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Quantum Mechanics and Atomic Physics 1

by: Mrs. Peter Toy

Quantum Mechanics and Atomic Physics 1 PHYS 3220

Marketplace > University of Colorado at Boulder > Physics 2 > PHYS 3220 > Quantum Mechanics and Atomic Physics 1
Mrs. Peter Toy

GPA 3.96

Michael Dubson

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Michael Dubson
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This 41 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 3220 at University of Colorado at Boulder taught by Michael Dubson in Fall. Since its upload, it has received 35 views. For similar materials see /class/232085/phys-3220-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.


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Date Created: 10/30/15
SJP QM 3220 Formalism l The Formalism 0f 5 Quantum Mechanics Our story so far I State of physical system normalizable I X t Observables operators 3c f7 if I Dynamics of I TDSE 171 W To solve lst solve TISE ll Ezp Solutions are stationary states nX En gt special solutions of TDSE 1 96 t wnxequotEquot39quot TDSE linear gt any linear combo of solutions is also a solution Discrete case l11x t Ec e39iEquot39h1p x n 123 1110c t fdk ke39iwk391w Continuum case k any real number 2 1271 A A Discrete Discrete Continuum Discrete Continuum n39s k39s form complete orthonormal sets fdx winJ 6m 9 1 dx 7 dxe k39kx 5 k k f w 2 f lt gt Page Fl M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 Notice similarity of 1 s to vectors Vector 7 complex function I x t Scalar real number a complex number c Any linear combination of vectors is a vector CaAbB 1Pa1111 l 2 Orthonormal basis vectors 221 amp 0 M J u wdx6m VVxxVyyVz TEcnwn Vx5cl7 c fdx1p111 Innerproduct 31 EAiBi f dwb HM fdxlgdmwmlilZW gagfmaw 9 The space of all complex squareintegrable functions I x is called Hilbert Space Norm flp111dxltoo Hilbert Space is an in nitedimensional vector space with complex scalars and normalizable vectors Page F2 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Forrnalism l 4905 tu a te 1 Every possible physical state of a system corresponds to a normed vector in Hilbert Space The correspondence is ltol except that vectors that differ by a phase factor scalar of modulus 1 corresponds to the same state I x t 4gt e16 I x t Dirac Notation defx gx fig complex number flg glfY is real non negative gt flc g Cflg vflg 5mg c any complex number 4905 tu a te 2 to be stated shortly associates with every observable a linear hermitean operator But first a little background De nition An operator Q is hermitean or hermitian both spellings are common if f Qg Qf Which can be written in position representation as fdx f Qg fdxQf g g for all f g in Hilbert space H space Question Is the operator Q di Hermitian The answer will be no x Let s see why 9 fdxf fdx g pam m d 9 df x x 7 xdx 7 xdx ffm f fg g So the answer is NO there s an extra unwanted minus sign that cropped up It is NOT the Qggt Qf g and that means Q is NOT hermitian case that for this particular operator that lt f g Instead we found here f ltQf By the way the surface turm in our integration by parts gave me zero because f and g belong to Hilbert space and thus should vanish off at infinity so they re normalizable Page F3 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 Question Is the operator Q hermitian Answer is Yes Let s see why 1 x hag ha 9 g quot ff Emir 4E llobelng gt f QgltQf g Question Is the operator Q C hermitian if c is some constant flcgcflg 0 cmg mg It depends This operator is hermitian only if c is Q Why are hermitean operators special Why only hermitean operators associated with physical observables Because hermitean operators produce Lal eigenvalues and measurements of observables always produce gal values Eigenvalue equation Q fx q fx eigenfunction K eigenvalue Z eorem The eigenvalues of a hermitean operator Q are real Proof Assume x is an eigenfunction Qf qf f gt where Q is hermitean gem 3 EH qltflfgt q f qltffgt gtqqflf0gtqq since flf0 Page F4 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism l Q eorem The eigenfunctions of a hermitean operator with distinct different eigenvalues are orthogonal Proof Given QfOC q39 fx Q gOC qquotgx With q g L f Qg ltQf g since Qhermitean a q f lg qf lg 61mg q real gt q39 qf g 0 gt 0 since q39 q 0 by assumption G os tu a ta 2 Operators Observables