Quant Chem & Stat Thermodyn I
Quant Chem & Stat Thermodyn I CEM 991
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HEB aw DJoNr ax ff 6 Am Wk 420 Anfc39jfd W39M 1 ZR fwfw aq E A NATL Aca39 fa aaw a FPG6lt 2 hof Hevm au A es AT mar 3 2 8 Cwm AN AAgtJE EC 6 1 1 f 7 C C7L pmgud 2 A M 4 1 e mac731 99 Aim0 7w Wm M ngpggfgt 3 fm WTBBfVLT yz 2 0 Aritea gff gowv ff 44 74397g 7L 7L 5 wch 2 Cw a C 424 p MW 0ch H Mm 0 5 gggt c2 A2DZgt c 0490 57 waxy 700 dfJAgtLgt 2 ltD gt 39 c c0pcgt 0 t if CDADCgt 1 0 2 2 7g C2 1 ZQCDDltgt 1 c o pltgt Mquot 5 Dz 7 Z D2gt 39L 2 1 w gt ltDZgt O 7114 czgt492gt ltmgt ltgt2 CZM D1L WINCD chunkMD quot pze 2 8 For the hydrogen atom H T V and hence En Te2ltr 1 r ince E equot 2an we nd by using 75 that e2 En ltTgtnlm a ltTgtnlm2Em 39henoe Eu 2 lt739gtnlm 39he relation between the mean value of the potential energy and that of the kinetic energy is 1115 ltVgtnlm 2ltTutm 81 5 9 Consider T to be normalized to unity By differentiating the de ning relation III3 e obtain dA 63F 63F 6A d ltat TAY A at EWgt 91 ut from the Schrodinger equation we have that 68quot 1 6 1quot 1 a a w am 1d thus from 91 equation 9a can be obtained If the observable A is not explicitly timedependent then 1 0 and equation 9a 15 the form dltAgt if d A H 92 lis equation expresses a very important fact namely that the mean value of an observable iuch commutes with the Hamiltonian is a constant of the motion In particular if H is not plicitly timedependent the total energy of the system is conserved and the system is said be conservative 10 Since the Hamiltonian of the particle is Pa H 2 Vr T Vr follows from 92 that 0510 iih rp HI 27 rvV d 39 long into account the fact that E041 0 the virial theorem fellows immediately 4x IA 7 itii x fxr I g g fx Pit um if isquot If the potential energy is proportional to rquot we have LVV nV and 10a becomes 2T nV Note that for the harmonic oscillator potential n 2 and for the Coulomb p n l the relations 46 and 81 follow directly from 101 11 Consider an arbitrary operator B The mean value in a state 1 of a physical c associated with the Hermitian operator BB will be 88 1 BB1p B p B p a 0 Now let C and D be two Hermitian operators and J a real number Taking B t we have B C ilD and o lt 33 OMMD 410 430 f say whenfunction f0 has no maximum and its minimum is given by the cc d dl 0 After some simple algebra we nd that CD DC 40 gt0 fmin 02gt whence ltC gtltD gt gt ltCDDCgt2 Let u and v be the dynamical variables associated with the Hermitian operators 6 of the uncertainty relation III7 The deviations 6U E U ltU 6V E V V of U and V from their mean values U gt and V will also be Hermitian operators sa the commutation rule USU 6V U V If we now take C 6U and D W 114 becomes lt6U gtlt6V2gtgt ltUV VUgt2 De ning the root mean square deviations AU b Ur and AV b m we obtain from 117 the uncertainty relation AUAVaa v MU V AXAPZ i i 11 flak 13f 12 Wng mwfmh m GILWM W igf rat t1 General miceptx with two arbitrary functions u and 139 both complex and only chosen 50 that the integrals exist With 2 the operator 73 and at real this yields h i Er h 711 E Lx39u 7 d3xu l a39 a 39 d 39 l a39 tun v 1 1 1 7x 3913 z 739quot quot 0x 393 or 39 lt7u d 39u 39 m ow w i lax 39 amquot If 0x 5X Reordcring leads on to a 3 d3xx m u 29 j d ru v nx or by partial integration on the lefthand side to j d xu v Zairl xu c which of course yields the result of a z Eq 75 again Problem 8 Derivatives of an operator Let fipx be an integer function ol the operators pkg Then the general relations 5 funk 31 and 39 ii M 82 Pk with the abbreviation 139 fr1 EU 991 shall be derived from the commutation rules Solution The commutation rules are t PiuPl 0 xxxi 0 Pin 1 611 83 From these we construct 81 and 82 in four consecutive steps I Let f p then we have 6f6Xk0 and afap 5 Hence Eq 81 and 82 become pkp0 and phxk 5 Le they are satis ed according to 83 in the same way for fax BfExk 6 amp 0 they may be proved to hold If x I 6th 7DEOH5 A 14 Qf z t 4 H V6 34 ll7 Iww aq dgfwW WA 9 WW HWA 9 WigW c39 f A E HAAHQfz 39gt ltLHAgt B 639 GigL ltHMgt 6M5 A A a EquotltCtgtlt3gt Veg mm Ma a lt 411w rm mom er glam an Ag KF1ojlk0f 7 5 103x Jun E Am I ij mejam 2 n7 5 39 4bgtltgt 49 33gt lt DH3 W digt 7 75gt M 2 A my QM L Vx91ogt vajJ Q04 72 Iakfl07Ed lt wegcr dam 4 1 dv dkgt so ago Avxxxgt F3 ltgt M m gt quotlt9HQkgt A47 We 039 ltM20 318 GENERAL ronMALisn PHYSICAL oonranr on vuz 12 system of equations 1 is formally identical to Hamilton s canonical equations of Classical Mechanics More generally a classical dynamical variable A Aq1 gig p1pvi obeys the equation of motion dAci DACL at 2 Acid 1903 y where Acid 1101 designates the Poisson bracket of Am and H91 according to the de nition A laz DA 2311 DA M1 59 323 529 3939 We see that the classical equation VIII42 is identical to the corresponding Heisenberg equation to the extent that one can identify the Poisson bracket A H with the commutator