace asummation signbyanintegral over x. Weshould mention onequaliﬁcation towhat wehave been saying. Any suitable setofbase states must becomplete ifitistobeused foranadequate ‘IForadiscussion ofprobability distributions seeVol.I,Section 6—4. 16-6 description ofwhat isgoing on.Foranelectron inonedimension itisnotreally sufficient tospecify only thebase states Ix),because foreach ofthese states the electron mayhave aspinwhich iseither upordown. Onewaybfgetting acomplete setistotaketwosetsofstates inx,oneforupspinandtheother fordown spin. Wewill,however, notworry about such complications forthetimebeing. 16-3 States ofdeﬁnite momentum Suppose wehave anelectron inastate I10)which isdescribed bytheprob- ability amplitude (xI1/)=1p(x). Weknow that thisrepresents astate inwhich theelectro"n isspread outalong thelineinacertain distribution sothattheprob- ability ofﬁnding theelectron inasmall interval dxatthelocation xisjust prob (x,dx)=I1/(x)I2 dx. What canwesayabout themomentum ofthiselectron? Wemight askwhat is theprobability that thiselectron hasthemomentum p?Let's start outbycal- culating theamplitude that thestate I1,b)isinanother state Imom p)which we deﬁne tobeastate with thedeﬁnite momentum p.Wecanﬁndthisamplitude by using ourbasic equation fortheresolution ofamplitudes, Eq.(16.20). Interms ofthestate Imom p) (mom pI1/1)=I“ (mom pIx)(xI1/1)dx. (16.21) And theprobability that theelectron willbefound with themomentum pshould begiven interms oftheabsolute square ofthis amplitude. Wehave again, however, asmall problem about thenormalizations. Ingeneral wecanonly askabout the probability ofﬁnding anelectron with amomentum inasmall range dpatthe momentum p.Theprobability thatthemomentum isexactly some value pmust be zero (unless thestate I1/1)happens tobeastate ofdeﬁnite momentum). Only ifwe askfortheprobability ofﬁnding themomentum inasmall range dpatthemo- mentum pwillwegetaﬁnite probability. There areseveral ways thenormalizations canbeadjusted. Wewillchoose oneofthem which wethink tobethemost convenient, although thatmaynotbeapparent toyoujustnow. Wetake ournormalizations sothattheprobability isrelated totheamplitude by 2dPprob(161111) =I<m<>mr>|~P>l 5;,‘ (16-22) With thisdeﬁnition thenormalization oftheamplitude (mom pIx)isdetermined. Theamplitude (momp Ix)is,ofcourse, justthecomplex conjugate oftheampli- tude (xImomp), which isjust theonewehave written down inEq.(16.15). With thenormalization wehave chosen, itturns outthat theproper constant of proportionality infront oftheexponential isjust 1.Namely, (momp Ix)I(xImom p)*=e_“""/'1. (16.23) Equation (16.21) then becomes (mom pI1/1)= eT'i"”/'l(x I1/)dx. (16.24) Thisequation together withEq.(16.22) allows ustoﬁndthemomentum distribu- tionforanystate I1;). Let’s look ataparticular example—for instance oneinwhich anelectron islocalized inacertain region around x=0.Suppose wetake awave function which hasthefollowing form: 1(1)=K6-1“/4"’. (16.25) Theprobability distribution inxforthiswave function istheabsolute square, or prob(>1,I/X)=P(x)1/X=K21-I’/2"’1/X. (16.26) 16-7 Fig. 16-1. The probability density forthewove function ofEq.(16.24).Theprobability density function P(x) istheGaussian curve shown inFig.16-1. Most oftheprobability isconcentrated between x=+0andx=-11. Wesay thatthe“half-width” ofthecurve is0'.(More precisily, 0isequal totheroot-mean- square ofthecoordinate xforsomething spread outaccording tothisdistribution.) Wewould normally choose theconstant Ksothattheprobability density P(x) isnotmerely proportional totheprobability perunitlength inxofﬁnding the electron, buthasascale such thatP(x)Axisequal totheprobability ofﬁnding theelectron inAxnearx.Theconstant Kwhich doesthiscanbefound byrequiring thatf_+,f,°P(x) dx=1,since there must beunitprobability thattheelectron is found somewhere. Here, wegetthatK=(21ro'2)_‘/4. [Wehave used thefact thatff;e“‘2dt=\/Tr; seeVol.I,page 40-6.] P(x) 0.4 0.3 .°M =1------—-— O2 qr--—-- ,,,_=1 -56-26 - 30''1’ Now let’sﬁndthedistribution inmomentum. Let’s let¢(p) stand forthe amplitude toﬁndtheelectron withthemomentum p, ¢>(.v)E(momp I1l/)- (16-27) Substituting Eq.(16.25) intoEq.(16.24) weget ¢(p)=Ll”6-"P="‘ -Ke_"2/“zdx. (16.28) theintregral canalsoberewritten as K6-Pi"/"’/+°° 2-<1/“’><*+2"'"*/">i1x. (16.29)—so Wecannowmake thesubstitution u=x+2iprr2/h, andtheintegral is /+°°e_“2’4"2du =261/71. (16.30) (The mathematicians would probably object tothewaywegotthere, buttheresult is,nevertheless, correct.) 1/>(p) =(81ra'2)1"‘e_"2"2/'12. (16.31) Wehave theinteresting result thattheamplitude function inphasprecisely thesame mathematical form astheamplitude function inx;only thewidth ofthe Gaussian isdifferent. Wecanwrite thisas ¢(.v)=(21r11"’)""“@"’2""’2, (16-32) where thehalf-width 1;ofthep-distribution function isrelated tothehalf-width tr ofthex-distribution by h.,_5- (16.33) 16-8 Ourresult says: ifwemake thewidth ofthedistribution inxvery small by makinga small, 17becomes large andthedistribution inpisvery much spread out. Or,conversely: ifwehave anarrow distribution inp,itmust correspond toa spread-out distribution inx.Wecan, ifwelike, consider 1;anda tobesome meas- ureoftheuncertainty inthelocalization ofthemomentum andoftheposition of theelectron inthestate wearestudying. Ifwecallthem ApandAxrespectively Eq.(16.33) becomes ApAx=Z’. (16.34)2 Interestingly enough, itispossible toprove that foranyother form ofa adistribution inxorinp,theproduct ApAxcannot besmaller than theone wehave found here. The Gaussian distribution gives thesmallest possible value fortheproduct oftheroot-mean-square widths. Ingeneral, wecansay ApAxZ (16.35) This isaquantatative statement oftheHeisenberg uncertainty principle, which we have discussed qualitatively many times before. Wehave usually made theap- proximate statement thattheminimum value oftheproduct ApAxisofthesame order ash. 16-4 Normalization ofthestates inx Wereturn now tothediscussion ofthemodiﬁcations ofourbasic equations which arerequired when wearedealing with acontinuum ofbase states. When wehave aﬁnite number ofdiscrete states, afundamental condition which must be satisﬁed bythesetofbase states is (iIj)=511- (16-36) lfaparticle isinonebase state, theamplitude tobeinanother base state is0.By choosing asuitable normalization, wehave deﬁned theamplitude (iI1')tobel. These twoconditions aredescribed byEq.(16.36). Wewant now toseehow this relation must bemodiﬁed when weusethebase states Ix)ofaparticle ona line. Iftheparticle isknown tobeinoneofthebase states Ix),what isthe amplitude thatitwillbeinanoIher base state Ix’)? Ifxandx’aretwo different locations along theline, thenjthe amplitude (xIx’)iscertainly O,sothat is consistent with Eq.(16.36). Butifxandx’areequal, theamplitude (xIx’)will notbe1,because ofthesame oldnormalization problem. Toseehow wehave to patch things up,wegoback toEq.(16.19), andapply thisequation tothespecial case inwhich thestate I1/5)isjustthebase state Ix’). Wewould have then <><'|i>=f<><'|><>to)dx. (16311 Now theamplitude (xI1/1)isjust what wehave been calling thefunction 1p(x). Similarly theamplitude (x’I1/1),since itrefers tothesame state 1/1,isthesame func- tion ofthevariable x’,namely 1/1(x’). Wecan, therefore, rewrite Eq.(16.37) as 1/(x')=/<x'|>1)1/1(x)dx. (16.32) This equation must betrueforanystate (I1and, therefore, foranyarbitrary function 1b(x). This requirement should completely determine thenature oftheamplitude (xIx’)—which is,ofcourse, justafunction thatdepends onxandx’. Our problem now istoﬁnd afunction f(x,x’)which when multiplied into 1/(x), andintegrated over allxgives justthequantity 1/1(x’). Itturns outthatthere isnomathematical function which willdothis! Atleast nothing likewhat we ordinarily mean bya“function.” 16-9 11‘(x) \\-.\.\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\®\\\\\\\\\\\\\\\\\\\\\\§\\\\\\\\\\\\R\\I1\\\\\\\\\.3\\\\\\\\\\\\\\\\\\‘_\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\bl I\) Q >< Fig. 16-2. Asetoffunctions, c|||of unit oreo, which look more and more like (f(x).Suppose wepick x’tobethespecial number 0anddeﬁne theamplitude (0|x)tobesome function ofx,let’ssayf(x). Then Eq.(16.36) would readas follows: ¢(o)=/f(x)1//(x) dx. (16.39) What kindoffunction f(x)could possibly satisfy thisequation? Since theintegral must notdepend onwhat values ¢(x)takes forvalues ofxother than 0,f(x) must clearly be0forallvalues ofxexcept 0.Butiff(x) is0everywhere, the integral willbe0,too, andEq.(16.39) willnotbesatisﬁed. Sowehave anim- possible situation: wewish afunction tobe0everywhere butatapoint, andstill togiveaﬁnite integral. Since wecan’t ﬁndafunction thatdoes this,theeasiest wayoutisjusttosaythatthefunction f(x) isdeﬁned byEq.(16.37). Namely, f(x) isthat function which makes (16.39) correct. Thefunction which does this wasﬁrstinvented byDirac andcarries hisname. Wewrite it5(x). Allwearesay- ingisthat thefunction 5(x) hasthestrange property that ifitissubstituted for f(x) intheEq.(16.39), theintegral picks outthevalue that(I/(x), takes onwhen xisequal 0;and, since theintegral must beindependent of1//(x) forallvalues ofxother thanO,thefunction 6(x)must be0everywhere except atx=0.Sum- marizing, wewrite (OIx)=6(x), (16.40) where 6(x)isdeﬁned by ¢(0)=/6(x)\[/(x) dx. (16.41) Notice what happens ifweusethespecial function “l”forthefunction 51/inEq. (16.41). Then wehave theresult 1=/6(x)dx. (16.42) That is,thefunction 6(x)hastheproperty thatitis0everywhere except atx=0 buthasaﬁnite integral equal tounity. Wemust imagine that thefunction 6(x)hassuch afantastic inﬁnity atonepoint thatthetotal area comes outequal toone. Onewayofimagining what theDirac 6-function islikeistothink ofasequence ofrectangles—or anyother peaked function youcare to—which gets narrower andnarrower andhigher andhigher, always keeping aunit area, assketched in Fig. 16-2. The integral ofthisfunction from —oe to—|—w isalways l.Ifyou multiply itbyanyfunction 1p(x) andintegrate theproduct, yougetsomething which isapproximately thevalue ofthefunction atx=0,theapproximation getting better andbetter asyouusethenarrower andnarrower rectangles. You canifyouwish, imagine the6-function interms ofthiskind oflimiting process. The only important thing, however, isthat the6-function isdeﬁned sothat Eq. (16.41) istrueforevery possible function ¢(x). That uniquely deﬁnes the6-function. Itsproperties arethen aswehave described. Ifwechange theargument ofthe6-function from xtox—x’,thecorre- sponding relations are 6(x—x’)=0, x’¢x, /.s(x-x’)\//(x)dx =tl/(x’). (16.43) Ifweuse6(x—x’)fortheamplitude (x]x’)inEq.(16.38), thatequation is satisﬁed. Ourresult then isthatforourbase states inx,thecondition corre- sponding to(16.36) is (x’Ix)=6(x—x’). (16.44) 16-10 Wehave now completed thenecessary modiﬁcations ofourbasic equations which arenecessary fordealing withthecontinuum ofbasestates corresponding tothepoints along aline. The extension tothree dimensions isfairly obvious; ﬁrstwereplace thecoordinate xbythevector r.Then integrals over xbecome re- placed byintegrals over x,y,andz.Inother words, they become volume integrals. Finally, theone-dimensional 6-function must bereplaced byjust theproduct of three 6-functions, oneinx,oneiny,and theother in2,6(x—x’)6(y—y’) 6(z—2'). Putting everything together wegetthefollowing setofequations for theamplitudes forparticle inthree dimensions: <¢1¢>=[<¢1»><r1¢>4v<>1. (16.45) <r111>=46). <11¢>=<46). <¢1¢>=[¢>*(r)¢<r)dv<>1. (16.47)(16.46) (r’II‘)=5(X—X’)<5(y-y’)5(Z—Z’), (16-43) What happens when there ismore than oneparticle? Wewilltellyouabout howtohandle twoparticles andyouwilleasily seewhat youmust doifyouwant todeal with alarger number. Suppose there aretwoparticles, which wecancall particle No.1andparticle No.2.What shall weuseforthebase states? One perfectly good setcanbedescribed bysaying that particle 1isatx1andparticle 2isatx2,which wecanwrite asI.X1Xg>. Notice that describing theposition of only oneparticle does notdeﬁne abase state. Each base state must deﬁne the condition oftheentire system. You must notthink thateach particle moves inde- pendently asawave inthree dimensions. Any physical state I1//)canbedeﬁned bygiving alloftheamplitudes (x1,x2I(0)toﬁndthetwoparticles atx1andx2. This generalized amplitude istherefore afunction ofthetwosetsofcoordinates x1andx2.You seethatsuch afunction isnotawave inthesense ofanoscillation thatmoves along inthree dimensions. Neither isitgenerally simply aproduct of twoindividual waves, oneforeachparticle. Itis,ingeneral, some kindofawave inthesixdimensions deﬁned byx1andx2. Ifthere aretwoparticles innature which areinteracting, there isnoway ofdescribing what happens tooneofthe particles bytrying towrite down awave function foritalone. Thefamous para- doxes that weconsidered inearlier chapters—where themeasurements made on oneparticle were claimed tobeabletotellwhat wasgoing tohappen toanother particle, orwere able todestroy aninterference—have caused people allsorts oftrouble because they have tried tothink ofthewave function ofoneparticle alone, rather than thecorrect wave function inthecoordinates ofboth particles. Thecomplete description canbegiven correctly only interms offunctions ofthe coordinates ofboth particles. l6—5 TheSchrodinger equation Sofarwehave just been worrying about how wecandescribe states which mayinvolve anelectron being anywhere atallinspace. Now wehave toworry about putting into ourdescription thephysics ofwhat canhappen invarious circumstances. Asbefore, wehave toworry about how states canchange with time. lfwehave astate I111)which goes over into another state |¢’) sometime later, wecandescribe thesituation foralltimes bymaking thewave function—which isjust theamplitude (rI1//)—a function oftime aswell asafunction oftheco- ordinate. Aparticle inagiven situation canthen bedescribed bygiving atime- varying wave function II/(r,t)=1//(x,y,z,r).This time-varying wave function describes theevolution ofsuccessive states that occur astime develops. This so-called “coordinate representation”—which gives theprojections ofthestate I1;»)intothebasestates Ir)maynotalways bethemost convenient onetouse- butwewillconsider itﬁrst. l6~ll InChapter 8wedescribed howstates varied intime interms oftheHamilto- nian H,-,-. Wesawthatthetime variation ofthevarious amplitudes wasgiven in terms ofthematrix equation .dC,-lhW=ZH,»,-C,-. (16.49)1‘ Thisequation saysthatthetimevariation ofeach amplitude C,isproportional to alloftheother amplitudes C,-,withthecoefﬁcients H,<,~. How would weexpect Eq.(16.49) tolook when weareusing thecontinuum ofbasestates Ix)?Let's ﬁrstremember thatEq.(16.49) canalsobewritten as .d. .-..143011») =;<1|H|1><11¢>- Now itisclear what weshould do.Forthex-representation wewould expect 1-4%(x|¢) =/aim x')(><'|¢)ax'. (16.50) Thesumover thebase states Ij),getsreplaced byanintegral over x’.Since (xIHIx’)should besome function ofxandx’,wecanwrite itasH(x, x’) which corresponds toH,-,4inEq.(16.49). Then Eq.(16.50) isthesame as ih%1//(x) =/H(x, x')¢(x’) dx’ with _ (16.51) H(x, x’)E(xIHIx’). According toEq.(16.51), therate ofchange ofthe11/atxwould depend onthe value oft}: atallother points x’;thefactor H(x, x’)istheamplitude perunittime that theelectron willjump from x’tox.Itturns outinnature, however, thatthis amplitude iszero except forpoints x’veryclose tox.This means—as wesawinthe example ofthechain ofatoms atthebeginning ofthechapter, Eq.(16.12)—that theright-hand sideofEq.(16.15) canbeexpressed completely interms of(Land thederivatives of1pwith respect tox,allevaluated attheposition x. Foraparticle moving freely inspace with noforces, nodisturbances, the correct lawofphysics is 22 /H(x, x’)¢(x’) dx’=—-3; ¢(x). Where didwegetthatfrom? Nowhere. It’snotpossible toderive itfrom anything youknow. Itcame outofthemind ofSchrodinger, invented inhisstruggle to ﬁndanunderstanding oftheexperimental observations oftherealworld. You can perhaps getsome clue ofwhy itshould bethatway bythinking ofourderivation ofEq.(16.12) which came from looking atthepropagation ofanelectron ina crystal. Ofcourse, freeparticles arenotvery exciting. What happens ifweputforces ontheparticle? Well, iftheforce ofaparticle canbedescribed interms ofascalar potential V(x)—which means wearethinking ofelectric forces butnotmagnetic forces—and ifwestick tolowenergies sothatwecanignore complexities which come from relativistic motions, then theHamiltonian which ﬁtstherealworld gives I / /H(x, x)¢(x )dx =~—5771-H;1,b(x) —I—V(x)1#(x). (16.52) Again, youcangetsome clue astotheorigin ofthisequation ifyougoback to themotion ofanelectron inacrystal, andseehow theequations would have to bemodiﬁed iftheenergy oftheelectron varied slowly from oneatomic siteto theother—as itmight doifthere were anelectric ﬁeld across thecrystal. Then 16-12 theterm E0inEq.(16.7) would vary slowly with position andwould correspond tothenewterm wehave added in(16.52). [You may bewondering whywewent straight from Eq.(16.51) toEq.(16.52) instead ofjust giving you thecorrect function fortheamplitude H(x, x’)= (xIllIx’). Wedidthat because H(x, x’)canonly bewritten interms ofstrange algebraic functions, although thewhole integral ontheright-hand side ofEq. (16.51) comes outinterms ofthings youareused to.Ifyouarereally curious, H(x, x’)canbewritten inthefollowing way: 2 H(x,x’) =—-217!6”(x —x’)+V(x) 6(x—-x’), where 6”means thesecond derivative ofthedelta function. This rather strange function canbereplaced byasomewhat more convenient algebraic differential operator, which iscompletely equivalent: H(x x’)=I—iiiL2—I—V(x)I 6(x—x).’ 2mdxz Wewillnotbeusing these forms, butwillwork directly with theform inEq. (l6.52).] Ifwenow usetheexpression wehave in(16.52) fortheintegral in(16.50) we getthefollowing differential equation for¢(x) :(xI50): 2 2 ih3;:=-2%%¢(><)+V(x)1//(x). (16.53) Itisfairly obvious what weshould useinstead ofEq.(16.53) ifweareinter- ested inmotion inthree dimensions. The only changes arethat d2/dxz gets replaced by 2_<12L21.2.V—6x‘~ 8y?+822 and V(x) gets replaced byV(x,y,z).The amplitude 1//(x,y,2)foranelectron moving inapotential V(x,y,z)obeys thedifferential equation 2 ihg=-£6V2111+V¢. (16.54) Itiscalled theSchrodinger equation, and was theﬁrst quantum-mechanical equation ever known. Itwaswritten down bySchrodinger before anyoftheother quantum equations wehave described inthisbook were discovered. Although wehave approached thesubject along acompletely different route, thegreat historical moment marking thebirth ofthequantum mechanical de- scription ofmatter occurred when Schrodinger ﬁrst wrote down hisequation in 1926. Formany years theinternal atomic structure ofmatter hadbeen agreat mystery. Noonehadbeen abletounderstand what heldmatter together, why there waschemical binding, andespecially how itcould bethat atoms could be stable. Although Bohr hadbeen able togiveadescription oftheinternal motion ofanelectron inahydrogen atom which seemed toexplain theobserved spectrum oflight emitted bythisatom, thereason thatelectrons moved inthisway remained amystery. Schrodinger’s discovery oftheproper equations ofmotion forelectrons onanatomic scale provided atheory from which atomic phenomena could be calculated quantitatively, accurately, and indetail. Inprinciple, Schrodinger’s equation iscapable ofexplaining allatomic phenomena except those involving magnetism and relativity. Itexplains theenergy levels ofanatom, andallthe facts ofchemical binding. This is,however, true only inprinciple—the mathe- matics soon becomes toocomplicated tosolve exactly anybutthesimplest prob- lems. Only thehydrogen andhelium atoms have been calculated toahigh accuracy. However, with various approximations, some fairly sloppy, many ofthefacts of more complicated atoms andofthechemical binding ofmolecules canbeunder- stood. Wehave shown yousome ofthese approximations inearlier chapters. l6—13 Fig. 16-3. Apotential well for o particle moving C110ng x.TheSchrodinger equation aswehave written itdoes nottakeintoaccount anymagnetic effects. Itispossible totakesucheffects intoaccount inanapproxi- mate waybyadding some more terms totheequation. However, aswehave seen inVolume II,magnetism isessentially arelativistic effect, andsoacorrect de- scription ofthemotion ofanelectron inanarbitrary electromagnetic ﬁeld can only bediscussed inaproper relativistic equation. Thecorrect relativistic equation forthemotion ofanelectron wasdiscovered byDirac ayear after Schrodinger brought forth hisequation, andtakes onquite adifferent form. Wewillnotbe abletodiscuss itatallhere. Before wegoontolook atsome oftheconsequences oftheSchrodinger equation, wewould liketoshow youwhat itlooks likeforasystem with alarge number ofparticles. Wewillnotbemaking anyuseoftheequation, butjust want toshow ittoyoutoemphasize that thewave function 1/1isnotsimply an ordinary wave inspace, butisafunction ofmany variables. Ifthere aremany particles, theequation becomes _.a¢(r,r2,r3, ...)_ 112I621!» aztp a2.//Ilh_‘-5T__ _ 5;+E+Ki+V(r1,r1, ...)¢.(16.55) Thepotential function Viswhat corresponds classically tothetotalpotential energy ofalltheparticles. Ifthere arenoexternal forces acting ontheparticles, the function Vissimply theelectrostatic energy ofinteraction ofalltheparticles. That is,iftheithparticle carries thecharge Z,-q,, thenthefunction Vissimply1' V(r1,r2,r3,...)=Z 22. (16.56)al1_ 1pairs 16-6 Quantized energy levels Inalaterchapter wewilllookindetail atasolution ofSchrodinger’s equation foraparticular example. Wewould likenow, however, toshow youhowoneof themost remarkable consequence ofSchrodinger’s equation comes about~name1y, thesurprising factthatadifferential equation involving only continuous functions ofcontinuous variables inspace cangiverisetoquantum effects such asthe discrete energy levels inanatom. Theessential facttounderstand ishowitcanbe thatanelectron which isconﬁned toacertain region ofspace bysome kind ofa potential “we11” must necessarily have only oneoranother ofacertain well- deﬁned setofdiscrete energies. Vlx) E-------------------------- --i ,‘______ ,1__toanY Suppose wethink ofanelectron inaone-dimensional situation inwhich its potential energy varies withxinawaydescribed bythegraph inFig.16-3. We willassume thatthispotential isstatic—it doesn’t varywithtime. Aswehave done somany times before, wewould liketolookforsolutions corresponding tostates ofdeﬁnite energy, which means, ofdeﬁnite frequency. Let’s tryasolution ofthe form if=a(x)e_"E‘/'1. (16.57) TWeareusing theconvention oftheearlier volumes according towhich e2Eqf/41re0. 16-14 Ifwesubstitute thisfunction intotheSchrodinger equation, weﬁndthatthe function a(x)must satisfy thefollowing differential equation: % =%[V(x)-E]a(x). (16.58) Thisequation saysthatateach xthesecond derivative ofa(x)withrespect tox isproportional toa(x), thecoefficient ofproportionality being given bythequan- tity(V—E).Thesecond derivative ofa(x)istherateofchange ofitsslope. If thepotential Visgreater than theenergy Eoftheparticle, therateofchange of theslope ofa(x) willhave thesame sign asa(x). That means that thecurve of a(x)willbeconcave away from theaxis. That is,itwillhave, more orless, the character ofthepositive ornegative exponential function, e*". This means that intheregion totheleftofx1,inFig. 16-3, where Visgreater than theassumed energy E,thefunction a(x) would have tolook likeoneoranother ofthecurves AV 0x a(x) i 12,-------- -- Y /~.K/Da(x)‘ >.5 ____________;~ >1: Y D(v>E v<E (0) lb) Fig.16-4. Possible shapes ofthe wave function a(x) forV>Eand for V<E. tive x. shown inpart (a)ofFig. 16-4. If,ontheother hand, thepotential function Vislessthan theenergy E,the second derivative ofa(x) with respect toxhastheopposite sign from a(x) itself, andthecurve ofa(x) willalways beconcave toward theaxislikeoneofthe pieces shown inpart (b)ofFig. 16-4. Thesolution insuch aregion has,piece-by- piece, roughly theform ofasinusoidal curve. Now let’sseeifwecanconstruct graphically asolution forthefunction a(x) which corresponds toaparticle ofenergy Eainthepotential Vshown inFig. 16-3. Since wearetrying todescribe asituation inwhich aparticle isbound inside thepotential well, wewant tolook forsolutions inwhich thewave amplitude takes onvery small values when xiswayoutside thepotential well. Wecaneasily imagine acurve liketheoneshown inFig. 16-5 which tends toward zero forlarge negative values ofx,andgrows smoothly asitapproaches x1.Since Visequal to Eaatx1,thecurvature ofthefunction becomes zero atthispoint. Between x1 andx2,thequantity V—Eaisalways anegative number, sothefunction a(x) isalways concave toward theaxis, andthecurvature islarger thelarger thediffer- ence between Eaand V.Ifwecontinue thecurve intotheregion between x1and x2,itshould gomore orlessasshown inFig. 16-5. Now let’scontinue thiscurve intotheregion totheright ofx2.There it curves away from theaxisandtakes offtoward large positive values, asdrawn in Fig. 16-6. Fortheenergy Eawehave chosen, thesolution fora(x)getslarger and larger with increasing x.lnfact, itscurvature isalso increasing (ifthepotential continues tostay ﬂat). The amplitude rapidly grows toimmense proportions. 16-15X Fig. 16-5. Awave function forthe energy E,which goes tozero fornega- a(x) X "nus;N *Y Fig. 16-6. Thewave function a(x) of Fig. 16-5 continued beyond x;. allFig. 16-7. The wave function a(x) foranenergy Ebgr V>E¢ V<Eceater than Eu. V>Ec . \_ ‘1 Fig. 16-8. Awave function forthe energy E,between*2 Eu(Ind Eb. E v\ —V\7t.~|-1501III//// Ai01X) “F 5/\ i%\ E3 E4A V |"ll"'lav11111111-1-1111111111-1- >1Y 11 °(X) What does thismean? Itsimply means that theparticle isnot“bound” inthe potential well. Itisinﬁnitely more likely tobefound outside ofthewell, than inside. Forthesolution wehave manufactured, theelectron ismore likely tobe found atx=+w than anywhere else. Wehave failed toﬁndasolution fora bound particle. Let’s tryanother energy, sayonealittle bithigher than E,,—say theenergy EbinFig.16-7. Ifwestart with thesame conditions ontheleft,wegetthesolution drawn inthelower halfofFig. 16-7. Itlooked atﬁrstasthough itwere going to bebetter, butitends upjustasbadasthesolution forE,,—except that now a(x) isgetting more andmore negative aswegotoward large values ofx. Maybe that’s theclue. Since changing theenergy alittle bitfrom EatoEb causes thecurve toﬁipfrom oneside oftheaxis totheother, perhaps there is some energy lying between EaandE1,forwhich thecurve willapproach zero for large values ofx.There is,indeed, andwehave sketched how thesolution might look inFig. 16-8. You should appreciate that thesolution wehave drawn intheﬁgure isa very special one. Ifwewere toraise orlower theenergy ever soslightly, thefunc- tionwould goover intocurves likeoneortheother ofthetwobroken-line curves shown inFig. 16-8, andwewould nothave theproper conditions forabound particle. Wehave obtained aresult thatifaparticle istobebound inapotential well, itcandosoonly ifithasavery deﬁnite energy. Does that mean that there isonly oneenergy foraparticle bound inapo- tential well? No. Other energies arepossible, butnotenergies tooclose toE,,. Notice that thewave function wehave drawn inFig. 16-8 crosses theaxisfour times intheregion between x1andx2. Ifwewere topick anenergy quite abit lower than EC,wecould have asolution which crosses theaxisonly three times, only twotimes, only once, ornotatall.The possible solutions aresketched in Fig. 16-9. (There may also beother solutions corresponding tovalues ofthe energy higher than theones shown.) Ourconclusion isthat ifaparticle isbound inapotential well, itsenergy cantakeononly thecertain special values inadiscrete energy spectrum. You seehow adifferential equation candescribe thebasic fact ofquantum physics. Wemight remark oneother thing. Iftheenergy Eisabove thetopofthe potential well, then there arenolonger anydiscrete solutions, andanypossible energy ispermitted. Such solutions correspond tothescattering offreeparticles byapotential well. Wehave seen anexample ofsuch solutions when weconsidered theeffects ofimpurity atoms inacrystal. Fig. 16-9. The function a(x) fortheﬁve lowest energy bound states. 16-16 I7 Symmetry and Conservation Laws 17-1 Symmetry Inclassical physics there areanumber ofquantities which areconserved- suchamomentum, energy, andangular momentum. Conservation theorems about corresponding quantities alsoexist inquantum mechanics. Themost beau- tifulthing ofquantum mechanics isthat theconservation theorems can, ina sense, bederived from something else, whereas inclassical mechanics they are practically thestarting points ofthelaws. (There areways inclassical mechanics todoananalogous thing towhat wewilldoinquantum mechanics, butitcanbe doneonlyatavery advanced level.) Inquantum mechanics, however, theconserva- tionlawsareverydeeply related totheprinciple ofsuperposition ofamplitudes, andtothesymmetry ofphysical systems under various changes. Thisisthesubject ofthepresent chapter. Although wewillapply these ideas mostly totheconserva- tionofangular momentum, theessential point isthatthetheorems about the conservation ofallkinds ofquantities are—in thequantum mechanics-related to thesymmetries ofthesystem. Webegin, therefore, bystudying thequestion ofsymmetries ofsystems. A verysimple example isthehydrogen molecular ion—we could equally welltakethe ammonia molecu1e—in which there aretwostates. Forthehydrogen molecular ionwetook asourbase states oneinwhich theelectron waslocated near proton number 1,andanother inwhich theelectron waslocated nearproton number 2. Thetwostates-which wecalled II)andI2)—are shown again inFig.l7—1(a). Now, solongasthetwonuclei areboth exactly thesame, thenthere isacertain symmetry inthisphysical system. That istosay,ifwewere toreﬂect thesystem intheplane halfway between thetwoprotons—by which wemean thateverything ononesideoftheplane getsmoved tothesymmetric position ontheother side- wewould getthesituations inFig. 17-l(b). Since theprotons areidentical, the operation ofreﬂection changes I1)intoI2)andI2)intoI1).We’ll callthisreﬂec- tionoperation Pandwrite ﬂo=mx Na=ui mu SoourPisanoperator inthesense thatit“does something” toastate tomake a newstate. Theinteresting thing isthatPoperating onanystate produces some other state ofthesystem. Now P,likeanyoftheother operators wehavedescribed, hasmatrix elements which canbedeﬁned bytheusual obvious notation. Namely, and arethematrix elements wegetifwemultiply PI1)andI3I2)ontheleftby(II. From Eq.(17.1) theyare <IlPl]>=P11: (1l2>=0,(17.2)mPM=Ps=um=1 Inthesame way wecangetP21andP22. Thematrix ofP-—with respect tothe basesystem II)andI2)-is 01P-(1 0). (1713) Weseeonce again thatthewords operator andmatrix inquantum mechanics are 17-144> ..17-1 Symmetry 17-2 Symmetry andconservation 17-3 Theconservation laws 17-4 Polarized light 17-5 Thedisintegration oftheA0 17-6 Summary oftherotation matrices Review: Chapter 52,Vol. I,Symmetry inPhysical Laws Reference: Angular Momentum in Quantum Mechanics: A.R.Edmonds, Princeton University Press, 1957 |P (bl PI|> 0P I l m\\ ‘P Fig. 17-1. Ifthestates I1)and I2) arereﬂected intheplane P-P,they go intoI2)andI1),respectively. PROS &\$ AFTER UME I l'>12> r_ (OV v _|> I2> ||> |2> AFTE1: TIME l|> 12>PROB W 1 (bl Fig. 17-2. Inasymmetric system, ifapure I1)state develops asshown in part lal,apure I2)state willdevelop asinpart (b). practically interchangeable. There areslight technical differences—like thediffer- ence between a“numeral” and a“number”-but thedistinction issomething pedantic thatwedon’t have toworry about. Sowhether Pdeﬁnes anoperation, orisactually used todeﬁne amatrix ofnumbers, wewillcallitinterchangeably anoperator oramatrix. Now wewould liketopoint outsomething. Wewillsuppose thatthephysics ofthewhole hydrogen molecular ionsystem issymmetrical. Itdoesn’t have tobe —itdepends, forinstance, onwhat elseisnearit.Butifthesystem issymmetrical, thefollowing ideashould certainly betrue. Suppose westart att=Owiththe system inthestate II)andﬁnd after aninterval oftime tthat thesystem turns outtobeinamore complicated situation-in some linear combination ofthetwo base states. Remember thatinChapter 8weused torepresent “going fora period oftime” bymultiplying bytheoperator U.That means thatthesystem would after awhi1e—say 15seconds tobedeﬁnite—be insome other state. For example, itmight be\/W parts ofthestate I1)andi\/W parts ofthestate I2), andwewould write 1.151115 sec)=U(15,0) I1)=4/2"/311) +i\/T/_3I2). (17.4) Now weaskwhat happens ifwestart thesystem inthesymmetric state I2)and wait for15seconds under thesame conditions? Itisclear that iftheworld is symmetric—as wearesupposing—we should getthestate symmetric to(17.4): I51/M15 sec) =U(15,0) I2)=\/2/3 I2)+ I1). (17.5) Thesame ideas aresketched diagrammatically inFig. 17-2. Soifthephysics ofa system issymmetrical with respect tosome plane, andwework outthebehavior ofaparticular state, wealsoknow thebehavior ofthestate wewould getby reflecting theoriginal state inthesymmetry plane. Wewould liketosaythesame things alittebitmore genera1ly—which means alittle more abstractly. LetQbeanyoneofanumber ofoperations thatyou could perform onasystem without changing thephysics. Forinstance, forQwe might bethinking ofP,theoperation ofareﬂection intheplane between thetwo atoms inthehydrogen molecule. Or,inasystem withtwoelectrons, wemight be thinking oftheoperation ofinterchanging thetwoelectrons. Another possibility would be,inaspherically symmetric system, theoperation ofarotation ofthe whole system through aﬁnite angle around some axis—which wouldn’t change thephysics. Ofcourse, wewould normally want togiveeach special casesome special notation for Speciﬁcally, wewillnormally deﬁne theR,,(0) tobethe operation “rotate thesystem about they-axis bytheangle 0”.ByQwemean justanyoneoftheoperators wehave described oranyother one—which leaves thebasic physical situation unchanged. Let’s think ofsome more examples. Ifwehave anatom with noexternal magnetic ﬁeld ornoexternal electric ﬁeld, andifwewere toturn thecoordinates around anyaxis, itwould bethesame physical system. Again, theammonia molecule issymmetrical withrespect toareflection inaplane parallel tothatof thethree hydrogens-so longasthere isnoelectric ﬁeld. When there isanelectric ﬁeld, when wemake areﬂection wewould have tochange theelectric ﬁeldalso, 17-2 andthatchanges thephysical problem. Butifwehave noexternal ﬁeld, the molecule issymmetrical. Now weconsider ageneral situation. Suppose westart with thestate I1/11) andafter some time orother under given physical conditions ithasbecome the state II02). Wecanwrite I‘//2) =(7I\P1>- (17-6) [You canbethinking ofEq.