This is a long postulate with 3 parts and many texts break this up into 2 or 3 postulates 1 For every physical observable Q X p E etc there corresponds a linear hermitean operator Q in the Hilbert Space which possesses a complete orthonormal set of eigenfunctions fn X and the corresponding eigenvalues qn Qf x q f x n could be discrete or continuous 2 The only possible results of a measurement of Q are one of the eigenvalues qnq1 qz q3 A h a 3 The momentum operator is p z x The position operator is 3c x Any function Q X p has operator Q Q0113 An example of Q Q0113 is the energy operator or Hamiltonian 71 gtvltxgtltgt A2 A I A h2 if 7 Vx 7 2m 2m Solutions of ll E zp form an orthonormal set since if is hermitian That wn39s form a complete set can be proven in some special cases like the infinite square well or SHO but in general completeness is taken as a postulate If energy is measured the only possible result is one of the Bus gtllt Page F5 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism l The momentum eigenstates are solutions of fp x p fp x wheref z x Eigenfunctions f p x Aeikx any constant A Eigenvalues p hk k any real number h d ikx ikx Proof faAe hkAe That s1t t In this case the eigenvalues of p form a continuum m real value of p hk is permitted and this leads to some mathematical subtleties Are the fp s orthonormal Ans kind of yes ffx fpoc dx 6p p39 quotDelta function orthogonalityquot fpxAei xh phk any realk or AdjustAsothat ltfpvl 3gt5pp39 fp39 ifp A 2 fdx expiry 61 p 7 Az Hm using Bcx x gt21f75PP39 c 1 6 112nm Question are the fp39s a complete set ipx7 Thus mg awrealpm orm Fourier Analysis Planchevel39s Theorem says that any fX can be written f x fdk Fk eW where H Fk fdxfx a Page F6 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Forrnalism 1 Now any I X t is a function of X at arbitrary t and thus from the previous page 1 1P xt 7 dk CI kt 2 where 17 f k ltIgtkk t fdx 11mm 6quot A quick change ofvariables k gt p hk dk dph leads us to de ne CI kt CI t ki where hk I7 4g I7 Putting it all together 1 At7 1P xt 7 d CI t e m f p p 1 CI t p m lPOM fdp 61gtp tfpx gt fp is are complete fdx lPc t Wm Note Previously we wrote similar relations when I X t was a free particle state V 0 But m function I X t can be Fourier analyzed In the special case of free particle then 2 CIgtkk t kequot quot where a 2m but this particular simple timedependence in Ik t is true only for the special case of a free particle Ip t is called the momentumspace wave function It is the Fourier transform of I Xt and tells quothow much p hk hAquot is in I Ip t contains all the same info as I Xt gtllt Page F7 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Forrnalism 1 We re ready now to restate Postulate 3 Previously Postulate 3 was stated as Prob nd position in X gt X dX l I 2 dX Our restatement will look very different but will be same 4905 tu a te 3 If a system is in state I Xt and a measurement of observable Q is made on the system where the corresponding operator Q has eigenfunctions fnX and eigenvalues q Qf bc q f x then the strongest predictive statement that can be made about the result of that measurement is Prob measure qn l lt fn l I gt 2 discrete spectrum If spectrum is continuous Q f4 x q f4 x with m real value of q then Probmeasureqn gtqdq lltfqlIgtlz dq Example of Postulate 3 Suppose a system has discrete energy eigenvalues ip E zp n 1 2 3 and your system is in a state that is a linear combo 45quot MM 2cm tux 2c e 411w 6 1 c e39iEquot3939 where c fdx 1011J then a measurement of energy will yield value En with probability lltnl 1 gtlzl0nlz Page F8 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 If the system is already in a particular single pure eigenstate n0 Wm mm 245 gt 45quot lI xt EC 2 A 1p x where c n 1 and all other cn s 0 then measurement of energy will yield Eno with probability Ic n 2 1 This means an eigenstate of energy is state of de nite energy Look back think about it convince yourself The formalism can look opaque but if you have an eigenstate of H there is only ONE TERM in our expansion and there is thus only one possible result when you measure energy A similar argument applies to any observable an eigenstate of Q is a state of de nite Q and the value of Q the eigenvalue of the eigenstate Previously we asserted that the expectation value of Q 11 We ve used this in the integral form