Ami1 Hi Making use of the fundamental commutation relations and of the similarity between the rules of commutator algebra and the rules of Poisson bracket algebra one can actually prove the identity of these two expressions provided one makes a suitabie choice of the order of the g s and the go s in the explicit expression of the Poisson bracket 12 Constants of the Motion The notion of constant of the motion is particularly simple to grasp in the Heisenberg representation A dynamical variable which does not depend upon the time explicitly is a constant of the motion if the observable CR representing it in the Heisenberg representation remains constant in time Consequently its system of eigenvectors remains stationary and the statistical distribution of the results of a possible measurement of this quantity is always independent of the time at which this measurement is undertaken According to the above de nition of the constant of the motion d lite 03 z 03 HH 0 The constants of the motion are thus represented by observables which commute with the Hamiltonian This result holds true equally well in the Schrodinger representation and in the Heisenberg representation on vni 13 GENERAL soansmsn PHYSICAL CONTENT 319 since the commutation relations are conserved in passing from one to the other f a on uiifxbl39u Moreover since 03 is timeindependent it is equal to its initial 95 5g 5 i l i value 05 Cum 0300 03 C If in particular the dynamical state of the system is represented by an eigenvector of C in the Heisenberg representation Olive ciao l the variable 0 keeps the same well defined value c in the course of time the eigenvalue 0 is thenwsg hdjp be a good quantum number As can easily be shown mommutes Heevi imnor Utto hence the hot low 0 e 0 r6 mger representation remains forever in the subspace of the eigenvalue 0 Clint 2 case also C U is 13 Equations of Motion for the Mean Values Time Energy Un 33 U 14gt certainty Relation 39 MM wimpy Starting from the Heisenberg representation it is particularly or simple to write down a differential equation for the mean value of a given observable AH Indeed since by is time independent dltA d dA 7 3 a ltwslAslegt on 3 also Using the Heisenberg equation we arrive at eq v72 once again gig A on VIII43 in particular one obtains the Ehrenfest equations Ch Vi 2 by carrying out this manipulation on system I As an application of eq VIIIA3 we shall give a precise statement of the timeenergy uncertainty relation of Ch IV 10 Consider a system whose Hamiltonian H does not explicitly depend upon the time and let A be another observable of this system which does not depend upon the time explicitly We consider the dynamical state of the system at a given time i Let lip be the vector representing 320 GENERAL FOBMALISM PHYSICAL CONTENT ClL Viu 13 that state Call AA AE39 the rootmeansquare deviations of A and of H respectively Applying the Schwarz inequality to the vectors A ltAgtl1pgt and H Hygt and carrying out the same manipu lations as in 4 we nd after some calculations AAAEgtltAHgt VIII44 the equality being realized when hp satis es the equation A 00W M11 lwgt where a y and s are arbitrarily real constants cf eq VIII10 However according to eq VIII43 ltAI1 lftdltAll thc inequality VIII44 may equally well be written AA W 39 M gt it or else rAAEgtM VIII4539 if one puts AA TA appears as a time characteristic of the evolution of the statistical distribution of A It is the time required for the center A of this distribution to be displaced by an amountequal to its width AA in other words it is the time necessary for this statistical distribution to be appreciably modi ed In this manner we can de ne a character istic evolution i739339quotf01 each dynamical variable of thesystem Let r be the shortest of the times thus de ned 1 may be considered as a characteristic time of evolution of the system itself whatever the measurement cari ii out on the system at an instant of time t the statistical distribution of the results is essentially the same as would be obtained at the instant t as long as the dili39erence t t39i is less than 7 According to the inequality VIII45 this time I and the energy spread AE39 satisfy the timeenergy uncertainty relation masts VIII47 If in particular 3 y53teiii is in a stationary state dA 2quotl0 no matter what 1 and consequently z is in nite however AEO in conformity with relation VllI47 cu V111 14 GENERAL FORMALISM PHYSICAL CONTENT 321 14 Intermediate Representations The Schrodinger and Heisenberg representations are not the only possible ones Any unitary transformation of thevectors and the observables of the Schrodinger or Heisenberg E representations de nes a new representation All these representations furnish strictly equivalent descriptions of quantum phenomena In practice one therefore adopts the representation which lends itself best to the solution of each particular problem ny problem