(l7.4).] Now imagine weperform theoperation Q onthewhole system. Thestate I$1)willbetransformed toastate Ii//1),which wecanalsowrite asQIII/1). Also thestate II//2)ischanged intoIil/§)=QIit/2). Now ifthephysics issymmetrical under Q(don’t forget theif;itisnotageneral property ofsystems), then, waiting forthesame time under thesame conditions, weshould have I¢a>=UI¢'i>. <17-7) [Like Eq.(17.5).] Butwecanwrite QI5&1)forI1//1)andQI$2)forI:14)so(17.7) canalsobewritten _ QI‘Pal= 1l/1>- (17-3) Ifwenowreplace I1172)byUIJ/1)~Eq. (17.6)—we getC$Q» CjlQ>AA QT/Ii!/i> '//1>- (17-9) It’snothard tounderstand what thismeans. Thinking ofthehydrogen ionit says that: “making areﬂection andwaiting awhile”—the expression onthe right ofEq.(17.9)—is thesame as“waiting awhile andthen making areﬂection”—— theexpression ontheleftof(17.9). These should bethesame solongasUdoesn’t change under thereﬂection. Since (17.9) istrueforanystarting state I11/1),itisreally anequation about theoperators: __ _AQU=UQ. (17.10) This iswhat wewanted toget—it isamathematical statement ofsymmetry. When Eq.(17.10) istrue, wesaythat theoperators Uand Qcommute. Wecanthen deﬁne “symmetry” inthefollowing way: Aphysical system issymmetric with respect totheoperation Qwhen Qcommutes with U,theoperation ofthepassage oftime. [Interms ofmatrices, theproduct oftwooperators isequivalent tothe matrix product, soEq.(17.10) alsoholds forthematrices QandUforasystem which issymmetric under thetransformation Q.] Incidentally, since forinﬁnitesimal times ewehave U=1—ibis/h—where Histheusual Hamiltonian (seeChapter 8)—you canseethatif(17.10) istrue, itisalsotruethat A_ QH= (17.11) is[Qt So(17.11) isthemathematical statement ofthecondition forthesymmetry ofa physical situation under theoperator Q.Itdeﬁnes asymmetry. 17-2 Symmetry andconservation Before applying theresult wehavejustfound, wewould liketodiscuss the ideaofsymmetry alittle more. Suppose thatwehave avery special situation: afterweoperate onastatewithQ,wegetthesame state. Thisisaveryspecial case, butlet’s suppose ithappens tobetrue forastate Iwo) that Iip’) =QIll/0)is physically thesame state asI1//0). That means thatIit’)isequal toIII/0)except forsome phase factor.'I How canthathappen? Forinstance, suppose thatwe Tlncidentally, youcanshow that Qtis necessarily aunitary 0peramr——which means thatifitoperates onIip)togivesome number times I\//),thenumber must beoftheform e“,where 6isreal. It’sasmall point, andtheproof rests onthefollowing observation. Anyoperation likeareﬂection orarotation doesn’t loseanyparticles, sothenormaliza- tionofIV)andI1/»)must bethesame; theycanonly diﬂ“er byapure imaginary phase factor. l7-3 Prob. | _______ __ II> |/2- - ___ -- 0 l'> 12> Prob I________ i’II> .,2__ ___ -_ O I'> I2> Fig. l7—3. The state II) and the state? II)obtained byreﬂecting II)in thecentral plane.have anHQ"ioninthestatewhich weoncecalled II).Forthisstate there isequal amplitude tobeinthebasestates II)andI2).Theprobabilities areshown asa bargraph inFig.17—3(a). Ifweoperate onII)with thereﬂection operator P,it ﬂipsthestate overchanging II)toI2)andI2)toI1)—we gettheprobabilities shown inFig.17—3(b). Butthat’s justthestate II)alloveragain. Ifwestartwith state I11)theprobabilities before andafterreﬂection lookjustthesame. However, there isadiﬂerence ifwelook attheamplitudes. Forthestate II)theamplitudes arethesame after thereﬂection, butforthestate I11)theamplitudes have the opposite sign. Inother words, P=pI1>+|2>=|2>+|1>= ,I1) I———\/5 ‘/5 II) (11.12) _1I1)—l2> _I2)—II)_ P|11)_PI~i-I_i_ -I11).W \/5 lfwe writeP I11/0)=emIII/0),wehavethate“ =1forthestateII)ande“;=—l forthestate III). Let’s look atanother example. Suppose wehave aRHC polarized photon propagating inthez-direction. Ifwedotheoperation ofarotation around the z-axis, weknow that thisjustmultiplies theamplitude bye“’when ¢istheangle oftherotation. Sofortherotation operation inthiscase, 6isjust equal tothe angle ofrotation. Now itisclear thatifithappens tobetruethatanoperator Qjustchanges the phase ofastate atsome time, sayt=0,itistrueforever. Inother words, ifthe state I11/1)goes over into thestate Iil/2)after atime I,or I7(l,0)I¢1) =Ill/2) (17-13) andifthesymmetry ofthesituation makes itsothat QII/1)=e“;II//1), (17-14) thenitisalsotruethat QI11/2)=81‘I¢2)- (17-15) Thisisclear, since QIIPZ) =QUI1//1): UQI‘//1), andifQII01)=e“3I1//1), then QII//2)=U916 I411) =ewﬂl 191)=etaI1P2)- [The sequence ofequalities follows from (17.13) and(17.10) forasymmetrical system, from (17.14), andfrom thefactthatanumber likee“Icommutes with an operator.] Sowith certain symmetries something which istrueinitially istrueforall times. Butisn’t thatjustaconservation law? Yes‘ Itsays thatifyoulook atthe original state andbymaking alittle computation onthesidediscover thatan operation which isasymmetry operation ofthesystem produces only amultiplica- tionbyacertain phase, thenyouknow thatthesame property willbetrueofthe ﬁnal state——the same operation multiplies theﬁnal state bythesame phase factor. This isalways true even though wemay notknow anything elseabout theinner mechanism oftheuniverse which changes asystem from theinitial totheﬁnal state. Even ifwedonotcaretolookatthedetails ofthemachinery bywhich the system getsfrom onestate toanother, wecanstillsaythatifathing 1Sinastate withacertain symmetry character originally, andiftheHamiltonian forthisthing issymmetrical under that symmetry operation, then thestate willhave thesame symmetry character foralltimes. That’s thebasis ofalltheconservation lawsof quantum mechanics. Let’s look ataspecial example. Let’s goback tothePoperator. Wewould likeﬁrst tomodify alittle ourdeﬁnition ofP.Wewant totake forPnotjusta 17-4 mirror reﬂection, because thatrequires deﬁning theplane inwhich weputthe mirror. There isaspecial kind ofareﬂection thatdoesn’t require thespeciﬁcation ofaplane. Suppose weredeﬁne theoperation Pthisway: First youreﬂect ina mirror inthez-plane sothatzgoesto—z,xstays x,andystays y;thenyouturn thesystem 180°about thez-axis sothatxismade togoto—xandyto—y.The whole thing iscalled aninversion. Every point isprojected through theorigin tothe diametrically opposite position. Allthecoordinates ofeverything arereversed. Wewillstillusethesymbol Pforthisoperation. Itisshown inFig.l7—4. Itisa littlemore convenient than asimple reﬂection because itdoesn't require thatyou specify which coordinate plane youusedforthereﬂection—you need specify only thepoint which isatthecenter ofsymmetry. Now let’ssuppose thatwehave astate Iit/0)which under theinversion opera- tiongoesintoe’6I‘//0)-—that is, It’/6)=Pliho =@’5Ii//o)- (17-16) Then suppose thatweinvert again. After twoinversions weareright back where westarted from——nothing ischanged atall.Wemust have that Plus=P-Pl¢..> =lm- FPI‘/’0) =P6’“3I'P0) =eI6PI¢0) =(@16)2I\l/0)-But Itfollows that(e16)2 = Soiftheinversion operator isasymmetry operation ofastate, there areonlytwo possibilities for6: e15 = $1, which means that PI¢@>=lt0> orPit/0>=—|¢0>, (17-17) Classically, ifastate issymmetric under aninversion, theoperation gives backthesame state. Inquantum mechanics, however, there arethetwopossibilities: wegetthesame state orminus thesame state. When wegetthesame state, PI$0)= Igl/0),wesaythatthestate Ii/10)hasevenparity. When thesignisreversed sothat PIil/0)=—I1&0), wesaythat thestate hasoddparity. (The inversion operator Pisalsoknown astheparity operator.) Thestate II)oftheH;ionhaseven parity; andthestate III)hasoddparity~see Eq.(17.12). There are,ofcourse, states which arenotsymmetric under theoperation P;these arestates with nodeﬁnite parity. Forinstance, intheHIsystem thestate II)hasevenparity, thestate III) hasoddparity, andthestate I1)hasnodeﬁnite parity. When wespeak ofanoperation likeinversion being performed “onaphysical system” wecanthink about itintwoways. Wecanthink ofphysically moving whatever isatrtotheinverse point at—r,orwecanthink oflooking atthesame system from anew frame ofreference x’,y’,z’related totheoldbyx’=-x, y’=—y,and2’=-—z. Similarly, when wethink ofrotations, wecanthink of rotating bodily aphysical system, orofrotating thecoordinate frame with respect towhich wemeasure thesystem, keeping the“system” ﬁxed inspace. Generally, thetwopoints ofview areessentially equivalent. Forrotation they areequivalent except thatrotating asystem bytheangle 0islikerotating thereference frame by thenegative of0.Inthese lectures wehave usually considered what happens when aprojection ismade intoanewsetofaxes. What yougetthatwayisthesame as what yougetifyouleave theaxes ﬁxed androtate thesystem backwards bythe same amount. When youdothat, thesigns oftheangles arereversed.I‘ IInother books youmayﬁndformulas with different signs; theyareprobably using adifferent deﬁnition oftheangles. 17-5Iz A (al %/ 7 X _U’%4,N \-\\\\ --\; /z ,Y x -r Al Fig.l7—4. The operation ofinver- sion, P.Whatever isatthepoint Acit (x,y,z)ismoved tothe point A’at (_X: —y: —Z)- Many ofthelaws ofphysics——but notall—are unchanged byareﬂection oran inversion ofthecoordinates. They aresymmetric with respect toaninversion. Thelawsofelectrodynamics, forinstance, areunchanged ifwechange xto-x, yto—y,andzto—zinalltheequations. Thesame istrueforthelawsofgravity, andforthestrong interactions ofnuclear physics. Only theweak interactions- responsible forB-decay—do nothave thissymmetry. (Wediscussed thisinsome detail inChapter 52,Vol.I.)Wewillfornowleave outanyconsideration ofthe /3-decays. Then inanyphysical system where B-decays arenotexpected toproduce anyappreciable eﬂect—an example would betheemission oflight byanatom— theHamiltonian Ii!andtheoperator Pwillcommute. Under these circumstances wehave thefollowing proposition. Ifastate originally hasevenparity, andifyou look atthephysical situation atsome later time, itwillagain have even parity. Forinstance, suppose anatom about toemitaphoton isinastate known tohave even parity. Youlook atthewhole thing—inc1uding thephoton—after theemis- sion; itwillagain have even parity (likewise ifyoustart withoddparity). This principle iscalled theconservation ofparity. You canseewhytheWords “conserva- tionofparity” and“reﬂection symmetry” areclosely intertwined inthequantum mechanics. Although until afewyears agoitwasthought thatnature always conserved parity, itisnowknown thatthisisnottrue. Ithasbeen discovered to befalsebecause theI6-decay reaction doesnothave theinversion symmetry which isfound intheother laws ofphysics. Now wecanprove aninteresting theorem (which istruesolong aswecan disregard weak interactions): Anystate ofdeﬁnite energy which isnotdegenerate must have adeﬁnite parity. Itmust have either even parity oroddparity. (Re- member thatwehavesometimes seensystems inwhich several states havethesame energy—-we saythatsuch states aredegenerate. Ourtheorem willnotapply to them.) Forastate I1//0)ofdeﬁnite energy, weknow that P71-00>=El-lo). <11-18> where Eisjustanumber~the energy ofthestate. Ifwehave anyoperator Q which isasymmetry operator ofthesystem wecanprove that Qlm=e"‘Ito (17.19) solong asI1//0)isaunique state ofdeﬁnite energy. Consider thenew state Iil/6) thatyougetfrom operating with Ifthephysics issymmetric, then Ii//(Q)must have thesame energy asI1//0). Butwehave taken asituation inwhich there is only onestate ofthat energy, namely III/0), soI¢{,) must bethesame state—it canonly diﬂer byaphase. That’s thephysical argument. Thesame thing comes outofourmathematics. Our deﬁnition ofsymmetry isEq-.(17.10) orEq.(17.11) (good foranystate ii/), 1=7QI\//) =Qﬁli//) (17-20) Butweareconsidering onlyastate II!/0)which isadeﬁnite energy state, sothat HI(lo)=EIIbo). Since Eisjustanumber thatﬂoats through Qifwewant, wehave QHI1/’0) = 1190) =EQI1/'0) So H{Q|-lo} =E{Q|v..>}- <11-21> SoIibé) =QI¢/0)isalsoadeﬁnite energy state ofH-—and with thesame E. Butbyourhypothesis, there isonlyonesuchstate; itmust bethatIil/5)=e“Iil/0). What wehavejustproved istrueforanyoperator Qthatisasymmetry opera- torofthephysical system. Therefore, inasituation inwhich weconsider only electrical forces andstrong interactions—and noB-decay——so thatinversion sym- metry isanallowed approximation, wehave that13IIP)=emI1;).Butwehave alsoseenthate“must beeither +1or—1.Soanystateofadeﬁnite energy (which isnotdegenerate) hasgoteither aneven parity oranoddparity. 17-6 17-3 Theconservation laws Weturn now toanother interesting example ofanoperation: arotation. Weconsider thespecial case ofanoperator thatrotates anatomic system byangle ¢>around thez-axis. Wewillcallthisoperatorj" R,(¢). Wearegoing tosuppose that wehave aphysical situation where wehave noinﬂuences lined upalong the x-andy-axes. Any electric ﬁeld ormagnetic ﬁeld lStaken tobeparallel tothe z-axisl sothat there willbenochange intheexternal conditions ifwerotate the whole physical system about thez-axis. Forexample, ifwehave anatom inempty space andweturntheatom around thez-axis byanangle ¢>,wehave thesame physical system. Now then, there arespecial states which have theproperty thatsuch anopera- tionproduces anewstate which istheoriginal state multiplied bysome phase factor. Letusmake aquick sideremark toshow youthatwhen thisistruethe phase change must always beproportional totheangle ¢.Suppose thatyouwould rotate twice bytheangle 4>.That’s thesame thing asrotating bytheangle 2¢.Ifa rotation by¢>hastheeffect ofmultiplying thestate I¢0) byaphase etasothat Rz(¢) I‘#0)=Q16I#0), to2twosuch rotations insuccession would multiply thestate bythefactor (e)= em, since Rz(¢)Rz(¢) I‘#0)=Rz(¢)eI6 I‘#0)=@“5Rz(¢) I‘#0)=916916 I#0)- Thephase change 5must beproportional to¢.‘IlWeareconsidering thenthose special states I1//0)forwhich Rz(¢) I\//0)=2"“I‘#0), (17-22) where missome realnumber. Wealsoknow theremarkable factthatifthesystem issymmetrical forarota- tionaround zandiftheoriginal state happens tohave theproperty that(17.22) istrue, then itwillalsohave thesame property later on.Sothisnumber misa veryimportant one. Ifweknow itsvalue initially, weknow itsvalue attheendof thegame. Itisanumber which isconserved—m isaconstant ofthemotion. The reason thatwepulloutmisbecause ithasn’t anything todowithanyspecial angle ¢,andalsobecause itcorresponds tosomething inclassical mechanics. Inquantum mechanics wechoose tocallmh——for such states asIi//0)—the angular momentum about thez-axis. Ifwedothat weﬁndthatinthelimit oflarge systems thesame quantity isequal tothez-component oftheangular momentum ofclassical me- chanics. Soifwehave astate forwhich arotation about thez-axis justproduces aphase factor e""", thenwehave astateofdeﬁnite angular momentum about that axis—-and theangular momentum isconserved. Itismfinow andforever. Of course, youcanrotate about anyaxis, andyougettheconservation ofangular momentum forthevarious axes. You seethat theconservation ofangular momentum isrelated tothefactthatwhen youturnasystem yougetthesame state withonlyanewphase factor. Wewould liketoshow youhowgeneral thisideais.Wewillapply ittotwo other conservation laws which have exact correspondence inthephysical ideas totheconservation ofangular momentum. Inclassical physics wealsohave conservation ofmomentum andconservation ofenergy, anditisinteresting to seethat both ofthese arerelated inthesame way tosome physical symmetry. IVery precisely, wewilldeﬁne l{,(4>) asarotation ofthephysical system by—¢about thez-axis, which isthesame asrotating thecoordinate frame by-I-¢. IWecanalways choose zalong thedirection oftheﬁeldprovided there isonly one ﬁeldatatime, anditsdirection doesn’t change. IIForafancier proof weshould make thisargument forsmall rotations eSince any angle ¢isthesumofasuitable nnumber ofthese,¢ =ne,R,(¢) =[R;(c)]" andthetotal phase change isntimes thatforthesmall angle e,andis,therefore, proportional to¢> 17-7 Suppose thatwehave aphysical system—an atom, some complicated nucleus, oramolecule, orsomething—and itdoesn’t make anydifference ifwetake the whole system andmove itover toadiﬂerent place. Sowehave aHamiltonian which hastheproperty that itdepends only ontheinternal coordinates insome sense, anddoes notdepend ontheabsolute position inspace. Under those cir- cumstances there isaspecial symmetry operation wecanperform which isa translation inspace. Let’s deﬁne D,(a) astheoperation ofadisplacement bythe distance aalong thex-axis. Then foranystate wecanmake thisoperation and getanewstate. Butagain there canbevery special states which have theproperty thatwhen youdisplace them byaalong thex-axis yougetthesame state except foraphase factor. It’salsopossible toprove, justaswedidabove, thatwhen this happens, thephase must beproportional toa.Sowecanwrite forthese special states I5&0) 5-.(a) I‘I/0)=elk”I‘I/0) (17-23) Thecoeﬂicient k,when multiplied byii,iscalled thex-component ofthemomentum. And thereason itiscalled thatisthatthisnumber isnumerically equal tothe classical momentum p,when wehave alarge system. Thegeneral statement is this: IftheHamiltonian isunchanged when thesystem isdisplaced, andifthe state starts with adeﬁnite momentum inthex-direction, thenthemomentum in thex-direction willremain thesame astimegoes on.Thetotal momentum ofa system before andafter collisions-—or after explosions orwhat not—will bethe same. There isanother operation thatisquite analogous tothedisplacement in space: adelay intime. Suppose thatwehave aphysical situation where there is nothing external thatdepends ontime, andwestart something oﬁ”atacertain moment inagiven state andletitroll. Now ifwewere tostart thesame thing oﬁ"again (inanother experiment) twoseconds later—or/say, delayed byatime 1'—and ifnothing intheexternal conditions depends ontheabsolute time, the development would bethesame andtheﬁnal state would bethesame asthe other ﬁnal state, except thatitwillgetthere later bythetime T.Under those circumstances wecanalsoﬁndspecial states which have theproperty thatthe development intime hasthespecial characteristic thatthedelayed state isjust theold,multiplied byaphase factor. Once more itisclear thatforthese special states thephase change must beproportional to1-.Wecanwrite at->Ito=ft"Iat (11.24) Itisconventional tousethenegative signindeﬁning to:with thisconvention whistheenergy ofthesystem, anditisconserved. Soasystem ofdeﬁnite energy is onewhich when displaced 1-intimereproduces itself multiplied bye"‘”. (That’s what wehave saidbefore when wedeﬁned aquantum state ofdeﬁnite energy, so we’re consistent with ourselves.) Itmeans thatifasystem isinastate ofdeﬁnite energy, andiftheHamiltonian doesn’t depend ont,thennomatter what goeson, thesystem willhave thesame energy atalllater times. You see,therefore, therelation between theconservation laws andthesym- metry oftheworld. Symmetry with respect todisplacements intime implies the conservation ofenergy; symmetry with respect toposition inx,y,orzimplies theconservation ofthat component ofmomentum. Symmetry with respect to rotations around thex-,y-,andz-axes implies theconservation ofthex-,y-,and z-components ofangular momentum. Symmetry with respect toreﬂection implies theconservation ofparity. Symmetry with respect totheinterchange oftwoelec- trons implies theconservation ofsomething wedon’t have aname for,andsoon. Some ofthese principles have classical analogs andothers donot. There aremore conservation laws inquantum mechanics than areuseful inclassical mechanics—- or,atleast, than areusually made useof. Inorder that youwillbeable toread other books onquantum mechanics, wemust make asmall technical aside—to describe thenotation that people use. Theoperation ofadisplacement with respect totime is,ofcourse, justtheopera- 17-8 tionUthatwetalked about before: 13,(-r)=on+1,t). (17.25) Most people liketodiscuss everything interms ofinﬁnitesimal displacements in time, orinterms ofinﬁnitesimal displacements inspace, orinterms ofrotations through inﬁnitesimal angles. Since anyﬁnite displacement orangle canbeac- cumulated byasuccession ofinﬁnitesimal displacements orangles, itisoften easier toanalyze ﬁrsttheinﬁnitesimal case. Theoperator ofaninﬁnitesimal displacement Atintimeis—as wehave deﬁned itinChapter 8— 1‘>,(Ai) =1-émﬁ. (17.26) Then Hisanalogous totheclassical quantity wecallenergy, because ifHII//) happens tobeaconstant times II0)namely, I?I1//)=EIIt/),then that constant istheenergy ofthesystem. Thesame thing isdone fortheother operations. Ifwemake asmall displace- ment inx,saybytheamount Ax,astate Iip)will, ingeneral, goover intosome other state Ii//).Wecanwrite Iil’)=f>.<A»<>I11/)=(1+Ax)Iii. (11.21) since asAxgoestozero, theIit’)should become justIip)orD,(0) =1,andfor small Axthechange ofD,,(Ax) from 1should beproportional toAx. Deﬁned this way, theoperator p,iscalled themomentum operator—for thex-component, of course. Foridentical reasons, people usually write forsmall rotations R.(A¢>)Iil) =(1+Ii.A-1)Ii> (11.28) andcallLtheoperator ofthez-component ofangular momentum. Forthose special states forwhich R,(¢) I1//0)=e"""Iit/0),wecanforanysmall angle—say A¢-—expand theright-hand sidetoﬁrstorder inA¢andget RAM) =@'7”“”I#0)=(1+imA¢) I‘#0)- Comparing thiswiththedeﬁnition ofLinEq.(17.28), wegetthat -itI‘#0)="171I‘#o)- (17-29) Inother words, ifyouoperate withI,onastate withadeﬁnite angular momentum about thez-axis, yougetmhtimes thesame state, where mhistheamount of z-component ofangular momentum. Itisquite analogous tooperating ona deﬁnite energy state withHtogetEI1//). Wewould nowliketomake some applications oftheideas oftheconservation ofangular momentum~—to show youhowtheywork. Thepoint isthattheyare really verysimple. Youknew before thatangular momentum isconserved. The onlything youreally have toremember from thischapter isthatifastate I1&0) hastheproperty thatupon arotation through anangle ¢about thez-axis, itbe- comes e”"‘I’I1//0);ithasaz-component ofangular momentum equal tomh.That’s allwewillneed todoanumber ofinteresting things. 17-4 Polarized light First ofallwewould liketocheck ononeidea. InSection 11-4 weshowed thatwhen RHC polarized light isviewed inaframe rotated bytheangle ¢about thez-axist itgetsmultiplied bye“1".Does thatmean thenthatthephotons oflight ISorry! Thisangle isthenegative oftheoneweused inSection 11-4. 17~9 Y (0)‘ii / Ily // / t if <-ELECTRON X -i»=-1--I», (bl Fig. 17-5. (clThe electric ﬁeld 8 inacircularly polarized light wave. (bl Themotion ofanelectron being driven bythecircularly polarized light.thatareright circularly polarized carry anangular momentum ofoneunitf along thez-axis? Indeed itdoes. Italsomeans thatifwehave abeam oflight containing alarge number ofphotons allcircularly polarized thesame way—as wewould have inaclassical beam——it willcarry angular momentum. Ifthetotal energy carried bythebeam inacertain time isW,then there areN=W/hm photons. Each onecarries theangular momentum ii,sothere isatotal angular momentum of J.=ivti= (17.30) Canweprove classically thatlight which isright circularly polarized carries anenergy andangular momentum inproportion toW/w? Thatshould beaclassical proposition ifeverything isright. Here wehave acasewhere wecangofrom the quantum thing totheclassical thing. Weshould seeiftheclassical physics checks. Itwillgiveusanideawhether wehave aright tocallmtheangular momentum. Remember what right circularly polarized light is,classically. It’sdescribed by anelectric ﬁeldwithanoscillating x-component andanoscillating y-component 90°outofphase sothattheresultant electric vector 8goes inacircle—-—as drawn in Fig. l7—5(a). Now suppose that such light shines onawall which isgoing to absorb it-——or atleast some ofit—-and consider anatom inthewall according to theclassical physics. Wehave often described themotion oftheelectron inthe atom asaharmonic oscillator which canbedriven into oscillation byanexternal electric ﬁeld. We’ll suppose that theatom isisotropic, sothat itcanoscillate equally well inthex-ory-directions. Then inthecircularly polarized light, the x-displacement andthey-displacement arethesame, butoneis90°behind the other. Thenetresult isthattheelectron moves inacircle, asshown inFig.l7—5(b). Theelectron isdisplaced atsome displacement rfrom itsequilibrium position atthe origin andgoes around with some phase lagwith respect tothevector 8.The relation between 8andrmight beasshown inFig. l7—5(b). Astime goes on,the electric ﬁeld rotates and thedisplacement rotates with thesame frequency, so their relative orientation stays thesame. Now let’s look atthework being done onthiselectron. Theratethatenergy isbeing putintothiselectron isv,itsvelocity, times thecomponent ofqt;parallel tothevelocity: dW—dT =q8,v. Butlook, there isangular momentum being poured intothiselectron, because there isalways atorque about theorigin. Thetorque isq8,r, which must be equal totherateofchange ofangular momentum d.l,/dt: Z-‘ff=qa,r. (17.32) Remembering thatv=wr,wehave that d],_I WVT5' Therefore, ifweintegrate thetotal angular momentum which isabsorbed, itis proportional tothetotal energy—the constant ofproportionality being 1/w, which agrees with Eq.(17.30). Light does carry angular momentum—l unit (times it)ifitisright circularly polarized along thez-axis, and—lunitalong the z-axis ifitisleftcircularly polarized. Now let’saskthefollowing question: Iflight islinearly polarized inthe x-direction, what isitsangular momentum? Light polarized inthex-direction canberepresented asthesuperposition ofRHC andLHC polarized light. There- fore, there isacertain amplitude that theangular momentum is-I—handanother ‘I’Itisusually veryconvenient tomeasure angular momentum ofatomic systems in units ofh.Then youcansaythataspinone-half particle hasangular momentum =1:1/2 with respect toanyaxis. Or,ingeneral. thatthez-component ofangular momentum ism.Youdon’t need torepeat theiiallthetime. 17-10 amplitude thattheangular momentum is-—h,soitdoesn’t have adeﬁnite angular momentum. Ithasanamplitude toappear with +handanequal amplitude to appear with —h. Theinterference ofthese twoamplitudes produces thelinear polarization, butithasequal probabilities toappear withplusorminus oneunit ofangular momentum. Macroscopic measurements made onabeam oflinearly polarized lightwillshow thatitcarries zeroangular momentum, because inalarge number ofphotons there arenearly equal numbers ofRHC andLHC photons contributing opposite amounts ofangular momentum—the average angular momentum iszero. And intheclassical theory youdon’t ﬁndtheangular mo- mentum unless there issome circular polarization. Wehave saidthatanyspin-one particle canhave three values of.l,,namely +1,O,—l(thethree states wesawintheStern-Gerlach experiment). Butlight is screwy; ithasonlytwostates. Itdoes nothave thezerocase. This strange lack isrelated tothefactthatlightcannot stand still. Foraparticle ofspinjwhich is standing still,there must bethe2j+1possible states withvalues ofj,going in steps of1from —jto+j.Butitturns outthatforsomething ofspinjwithzero mass onlythestates withthecomponents +jand—jalong thedirection ofmotion exist. Forexample, light does nothave three states, butonly two—although a photon isstillanobject ofspinone.How isthisconsistent withourearlier proofs- based onwhat happens under rotations inspace—that forspin-one particles three states arenecessary” Foraparticle atrest, rotations canbemade about any axis without changing themomentum state. Particles with zero restmass (like photons andneutrinos) cannot beatrest; only rotations about theaxisalong the direction ofmotion donotchange themomentum state. Arguments about rota- tions around oneaxisonly areinsuﬁicient toprove thatthree states arerequired, given that oneofthem varies ase“under rotations bytheangle ¢.'I One further sideremark. Forazero restmass particle, ingeneral, only one ofthetwospinstates with respect tothelineofmotion (+j, —j)isreally necessary. Forneutrinos—which arespin one-half particlesI—only thestates with theconi- ponqnt ofangular momentum opposite tothedirection ofmotion (-ii/2) exist innature [and only along themotion (—I—h/2)forantineutrinos]. When asystem has inversion symmetry (sothatparity isconserved, asitisforlight) bothcomponents (+j,and—j)arerequired. 17-sThedisintegration oftheA0 Now wewant togiveanexample ofhowweusethetheorem ofconservation ofangular momentum inaspeciﬁcally quantum physical problem. Welook at break-up ofthelambda particle (A0), which disintegrates intoaproton anda1r’ meson bya“weak” interaction: A°—>p+7r“. Assume weknow thatthepion hasspinzero, thattheproton hasspinone-half, andthattheA°hasspinone-half. Wewould liketosolve thefollowing problem: Suppose thataA0were tobeproduced inawaythatcaused ittobecompletely polarized—by which wemean thatitsspinis,say“up,” withrespect tosome suit- ablychosen z-axis——see Fig.17—6(a). Thequestion is,withwhat probability willit disintegrate sothattheproton goesoffatanangje 6‘withrespect tothez-axis—as inFig.l7~6(b)? Inother words, what istheanglardistribution ofthedisintegra- tions? Wewilllook atthedisintegration inthecoordinate system inwhich the A°isatrest—we willmeasure theangles inthisrestframe; thentheycanalways betransformed toanother frame ifwewant. TWehave tried toﬁndatleast aproof thatthecomponent ofangular momentum along thedirection ofmotion must forazero mass particle beanintegral multiple of it/2——and notsomething likeit/3. Even using allsorts ofproperties oftheLorentz transformation andwhat not,wefailed. Maybe it’snottrue. We’ll have totalkabout itwith Prof. Wigner, whoknows allabout suchthings. 17-llBEFORE AFTER Iz >O _______.,>iiii__N \O g____¢____’\\\Q §,\U //_ // 7r/ V1 (11) Fig.17-6. A11°with spin "up" decays into ciproton and apion (inthe CMsystem]. What istheprobability that theproton willgooffattheangle 6?5,‘. BEFORE AFTER >- ——————-e->———————-:#‘U -o---:--o>»<N=1O“K =1 ——O—————C/————<Q-<<qUIZ P II- YES NO (<1) (bl (cl Fig. l7—7. Two possibilities for the decoy ofaspin "up" A0with theproton going along the —I—z-axis. Only (bl conserves angular momentum. BEFORE AFTER >- _/\____.~_< ‘U O-\ _(:___...Qu-< U-1-I =l ___O_____<# =1 __O..___<1 no YES (0) lb) (C) Fig. l7—8. The decay along the z-axis foraA0with spin “down.”Webegin bylooking atthespecial circumstance inwhich theproton isemitted into asmall solid angle AS2along thez-axis (Fig. 17-7). Before thedisintegration wehaveaA0withitsspin“up,” asinpart(a)oftheﬁgure. After ashort time—for reasons unknown tothisday,except thattheyareconnected withtheweak decays— theA0explodes intoaproton andapion. Suppose theproton goesupalong the +2-axis. Then, from theconservation ofmomentum, thepion must godown. Since theproton isaspinone-half particle, itsspinmust beeither “up” or“down”— there are,inprinciple, thetwopossibilities shown inparts (b)and(c)oftheﬁgure. Theconservation ofangular momentum, however, requires thattheproton have spin“up.” This ismost easily seen from thefollowing argument. Aparticle moving along thez-axis cannot contribute anyangular momentum about thisaxisbyvirtue ofitsmotion; therefore, only thespins cancontribute toJ2.The spin angular momentum about thez-axis is—I-ii/2 before thedisintegration, soitmust alsobe —I—h/2 afterward. Wecansaythat since thepion hasnospin, theproton spin must be“up.” Ifyouareworried thatarguments ofthiskindmaynotbevalid inquantum mechanics, wecantakeamoment toshow youthattheyar‘e. Theinitial state (before thedisintegration), which wecancallIA0,spin+2)hastheproperty that ifitisrotated about thez-axis bytheangle 4:,thestate vector getsmultiplied by thephase factor e“1’/2. (Intherotated system thestatevector ise“"’2IA0,spin+z).) That’s what wemean byspin “up” foraspin one-half particle. Since nature’s behavior doesn’t depend onourchoice ofaxes, theﬁnal state (the proton plus pion) must have thesame property. Wecould write theﬁnal state as,say, Iproton going +2,spin-l—z;piongoing -2). Butwereally donotneed tospecify thepion motion, since intheframe wehave chosen thepion always moves opposite theproton; wecansimplify ourdescription oftheﬁnalstate to Iproton going +2,spin+2). Now what happens tothisstate vector ifwerotate thecoordinates about the z-axis bytheangle ¢? Since theproton andpion aremoving along thez-axis, their motion isn’t changed bytherotation. (That’s why wepicked thisspecial case; wecouldn’t make theargument otherwise.) Also, nothing happens tothepion, because itis spinzero. Theproton, however, hasspinone-half. Ifitsspinis“up” itwillcon- tribute aphase change ofe“”/2 inresponse totherotation. (Ifitsspin were “down” thephase change duetotheproton would bee_‘l’I 2.)Butthephase change with rotation before andafter theexcitement must bethesame ifangular mo- mentum istobeconserved. (And itwillbe,since there arenooutside inﬂuences in theHamiltonian.) Sotheonlypossibility isthattheproton spinwillbe“up.” Iftheproton goesup,itsspinmust alsobe“up.” Weconclude, then, that theconservation ofangular momentum permits the process shown inpart(b)ofFig.17-7, butdoes notpermit theprocess shown in part (c). Since weknow that thedisintegration occurs, there issome amplitude forprocess (b)*proton going upwith spin“up.” We’ll letastand fortheamplitude thatthedisintegration occurs inthiswayinanyinﬁnitesimal interval oftime.I Now let’sseewhat would happen iftheA0spinwere initially “down.” Again weaskabout thedecays inwhich theproton goesupalong thez-axis, asshown in Fig.17-8. Youwillappreciate thatinthiscasetheproton must havespin“down” ifangular momentum isconserved. Let’s saythattheamplitude forsuch adis- integration isb. Wecan’t sayanything more about thetwoamplitudes aandb.They depend ontheinner machinery ofA0,andtheweak decays, andnobody yetknows howto IWearenowassuming thatthemachinery ofthequantum mechanics issufﬁciently familiar toyouthat wecanspeak about things inaphysical way without taking thetime towrite down allthemathematical details. Incase what wearesaying here isnotclear toyou, wehave putsome ofthemissing details inanoteattheendofthesection. 17-12 calculate them. We’ll have togetthem from experiment. Butwith justthese twoamplitudes wecanﬁndoutallwewant toknow about theangular distribution ofthedisintegration. Weonlyhave tobecareful always todeﬁne completely the states wearetalking about. Wewant toknow theprobability thattheproton willgooﬁattheangle 0 with respect tothez-axis (into asmall solid angle A9)asdrawn inFig. 17-6. Let’s putanewz-axis inthisdirection andcallitthez’-axis. Weknow howto analyze what happens along thisaxis. With respect tothisnewaxis, theA°no longer hasitsspin“up,” buthasacertain amplitude tohave itsspin“up” and another amplitude tohave itsspin“down.” Wehave already worked these out inChapter 6,andagain inChapter 10,Eq.(10.30). The amplitude tobespin “up” iscos0/2, andtheamplitude tobespin “down” isI—sin 0/2. When the A°spin is“up” along thez’-axis itwillemit aproton inthe+2’-direction with the amplitude a.Sotheamplitude toﬁndan“up”-spinning proton coming outalong thez’-direction is acosg- (17.33) Similarly, theamplitude toﬁnda“down”-spinning proton coming along theposi- tivez’-axis is -bsing- (17.34) Thetwoprocesses thatthese amplitudes refer toareshown inFig.17-9. >5 .