many times Wgt q gem quot c qnzqn We can now show this follows from Postulate3 1P Hermiticity completeness M1 1 em lew egg 661 n n Eq IC 2 Eq Prob q n weighted average of qn39s This is what you would mean by the expectation value of measurements of Q Page F9 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism l Eigenstates of 1 states of defmite energy Eigenstates of f7 states of def39mite momentum Eigenstates of 2 states of definite position Let s look at these position eigenstates ng x P Must be delta function Ing 00 xogxu 6 variable X particular X X0 fcx gtxgxx0gxgtx x0gx0 X0 gt gX is zero everywhere except at X X0 gt ng x 5 x x0 Notation gxoX Postulate 3 says Prob X0 gt X0 dX l ltX0 l 1 Kat gtl2 dX fdx 5x x01pxtl2dx l1Px0 tl2dx agrees with our previous version of Postulate 3 what we ve been using all along G os tu a ta 4 Wave function collapse If a measurement of observable Q gives result q then the wavefunction instantly collapses into the corresponding eigenfunction of Q fn X Discrete spectrum eXample l11x t EC t 1p x EC e39iEquot39h1p x where 1p x is an eigenstate of 1 If measure energy and if nd E E n gt 1p 3 WAX and the new lIJXt g iEm39p39 collapse w u x Page FlO M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 Continuous spectrum example 6 ipxh L11xrfdplt1gtpr m fdpcmmmm x momentum erstates lCIgtp0lA2 is the probability density telling you the probability if you measure momentum that you will get p within p0 gt p0 Ap No measurement of continuous variable has in nite precision Precision Ap depends on measurement This means that in practice collapse is to a normalizable I that is almost an eigenstate fpn x MILO After collapse gt AP M P0A JAAVAVWVWV AxApn uncertainty principle Probp gt p dp lltfhxwxtgt2dp lltIgtpot2dp Page Fll M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 all other Hilbert Space Function nonnormalizable fcns Normalizable fcns Space igenfunctions of Hermitean operators w continuous eigenvalues eg X sol ns of SE that are non normalizable but useful quotSuburbs of Hspacequot quotNearly normalizable 8function normalizablequot To make our list of postulates complete Q os tu a te 4 Schrodinger Equation The time evolution of the wavefunction I Xt is determined by the TDSE 93m 171 31 A 2 H 177 Vx To solve TDSE Separation of variables gt special solutions lI x t Edh 1P x E 39s 1pquot X s are from TISE if x E 1Px General solution to TDSE tux r Eamon Ege mn moo c c t 0 f 1pIIJx0dx 1 1px0 n39s form complete orthonormal set since 1 is hermitean Any hermitean operator associated with an observable has a complete orthonormal set of eigenfunctions but the energy eigenfunctions are special in that they provide the time dependence of I Xt Page F12 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Forrnalism 1 Comment about probabilities and normalization consider normalized basis states nX and wnormalized l c11 c2 z Note that lt I I gt l c1 2 lcg 2 i 1 In this case postulate 3 should read 2 2 lql WW Prob ndq 1 l61l2lczl2 W You must diVide by lt I I gt for probabilities to add up to l E Kw lz E E Prob ndq quot n721 W E c n n 2 C n Review to this point System I Xt Measure Q Qf x q f x Post3 Find qn with Prob f xl1P 2 Post4 I Xt gt fn X collapse ltf xl1P is quotprojection of I ontofn Xquot Euclidean space R 501 projection of I along 3c Page Fl3 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Forrnalism l Hilbert Space is a complex in nitedimensional vector space Basis states n from y E zp Any wavefunction 1P 26311 ll W3 11 W2 complex numbers 1 V all perpendicular lt I n I m gt 5am 4 2 A 1 1 1 02 W2 02 a quotpoetic representationquot ltIz l Tgt gt 1 H c1 ltw1 l Pgt 1P 2631 x cn tells how much of I is along w aXis in Hspace 1P 1p x 01 Page Fl4 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 We have now seen 3 different equivalent ways to represent the wavefunctions 1mm ltIgtpt 6 11mm xl P ltIgtpr pl c WM cn looks different than functions I and I but not really cn is an in nite set of numbers that associate a number cn with a quotcoordinatequot n Likewise I Xt is 00 set of numbers that associate a number I X with a coordinate X Multiple ways to represent the quotstatequot of the system like multiple ways to represent an ordinary vector V VxVyVz VxVyVz VV9Vq There is a vector 7 which eXists independent of its representation in any particular basis Likewise there is a