of Quantum Mechanics essentially consists of a more or hss wiliphH and more or less precise determination of the properties of llu unimr x39 upt39l utm if I indeed all the predictions of the theory irv gix wn hr unit rix elements of UU I stch as the one occurring in H 39l limo 39l39ln solution of equation Vll 132 is thus the central ll39Ull39lil nl 1hr llivni 39hn one knows an approximate solution l39 t I of this l qlmilon it is often convenient to set U Uioiu39 VIII48 Substituting this expression in eq VlII3 2and multiplying both sides from the left by the unitary operator Um we obtain the differential equation d lU l39 39 O 0 17L dt U UH ltHU iii dt gt U VIII49 U is the solution of this equation satisfying the initial condition Uluii 0 1 If the apprmiination is justified U is an operator changing slowly as a function of time this is actually quite evident from eq VIIIAQ sinCc in that case the operatorHUl0 ifidU quotdl almost vanishes Equation 39lll49 therefore lends itself better than equation Vlll32 to an approximate solution 1 Since UN is unitary the operator 11w t 2 iii Ult0gti in Ult012t0 1 The procedure discussed here is the generalization to the differential equations between operators of the method of variation of Constants of the elementary theory of differential equations CDR J6 414 7L C 39 ELHE lfb 39 4 e Elf5 C x f a 2 2 56 wm t emf 39 was El 5M mz x 6v cfLk39 654 C Y7J7L 722r a aimr anew 727k V f g Am My orcu39 Qt one 004 v 7 071194 2 fadHard 7 139 2776 0311 of fa e u Wbm aw W 749 luvu 12L ohmarm 104 7 am w od a u mmr my A W A71 2760 8m m zRfhhl39 a loamg r57 W e MW w Mix 2 A wag m m mea WNW ww 94 11k 57 irWv dbJM vf39gx ZZA balmL 713mm my A 1 7L u m ww oxa Z A APPENDIX Review of pertinent quantum mechanics Ai The standard twostate system This review is intended to aid the dispersion to experimental workers in the biological eld of the pertinent present knowledge ofquan tum mechanics which has grown up in the elds of physics and chemistry For many readers this appendix is unnecessary but my Own experience and the lessons of the Philadelphia Tunnelling Conference November 1977 Chance et al eds 19791 show that many of us need at least parts of it I assume the reader has an elementary knowledge of quantum mechanics at least as taught in general physics and chemistry courses uncertainty principle Schroedinger equation etc I suppose not all readers are familiar with transformation theory Dirac 1927 935 Jordan 1927a b how to change from one representation of a system to another and some other concepts very pertinent to tunnelling discussion which 1 will try to define and explain in elementary terms An excellent beginning text for further reading is the Feynman lectures Feynman Leighton amp Sands 1965 A few notions which the reader undoubtedly knows already will be repeated anyway for emphasis and completeness Quantum mechanics attempts to tell us what we can know about the future of a system given what we know about it now Since scienti c knOWledge is always ofa statistical nature and never complete quantum mechanics quite properly deals with probabilities However it does not deal with the probabilities directly It deals with an underlying structure of probability amplitudes from which probabilities can be calculated by taking the squares of the absolute values of the probability amplitudes The amplitudes themselves are in general complex numbers but the probabilities ofcourse must be real and positive One describes a system in terms of a set of basestarcs The set must J A to standard tn tJslttlt39 8l ft lll i be complete in that any state of the system to be contemplated must be describable as a combination of the basestates Also the base states must be orthogonal and normalized that is if the system is kn0wn to be in a single basestate it must have zero amplitude to be in any other and unit amplitude to be in itself Different sets of basestates may be used to describe the same system Sometimes one set or representation is more convenient and sometimes another lig Al Tunnelling between two boxes A C Energy v patticle coordinate as in Fig 58 Boundaries ofsltaded areas are potential energy Solid horizontal lines within boxes represent the energy eigenvalues dashed lines the values of I and H A Box a has lower energy than b B Boxes have equal energy C Box a has higher energy than I D E and F Graphs of occupation of states correSponding to the polentinlencrgy diagrams A H and C respectively The occupations or probabilities are showu partitioned between basestates I and h and between eigenstates l and 2 according to equations Al I G A plot of the energies t39 a continuous scale of separation between HM and If The abscissa is r tHW limo2 Huh See equations A 0 E nergy Occupation a E2 ampt gt on m i 5 x EKJ H w uh V I F L l 0 G last nppuiam i usumu quantum mumtit Physical quantities are represented as operators which affect the