__..___-+__-—>N---——>N ‘T7_vs P‘5 —-———>N ————>N ’$1-=2 \\ -t\\__}~_'\/Q\\ .9»:> —.> \()=l\‘R,,\\\\ n<\/ :"O/ v Amplitude 0cos6/2 Amplitude -bcos6/2 Fig. 17-9. Twopossible decay states fortheA0. Let’s nowaskthefollowing easyquestion. IftheA°hasspinupalong the z-axis, what istheprobability thatthedecay proton willgooffattheangle 0? The twospin states (“up" or“down” along 2')aredistinguishable even though wearenotgoing tolookatthem. Sotogettheprobability wesquare theamplitudes andadd. Theprobability f(0)ofﬁnding aproton inasmall solid angle A9at6is {(0)=IaI2cos2 g+|bI2sin2;- (17.35) Remembering thatsin?6/2=§(1—cos0)andthatcos’0/2=-}(1—I—cos0), wecanwrite f(0)as /(0)= + coso. (17.36) IWehave chosen toletz’beinthexz-plane andusethematrix elements forR,,(0). Youwould getthesame answer foranyother choice. 17-13I P9 /,1 $1-='7z Theangular distribution hastheform f(0)=ﬂ(1+otCOS 0). (17.37) Theprobability hasonepartthatisindependent of0andonepartthatvaries linearly withcos6.From measuring theangular distribution wecanget(XandB, andtherefore, IaIandIbI. Now there aremany other questions wecananswer. Areweinterested only inprotons withspin“up” along theoldz-axis? Each oftheterms in(17-33) and (17-34) willgiveanamplitude toﬁndaproton with spin“up” andwith spin “down” withrespect tothez’-axis (+z' and—-z’). Spin “up” withrespect tothe oldaxisI+2)canbeexpressed interms ofthebase states I+2’) andI—z'). Wecanthen combine thetwoamplitudes (17.33) and(17.34) with theproper coefﬁcients (cos0/2and-sin 9/2)togetthetotal amplitude (acoszg+bsin? - Itssquare istheprobability thattheproton comes outattheangle 0withitsspin thesame astheA0(“up” along thez-axis). Ifparity were conserved, wecould sayonemore thing. Thedisintegration ofFig.17-8isjustthereﬂection—in say,theyz-plane ofthedisintegration of Fig.l7—7.I Ifparity were conserved. bwould have tobeequal toaorto—a. Then thecoeﬂicient ozof(17.37) would bezero, andthedisintegration would be equally likely tooccur inalldirections. Theexperimental results show, however, thatthere isanasymmetry inthe disintegration. Themeasured angular distribution doesgoascos0aswepredict- andnotascoszI9oranyother power. Infact,since theangular distribution has thisform, wecandeduce from these measurements thatthespinoftheA0is1/2. Also, weseethatparity isnotconserved. Infact,thecoefficient aisfound experi- mentally tobe-0.62 i0.05, sobisabout twice aslarge asa.Thelackofsym- metry under areﬂection isquite clear. Youseehowmuch wecangetfrom theconservation ofangular momentum. Wewillgivesome more examples inthenextchapter. Parenthetical note. Bytheamplitude ainthissection wemean theamplitude thatthe state Iproton going +2,spin-I-z)isgenerated inaninﬁnitesimal timedtfrom thestate IA,spin-I-z), or,inother words, that (proton going +2,spin+zIHIA,spin+2) =iha, (17.38) where HistheHamiltonian oftheworld—or, atleast, ofwhatever isresponsible forthe A-decay. Theconservation ofangular momentum means thattheHamiltonian must have theproperty that (proton going +2,spin-zIHIA,spin+2) =0. (17.39) Bytheamplitude bwemean that (proton going +z,spin—zIHIA,spin—z) =ihb. (17.40) Conservation ofangular momentum implies that (proton going +z,spin+zIHIA,spin—z) =0. (17.41) Iftheamplitudes written in(17.33) and(17.34) arenotclear, wecanexpress them more mathematically asfollows. By(17.33) weintend theamplitude thattheAwith spinalong +zwilldisintegrate intoaproton moving along the-I-z’-direction with its spinalsointhe-I-2’-direction, namely theamplitude (proton going -I-z’, spin+2’IHIA,spin+2). (17.42) Bythegeneral theorems ofquantum mechanics, thisamplitude canbewritten as Z(proton going +z’, spin+2’IHIA,i)(A, iIA, spin+z), (17.43) IRemembering thatthespinisanaxial vector andﬂipsoverinthereﬂection. 17-14 where thesumistobetaken overthebasestates IA,i)oftheA-particle atrest. Since the A-particle isspin one-half, there aretwosuch base states which canbeinanyreference base wewish. Ifweuseforbase states spin“up” andspin“down” withrespect toz’ (+z', —z’), theamplitude of(17.43) isequal tothesum (proton going +z’, spin+z’IHIA,+z’)(A, +z’IA,+2) +(proton going +2’, spin+z’IHIA,—z’)(A, -2’IA,+z). (17.44) Theﬁrstfactor oftheﬁrstterm isa,andtheﬁrstfactor ofthesecond term iszero—from thedeﬁnition of(17.38), andfrom (17.41), which inturnfollows from angular momentum conservation. Theremaining factor (A,+2’IA,+2)oftheﬁrstterm isjusttheamplitude thataspinone-half particle which hasspin“up” along oneaxiswillalsohave spin“up” along anaxistilted attheangle 0,which iscos0/2—see Table 6-2. So(17.44) isJust acos0/2,aswewrote in(17.33). Theamplitude of(17.34) follows from thesame kind ofarguments foraspin“down” A-particle. 17-6 Summary oftherotation matrices Wewould likenowtobring together inoneplace thevarious things wehave learned about therotations forparticles ofspinone-half andspinone—so theywill beconvenient forfuture reference. Onthenextpageyouwillﬁndtables ofthetwo rotation matrices R,(¢) andR,,(6) forspinone-half particles, forspin-one particles, andforphotons (spin-one particles withzerorestmass). Foreach spinwewill givetheterms ofthematrix (jIRIi)forrotations about thez-axis orthey-axis. They are,ofcourse, exactly equivalent totheamplitudes like(+T I0S)wehave used inearlier chapters. Wemean byR,(4>) thatthestate 1Sp1'O_]CClCd intoanew coordinate system which isrotated through theangle ¢about thez-axis—using always theright-hand ruletodeﬁne thepositive sense oftherotation. ByR,,(9) wemean thatthereference axesarerotated bytheangle 0about they-axis. Know- ingthese tworotations, youcan,ofcourse, work outanyarbitrary rotation. As usual, wewrite thematrix elements sothatthestate ontheleftisabasestate of thenew(rotated) frame andthestate ontheright isabase state oftheold(un- rotated) frame. Youcaninterpret theentries inthetables inmany ways. For instance, theentry e_“/2inTable 17-1means thatthematrix element (—IRI—)= e““"/2. Italsomeans thatRI—)=e_‘¢'2 I—),orthat(—IR=(—Ie_“l’/2. It’sallthesame thing. 17-15 Table 17-1 Rotation matrices forspinone-half Twostates: I+),“up” along thez-axis, m=-I-1/2 I—),“down" along thez-axis, m=-1/2 <_I_I e+t¢/2 0 (—I 0e“'W/2 RII(9) I+) I-I (+I cos0/2 (—I —sin 0/2sin0/2 cos0/2 Table 17-2 Rotation matrices forspinone Three states: I+),ta= I0),m l_>s m=+1 =0 -1 R.(¢) I+) I0) |—> (+I (OI (-Ie+“’ 0 0 1 0 O0 0 e“‘¢ Ru(9) l+> I0) I—) (+I (OI (—I§(l—I—cos0) +\-;—isin0 —\%sin0 cos0 §(1—cos0) —%sin0§(l—cos0) 1.—I—72sin0 §(l—I—cos0) Table 17-3 Photons Twostates: IR)=I}?(Ix) —I—iIy)), m=+1(RHC polarized) 17-16IL)=‘+2 (Ix)—iIy)),m=-1(LHC polarized) R.(¢) IR> IL) (RI e+e'¢ 0 (LI 0 e“¢ I8 Angular Momentum 18-1 Electric dipole radiation Inthelastchapter wedeveloped theidea oftheconservation ofangular momentum inquantum mechanics, andshowed how itmight beused topredict theangular distribution oftheproton from thedisintegration oftheA-particle Wewant now togive you anumber ofother, similar, illustrations ofthecon- sequences ofmomentum conservation inatomic systems Our ﬁrst example is theradiation oflight from anatom. The conservation ofangular momentum (among other things) will determine thepolarization and angular distribution oftheemitted photons. Suppose wehave anatom which isinanexcited state ofdeﬁnite angular momentum—say with aspinofone——-and itmakes atransition toastate ofangular momentum zero atalower energy, emitting aphoton. The problem istoﬁgure outtheangular distribution and polarization ofthephotons. (This problem is almost exactly thesame astheA0disintegration, except that wehave spin-one instead ofspin one-half particles.) Since theupper state oftheatom isspin one, there arethree possibilities foritsz-component ofangular momentum. Thevalue ofmcould be+1, or0,or-1. Wewilltake m=+1forourexample. Once youseehow itgoes, youcanwork outtheother cases. Wesuppose thattheatom issitting with itsangular momentum along the+2-axis—as inFig. l8—l(a)—and askwith what amplitude itwillemit right circularly polarized light upward along thez-axis, sothat theatom ends upwith zero angular momentum—as shown in part (b)oftheﬁgure. Well, wedon't know theanswer tothat. Butwedoknow that right circularly polarized light hasoneunit ofangular momentum about its direction ofpropagation. Soafter thephoton isemitted, thesituation would have tobeasshown inFig. l8—l(b)—the atom isleftwith zero angular momentum N AN N nncPHOTON ' j=I ATOMIN i=0 ATOM IN 1,,m=, EXCITED m= cnouuo __sure sure "‘'' AMPLITUDE I O12> BEFORE AFTER18-1 Electric dipole radiation 18-2 Light scattering 18-3 Theannihilation ofpositronium 18-4 Rotation matrix foranyspin 18-5 Measuring anuclear spin 18-6 Composition ofangular mo- mentum Added Note 1:Derivation oftherota- tionmatrix Added Note 2:Conservation ofparity inphoton emission i>N ‘DI’IIC OTON 3..OO I AMPLITUDE »I1,BEFORE AFTE R (0) (bl (O) lb) Fig. l8—l. Anatom with m=+1 emits 0RHC photon along the—I—z-cixis. 18-1Fig. 18-2. Anatom with mI—l emits ciLHC photon along the+2-axis. is -e (b) __"® Fig. 18-3. Ifthe process of(ci)is transformed byoninversion through the center oftheatom, itappears usinlb).about thez-axis, since wehave assumed anatom whose lower state isspin zero. Wewillletastand fortheamplitude forsuch anevent. More precisely, weleta betheamplitude toemit aphoton intoacertain small solid angle A9,centered onthez-axis, during atimedt.Notice thattheamplitude toemitaLHC photon inthesame direction iszero. Thenetangular momentum about thez-axis would be-1forsuch aphoton andzerofortheatom foratotal of-1,which would notconserve angular momentum. Similarly, ifthespinoftheatom isinitially “down” (-1along thez-axis), itcanemitonlyaLHC polarized photon inthedirection ofthe+2-axis, asshown inFig.18-2. Wewillletbstand fortheamplitude forthisevent—meaning again theamplitude thatthephoton goesintoacertain solid angle A9. Ontheother hand, iftheatom isinthem=0state, itcannot emitaphoton inthe+z-direction atall,because aphoton canhave onlytheangular momentum +1or-1along itsdirection ofmotion. Next, wecanshow thatbisrelated toa.Suppose weperform aninversion of thesituation inFig.18-1, which means thatweshould imagine what thesystem would looklikeifwewere tomove eachpartofthesystem toanequivalent point ontheopposite sideoftheorigin. Thisdoes notmean thatweshould reﬂect the angular momentum vectors, because theyareartiﬁcial. Weshould, rather, invert theactual character ofthemotion thatwould correspond tosuch anangular momentum. InFig.l8—3(a) and(b)weshow what theprocess ofFig.18-1looks likebefore andafter aninversion withrespect tothecenter oftheatom. Notice thatthesense ofrotation oftheatom isunchanged."I' Intheinverted system of Fig.l8—3(b) wehave anatom withm=+1emitting aLHC photon downward. Ifwenowrotate thesystem ofFig.l8—3(b) by180°about thex-ory-axis, it becomes identical toFig.18-2. Thecombination oftheinversion androtation turns thesecond process intotheﬁrst. Using Table 17-2, weseethatarotation of180°about they-axis justthrows anm=-1state intoanm=+1state, sotheamplitude bmust beequal totheamplitude aexcept forapossible sign change duetotheinversion. Thesignchange intheinversion willdepend onthe parities oftheinitial andﬁnalstate oftheatom. Inatomic processes, parity isconserved, sotheparity ofthewhole system must bethesame before andafterthephoton emission. What happens willdepend onwhether theparities oftheinitial andﬁnalstates oftheatom areeven orodd— theangular distribution oftheradiation willbedifferent fordifferent cases. We willtakethecommon caseofoddparity fortheinitial state andevenparity forthe ﬁnal state; itwillgivewhat iscalled “electric dipole radiation.” (Iftheinitial andﬁnalstates have thesame parity wesaythere is“magnetic dipole radiation,” which hasthecharacter oftheradiation from anoscillating current inaloop.) Iftheparity oftheinitial stateisodd,itsamplitude reverses itssignintheinversion which takes thesystem from (a)to(b)ofFig.18-3. Theﬁnalstate oftheatom haseven parity, soitsamplitude doesn’t change sign. Ifthereaction isgoing to conserve parity, theamplitude bmust beequal toainmagnitude butofthe opposite sign. Weconclude thatiftheamplitude isathatanm=+1state willemit a photon upward, then fortheassumed parities oftheinitial andﬁnal states the amplitude thatanm=-1state willemitaLHC photon upward is-a.I Wehaveallweneed toknow toﬁndtheamplitude foraphoton tobeemitted atanyangle 0with respect tothez-axis. Suppose wehave anatom originally polarized with m=+1. Wecanresolve thisstate into+1,0,and-1states withrespect toanewz’-axis inthedirection ofthephoton emission. Theampli- tudes forthese three states arejusttheonesgiven inthelower halfofTable 17-2. TWhen wechange x,y,zinto-x,-y,—z,youmight think thatallvectors getre- versed. That istrueforpolar vectors likedisplacements andvelocities, butnotforan axial vector likeangular momentum—or anyvector which 1Sderived from across product oftwopolar vectors. Axial vectors have thesame components after aninversion. ISome ofyoumayobject totheargument wehavejustmade, onthebasis thattheﬁnal states wehave been considering donothave adeﬁnite parity. You willﬁndinAdded Note 2attheendofthischapter another demonstration, which youmayprefer. 18-2 Theamplitude thataRHC photon isemitted inthedirection 0isthen atimes the amplitude tohave m=+1inthatdirection, namely, a(+IRy(0)] +)=2?(1+cose). (18.1) Theamplitude thataLHC photon isemitted inthesame direction is-atimes the amplitude tohave m=-1inthenewdirection. Using Table 17-2, itis -a(— IR,,(0)I +)={L2(1-cos0). (18.2) Ifyouareinterested inother polarizations youcanﬁndouttheamplitude forthem from thesuperposition ofthese two amplitudes Togettheintensity ofany component asafunction ofangle, youmust, ofcourse, take theabsolute square oftheamplitudes. 18-2 Light scattering Let’s usethese results tosolve asomewhat more complicated problem- butalsoonewhich issomewhat more real. Wesuppose thatthesame atoms are sitting intheir ground state (j=O),and scatter anincoming beam oflight. Let’s saythatthelightisgoing initially inthe+z-direction, sothatwehavephotons coming uptotheatom from the—z-direction, asshown inFig. 18-4(a). Wecan consider thescattering oflight asatwo-step process: The photon isabsorbed, andthen isre-emitted. Ifwestart with aRHC photon asinFig.l8—4(a), and angular momentum isconserved, theatom willbeinanm=+1state after the absorption—as shown inFig. l8—4(b). Wecalltheamplitude forthisprocess c. The atom canthen emit aRHC photon inthedirection 0—as inFig. l8—4(c). The total amplitude that aRHC photon isscattered inthedirection 0isjust ctimes (18.1). Let’s callthisscattering amplitude (R’ISIR);wehave (R’IsIR)=925(1+cos0). (18.3) There isalsoanamplitude thataRHC photon willbeabsorbed andthat aLHC photon willbeemitted. Theproduct ofthetwoamplitudes istheamplitude (L’ISIR)thataRHC photon isscattered asaLHC photon. Using (18.2), wehave (L’ISIR)=-925(1-cose). (18.4) Now let’saskabout what happens ifaLHC photon comes in.When itis absorbed, theatom willgointoanm=-1state. Bythesame kind ofarguments weused inthepreceding section, wecanshow that thisamplitude must be—c. Theamplitude thatanatom inthem=-1state willemit aRHC photon atthe angle 0isatimes theamplitude (+IR,,(0) I-),which is%(l—cos0).Sowehave (R’|s]L)=-%(1-cose). (18.5) Finally, theamplitude foraLHC photon tobescattered asaLHC photon is (L’|s|L)=gf(1+cose). (18.6) (There aretwominus signs which cancel.) Ifwemake ameasurement ofthescattered intensity foranygiven combina- tion ofcircular polarizations itwillbeproportional tothesquare ofoneofourfour amplitudes. Forinstance, with anincoming beam ofRHC light theintensity of theRHC light inthescattered radiation willvary as(1+cos0)2. That’s allvery well, butsuppose westart outwith linearly polarized light. What then? Ifwehave x-polarized light, itcanberepresented asasuperposition l8-3Z 1 ill / I>1Q5 gga._ \/\/\(>/\/\.> Fig. 18-4. Thescattering oflight by onatom seen Clsutwo-step process. ofRHC andLHC light. Wewrite (seeSection ll-4) l IX)='—(IR)-1-IL))- (13-7)\/5 Or,ifwehave y-polarized light, wewould have 1))=-§i<1R>— lL>)- (188) Now what doyouwant toknow? Doyouwant theamplitude thatanx-polarized photon willscatter intoaRHC photon attheangle 0?You cangetitbytheusual ruleforcombining amplitudes. First, multiply (18.7) by(R'IStoget 1R’S =— R’SR R’SL, 18.9 (IIX) \/i(( II)+(1I)) () andthenuse(18.3) and(18.5) forthetwoamplitudes. Youget (R'ISIx)=$2cos6. (18.10) Ifyouwanted theamplitude that anx-photon would scatter intoaLHC photon, youwould get (L’ISIx) =Ex/638888. (18.11) Finally, suppose youwanted toknow theamplitude thatanx-polarized photon willscatter while keeping itsx-polarization. What youwant is(x’ISIx).This canbewritten as <1’ISIX)=(><’1R’)(R’ 1SIX)+(><’1L’)(L' I$1><)- (1312) Ifyouthenusetherelations IR’)=—(IX’)+ily’)), (18-13)3.. 1IL’)=—(IX’)—1|)/)), (13-14)\/i itfollows that §_(x'IR’)=—, (18.15) (x’IL’) = (18.16) S0yougetthat (x’ISIx)=accos0. (18.17) Theanswer isthatabeam ofx-polarized light willbescattered atthedirection 0 (inthexz-plane) with anintensity proportional tocos2 0.Ifyouaskabout y-polar- izedlight, youﬁndthat (y'ISIx) =0. (18.18) Sothescattered lightiscompletely polarized inthex-direction. Now wenotice something interesting. Theresults (18.17) and(18.18) corre- spond exactly totheclassical theory oflight scattering wegave inVol.l,Section 32-6, where weimagined thattheelectron wasbound totheatom byalinear restoring force—so thatitacted likeaclassical oscillator. Perhaps youarethink- 1ng:“It’s somuch easier intheclassical theory; 1fitgives theright answer why bother with thequantum theory?” Foronething, wehave considered sofar onlythespecial—though common-—case ofanatom with aj=1excited state andaj=0ground state. Iftheexcited state hadspintwo,youwould getadiffer- entresult. Also, there isnoreason whythemodel ofanelectron attached toa 18-4 spring anddriven byanoscillating electric ﬁeldshould work forasingle photon. Butwehave found thatitdoesinfactwork, andthatthepolarization andintensi- tiescome outright. Soinacertain sense wearebringing thewhole course around totherealtruth. Whereas wehave, inVol.I,done thetheory oftheindex of refraction, andoflight scattering, bytheclassical theory, wehave now shown that thequantum theory gives thesame result forthemost common case. Ineffect wehave nowdone thepolarization ofskylight, forinstance, byquantum me- chanical arguments, which istheonlytruly legitimate way. Itshould be,ofcourse, thatalltheclassical theories which work aresup- ported ultimately bylegitimate quantum arguments. Naturally, those things which wehave spent agreat dealoftimeinexplaining toyouwere selected from Just those parts ofclassical physics which still maintain validity inquantum mechanics. You’ll notice thatwedidnotdiscuss ingreat detail anymodel ofthe atom which haselectrons going around inorbits. That’s because such amodel doesn’t giveresults which agree with thequantum mechanics. Buttheelectron onaspring—which isnot,inasense, atallthewayanatom “looks”—does work, andsoWeused thatmodel forthetheory oftheindex ofrefraction. 18-3 Theannihilation ofpositronium Wewould likenexttotakeanexample which isverypretty. Itisquite inter- esting and, although somewhat complicated, wehope nottoomuch so.Our example isthesystem called positronium, which isan“atom” made upofanelec- tron andapositron—a bound state ofane+andane_. Itislikeahydrogen atom, except thatapositron replaces theproton. This object has—like thehydro- genatom—many states. Also likethehydrogen, theground state issplit into a “hyperﬁne structure” bytheinteraction ofthemagnetic moments. Thespins of theelectron andpositron areeach one-half, andthey canbeeither parallel or antiparallel toanygiven axis. (Intheground state there isnoother angular momentum duetoorbital motion.) Sothere arefour states: three arethesub- states ofaspin-one system, allwith thesame energy; andoneisastate ofspin zero with adiﬁerent energy. The energy splitting is,however, much larger than the1420 megacycles ofhydrogen because thepositron magnetic moment isso much stronger—l00O times stronger-—than theproton moment. The most important difference, however, isthat positronium cannot last forever. Theposition istheantiparticle oftheelectron; they canannihilate each other. The twoparticles disappear completely-converting their restenergy into radiation, which appears asY-rays (photons). Inthedisintegration, twoparticles with aﬁnite restmass gointotwoormore objects which have zero restmass.'I' Webegin byanalyzing thedisintegration ofthespin-zero state oftheposi- tronium. Itdisintegrates into two ‘Y-rays with alifetime ofabout l0‘1° second. Initially, wehave apositron andanelectron close together andwith spins anti- parallel, making thepositronium system. After thedisintegration there aretwo photons going outwith equal andopposite momenta (Fig. 18-5). Themomenta must beequal andopposite, because thetotal momentum after thedisintegration must bezero, asitwasbefore, ifwearetaking thecase ofannihilation atrest. Ifthepositronium isnotatrest,wecanridewithit,solve theproblem, andthen transform everything back tothelabsystem. (See, wecandoanything now; wehave allthetools.) First, wenote thattheangular distribution isnotvery interesting. Since theinitial state hasspinzero, ithasnospecial axis itissymmetric under all rotations. Theﬁnal state must then also besymmetric under allrotations. That means that allangles forthedisintegration areequally likely—-the amplitude is thesame foraphoton togoinanydirection. Ofcourse, once weﬁndoneof thephotons insome direction theother must beopposite. TInthedeeper understanding oftheworld today, wedonothave aneasy wayto distinguish whether theenergy ofaphoton 1Sless“matter” thantheenergy ofanelectron, because asyouremember alltheparticles behave verysimilarly. Theonlydistinction is thatthephoton haszerorestmass. 18-5POSITRONIUM 1‘, \/ ete- BEFORE AFTER (0) (b) Fig. 18-5. Thetwo-photon cinnihila tion ofpositronium. Z Z m=+| ($3 RHC I POSITRONIUM I /\ i=0@ r1m=Q \_/ EG- | m=-|<<7RHc Fig. l8—6. One possibility forposi- tronium annihilation along thez-axis.Theonly remaining question, which wenow want tolook at,isabout the polarization ofthephotons. Let’s callthedirections ofmotion ofthetwophotons theplusandminus z-axes. Wecanuseanyrepresentations wewant forthepolar- ization states ofthephotons; wewillchoose forourdescription right andleft circular polarization—always with respect tothedirections ofmotion. Right away, wecanseethatifthephoton going upward isRHC, then angular momentum willbeconserved ifthedownward going photon isalsoRHC. Each willcarry +1 unitofangular momentum withrespect toitsmomentum direction, which means plusandminus oneunitabout thez-axis. Thetotal willbezero, andtheangular momentum after thedisintegration willbethesame asbefore. SeeFig.18-6. The same arguments show that iftheupward going photon isRHC, the downward cannot beLHC. Then theﬁnalstate would have twounits ofangular momentum. This isnotpermitted iftheinitial state hasspinzero. Note that such aﬁnal state isalsonotpossible fortheother positronium ground state of spinone,because itcanhave amaximum ofoneunitofangular momentum in anydirection. Now wewant toshow thattwo-photon annihilation isnotpossible atall from thespin-one state. Youmight think thatifwetookthej=1,m=0state—- which haszeroangular momentum about thez-axis-—it should belikethespin-zero state, andcould disintegrate intotwoRHC photons. Certainly, thedisintegration sketched inFig. l8—7(a) conserves angular momentum about thez-axis. Butnow lookwhat happens ifwerotate thissystem around they-axis by180°; wegetthe picture shown inFig.l8—7(b). Itisexactly thesame asinpart(a)oftheﬁgure. Allwehavedone isinterchange thetwophotons. Now photons areBose particles; ifweinterchange them, theamplitude hasthesame sign, sotheamplitude forthe disintegration inpart(b)must bethesame asinpart(a).Butwehave assumed thattheinitial object isspinone. Andwhen werotate aspin-one object inastate withm=0by180°about they-axis, itsamplitudes change sign(seeTable 17-2 for6=1r).Sotheamplitudes for(a)and(b)inFig.18-7should have opposite signs; thespin-one state cannot disintegrate intotwophotons. When positronium isformed youwould expect ittoendupinthespin-zero state 1/4ofthetimeandinthespin-one state (with m=—1,0,or+l)3/4 ofthe time. So1/4ofthetimeyouwould gettwo-photon annihilations. Theother 3/4 1 “f/\ /T 11--to \ (cl m= \/ I) m=O \\/ .1 +_ I iFig. 18-7. FortheI=lstate ofpositronium, theprocess Ia)and its180° rotation about y(b)areexactly thesame. TNote thatwealways analyze theangular momentum about thedirection ofmotion of theparticle. Ifwewere toaskabout theangular momentum about anyother axis, we would have toworry about thepossibility of“orbital” angular momentum—from a pXrterm. Forinstance, wecan’t saythatthephotons leave exactly from thecenter ofthepositronium. They could leave liketwothings shotoutfrom therimofaspinning wheel. Wedon’t have toworry about such possibilities when wetakeouraxisalong the direction ofmotion. 18-6 ofthetimethere canbenotwo-photon annihilations. There isstillanannihilation, butithastogowiththree photons. Itisharder forittodothatandthelifetime is1000times longer—about l0_7 second. Thisiswhat isobserved experimentally. Wewillnotgointoanymore ofthedetails ofthespin-one annihilation. Sofarwehave thatifweonlyworry about angular momentum, thespin-zero state ofthepositronium cangointotwoRHC photons. There isalsoanother possibility: itcangointotwoLHC photons asshown inFig.18-8. Thenext question is,what istherelation between theamplitudes forthese twopossible decay modes? Wecanﬁndoutfrom theconservation ofparity. Todothat, however, weneed toknow theparity ofthepositronium. Now theoretical physicists have shown inaway that isnoteasy toexplain that the parity oftheelectron and thepositron—-its antiparticle—must beopposite, so thatthespin-zero ground state ofpositronium must beodd. Wewilljustassume thatitisodd, andsince wewillgetagreement with experiment, wecantake that assuﬁicient proof. Let’s seethen what happens ifwemake aninversion oftheprocess inFig. 18-6. When wedothat, thetwophotons reverse directions andpolarizations. The inverted picture looks just likeFig. 18-8. Assuming that theparity ofthe positronium isodd, theamplitudes forthetwoprocesses inFigs. 18-6 and 18-8 must have theopposite sign. Let’s letIR1R2) stand fortheﬁnal state ofFig. 18-6 inwhich both photons areRHC, andletIL1L2) stand fortheﬁnalstate of Fig.18-8, inwhich both photons areLHC. Thetrueﬁnalstate—let’s callitIF)—— must be IF)=IR1R2)_ lL1L2)- (13-19) Then aninversion changes theR’sintoL’sandgives thestate PIF) =1lL1L2)— lR1R2)= —IF), (18-20) which isthenegative of(18.19). Sotheﬁnalstate IF)hasnegative parity, which isthesame astheinitial spin-zero state ofthepositronium. Thisistheonlyﬁnal statethatconserves bothangular momentum andparity. There issome amplitude that thedisintegration into thisstate willoccur, which wedon’t need toworry about now, however, since weareonlyinterested inquestions about thepolariza- tion. What does theﬁnalstate of(18.19) mean physically? Onething itmeans is thefollowing: Ifweobserve thetwophotons intwodetectors which canbeset tocount separately theRHC orLHC photons, wewillalways seetwoRHC photons together, ortwoLHC photons together. That is,ifyoustand ononeside ofthepositronium andsomeone elsestands ontheopposite side,youcanmeasure thepolarization andtelltheother guywhat polarization hewillget.Youhave a 50-50 chance ofcatching aRHC photon oraLHC photon; whichever oneyouget, youcanpredict thathewillgetthesame. Since there isa50-50 chance forRHC orLHC polarization, itsounds as though itmight belikelinear polarization. Let’s askwhat happens ifweobserve thephoton incounters thataccept onlylinearly polarized light. For“r-rays itis notaseasytomeasure thepolarization asitisforlight; there isnopolarizer which works wellforsuchshort wavelengths. Butlet’simagine thatthere is,tomake the discussion easier. Suppose that youhave acounter that only accepts light with x-polarization, andthatthere isaguyontheother sidethatalsolooks forlinear polarized light with, say,y-polarization. What isthechance youwillpick upthe twophotons from anannihilation? What weneed toaskistheamplitude that IF)willbeinthestate Ixlyz). Inother words, wewant theamplitude (X012 IF)» which is,ofcourse, just (X11/2 IRIR2) _(X1)/2 lL1L2)- (18-21) Now although weareworking with two-particle amplitudes forthetwo photons, wecanhandle them justaswedidthesingle particle amplitudes, since 18-72....s OO@m= \\/ ete- c%>u-cc Fig. l8-8. Another possible process forpositronium annihilation. each particle actsindependently oftheother. That means thattheamplitude (x1y2IR1R2) isjust theproduct ofthetwoindependent amplitudes (x1IR1) and(jigIR2). Using Table 17-3, these twoamplitudes are1/\/2 andi/\/5, so (x1.V2I-RIR2) =+5 Similarly, weﬁndthat (X1112 lLiL2) =— Subtracting these twoamplitudes according to(18.21), wegetthat (X1)/2 IF)=+i. (18.22) Sothere isaunitprobability'I" thatifyougetaphoton inyour x-polarized detector, theother guywillgetaphoton inhisy-polarized detector. Now suppose thattheother guysetshiscounter forx-polarization thesame asyours. Hewould never getacount when yougotone. Ifyouwork itthrough, youwillﬁndthat <X1X2 IF)=0. (18.23) Itwill,naturally, alsowork outthatifyousetyour counter fory-polarization he willgetcoincident counts onlyifheissetforx-polarization. Now thisallleads toaninteresting situation. Suppose youwere tosetup something likeapiece ofcalcite which separated thephotons intox-polarized andy-polarized beams, andputacounter ineachbeam. Let’s callonethex-counter andtheother they-counter. Iftheguyontheother sidedoes thesame thing, youcanalways tellhimwhich beam hisphoton isgoing togointo. Whenever youandhegetsimultaneous counts, youcanseewhich ofyour detectors caught thephoton andthentellhimwhich ofhiscounters hadaphoton. Let’s saythat inacertain disintegration youﬁndthataphoton went intoyour x-counter; you cantellhimthathemust have hadacount inhisy-counter. Now many people wholearn quantum mechanics intheusual (old-fashioned) wayﬁndthisdisturbing. They would liketothink thatonce thephotons areemitted itgoes along asawave with adeﬁnite character. They would think thatsince “any given photon” hassome “amplitude” tobex-polarized ortobey-polarized, there should besome chance ofpicking itupineither thex-ory-counter andthat thischance shouldn’t depend onwhat some other person ﬁnds outabout acom- pletely different photon. They argue that“someone elsemaking ameasurement shouldn’t beable tochange theprobability thatIwillﬁndsomething.” Our quantum mechanics says, however, thatbymaking ameasurement onphoton number one,youcanpredict precisely what thepolarization ofphoton number twoisgoing tobewhen itisdetected. Thispoint wasnever accepted byEinstein, andheworried about itagreat deal-it became known asthe“Einstein-Poda1sky- Rosen paradox.” Butwhen thesituation 1Sdescribed aswehave done ithere, there doesn’t seem tobeanyparadox atall;itcomes outquite naturally thatwhat ismeasured inoneplace iscorrelated with what ismeasured somewhere else. The argument thattheresult isparadoxical runssomething likethis: (1)Ifyouhave acounter which tellsyouwhether your photon isRHC orLHC, youcanpredict exactly what kindofaphoton (RHC orLHC) hewillﬁnd. (2)Thephotons hereceives must, therefore, eachbepurely RHC orpurely LHC, some ofonekind andsome oftheother. (3)Surely youcannot alter thephysical nature ofhisphotons bychanging the kind ofobservation youmake onyour photons. Nomatter what measure- ments youmake onyours, hismust stillbeeither RHC orLHC. ‘I’Wehave notnormalized ouramplitudes, ormultiplied them bytheamplitude for thedisintegration intoanyparticular ﬁnalstate, butwecanseethatthisresult iscorrect because wegetzeroprobability when welook attheother alternative—see Eq.(18.23). 18-8 (4)Now suppose hechanges hisapparatus tosplithisphotons intotwolinearly polarized beams with apiece ofcalcite sothatallofhisphotons goeither intoanx-polarized beam orintoay-polarized beam. There isabsolutely no way, according toquantum mechanics, totellintowhich beam anypar- ticular RHC photon willgo.There isa50% probability itwillgointo the x-beam anda50% probability itwillgointothey-beam. And thesame goesforaLHC photon. (5)Since each photon isRHC orLHC—according to(2)and(3)—each one must have a50-50 chance ofgoing intothex-beam orthey-beam andthere isnowaytopredict which wayitwillgo. (6)Yetthetheory predicts thatifyou seeyour photon gothrough anx-polarizer youcanpredict with certainty that hisphoton willgointo hisy-polarized beam. This isincontradiction to(5)sothere isaparadox. Nature apparently doesn’t seethe“paradox,” however, because experiment shows thattheprediction in(6)is,infact,true. Wehave already discussed thekey tothis“paradox” inourvery ﬁrst lecture onquantum mechanical behavior in Chapter 35,Vol. I.Intheargument above, steps (1),(2),(4),and(6)areall correct, but(3),anditsconsequence (5),arewrong; theyarenotatruedescription ofnature. Argument (3)saysthatbyyourmeasurement (seeing aRHC oraLHC photon) youcandetermine which oftwoalternative events occurs forhim(seeing aRHC oraLHC photon), andthateven ifyoudonotmake your measurement youcanstillsaythathisevent willoccur either byonealternative ortheother. Butitwasprecisely thepoint ofChapter 35,Vol.I,topoint outright atthebegin- ning thatthisisnotsoinNature. Herwayrequires adescription interms ofinter- fering amplitudes, oneamplitude foreach alternative. Ameasurement ofwhich alternative actually occurs destroys theinterference, butifameasurement is notmade youcannot stillsaythat“one alternative ortheother isstilloccurring.” Ifyoucould determine foreach oneofyour photons whether itwasRHC and LHC, andalsowhether itwasx-polarized (allforthesame photon) there would indeed beaparadox. Butyoucannot dothat—it isanexample oftheuncertainty principle. Doyoustillthink there isa“paradox”? Make surethatitis,infact,aparadox about thebehavior ofNature, bysetting upanimaginary experiment forwhich thetheory ofquantum mechanics would predict inconsistent results viatwo different arguments. Otherwise the“paradox” isonlyaconﬂict between reality andyour feeling ofwhat reality “ought tobe.” Doyouthink thatitisnota“paradox,” butthatitisstillvery peculiar? Onthatwecanallagree. Itiswhat makes physics fascinating. 