quotstate vectorquot l S gt or l I gt which eXists quotout therequot in an abstract Hilbert space independent of any representation Dirac39s Notation abstract state vector l I gt or l S gt to avoid confusion with I X l I gt is called a quotketquot because it is the right hand side ofa quotbracketquot lt w l I gt When can two different operators have simultaneous eigenfunctions Answer to be shown When they commute Recall Definition Commutator of operators A and I A E an operator So 2 operators commute if their commutator is zero A3 ltgt 1 0 Why would we care if there are states that are simultaneously eigenfunctions of 2 operators A and B Recall eigenfunction of A is a state of definite A so eigenfunction of both A and I state of definite A and B Example 13x z Bx Bx z muf lxi 1xfjf B xi xi Bx Bx 3x HP 1 Operate on arbitrary state fX Z This is true for any f so 13x if l Page F15 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 But 1 0 213o 1313o So it is possible to have a state that is simultaneously a state of de nite X and de nite py W gt V AND X both allowed simultaneously subject to usual caveat about nonrenormalizable states But it is NOT possible to have simultaneous eigenstates of 3c and x X M This is veg different from the classical situation X px m vx lt can have wellde ned precise values of X AND px In QM if we start with a state of de nite px 1y W and we measure X then 11 collapses to a state of de nite X 111 h and the momentum information is destroyed Page Fl6 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 Theorem If A E 0 then there exist simultaneous eigenfunctions of A and 1A3 Proof Given 11 such that At mp 31p bzp same 1p assume that A w a w assume that w is a nondegenerate eigenfunction of A We39ll relax this condition later 11 nondegenerate eigenfunciton of A means that only 11 and multiples of w c w are eigenfunctions No other linearly independent eigenstates eXist Now operate with 13 on both sides ofA 1y a w lg A 1pl a 1p al 1psincel islinearop Asz al 1p gt 1 1p is also eigenstate ofA But assumed eigenstate of A nondegenerate gt lg zp is a multiple of 1p gt 31p bzp for some b Done So 11 is a state of definite A value a and a state of definite B value b It can be shown that 13x 139th x this would be a straightforward HW problem 51gt ifV 0 constant free particle then 3 V 3 X 0 and in this case if px 0 gt it IS possible to have states of definite energy and definite momentum Easy we ve seen this 1px Ae lkx39w39 A A 21sz pwhkw WW 7112 2m But only true for fr particle If any VX i const present then eigenstates of 1 are not p eigenstates Page Fl7 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 Now that we have some familiarity with commutation relations we can show how expectation values change with time T eorem For any linear hermitean operator Q that does not depend on time d i A A ngglt I Q gt QIIIgt lt1P Proof d d A 011 39 Equ QT lt3 gm Now since Q assumed t independent 01 at d 311 A A 311 gt 7 7 1P 1P 7 dtltQgt ltat Q gt lt Qatgt On the other hand we know 3371p 1P this is the TDSE gt 1 Note not do you see why ilt 1 h dt Q lpgt III 1mm K since if is hermitean Q 0111 As we claimed i h So any observable Q whose operator Q commutes with the Hamiltonian if has ltQgt I 0 gt Q const ltgt Q is conserved Q I gt constant in time for m time I Xt In classical mechanics conservation of Q means Q constant for isolated system In QM conservation of Q means lt Q gt constant Classically measured conserved Q gt get same Q every time But in QM if you measure a conserved Q get one of the qn39s f qnfu In QM conservation on is only evident if you make many measurements on an ensemble of identical systems Page F18 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 Examples A A d o 0 gtEltE0gt ltEgtconstant We really already know this ltEgt EEn culz is time independent A A d O 0 gt 2x 0 gt lt X gt changes w1th t1me 1n general I In fact we can work out the commutator on the left Em A A A A A A if A p 17 x 17214 p Whlch means 1 l dx I3 dt m g V22 l i E We ve seen this many times before but now it s formally derived It s Ehrenfest s theorem and tells us that expectation values obey Classical Laws 0 You showed in a HW E ihLZ V X dt 6x Ehrenfest again It s Newton s 2quotd law Classically 1 Fm 0Vax Page Fl9 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Forrnalism l The Heisenberg Uncertainty Principle