probability amplitudes by operating on them Here we will be concerned mainly with the energy operator which is also called the Hamiltonian We will discuss how this works in the ease of a system which has only two basestates This may sound like an oversimpli ed system but actually a large number of physical problems can be discussed in terms of the standard twostate system Feynman et al 1965 including many of the tunnelling concepts Consider a system that consists of two boxes with one particle which can be in either box There is a wall between the boxes that the particle cannot penetrate in a classical manner because it does not have enough kinetic energy This wall is not a real one penetration of which requires loss of energy to heat through inelastic collision with its material but is rather a force eld which classically slows the particle on entering to stopping if the particle does not have enough momentum to get over the hump and speeds it on leaving The particle can however get from one side to the other by quantummechanical tunnelling Fig AlB shows the situation On an energy v particlecoordinate diagram All The Hamiltonian Let Ca be the probability amplitude for the system to be in base state a described as particle in box a and Cb the amplitude to be in base state b described as particle in box b39 The states a and 7 form a complete set of basestates for this twostate system Ca and Cb can be complex numbers The time operator ihddt and the energy operator H act on Ca and Cb as follows HaaCa Habe ih dCbdt HbaCa beCb 39 The Hs can be set forth in a square array H H b H no a g Hba be which is called the Hamiltonian matrix39 The Hs are the matrix elements They apply strictly to the basestates a and b The matrix A2 is the Hamiltonian of the system in the a b representation39 Another set of basestates will take a different set of matrix elements It is seen from equations A that H is the effect of Ca itself on dCadt H is the energy ofthe system in state a if there were no influence on it from state b that is no possibility of the particle tunnelling from one box to the other Likewise be is the energy of the system in state b if there were no effect ll ll ilttlttlultl iltu lfll 39llthl it from u Hub is the effect of 39h on d39quotdt and HM is the effect of Cquot on debtit They represent the rate at which probability amplitude ows from one state to the other lfeqntrtions Al were divided through by it they would have the appearance of chemical rate equations with first order rate constants Hh except for the presence of the imaginary root i I in order that theenergy be real it is necessary that thediagonal elements HM and HM be real and that Huh H u where 39 means complex c0njugate39 Thus i1ulli i hrti39 A l 2 Eigenstutes and eigenvalues u Symmetrical mse Equations Al can be solved easily if the Us do not change with time To simplify further let HM HM E The particle has the same energy in either box Let also Huh H A Then ifwe add the cqttationstA l l we get it d a Chiquottit E0 IIHC39H 39b tAJi Subtracting the second ofA l from the first gives it d a hdt E0 I 1 a Cb A4 The form of these equations all0ws a simpler description of the states suppose we let I Ca C39bh39VZ be the amplitude to be in a state we ll call 139 and C2 C zI Cb2 be the amplitude to be in a state called 2 The 2 is necessary for proper normalization Then equations A3 and A4 become ithldt EICI if dCzdt EZCZi where El 2 Eo A and E2 130 A States I and 2 may be taken as a new set of basestates in this re presentation equations A5 sh0w that the Hamiltonian matrix is H 0 39 Hlt0 H Am 22 where Iill 13l and Huzlil We have diagonttlizcd the matrix The eqL39Iations A5 are seen to be independent of each other and solutions are 39n 0e quotquot3quotquot39 and lm Zion quotgt3quot quot A7 A5 39I and C2 oscillate in the complex plane with radian frequencies Iiih and Ezh respectively t 126 Appendix Pertinent quantum mechanics Their probabilities C2 and C22 stay cOnstant with time States 1 and 2 are thus stationary states or eigenstates Their energy eigenvalues are El and E2 The splitting between the two energies is seen to be El E1 2A 2Hub This splitting is the result of the particle s ability to tunnel from one box to the other It is shown in Fig Al B Note other phase factors than 1 could have been used for Hquot and HM but the results would be practically the same The representation is in terms of eigenstates whenever the matrix is diagonalized On the other hand if off diagonal elements had finite values this would indicate that the basestate amplitudes would be subject to change over a period of time and so such basestates are not stationary states If the particle is initially in a how fast does it get into b and what is the future time course of the system This is answered by noting that from the de nition of CI and C2 if Cn0 0 then Cl0 20 lVZ Also note from the de nition that