18-4 Rotation matrix foranyspin Bynowyoucansee,wehope, howimportant theideaoftheangular mo- mentum isinunderstanding atomic processes. Sofar,wehave considered only systems with spins—or “total angular momentum”—of zero, one-half, orone. There are,ofcourse, atomic systems withhigher angular momenta. Foranalyzing such systems wewould need tohave tables ofrotation amplitudes likethose in Section 17-6. That is,wewould need thematrix ofamplitudes forspin %,2, it,3,etc. Although wewillnotwork outthese tables indetail, wewould like toshow youhowitisdone, sothatyoucandoitifyoueverneed to. Aswehave seenearlier, anysystem which hasthespinor“total angular mo- mentum” jcanexist inanyoneof(2j-1-1)states forwhich thez-component of angular momentum canhave anyoneofthediscrete values inthesequence j, j-l,j—2,...,-(j —1),-1(allinunits ofh).Calling thez-component of angular momentum ofanyparticular state mh,wecandeﬁne aparticular angular momentum state bygiving thenumerical values ofthetwo“angular momentum quantum numbers” _]andm.Wecanindicate such astate bythestate vector Ij,m). Inthecase ofaspinone-half particle, thetwostates arethen I%,%)andI%,—%); orforaspin-one system, thestates would bewritten inthis notation asI1,+1), I1,0),I1,-1). Aspin-zero particle has,ofcourse, only the onestate I0,0). 189 Now wewant toknow what happens when weproject thegeneral state I1',m) intoarepresentation withrespect toarotated setofaxes. First. weknow thatj isanumber which characterizes thesystem, soitdoesn’t change. Ifwerotate the axes, allwedoisgetamixture ofthevarious m-values forthesame j.Ingeneral, there willbesome amplitude thatintherotated frame thesystem willbeinthe state Ij,in’),where m’gives thenewz-component ofangular momentum. Sowhat wewant areallthematrix elements (j,m’IRIj,m)forvarious rotations. We already know what happens ifwerotate byanangle 4)about thez-axis. Thenew state isjusttheoldonemultiplied bye’"“’—it stillhasthesame m-value. Wecan write thisby R.(¢) I111") =8"“IJ'-m)- (13-24) Or,ifyouprefer, (I-1"’ IRz(¢) I11m)=5m.~.1@””°’ (13-25) (where 6,,,,,,,/ is1ifm’=in,orzerootherwise). Forarotation about anyother axisthere willbeamixing ofthevarious m-states. Wecould, ofcourse, trytowork outthematrix elements foranarbitrary rotation described bytheEuler angles /3,OZ,and7.Butitiseasier toremember that themost general such rotation canbemade upofthethree rotations R,(W), R,,(a), R,(I6); soifweknow thematrix elements forarotation about they-axis, wewillhave allweneed. How canweﬁndtherotation matrix forarotation bytheangle 6about the y-axis foraparticle ofspinj?Wecan't tellyouhow todoitinabasic way (with what wehave had). Wediditforspin one-half byacomplicated symmetry argu- ment. Wethen diditforspin onebytaking thespecial case ofaspin-one system which wasmade upoftwospinone-half particles. Ifyouwillgoalong with usand accept thefactthatinthegeneral case theanswers depend only onthespinj,and areindependent ofhow theinner guts oftheobject ofspinjareputtogether, we canextend thespin-one argument toanarbitrary spin. Wecan, forexample, cook upanartiﬁcial system ofspin %outofthree spin one-half objects. Wecan even avoid complications byimagining thattheyarealldistinct particles—like a proton, anelectron, andamuon. Bytransforming each spinone-half object, we canseewhat happens tothewhole system——remembering thatthethree amplitudes aremultiplied forthecombined state. Let’s seehow itgoes inthiscase. Suppose wetake thethree spin one-half objects allwith spins “up”; wecan indicate thisstate byI+++). Ifwelook atthissystem inaframe rotated about thez-axis bytheangle ¢,each plus stays aplus, butgets multiplied bye“’/2. Wehave three such factors, so R.<¢>|+++>=e"“’*‘ 1+++). (18.26) Evidently thestate I+++)isjustwhat wemean bythem=+%state, or thestate I%,-I-%). Ifwenowrotate thissystem about they-axis, eachofthespinone-half objects willhave some amplitude tobeplusortobeminus, sothesystem willnowbea mixture oftheeight possible combinations I+++),I‘l'+")>I*1‘—+), I—++),I+——),I—+—),I——+),orI———).Itisclear, however, thatthese canbebroken upintofoursets,each setcorresponding toaparticular value ofm.First, wehave I++—I—),forwhich m=%.Then there arethe three states I+—I——).I-lr-'+),andI—++)-—each with twoplusses and oneminus. Since each spinone-half object hasthesame chance ofcoming out minus under therotation, theamounts ofeachofthese three combinations should beequal. Solet’stakethecombination 1—I->l->+|—++> (18.27) X/§{++ +++ } withthefactor 1/\/3 putintonormalize thestate. Ifwerotate thisstate about thez-axis, wegetafactor e”/2 foreach plus, ande_‘°'2 foreach minus. Each term in(18.27) ismultiplied bye“"/2, sothere isthecommon factor e‘°/2.This 18-10 one“—" pieces. Forinstance, =a2cI+!+!+/>_I_a2dI+l+/_/>+abcI_I_!_/_I__/> +bacI-'+'+')+abdI+'_'-')+bad]-'+'-') +8%]-'-'+')+b2dI-'-'-'). (18.33) Adding twosimilar expressions forI+—+)andI—++)anddividing by \/3,weﬁnd _ I%,+%,S) =\/3a2cI%,+%,T) +(a2d +211116)I%,+%,T) +(2bad +b2c)I%,—%,T) +\/3b2a'I %,—%,T). (18.34) Continuing theprocess weﬁndalltheelements (jTIiS)ofthetransformation ma- trixasgiven inTable 18-2. Theﬁrstcolumn comes from Eq.(18.32); thesecond from (18.34). Thelasttwocolumns were worked outinthesame way. Table 18-2 Rotation matrix foraspin%particle I%,+%,-9) I%»~%»5) 1%-~%»$) (%.+%.TIa3x/3a2c \/3ac‘) c3 <%,+%-T I \/3a2b a2d—I—Zabc c2b+Zdac \/302d (§,_l’TI \/3abz 2bad —I—b2c 2cdb +d2a '\/3cdz <%,-%-Tl (/5824 \/3‘8112(The coefﬁcients a,b,c,anddaregiven inTable 12-4.) (ITIts) I8.+-as) 22 b3 d3 Now suppose theT-frame were rotated withrespect toSbytheangle I9about their y-axes. Then a,b,c,anddhave thevalues [see(l2.54)] a=d=cos0/2, andc=—b=sin6/2. Using these values inTable 18-2 wegettheforms which correspond tothesecond part ofTable 17-2, butnow foraspin £1system. Thearguments wehave justgone through arereadily generalized toasystem ofanyspinj.Thestates Ij,m)canbeputtogether from 2jparticles, each of spin one-half. (There arej —I—mofthem intheI+)state andj —mintheI—) state.) Sums aretaken over allthepossible ways thiscanbedone, andthestate isnormalized bymultiplying byasuitable constant. Those ofyouwho aremathe- matically inclined may beable toshow that thefollowing result comes out'I: <1.m’IRlI(0)I1.m>=to+mm"—m>Io'+m’)1(j-m')1]”2 (-l)k(cos 0/2)*1+'"'—"*-“(sin 0/2)’"-'”'+2" X2)(m-m’+k)!(j+m’-k)!(j-m-k)!k!’ (1835) where kistogooverallvalues which giveterms Z0inallthefactorials. Thisisquite amessy formula, butwithityoucancheck Table 17-2forj=1 andprepare tables ofyour ownforlarger j.Several special matrix elements areof extra importance and have been given special names. Forexample thematrix elements form=m’=0andintegral jareknown astheLegendre polynomials andarecalled P,(cos0): (j,0IR,,(0)I1'.0)=P,(eos 0). (18.36) TIfyouwant details, they.aregiven inanappendix tothischapter. 18-12 Theﬁrstfewofthese polynomials are: P0(cosI9)=1, (18.37) P;(cos0)=cos0, (18.38) P2(cos0)=-§(3cos20—1), (18.39) P3(cos0)==}(5cos30—3cos6). (18.40) 18-5 Measuring anuclear spin Wewould liketoshow youoneexample oftheapplication ofthecoeﬁicients wehavejustdescribed. Ithastodowith arecent, interesting experiment which youwillnowbeabletounderstand. Some physicists wanted toﬁndoutthespin ofacertain excited state oftheNe2° nucleus. Todothis, they bombarded a carbon target with abeam ofaccelerated carbon ions, andproduced thedesired excited state ofNe2°—called Ne2°*—in thereaction C12 +C12 _,Ne20* +ah where <11isthe11-particle, orHe‘. Several oftheexcited states ofNe2° produced thiswayareunstable anddisintegrate inthereaction Ne2°* —>O16 —I—a2. Soexperimentally there aretwoa-particles which come outofthereaction. We callthem a1and a2;since they come offwith different energies, they canbe distinguished from each other. Also, bypicking aparticular energy for111we canpickoutanyparticular excited state oftheNe2°. Theexperiment wassetupasshown inFig.18-9. Abeam of16-Mev carbon ionswasdirected onto athinfoilofcarbon. Theﬁrstat-particle wascounted ina silicon diffused junction detector marked a,—set toaccept or-particles ofthe proper energy moving intheforward direction (with respect totheincident C12 beam). Thesecond a-particle waspicked upinthecounter 012attheangle 0 withrespect to(X1.Thecounting rateofcoincidence signals from I11and012were measured asafunction oftheangle 0. Theideaoftheexperiment isthefollowing. First, youneed toknow thatthe spins ofC12,01°,andthea-particle areallzero. Ifwecallthedirection ofmotion oftheinitial C12the+z-direction, thenweknow thattheNe2°* must have zero angular momentum about thez-axis. None oftheother particles hasanyspin; theC12arrives along thez-axis andthe(X1leaves along thez-axis sotheycan’t have anyangular momentum about it.Sowhatever thespinjoftheNe2°* is, weknow thatitisinthestate Ij,0).Now what willhappen when theNe2°* disintegrates intoanO12andthesecond at-particle? Well, theat-particle ispicked upinthecounter 012andtoconserve momentum theO16must gooffintheop- posite direction.'I About thenewaxisthrough (12,there canbenocomponent of angular momentum. Theﬁnal state haszeroangular momentum about thenew axis,sotheNe2°* candisintegrate thiswayonlyifithassome amplitude tohave m’equal tozero, where m’isthequantum number ofthecomponent ofangular momentum about thenewaxis. Infact,theprobability ofobserving 0:2attheangle 0isjustthesquare oftheamplitude (ormatrix element) (j,0IR,,(I9) Ij,0). (18.41) ToﬁndthespinoftheNe2°* state inquestion, theintensity ofthesecond a-particle wasplotted asafunction ofangle andcompared with thetheoretical ‘IWecanneglect therecoil given totheNe2°"' intheﬁrstcollision. Orbetter still, wecancalculate what itisandmake acorrection forit. 18-13SILICON JUMITION DETECTORS 02,, ’ / CIZBEAM _’_ __ U16Mev —-v —— / °'CARBON FQIL 30j.i.q /cm Fig. 18-9. Experimental arrange- ment fordetermining thespin ofcertain states ofNew. ..- I 5.80MnSTATE J-I oiz- 0.61-§,IP,1colO)]2‘ __I (I-RADANE’0 5 PERsrsOO28 11¢»:OO|\’ EC E/DR5.53 Mev STATE J13 036'}y[P3(cos0l]' - /- 2040 so so iooI20140 160 CENTER-OF-MASS ANGLE INDEGREEScozucO2.051 [coNOO'§’ Fig. 18-10. Experimental results for theangular distribution oftheat-particles from two excited states ofNe2° pro- duced inthesetup ofFig.18-9. [From J.A.Kuehner, Physical Review, Vol. 125, p.1853, 1982.1curves forvarious values ofj.Aswesaidinthelastsection, theamplitudes (j,0IR,,(0) Ij,0)arejustthefunctions P,(cos0).Sothepossible angular distribu- tions arecurves of[P,(cos0)]2. Theexperimental results areshown inFig.18-10 fortwooftheexcited states. You canseethattheangular distribution forthe 5.80-Mev state ﬁtsverywellthecurve for[P1(cos(9)12,andsoitmust beaspin-one state. Thedata forthe5.63—Mev state, ontheother hand, arequite different; theyﬁtthecurve [P3(cos0)]2. Thestate hasaspinof3. From thisexperiment wehave been abletoﬁndouttheangular momentum of twooftheexcited states ofNe2°*. This information canthen beused fortrying tounderstand what theconﬁguration ofprotons andneutrons isinside this nucleus—one more piece ofinformation about themysterious nuclear forces. 18-6 Composition ofangular momentum When westudied thehyperﬁne structure ofthehydrogen atom inChapter 12 wehadtowork outtheinternal states ofasystem composed oftwoparticles- theelectron andtheproton-—each withaspinofone-half. Wefound thatthefour possible spin states ofsuch asystem could beputtogether intptwogroups—a group with oneenergy thatlooked totheexternal world likeaspin-one particle, andoneremaining state thatbehaved likeaparticle ofzerospin. That is,putting together twospin one-half particles wecanform asystem whose “total spin” isone,orzero. Inthissection wewant todiscuss inmore general terms thespin states ofasystem which ismade upoftwoparticles ofarbitrary spin. Itisanother important problem about angular momentum inquantum mechanical systems. Let’s ﬁrstrewrite theresults ofChapter 12forthehydrogen atom inaform thatwillbeeasier toextend tothemore general case. Webegan withtwoparticles which wewillnowcallparticle a(theelectron) andparticle b(theproton). Particle ahadthespinj,,(=§-), anditsz-component ofangular momentum macould have oneofseveral values (actually 2,namely m,,=+%orm,,=—-}).Similarly, thespinstate ofparticle bisdescribed byitsspinji,anditsz-component ofangular momentum mh. Various combinations ofthespin states ofthetwoparticles could beformed. Forinstance, wecould have particle awithma=1}andparticle bwith mi,=—%,tomake astate Ia,+%; b,-1}). Ingeneral, thecombined states formed asystem whose “system spin,” or“total spin,” or“tota1 angular momentum” Jcould be1,or0.And thesystem could have az-component of angular momentum M,which was+1,0,or-1when J=1,or0when J=0. Inthisnewlanguage wecanrewrite theformulas in(12.41) and(12.42) asshown inTable 18-3. Inthetable theleft-hand column describes thecompound state interms of itstotal angular momentum Jandthez-component M.Theright-hand column shows howthese states aremade upinterms ofthem-values ofthetwoparticles aandb. Wewant nowtogeneralize thisresult tostates made upoftwoobjects aand bofarbitrary spins j,,andjb.Westartbyconsidering anexample forwhich j,,=§ Table 18-3 Composition ofangular momenta fortwo spinitparticles (in=1},jb= I1-1.M-+1>-Ia.+8;1>,+8> IJ=1.M=11>-\%{|a.+%:b,—%)+|11.—%;b.+=1)} IJ=1-M =-1)=I11.-‘1“;l>,—%) 1 IJ =0:M = =_\/-2: a:+%; _IaI_%; 18-14 andjb=1,namely, thedeuterium atom inwhich particle aisanelectron (e)and particle bisthenucleus—a deuteron (d). Wehave then thatjg=je=5The deuteron isformed ofoneproton andoneneutron inastate whose total spin is one,sojl,=jd=1.Wewant todiscuss thehyperﬁne states ofdeuterium—just thewaywedidforhydrogen. Since thedeuteron hasthree possible states mb= m,1=+1, 0,-1, and theelectron hastwo, ma=me=+%, —%, there are sixpossible states asfollows (using thenotation Ie,me;d,md)): I@,+%;d,+1>, |e,+»1=; d,0>;I¢,—%; d,+l>, I¢,+%; d,—1>; |¢,~%; 11,0), Ie,—%; d,—l).(18.42) Youwillnotice thatwehave grouped thestates according tothevalues ofthesum ofmeandmd—arranged indescending order. Now weask: What happens tothese states ifweproject into adifferent coordinate system‘? Ifthenewsystem isjustrotated about thez-axis bytheangle ¢,then thestate Ie,mu;d,ma)getsmultiplied by eimeeeimdo =ei(me+md)¢>_ (The state maybethought ofastheproduct Ie,me)Id,md),andeach state vector contributes independently itsownexponential factor.) Thefactor (18.43) isofthe form e‘M", sothestate Ie,me;d,md)hasaz-component ofangular momentum equal to M=me—I—md. (18144) Thez-component ofthetotal angular momentum isthesum ofthez-components of angular momentum oftheparts. Inthelistof(18.42), therefore, thestate inthetoplinehasM=+%, the twointhesecond linehave M=+%, thenext twohave M=~%,and the laststate hasM=—%. Weseeimmediately onepossibility forthespinJofthe combined state (thetotal angular momentum) must be%,andthiswillrequire four states with M=-I-%,+5,—%,and—%. There isonly onecandidate forM=%,soweknow already that 11=%,M=+%>=|<=,+%;<1,+1>. <18-45> Butwhat isthestate IJ=%,M=%)?Wehave twocandidates inthesecond line of(18.42), and, infact, anylinear combination ofthem would also have M= So,ingeneral, wemust expect toﬁndthat IJ=%,M=+%>=v<l¢,+%;d,0> +BI¢,—%;d,+1>, (18-46) where ozandBaretwonumbers. They arecalled theClebsch-Gordon coeﬁicients. Ournextproblem istoﬁndoutwhat theyare. Wecanﬁndouteasily ifwejustremember thatthedeuteron ismade upofa neutron andaproton, andWrite thedeuteron states outmore explicitly using the rules ofTable 18-3. Ifwedothat, thestates listed in(18.42) thenlookasshown in Table 18-4. Wewant toform thefour states ofJ=%,using thestates inthetable. Butwealready know theanswer, because inTable 18-1 wehave states ofspin %formed from three spin one-half particles. The ﬁrst state inTable l8—1 has IJ=%,M=+%)anditisI+++),which—in ourpresent notation—is the same asIe,—I-71;; n,—|—§, p,+%), ortheﬁrststate inTable 18-4. Butthisstate is alsothesame astheﬁrstinthelistof(18.42), conﬁrming ourstatement in(18.45). Thesecond lineofTable 18-1says—changing toourpresent notation—that I1=s;M=+s>={I<=,+an,+%; p,-s> +I¢,+%;-n,—%; p,+%) +I¢,—%; n,+%; r>,+%>}. (18.47) 18-15 Table 18-4 Angular momentum states foradeuterium atom m=% |e.+=1¢;d.+1>=I<=.+%;n,+&;p.+%> "1==12 Ie,+%; 11.0)=5-iii e,+1};n,+%;p.—%) +Ie.+%; 11.—%;r>,+%)} I°,—%; d,+1) =I°:—‘l"; n.+%; t>,+%) "1=—% I¢,+%;d,—1) =Ie,+%: r1,—%;r>.—%) Ies_%; =8&5 e>_"%; ns+%; +IeI_%; 111-2; "1=-% I=,—%; d,—1) =I=,—%; r1,—%;r>,—%) Theright sidecanevidently beputtogether from thetwoentries inthesecond line ofTable 18-4 bytaking \/2/3oftheﬁrstterm with \/1/3ofthesecond. That is, Eq.(18-47) isequivalent to IJ=8,M=%)=\/2/3 I¢,+%; d,0)+\/1/3 I¢,—%; d,1)- (18-48) Wehave found ourtwoClebsch-Gordon coefﬁcients atand,8inEq.(18.46): a=\/2/3, 6=\/1/3. (18.49) Following thesame procedure wecanﬁndthat IJ=8.114=-é)=V1/3I¢,+%;<1,—1)+\/2/3I¢,-'lt;d,0)- (18-50) And, also, ofcourse, IJ=%.M=—%>=le,—%;<1.-1>. (18.51) These aretherules forthecomposition ofspinlandspin1}tomake atotal J=§. Wesummarize (18.45), (18.48), and(18.50) inTable 18-5. Wehave, however, onlyfourstates herewhile thesystem weareconsidering hassixpossible states. Ofthetwostates inthesecond lineof(18.42) wehave used only onelinear combination toform IJ=1%,M=+-5). There isanother linear combination orthogonal totheonewehave taken which also hasM=+-}, namely V1/3I¢,+%;d,0) -V2/3 I¢,—%; d,+1)- (13-52) Table 18-5 TheJ--3-states ofthedeuterium atom I1=isM=+%)=I=.+%;d,+1) IJ=g,M=+%>=\/2/_3Ie,+§;d,0) +\/U§|e.—%;d.1> |J=gM=-§>=\/T/?|e.+%;d.—1>+ \/Y/3|-=.—1.\;<1.<>> IJ=%,M=—%)=I¢,—%;d.—1> 18-16 Similarly, thetwostates inthethird lineof(18.42) canbecombined togivetwo orthogonal states, each with M=~%. Theoneorthogonal to(18.52) is \/'2? Ie>+%; — V I€s—%; d>O>' These arethetworemaining states. They have M=me+md==I=%; and must bethetwostates corresponding toJ=%.Sowehave IJ=s,M=s>=\/1/3Ie,+e;d,0> —\/i7§I<>,—%; d,+1>.(18.54) I1=%.M=—%>=~/2/3Ie.+s;<1.-1>—v1/3Ie.-=1;d.0>- Wecanverify thatthese twostates doindeed behave likethestates ofaspin one-half object bywriting outthedeuterium parts interms oftheneutron and proton states——using Table 18-3. Theﬁrststate in(18.53) is \/1/_6{I ¢,+%;n,+%;i>,—%) +Ie,+%;I1,—%;P,+%>} —\/2/3I¢,—%;n,+%; p,+%>, (13-55) Wl’11C1'1 canalsobewritten \/1/3I\/Vi {I@,+%;n,+%;p,—%>—I¢,—%sn,+%;p,+%>} +\/W (Ie,+%;n,—%;p,+%) —I=,—%1n,+%;p,+%)}I-(18.56) Now look attheterms intheﬁrstcurly brackets, andthink oftheeandptaken together. Together theyform aspin-zero state (seethebottom lineofTable 18-3), andcontribute noangular momentum. Only theneutron isleft,sothewhole of theﬁrst curly bracket of(18.56) behaves under rotations likeaneutron, namely asastate with J=%,M=+5 Following thesame reasoning, weseethat inthesecond curly bracket of(18.56) theelectron andneutron team uptoproduce zero angular momentum, andonly theproton contribution—with mp=§—is left. Theterms behave likeanobject with J=%,M=+%. Sothewhole ex- pression of(18.56) transforms likeIJ=+%, M=+%) asitshould. The M=—%state which corresponds to(18.57) canbewritten down (bychanging theproper +§’s to——%’s) toget \/V-’7[\/1/2 {Ie,+%;n,—%;p,—%> —I¢,—%;11,—%;p,+%>} +\/1/2 {Ie,+%;I1’-%;i>,—%) —le,—%;n,+%;p,-%)}I' (18.57) Youcaneasily check thatthisisequal tothesecond lineof(18.54), asitshould be ifthetwoterms ofthatpairaretobethetwostates ofaspinone-half system. Soour results areconﬁrmed. Adeuteron andanelectron canexist insixspinstates, four ofwhich actlikethestates ofaspin5%object (Table 18-5) andtwoofwhich act likeanobject ofspinone-half (18.54). Theresults ofTable 18-5andofEq.(18.54) were obtained bymaking useof thefactthatthedeuteron ismade upofaneutron andaproton. Thetruth ofthe equations does notdepend onthatspecial circumstance. Foranyspin-one object puttogether with anyspinone-half object thecomposition laws (and thecoefﬁ- cients) arethesame. Thesetofequations inTable 18-5 means thatiftheco- ordinates arerotated about, say,they-axis—so thatthestates ofthespinone-half particle andofthespin-one particle change according toTable 18-1/and Table l8—2—the linear combinations ontheright-hand sidewillchange intheproper way foraspin %object. Under thesame rotation thestates of(18.54) will change asthestates ofaspinone-half object. Theresults depend only onthe 18—17 Table 18-6 Composition ofaspinone-half particle (ja=h‘) andaspin-one particle (jb=1). IJ= = =Ia,+§; b,-I-1) IJ= = =\/7/3| a.+%;b,0 >+\/1*/51 a,—%;1>,+1) IJ= =—=\/ﬁIa.+%;1>.—1> +\/%Ia,—%;1>.0> IJ= =— =Ia»—%;b,—1) :;>I==gmr_~_=k-=gmKRRK~>-wIw NIQl\>L-1\/\/\/\/ IJ= =+t>=\F1/_3|a.+s;1>.<>> —~/iFIa.—%;b.+1> I]= =—%> =\/ﬁIa’+%;b>'_1> _\/U§Ia’_%;b’O> weweKK rotation properties (that is,thespinstates) ofthetwooriginal particles butnot inanywayontheorigins oftheir angular momenta. Wehave only made useof thisfacttowork outtheformulas bychoosing aspecial caseinwhich oneofthe component parts isitself made upoftwospin one-half particles inasymmetric state. Wehave putallourresults together inTable 18-6, changing thenotation “e”and“d”to“a”and“b”toemphasize thegenerality oftheconclusions. Suppose wehave thegeneral problem ofﬁnding thestates which canbe formed when twoobjects ofarbitrary spins arecombined. Sayonehasj,,(soits z-component maruns overthe2j,,+1values from —j,,to+111)andtheother has jb(with z-component ml,running overthevalues from —jbto+j,,). Thecombined states areIa,ma;b,ma), andthere are(2j,,+l)(2jb —I—1)diiierent ones. Now what states oftotal spinJcanbefound? Thetotal z-component ofangular momentum Misequal toma+mb,and thestates canallbelisted according toM[asin(l8.42)]. Thelargest Misunique; itcorresponds toma=j,,andmi,=jb,andis,therefore, justja+jb.That means thatthelargest total spinJisalsoequal tothesumja+jb: J=(M)max =ja FortheﬁrstMvalue smaller than (M)m“, there aretwostates (either maormi, isoneunitlessthan itsmaximum). They must contribute onestate tothesetthat goes with J=ja+jb,andtheoneleftover willbelong toanewsetwith J= j,,+jb—1.ThenextM-va1ue—the third from thetopofthelist-—can beformed inthree ways. (From ma=j,,—2,mb=j,,;from m,,=j,,—1,m1,=jb—1; andfrom ma=ja,m1,=jb—2.)Two ofthese belong togroups already started above; thethird tellsusthatstates ofJ=j,,+jb—2must alsobeincluded. This argument continues until wereach astage where inourlistwecannolonger goonemore stepdown inoneofthem’stomake newstates. Letjbbethesmaller ofjaandj;,(iftheyareequal takeeither one); then only 2j,,values ofJarerequired—going ininteger steps from j,,+jl,down toj,—jb. That is,when twoobjects ofspinj,,andjbarecombined, thesystem canhave a total angular momentum Jequal toanyoneofthevalues ja. ju _l J= ja-1-jb —2 (18.58) Ija. (Bywriting Ij,,—jbIinstead ofja—jbwecanavoid theextra admonition that ja2 Foreachofthese Jvalues there arethe2J+1states ofdiﬁerent M-values-— with Mgoing from +Jto—J. Each ofthese isformed from linear combinations oftheoriginal states Ia,ma;b,mb)withappropriate factors—the Clebsch-Gordon 18—18 coefﬁcients foreach particular term. Wecanconsider thatthese coefﬁcients give the“amount” ofthestate Ij,,,ma;j1,, mb)which appears inthestate IJ,M). So each oftheClebsch-Gordon coeﬂicients has,ifyouwish, sixindices identifying itsposition intheformulas likethose ofTables 18-3 and 18-6. That is,calling these coefﬁcients C(J,M;ja,ma;jb,mb), wecould express theequality ofthe second lineofTable 18-6 bywriting C(§-,+%; %,+112“;1,0)=V2/3, 2»_%; =V Wewillnotcalculate here thecoefﬁcients foranyother special cases.'I' You can,however, ﬁndtables inmany books. You might wish totryanother special caseforyourself. Thenextonetodowould bethecomposition oftwospin-one particles. Wegivejusttheﬁnal result inTable 18-7. These laws ofthecomposition ofangular momenta arevery important in particle physics—where they have innumerable applications. Unfortunately, we have notime tolook atmore examples here. Table 18-7 Composition oftwospin-one particles (j,,=1,jb=1) I1=2.M=+2)=Ia,+1;b.+1> ée3*I1=2.M=+1)=—|a,+1;b,0) +%Ia.0;1>.+1> 1 2a ‘/6 a ‘/6 a ) I1=2.M=—1>-\-;—5l11,0;b,—1) +$11.-1;1>.o> IJ=2,M=-2)=Ia,—1;b,—1) 1 I a \/2a 3,- 1IJ=1,M =0)=—a,+1;b,—1)- —Ia,—1;b,+1)\/E 3: 1 1 IJ :1!M = :T_—Ia20;b:_1>— iIa:—1;b!0> \/2 \/2 I1=0.M-<>>=L{la.+1;b.—1)+|a,—1;b.+1>—|a.0;b.0)}(/5 Added Note 1:Derivation oftherotation matrixft Forthose who would liketoseethedetails, wework outhere thegeneral rotation matrix forasystem withspin(total angular momentum)j. Itisreally not veryimportant towork outthegeneral case; once youhave theidea, youcanﬁnd thegeneral results intables inmany books. Ontheother hand, after coming thisfaryoumight liketoseethatyoucanindeed understand even theverycom- plicated formulas ofquantum mechanics, such asEq.(18.35), thatcome intothe description ofangular momentum. IAlarge partofthework isdone nowthatwehave thegeneral rotation matrix Eq. (18.35). IThe material ofthisappendix wasoriginally included inthebody ofthelecture. Wenowfeelthatitisunnecessary toinclude suchadetailed treatment ofthegeneral case. 18-19 Weextend thearguments ofSection 18-4 toasystem with spinj,which we consider tobemade upof2jspinone-half objects. Thestate with m=jwould beI+++-~-+)(withj plussigns). Form=j—1,there willbe2jterms like I++~--++—),I++---+—+),and soon. Let’s consider the general caseinwhich there arerplusses andsminuses—with r+s=2j.Under arotation about thez-axis eachoftherplusses willcontribute e+‘°/2.Theresult isaphase change ofi(r/2 —s/2)¢. You seethat ___?‘-S m- 2- (18.59) JustasforJ=%,each state ofdeﬁnite mmust bethelinear combination with plussigns ofallthestates with thesame rands-that is,states corresponding to every possible arrangement which hasrplusses andsminuses. Weassume that youcanﬁgure outthatthere are(r+s)!/r!s! such arrangements. Tonormalize each state, weshould divide thesumbythesquare root ofthisnumber. Wecan write -1/2 :{I_I__I_+(_..._I__I::.____.Y..__Z +(allrearrangements oforder)} =Ij,m) (18.60) with ._r+s __r—s_)__--2 ,111-12 (18.61) Itwillhelp ourwork ifwenow gotostillanother notation. Once wehave deﬁned thestates byEq.(18.60), thetwonumbers randsdeﬁne astate justas wellasjandm.Itwillhelpuskeep track ofthings ifwewrite I111") =II). (13-62) where, using theequalities of(18.67) r=j+m, s=j—m. Next, wewould liketowrite Eq.(18.60) with anewspecial notation as - '1 + +1/2 r s I1.m>=I.>= 1|+>I->1...... (18.63) Note thatwehave changed theexponent ofthefactor infront toplusQ-.Wedo thatbecause there arejustN=(r+s)!/r!s! terms inside thecurly brackets. Comparing (18.63) with (18.60) itisclear that {I+)'I—)“}perm isjustashorthand wayofwriting {I—I——I—---——)+allrearrangements}N . where Nisthenumber ofdiﬁferent terms inthebracket. Thereason thatthis notation isconvenient isthateach time wemake arotation, alloftheplussigns contribute thesame factor, sowegetthisfactor totherthpower. Similarly, all together thesminus terms contribute afactor tothesthpower nomatter what the sequence oftheterms is. Now suppose werotate oursystem bytheangle 0about they-axis. What we want isR,,(0) IQ).When R,,(0) operates oneach I+)itgives R1/(9)I+)=I+)C+ I—)$. (13.64) where C=cos0/2andS=sin0/2. When R,,(0) operates oneach I—}itgives Ry(9)I—)=I-)C— I+)$- 18-20 Sowhat wewant is 121(0)I:>= f>‘I"2R.<@>{I+>’I->‘1...... Z {<R.<1>I+>>'<R.<1> I—>>”} = 11+>c+I—>S)’(I—>c-I+>s*1...... (18.65) Now each binomial hastobeexpanded outtoitsappropriate power andthetwo expressions multiplied together. There willbeterms with I—I—)toallpowers from zero to(r+s).Let’s look atalloftheterms which have I+)tother’power. They willappear always multiplied with I—)tothes’power, where s’=2j—r’. Suppose wecollect allsuch terms. Foreach permutation they willhave some numerical COClTlClCI1t involving thefactors ofthebinomial expansion aswellas thefactors CandS.Suppose wecallthatfactor A,1.Then Eq.(18.65) willlooklike r-I-.1 R1109) ID=Z {AV I+>T I_>8lp<*r1n' (18-66) 1"=0 Now let’ssaythatwedivide A,1bythefactor [(r’—I—s’)!/r’Is’!]1/2 andcallthe quotient B,1. Equation (18.66) isthenequivalent to r+s / /1/2 Ryw) I =2 Br’I:rr/—|I;/|s:I +>T) I_>8’}lJ€l'n1- (18-67) r’=0 '' (Wecould justsaythatthisequation deﬁnes B,1bytherequirement that(18.67) gives thesame expression thatappears in(18.65).) With thisdeﬁnition ofB,»theremaining factors ontheright-hand sideof Eq.(18.67) arejustthestates IQ). Sowehave that RttoI;>=fjB.1|;i>. (18.68)rD1 with s’always equal tor+s—r’.This means, ofcourse, thatthecoeﬂicients B,-arejustthematrix elements wewant, namely (iiIR11(@)I §)=B~- (18-69) Now wejusthave topush through thealgebra toﬁndthevarious B,1. Com- paring (18.39) with (l8.37)—-and remembering that r’—I—s’=r—I—s—-we see thatB,1isjustthecoefficient ofa"bs' inthefollowing expression: r/Is/I 1/2 (aC+bS)’(bC -as)". (18.70) ltisnow only adirty jobtomake theexpansions bythebinomial theorem, and collect theterms withthegiven power ofaandb.Ifyouwork itallout,youﬁnd thatthecoefficient ofa"b“" in(18.70) is r’!s’! 1/2 1T—T'+2lC .+w_21 r! S! Ir!s!I (-1)S C '(r-r’+k)!(r’ -k)!'(S-k)!k!' (18.71) Thesumistobetaken overallintegers kwhich giveterms ofzeroorgreater inthe factorials. This expression isthen thematrix element wewanted. Finally, wecanreturn toouroriginal notation interms ofj,m,andm’using r=j+m, r’=j+m', s=j—m, s’=j—m’. Making these substitutions, wegetEq.(18.34) inSection 18-4. 18-21 Added Note 2:Conservation ofparity inphoton emission lnSection 1ofthischapter weconsidered theemission oflight byanatom thatgoesfrom anexcited state ofspin1toaground state ofspin0.Iftheexcited state hasitsspinup(m=+1), itcanemit aRHC photon along the+2-axis or aLHC photon along the-2-axis. Let’s callthese twostates ofthephoton IR.,,,) andILg“); Neither ofthese states hasadeﬁnite parity. Letting Pbetheparity operator, PIR,,,,) =IL,i,,)andPIL4“) =IR,,,,). What about ourearlier proof thatanatom inastate ofdeﬁnite energy must have adeﬁnite parity, andourstatement thatparity isconserved inatomic proc- esses? Shouldn’t theﬁnalstate inthisproblem (thestate after theemission ofa photon) have adeﬁnite parity? Itdoes ifweconsider thecomplete ﬁnal state which contains amplitudes fortheemission photons intoallsorts ofangles. In Section 1wechose toconsider onlyapartofthecomplete ﬁnal state. Ifwewish wecanlook onlyatﬁnalstates thatdohave adeﬁnite parity. For example, consider aﬁnal state It!/F) which hassome amplitude oztobeaRHC photon going along +2andsome amplitude BtobeaLHC photon going along —z. Wecanwrite IIPF) =11IRup) —I—5ILdn>- (13-72) Theparity operation onthisstate gives PI11>=61IL..>+11IR...) (18-13) Thisstate willbe===I(I/F)ifti=aorif[3=—a. Soaﬁnalstate ofeven parity 1S I =a{R11p> + ILdnlls andastate ofoddparity is III/F) =<1{IR1111)-IL<ii.)}- (13-75) Next, wewish toconsider thedecay ofanexcited state ofoddparity toa ground state ofeven parity. Ifparity istobeconserved, theﬁnal state ofthe photon must have oddparity. Itmust bethestate in(18.75). Iftheamplitude to ﬁndIRup> isa,theamplitude toﬁndILd,,) is—-a. Now notice what happens when weperform arotation of180° about the y-axis. Theinitial excited state oftheatom becomes anm=——lstate (with no change insign, according toTable 17-2). Andtherotation oftheﬁnalstate gives R.<180°) I1;)=<1{IR...)-IL..>I- (18.16) Comparing thisequation with (18.75), youseethatfortheassumed parity ofthe ﬁnal state, theamplitude togetaLHC photon along —I—zfrom them=-1 initial state isthenegative oftheamplitude togetaRHC photon from them=+1 initial state. This agrees withtheresult wefound inSection 1. 18-22 I9 The Hydrogen Atom and The Periodic Table 19-1 Schrodinger’s equation forthehydrogen atom Themost dramatic success inthehistory ofthequantum mechanics wasthe understanding ofthedetails ofthespectra ofsome simple atoms andtheunder- standing oftheperiodicities which arefound inthetable ofchemical elements. Inthischapter wewillatlastbring ourquantum mechanics tothepoint ofthis important achievement, speciﬁcally toanunderstanding ofthespectrum ofthe hydrogen atom. Wewillatthesame timearrive ataqualitative explanation ofthe mysterious properties ofthechemical elements. Wewilldothisbystudying in detail thebehavior oftheelectron inahydrogen atom—for theﬁrsttime making adetailed calculation ofadistribution-in-space according totheideas wedeveloped inChapter 16. Foracomplete description ofthehydrogen atom weshould describe themo- tions ofboth theproton andtheelectron. ltispossible todothisinquantum mechanics inawaythatisanalogous totheclassical ideaofdescribing themotion ofeach particle relative tothecenter ofgravity, butwewillnotdoso.Wewill justdiscuss anapproximation inwhich weconsider theproton tobevery heavy, sowecanthink ofitasﬁxed atthecenter oftheatom. Wewillmake another approximation byforgetting thattheelectron hasa spinandshould bedescribed byrelativistic laws ofmechanics. Some small cor- rections toourtreatment willberequired since wewillbeusing thenonrelativistic Schrodinger equation andwilldisregard magnetic effects. Small magnetic effects occur because from thee1ectron’s point-of-view theproton isacirculating charge which produces amagnetic ﬁeld. Inthisﬁeld theelectron willhave adifferent energy withitsspinupthan withitdown. Theenergy oftheatom willbeshifted alittle bitfrom what wewillcalculate. Wewillignore thissmall energy shift. Also wewillimagine thattheelectron isjustlikeagyroscope moving around in space always keeping thesame direction ofspin. Since wewillbeconsidering a freeatom inspace thetotal angular momentum willbeconserved. lnourapproxi- mation wewillassume thattheangular momentum oftheelectron spinstays con- stant, soalltherestoftheangular momentum oftheatom-—what isusually called “orbital” angular momentum—will alsobeconserved. Toanexcellent approxi- mation theelectron moves inthehydrogen atom likeaparticle without spin—the angular momentum ofthemotion isaconstant. With these approximations theamplitude toﬁndtheelectron atdifferent places inspace canberepresented byafunction ofposition inspace andtime. Welet¢(x,y,z,t)betheamplitude toﬁndtheelectron somewhere atthetime t. According tothequantum mechanics therateofchange ofthisamplitude with timeisgiven bytheHamiltonian operator working onthesame function. From Chapter 16, ihgif=so/,, (19.1) with A) 1122at=-2-iv +1/(1). (19.2) Here, mistheelectron mass, andV(r)isthepotential energy oftheelectron inthe 19-119-1 Schrodinger’s equation forthe hydrogen atom 19-2 Spherically symmetric solutions 19-3 States with anangular dependence 19-4 Thegeneral solution for hydrogen 19-5 Thehydrogen wave functions 19-6 Theperiodic table ‘Z ~\\\\\\ \ P IIIIoI6I/I/I/qI1/l/¥\\\\\--——;i> // y X Fig. 19-1. The spherical polar co- ordinates r,0,¢>ofthepoint P.electrostatic ﬁeldoftheproton. Taking V=0atlarge distances from theproton wecanwrite'I' 2V=_e_.r Thewave function ipmust then satisfy theequation . __ hz 2 32 Wewant tolook fordeﬁnite energy states, sowetrytoﬁndsolutions which have theform wt-.1)=e-"'"""=1<-)- (19.4) Thefunction ib(r)must then beasolution of 2 2 -%V21=(E+2)1. <19-5) where Eissome constant—the energy oftheatom. Since thepotential energy term depends only ontheradius, itturns outto bemuch more convenient tosolve thisequation inpolar coordinates rather than rectangular coordinates. TheLaplacian isdeﬁned inrectangular coordinates by 2 2 2 2_L L L.V_8x2+6y‘1+6z2 Wewant touseinstead thecoordinates r,0,¢shown inFig. 