Recall standard deviation 09 J0 Classically for any random variable X A gtlt xavg fraqX x x deviation a x avg deviation 0 lt 16 x2gt avg deviation2 0 1 ltx x2gt rms deviation 5 Ispread about avg x I eorem proof later For any two linear hermitean operators 2 1 A 0394 0393 2 B This is called a quotGeneralized Uncertainty Principlequot 1 Example 2xl x 3111 if gt 0A 032 4 That s the original Heisenberg Uncertainty Principle derived now Often written sloppily AX Ap E h gt if X known precisely AX Z 0 Ap very large if p known precisely Ap Z 0 AX very large Note large Ap implies large p since if p known small gt Ap small But if p large then KE pzZm large So if AX small particle con ned to small space then Ap s hAx at p is large gt energy is large 2 A 2 2 KB I 2 m 5 L2 2m 2m 2mAx We saw this in ground state of particle in in nite square well E3quot hz zmaz gt it always takes a big energy to confine particle to a small space Page F20 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Formalism 1 Note The Uncertainty Principle does not refer to uncertainty in the mind or even just the apparatus of the observer The uncertainty is in Nature itself If the particle has welldefined momentum then is does not can not have a welldefined position Proof of Generalized Uncertainty Principle same as in text A ltAgt hermitean defines f 2 A 2 K A A K 0A 111 A A 1P A A1PA A1Pgtflf Similarly a glg where g E B1P 2 010 f leglggt Z Kf lg This is called the Schwartz Inequality proven in homework Translating into more conventional vector notation it s equivalent to 2 E2 which is just AZEztasiQ lt l Now ltfl g gt is some compleX number z and le Rez2 may 2 hwy Z 2 2 Whmwwmz mUWWMWWWW ALvgtlt2sgt A AXE My notation here is that A 11 A AXE etc I know that ltAgt is real so it comes out of since A11J 31p AXIIJ the ket Thus f lg AXE AXE AXE A AXE Likewise A EXA wwwM MWWH Putting it all together then gt 0 1 7 Which is the generalized uncertainty principle done gtllt Page F21 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 Forrnalism 1 In addition to the positionmomentum UP Ax Ap 2 there is the timeenergy UP At AE 2 This looks similar but is guite different In QM time t is a parameter not an observable You don t measure quotthe time of a particlequot There is no observable corresponding to time in nonrelativistic QM At i uncertainty in time measurement there is no quotexpectation value of timequot At time interval for system quotto change significantly made precise below d M ltA gt Recall QLlt gt and 710 2 dt h 2quot A A A A 1 A A 2 TakeA H B Qgt ago 2 f Q 1 h dltQgtj2 Zii dt 6 diff gt O HO Q 2 dt De ne A E 6H l sigma uncertainty in energy At is time required for lt Q gt to change by 1 standard deviation 6 d De ne At 0Q 39 Al and we have our Energytime UP dt Examples 0 If I is energy eigenstate E known exactly gt AE 0 gt At 00 It takes forever for a stationary state to change 0 If I is superposition of Eeigenstates E1 amp E2 say then AE Z l E2 El and At a hIE2 E1 This is consistent with what we ve seen before on homeworks 2 2 Mr 2Re1p11p2cos E2 Eljt Page F22 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom l Angular Momentum warmup for Hatom Classically angular momentum defined as for a 1particle system ZFxf y m x y z 7 mv x y z X I X 1 Py p 0 Note Z defined wrt an origin of coords 1 J cyp zp5 zp xp5xpy yp Tietc In QM the operator corresponding to LX is Z 513 2 y p1 h I I 2 according to prescription of Postulate 2 part 3 Classically torque defined as 1 E FxF39 and iquot rotational version of F m5 If the force is radial central force then 1 FxF39 0 gt Z const HatOIIll electron lquot7 F Coulomb force proton at origin In a multiparticle system total average momentum im is conserved for system k sum over particles isolated from external torques Internal torques can cause exchange of average momentum among particles but Ltot remains constant In classical and quantum mechanics only 4 things are conserved I energy linear momentum angular momentum I electric charge Page Hl M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 Back to QM Define vector operator L operator unit vector Li L35 1 2 Esta 2w1gt dltLgtiltWlgt dt h dltLxgtA dltLygt A 7x 7 t dt Claim for a central force such as in Hatom V Vr k62 r then 0 will show this later This implies 0 just like in classical mechanics Angular momentum of electron is Hatom is constant so long