Ca c1 C2 and C Cl C22 If we substitute the solutions A7 for C1 and C2 into these expressions we get Cam eiElth elEzth eIEotIt COSMth ch e tEgt eiEztlt ie t Eotlh SinU Hh lat cos Arm g cos2Ath lCil2 sinztAth t J5 cos2Ath Equations A8 show that the probability of nding the particle oscillates back and forth between the two boxes with a frequency equal to the splitting E2 5 divided by it This is the resonance phenomenon familiar to chemists occurring in molecules with two equivalent electronic structures The lowering of El below E0 is the resonance lowering of energy and it results from the ability of the particle electron to go back and forth between the two structures It is always accompanied by the raising of a companion state to a higher energy The timecourse of Cbz is plotted in Fig A2 A The system must be in state 1 to be in the lower energy state How are states 1 and 2 to be described If the systemis known to be in state 1 then C1 land C2 0 From the de nitions we then have C C2 and c C0 Thus in state 1 we have Ca Cb l2 The particle has equal probability to be in both boxes in each If the system is in state 2 then Cl 0 Mi A The standard twostate system 39 l27 and C2 1 and Ca Cb0 and Ca Ch 2 Thus in state 2 C Cb l2 Again we have equal probability G each to be in both a and b but the amplitudes to be in a and in b are symmetrical in state and antisymmetrical in state 2 Note that one cannot observe both the going back and forth and the splitting ofthe energy at the same time To observe the resonance oscilla tion one should make observations at intervals of time considerably less than one period which is hZA The uncertainty in the time measure ments must be considerably less than li3921 By the uncertainty principle the uncertainty in energy must then be considerably larger than 2A which is the splitting Thus the splitting would be unobservable To be able to tell in which of the two energy eigenstates the System is one would have to observe it Over a period of time long enough to reduce the un certainty in energy to considerably less than 2A By the uncertainty principle the observation would require a time considerably longer than hZA the period of the oscillation Therefore if one Ctttlit39l measure the Fig A2 Timecourse ofICblZ the occupancy of box 5 starting with the particle in box a at time 0 A H Hquot B H H i 2 Haul C HM 1 4H bl D Distributed H p 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wt C 77 I J 1 7 Jane K V2 L manman 27 V3 w 3 O i 39Qquot f L x J F Jquot M y 01m r IND EPEN DENT Z 77 mime 71m u Lue Av V OMA H r ii l jgi 3973W44 lj ya hg gm haggzg 4A 4 atwgktf quotW w Vaquot 5 425 Jim 3974 4p m WM M Fritz5t 50w Mon 140 f gt H I LANL X V39n 0 71m 0 00 Exau lu Lam ma 39HD97015 N H77577 lm WWW V oc m 3 j afg39ijigg i quotf ffL 4M u 44 w 5 M W Ci 5 ay a Va r1 71 V J OCCUVJ at w k loucnmc 9 nAf n 7 a 7kraac nd mvoampd g 1 06w 39 cfbhs 39 7 15 7 395 a 7 7 VAL E 39 ELM a fog7 A E 395 1 I J M m v n V x 10 5 177 7 7 457 A 04quot QZ39LJ C ha V 71 a 04atu4p wg 4 mil3 gt 0 L RVQ Man i 9 6475 i 39 1 1a 1 Li r t quot194 quot1quot quotm f a AJ V 7 JM 4 5 kOK all 84 Llt4 quot Iiigame 14 SJI124L Kt71f2k 1 WW J WNW M DI POLE peknroe r r v A r L m 94391 3 l kgW Lwa 117W Way 7 3409 33er gt a ii Y 393 Nquot magazg asquot W MrwI39 tl uv XTVAWLQCOJ P u V m a tc lfh9Slfn EX WM Lu wovdvuJ m Hz wrote k IrrRVRN vrquotalv 39 i i 11T I 39 T I Ih 1 Hv X s 7 L 0139 quotJY JS i 39h w49 PS 0595w29 do Ix LINn 397 n n w WMMO 4 J m 75 E 37 0 49 J QJQJS 2ef w9 AM k j 2mm 0 pgmzw rm meagedexfrz m V 131421 411 AN rortlj t1 13 J i 1 I I j IH391 39HI IJ 3mm A 4 mm 39 n I2J1 72 g M quot quot 739 MH 39 m 29 J Iquot 7 5 7m 39 quot WrM 39 1114 53 9 K I f 66 fdVzi 196650 2 M 5fo 06 AM M W3m quot A 6 1w Rama 7 f F 1 7777 mm WLJMWQ VVVVVVVVVVVV W Agczawauc PM arAIzf ef W mm my be MWan warty V w f 1 wre hcrlr ohs L m 0 gm 5 5 V XB u a f wax F02 Zo a or i I lt V L L 2 2 739 4o 7 3c v v 3 M79mKM L M p Mme ggagwm Mm J c quot V K t J c7 an f v 139 Vx E a 142 kz 2 wa 39Wu zgdmg g I 3 1i fillX KA h 7 a i39Vf q 7 A dls pu 23 W Nah w ug y quotin w F h N Ha 7 wwwx W 737x t Jail 1 A Q ELQ A f a 39dim 39fLIif 739quot Cg T T 377 x i E39 L39meif f h n i R m 2 akfr7oeg 77 77m o aif223331W 39 f L 7 x ijk 4 aw 4 2322L 71173 14739 5m2g 5AM 39 m mw V quot n A M H 7 quot175 1536 Mx2f 11 2c 7 51quot 757 49 gg f a J m f 5 g 7quot g g m ww wvmmm MMHW 1 f E 6M V 7r 39 742wVW hu riv A 7 7 7 r v I F V quota E r 1 V 7n r g a J x x K ML H 7g c I M A fl quot m 7 K A m m M1915 79 W aftquot 0N 44 L I a CPM 119Ny 19164 A ua fitLo WV 17 7523 aid mtnzmg 15 277 Mar S j f 39 39 39 39 ESL QM k A W D A ma 943 gijgii39JTIQLJLQ m A If W 7 0WAm7 j V 6 39 339 if 39 M quot 7 HQquot 51 sinLa 444w A I 5 quot A x A mum s a 7 1i quot 39 E 5 HOW THE E W 5 M 02 awe a 4 L 27cxs o 2 A m 39 W v39 gs f t Fri r mfg f 7 quot 39 r L ClawA146 to we Mmm4 6 9quot Q 4 cos NIP15 5723 91 U121 W a Q3112 9N 2amp m Mam 39 39 