19-1. These coordinates arerelated tox,y,zby x=rsin6cos¢; y=rsin0sin¢; z=rcos9. lt‘sarather tedious mess towork through thealgebra, butyoucaneventually show thatforanyfunction f(r) =f(r,0,¢), 2 2. l6 l l d . 3 1 3 V2](r’0’lb)=;M7(rf)+F5Isin060(Sm068>+sin?06¢f‘I' (H6) Sointerms ofthepolar coordinates, theequation which istobesatisﬁed by ¢(r.18¢)is 1a2 11a.aip 1a’¢ 2 2 ;19?(Hp)+F5Isin058(Sm080)+sin?084>? =Qfig} E+e7lb" (19.7) 19-2 Spherically symmetric solutions Let’s ﬁrsttrytoﬁndsome very simple function that satisﬁes thehorrible equation in(19.7). Although thewave function I11will,ingeneral, depend onthe angles I9and¢aswellasontheradius r,wecanseewhether there might beaspecial situation inwhich 1/1does notdepend ontheangles. Forawave function that doesn't depend ontheangles, none oftheamplitudes willchange inanywayif yourotate thecoordinate system. That means thatallofthecomponents ofthe angular momentum arezero. Such aipmust correspond toastate whose total angular momentum iszero. (Actually, itisonly theorbital angular momentum which iszero because westillhave thespinoftheelectron, butweareignoring thatpart.) Astate withzeroorbital angular momentum iscalled byaspecial name. ltiscalled an“s-state”—you canremember “sforspherically symmetric."I Asusual, e2=:12/41re().-e O - 1Since these special names arepartofthecommon vocabulary ofatomic physics, you willjusthave tolearn them. Wewillhelpoutbyputting them together inashort “dic- tionary" later inthechapter. 19-2 Now ifipisnotgoing todepend on0andqtthentheentire Laplacian contains onlytheﬁrstterm andEq.(19.7) becomes much simpler: 1d2 2m e2 ;Jr;("t/') =-Z5(E-1-rr-)1/h (19-8) Before youstart towork onsolving anequation likethis,it’sagood ideatoget ridofallexcess constants likee2,m,andh,bymaking some scale changes. Then thealgebra willbeeasier. Ifwemake thefollowing substitutions: h2 I‘=;n—‘;§p, and H1€4 E= 6, (19.10) then Eq.(19.8) becomes (after multiplying through byp) d2(P¢) 2 These scale changes mean thatwearemeasuring thedistance randenergy Eas multiples of“natural” atomic units. That is,p=r/rB, where rB=112/me2, iscalled the“Bohr radius” andisabout 0.528 angstroms. Similarly, e=E/ER, with ER=me‘/2h2. This energy iscalled the“Rydberg” andisabout 13.6 electron volts. Since theproduct pibappears onboth sides, itisconvenient towork with it rather than with1pitself. Letting pib=f, (19.12) wehave themore simple-looking equation 2 =——(6+ (19.13) Now wehave toﬁndsome function fwhich satisﬁes Eq.(19.13)—in other words, wejust have tosolve adifferential equation. Unfortunately, there isno veryuseful, general method forsolving anygiven diﬁerential equation. YouJust have toﬁddle around. Our equation isnoteasy, butpeople have found that it canbesolved bythefollowing procedure. First, youreplace f,which issome function ofp,byaproduct oftwofunctions f(P)=@“°“’g(p)- (19-14) Thisjustmeans thatyouarefactoring e_°"’ outoff(p). Youcancertainly dothat foranyf(p) atall.This_]US1lshifts ourproblem toﬁnding theright function g(p). Sticking (19.14) into(19.13), wegetthefollowing equation forg: d2g dg 2HF-2a;]3+<;+e+<>F)g=0. (19.15) Since wearefreetochoose 01,let’smake 042=-6, (19.16) andget d2g dg 2_8? -—20¢2;)-1-Bg-O. (19.17) You may think wearenobetter oﬁ"than wewere atEq.(19.13), butthehappy thing about ournewequation isthatitcanbesolved easily interms ofapower series inp(ltispossible, inprinciple, tosolve (19.13) that Way too, butitis 19-3 much harder.) Wearesaying thatEq.(19.17) canbesatisﬁed bysome g(p)which canbewritten asaseries, 017 g(p)=2aw", (19-18)k=1 inwhich theakareconstant coefﬁcients. Now allwehave todoisﬁndasuitable inﬁnite setofcoefﬁcients! Let’s check toseethatsuch asolution willwork. The ﬁrstderivative ofthisg(p)is dg_w k-1F15 —IE1 akkp , andthesecond derivative is “'28 _W k-2Zip -—- —' . Using these expressions in(19.17) wehave Zk(k-l)a,,pk_2 -Z2akakpk_1 +Z2a,.p’“—‘ =0.(19.19)i=1 (=1 Ic=1 lt’snotobvious thatwehave succeeded; butweforge onward. ltwillalllook better ifwereplace theﬁrstsumbyanequivalent. Since theﬁrstterm ofthesum iszero, wecanreplace each kbyk+1without changing anything intheinﬁnite series; with thischange theﬁrstsumcanequally wellbewritten as E(k-1"l)ka1¢+1Pk*1- k=1 Now wecanputallthesums together toget Z[(/<+l)kak+1 -Zakak+2a,.]p’~—‘ =0. (19.20)k.-=1 This power series must vanish forallpossible values ofp.Itcandothatonly ifthecoefﬁcient ofeach power ofpisseparately zero. Wewillhave asolution forthehydrogen atom ifwecanﬁndasetakforwhich (k+l)kak+1 —2(ak —l)a;, =0 (19.21) forallk>1.That iscertainly easy toarrange. Pick anya1youlike. Then generate alloftheother coeﬁicients from G;;+1 = Gk. With thisyouwillgeta2,a3,a4,andsoon,andeach pairwillcertainly satisfy (19.21). Wegetaseries forg(p)which satisﬁes (19.17). With itwecanmake a (L,thatsatisﬁes Schrodinger’s equation. Notice thatthesolutions depend onthe assumed energy (through oz),butforeachvalue ofe,there isacorresponding series. Wehave asolution, butwhat does itrepresent physically? Wecangetan ideabyseeing what happens farfrom theproton—for large values ofp.Outthere, thehigh-order terms oftheseries arethemost important, soweshould look at what happens forlarge k.When k>>1,Eq.(19.22) isapproximately thesame as 201ilk-|-1 =-k-ak, which means that 2 k 111+.~§-ki,)-- (19.23) Butthese arejustthecoelﬁcients oftheseries fore+2"“’. Thefunction ofgisa rapidly increasing exponential. Even coupled with e_“" toproduce f(p)——see 19-4 Eq.(19.l4)—it stillgives asolution forf(p)which goes likeea”forlarge p.We have found amathematical solution butnotaphysical one. 1trepresents asitua- tion inwhich theelectron isleast likely tobenear theproton! Itisalways more likely tobefound ataverylarge radius p.Awave function forabound electron must gotozeroforlarge p. Wehave tothink whether there issome waytobeatthegame, andthere is. Observe! Ifitjusthappened byluck thatozwere equal to1/n,where nisany integer, then Eq.(19.22) would make a,,+1 =O.Allhigher terms would alsobe zero. Wewouldn’t have aninﬁnite series butaﬁnite polynomial. Anypolynomial increases more slowly than e°"’,sotheterm e_°"’willeventually beatitdown, and thefunction fwillgotozeroforlarge p.Theonlybound-state solutions arethose forwhich or=1/n,with n=1,2,3,4,andsoon. Looking back toEq.(19.16), weseethatthebound-state solutions tothe spherically symmetric wave equation canexist only when --ezllslsl 1 ’4916 n2 Theallowed energies arejustthese fractions times theRydberg, ER=me“/2fi2, ortheenergy ofthenthenergy level is 1E,,=—ER (19.24) There is,incidentally, nothing mysterious about negative numbers fortheenergy. Theenergies arenegative because when wechose towrite V=—e2/r, wepicked ourzeropoint astheenergy ofanelectron located farfrom theproton. When it isclose totheproton, itsenergy isless,sosomewhat below zero. Theenergy is lowest (most negative) forn=1,andincreases toward zerowithincreasing n. Before thediscovery ofquantum mechanics, itwasknown from experimental studies ofthespectrum ofhydrogen thattheenergy levels could bedescribed by Eq.(19.24), where ERwasfound from theobservations tobeabout 13.6electron volts. Bohr then devised amodel which gave thesame equation andpredicted thatERshould beme‘/Zhz. Butitwastheﬁrstgreat success oftheSchrodinger theory thatitcould reproduce thisresult from abasic equation ofmotion forthe electron. Now thatwehave solved ourﬁrstatom, let’slook atthenature ofthesolution wegot. Pulling allthepieces together, each solution looks likethis: 11..=f—",-ﬁi)=gap). (19.25)where gap)=ZYatpk (19.26)k=l and a),+1= a,,. (19.27) Solong aswearemainly interested intherelative probabilities ofﬁnding the electron atvarious places wecanpickanynumber wewishfora1.Wemayaswell seta1=1.(People often choose a1sothatthewave function is“normalized,” thatis,sothattheintegrated probability ofﬁnding theelectron anywhere inthe atom isequal to1.Wehave noneed todothatjustnow.) Forthelowest energy state, n=1,and 1,!/1(p) =e_". (19.28) Forahydrogen atom initsground (lowest-energy) state, theamplitude toﬁndthe electron atanypoint drops offexponentially with thedistance from theproton. Itismost likely tobefound right attheproton, andthecharacteristic spreading distance isabout oneunitinp,orabout oneBohr radius, rB. l9—5 Fig. 19-2. The wave functions for theﬁrst three I=0states ofthehydro- gen atom. (The scales arechosen sothat thetotal probabilities areequal.)Putting n=2gives thenext higher level. Thewave function forthisstate willhave twoterms. Itis ¢3(p)=(1-g)er”? (19.29) Thewave function forthenextlevel is mp)=(1—33‘!+,3,p2)@—"’*“‘- (19.10) The wave functions forthese ﬁrst three levels areplotted inFig. 19-2. You can seethegeneral trend Allofthewave functions approach zerorapidly forlarge pafter oscillating afewtimes. Infact, thenumber of“bumps” isjust equal to ii——-or. ifyouprefer, thenumber ofZC1'O-Cl‘OSS111gS ofil/,,isii—1. ‘ill n=3 \i _T; ': n=2 19-3 States withanangular dependence lnthestates described bythe¢,,(r) wehave found thattheprobability ampli- tude forﬁnding theelectron isspherically symmetric—depending only onr,the distance fortheproton. Such states have zero orbital angular momentum. We should nowinquire about states which mayhave some angular dependences. Wecould, ifwewished, just investigate thestrictly mathematical problem of ﬁnding thefunctions ofr,6,and¢which satisfy theditferential equation (19.7)- putting intheadditional physical conditions that theonly acceptable functions areones which gotozero forlarge r.You willﬁndthisdone inmany books. Wearegoing totake ashort cutbyusing theknowledge wealready have about howamplitudes depend onangles inspace. The hydrogen atom inanyparticular state isaparticle with acertain “spin” j—the quantum number ofthetotal angular momentum. Partofthisspincomes from theelectron’s intrinsic spin, and part from theelectron’s motion. Since each ofthese twocomponents actsindependently (toanexcellent approximation) wewill again ignore thespin part and think only about the“orbital” angular momentum. This orbital motion behaves, however, justlikeaspin. Forexample, iftheorbital quantum number isI,thez-component ofangular momentum can beI,I—1,I—2,...,—l. (Weare,asusual, measuring inunits ofh.)Also. alltherotation matrices and other properties wehave worked outstill apply (From nowonwewillreally ignore thee1ectron’s spin; when wespeak of“angular momentum” wewillmean only theorbital part.) Since thepotential Vinwhich theelectron moves depends onlyonrandnot on0or¢,theHamiltonian issymmetric under allrotations. Itfollows that the angular momentum andallitscomponents areconserved. (This istrueformotion inany“central ﬁeld”—one which depends only onr—so isnotaspecial feature of theCoulomb e2/rpotential) 19-6 Now let'sthink ofsome possible state oftheelectron; itsinternal angular structure willbecharacterized bythequantum number I.Depending onthe “orientation” ofthetotal angular momentum with respect tothez-axis, the z-component ofangular momentum willbem,which isoneofthe2I+1possi- bilities between +Iand—I.Let’s saym=1.With what amplitude willtheelec- tronbefound onthez-axis atsome distance r?Zero. Anelectron onthez-axis cannot have anyorbital angular momentum around thataxis. Alright, suppose miszero, then there canbesome nonzero amplitude toﬁndtheelectron ateach distance from theproton. We’ll callthisamplitude F)(r). Itistheamplitude to ﬁndtheelectron atthedistance rupalong thez-axis, when theatom isinthe state II,0),bywhich wemean orbital spinIandz-component m=0. lfweknow F)(r)everything isknown. Foranystate II,m),weknow the amplitude i/q_,,,(r) toﬁndtheelectron anywhere intheatom. How? Watch. Suppose wehave theatom inthestate II,m),what istheamplitude toﬁndtheelectron at theangle 0,4>andthedistance rfrom theorigin? Putanewz-axis, sayz’,atthat angle (seeFig.19-3), andaskwhat istheamplitude thattheelectron willbeat thedistance ralong thenewaxisz’?Weknow thatitcannot befound along z’ unless itsz’-component ofangular momentum, saym’,iszero. When m’iszero, however, theamplitude toﬁndtheelectron along z’isF)(r). Therefore, theresult istheproduct oftwofactors. Theﬁrstistheamplitude thatanatom inthestate II,m)along thez-axis willbeinthestate II,m’=0)withrespect tothez’-axis. Multiply thatamplitude byF1(r) andyouhave theamplitude 1//;_m(r) toﬁndthe electron at(r,0,¢)with respect totheoriginal axes. Let’s write itout. Wehave worked outearlier thetransformation matrices forrotations. Togofrom theframe x,y,ztotheframe x’,y’,z’ofFig.19-3, wecanrotate ﬁrstaround thez-axis bytheangle ¢,andthenrotate about thenew y-axis (y’)bytheangle 0.This combined rotation istheproduct R1(@)R.(¢)- Theamplitude toﬁndthestate I,m’=0after therotation is (1.0lR11(¢)R,(¢) l1,m)- (19-31) Ourresult, then, is ¢),,,,(r) =(I,OIRy(0)Rz(¢) II,m)F)(r). (19.32) Theorbital motion canhave onlyintegral values ofI.(lftheelectron canbe found anywhere atr;£0,there issome amplitude tohave m=0inthatdirection. And m=0states exist only forintegral spins.) Therotation matrices forI=l aregiven inTable 17-2. Forlarger Iyou canusethegeneral formulas weworked outinChapter 18.Thematrices forR,(¢) aridR,,(0) appear separately, butyou know howtocombine them. Forthegeneral caseyouwould start with thestate II,m)andoperate withR,(¢) togetthenewstate R,(¢) II,m).Then youoperate onthisstate withR,,(0) togetthestate R,,(0)R,(¢) II,m)(which isjuste“"°‘II,m)). Multiplying by(I,0Igives thematrix element (19.31). Thematrix elements oftherotation operation arealgebraic functions of0 and4>.Theparticular functions which appear in(19.31) alsoshow upinmany kinds ofproblems which involve waves inspherical geometries andsohasbeen given aspecial name. Noteveryone usesthesame convention; butoneofthemost common ones is (1,0lR1/(1-9)R=() lLm)EaYi,m(@, ¢)- (19-33) Thefunctions Y;,,,,(0, ¢)arecalled thespherical harmonics, andaisjustanumerical factor which depends onthedeﬁnition chosen forY)_,,,. Fortheusual deﬁnition a=\/“T2I+1 With thisnotation, thehydrogen wave functions canbewritten 1/’Z,m(") =Yl,m(0> 4>)Fi(r)- (19-35) 19—79 (19.34)1z 2, (r,0,1») Illll I/i/9|/QI/I// |/ -§__-______________________\\\I Y 4 /// y,1 X II Fig. 19-3. The point (r,0,11>)ison thez’-axis ofthex'y'z' coordinate frame. Hz 112 G3Q 4Ne2O X Xli,m> _ /' 9 ¢\j y O16 Fig, 19-4. The decay ofanexcited state ofNew.The angle functions Y;,,,,(0, ¢)areimportant notonly inmany quantum- mechanical problems, butalso inmany areas ofclassical physics inwhich theV2 operator appears, such aselectromagnetism. Asanother example oftheir usein quantum mechanics, consider thedisintegration ofanexcited state ofNew (such aswediscussed inthelastchapter) which decays byemitting ana-particle andgoing intoO1“: Ne2U* __,0111+ H64‘ Suppose thattheexcited state hassome spinI(necessarily aninteger) andthatthe z-component ofangular momentum ism. Wemight now askthefollowing' given Iandm,what istheamplitude thatwewillﬁndthea-particle going oﬁ"ina direction which makes theangle 6with respect tothez-axis andtheangle <1)with respect tothexz-plane—as shown inFig.19-4. Tosolve thisproblem wemake, ﬁrst, thefollowing observation. Adecay in which thea-particle goesstraight upalong zmust come from astate with m=0. Thisissobecause both O1"andthea-particle have spinzero, andbecause their motion cannot have anyangular momentum about thez-axis Let’s callthis amplitude a(perunitsolid angle). Then, toﬁndtheamplitude foradecay atthe arbitrary angle ofFig.19-4, allweneed toknow iswhat amplitude thegiven initial state haszero angular momentum about thedecay direction. The amplitude for thedecay at0and¢isthen atimes theamplitude thatastate II,m)withrespect tothez-axis willbeinthestate II,O)withrespect toz’—the decay direction. This latter amplitude isjustwhat wehave written in(19.31). Theprobability toseethe a-particle at6,¢is P(9,11>)=112|(I,0IRy(6)Rz(¢) l1,m)l2- Asanexample, consider aninitial state with I=1andvarious values ofm. From Table 17-2 weknow thenecessary amplitudes. They are <1.01R1<11>R.(¢)1 1.+1>=~sin116"”. (1,0|R,(e)R,(¢)I 1,0)=cos(1, (19.36) (1,0IR,,(6)R,(¢)I1,~1)= --bsinea-"”. These arethethree possible angular distribution amplitudes—depending onthe m-value oftheinitial nucleus. Amplitudes such astheones in(19.36) appear sooften andaresufficiently important thattheyaregiven several names. Iftheangular distribution amplitude isproportional toanyoneofthethree functions oranylinear combination ofthem, wesay,“The system hasanorbital angular momentum ofone.” Orwemaysay, “The Ne2°* emits ap-wave a-particle.” Orwesay, “The 01-p3ftlCl€ isemitted in anI=1state.” Because there aresomany ways ofsaying thesame thing itis useful tohave adictionary. Ifyouaregoing tounderstand what other physicists aretalking about, youwilljusthave tomemorize thelanguage. InTable 19-1 wegive adictionary oforbital angular momentum. lftheorbital angular momentum iszero, then there isnochange when you rotate thecoordinate system andthere isnovariation withangle-—the “dependence” onangle isasaconstant, say1.This isalsocalled an“s-state”, andthere isonly onesuch state—as farasangular dependence isconcerned. Iftheorbital angular momentum is1,thentheamplitude oftheangular variation maybeanyoneofthe three functions given—depending onthevalue ofm——or itmay bealinear combina- tion. These arecalled “p-states,” andthere arethree ofthem. lfthe orbital angular momentum is2then there aretheﬁvefunctions shown. Any linear combination iscalled an“I=2,”ora“d-wave” amplitude. Now youcanimmediately guess what thenext letter is—what should come after s,p,d?Well, ofcourse,f, g,h, andsoondown thealphabet! Theletters don’t mean anything. (They didonce mean something—they meant “sharp” lines, “principal” lines, “diffuse” lines and l9—8 Table 19-1 Dictionary oforbital angular momentum (I=j=aninteger) Orbital angular Z‘ Angular dependence Number of Orbital momentum, compinem’ ofamplitudes Name states parity I1 O O I1 s 1 + I 11 1 +1 ——~A;sin6e“”] \/21 .O T p 3 — 1-1 ~—:sin0e—‘¢ \/2 1cos0 1 t I, , +2 gsin‘0e2'¢ '4 I_ 1 +1 2225111660898” 2 J0gmosze-1) Kd 5 + -1 —-E/2—6Sll10C0S0€_“b -2 sin219e"2'¢ 3 l 1,9IRi<11)R.<¢>| I.m /' 4 =Yl,m(6a¢) 1: 21+ 1 (_j)1 5 l =PI"(cos 0)e”"¢ , Ii I\ Y___/ “fundamental” lines oftheoptical spectra ofatoms. Butthose were inthedays when people didnotknow where thelines came from. After fthere were no special names, sowenowjustcontinue withg,h,andsoon.) The angular functions inthetable gobyseveral names—and aresometimes deﬁned withslightly diﬁerent conventions about thenumerical factors thatappear outinfront. Sometimes they arecalled “spherical harmonics,” andwritten as Y)_,,,(0, qb).Sometimes they arewritten P,'"(cos 0)e"”'1’, andifm=0,simply as P,(cos 6).The functions P)(cos 0)arecalled the“Legendre polynomials“ in cos0,andthefunctions P,"‘(cos 6)arecalled the“associated Legendre functions.“ You willﬁnd tables ofthese functions inmany books. Notice, incidentally, thatallthefunctions foragiven Ihave theproperty that thattheyhave thesame parity—for oddItheychange signunder aninversion and foreven Itheydon’t change. Sowecanwrite thattheparity ofastate oforbital angular momentum Iis(—1)'. Aswehave seen, these angular distributions mayrefer toanuclear disintegra- tion orsome other process, ortothedistribution oftheamplitude toﬁnd anelec- tronatsome place inthehydrogen atom. Forinstance, ifanelectron isinap-state (I=1)theamplitude toﬁnditcandepend ontheangle inmany possible ways— butallarelinear combinations ofthethree functions forI=1inTable 19-1. Let’s take thecase cos0.That’s interesting. That means that theamplitude is positive, say,intheupper part(0<1r/2), isnegative inthelower part(0>1r/2), andiszerowhen 6is90°. Squaring thisamplitude weseethattheprobability of ﬁnding theelectron varies with 0asshown inFig.19-5-and isindependent of¢ This angular distribution isresponsible forthefactthatinmolecular binding the attraction ofanelectron inanl=1state foranother atom depends ondirection- itistheorigin ofthedirected valences ofchemical attraction. 19-9PROBABILITY Fig. 19-5. Apolar graph ofcosz 9, which istherelative probability ofﬁnding anelectron atvarious angles from the z-axis (foragiven r)inanatomic state with1= landm =O. 19-4 Thegeneral solution forhydrogen InEq.(19.35) wehave written thewave functions forthehydrogen atom as lf’l,m(") =Yi.m(9, <l>)Fi(')- (19-37) These wave functions must besolutions ofthedifferential equation (19.7). Let’s seewhat thatmeans. Put(19.37) into(19.7); youget Ya’ E,a aY,,, F,62Y l,m_ _ - 1. 1.». r672OF’)+r2sin0as(ma aa)+r2sin2 0a¢2 2 =-3hi,T’-(E+5;)Y;_,,,F). (19.38) Now multiply through byr2/F) andrearrange terms. Theresult is 1 3 . 6Yjnn) 1 822’Lm sin05?(Sm0as+S11120a¢2 _ r21d2 2m e2_-IE1; w(rm)+-,7(E+ rt... (19.39) Theleft-hand sideofthisequation depends on19and¢,butnotonr.Nomatter what value wechoose forr,theleftsidedoesn’t change. Thismust alsobetrue fortheright-hand side. Although thequantity inthesquare brackets hasr’sall over theplace, thewhole quantity cannot depend onr,otherwise wewouldn’t have anequation good forallr.Asyoucansee,thebracket alsodoes notdepend on19or11>.Itmust besome constant. Itsvalue maywelldepend ontheI-value of thestate wearestudying, since thefunction F1must betheoneappropriate tothat state; we’ll calltheconstant K).Equation (19.35) istherefore equivalent totwo equations: 1a.aY 1a2 siiiéea(Sm9aim) +sin?0.902=_K‘Y""" (1940) 2 z %%(rF;)+3+(E+F)=K, (19.41) Now look atwhat we’ve done. Foranystate described byIandm,weknow thefunctions Y;_,,,; wecanuseEq.(19.40) todetermine theconstant K1.Putting KiintoEq.(19.41) wehave adifferential equation forthefunction F¢(r). Ifwe cansolve thatequation forF;(r), wehave allofthepieces toputinto(19.37) to give1/»(r). What isKI?First, notice thatitmust bethesame forallm(which gowith a particular I),sowecanpickanymwewant forYl,,,,andplug itinto(19.40) to solve forK).Perhaps theeasiest onetouseisI/1,1. From Eq.(18.24), R,(¢) II,I)=e"‘1’II,I). (19.42) Thematrix element forR,,(0) isalsoquite simple: (I,OIR,,(6) II,I)=b(sin0)’, (19.43) where bissome number.'I' Combining thetwo,weobtain Y)’;cce"‘1*sin’0. (19.44) I‘Youcanwith some work show thatthiscomes outofEq.(18.35), butitisalsoeasy towork outfrom ﬁrstprinciples following theideas ofSection 18-4. Astate I1,1)can bemade outof2!spinone-half particles allwith spins up;while thestate II,0)would have IupandIdown. Under therotation theamplitude thatanup-spin remains up iscos0/2,andthatanup-spin goes down issin6/2. Weareasking fortheamplitude thatIup-spins stayup,while theother Iup-spins godown. Theamplitude forthatis (cos0/2sin0/2)‘ which 1Sthesame assin‘0. 19-10 Putting thisfunction into(19.40) gives K,=l(l+1). (19.45) Now thatwehave determined K),Eq.(19.41) tellsusabout theradial function F1(r). Itis,ofcourse, justtheSchrodinger equation withtheangular partreplaced byitsequivalent KZF)/r2. Let's rewrite (1941)intheform wehadinEq(19.8), asfollows: 1d2 2 2111112;W(rF))=-Z’?IE+§- F). (19.46) Amysterious term hasbeen added tothepotential energy. Although wegotthis term bysome mathematical shenanigan, ithasasimple physical origin. Wecan give youanidea about where itcomes from interms ofasemi-classical argument. Then perhaps youwillnotﬁnditquite somysterious. Think ofaclassical particle moving around some center offorce. The total energy isconserved andisthesumofthepotential andkinetic energies U=V(r)-1-%mv2 =constant. Ingeneral, 11canberesolved into aradial component 1',andatangential compo- nent r0;then 112= +(r6)2. Now theangular momentum mr20 isalsoconserved; sayitisequal toL.Wecan then write H1726 =L, or r19=-Q=mr andtheenergy is Lu u=%mv?+V(r)+517,-- Ifthere were noangular momentum wewould have justtheﬁrsttwoterms. Adding theangular momentum Ldoes totheenergy justwhat adding aterm L2/2m/'2 tothepotential energy would do.Butthisisalmost exactly theextra term in(1946) The only difference isthat I(l—I—l)happears fortheangular momentum instead of12112aswemight expect. Butwehave seen before (forex- ample, Volume II,Section 34-7)'I thatthisisjustthesubstitution thatisusually required tomake aquasi-classical argument agree with acorrect quantum- mechanical calculation. Wecan, then, understand thenew term asa“pseudo- potential" which gives the“centrifugal force” term thatappears intheequations ofradial motion forarotating system. (See thediscussion of“pseudo-forces” in Volume I,Section 12-5.) Wearenow ready tosolve Eq.(19.46) forF)(r). Itisvery much likeEq. (19.8), sothesame technique willwork again. Everything goes asbefore until yougettoEq.(19.19) which willhave theadditional term —l(1+1)fla1p‘_2- (19.47)!c=1 This term canalsobewritten as —I(I -1-l){ip1 —£3Gk-1-iPk‘1}' (19.48) k=l (We have taken outtheﬁrstterm andthen shifted therunning index kdown by1.)Instead ofEq.(19.20) wehave 2[{1<(1<+1)-1(1+1)}a,,+, ~2(al(-i)a,]p’~-1 k:1 _ =0 (I949) p . . I‘SeeAppendix tothisvolume. 19-11 Al ‘ r ' ‘s""'° 25,11-= IA) Lb) *2 Al 1 ‘I1: 1 WW434p,mn0 Le) (at) .. 211.1--n 1 AP. -1- _ ‘1.4; (4,'4 WW 41,-no H)1' Id-,iii»: (E) Fig. 19-6. Rough sketches showing thegeneral nature ofsome ofthehydro- gen wave functions. Theshaded regions show where theamplitudes are large. Theplus and minus signs show therelative sign oftheamplitude ineach region.There isonly oneterm inp_‘, soitmust bezero. Thecoeﬁicient a1must bezero (unless I=0aridwehave ourprevious solution). Each oftheother terms is made zerobyhaving thesquare bracket come outzeroforevery k.Thiscondition replaces Eq.(19.21) by ai,+1 = Gk. This istheonlysigniﬁcant change from thespherically symmetric case. Asbefore theseries must terminate ifwearetohave solutions which can represent bound electrons. Theseries willendatk=nifan=1.Wegetagain thesame condition ona,thatitmust beequal to1/n,where nissome integer. However, Eq.(19.50) alsogives anewrestriction. Theindex kcannot beequal to I,thedenominator becomes zeroanda1+1 isinﬁnite. That is,since a1=0,Eq. (19.50) implies thatallsuccessive aharezero until wegettoai+1, which canbe nonzero. This means thatkmust start atI—I—1andendatn. Ourﬁnalresult isthatforanyIthere aremany possible solutions which we cancallFm)where nZI+1.Each solution hastheenergy me4 1E11='_-2713 (19.51) Thewave function forthestate ofthisenergy with theangular quantum numbers Iandmis I/’1i,l,1n = Yl,m(0: ¢)Fn,I(p)i with PFn,i(P) =£7” 2 fl1¢Pk- (19-53) k=l-I-1 Thecoefficients a),areobtained from (19.50). Wehave, ﬁnally, acomplete de- scription ofthestates ofahydrogen atom. 19-5 Thehydrogen wave functions Let’s review what wehave discovered. Thestates which satisfy Schrodinger’s equation foranelectron inaCoulomb ﬁeld arecharacterized bythree quantum numbers n,I,m,allintegers. Theangular distribution oftheelectron amplitude canhave onlycertain forms which wecallY),,,.They arelabeled byI,thequantum number oftotal angular momentum, and m,the“magnetic” quantum number, which canrange from —lto+I.Foreach angular conﬁguration, various possible radial distributions F,,,)(r) oftheelectron amplitude arepossible; theyarelabeled bytheprinciple quantum number n—-which canrange from I+1tocc.Theenergy ofthestate depends only onn,andincreases withincreasing n. Thelowest energy, orground, state isans-state. IthasI=0,n=0,and m=0.Itisa“nondegenerate” state—there isonly onewith thisenergy, andits wave function isspherically symmetric. Theamplitude toﬁndtheelectron isa maximum atthecenter, andfalls offmonatonically with increasing distance from thecenter. Wecanvisualize theelectron amplitude asablob asshown inFig. 19-6(a). There areother s-states with higher energies, forn=2,3,4, ...Foreach energy there isonlyoneversion (m=0),andtheyareallspherically symmetric. These states have amplitudes which alternate insign oneormore times with increasing r.There aren—1spherical nodal surfaces—the places where ipgoes through zero. The2s-state (I=0,n=2),forexample, willlook assketched in Fig.19-6(b). (The dark areas indicate regions where theamplitude islarge, and theplusandminus signs indicate therelative phases oftheamplitude.) Theenergy levels ofthes-states areshown intheﬁrstcolumn ofFig.19-7. Then there arethep-states-with I=1.Foreach n,which must be2or greater, there arethree states ofthesame energy, oneeach form=+1,m=0, andm=——1.Theenergy levels areasshown inFig.19-7. Theangular de- pendences ofthese states aregiven inTable 19-1. Forinstance, form=0,ifthe 19-12 amplitude ispositive for0near zero, itwillbenegative for0near 180°. There is anodal plane coincident with thexy-plane. Forn>2there arealsospherical nodes. The n=2,m=Oamplitude issketched inFig. l9—6(c), andthen=3, m=0wave function issketched inFig.l9—6(d). You might think that since mrepresents akind of“orientation” inspace, there should besimilar distributions withthepeaks ofamplitude along thex-axis oralong they-axis. Arethese perhaps them=+1andm=——lstates? No. Butsince wehave three states withequal energies, anylinear combinations ofthe three willalsobestationary states ofthesame energy. ltturns outthatthe“x”- state—which corresponds tothe“z”-state, orm=0state, ofFig. l9—6(c)——-is alinear combination ofthem=+1andm=—lstates. Thecorresponding “y”-state isanother combination. Speciﬁcally, wemean that “Zn = I1,0), “X9! Z +_i1ﬂ l>, \/2 |l,+l) —[1,-1).V A/5 These states alllook thesame when referred totheir particular axes. Thed-states (I=2)have ﬁvepossible values ofniforeach energy, thelowest energy hasn=3.Thelevels goasshown inFig.19-7. Theangular dependences getmore complicated. Forinstance them=0states have twoconical nodes, so thewave function reverses phase from +,to—,to+asyougoaround from the north poletothesouth pole. Therough form oftheamplitude issketched in(e) and(f)ofFig. 19-6 forthem=0states with n=3andn=4.Again, the larger n’shave spherical nodes. Wewillnottrytodescribe anymore ofthepossible states. Youwillﬁndthe hydrogen wave functions described inmore detail inmany books. Two good references areL.Pauling andE.B.Wilson, Introduction toQuantum Mechanics, McGraw-Hill (1935); andR.B.Leighton, Principles ofModern Physics, McGraw- Hill(1959). You willﬁndinthem graphs ofsome ofthefunctions andpictorial representations ofmany states. Wewould liketomention oneparticular feature ofthewave functions for higher I:forl>0theamplitudes arezero atthecenter. That isnotsurprising, since it’shard foranelectron tohave angular momentum when itsradius armis verysmall. Forthisreason, thehigher thel,themore theamplitudes are“pushed away” from thecenter. Ifyoulook atthewaytheradial functions F(r)vary for small r,youﬁndfrom (19.53) that F,,';(r) zrl. Such adependence onrmeans thatforlarger l’syouhave togofarther from r=O before yougetanappreciable amplitude. Thisbehavior is,incidentally, determined bythecentrifugal force term intheradial equation, sothesame thing willapply foranypotential thatvaries slower than l/r2 forsmall r—~which most atomic potentials do. 19-6 The periodic table Wewould likenowtoapply thetheory ofthehydrogen atom inanapproxi- mate waytogetsome understanding ofthechemist’s periodic table oftheelements. Foranelement with atomic number Zthere areZelectrons heldtogether bythe electric attraction ofthenucleus butwith mutual repulsion oftheelectrons. To getanexact solution wewould have tosolve Schrodinger’s equation forZelectrons inaCoulomb ﬁeld. Forhelium theequation is -@"’i”= -712-<v2¢+v”ia+(-35-@+i2)¢. F1 F2 16t 2m 1 2 rm 19-13we Q________ __ 4s 3s 2s -I36evi ---------------- -- s p d 1And SO Oh _4L _3P_ ZP4d in-_’3L_- 9=l 2 3 4 Fig. 19-7. Theenergy level diagram forhydrogen.J:0-av(D63 3 2 where VfisaLaplacian which operates onr1,thecoordinate ofoneelectron; Vioperates onr2;andr12=Ii-1—r2|. (Weareagain neglecting thespinofthe electrons.) Toﬁndthestationary states andenergy levels wewould have toﬁnd solutions oftheform ¢=/‘('1 r2)e—(i/ﬁ)Ii't, . Thegeometrical dependence iscontained inf,which isafunction ofsixvariables —the simultaneous positions ofthetwoelectrons. Noonehasfound ananalytic solution, although solutions forthelowest energy states have been obtained by numerical methods. With 3,4,or5electrons itishopeless totrytoobtain exact solutions, andit1S going toofartosaythatquantum mechanics hasgiven aprecise understanding of theperiodic table. Itispossible, however, even with asloppy approximation~and some ﬁxing—to understand, atleast qualitatively, many chemical properties which show upintheperiodic table. Thechemical properties ofatoms aredetermined primarily bytheir lowest energy states. Wecanusethefollowing approximate theory toﬁndthese states andtheir energies. First, weneglect theelectron spin, except thatweadopt the exclusion principle andsaythatanyparticular electronic state canbeoccupied byonly oneelectron. This means thatanyparticular orbital conﬁguration can have uptotwoelectrons—one with spinup,theother with spindown. Next we disregard thedetails oftheinteractions between theelectrons inourﬁrstapproxi- mation, andsaythateach electron moves inacentral ﬁeld which isthecombined ﬁeldofthenucleus andalltheother electrons. Forneon, which has10electrons, wesaythatoneelectron seesanaverage potential duetothenucleus plustheother nineelectrons. Weimagine thenthatintheSchrodinger equation foreachelectron weputaV(r)which isal/rﬁeld modiﬁed byaspherically symmetric charge density coming from theother electrons. Inthismodel each electron actslikeanindependent particle. Theangular dependence ofitswave function willbejustthesame astheones wehadforthe hydrogen atom. There willbes-states, p-states, andsoon;andtheywillhave the various possible m-values. Since V(r)nolonger goesasl/r,theradial partofthe wave functions willbesomewhat different, butitwillbequalitatively thesame, so wewillhave thesame radial quantum numbers, n.Theenergies ofthestates will also besomewhat different. H With these ideas, let’sseewhat weget. Theground state ofhydrogen has I=m=0andn=l;wesaytheelectron conﬁguration isls.Theenergy is -13.6 ev.This means thatittakes 13.6electron volts topulltheelectron offthe atom. Wecallthisthe“ionization energy”, W1.Alarge ionization energy means thatitisharder topulltheelectron oﬁand,ingeneral, thatthematerial ischem- ically lessactive. He Now takehelium. Both electrons canbeinthesame lowest state (one spin upandtheother spin down). Inthislowest state theelectron moves inapotential which isforsmall rlikeaCoulomb ﬁeldforz=2andforlarge rlikeaCoulomb ﬁeld forz=1.Theresult isa“hydrogen-like” lsstate with asomewhat lower energy. Both electrons occupy identical lsstates (I=0,m=0).Theobserved ionization energy (toremove oneelectron) is24.6 electron volts. Since thels “shell” isnowﬁlled—we allow onlytwoelectrons—there ispractically notendency foranelectron tobeattracted from another atom. Helium ischemically inert. Li Thelithium nucleus hasacharge of3.Theelectron states willagain behy- drogen-like, and thethree electrons will occupy thelowest three energy levels. Two willgointolsstates andthethird willgointoann=2state. ButwithI=0 orl==1?Inhydrogen these states have thesame energy, butinother atoms they l9—l4 don’t, forthefollowing reason. Remember thata2sstate hassome amplitude to benearthenucleus while the2pstate does not. That means thata2selectron will feelsome ofthetriple electric charge oftheLinucleus, butthata2pelectron will stayoutwhere theﬁeldlooks liketheCoulomb ﬁeldofasingle charge. Theextra attraction lowers theenergy ofthe2sstate relative tothe2pstate. Theenergy levels willberoughly asshown inFig.19—8—which youshould compare with the corresponding diagram forhydrogen inFig.19-7. Sothelithium atom willhave twoelectrons inlsstates andoneina2s.Since the2selectron hasahigher energy than alselectron itisrelatively easily removed. Theionization energy oflithium isonly5.4electron volts, anditisquite active chemically. Soyoucanseethepatterns which develop; wehave given inTable 19-2 a listoftheﬁrst36elements, showing thestates occupied bytheelectrons inthe ground state ofeach atom. TheTable gives theionization energy forthemost loosely bound electron, andthenumber ofelectrons occupying each “shell”—— bywhich wemean states withthesame n.Since thedifferent I-states have different Theelectron conﬁgurations oftheﬁrst36elementsTable 19-2 Z Element W1(ev)Electron Conﬁguration ls 2s2p 3s3p3d 4s4p4d4f 1H 2Hehydrogen helium13.6 24.61 2 I-lO\OO0~lO\U|J>u-\"l'lOZGLilithium Beberyllium Bboron carbon nitrogen oxygen ﬂuorine Neneon5.4 9.3 8.3 11.3 14.5 13.6 17.4 21.6FILLED (2 IQNNNDNNNF‘ O\L!