as it does not absorb or emit photon Throughout present discussion we ignore interaction of Hatom wphotons Will show that for Hatom or for any atom molecule solid any collection of atoms the angular momentum is quantized in units ofh Z can only change by integer number ofh39s Units of L L NoteLrp p sincep hk L Bldg h Claim Exiyihi i j k cyclic X y z or y z X or Z X y Page H2 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 To prove need two very useful identities EA B C A C B C ABC ABC ACB Proof L1Ly 117 ZPZPx x171 I39msz Iprp 212 sz Epywpz y x 0 0 x M ih z39h all other terms like y px 0 7 10chy ypx 171 Have used xp ih xy 0 Egpy 0 I pxpy 0 etc I39m dropping the over operators when no danger of confusion Since LX Ly 0 cannot have simultaneous eigenstates of I and iy 2 2 2 1 2 h 2 0 onth 2 lt Ex Lygt ltLzgt ihltizgt However L2 L L L does commute with L2 Claim EAL 0 iXyorz Proof L2Lz2L Eryzalqll39 12142 I 0 LxLx Lz Lx Lzlx Ly Ly LZ Ly Lzly ihLy ihLy ihL H39th 0 Note cancellations L2 L2 0 gt can have simultaneous eigenstates of 221 orl 2 ii any i 1 Page H3 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 Looking forward to Hatom 1V2Vr 2m We will show that ELI 0 Eli 0 gt simultaneous eigenstates of if E L 2 energy qnbr 1 Ip n l m L2 7 L2 qnbr qnbr When we solve the TISE D 11 E 11 for the Hatom the natural coordinates to use will be spherical coordinates r 6 p not X y z Xrsin6 COStp yrsin6 sintp zrcos6 a2 a2 62 Iust rewr1t1ng V2 72 7 7 1n spher1cal coord1nates1s gawdawful But 0x ay2 922 separation ofvariables will give special solutions energy eigenstates of form 0659 Rr Y9 Rr99lt1gtltP The angular part of the solution Y 6 p will turn out to be eigenstates of L2 LZ and will have form completely independent of the potential V r a Page H4 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 Given only L2 L2 0 and 7 it hermitean we know there must exist simultaneous eigenstates f which will turn out to be the Y 6 p mentioned above such that Z fwlf ifuf A will be related to I and uwill be related to m We will show thatfwill depend on quantumnumbers I m so we write it asfzm and that sz h21l1f quot szl hmfquot39 wherel0l l l1l 1l fquot Yquot396tp will be determined later Notice maX eigenvalue of LZ h is smaller than square root of eigenvalue of L2 hllil1 So in QM Lzlt L Odd Also notice I 0 m 0 state has zero angular momentum L2 0 L2 0 so unlike Bohr model can have electron in state that is quotjust sitting therequot rather than revolving about proton in Hatom Proof ofboxed formulae This proof takes 2 12 pages Define L Lx iLy quotraising operatorquot L LX iLy quotlowering operatorquot Note L L L L Ai hermitean adjoint ofA Neither L or L are hermitean selfadjoint Note LZL0 I242Lamp2aLx iEZ LyO 0 0 gtConsiderf L2fl39f szlVf Page H5 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 Claim g Lf is an eigenfunction of LZ with eigenvalue u h So L operator raises eigenvalue of LZ by 1 h Proof ng L2LfLL2fALfAg To prove ng u h g need to show that L2 L h L LzsLxiLy Lz Lx iLz Ly hLxiLy H H hLy ith L Now Lg LLf LLf hLf nhLf w r f L 39lhL hk z So operating onfwith raising operator L raises eigenvalues of L2 by 1h but keeps eigenvalue of L2 unchanged Similarly L lowers eigenvalue of LZ by 1h Operating repeatedly with L raises eigenvalue of LZ by h each time L Lf has p 2h etc But eigenvalue of LZ cannot increase without limit since L1 cannot exceed L2 5gt lt5gt sew lt1 M2 2 0 20 A gt M There is only one way out There must be for a given A a quottop statequot for which Lft Likewise there must be for a given A a quotbottom statequotfL for which L 0 Lz all with same A eigenvalue of L2 lt Page H6 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 Write sz In E f In changes by integers only Lz PIft fmvalueofm Lz Want to write L2 in terms of L L2 2 2 LL Lx zLyLx zLy L Ly zLxLy EH Lz Lg 1711 EH hL 1 gt L2 LL Li hL Also L2 LL Li hLl gt L2 LLf Lift 71sz h22 1f HF HF 0 hzlzf We So sz h2 1f where 1 max In samelfor all m39s W A Repeatfor fb szb hZfb 2 minvalue ofm szb LLf L fb hszb h222 1f EI t w W5 H J 0 my h2efb A AA 221 22 1 gt 2 z tryit So mmin mmax and In changes only in units of 1 gtmf1 1 1 2f11 N integer steps gt21 N 1 N2 gt 1 012132 252 End ofproof of Page H7 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom l m 2 1 1 0 0 0 1 1 2 8 0 1 2 3 We39ll see later that there are 2 avors of angular momentum 3 1 Orbital 2 Spin