TM an in 0 LOW Qm ml 6 r 7 397quot394 7quotL quotgt j 339 i l gt 39 iii1 M quot FL1 M154ampMb l 7 JJgt i3iIquotATZ 22 M 7 i k ergac 43 757 1 quot aa 1apwzgm mgr L A 5m K J f m Ef 1i FI Cloggirma quotP a 3 w m 9 i 23L3gt32gt 1 s oomlg 7fj ffj A v gbig en 1 quotA V 1c b E SEdie EEeS JF39T7L ngw22 31 d7 2 M77 A fv 31 z f q 4 g r t quot ff an LMigf7 Mf j L 39 n 3quot Eras 1 jmi22fxfu uoz39ua in if Wing 6Qrafjnmom A 7 W K adov VAAA V AMEM a T AMMA V 7 L Follow Qua 7 639 ZIP MIJm 63 B6M b w 3 V WI f WAN 4L aflon w I A vaglpngnMmmmlagglt avg 39 low glo wc 7 quot I W ii Vdi 7 LQQL M Jg wsjw 39 139 m 4MM7AJ 1 3 1 ML N Mg m WWW Q if39ie ulttamp t 1 7 N Y x Canr 39939 P9 as x mm 71 Ag V I T A33nh r V f EfZLZE73 g i Va 3937 I I 5191993 VAT Pb JW 991 E FKFMI39W H W A 7 a 7 an v 39 w Awn Wall9m fragufu o 5213 39 3 39 quot u 77 7 2 7 z I lt quotEi 60wa 1 5 VELOCITY DEPENDENT POTENTIALS 19 1 5 Velocitydependent potentials and the dissipation function La grange s equations can be put in the form 1 53 even if the system is not conservative in the usual sense providing the generalized forces are ob tained from a function Uq 1 by the prescription 1 L Q aqdzag In such case Eqs 1 53 still follow from Eqs 150 with the Lagrangian given by 1 54 LT U U may be called a generalized potential or velocity dependent po tential The possibility of using such a potential is not academic it applies to one very important type of force eld namely the electro magnetic forces on moving charges Considering its importance a diversion on this subject is well worthwhile In Gaussian units the Maxwell equations are 1013 VXEfcat O VD 47rp 162mm VXH COt C VB O 155 The force on a charge q is not given entirely by the electric force so that the system is not conservative in this sense Instead the complete force is 39 FqEva we The history of the designation given to such a potential is curious Appar ently spurred by Weber s early and erroneous classical clectrodynmnics which postulated velocitydependent forces the German mathematician E Schering seems to have been the rst to attempt seriously to include such forces in the frame work of mechanics cf Gott Abh 18 3 1873 The rst edition of Whittaker s Analytical Dynamics 1904 thus refers to the potential as Schering s potential function but the name apparently did not stick for the title was dropped in later editions We shall prefembly use the name generalized potential including Within this designation also the ordinary potential energy a function of position Only 1 5239 I 5M seem 133 V XML 39 mg an 59 xAH 8 WM r Ux ax 8 is ega l 20 SURVEY OF THE ELEMENTARY PRINCIPLES lcmv I E is not the gradient cf39a scalar function since V x E y 0 but from V B 0 it follows that B can be represented as the curl of a vector BVxA where A is called the magnetic vector potential becomes Hence we can set 16A Mv Ec6 d or 39 16A EV SZ VxE vavxE 18A 157 Then the curl E equation 0 1 58 In terms of the potentials lt6 and A the socalled Lorentz force 13956 becomes 1 6A Fq V EVXVXA39 1 59 The terms of Eq 1159 can be rewritten in a more convenient form As an example consider the 2 component gr 81 and 8A 5245 a 6A 6A WXVXALW mamp 8A 61 Q43 m v 61 3 61 91 6x where we have added and subtracted the term 82 Now the total time derivative of A is M 6A 6A as v 62 U It a quot8y I 32 3A 6A quot is le 6 where the rst term arises from the explicit variation of A with time and the secorrd term results from the motion of the particle with time which HIM dt if keg 3 J8 Di ifig2l 33 31 a r J 4quot 9 a 91 ilk L 3 ivy 3g fat V r q lv IthqWIW 1 5 VELOCITYDEPENDENT POTENTIALS 2 changes the spatial point at which AI is evaluated The x component of v x V xA can therefore be written as 6 mice VXVXA axv39A 016 at If With these substitutions 1 59 becomes A x 576 L aq iG gvAf iec Since the scalar potential is independent of velocity this expression is equivalent to 6U cilaU v l P 62 39 dt 61 where U git EA 1 60 is a ner 39 39 fE 1 54 and the Lagrangian for a charged particle in an electromagnetic eld can be written LT q C ZAv 1 61 It should be noted that if only some of the forces acting on the system are derivable from a potential then Lagrange s equations can always be written in the form aggaq It ad 61139 n where L contains the potential of the conservative forces as before and Q represents the forces not arising from a potential Such a situation often occurs when there are frictional forces present It frequently happens that the frictional force is proportional to the velocity of the particle so that its a component has the form Ff lcvz Frictional forces of this type may be derived in terms of a function 5 known as Rayleigh s dissipation function and de ned as 1 lt3 E 2 mg km kzv 3 39V 5 T Axverqufr e x i 4xe 39 A quotIi 1 U 09 2r maul ow fora7522 CHAPTER VIII TIME DEPENDENT PERTURBATIONS RADIATION THEORY 8a TimeDependent Perturbations In the previous chapter we discussed the problem of determining the new energy levels and wave functions for a system subjected to a perturbation which depended only on the space coordinates of the system For certain problems particu larly those dealing with the emission and absorption of radiation we need to calculate the effects produced by a perturbation which is a func tion of the time therefore we shall now develop the timedependent perturbation theory quot The wave equation in the form which expresses the manner in which V the complete wave function 139 changes with time is r v 39 h 0 11 as Hip339 2mm at 81 h where ft 3 Let us now write the Hamiltonian Operator as uW39 H 2 Ho H where H0 is independent of time and H is a time dependent perturbation The unperturbed eigenfunctions IIO satisfy the equation HO I O z 8 2 E n and are of the form 301 t r3ge 7 In order to obtain a solu tion of 81 we expand the function I in terms of the unperturbed eigen functions Wg with coefficients which are functions of the time that is we write so a Emerge t 83 Substituting this expression for 139 in equation 81 gives 36 anquot ZCRHO Ig ZCan Ig Et I g ZhZCn F7 n n n n G Since the unperturbed eigenfunctions 1 satisfy equation 82 the above 107 I 108 TIMEDEPENDENT PER i URBATIONS RADIATION THEORY equation immediately reduces to d n goal1 92 iquot r n n dt Let us now multiply both sides of this equation by If and integrate over coordinate space Then of A 85 d 11 dcm Zen farm39ng a a 3 fxpgj tg dr m 86 n n dam f 0 I 0 39 3 quot39 n m H I n d quotF or di h 1 139 139 8 l 39 In any particular problem we will thus have a set of simultaneous di er ential equations which can be solved to give explicit expressions for the en s 8b The Wave Equation for a System of Charged Particles under the In uence of an External Electric or Magnetic Field The most important problem to which the time dependent perturbation theory will be applied is that of radiation In order to discuss radiation theory we need the Hamiltonian Operator for a charged particle in an electro magnetic eld In deriving the classical Hamiltonian39function it is more convenient to use the vector potential A and the scalar potential 0 rather than the electric and magnetic eld strengths E and H The relations between these quantities are given by the equations 1 a H FXA39 E m A w c a 0 8 8 Where 0 is the velocity of light A particle of mass m and charge e moving with a velocity v in an elec tromagnetic eld is subjected to a force F so ileD The equations of motion are therefore 12 5 6 8A 6 1 dz m 85quot5quot5f shim 32H 8399 62y 22 With Similar expresswns for 2172 and Usmg the relatlon H WA these equations become mfg agan dz ea c a 0 dt ea 63 dz 6A 6A diltax 82 810 THE WAVE EQUATION 109 etc It is not dif cult to demonstrate that these equations of motion are derivable from the Lagrangian function 39 L 22gt2i I 8 dx dy dz A A A2 391 Ila cdt dt ydt 6amp0 81 am 6L 4 From the de nition of generalized momentum p g we see that dx 6 39pI m E EA With analogous values for 12 and 12 The Hamil 39 tonian fimction is therefore 7 div dy dz quot e sm le e z z L quotquot 2 T y H pdtpydtpdt m d2 2 dy 2 dz 2 7 22 allW 83912 D x 14 n or in terms of coordinates and momenta LR W2 302 all 813 pas c c py c 1 pz 0 2 8 39 The procedure for constructing the Hamiltonian operator is identical with that followed before in the classical Hamiltonian function the 6 wave function 11 the rst term in the Hamiltonian gives a 2 a 39 momentum p 1s replaced by z a etc Operating on a 2m 6 g 1 62 e 6 e 6 e2 v quothz hquot Aa ih Azquot A22 2m 6x22 662 20 ax02l I 1 62 11 eaA h2 r 2m 622 2hc 62 as e aw e2 ihEAI m Ax leJZ I 814 c 62 6 62 c After collecting terms and expressing our results in vector notation we see that the Hamiltonian operator is 2 ih2v2ih9vA2ih3AvIAJZ6 815 2m c c c 110 TlMEwDEPENDENT PERTURBA I IONS 39RADIATION THEORY For an electromagnetic eld such as that associated with a light wave V 0A 0 and p 0 Since the perturbation of a system by a light 2 wave will be a small perturbation the term A12 may be neglected in discussing radiation although it must be retained when discussing per turbations due to strong magnetic elds For a system of charged particles with an internal potential energy V we will therefore have the Hamiltonian operator H H0 H where it 2 e no Z v V H immac 810 239m5 5 77er The operator BI0 is just the operator for the system in the absence of an electromagnetic eld the perturbation H may equally well be written e as H mjc 3 9 8c Induced Emission and Absorption of Radiation Let us con sider an atomic or molecular system subjected to the perturbation H of an electromagnetic eld For simplicity we rst will assume that the field is that of a plane polarized light wave with AI and AZ equal to zerofquot We shall need to calculate the values of such matrix elements as foiif n e d1 rgle lxrg o2 e2 so e 39 Z Ana39ij my 1 Since molecular dii nensions are of the order of 