|-bbJl\It—) 11Nasodium 12 Mgmagnesium 13Alaluminum 14Sisilicon 15Pphosphorus 16 Ssulfur 17Clchlorine 18A argon5.1 7.6 6.0 8.1 10.5 10.4 13.0 15.8 19K 20Ca 21 Sc 22Ti 23 Vpotassium calcium scandium titanium vanadium 24 Crchromium 25 Mnmanganese 26 Feiron 27 Cocobalt 28 Ninickel 29 Cucopper 30 Znzinc4.3 6.1 6.5 6.8 6.7 6.8 7.4 7.9 7.9 7.6 7.7 9.4(2) (2) (3) (3)/\Xv l\IP\JlQl\)PQI\!|\)"* O\tJ|&U)l\)>-1FILLED—— I-1:-1_——FILLED—— 31 Gagallium 32 Gegermanium 33 Asarsenic 34 Seselenium 35 Brbromine 36 Krkrypton6.0 7.9 9.8 9.7 11.8 14.0iFlLLEDi (2) (8) (13) l~)l\)l\)l\)l\)lQPOP-"I\Jl\)I\)|\)\-*I\)|\)|~JI~)t—lNumber ofelectrons ineachstate O\LII-Pbdlqii-A 19-15we o---------------------------- -5 l-- 3:5 _ ___- "-2-§C‘_'_:3 j ’ ”-.74: 4s______ ''‘F;,4-— /3P,~~'\ 0|ab\\\\\\\_______-= 3s,» /.2 2', "’;’i' ___,.¢; Z! -~’._.; I8 . 8 p d f Fig. 19-8. Schematic energy level diagram foranatomic electron withother electrons present. (The scale isnotthe same asFig.19-7.) energies, each I-value corresponds toasub-shell of2(2l+1)possible states (of different mandelectron spin). These allhave thesame energy—except forsome verysmall effects weareneglecting. Be Beryllium islikelithium excepethat ithastwoelectrons inthe2sstate as wellastwointheﬁlled lsshell. BtoNe Boron has5electrons. Theﬁfthmust gointoa2pstate. There are2X3=6 different 2pstates, sowecankeep adding electrons until wegettoatotal of8. This takes ustoneon. Asweaddthese electrons wearealsoincreasing Z,sothe whole electron distribution getspulled incloser andcloser tothenucleus andthe energy ofthe2pstates goesdown. Bythe_time wegettoneon theionization energy isupto21.6volts. Neon does noteasily giveupanelectron. Also there areno more low-energy slots tobeﬁlled, soitwon’t trytograb anextra electron. Neon ischemically inert. Fluorine, ontheother hand, doeshave anempty position where anelectron candrop intoastate oflowenergy, soitisquite active inchemical reactions. Nat0A With sodium theeleventh electron must start anewshell—going intoa3s state. Theenergy level ofthisstate ismuch higher; theionization energy jumps down; andsodium isanactive chemical. From sodium toargon thesandpstates with n=3areoccupied inexactly thesame sequence asforlithium toneon. Angular conﬁgurations oftheelectrons intheouter unﬁlled shell have thesame sequence, andtheprogression ofionization energies isquite similar. You cansee whythechemical properties repeat with increasing atomic number. Magnesium actschemically much likeberyllium, silicon likecarbon, andchlorine likeﬂuorine. Argon isinert likeneon. You may have noticed thatthere isaslight peculiarity inthesequence of ionization energies between lithium andneon, andasimilar onebetween sodium and argon. The lastelectron isbound totheoxygen atom somewhat lessthan wemight expect. And sulphur issimilar. Why should thatbe? Wecanunder- stand itifweputinjust alittle bitoftheeffects oftheinteractions between in- dividual electrons. Think ofwhat happens when weputtheﬁrst2pelectron onto theboron atom. Ithassixpossibilities—-three possible p-states, each with two spins. Imagine thattheelectron goes with spinupintothem=0state, which wehave alsocalled the“z”state because ithugs thez-axis. Now what willhappen incarbon? Therei arenow two2pelectrons. Ifoneofthem goes intothe“z” state, where willthesecond onego?Itwillhave lower energy ifitstays away from theﬁrstelectron, which itcandobygoing into, say,the“x”state ofthe2pshell. (This state is,remember, justalinear combination ofthem=+1andm=—l states.) Next, when wegotonitrogen, thethree 2pelectrons willhave thelowest energy ofmutual repulsion ifthey gooneeach into the“x,” “y,” and“z”con- ﬁgurations. Foroxygen, however, thejigisup.Thefourth electron must gointo oneoftheﬁlled states——with opposite spin. Itisstrongly repelled bytheelectron already inthatstate, soitsenergy willnotbeaslowasitmight otherwise be,and itismore easily removed. That explains thebreak inthesequence ofbinding energies which appears between nitrogen andoxygen, andbetween phosphorus andsilicon. Kto Zn After argon, youwould, atﬁrst, think thatthenewelectrons would start to ﬁllupthe3dstates Buttheydon’t. Aswedescribed earlier—and illustrated in Fig.19—7——the higher angular momentum states getpushed upinenergy. Bythe time wegettothe3dstates they arepushed toanenergy alittle bitabove theenergy ofthe4sstate. Soinpotassium thelastelectron goesintothe4sstate. After this 19-16 shell isﬁlled (with twoelectrons) atcalcium, the3dstates begin tobeﬁlled for scandium, titanium, andvanadium. The energies ofthe3pand 4sstates aresoclose together that small effects canshiftthebalance either way. Bythetime wegettoputfourelectrons intothe 3dstates, their repulsion raises theenergy ofthe4sstate justenough thatitsenergy isslightly above the3denergy, sooneelectron shifts over. Forchromium wedon’t geta4,2combination aswewould have expected, butinstead a5,1combination. Thenewelectron added togetmanganese ﬁllsupthe4sshell again, andthestates ofthe3dshell arethen occupied onebyoneuntil wereach copper. Since theoutermost shell ofmanganese, iron, cobalt, andnickel have thesame conﬁgurations, however, they alltend tohave similar chemical properties. (This effect ismuch more pronounced intherare-earth elements which allha_ve thesame outer shell butaprogressively ﬁlling inner shell which hasmuch lessinﬂuence on their chemical properties.) Incopper anelectron isrobbed from the4sshell, ﬁnally completing the3d shell. Theenergy ofthe10,lcombination is,however, soclose tothe9,2con- ﬁguration forcopper thatjust thepresence ofanother atom nearby canshift the balance. Forthisreason thetwolastelectrons ofcopper arenearly equivalent, andcopper canhave avalence ofeither lor2(ltsometimes acts asthough its electrons were inthe9,2combination.) Similar things happen atother places and account forthefactthatother metals, such asiron, combine chemically witheither oftwovalences. Byzinc, both the3dand4sshells areﬁlled once andforall. GatoKr From gallium tokrypton thesequence proceeds normally again, ﬁlling the 4pshell. Theouter shells, theenergies, andthechemical properties repeat the pattern ofboron toneon andaluminum toargon. Krypton, likeargon andneon, isknown as“noble” gas. Allthree arecheni- ically “inert.” This means only that, having ﬁlled shells ofrelatively lowenergy, there arefewsituations inwhich there isanenergy advantage forthem to_]Ol[1ina simple combination with other elements. Having aﬁlled shell isnotenough. Beryllium andmagnesium h_gve ﬁlled s-shells, buttheenergy ofthese shells istoo high tolead tostability. Similarly, onewould have expected another “noble” element atnickel, iftheenergy ofthe3dshell hadbeen lower (orthe4s,higher). Ontheother hand, krypton isnotcompletely inert; itwillform aweakly-bound compound withchlorine. Since oursample hasturned upmost ofthemain features oftheperiodic table, westop ourexamination atelement number 36—there arestillseventy or somore! Wewould liketobring uponlyonemore point—that wenotonlycanunder- stand thevalences tosome extent butalsocansaysomething about thedirectional properties ofthechemical bonds. Take anatom likeoxygen which hasfour 2p electrons. The ﬁrst three gointo “x,” “y,” and “z”states and thefourth will double oneofthese states, leaving two——say “x”and“y”—vacant. Consider then what happens inH20. Each ofthetwohydrogens arewilling toshare anelectron with theoxygen, helping theoxygen toﬁllashell. These electrons willtendtogo intothe“x”and“y”vacancies. Sothewater molecule should have thetwohy- drogen atoms making aright angle withrespect tothecenter oftheoxygen. The angle isactually 105°. Wecaneven understand why theangle islarger than 90°. Insharing their electrons thehydrogens endupwith anetpositive charge. The electric repulsion “strains” thewave functions andpushes theangle outto105°. Thesame situation occurs inHZS. Butbecause thesulphur atom islarger, the two hydrogen atoms arefarther apart, there islessrepulsion, and theangle is only pushed outtoabout 93°. Selenium iseven larger, soinH2Se theangle is very nearly 90°. Wecanusethesame arguments tounderstand thegeometry ofammonia. H3N. Nitrogen hasroom forthree more 2pelectrons, oneach forthe“x,” “y,” and"z"typestates. Thethree hydrogens should joinonatright angles toeach other. Theangles come outalittle larger than90°—-again from theelectric repul- l9—l7 sion—but atleastweseewhythemolecule ofH3N isnotﬂat. Theangles in phosphene, H3P,areclose to90°,andinH3As arestillcloser. Weassumed that NH3 wasnotﬁatwhen wedescribed itasatwo-state system. Andthenonﬁatness iswhat makes theammonia maser possible. Now weseethatalsothatshape can beunderstood from ourquantum mechanics. TheSchrodinger equation hasbeen oneofthegreat triumphs ofp/hysics. By providing thekeytotheunderlying machinery ofatomic structure ithasgiven anexplanation foratomic spectra, forchemistry, andforthenature ofmatter. 19-18 20 llperators 20-1 Operations andoperators Allthethings wehave done sofarinquantum mechanics could behandled with ordinary algebra, although wedidfrom time totime show yousome special ways ofwriting quantum-mechanical quantities andequations. Wewould like now totalksome more about some interesting anduseful mathematical ways of describing quantum-mechanical things. There aremany ways ofapproaching the subject ofquantum mechanics, andmost books useadifferent approach from the onewehave taken. Asyougoontoread other books youmight notseeright away theconnections ofwhat youwillﬁndinthem towhat wehave been doing. Although wewillalsobeabletogetafewuseful results, themain purpose ofthis chapter istotellyouabout some ofthedifferent ways ofwriting thesame physics. Knowing them youshould beable tounderstand better what other people are saying. When people were ﬁrstworking outclassical mechanics theyalways wrote alltheequations interms ofx-,y-,andz-components. Then someone came along andpointed outthatallofthewriting could bemade much simpler byinventing thevector notation. It’struethatwhen youcome down toﬁguring something outyouoften have toconvert thevectors back totheir components. Butit’s generally much easier toseewhat's going onwhen youwork withvectors andalso easier todomany ofthecalculations. Inquantum mechanics wewere able to write many things inasimpler waybyusing theideaofthe“state vector.” The state vector Ill!)has, ofcourse, nothing todowith geometric vectors inthree dimensions butisanabstract symbol thatstands foraphysical state, identiﬁed bythe“label,” or“name,” 1//.Theidea isuseful because thelaws ofquantum mechanics canbewritten asa_lgebraic equations interms ofthese symbols. For instance, ourfundamental lawthatanystate canbemade upfrom alinear com- bination ofbase states iswritten as |o=Zcm, mo where theC,areasetofordinary (complex) numbers—the amplitudes C,=(iIil) ——while I1),I2),I3),andsoon,stand forthebase states insome base, orrepre- sentation. lfyoutake some physical state anddosomething toit—like rotating it,or likewaiting forthetime At—you getadifferent state. Wesay, “performing anoperation onastate produces anewstate." Wecanexpress thesame ideaby anequation: l¢)=/flit) (20-2) Anoperation onastate produces another state. Theoperator /fstands forsome particular operation. When thisoperation isperformed onanystate, sayI11),it produces some other state Iqs). What does Eq.(20.2) mean? Wedeﬁne itthisway. Ifyoumultiply the equation by(iIandexpand I¢)according toEq.(20.1), youget mo=Zm1mmo- mm (The states Ij)arefrom thesame setasIi).)Thisisnowjustanalgebraic equation. Thenumbers (iI¢)givetheamount ofeach base state youwillﬁndinI¢),and itisgiven interms ofalinear superposition oftheamplitudes (jI1//)thatyouﬁnd 20-120-1 Operations andoperators 20-2 Average energies 20-3 Theaverage energy ofan atom 20-4 Theposition operator 20-S The momentum operator 20-6 Angular momentum 20-7 Thechange ofaverages withtime II//>ineach base state. Thenumbers (iI/fIj)arejustthecoefficients which tell howmuch of(jIIll)goesintoeachsum. Theoperator /fisdescribed numerically bythesetofnumbers, or“matrix,” /1,,E<1I.4|1). (20.4) SoEq.(20.2) isahigh-class wayofwriting Eq.(20.3). Actually itisalittle more than that; something more isimplied. InEq.(20.2) wedonotmake any reference toasetofbase states. Equation (20.3) isanimage ofEq.(20.2) in terms ofsome setofbasestates. But,asyouknow, youmayuseanysetyouwish. Andthisideaisimplied inEq.(20.2). Theoperator wayofwriting avoids making anyparticular choice. Ofcourse, when youwant togetdeﬁnite youhave tochoose some set. When youmake your choice, youuseEq.(20.3). Sotheoperator equation (20.2) isamore abstract wayofwriting thealgebraic equation (20.3). It’ssimilar tothedifference between writing c=aXb instead of c,=a,,b,—a,b,,, cg=a,b,—a,,bU, c,=a,b,, —a,,b,. Theﬁrstwayismuch handier. When youwant results, however, youwilleventually have togivethecomponents with respect tosome setofaxes. Similarly, ifyou want tobeabletosaywhat youreally mean by/f,youwillhave tobeready to givethematrix AUinterms ofsome setofbase states. Solong asyouhave in mind some setA,~,,Eq.(20.2) means justthesame asEq.(20.3). (You should remember alsothatonce youknow amatrix foroneparticular setofbase states youcanalways calculate thecorresponding matrix thatgoeswith anyother base. Youcantransform thematrix from one“representation” toanother.) Theoperator equation in(20.2) alsoallows anewwayofthinking. Ifwe imagine some operator /f,wecanuseitwith anystate Iip)tocreate anewstate /fI¢).Sometimes a“state” wegetthiswaymay bevery peculiar—it may not represent anyphysical situation wearelikely toencounter innature. (Forinstance, wemay getastate thatisnotnormalized torepresent oneelectron.) Inother words, wemay attimes getI‘states" that aremathematically artiﬁcial. Such artiﬁcial “states” maystillbeuseful, perhaps asthemid-point ofsome calculation. Wehave already shown youmany examples ofquantum-mechanical op- erators. Wehave hadtherotation operator R,,(0) which takes astate Ip)and produces anewstate, which istheoldstate asseeninarotated coordinate system. Wehave hadtheparity (orinversion) operator P,which makes anewstate by reversing allcoordinates. Wehave hadtheoperators 6,,6,,and6,forspinone- halfparticles. Theoperator J,wasdeﬁned inChapter 17interms oftherotation operator forasmall angle e. R.(@)=1+I5L. (20.5) Thisjustmeans, ofcourse, that Rxolth)=Ii>+I6J.l¢)- (20.6) Inthisexample, J,II//>ish/ietimes thestate yougetifyou rotate IIL)bythesmall angle eandthen subtract theoriginal state. Itrepresents a“state” which isthe difference oftwostates. Onemore example. Wehadanoperator p,—called themomentum operator (x-component) deﬁned inanequation like(20.6). IfD,,(L) istheoperator which 20—2 displaces astate along xbythedistance L,then13,,isdeﬁned by 15.0)=1+Itr... (20.1) where 6isasmall displacement. Displacing thestate Iih)along xbyasmall dis- tance 6gives anew state Ith’). Wearesaying that thisnew state istheoldstate plus asmall new piece 1 A 715.01 l Theoperators wearetalking about work onastate vector likeI1/»),which is anabstract description ofaphysical situation. They arequite different from algebraic operators which work onmathematical functions. Forinstance, d/dx isan“operator” that works onf(x) bychanging ittoanew function f’(x) = df/dx. Another example isthealgebraic operator V2. You canseewhy thesame word isused inboth cases, butyoushould keep inmind thatthetwokinds of operators aredifferent. Aquantum-mechanical operator /fdoes notwork onan algebraic function, butonastate vector likeIip). Both kinds ofoperators are used inquantum mechanics andoften insimilar kinds ofequations, asyou will seealittle later. When youareﬁrstlearning thesubject itiswelltokeep the distinction always inmind. Later on,when youaremore familiar with thesubject, youwillﬁndthatitislessimportant tokeep anysharp distinction between the twokinds ofoperators. Youwill,indeed, ﬁndthatmost books generally usethe same notation forboth! We’ll goonnow andlook atsome useful things youcandowith operators. Butﬁrst, onespecial remark. Suppose wehave anoperator /fwhose matrix in some base isA,,E(iI/fIj).Theamplitude thatthestate /fIi//)1Salsoinsome other state I¢>)is(¢>I/fII//>.lsthere some meaning tothecomplex conjugate of thisamplitude? Youshould beabletoshow that (¢>I/fIi>*=0I/T‘Ii>>. (20.8) where /fl(read “Adagger”) isanoperator whose matrix elements are Al,=(A,,)*. (20.9) Togetthei,jelement ofAlyougotothej,ielement of/f(theindexes arereversed) andtakeitscomplex conjugate. Theamplitude thatthestate /flIqs)isinI(1))is thecomplex conjugate oftheamplitude that/fIi//)isinI¢).Theoperator /flis called the“Hermitian adjoint” of/f.Many important operators ofquantum mechanics have thespecial property thatwhen youtake theHermitian adjoint, yougetthesame operator back. IfBissuch anoperator, then til=E, anditiscalled a“self-adjoint” or“Hermitian,” operator. 20-2 Average energies Sofarwehave reminded youmainly ofwhat you already know. Now we would liketodiscuss anewquestion. How would youﬁndtheaverage energy of asystem—say, anatom‘? Ifanatom isinaparticular state ofdeﬁnite energy and youmeasure theenergy, youwillﬁnd acertain energy E.lfyou keep repeating themeasurement oneach oneofawhole series ofatoms which areallselected to beinthesame state, allthemeasurements willgiveE,andthe“average” ofyour measurements will,ofcourse, bejustE. Now, however, what happens ifyoumake themeasurement onsome state It//>which isnotastationary state? Since thesystem does nothave adeﬁnite energy, onemeasurement would giveoneenergy, thesame measurement onanother atom inthesame state would giveadifferent energy, andsoon.What would you getfortheaverage ofawhole series ofenergy measurements? 20-3 Wecananswer thequestion byprojecting thestate I¢)onto thesetofstates ofdeﬁnite energy. Toremind youthatthisisaspecial baseset,we’ll callthestates I17,). Each ofthestates I1),)hasadeﬁnite energy E,.Inthisrepresentation, IM=ZQm) mm When youmake anenergy measurement andgetsome number E,,youhave found thatthesystem wasinthestate 11,.Butyoumaygetadifferent number foreach measurement Sometimes youwillgetE1,sometimes E2,sometimes E2,andso on.Theprobability thatyouobserve theenergy E1isjusttheprobability ofﬁnding thesystem inthestate In1),which is,ofcourse, justtheabsolute square ofthe amplitude C1=(11,III). Theprobability ofﬁnding each ofthepossible energies E,is P,=IC.I“. (20.11) How arethese probabilities related tothemean value ofawhole sequence ofenergy measurements? Let's imagine thatwegetaseries ofmeasurements like this: E1,E7,E11,E2,E1,E10,E7,E2,E3,E9,E6,E4,andsoon.Wecontinue for,say,athousand measurements. When weareﬁnished weaddalltheenergies anddivide byonethousand. That’s what wemean bytheaverage. There’s also ashort-cut toadding allthenumbers. Youcancount uphowmany times youget E1,saythatisN1,andthencount upthenumber oftimes yougetE2,callthat N2,andsoon.Thesumofalltheenergies iscertainly just mn+mo+mn+m=Zma Theaverage energy isthissumdivided bythetotal number ofmeasurements which isjust thesumofalltheN,’s, which wecancallN; =24,1-I’-*5 (20.12) Wearealmost there. What wemean bytheprobability ofsomething happen- ingisjustthenumber oftimes weexpect ittohappen divided bythetotal number oftries. Theratio N,/Nshould—for large N—-be veryneartoP,,theprobability ofﬁnding thestate I1),),although itwillnotbeexactly P,because ofthestatistical ﬂuctuations. Let’s write thepredicted (or“expected”) average energy as(E),,,.; thenwecansaythat (5),,=ZP,E,. (20.13) Thesame arguments apply foranymeasurement. Theaverage value ofameasured quantity Ashould beequal to mm=Zem where A,arethevarious possible values oftheobserved quantity, andP,isthe probability ofgetting thatvalue. Let‘s goback toourquantum-mechanical state Iip).It’saverage energy is (15),,=ZIC,I2E, =Zcfc,E,. (20.14) Now watch thistrickery! First, wewrite thesumas Next wetreat theleft-hand (uhIasacommon “factor.” Wecantakethisfactor outofthesum, andwrite itas oq;mwowI 20-4 This expression hastheform (1/1I¢>, where I¢)issome “cooked-up” state deﬁned by |¢>=Z|»t>E1<»11l¢>. (20.16) Itis,inother words, thestate yougetifyoutakeeach basestate I11,)intheamount E1<'71 I Now remember what wemean bythestates I-4,). They aresupposed tobe thestationary states—-by which wemean thatforeach one, HITh> =EtI711>- Since E,is_]USlanumber, theright-hand sideisthesame asI1;,)E,, andthesum inEq.(20.16) isthesame as ZHIm)(m I¢)- Now Iappears only inthefamous combination thatcontracts tounity, so Em m)<m|V/) =HZ:|m>(m|~l/) =HIM- Magic! Equation (20.16) isthesame as 1¢>=H|¢>- <20-11> Theaverage energy ofthestate I11/)canbewritten veryprettily as <E>&V=<¢11¥|¢>- <20-18> Togettheaverage energy youoperate on|¢)with ii,andthen multiply by(wkI. Asimple result. Ournewformula fortheaverage energy isnotonly pretty. ltisalsouseful, because now wedon’t need tosayanything about anyparticular setofbase states. Wedon’t even have toknow allofthepossible energy levels. When wego tocalculate, we'll need todescribe ourstate interms ofsome setofbase states, butifweknow theHamiltonian matrix H”forthatsetwecangettheaverage energy. Equation (19.18) says that foranysetofbase states Ii),theaverage energy canbecalculated from (E>n\'=Z<¢|1><i|HInoI¢>, <20-19> where theamplitudes (iIHIj)arejusttheelements ofthematrix H”. Let’s check thisresult forthespecial casethatthestates Ii)arethedeﬁnite energy states. Forthem, HIj)=E,Ij),so(1IHIj)=E,6,,and <E>11v=Z<¢|1>E,at,</|¢> =ZE1<¢Ii><il¢), which isright. Equation (20.19) can, incidentally, beextended toother physical measure- ments which youcanexpress asanoperator. Forinstance, 1:,istheoperator of thez-component oftheangular momentum L.Theaverage ofthez-component forthestate I1//)is <LZ)ﬂv=<¢|1-11¢>. Onewaytoprove itistothink ofsome situation inwhich theenergy isproportional totheangular momentum. Then allthearguments gothrough inthesame way. 20—S Insummary, ifaphysical observable Aisrelated toasuitable quantum- mechanical operator A,theaverage value ofAforthestate I¢)isgiven by <A>..»=<iI/5I¢>. (20.20)Bythiswemean that AILV=(itI¢), (2021)with I¢>=1|¢>- (20.22) Z0-3 The average energy ofanatom Suppose wewant theaverage energy ofanatom inastate described bya wave function i//(r); How doweﬁndit?Let’s ﬁrstthink ofaone-dimensional situation withastate Iip)deﬁned bytheamplitude (xIip)=¢(x). Weareasking forthespecial caseofEq.(20.19) applied tothecoordinate representation. Follow- ingourusual procedure, wereplace thestates I1)andIj)byIx)andIx’),and change thesums tointegrals. Weget <E>,,=I/<IiIx)(xIHIx’)(x’IIt)dxdx’. (20.23) This integral can, ifwewish, bewritten inthefollowing way: [<¢Ix><xI¢>dx, (20.24)with (XI¢>=I(xIHIx’)(x’I¢>dx’. (20.25) Theintegral over x’in(20.25) isthesame onewehadinChapter l6—-—see Eq. (16.50) andEq.(l6.52)—and isequal to 2 2 ~2";;,d~x—,two+V<><w<»<>. Wecantherefore write ifd2(xI4,)=I-:27‘2?;+V(x)I ¢(x). (20.26) Remember that<1//Ix)=(xI1//>*=¢*(x); using thisequality, theaverage energy inEq.(20.23) canbewritten as ,, if11’<E>B.V=-P(X)I—E3;+VI¢»(><)dx- (20-27) Given awave function I//(X), youcangettheaverage energy bydoing thisintegral. You canbegin toseehowwecangoback andforth from thestate-vector ideas tothewave-function ideas. The quantity inthebraces ofEq.(20.27) isanalgebraic operator.jI§ Wewill write itas3'0 2.]___h2 (L %— 2E?1'x2+V' With thisnotation Eq.(20.23) becomes <E>..'=j¢*<x>:t¢<x>dx. <20-28> Thealgebraic operator 3'6deﬁned here is,ofcourse, notidentical tothe quantum-mechanical operator H.Thenew operator works onafunction of position ¢(x) =(xI1//)togive anew function ofx,¢(x) =(xI¢);while H IThe“operator” V(x)means “multiply byV(x)." 20-6 operates onastate vector Iil/)togiveanother state vector I¢),without implying thecoordinate representation oranyparticular representation atall.Nor IS31‘ strictly thesame asHeven inthecoordinate representation. lfwechoose to work inthecoordinate representation, wewould interpret £7interms ofamatrix (xI1-7Ix’) which depends somehow onthetwo"indices" xandx’;thatis,we expect—according toEq. (20.25)——that (xI¢)isrelated toalltheamplitudes (xII11)byanintegration. Ontheother hand. weﬁndthatUPCisadifferential op- erator. Wehave already worked outinSection 16-5 theconnection between (xIHIx’)andthealgebraic operator 3'6. Weshould make onequaliﬁcation onourresults. Wehave been assuming thattheamplitude \//(X) =(xI¢)isnormalized. Bythiswemean thatthescale hasbeen chosen sothat /|¢<x>|2d><= 1; sotheprobability ofﬁnding theelectron somewhere isunity. Ifyoushould choose towork with a-//(x) which isnotnormalized youshould write ft/*<><>:w<»<> dx<E>:\v ="‘—*_t" f¢*<w<><> dx lt’sthesame thing. Notice thesimilarity inform between Eq.(20.28) andEq.(20.18). These twoways ofwriting thesame result appear often when youwork withthex-repre- sentation. You cangofrom theﬁrstform tothesecond with any/ii which isa localoperator, where alocal operator isonewhich intheintegral /<xIAIx’)(x’IIt)dx’ canbewritten asft\//(X), where tiisad1lTCI‘€l’lIl3.l algebraic operator. There are, however, operators forwhich thisisnottrue. Forthem youmust work with thebasic equations in(20.21) and(20.22). Youcaneasily extend thederivation tothree dimensions. Theresult isthatl (15),.=/¢(r)s‘c(t(r)dv<>i, (20.30) with - hz2ac=-Hv+V(r), (20.31) andwith theunderstanding that II¢I2dvo1 =i. (20.32) Thesame equations canbeextended tosystems with several electrons inafairly obvious way, butwewon’t bother towrite down theresults. With Eq.(20.30) wecancalculate theaverage energy ofanatomic state even without knowing itsenergy levels. Allweneed isthewave function. It's animportant law. We’ll tellyouabout oneinteresting application. Suppose you want toknow theground-state energy ofsome system——say thehelium atom, but it’stoohard tosolve Schrodinger’s equation forthewave function, because there aretoomany variables. Suppose, however, that you take aguess atthewave function——pick anyfunction youlike~—and calculate theaverage energy. That is, youuseEq.(20.29)—generalized tothree dimensions—to ﬁndwhat theaverage energy would beiftheatom were really inthestatedescribed bythiswave function. Thisenergy willcertainly behigher thantheground-state energy which isthelowest 1We write dVolfortheelement ofvolume. Itis,ofcourse, Justdxdydz.andthe integral goesfrom —1:to+1> inallthree coordinates. 20-7 iPix) > X Fig.20-1. Acurve ofprobability density representing 0localized particle.possible energy theatom canhave.I Now pickanother function andcalculate its average energy. lfitislower thanyour ﬁrstchoice youaregetting closer tothe trueground-state energy. Ifyoukeep ontrying allsorts ofartiﬁcial states you willbeabletogetlower andlower energies, which come closer andcloser tothe ground-state energy. Ifyouareclever, youwilltrysome functions which have a fewadjustable parameters. When youcalculate theenergy itwillbeexpressed interms ofthese parameters. Byvarying theparameters togivethelowest possible energy, youaretrying outawhole class offunctions atonce. Eventually youwill ﬁndthatitisharder andharder togetlower energies andyouwillbegin tobe convinced thatyouarefairly close tothelowest possible energy. Thehelium atom hasbeen solved injustthisway—not bysolving adifferential equation, butby making upaspecial function withalotofadjustable parameters which areeventu- allychosen togivethelowest possible value fortheaverage energy. 20-4 Theposition operator What istheaverage value oftheposition ofanelectron inanatom? Forany particular state I1/»)what istheaverage value ofthecoordinate x‘?We’ll work in onedimension andletyouextend theideas tothree dimensions ortosystems with more than oneparticie. 'wenave astatedescribed byIt/(x), andwekeep measuring xoverandoveragain. What istheaverage? Itis fxP(x) dx, where P(x) istheprobability ofﬁnding theelectron inalittle element dxatx. Suppose theprobability density P(x) varies with xasshown inFig.20-1. The electron ismost likely tobefound nearthepeak ofthecurve. Theaverage value ofxisalsosomewhere near thepeak. Itis,infact,justthecenter ofgravity of theareaunder thecurve. Wehave seenearlier thatP(x) isjustIll/(X) I2=(I/*(x)¢(x), sowecanwrite theaverage ofxas <2)...=/¢*(x)»i(x)d» (20.33) Ourequation for(x),,., hasthesame form asEq.(20.33). Fortheaverage energy, theenergy operator 5Cappears between thetwoi//’s,fortheaverage position there isjustx.(Ifyouwishyoucanconsider xtobethealgebraic operator “multi- plybyx.”) Wecancarry theparallelism stillfurther, expressing theaverage posi- tioninaform which corresponds toEq.(20.18). Suppose wejustwrite <2)...=<010> (20-34) Ia)=2It), (20.35)with andthen seeifwecanﬁndtheoperator itwhich generates thestate Ioz),which willmake Eq.(20.34) agree withEq.(20.33). That is,wemust ﬁndaIa),sothat <0I0)=(x>.v=/<0Ix)x<xII)dx. (20.30) First, let’sexpand (I/1I4))inthex-representation. Itis <0I02>=/<0Ix)(xI.).)dx. (20.37) Now compare theintegrals inthelasttwoequations. Youseethatinthex-repre- sentatiori (xIat)=x(xI¢>. (20.38) IYou canalsolook atitthisway. Any function (that is,state) youchoose canbe written asalinear combination ofthebase states which aredeﬁnite energy states. Since inthiscombination there isamixture ofhigher energy states inwith thelowest energy state, theaverage energy willbehigher than theground-state energy. 20-8 Operating onIt//)with ittogetIa)isequivalent tomultiplying ¢(x) =(xIip) byxtogeta(x) =(xI(1).Wehave adeﬁnition of)2inthecoordinate representa- tion.1 [Wehave notbothered totrytogetthex-representation ofthematrix ofthe operator st.Ifyouareambitious youcantrytoshow that (xIxIx’)=x6(x—x’). (20.39) Youcanthen work outtheamusing result that itIx)=xIx). (20.40) Theoperator ithastheinteresting property thatwhen itworks onthebase states Ix)itisequivalent tomultiplying byx.] Doyouwant toknow theaverage value ofx2? Itis wn=ﬂMmMoa (mm Or,ifyoupreferyoucanwrite wm=0w> Ia’)=22I0). (20.42)with By>22wemean xx—the twooperators areused oneafter theother. With the second form you cancalculate <X2>av using anyrepresentation (base-states) you wish. Ifyouwant theaverage ofx",orofanypolynomial inx,youcanseehow togetit. 20-5 Themomentum operator Now wewould liketocalculate themean momentum ofanelectron—again, we'll stick toonedimension. LetP(p)dpbetheprobability thatameasurement willgiveamomentum between pandp—I—dp.Then ou=pHm@ own Now welet(pIt//)betheamplitude thatthestate Iup)isinadeﬁnite momentum state Ip).This isthesame amplitude wecalled (mom pIt//)inSection 16-3 and isafunction ofpjustas(xII0)isafunction ofx.There wechose tonormalize theamplitude sothat 1P(12)=HI<pI¢)I2- (20.44) Wehave, then, <11)”=/(ItIP>P<P Iit)2%; (20-45) Theform isquite similar towhat wehadfor(x)...-. Ifwewant, wecanplayexactly thesame game wedidwith (x),w. First, we canwrite theintegral above as /mnmogg ma) Youshould nowrecognize thisequation asjusttheexpanded form oftheamplitude (I0IB)—expanded interms ofthebase states ofdeﬁnite momentum. From Eq. 1Equation (20.38) does notmean thatIa)=xI¢). You cannot “factor out” the (xI,because themultiplier xinfront of(xI¢)isanumber which isdifferent foreach state (x Itisthevalue ofthecoordinate oftheelectron inthestate Ix).SeeEq.(20.40). 20-9 (20.45) thestate II8)isdeﬁned inthemomentum representation by (12I/3)=P(1>IIt) (20-47) That is,wecannowwrite (12)..=()0I6) (20-43) with I6)=15IIP)» (20-49) where theoperator pisdeﬁned interms ofthep-representation byEq.(20.47). [Again, youcanifyouwish show thatthematrix form ofﬁis (11I15I11’)=P5(1)—P’), (20-50) andthat 13IP)=12IP)- (20-51) ltworks outthesame asforx.] Now comes aninteresting question. Wecanwrite ([2)... aswehave done in Eqs. (20.45) and(20.48), andweknow themeaning oftheoperator 13mthemo- mentum representation. Buthowshould weinterpret pinthecoordinate representa- tion? That iswhat wewillneed toknow ifwehave some wave function 1//(x), andwewant tocompute itsaverage momentum. Let's make clear what wemean. Ifwestartbysaying that(p),,, isgiven byEq.(20.48), wecanexpand thatequation interms ofthep-representation togetback toEq.(20.45). Ifwearegiven the p-description ofthestate——namely theamplitude (pIip),which isanalgebraic function ofthemomentum p—we canget(pI¢)from Eq.(20.47) andproceed toevaluate theintegral. Thequestion now is:What dowedoifwearegiven a description ofthestate inthex-representation, namely thewave function ¢(x) = (XIt/)2 Well, let’sstart byexpanding Eq.(20.48) inthex-representation. Itis <1»)...=I(¢|><)<><I0) 21» (20.52) Now, however, weneed toknow what thestate IB)isinthex-representation. Ifwecanﬁndit,wecancarry outtheintegral. Soourproblem istoﬁndthe function (3(x) =(xIB). Wecanﬁnditinthefollowing way. InSection l6—3wesawhow(pI(3)was related to(xIB).According toEq.(16.24), (pI0)=/e-""’"(x I(2)213. (20.52) Ifweknow (pI(3)wecansolve thisequation for(xI5).What wewant, ofcourse, istoexpress theresult somehow interms ofIp(x) =(xIll/>,which weareassuming tobeknown. Suppose westart withEq.(20.47) andagain useEq.(16.24) towrite <2I0)=p<12I¢> =pIe"'*'*I"(i(>() 21» (20.54) Since theintegral isover xwecanputthepinside theintegral andwrite <2I0)=[@"""'"p((x) dx. (20.55) Compare thiswith (20.53). Youwould saythat(xIB)isequal top\l/(X). No,No! Thewave function (xI,6)=B(x) candepend only onx—not onp.That‘s the whole problem. However, some ingenious fellow discovered thattheintegral in(20.55) could beintegrated byparts. Thederivative ofe_””" withrespect toxis(-1/h)pe'””‘”‘, sotheintegral in(20.55) isequivalent to ftd_.I.