Ang Mom Ang Mom Integer 1 3 ml integer g integer 0K Page H8 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom l 0 me mp gtgt Ine gt proton nearly 9 mp z 1840 me statlonary A A 2 Hamiltonian of electron if 2 Vr m e2 kZe2 r k o orVr r 3713 h2V2 m 2m Vr TISE 7an El 1 gt special solutions stationary states 45 wnm wnxt wnxe 4 General Solutlon to TDSE lI xt Ecne nx Spherical Coordinate System z r cos 6 Xrsin6 COStp yrsin6 sintp w w r39 639 p volume Normalization fdw l 1 ea Ir 2 2 1 dr 010 1 dtpr smelld 1 Need V2 in spherical coordinates 2 62f HardWay V fVVfaizm x Page H9 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom l ii i i 6x 6r 6x 60 6x atp 6x 62 f 6g 0g 6r i7i7m n1 tmare 6x2 6x 6r 6 gh Also need 9 derivatives 1 6 6y 6x Easier Way Curvilinear coordinates See Boas path element d idx fidy Edz 9dr rd Jrsianxp 1hl dxl zhzdxz39l39 aha dxs 3 2 ihidxi hi quotsca1e factorquot ii W39qum W 1 6mvaww 6x1 hl 9x1 63 2 hz 6quot2 n 6 ea Sphericalcoordinates i I 0 X r rsm gtV2f 2 izirzi2139isin6 239l2 L rar 6r rsm660 60 rsm 611 radial i2 angular r In Classical Mechanics CM KE p2 2m KE i l7xmi139l mrvl radial motion KE angular aXial motion KE L VJ 1quot 1 2 m 2 2 P2 L2 KEimv 7v v 2 2 r i m 2mr2 Hg 2 radial 1 7 x angular r2 Page H10 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 Same splitting in QM h 2 1 a a 1 a2 L2 inV h2 if 307 2772 i s0 60 60 s 6 61 Notice 1 depends only on 6 p and not r 2 h V21 Vrl1 E1p 2m an 4121 261 2m r2 Br 6r 32 72 Vow my 2mr Separation of Variables as usual Seek special solution of form Wr stp Rr39Y0 P Rr399 1 Normalization I dV I 2 j drrlelz 5210 7d sm0lyl2 1 0 0 0 1 1 Convention normalize radial angular parts individually Plugll RYinto TISE gt 2 h Yi 2am R I7YVR39YERY 2m err dr 2mr2 Multiply thru by 39 hz RY 39 2 ii z 2 392 VE 7 LZY R dr dr h 71 Y 4 W l fr g0 gt r g 6 p constantC 19 1 LAZY hZCY h22e1Y PageHS Page H11 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 Have separated TISE into radial part r 19 1 involving V r and angular part g 6 p 19 1 which is independent ofV r gt All problems with spherically symmetric potential V V r have exactly same angular part of solution Y Y6 p called quotspherical harmonicsquot We39ll look at angular part later Now let39s examine hZ Radial SE x R 2mr 2 2 quoth i rz rRV E 4719am 2mrdr dr 2mr Change ofvariable u r r Rr idrlulz 1 2 Can showthat r dr dr dr2 du dR dzu dR dR dzR 7 r7 7z77r 2 dr dr dr dr dr dr same 1d 2dR 1 dR 2c121 dR PR 77 r 7 7 2r7r 7 27r rdr dr r dr dr2 dr dr2 h2d2u 7 2m dr2 2m r2 2 Notice identical to 1D TISE h2 dzzp 2m dr2 V II E1 except r 0 gt 00 instead ofx 00 gt 00 and hZ Vx replaced with Vff V 2 2 mr at 1 Veff quoteffective potentialquot Boundary conditions Page H12 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom l u r 00 0 from normalization I dr u 2 1 u r 0 0 otherwise R 3 blows up at r0 subtle r A A B Vr 7 Veff 772 r Vquot 39 11 Notice that energy 1 eigenvalues given by solution to radial 2 r equation alone Seek bound I state solutions E lt 0 i0 E gt 0 solutions are unbound states 39 1Ir scattering solutions Full solution of radial SE is very messy even though it is effectively a 1D problem different problem for each 1 Power series solution see text for details Solutions depend on 2 quantum numbers n and 1 for each effective potential 1 0 1 2 have a set ofsolutions labeled by index 11 Solutions 11 1 2 3 for given 11 1 01 nl maxn 1 n quotprincipal quantum numberquot energy eigenvalues depend on 11 only it turns out 22 En 21 E1 mke n 2T12 independent of 1 same as Bohr model agrees with experiment Page H13 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 First few solutions Rn r rormalization quotBohr radiusquot 0 A47 4mth R10 Aloe a0 me mez r r R20 A201Te All 0 r 739 R21 A21e Allquot 0 NOTE for 1 0 s states R r 0 0 gt wavefunction 11 quottouchesquot nucleus for 1 0 R r 0 0 gt 11 does not touch nucleus 1 0 gt electron has angular momentum Same as classical behavior particle with nonzero L cannot pass thru origin 1 Fxp r 0gt p on Can also see this in QM for 1 0 VeffhaS infinite barrier at origin gt ur must decay to zero at r0 exponentially V r gt exponential decay in Rr u r as well r Page H14 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 Back to angular equation BY if 1Ylquot39 Want to solve for the Yquotquots quotspherical harmonicsquot Before started with commutation relations ii ij mil a LZ LI39 0 and using operator algebra solved for the eigenvalues of L2 L2 We found 2 m 2 m LYI 39hza39l39l l wheref012132 LZY quotth quot m 1 f1 In the process we defined raising and lowering operators A A L Lxf4y LYquot cht39 1 formltmm l LY l chl39H form gt mmill Z Cm is some constant if iYl 0 and iY 0 So if we can find for a given 1 a single eigenstate Yl then we can generate all the others other m39s by repeated application of A or i Y1quot YZquot0ltP It39s easy to find the pdependence don39t need the 1 business yet iz showed in HW l Mp izY hmY and you can cancel the h l P Assume YH p 80 ltP017 gt im39 Igt gt ltIgtp 6quot dtp If we assume postulate that 11 is singlevalued than t1gt p t ltIgtp gt emquot 1 gt m0r1r2 Butm1 1 So for orbital angular momentum 1 must be integer only 1 0 1 2 throw out 12 integer values Page H15 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom l algebra 6 a LLxzL he izcot07 y 60 6 L Lx iL he iicott9i y 60 640 L LL anoint fIA39g Afggt Can deduce Yf from LY 0 1 z gt i icotl9 0 60 a Ytt0p sine eiltp Solutlon unnormallzed Checks Plug back in sin0quot cos 0 W icot0sin WagW 0 l cosH sin0 cote Now can get other Ylm39s by repeated application of i Somewhat messy HW Normalization from fdefdtp sin 0 Ylm 0 Notice case1 0 Yo 3 mSty41z It 21 39 d0 1 39 0 dQ4 smce g39 qusm f It Example 1 3 i Y1 isln e 81139 sine cot0sin0 0 2 If1 1 1 sinHequot 81 Convention on r sign Yf39quot 1quotl Yf39 Page H16 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom l The spherical harmonics form a complete orthonormal set since eigenfunctions of hermitean operators fdsz Yf39Y quot39 a 5m Any function of anglesff6 p can be written as linear combo of Ylm39s ea Z Y Likewise 1 dr r2RM39Rnl and gt Hatom mg eigenstates are mam anrYZquot0qJ Ruezmem n12 1 01 n1 m1 1 Arbitrary bound state is w n m cm 39wntm c39s are any compleX constants energy of state 11 1 m depends only on 11 En constantn2 states 1 m with same 11 are degenerate n 4 A 1 4p 3 4d 5 4r 7 3 3 3p 3d 2 2 29 1 1 Degeneracy of 11th level is 2 2n2 ifyou include spin Page H17 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom l delxle 1 Prob find particle in dV about 7 0 de If 011 1pm then delcle fdr4m2l rl2 Prob find in r r dr Prdr 4m2ltpr2dr Pr radial probability density Pr Ground state 1pm Ae 2r Pr A2 4mze A Notice Pr very different from IMr P f If1 0 11 11 r 6 p Rr Y6 p then fanW dr r2 R 2 draw 1 1 Prob find in r r dr r2 R2 dr Prr2 IRIZ evenif1 0 Note 1 1prR39YR39gtlRl2 4 rl pl2 if 7 r so Prr2l1al2 4 rzw quotsolid angle quot Hatom and emissionabsorption of radiation If Hatom is in excited state 11 2 1 1 In 0 then it is in energy eigenstate stationary state If atom is isolated then atom should remain in state 111210 forever since stationary state has simple time dependence W w2ore39E Page H18 M Dubson typeset by J Anderson Mods by S Pollock Fall 20 SJP QM 3220 HAtom 1 But experimentally we find that Hatom emits photon and deexcites 111210 111100 in 10397 s gt 10399 s E 2p W EA hfAE 1s The reason that the atom does not remain in stationary state is that it is not truly isolated The atom feels a uctuating EM field due to quotvacuum uctuationsquot Quantum Electrodynamics is a relativistic theory of the QM interaction of matter and light It predicts that the quotvacuumquot is not quotemptyquot or quotnothingquot as previously supposed but is instead a seething foam ofvirtual photons and other particles These vacuum uctuations interact with the electron in the Hatom and slightly alter the potential Vr So eigenstates of the coulomb potential are not eigenstates of the actual potential Vcoulomb Vvacuum Photons possess an intrinsic angular momentum spin of1 h meaning e1lil hle1 h and L 739 So when an atom absorbs or emits a single photon its angular momentum must change by 1 h by Conservation of Angular Momentum so the orbital angular momentum quantum number 1 must change by 1 quotSelection Rulequot A1 r 1 in any process involving emission or absorption of1 photon gt allowed transitions are s p cl n1 If an Hatom is in state 2s 11 2 1 0 then it cannot deexcite to ground state by emission ofa photon since this would violate the selection rule It can only lose its energy deexcite by collision with another atom or via a rare 2photon process Page H19 M Dubson typeset by J Anderson Mods by S Pollock Fall 20


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