-TIE (eP”)¢()<)d)<. 20-10 Ifweintegrate byparts, itbecomes h -11):/ii +°° -11):/it—' I8 ¢(X)I_w +7 e g dx. Solongasweareconsidering bound states, sothatil(x)goestozeroatx==I=~/., thebracket iszero andwehave _h -—! I/ﬁt(pI6)-?/e "5dx. (20.56) Now compare thisresult with Eq.(20.53). You seethat hd(XIB)=7EI//(X) (20-57) Wehave thenecessary piece tobeabletocomplete Eq.(20.52). Theanswer is (10..=/(*0)? %tax)ax. (20-58) Wehave found howEq.(20.48) looks inthecoordinate representation. Now youshould begin toseeaninteresting pattern developing. When we asked fortheaverage energy ofthestate I1//)wesaiditwas (E)...=<0I<0),withI01>=HIti)- Thesame thing iswritten inthecoordinate world as (E)...=ft/»*(>t)¢(>t)t1x withtot)=:t0i(>t)- Here 30isanalgebraic operator which works afunction ofx.When weasked about theaverage value ofx,wefound thatitcould alsobewritten <x>av=(IIIIct),with I0t)=8I=//)- Inthecoordinate world thecorresponding equations are (X)...=]¢*(>t)h(>t)d»t. witha(x)=a(x)- When weasked about theaverage value ofp,wewrote (1>>..v=(ItI6),with I6)=ﬁlth)- lnthecoordinate world theequivalent equations were <12)...=/¢(x)0(»t)d»t. with(tot)=gtot). lneach ofourthree examples westart with thestate Iip)andproduce another (hypothetical) state byaquantum-mechanical operator. Inthecoordinate repre- sentation wegenerate thecorresponding wave function byoperating onthewave function (1/(x) with analgebraic operator. There arethefollowing one-to-one correspondences (forone-dimensional problems): h2d2I/(X), 2->X, (20.59) .-_h0"*“"ra 20-1i Table 20-1 Physical Quantity Operator Coordinate Form Energy 1-7 GAC=—ZimV2+V(r) N)‘<9>9Position x J’ Z A ft6Momentum 1),, (P,—IT5; it6 7(T) ft6 i5}pl! (;)y : pz (P: = Inthislist,wehave introduced thesymbol (P,forthealgebraic operator (h/i)6/6x: - h(3 andwehave inserted thexsubscript on0’toremind youthatWehave been working onlywiththex-component ofmomentum. You caneasily extend theresults tothree dimensions. Fortheother com- ponents 'ofthemomentum, . - I18 Pii_’(Pi/=75’ . »~ hi) pz—)(Pz=_l-62 Ifyouwant, youcaneventhink ofanoperator ofthevector momentum andwrite - A h 6 8 8 P'_*(P _ ey$+ez&)’ where et,e,_,,ande,aretheunitvectors inthethree directions. ltlooks even more elegant ifwewrite - ~hp—>(P=7V. (20.61) Ourgeneral resulttis thatforatleast some quantum-mechanical operators, there arecorresponding algebraic operators inthecoordinate representation. Wesummarize ourresults sofar-—extended tothree dimensions—in Table 20—l. Foreach operator wehave thetwoequivalent forms:I I0)=1Iti) (20-02)OI‘ A <p(r) =(it//(r). (20.63) Wewillnowgiveafewillustrations oftheuseofthese ideas. Theﬁrstoneis justtopoint outtherelation between (PandIRI.lfweuseti’,twice. weget i-_hag? 6”‘0)”__ 0x1 IInmany books thesame symbol isusedfor/iand(3,because theyboth stand forthe same physics, andbecause itisconvenient nottohave towrite different kinds ofletters. You canusually tellwhich oneisintended bythecontext 20-l2 Thismeans thatwecanwrite theequality a=5ﬂa@+aa+a@+v@ Or,using thevector notation, 50=%0°:-6-+V(r). (20.64) (Inanalgebraic operator, anytermwithout theoperator symbol (0)means justa straight multiplication.) This equation isnicebecause it’seasy toremember if youhaven’t forgotten your classical physics. Everyone knows thattheenergy is (nonrelativistically) justthekinetic energy p2/2m plusthepotential energy, and 3'0istheoperator ofthetotal energy. This result hasimpressed people somuch thattheytrytoteach students all about classical physics before quantum mechanics. (We think differently!) But such parallels areoften misleading. Foronething, when youhave operators, the order ofvarious factors isimportant; butthatisnottrueforthefactors ina classical equation. InChapter 17wedeﬁned anoperator ﬁ,interms ofthedisplacement operator D,by[seeEq.(l7.27)] Iw=0mno=0+;aQw) own where 6isasmall displacement. Weshould show youthatthisisequivalent to ournew deﬁnition. According towhat wehave justworked out,thisequation should mean thesame as wm=no+§t Buttheright-hand sideisjusttheTaylor expansion of¢(x+6),which iscertainly whatyougetifyoudisplace thestatetotheleftby6(orshiftthecoordinates to theright bythesame amount). Ourtwodeﬁnitions ofpagree! Let’s usethisfacttoshow something else. Suppose wehave abunch ofparti- cleswhich welabel 1,2,3,...,insome complicated system. (Tokeep things simple we’ll stick toonedimension.) Thewave function describing thestate isafunction ofallthecoordinates x1,x2,x3,...Wecanwrite itas¢(x1, x2,x3,...).Now displace thesystem (totheleft)by8.Thenewwave function ¢’(x1,x2,x3, ...)=¢(x1 +¢S,x2 +6,x3 +6,...) canbewritten as \0’(x1, x2,x3,...)=¢(x1, X2,x3,...) I.~.+a3‘i+ai+i)i+---I- (20.66)X1 6x2 6X3 According toEq.(20.65) theoperator ofthemomentum ofthestate IIb)(let’s callitthetotal momentum) isequal to - h 8 8 6 0)toiiii=?I'5Z+TQ+53g+"'I' Butthisisjustthesame as (ti...)=(fa.+6*.)+(5.3+---. (20.01) Theoperators ofmomentum obey therulethatthetotal momentum isthesumof themomenta ofalltheparts. Everything holds together nicely, andmany ofthe things wehavebeensaying areconsistent witheachother. 20-l3 y’y Isl P ___,-vx’I,__I\lI Fig. oround20-2. Rotcition of the cixes thez-cixis bythesmoll ongle €.le >X20-6 Angular momentum Let’s forfunlook atanother operation-the operation oforbital angular momentum. lnChapter 17wedeﬁned anoperator .7,interms of102(0)), theoperator ofarotation bytheangle (0about thez-axis. Weconsider hereasystem described simply byasingle wave function ll/(V), which isafunction ofcoordinates only, anddoes nottakeintoaccount thefactthattheelectron mayhave itsspineither upordown. That is,wewant forthemoment todisregard intrinsic angular momentum andthink about only theorbital part. Tokeep thedistinction clear, we’ll calltheorbital operator inanddeﬁne itinterms ofthe operator ofarotation byaninﬁnitesimal angle eby R.(t)Ii> =(1+6L.)Ii>- (Remember, thisdeﬁnition applies onlytoastate I(,0)which hasnointernal spin variables, butdepends only onthecoordinates r=x,y,x)Ifwelook atthe state I\l/>inanewcoordinate system, rotated about thez-axis bythesmall angle e,weseeanewstate IIV)=R.-(¢)I Ir)- lfwechoose todescribe thestate I(L)inthecoordinate representation——that is,byitswave function ¢(r), wewould expect tobeable towrite I//(r)=l+i (pot). (20.08)/'\ §-at£1» g What is,0? Well, apoint Patxandyinthenewcoordinate system (really x’ andy’,butwewilldrop theprimes) wasformerly atx—eyandy—I—ex,asyou canseefrom Fig.20-2. Since theamplitude fortheelectron tobeatPisn’tchanged bytherotation ofthecoordinates wecanwrite 6 6 tl’(X,y, Z)=¢(x+6%)’-ex.z)=t//(x,y,2) +ey5‘-I—EX5%J’ (remembering thateisasmall angle). This means that 5,= ~y (20.60) That’s ouranswer. Butnotice. Itisequivalent to .0.=Xe,—)0... (20.70) Returning toourquantum-mechanical operators, wecanwrite L‘.=xp,-yp, (20.71) This formula iseasy toremember because itlooks likethefamiliar formula of classical mechanics; itisthez-component of L=r><p. (20.72) One ofthefunparts ofthisoperator business isthatmany classical equations getcarried over into aquantum-mechanical form. Which ones don’t? There hadbetter besome thatdon’t come outright, because ifeverything did,then there would benothing diﬂerent about quantum mechanics. There would beno newphysics. Here isoneequation which isdifferent. lnclassical physics xp,—prx =0. What isitinquantum mechanics? X./5:0 _1071-if =(2 20-14 Let’s work itoutinthex-representation. Sothatwe’ll know what wearedoing weputinsome wave function ¢(x). Wehave x(P:r¢’(-X) _63::-x‘l’(x)> OI‘ h6 h6X?I//'/‘(X) _g5;20/‘(Xi Remember now thatthederivatives operate oneverything totheright. Weget h6¢ h h(N_ h Theanswer isnotzero. Thewhole operation isequivalent simply tomultiplication by—h/i: ., h32,),~1),):=-7- (20.74) IfPlank’s constant were zero, theclassical andquantum results would bethesame, andthere would benoquantum mechanics tolearn! Incidentally, ifanytwooperators AandB,when taken together likethis: /TE—B/i, donotgivezero, wesaythat“theoperators donotcommute.” And anequation such as(20.74) iscalled a“commutation rule.” Youcanseethatthecommutation ruleforp,andyis ﬁx?-I113;=0- There isanother veryimportant commutation rulethathastodowith angular momenta. Itis 13,15,-13,,£,,=#113,. (20.75) You cangetsome practice with5candj)operators byproving itforyourself. Itisinteresting tonotice thatoperators which donotcommute canalsooccur inclassical physics. Wehave already seenthiswhen wehave talked about rotation inspace. Ifyourotate something, such asabook, by90°around xandthen90° around y,yougetsomething different from rotating ﬁrstby90°around yandthen by90°around x.Itis,infact,justthisproperty ofspace thatisresponsible for Eq.(20.75). 20-7 Thechange ofaverages withtime Now wewant toshow yousomething else. How doaverages change with time? Suppose forthemoment thatwehave anoperator 24‘,which does notitself have timeinitinanyobvious way. Wemean anoperator like5:orp.(Weexclude things like,say,theoperator ofsome external potential thatwasbeing varied with time, such asV(x,t).)Now suppose wecalculate (A)..v, insome state IIt/),which is <4)...=(I1/IMi). (20-70) How will(A),,v depend ontime? Why should it?Onereason might bethatthe operator itself depended explicitly ontime—for instance, ifithadtodowith a time-varying potential likeV(x,t).Buteven iftheoperator does notdepend on t,say,forexample, theoperator /i=)2,thecorresponding average may depend ontime. Certainly theaverage position ofaparticle could bemoving. How does such amotion come outofEq.(20.76) if/ihasnotime dependence? Well, the state IIk)might bechanging with time. Fornonstationary states wehave often shown atimedependence explicitly bywriting astate asI¢(t)). Wewa_nt toshow thattherateofchange of(A),,v isgiven byanewoperator wewillcall/i.Remem- berthatiiisanoperator, sothatputting adotovertheAdoesnotheremean taking 20-!5 thetimederivative, butisjustawayofwriting anewoperator which isdeﬁned by §,<4)...=(iIATIi). (20-77) Ourproblem istoﬁndtheoperator First, weknow thattherateofchange ofastate isgiven bytheHamiltonian. Speciﬁcally, thIi(t)>=HI((0). (20-78) This isjusttheabstract wayofwriting ouroriginal deﬁnition oftheHamiltonian: .dC,I/‘IT,=H.,c,. (20.70) Ifwetakethecomplex conjugate ofthisequation, itisequivalent to -it%<i(t)I=<70)Iit (20-80) Next, seewhat happens ifwetakethederivatives with respect totofEq.(20.76). Since each ipdepends ont,wehave ;§</1)..=(;§iI<iI)1Ii +<iI4(§IIi>)- (20-81) Finally, using thetwoequations in(20.78) and(20.79) toreplace thederivatives, weget §,<A)..= £{(¢I1‘7/fI¢)— (¢I/fP7I\t)]- This equation isthesame as d 1 AA AA 3;<4)...-Z(tiI(HA-4H)I=i>- Comparing thisequation with Eq.(20.77), youseethat ¢ A/\ AA A_,1(HA-AH). (20.22) That isourinteresting proposition, anditistrueforanyoperator /i. Incidentally, iftheoperator /fshould itsey betimedependent, wewould have had ¢1'-A A 6AA-—z(HA —AH) —I—-5- (20.83) LetustryoutEq.(20.82) onsome example toseewhether itreally makes sense. Forinstance, what operator corresponds to3??Wesayitshould be i=%(Hx -xii). (20.84) What isthis? Onewaytoﬁndoutistowork itthrough inthecoordinate repre- sentation using thealgebraic operator forIRZ.Inthisrepresentation thecommutator is -1 A h2dz hzdzJLX —Xi“: = V(X)}X — ' Ifyouoperate with thisoranywave function ¢(x) andwork outallofthede- rivatives where youcan,youendupafter alittle work with it’dii 2mdx 20-16 Butthisisjustthesame as .hA—lﬁ(l)I¢, soweﬁndthat 172=2H=-1%p, (20.85) orthat 4 "_122.x-m (20.86) Apretty result. Itmeans that ifthemean value ofxischanging with time the drift ofthecenter ofgravity isthesame asthemean momentum divided bym. Exactly likeclassical mechanics. Another example. What istherateofchange oftheaverage momentum ofa state? Same game. Itsoperator is ,3=;I(i7p ~pH). (20.87) Again youcanwork itoutinthexrepresentation. Remember thatp‘becomes d/dx, andthismeans thatyouwillbetaking thederivative ofthepotential energy V(inthe30(1)-but only inthesecond term. Itturns outthatitistheonly term which does notcancel, andyouﬁndthat AA A» 506’ —-(P30 —-ih d—Vdx orthat - avp_--5- (20.88) Again theclassical result. Theright-hand sideistheforce, sowehave derived Newton’s law! Butremember—these arethelaws fortheoperators which give theaverage quantities. They donotdescribe what goes onindetail inside an atom. Quantum mechanics hastheessential difference thatﬁxisnotequal tosp. They differ byalittle bit-by thesmall number h.Butthewhole wondrous compli- cations ofinterference, waves, andall,result from thelittle factthatxp-pxis notquite zero. Thehistory ofthisideaisalsointeresting. Within aperiod ofafewmonths in 1926, Heisenberg andSchrtidinger independently found correct laws todescribe atomic mechanics. Schrddinger invented hiswave function I//(x) andfound his equation. Heisenberg, ontheother hand, found thatnature could bedescribed byclassical equations, except thatxp—pxshould beequal toh/i,which hecould make happen bydeﬁning them interms ofspecial kinds ofmatrices. Inourlan- guage hewasusing theenergy-representation, withitsmatrices. Both Heisenberg’s matrix algebra andSchr6dinger’s differential equation explained thehydrogen atom. Afewmonths later Schrtidinger wasabletoshow thatthetwotheories were equivalent-—as wehave seenhere. Butthetwodifferent mathematical forms ofquantum mechanics were discovered independently. 20-17 21 The Schrodinger Equntion inuClassical Context: ASeminar onSuperconductivity 21-1 Schriidinger’s equation inamagnetic ﬁeld This lecture isonly forentertainment. Iwould liketogivethelecture ina somewhat different style—-just toseehowitworks out.It’snotapartofthecourse —inthesense thatitisnotsupposed tobealastminute effort toteach yousome- thing new. But, rather, Iimagine thatl’mgiving aseminar orresearch report on thesubject toamore advanced audience, topeople whohavealready beeneducated inquantum mechanics. Themain difference between aseminar andaregular lecture isthattheseminar speaker does notcarry outallthesteps, orallthe algebra. Hesays: “Ifyoudosuch andsuch, thisiswhat comes out,” instead ofshowing allofthedetails. Sointhislecture I’lldescribe theideas alltheway along butjustgiveyoutheresults ofthecomputations. Youshould realize that you’re notsupposed tounderstand everything immediately, butbelieve (more or less)thatthings would come outifyouwent through thesteps. Allthat aside, thisisasubject Iwant totalkabout. Itisrecent andmodern andwould beaperfectly legitimate talktogiveataresearch seminar. Mysubject istheSchrodinger equation inaclassical setting—the caseofsuperconductivity. Ordinarily, thewave function which appears intheSchrodinger equation applies toonly oneortwoparticles. Andthewave function itself isnotsome- thing thathasaclassical meaning—unlike theelectric ﬁeld, orthevector potential, orthings ofthatkind. Thewave function forasingle particle isa“ﬁeld”-—in thesense thatitisafunction ofposition—but itdoesnotgenerally have aclassical signiﬁcance. Nevertheless, there aresome situations inwhich aquantum me- chanical wave function doeshave classical signiﬁcance, andtheyaretheones I would liketotakeup.Thepeculiar quantum mechanical behavior ofmatter on asmall scale doesn’t usually make itself feltonalarge scale except inthestandard waythatitproduces NeWton’s laws—the laws oftheso-called classical mechanics. Butthere arecertain situations inwhich thepeculiarities ofquantum mechanics cancome outinaspecial wayonalarge scale. Atlowtemperatures, when theenergy ofasystem hasbeen reduced very, verylow,instead ofalarge number ofstates being involved, only avery, very small number ofstates neartheground state areinvolved. Under those circum- stances thequantum mechanical character ofthatground state canappear ona macroscopic scale. ltisthepurpose ofthislecture toshow aconnection between quantum mechanics andlarge-scale effects—not theusual discussion oftheway thatquantum mechanics reproduces Newtonian mechanics ontheaverage, buta special situation inwhich quantum mechanics willproduce itsowncharacteristic effects onalarge or“macroscopic” scale. Iwillbegin byreminding youofsome oftheproperties oftheSchrodinger equation.'I' Iwant todescribe thebehavior ofaparticle inamagnetic ﬁeldusing theSchrodinger equation, because thesupercolnductive phenomena areinvolved withmagnetic ﬁelds. Anexternal magnetic ﬁeldisdescribed byavector potential, andtheproblem is:what arethelawsofquantum mechanics inavector potential? Theprinciple that describes thebehavior ofquantum mechanics inavector potential isvery simple. Theamplitude thataparticle goes from oneplace to another along acertain route when there’s aﬁeldpresent isthesame astheampli- TI’mnotreally reminding you,because Ihaven’t shown yousome ofthese equations before; butremember thespirit ofthisseminar. 21-121-1 21-2 21-3 21-4 21-5 21-6 21-7 21-8 21-9Schr6dinger’s equation ina magnetic ﬁeld Theequation ofcontinuity ft probabilities Twokinds ofmomentum Themeaning ofthewave function Superconductivity TheMeissner effect Flux quantization Thedynamics of superconductivity TheJosephson junction b r O Fig. 2l»~l. Theamplitude togofrom 0tobalong thepcith I‘isproportioncil to EXPliq/ﬁ) EA-ds.tude thatitwould goalong thesame route when there's noﬁeld, multiplied bythe exponential ofthelineintegral ofthevector potential, times theelectric charge divided byPlanck's constant‘ (seeFig.21-1): ~l> (b|a),,,_4 =<bia>,,=,,-expllzl A-as)» (21.1) ltisabasic statement ofquantum mechanics. Now without thevector potential theSchrodinger equation ofacharged particle (nonrelativistic, nospin) is ha¢_,_1<h)(h> 1275-3“//-5277 7V1P+q¢lP, where 41>istheelectric potential sothatq¢-isthepotential energyfi Equation (21.1) isequivalent tothestatement thatinamagnetic ﬁeld thegradients intheHamilton- ianarereplaced ineach case bythegradient minus qA,sothatEq.(21.2) becomes —€l%l€=J11//=ﬁ<?V—q/1)-<?V~qA>¢—l—q¢¢. (21.3) This istheSchrodinger equation foraparticle with charge qmoving inanelec- tromagnetic ﬁeld A,¢(nonrelativistic, nospin). Toshow thatthisistrue l’dliketoillustrate byasimple example inwhich instead ofhaving acontinuous situation wehave alineofatoms along thex-axis with thespacing bandwehave anamplitude —Kforanelectron tojump from oneatom toanother when there isnoﬁeld.I Now according toEq. (21.1) if there's avector potential inthex-direction A,(x, I),theamplitude tojump will bealtered from what itwasbefore byafactor exp(zq/hA,,b), theexponent being iq/htimes thevector potential integrated from oneatom tothenext. Forsimplicity weWlllwrite (q/h)AI Ef(x), since A,will, ingeneral, depend onx.lftheampli- tude toﬁnd theelectron attheatom “n”located atxiscalled C(x) EC,,,then therateofchange ofthatamplitude isgiven bythefollowing equation‘ -if5;co-)=E,,C(x) -Ke—‘l’/“+"/2)C(,t‘ +1») -Ke+'l’/”—[’l2)C(x -b). (21.4) There arethree pieces. First, there’s some energy E‘,iftheelectron 1Slocated atx.Asusual, thatgives theterm E0C(x). Next, there istheterm —KC(x +b), which 1Stheamplitude fortheelectron tohave jumped backwards onestepfrom atom “n-1-1,”located atx+b.However, indoing soinavector potential, the phase oftheamplitude must beshifted according totheruleinEq.(21.1). lfA, isnotchanging appreciably inoneatomic spacing, theintegral canbewritten as justthevalue ofA,atthemidpoint, times thespacing b.So(iq/h) times theintegral isjust bf(x +b/2). Since theelectron isjumping backwards, 1showed this phase shift with aminus sign. That gives thesecond piece. lnthesame manner there’s acertain amplitude tohave jumped from theother side, butthistimewe need thevector potential atadistance (b/2) ontheother sideofx,times thedis- tance b.That gives thethird piece. Thesum gives theequation fortheamplitude tobeatxinavector potential Now weknow that ifthefunction C(x) issmooth enough (wetake thelong wavelength limit), and ifwelettheatoms getcloser together, Eq.(164) will approach thebehavior ofanelectron infreespace. Sothenext stepistoexpand both sides of(21.4) inpowers ofb,assuming bisvery small. Forexample, ifb iszero theright-hand sideisjust(E0—2K)C(x). sointhezeroth approximation 1Volume II,Section 15-5. TNot tobeconfused with ourearlier use0f¢ forastate label‘ IKisthesame quantity thatwascalled Aintheproblem ofalinear lattice with no magnetic ﬁeld SeeChapter 13. Zl—2 theenergy isE0—2K. Next comes theterms inb.Butbecause thetwoex- ponentials have opposite signs, onlyeven powers ofbremain. Soifyoumake a Taylor expansion ofC(x), off(x), andoftheexponentials, andthen collect the terms inb2,youget — =E0C(x) —2KC(x) -K52{C”(X) —ZIf(X)C’(X) —If’(X)C(X) —f2(X)C(X)}- (Z1-5) (The “primes” mean diﬂerentiation withrespect tox.) Now thishorrible combination ofthings looks quite complicated. But mathematically it’sexactly thesame as _% =(E0-2K)C(x) ~102% -if(x)H% -[f(x)]C(x). (21.6) Thesecond bracket operating onC(x)gives C’(x) plusif(x)C(x). Theﬁrstbracket operating onthese twoterms gives theC”term andterms intheﬁrstderivative off(x) andtheﬁrstderivative ofC(x). Now remember thatthesolutions forzero magnetic ﬁeld” represent aparticle withaneﬂective mass mp“given by Kb2=1-mcff Ifyouthen setE0=—-2K, andputback f(x) =(q/h)A,, youcaneasily check thatEq.(21.6) isthesame astheﬁrstpartofEq.(21.3). (Theorigin ofthepotential energy term iswellknown, soIhaven’t bothered toinclude itinthisdiscussion.) Theproposition ofEq.(21.1) thatthevector potential changes alltheamplitudes bytheexponential factor isthesame astherulethatthemomentum operator, (h/i)V getsreplaced by éV—qA, asyouseeintheSchrodinger equation of(21.3). 21-2 Theequation ofcontinuity forprobabilities Now Iturntoasecond point. Animportant partoftheSchrodinger equation forasingle particle istheideathattheprobability toﬁndtheparticle ataposition isgiven bytheabsolute square ofthewave function. Itisalsocharacteristic of thequantum mechanics thatprobability isconserved inalocal sense. When the probability ofﬁnding theelectron somewhere decreases, while theprobability of theelectron being elsewhere increases (keeping thetotal probability unchanged), something must begoing oninbetween. Inother words, theelectron hasacon- tinuity inthesense that iftheprobability decreases atoneplace andbuilds up atanother place, there must besome kindofﬂowbetween. Ifyouputawall, for example, intheway, itwillhave aninﬂuence andtheprobabilities willnotbethe same. Sotheconservation ofprobability alone isnotthecomplete statement of theconservation law,justastheconservation ofenergy alone isnotasdeep and important asthelocal conservation ofenergy.3 Ifenergy isdisappearing, there must beaﬂowofenergy tocorrespond. Inthesame way, wewould liketoﬁnda “current” ofprobability suchthatifthere isanychange intheprobability density (theprobability ofbeing found inaunitvolume), itcanbeconsidered ascoming from aninﬂow oranoutﬂow duetosome current. Thiscurrent would beavector which could beinterpreted thisway—the xcomponent would bethenetprob- ability persecond andperunitareathataparticle passes inthexdirection across aplane parallel tothey-zplane. Passage toward +xisconsidered apositive ﬂow, andpassage intheopposite direction, anegative ﬂow. 2Section 13-3. 3Volume II,Section 27~l. 21-3 lsthere such acurrent? Well, youknow thattheprobability density P(r,t) isgiven interms ofthewave function by P(r,t)=jl/*(r, r)i//(r, I). (21.7) lamasking: lsthere acurrent Jsuch that 6P_ _95;- VJ. (21.8) lfltake thetime derivative ofEq.(21.7), Igettwoterms: aP_,.at aw 57"‘F5+‘fat <2”) Now usetheSchrodinger equation—Eq. (2l.3)—for 611//61; andtakethecomplex conjugate ofittoget610*/6t—each 1'getsitssignreversed. Youget %’€=—i’.-*@%(?v —1/1)-(iv—1A)~*+~>~i*~* —112';-1 v+qA)-<§ v+qA)¢* —@¢w*- (M0) Thepotential terms andalotofother stuff cancel out. And itturns outthatwhat isleftcanindeed bewritten asaperfect divergence. Thewhole equation isequiva- lentto %§=»-ia-<%v~a>i+1<~1~A>i*i» W»Itisreally notascomplicated asitseems. Itisasymmetrical combination of )1/*times acertain operation onip,plusjb*times thecomplex conjugate operation onwk.Itissome quantity plusitsowncomplex conjugate, sothewhole thing is real—as itought tobe.Theoperation canberemembered thisway: itisjustthe momentum operator fl’minus qA. Icould write thecurrent inEq.(21.8) as _1 ]*.[@:;q/1] ]. 1-ill mii¢+¢ m~»i <21-12> There isthen acurrent Jwhich completes Eq.(21.8). Equation (21.10) shows thattheprobability isconserved locally. Ifaparticle disappears from oneregion itcannot appear inanother without something going oninbetween. Imagine thattheﬁrstregion issurrounded byaclosed surface far enough outthat there iszero probability toﬁndtheelectron atthesurface The total probability toﬁnd theelectron somewhere inside thesurface isthevolume integral ofP.Butaccording toGauss’s theorem thevolume integral ofthedi- vergence Jisequal tothesurface integral ofJ.Ifjbiszeroatthesurface, Eq. (21.10) says thatJiszero, sothetotal probability toﬁndtheparticle inside can’t change. Only ifsome oftheprobability approaches theboundary cansome ofit leak out. Wecansaythat itonly getsoutbymoving through thesurface—and thatislocal conservation. 21-3 Two kinds ofmomentum The equation forthecurrent israther interesting, and sometimes causes a certain amount ofworry. Youwould think thecurrent would besomething like thedensity ofparticles times thevelocity. Thedensity should besomething like it/1//*,which iso.k.AndeachterminEq.(21.12) looks likethetypical form forthe average-value oftheoperatorA (P—qA~W— (21.13) 21-4 somaybe weshould think ofitasthevelocity ofﬂow. Itlooks asthough wehave twosuggestions forrelations ofvelocity tomomentum, because wewould also think thatmomentum divided bymass, 05/m, should beavelocity. Thetwopossi- bilities differ bythevector potential. Ithappens thatthese twopossibilities were alsodiscovered inclassical physics, when itwasfound thatmomentum could bedeﬁned intwoways.‘* Oneofthem iscalled “kinematic momentum," butforabsolute clarity Iwillinthislecture call itthe“mu-momentum.” This isthemomentum obtained bymultiplying mass byvelocity. Theother isamore mathematical, more abstract momentum, some- times called the“dynamical momentum,” which I’llcall“p-momentum.” The twopossibilities are mzi-momentum =mv, (21.14) p-momentum =mu+qA. (21.15) Itturns outthatinquantum mechanics with magnetic ﬁelds itisthep-momentum which isconnected tothegradient operator 5’,soitfollows that (21.13) isthe operator ofavelocity. I'dliketomake abrief digression toshow youwhat thisisallabout—why there must besomething likeEq.(21.15) inthequantum mechanics. The wave function changes with time according totheSchrodinger equation inEq.(21.3). IfIwould suddenly change thevector potential, thewave function wouldn't change attheﬁrst instant; only itsrateofchange changes. Now think ofwhat would happen inthefollowing circumstance. Suppose Ihave along solenoid, in which Icanproduce aﬂuxofmagnetic ﬁeld (B-ﬁeld), asshown inFig.21-2. And there isacharged particle sitting nearby. Suppose thisﬂuxnearly instantaneously builds upfrom zero tosomething. Istart with zero vector potential andthen I turn onavector potential. That means thatIproduce suddenly acircumferential vector potential A.You’ll remember that thelineintegral ofAaround aloop is thesame astheﬂuxofBthrough theloop.5 Now what happens ifIsuddenly turn onavector potential? According tothequantum mechanical equation thesudden change ofAdoes notmake asudden change ofip;thewave function isstillthe same. Sothegradient isalsounchanged. Butremember what happens electrically when lsuddenly turn onaﬂux. During theshort time that theﬂux isrising, there’s anelectric ﬁeld generated whose lineintegral istherateofchange oftheﬂuxwith time: a/1E_-5- (21.16) That electric ﬁeld isenormous iftheﬂuxischanging rapidly, anditgives aforce ontheparticle. Theforce isthecharge times theelectric ﬁeld, andsoduring the build upoftheﬂuxtheparticle obtains atotal impulse (that is,achange inmv) equal to—qA. Inother words, ifyousuddenly turn onavector potential ata charge, thischarge immediately picks upan“mu” momentum equal to—qA. Butthere issomething thatisn’tchanged immediately andthat’s thedifference between mvand—qA. And sothesump =mv+qAissomething which isnot changed when youmake asudden change inthevector potential. This quantity piswhat wehave called thep-momentum andisofimportance inclassical me- chanics inthetheory ofdynamics, butitalsohasadirect signiﬁcance inquantum mechanics. Itdepends onthecharacter ofthewave function, anditistheoneto beidentiﬁed with theoperator 63=£lv.l “See,forexample, J.DJackson, Classical Electrodynamics, John Wiley andSons, Inc. New York (1962), p.408. 5Volume II,Chapter 14,Section 14-1. 21-5B Ee;{;Z€§titi(f(E-///~gig \-/1 Fig. 21-2. The electric ﬁeld outside osolenoid with onincreasing current. 21-4 Themeaning ofthewave function When Schrodinger ﬁrstdiscovered hisequation hediscovered theconservation lawofEq.(21.9) asaconsequence ofhisequation. Butheimagined incorrectly thatPwastheelectric charge density oftheelectron andthatJwastheelectric current density, sohethought thattheelectrons interacted withtheelectromagnetic ﬁeld through these charges andcurrents. When hesolved hisequations forthe hydrogen atom andcalculated 11/,hewasn’t calculating theprobability ofanything —there were noamplitudes atthattime——the interpretation wascompletely (llil8l'- ent. Theatomic nucleus wasstationary butthere were currents moving around; thecharges Pandcurrents Jwould generate electromagnetic ﬁelds andthething would radiate light. Hesoon found ondoing anumber ofproblems thatitdidn’t work outquite right. Itwasatthispoint thatBorn made anessential contribution toourideas regarding quantum mechanics. ltwasBorn whocorrectly (asfar asweknow) interpreted the\[/oftheSchrodinger equation interms ofaprobability amplitude——that very difficult ideathatthesquare oftheamplitude isnotthe charge density butisonlytheprobability perunitvolume ofﬁnding anelectron there, andthatwhen youdoﬁndtheelectron some place theentire charge isthere. That whole ideaisduetoBorn. Thewave function \b(r)foranelectron inanatom does not,then, describe asmeared-out electron with asmooth charge density. Theelectron iseither here, orthere, orsomewhere else,butwherever itis,itisapoint charge. Ontheother hand, think ofasituation inwhich there areanenormous number ofparticles in exactly thesame state, averylarge number ofthem with'exactly thesame wave function. Then what? Oneofthem ishereandoneofthem isthere, andthe probability ofﬁnding anyoneofthem atagiven place isproportional to¢¢*. Butsince there aresomany particles, ifIlook inanyvolume dxdydzIwill generally ﬁndanumber close toW/*dxdydz.Soinasituation inwhich itisthe wave function foreach ofanenormous number ofparticles which areallinthe same state. w*canbeinterpreted asthedensity ofparticles. If,under these circumstances, each particle carries thesame charge q,wecan, infact, gofurther andinterpret 11/*¢asthedensity ofelectricity. Normally, W*isgiven thedimen- sions ofaprobability density, then ipshould bemultiplied byqtogivethedimen- sions ofacharge density. Forourpresent purposes wecanputthisconstant factor into KP,andtake 1)/41*itself astheelectric charge density. With thisunder- standing, J(the current ofprobability Ihave calculated) becomes directly the electric current density. Sointhesituation inwhich wecanhave very many particles inexactly the same state, there ispossible anew physical interpretation ofthewave functions. Thecharge density andtheelectric current canbecalculated directly from the wave functions andthewave functions takeonaphysical meaning which extends intoclassical, macroscopic situations. Something similar canhappen with neutral particles. When wehave the wave function ofasingle photon, itistheamplitude toﬁndaphoton somewhere. Although wehaven‘t everwritten itdown there isanequation forthephoton wave function analogous totheSchrodinger equation fortheelectron. Thephoton equation isjust thesame asMaxwell’s equations fortheelectromagnetic ﬁeld, andthewave function isthesame asthevector potential A.Thewave function turns outtobejustthevector potential. Thequantum physics isthesame thing astheclassical physics because photons arenoninteracting Bose particles and many ofthem canbeinthesame state—as youknow, they liketobeinthesame state. Themoment thatyouhave billions inthesame state (that is.inthesame electromagnetic wave), youcanmeasure thewave function, which isthevector potential, directly. Ofcourse, itworked historically theother way. Theﬁrstob- servations wereonsituations withmany photons inthesame state, andsowewere abletodiscover thecorrect equation forasingle photon byobserving directly withourhands onamacroscopic levelthenature ofwave function. Now thetrouble with theelectron isthat youcannot putmore than onein thesame state. Therefore, itwas long believed thatthewave function ofthe 21-6 Schrodinger equation would never have amacroscopic representation analogous tothemacroscopic representation oftheamplitude forphotons. Ontheother hand, itisnow realized thatthephenomena ofsuperconductivity presents uswith justthissituation. 21-5 Superconductivity Asyou know, very many metals become superconducting below acertain Iemperature"—the temperature isdifferent fordifferent metals. When youreduce thetemperature sufficiently themetals conduct electricity without anyresistance This phenomenon hasbeen observed foravery large number ofmetals butnotfor all,andthetheory ofthisphenomenon hascaused agreat deal ofdifficulty. It took avery long time tounderstand what wasgoing oninside ofsuperconductors, andIwillonly describe enough ofitforourpresent purposes. Itturns outthat duetotheinteractions oftheelectrons with thevibrations oftheatoms inthe lattice, there isasmall neteffective attraction between theelectrons. The result isthattheelectrons form together, ifImay speak very qualitatively andcrudely, bound pairs. Now youknow thatasingle electron isaFermi particle. Butabound pair would actasaBose particle, because ifIexchange both electrons inapairIchange thesign ofthewave function twice, andthatmeans thatIdon’t change anything. ApairisaBose particle. Theenergy ofpairing—-that is,thenetattraction—is very, very weak. Only atiny temperature isneeded tothrow theelectrons apart bythermal agitation, andconvert them back to“normal” electrons. Butwhen youmake thetempera- turesuﬁiciently lowthatthey have todotheir very besttogetinto theabsolutely lowest state; then they docollect inpairs. Idon’t wish youtoimagine thatthepairs arereally held together very closely likeapoint particle. Asamatter offact, oneofthegreat difﬁculties ofunder- standing thisphenomena originally wasthat that isnottheway things are. The twoelectrons which form thepair arereally spread over aconsiderable distance; andthemean distance between pairs isrelatively smaller than thesizeofasingle pair. Several pairs areoccupying thesame space atthesame time. Both thereason why electrons inametal form pairs andanestimate oftheenergy given upin forming apairhave been atriumph ofrecent times. This fundamental point inthe theory ofsuperconductivity wasﬁrstexplained inthetheory ofBardeen, Cooper, andSchrietTer,7 butthat1SnottheSL1b_|CCI ofthisseminar. Wewillaccept, however, theidea that theelectrons do,insome manner orother, work inpairs, that we canthink ofthese pairs asbehaving more orlesslikeparticles, andthat wecan therefore talkabout thewave function fora“pair.” Now theSchrodinger equation forthepairWlllbemore orlesslikeEq.(21.3). There willbeonedifference inthatthecharge qwillbetwice thecharge ofanelec- tron. Also, wedon't know theinertia—or effective mass——for thepairinthecrystal lattice, sowedon’t know what number toputinform.Norshould wethink that ifwegotovery high frequencies (orshort wavelengths), thisisexactly theright form, because thekinetic energy that corresponds tovery rapidly varying wave functions may besogreat astobreak upthepairs. Atﬁnite temperatures there arealways afewpairs which arebroken upaccording totheusual Boltzmann theory. Theprobability thatapairisbroken isproportional toexp(—E,,.,,,/kT). Theelectrons that arenotbound inpairs arecalled “normal” electrons andwill move around inthecrystal intheordinary way. Iwill, however, consider only thesituation atessentially zero temperature—or, inanycase, Iwilldisregard the complications produced bythose electrons which arenotinpairs. f’First discovered byOnnes in1911; H.K.Onnes, Comm. Phys. Lab,Univ. Leyden, Nos. 119,120,122(1911). Youwillﬁndaniceup-to-date discussion oftheSl1l)]ECl. in E.A.Lynton, Superconductivity, John Wiley andSons, Inc.,New York, 1962. 7J.Bardeen, L.N.Cooper, andJ.R.Schrietfer, Phys. Rev.108,1175 (1957). 2l~7 Since electron pairs arebosons, when there arealotofthem inagiven state there isanespecially large amplitude forother pairs togotothesame state. So nearly allofthepairs willbelocked down atthelowest energy inexactly t/zesame .YI(1[€—l[ won’t beeasytogetoneofthem intoanother state. There’s more ampli- tudetogointothesame state thanintoanunoccupied state bythefamous factor \/H,where nistheoccupancy ofthelowest state. Sowewould expect allthepairs tobemoving inthesame state. What thenwillourtheory looklike? I'llcall11/thewave function ofapair inthelowest energy state. However, since W/*isgoing tobeproportional tothe charge density p,Icanjustaswellwrite ¢asthesquare root ofthecharge density times some phase factor: Mr)=t>(t')¢"”('), (21-17) where pand0arerealfunctions ofr.(Any complex function can,ofcourse, be written thisway.) It’sclear what‘ wemean when wetalkabout thecharge density, butwhat isthephysical meaning ofthephase 0ofthewave function? Well, let’s seewhat happens ifwesubstitute ip(r)intoEq.(21.12), andexpress thecurrent density interms ofthese newvariables pand0.It’sjustachange ofvariables and Iwon't gothrough allthealgebra, butitcomes out hJ=7n<va-gA)p- (21.18) Since both thecurrent density andthecharge density have adirect physical meaning forthesuperconducting electron gas,both pand9arerealthings. Thephase is justasobservable asp;itisapiece ofthecurrent density J.Theabsolute phase is notobservable, butifthegradient ofthephase isknown everywhere, thephase is known except foraconstant. Youcandeﬁne thephase atonepoint, andthenthe phase everywhere isdetermined. Incidentally, theequation forthecurrent canbeanalyzed alittle nicer, when youthink thatthecurrent density Jisinfactthecharge density times thevelocity ofmotion oftheﬂuid ofelectrons, orpv.Equation (21.18) isthen equivalent to mv=hV0 —qA. (21.19) Notice thatthere aretwopieces inthemu-momentum; oneisacontribution from thevector potential, and theother, acontribution from thebehavior ofthe wave function. Inother words, thequantity hV0isjustwhat wehave called the p-momentum. 21-6 TheMeissner effect Now wecandescribe some ofthephenomena ofsuperconductivity. First, there isnoelectrical resistance. There’s noresistance because alltheelectrons are collectively inthesame state. Intheordinary ﬂow ofcurrent you knock one electron ortheother outoftheregular ﬂow, gradually deteriorating thegeneral momentum. Buthere togetoneelectron away from what alltheothers aredoing isvery hard because ofthetendency ofallBose particles togointhesame state. Acurrent once started, justkeeps ongoing forever. It’salso easy tounderstand that ifyouhave apiece ofmetal inthesuper- conducting state andturn onamagnetic ﬁeld which isn’t toostrong (wewon’t gointothedetails ofhowstrong), themagnetic ﬁeldcan’t penetrate themetal. If,as youbuild upthemagnetic ﬁeld, anyofitwere tobuild upinside themetal, there would bearateofchange ofﬂuxwhich would produce anelectric ﬁeld, andan electric ﬁeld would immediately generate acurrent which, byLenz‘s law, would oppose theﬂux. Since alltheelectrons willmove together, aninﬁnitesimal electric ﬁeld willgenerate enough current tooppose completely anyapplied magnetic ﬁeld. Soifyouturn theﬁeld onafter you‘ve cooled ametal tothesuperconducting state, itwillbeexcluded. 21-8 Even more interesting isarelated phenomenon discovered experimentally byMeissner.8 Ifyouhave apiece ofthemetal atahigh temperature (sothatitisa normal conductor) andestablish amagnetic ﬁeld through it,andthen youlower thetemperature below thecritical temperature (where themetal becomes asuper- conductor), theﬁeld ISexpelled. Inother words, itstarts upitsown current——and injusttheright amount topush theﬁeld out. Wecanseethereason forthatintheequations, andI'dliketoexplain how. Suppose that wetake apiece ofsuperconducting material which isinonelump. Then inasteady situation ofanykind thedivergence ofthecurrent must bezero because there’s noplace forittogo. Itisconvenient tochoose tomake the divergence ofAequal tozero. (Ishould explain why choosing thisconvention doesn't mean anylossofgenerality, butIdon’t want totake thetime.) Taking thedivergence ofEq.(21.18), then gives that theLaplacian of0isequal tozero. One moment. What about thevariation ofp? Iforgot tomention animportant point. There isabackground ofpositive charge inthismetal duetotheatomic ions ofthelattice. Ifthecharge density pisuniform there isnonetcharge andno electric ﬁeld. Ifthere would beanyaccumulation ofelectrons inoneregion the charge wouldn’t beneutralized andthere would beaterriﬁc repulsion pushing the electrons apart.T Soinordinary circumstances thecharge density oftheelectrons inthesuperconductor isalmost perfectly uniform—I cantake pasaconstant. Now theonly way that V26canbezero everywhere inside thelump ofmetal is for9tobeaconstant. And that means that there isnocontribution toJfrom p-momentum. Equation (21.18) then says that thecurrent isproportional top times A.Soeverywhere inalump ofsuperconducting material thecurrent is necessarily proportional tothevector potential: J=—p%A. (21.20) Since pandqhave thesame (negative) sign, andsince pisaconstant, Icanset pq/m =-(some constant); then J=—(some constant)A. (21.21) This equation wasoriginally proposed byLondon andLondon“ toexplain the experimental observations ofsuperconductivity—long before thequantum me- chanical origin oftheeffect wasunderstood. Now wecanuseEq.(21.20) intheequations ofelectromagnetism tosolve fortheﬁelds. Thevector potential isrelated tothecurrent density by 2__L vA_600,J. (21.22) lfluseEq.(21.21) forJ,Ihave VZA=VA, (21.23) where >12isjustanewconstant; 2= ‘I_X pGomez (21.24) Wecannow trytosolve thisequation forAandseewhat happens indetail. Forexample, inonedimension Eq.(21.23) hasexponential solutions oftheform e_)"‘ and 12+“. These solutions mean that thevector potential must decrease exponentially asyou gofrom thesurface into thematerial. (Itcan’t increase 8W.Meissner andR.Ochsenfeld, Naturwiss. 21,787(1933). 9H.London andF.London, Proc. Roy. Soc (London) A149, 71(1935); Physica 2, 341(1935). TActually iftheelectric ﬁeld were toostrong, pairs would bebroken upandthe “normal” electrons created would move intohelpneutralize anyexcess ofpositive charge. Still, ittakes energy tomake these normal electrons, sothemain point isthatanearly uniform density pishighly favored energetically. 21-9 >2(u (b)B Q B I l I I I -I Fig 21-3 (0)Asuperconducting cyl- inder ISomcignetic ﬁeld; (blthemagnetic field Bosufunction ofr.>because there would beablow up.) lfthepiece ofmetal isvery large compared to1/>\,theﬁeldonlypenetrates toathinlayer atthesurface—a layer about 1/it inthickness. Theentire remainder oftheinterior isfreeofﬁeld, assketched in Fig.21—3. Thisistheexplanation oftheMeissner effect. How bigisthedistance >1?Well. remember that r0,the“electromagnetic radius” oftheelectron (2.8 X10_] 3cm), isgiven by 2 1 _.& _n 4'rre0r(, Also, remember thatqinEq.(21.24) istwice thecharge onanelectron, so "o _q__81come’ q,. Writing pasq,N, where Nisthenumber ofelectrons percubic centimeter, wehave X2=81rNr0. (21.25) Forametal suchasleadthere areabout 3X1022atoms percm“, soifeach one contributed only oneconduction electron, 1/)1would beabout 2X10‘5 cm. That gives youtheorder ofmagnitude 21-7 Flux quantization TheLondon equation (21.21) wasproposed toaccount fortheobserved facts ofsuperconductivity including theMeissner effect. Inrecent times, however, there have been some even more dramatic predictions. Oneprediction made by London wassopeculiar thatnobody paid much attention toituntil recently. 1Wlllnowdiscuss it.Thistimeinstead oftaking asingle lump, suppose wetake aringwhose thickness islarge compared to1/>\,andtrytoseewhat would happen ifwestarted withamagnetic ﬁeldthrough thering, thencooled ittothesuper- conducting state, andafterward removed theoriginal source ofB.Thesequence of events issketched inFig21—4. lnthenormal statethere willbeaﬁeldinthebody oftheringassketched inpart(a)oftheﬁgure. When theringismade super- conducting, theﬁeld isforced outside ofthematerial (aswehave just seen). There willthen besome ﬂuxthrough thehole oftheringassketched inpart (b). Iftheexternal ﬁeldisnowremoved, thelines ofﬁeldgoing through theholeare “trapped” asshown inpart (c). The ﬂux<1>through thecenter can‘t decrease because 6/61 must beequal tothelineintegral ofEaround thering, which is zeroinasuperconductor. Astheexternal ﬁeldisremoved asuper current starts ﬂowing around thering tokeep theﬂuxthrough thering aconstant. (lt’s the oldeddy-current idea, onlywithzero resistance.) These currents will,however, allﬂow near thesurface (down toadepth 1/)\), ascanbeshown bythesame kind ofanalysis thatImade forthesolid block. These currents cankeep themagnetic ﬁeld outofthebody ofthering, andproduce thepermanently trapped magnetic ﬁeldaswell. Now, however, there isanessential difference, andourequations predict a surprising effect. Theargument 1made above that0must beaconstant inasolid block does notapplyfor aring, asyoucanseefrom thefollowing arguments. Well inside thebody oftheringthecurrent density J1Szero; soEq.(21.18) gives hve=qA. (21.26) Now consider what wegetifwetake thelineintegral ofAaround acurve I‘, which goes around thering near thecenter ofitscross-section sothat itnever getsnear thesurface. asdrawn inFig.21-5. From Eq.(21.26), h%V0-ds =q%A't/s (21.27) 21-10 Now youknow thatthelineintegral ofAaround anyloop isequal totheﬂux ofBthrough theloop f-A~ds=. yfve-as =§<1>. (21.28)Equation (21.27) thebecomes Thelineintegral ofagradient from onepoint toanother (sayfrom point 1topoint 2)isthedifference ofthevalues ofthefunction atthetwopoints. Namely, 2 [V@‘dS=0g'"01. l Ifweletthetwoendpoints 1and2come together tomake aclosed loopyoumight atﬁrstthink that02would equal 01,sothattheintegral inEq.(21.28) would be zero. That would betrueforaclosed loopinasimply-connected piece ofsuper- conductor, butitisnotnecessarily trueforaring-shaped piece. Theonlyphysical requirement wecanmake isthatthere canbeonlyonevalue ofthewavefunction foreachpoint. Whatever 0doesasyougoaround thering, when yougetback to thestarting point the0yougetmust givethesame value forthewave function 11/=~/5e"’- Thiswillhappen if9changes by21m,where nisanyinteger. Soifwemake one complete turnaround theringtheleft-hand sideofEq.(21.27) must beh-21rn. Using Eq.(21.28), Igetthat 21rnh=q. (21.29) Thetrapped ﬂuxmust always beaninteger times 21rh/q! Ifyouwould think ofthe ringasaclassical object with anideally perfect (that is,inﬁnite) conductivity, youwould think thatwhatever ﬂuxwasinitially found through itwould juststay there——any amount ofﬂuxatallcould betrapped. Butthequantum-mechanical theory ofsuperconductivity saysthattheﬂuxcanbezero, or21rh/q, or41rh/q, or61rh/q, andsoon,butnovalue inbetween. Itmust beamultiple ofabasic quantum mechanical unit. London“) predicted thattheﬂuxtrapped byasuperconducting ringwould bequantized andsaidthatthepossible values ofthefluxwould begiven byEq. (21.29) with qequal totheelectronic charge. According toLondon thebasic unitofﬂuxshould be21rh/q,, which isabout 4X10" gauss =cm2. Tovisual- izesuch aﬂux, think ofatinycylinder atenth ofamillimeter indiameter; the magnetic ﬁeldinside itwhen itcontains thisamount ofﬂuxisabout onepercent oftheearth’s magnetic ﬁeld. Itshould bepossible toobserve such aﬂuxbya sensitive magnetic measurement. In1961 such aquantized ﬂuxwaslooked forandfound byDeaver and Fairbank“ atStanford University andatabout thesame time byDoll and Nabauer” inGermany. Intheexperiment ofDeaver andFairbank, atinycylinder ofsuperconductor wasmade byelectroplating athinlayer oftinonaone-centimeter length ofNo. 56(1.3X10'3 cmdiameter) copper wire. Thetinbecomes superconducting below 3.8°K, while thecopper remains anormal metal. Thewirewasputina small controlled magnetic ﬁeld, andthetemperature reduced until thetinbecame superconducting. Then theexeternal source ofﬁeldwasremoved. You would 1°F.London, Superﬂui'd.r; John Wiley andSons, Inc.,New York, 1950, Vol.I,p.152 11B.S.Deaver, Jr.,andW.M.Fairbank, Phys. Rev.Letters 7,43(1961). 12R.DollandM.Nabauer, Phys. Rev.Letters 7,51(1961). 21-11B ii 1-1 l-1 H Ll h . (<1) ‘ﬁll l M B H 1 i ,ill‘ ii (b) ‘/ l 1l 1 l i ,Hi1ij)lH~, B i ,,, (c) Fig. 21-4. Aring inqmqgnetig ﬁeld: (ct)inthenormal state; (b)inthe superconducting state; (cloffer theex- ternol ﬁeld isremoved.—~s —\ /55?\\\ ii§§ls Fig. 21-5. The curve Finside o superconducting ring expect thistogenerate acurrent byLenz’s lawsothattheﬂuxinside would not change. Thelittlecylinder should nowhavemagnetic moment proportional tothe ﬂuxinside. Themagnetic moment wasmeasured byjiggling thewire upanddown (like theneedle onasewing machine, butattherate of100cycles persecond) inside apairoflittle coils attheends ofthetincylinder. Theinduced voltage in thecoils wasthen ameasure ofthemagnetic moment. When theexperiment wasdone byDeaver andFairbank, they found thatthe ﬂux wasquantized, butthat thebasic unitwasonly one-half aslarge asLondon hadpredicted. Doll andNabauer gotthesame result. Atﬁrstthiswasquite mys- terious,‘|' butwenow understand whyitshould beso.According totheBardeen, Cooper, and Schrielfer theory ofsuperconductivity, theqwhich appears inEq. (21.29) isthecharge ofapair ofelectrons andsoisequal to2q,. Thebasic ﬂux unitis <1»,=3;’~2><10-’gauss-cm (21.30) orone-half theamount predicted byLondon. Everything, now ﬁtstogether, and themeasurements show theexistence ofthepredicted purely quantum-mechanical effect onalarge scale. 21-8 Thedynamics ofsuperconductivity TheMeissner effect andtheﬂuxquantization aretwoconﬁrmations ofour general ideas. Justforthesake ofcompleteness Iwould liketoshow youwhat thecomplete equations ofasuperconducting ﬂuid would befrom thispoint of view—it israther interesting. Uptothispoint Ihave onlyputtheexpression for upintoequations forcharge density andcurrent. IfIputitintothecomplete Schrodinger equation Igetequations forpand0.ltshould beinteresting tosee what develops, because herewehave a“ﬂuid” ofelectron pairs with acharge density pandamysterious 0—we cantrytoseewhat kindofequations wegetfor sucha“ﬂuid”! Sowesubstitute thewave function ofEq.(21.17) intotheSchro- dinger equation (21.3) andremember thatpand0arerealfunctions ofx,y,and z.Ifweseparate realandimaginary parts weobtain then twoequations. To write them inashorter form Iwill—fol1owing Eq.(21.19)——write -'1v0-flA=v. (21.31)m m Oneoftheequations 1getisthen 6p_ _It-Vpv. (21.32) Since pvisﬁrstJ,thisisjustthecontinuity equation once more. Theother equation Iobtain tellshow0varies; itis 66_ 9'1 2 _h2 VL 2 _ 115- *fl’ ‘l'q¢ 271{\/;V(\/.5): (21-33) Those who arethoroughly familiar with hydrodynamics (ofwhich l’msure few ofyouare)willrecognize thisastheequation ofmotion foranelectrically charged ﬂuid ifweidentify hﬂasthe“velocity potential”—except thatthelastterm, which should betheenergy ofcompression oftheﬂuid, hasarather strange dependence onthedensity p.Inanycase, theequation saysthatrateofchange ofthequantity h6isgiven byakinetic energy term, %mv2, plus apotential energy term, q¢,with anadditional term, containing thefactor hz,which wecould calla“quantum mechanical energy.” Wehave seen that inside asuperconductor piskept very TIthasonce been suggested byOnsager thatthismight happen (seeF.London, Ref. 10),although nooneelseeverunderstood why. 21-12 uniform bytheelectrostatic forces, sothisterm canalmost certainly beneglected inevery practical application provided wehave only onesuperconducting region. Ifwehave aboundary between twosuperconductors (orother circumstances in which thevalue ofpmay change rapidly) thisterm canbecome important. Forthose who arenotsofamiliar with theequations ofhydrodynamics, Icanrewrite Eq.(21.33) inaform thatmakes thephysics more apparent byusing Eq.(21.31) toexpress Binterms ofv.Taking thegradient ofthewhole ofEq. (21.33) andexpressing V0interms ofAandvbyusing (21.31), Iget aA hz 9-g'=%<—V¢—w)-v><(V><v)—(v><V)u— Vﬁ(\/L5v—\/p). (21.34) What does thisequation mean? First, remember that -v¢‘- %§=E. (2L35) Next, notice thatifItakethecurlofEq.(21.19), Iget vXv=—%vXA cum since thecurlofagradient isalways zero. ButVXAisthemagnetic ﬁeldB, sotheﬁrsttwoterms canbewritten as 1m(E+vXB). Finally, you should understand that 6v/6t stands fortherate ofchange ofthe velocity oftheﬂuid atapoint. lfyouconcentrate onaparticular particle, its acceleration isthetotal derivative ofv(or,asitissometimes called inﬂuid dy- namics, the“comoving acceleration”), which isrelated toav/6t by” d a%mmm=£+@yp amp This extra term alsoappears asthethird term ontheright sideofEq.(21.25). Taking ittotheleftside,Icanwrite Eq.(21.25) inthefollowing way: mg =qw+vxBy-v@<#vW@- ammdt COm0vii|g 2 Wealsohave from Eq.(21.36) that vxv=-in am»m These twoequations aretheequations ofmotion ofthesuperconducting electron ﬂuid. Theﬁrstequation isjustNewton’s lawforacharged ﬂuid inan electromagnetic ﬁeld. Itsaysthattheacceleration ofeach particle oftheﬂuid whose charge isqcomes from theordinary Lorentz force q(E+vXB)plus an additional force, which isthegradient ofsome mystical quantum mechanical potentia1——a force which isnotverybigexcept atthejunction between twosuper- conductors. Thesecond equation saysthattheﬂuidis“ideal"—the curlofvhas zerodivergence (thedivergence ofBisalways zero). That means thatthevelocity canbeexpressed interms ofvelocity potential. Ordinarily onewrites that VX v=0foranideal ﬂuid, butforanidealcharged ﬂuid inamagnetic ﬁeld, thisgets modiﬁed toEq.(21.40). ' So,Schrodinger’s equation fortheelectron pairs inasuperconductor gives ustheequations ofmotion ofanelectrically charged idealﬂuid. Superconductivity isthesame astheproblem ofthehydrodynamics ofacharged liquid. Ifyouwant 13SeeVolume II,Section 40—2. 21-13 INSULATOR/ >\§‘“\*§\ \pp \. \ 2 ( .\ sf: \\\\<\\»~s ssx 2.>;55#rs ss; \\§si\Q\ .;.\§\\,\, \\\ T;.5. \&i\\\%;l\\ \€>J 23422’. \ /SUPERCONDUCTOR Fig.21-6. Twosuperconductors sep- arated byathininsulator.tosolve anyproblem about superconductors youtake these equations forthe ﬂuid [ortheequivalent pair, Eqs. (21.32) and(2l.33)], andcombine them with Maxwell's equations togettheﬁelds. (The charges andcurrents youusetoget theﬁelds must, ofcourse, include theones from thesuperconductor aswellas from theexternal sources.) Incidentally, Ibelieve thatEq.(21.38) isnotquite correct, butought tohave anadditional term involving thedensity. This newterm does notdepend on quantum mechanics, butcomes from theordinary energy associated with varia- tions ofdensity. Justasinanordinary ﬂuid there should beapotential energy density proportional tothesquare ofthedeviation ofpfrom po,theundisturbed density (which is,here, alsoequal tothecharge density ofthecrystal lattice). Since there willbeforces proportional tothegradient ofthisenergy, there should beanother term inEq.(21.38) oftheform: (const) V(p—p0)2. Thisterm did notappear from theanalysis because itcomes from theinteractions between parti- cles, which Ineglected inusing anindependent-particle approximation. Itis, however, justtheforce Ireferred towhen Imade thequalitative statement that electrostatic forces would tendtokeeppnearly constant inside asuperconductor. 21-9 TheJosephson junction Iwould liketodiscuss next avery interesting situation that wasnoticed by Josephson 14while analyzing what might happen atajunction between twosuper- conductors. Suppose wehave two superconductors which areconnected bya thin layer ofinsulating material asinFig. 21-6. Such anarrangement isnow called a“Josephson junction.” Iftheinsulating layer isthick, theelectrons can’t getthrough; butifthelayer isthinenough, there canbeanappreciable quantum mechanical amplitude forelectrons tojump across. This isjustanother example ofthequantum-mechanical penetration ofabarrier. Josephson analyzed this situation anddiscovered thatanumber ofstrange phenomenon should occur. Inorder toanalyze such ajunction I’llcalltheamplitude toﬁndanelectron ononeside, (01,andtheamplitude toﬁnditontheother, 11/2.Inthesuperconduct- ingstate thewave function, thyisthecommon wave function ofalltheelectrons ononeside, andt//2isthecorresponding function ontheother side. Icould do thisproblem fordifferent kinds ofsuperconductors, butletustake avery simple situation inwhich thematerial isthesame onboth sides sothat thejunction is symmetrical andsimple. Also, foramoment letthere benomagnetic ﬁeld. Then thetwoamplitudes should berelated inthefollowing way: it95=I/it/1+Kn, dt 9:72 = I/21,02 + Theconstant Kisacharacteristic ofthejunction. IfKwere zero, these two equations would justdescribe thelowest energy state——with energy U—-of each superconductor. Butthere iscoupling between thetwosides bytheamplitude K thatthere maybeleakage from onesidetotheother. (Itisjustthe“ﬂip-ﬂop” amplitude ofatwo-state system.) Ifthetwosides areidentical, U1would equal U2andIcould justsubtract them off.Butnowsuppose thatweconnect thetwo superconducting regions tothetwoterminals ofabattery sothatthere isapo- tential difference Vacross thejunction. Then U1——U2=qV.Ican,forcon- venience, dcﬁne thezeroofenergy tobehalfway between, thenthetwoequations are lh%1 =gzillti -1-K11/2, (2r40) .% __qV 171at— 7IP2-1-Kiln- 14B.D.Josephson, Physics Letters 1,251(1962). 21-14 These arethestandard equations fortwoquantum mechanical states coupled together. Thistime, let’sanalyze these equations inanother way. Let’s make the substitutions ‘pl : \/;Te10‘s d/2 : VP261029 where 6land02arethephases onthetwosides oftheJunction andplandpg arethedensity ofelectrons atthose twopoints. Remember thatinactual practice plandp2arealmost exactly thesame andareequal topo,thenormal density of electrons inthesuperconducting material. Now ifyousubstitute these equations for1//land1&2into(21.40), yougetfourequations byequating therealandimaginary parts ineach case. Letting (02—6l)=5,forshort, theresult is(21.41) Z . £51=+EKVP291 $1115, (21.42) 2 . $52=""§K\/P2915195, K I V61=‘|-'5 %COS5—%T . KT Ve2=+-,;\/%¢osa+-‘;7-(21.43) Theﬁrsttwoequations saythatpl=—p2. “But,” yousay,“they must both bezeroifplandp2arebothconstant andequal tozero.” Notquite. These equations arenotthewhole story. They saywhat plandpgwould bezfthere were noextra electric forces duetoanunbalance between theelectron ﬂuid and thebackground ofpositive ions. They tellhowthedensities would starttochange, andtherefore describe thekindofcurrent thatwould begin toﬂow. Thiscurrent from side1toside2would bejustpl(or —p2), or 21<4_.J=7;\/plpz Slna. (21.44) Such acurrent would soon charge upside2,except thatwehave forgotten that thetwosides areconnected bywires tothebattery. Thecurrent thatﬁows will notcharge upregion 2(ordischarge region 1)because currents willﬂow tokeep thepotential constant. These currents from thebattery have notbeen included inourequations. When theyareincluded, plandp2donotinfactchange, but thecurrent across thejunction isstillgiven byEq.(21.44). Since plandp2doremain constant andequal top0,let’sset2Kp(,/h =J0, andwrite .1=J0sin6. (21.45) J0,likeK,isthen anumber which isacharacteristic oftheparticular _]Lll1Cl£10I1. Theother pairofequations (21.43) tellsusabout Bland62.Weareinterested inthedifference 5=02—BltouseEq.(21.45); what wegetis That means thatwecanwrite 6(1)=al,+gfV(t)dr, (21.47) where 60isthevalue of6att=0.Remember alsothatqisthecharge ofapair, namely, q=2q,,. InEqs. (21.45) and(21.47) wehave animportant result, the general theory oftheJosephson junction. 21-15 Now what aretheconsequences? First, putonadcvoltage. Ifyouputona dcvoltage, V0,theargument ofthesinebecomes (50+(q/h)V (,1)Since hisa small number (compared toordinary voltage andtimes), thesineoscillates rather rapidly andthenetcurrent isnothing. (Inpractice, since thetemperature isnot zero, youwould getasmall current duetotheconduction by“normal” electrons.) Ontheother hand ifyouhave zerovoltage across thejunction, youcangetacur- rent! With novoltage thecurrent canbeanyamount between +J,, and —Jl, (depending onthevalue of60).Buttrytoputavoltage across itandthecurrent goes tozero. This strange behavior hasrecently been observed experimentally.“ There isanother wayofgetting acurrent——by applying avoltage atavery highfrequency inaddition toadcvoltage. Let V=Vt)+1/coswt, where v<<V.Then 5(2)is 61) + glV“, + £511] (DI. Now forAxsmall, sin(x-1-Ax) zsinx +Axcosx. Using thisapproximation forsin6,lget J=J0[sin(60-1-2Vot) +g sinwtcos(6,,+gVlltn» Theﬁrstterm iszero ontheaverage, butthesecond term isnotif =QOJ There should beacurrent iftheacvoltage hasjust thisfrequency. Shapiro“ claims tohave observed sucharesonance effect. Ifyoulook uppapers onthesubject youwillﬁndthat they often write the formula forthecurrent as J= 1./Q SlI1<6() + '2%£ '0,3)’ where theintegral istobetaken across thejunction. Thereason forthisisthat when there's avector potential across thejunction theﬁip-ﬂop amplitude is modiﬁed inphase intheway that weexplained earlier Ifyouchase that extra phase through, itcomes outasgiven above. Finally, Iwould liketodescribe avery dramatic andinteresting experiment which hasrecently been made ontheinterference ofthecurrents from each of twojunctions. Inquantum mechanics we’re used totheinterference between amplitudes from twodifferent slits. Now we’re going todotheinterference be- tween twojunctions caused bythediflerence inthephase ofthearrival ofthe currents through twodifferent paths. InFig.21-7, Ishow twodiﬁerent junctions, “a”and“b”, connected inparallel. Theends, PandQ,areconnected toourelec- trical intruments which measure anycurrent ﬂow. Theexternal current, Jlolll, willbethesumofthecurrents through thetwojunctions. LetJ,andJl,bethe currents through thetwojunctions, andlettheir phases be6,,and5h.Now the phase difference ofthewave functions between PandQmust bethesame whether yougoononeroute ortheother. Along theroute through junction “a”,thephase diﬂerence between PandQis6,plusthelineintegral ofthevector potential along theupper route: APhasep.,Q =5..+3,?IA-ds. (21.49)upper 15P.W.Anderson andJ.M.Rowell, Phys RevLetters 10,230(1963). ‘GS.Shapiro, Phys. Rev. Letters 11,80(1963). 21-16 LOOP I" 0/INSULATOR / ////// ///,,(_//.//fj/f.4,/L///rt//tAI/4’'.7//I/--v/areJTOTAL §'—~P. —~ SUPERCONDUC OR Why‘? Because thephase 6isrelated toAbyEq.(21.26). Ifyouintegrate that equation along some path, theleft-hand sidegives thephase change, which isthen justproportional tothelineintegral ofA,aswehave written here. The phase change along thelower route canbewritten similarly APhﬂSCp_,Q =al.+Q A~ds. (21.50)h lower These twomust beequal; andiflsubtract them Igetthat thedifference ofthe deltas must bethelineintegral ofAaround thecircuit: Zt5,4a,=-%}€_A-(ts. I-Iere theintegral isaround theclosed loop I‘ofFig.21-7 which circles through both junctions. Theintegral over Aisthemagnetic ﬂuxthrough theloop. So thetwo 5’saregoing todiffer by2q,_,//t times themagnetic ﬂux¢I>which passes between thetwobranches ofthecircuit: ab-a,=2%<1». (21.51) 1c'ancontrol thisphase diﬁerence bychanging themagnetic ﬁeld onthecircuit, soIcanadjust thedifferences inphases andseewhether ornotthetotal current that ﬂows through thetwojunctions shows anyinterference ofthetwo parts. Thetotal current willbethesum ofJ,,and.11..Forconvenience, IwillWrite a,=a,l+9£"<r>, ab=5.,-%<t>. Then, J(,,l,,1 =./(j{S1f1(t§() + Q)-1-sin(60—E25)} . =J0sin60cos31;! (21.52) Now wedon’t know anything about 60,andnature canadjust that anyway shewants depending onthecircumstances. Inparticular, itwilldepend onthe external voltage weapply tothejunction. Nomatter what wedo,however, sinél, cannever getbigger than l.Sothemaximum current foranygiven isgiven by Jiiiax : JO COS gil- This maximum current willvary with andwillitself have maxima whenever lznﬂ.‘It with nsome integer. That istosaythatthecurrent takes onitsmaximum values where theﬂuxlinkage hasjustthose quantized values wefound inEq.(2l.30)' 21-17Fig. 21-7. Two Josephson junctions inporollel. Fig. 21-8. Arecording ofthecurrent through cipair ofJosephson junctions aso function ofthemagnetic ﬁeld intheregion between thetwojunctions (see Fig.21-7). [This recording was provided byR.C. Jaklevic, J.Lambe, A.H.Silver, and J.E. Mercereou oftheScientiﬁc Laboratory, Ford Motor Company]JOSEPHSONCURRENTARBTARYUNT)S R §-I I l I l l 1 I l I I l -500 -400 -300 -200 -IOO O IOO ZOO 300 400 500 MAGNETIC FIELD (MILLIGAUSSI The Josephson current through adouble junction wasrecently measurcdlf asafunction ofthemagnetic ﬁeld inthearea between thejunctions. Theresults areshown inFig. 21-8. There isageneral background ofcurrent from various effects wehave neglected, buttherapid oscillations ofthecurrent with changes in themagnetic ﬁeld areduetotheinterference term cosqe4>/h ofEq.(21.52). One oftheintriguing questions about quantum mechanics isthequestion of whether thevector potential exists inaplace where there’s noﬁeld. 18This experi- ment Ihave just described hasalso been done with atiny solenoid between the twojunctions sothat theonly signiﬁcant magnetic Bﬁeld isinside thesolenoid andanegligible amount isonthesuperconducting wires themselves. Yetitis reported thattheamount ofcurrent depends oscillatorily ontheﬁuxofmagnetic ﬁeldinside thatsolenoid even though thatﬁeldnever touches thewires—another demonstration ofthe“physical reality” ofthevector pOI€I'1[lHl.]9 Idon’t know what willcome next. Butlook what canbedone. First, notice thattheinterference between twojunctions canbeused tomake asensitive mag- netometer. Ifapair ofjunctions ismade with anenclosed area of,say, lmmz, themaxima inthecurve ofFig.21-8 would beseparated by2Xl0_" gauss. It iscertainly possible totellwhen youarel/10 oftheway between twopeaks; so itshould bepossible tousesuch ajunction tomeasure magnetic ﬁelds assmall as 2><IOT7 gauss—or tomeasure larger ﬁelds tosuch aprecision. One should be able togoeven farther. Suppose forexample weputasetof10or20junctions close together andequally spaced. Then wecanhave theinterference between 10or20slitsandaswechange themagnetic ﬁeld wewillgetvery sharp maxima andminima. Instead ofa2-slit interference wecanhave a20-orperhaps even a 100-slit interferometer formeasuring themagnetic ﬁeld. Perhaps wecanpredict that themeasurement ofmagnetic ﬁelds will-by using theeffects ofquantum- mechanical interference—eventually become almost asprecise asthemeasurement ofwavelength oflight. These then aresome illustrations ofthings that arehappening inmodern times—the transistor, thelaser, andnow these junctions, whose ultimate practical applications arestillnotknown. Thequantum mechanics which wasdiscovered in1926 hashadnearly 40years ofdevelopment, andrather suddenly ithasbegun tobeexploited inmany practical andrealways. Wearereally getting control of nature onavery delicate andbeautiful level. Iamsorry tosay,gentlemen, thattoparticipate inthisadventure itisab- solutely imperative thatyoulearn quantum mechanics assoon aspossible. Itwas ourhope thatinthiscourse wewould ﬁndawaytomake comprehensible toyou attheearliest possible moment themysteries ofthispartofphysics 17Jaklevic, Lambe, Silver, andMercereau, Phys. Rev Letters 12,159(1964). 18Jaklevic, Lambe, Silva. andMercereau, Phys Rev.Letters 12,274(1964). 19SeeVolume II,Chapter 15,Section 15-5. 21-18 Feynman ’sEpilogue Well, I’vebeen talking toyoufortwoyears andnow I’mgoing toquit. In some ways Iwould liketoapologize, andother ways not. Ihope-—in fact, Iknow- that two orthree dozen ofyouhave been able tofollow everything with great excitement, andhave hadagood time with it.ButIalsoknow that“the powers of instruction areofvery little etﬁcacy except inthose happy circumstances inwhich they arepractically superﬂuous.” So,forthetwoorthree dozen who have under- stood everything, may IsayIhave done nothing butshown youthethings. For theothers, ifIhavemade youhatethesubject, I’msorry. Inever taught elementary physics before, andIapologize. Ijusthope thatIhaven’t caused aserious trouble toyou,andthatyoudonotleave thisexciting business. Ihope thatsomeone else canteach ittoyouinawaythatdoesn’t giveyouindigestion, andthatyouwill ﬁndsomeday that, after all,itisn’tashorrible asitlooks. Finally, may Iaddthatthemain purpose ofmyteaching hasnotbeen to prepare youforsome examination—it wasnoteven toprepare youtoserve in- dustry orthemilitary. Iwanted most togiveyousome appreciation ofthewonder- fulworld andthephysicist’s wayoflooking atit,which, Ibelieve, isamajor part ofthetrueculture ofmodern times. (There areprobably professors ofother sub- jects who would object, butIbelieve thatthey arecompletely wrong.) Perhaps youwillnotonlyhave some appreciation ofthisculture; itiseven possible thatyoumaywant tojoininthegreatest adventure thatthehuman mind haseverbegun. 21-19