11000 the wavelengths of Visible light a suf ciently good approximation for our present purposes will be to take A as constant over the molecule The matrix elements That this situation represents a planepolarized wave may be seen as follows We take A mg with A Agcos 2wt This represents a wave moving in the z direction with a velocity equalto c The associated electric and magnetic elds are then 2 H VxA 3 Agsin2wltt The electric and magnetic elds therefore have equal amplitudes and are at right angles to each other as well as to the direction of prepngation and hence represent a planepolarized light wave moving along the z axis with a velocity 0 w n wmhn INDUCED EMISSION AND ABSORPTION 111 x19 h Emquot39En TAxe 21 139 mi can then be written as I 6 013339 GIG e 1 AI 19 5 ng a 12m IPA j m 1 8 C 3A32i 2quot Z 39l itpquot 818 956 n In order to obtain usable results we wish to express the matrix elements 6 of x in terms of those of c Let us for the moment39consider that the j 50 s are functions only of one coordinate 3 Then 31an and ybg satisfy the equations dag 2m dx2 gill7quot Vrcl 32quot 0 8182 d2 9 2 33 En Vx 2 0 818b If we multiply the rst of these equations by scybg and the second by 230 and subtract we obtain upon integrating the resulting equation 00 d2 0 d2 0 00 f all x 9 an o dx dz hquot 00 The rst two terms may now be integrated by parts since the wave functions vanish at in nity we have d 0 dag d 0 dag wd nE 39Zzx m 173 2 g ZltEm Engt f 3c 3dr 818d 0139 r of we integrate the rst term by parts we see that it is equal to the second For this one dimensional example we have thus obtained the result oaogtn1 0 o m 6514 h2Em En mlfvl 818 Thisvresult can be generalized The matrix element 11 IH39 MS may therefore be written as fig dz 21E m En f 429be dz 8186 d3 71 oo 0 o 1 239 Iml qun AIEm EnXme 818g L 112 TIME DEPENDENT PERTURBATIONS RADIATION THEORY 39 where an ex 2 is the matrix element for the a compo J nent of the dipole moment If we now assume that the system was originally in the state 12 so that on 1 and all the other c s are zero at the time t 0 we have for times suf ciently small so that all the 0 s are negligible except on dam 1 Em En E AsznEm En 5 If the light has the frequency v the time39 dependence of A may be expressed as A3 A2 cos 27m 2022 6 2quotquot Then dcm 1 0 dt a 2ph2AmenEm En e Em Enhr Em En hv 39 15 8 IS 820 Integrating and choosing the constant of integration so that cm equal zero when t 0 we have Em Enhv Em En hu e 5 1 e 5 1 821 Em En L o c MAquot XmquotEquot Equot Em En hu Let us consider that E39m gt En so that the transition corresponds to absorption cm will be large only when the denominator of the second term in brackets is nearly zero that is when Em Eng hr The probability that the system will be in the state m at time 15 will be the value of the product cfgcm The rst term in brackets being neg lected this product is 2 Em En hy Sln 2ht A0 2 mn 2 m 39 En 2 V lawlw ampmg 25 i AM t2 I cmcm 40 822 So far we have considered only a single frequency v To obtain the correct value of 620quot we will have to integrate over a range of fre quencies Since 00quot is Very small except for frequencies such that Em En 1 hr it will be satisfactory to integrate from co to 00 and regard Ag as constant and equal to A205 This integration gives 2 7r2umn th2 823 A2an2anl2t cmcm 5 1 Where Em En has been replaced by hymn INDUCED EMISSION AND ABSORPTION 113 Equation 823 has been derived upon the assumption that the light was plane polarized In the general case in which A and Az are not 39zero equation 823 will of course contain additional terms in A and AZ In calculating 00 the cross terms will vanish because of the randomness of the phase differences between the various components of A the nal expression for 020m will therefore be Cakcm 7723 m c h2 IASemollemlz A3ltangtzlrmnl2 lA3anl2lZmn2t 824 If the radiation is isotropic then IA2ltangtlz IA2ltVMgtIZ IA2an2 3 A um2 825 We may express IA0an2 in terms of the radiation density pumn From Equation 88 21ran Ean TA0ymn sin 2711113 Then n 2 2 2 E2an 16112 A0anl2 since the average value of sin2 2m is In electromagnetic theory it l 2 0 IS shown that pan E20 so that A2vmnl2 T pan 41r 3 minquot The product cfncm may now be written in terms of the radiation den sity as 2 quot I 016quot 2 5X2 ilenlz lYmnlz quotlquot Izmnlzlpymnt Since the probability that the system will be in the state m is zero at time t O and is the value given by 826 at time t the probability that a transition from the state 72 to the state m will take place in unit time resulting in absorption of energy from the39electromagnetic eld is 2w BnquO mn 23quot lRmnl2Pan I 83927 Where 11m2 IanI2 KM2 z2 828 If the system is originally in the state m then the same treatment shows that the probability of transition to the state 71 resulting in the emission of energy due to the perturbing effect of the electromagnetic 114 TIME DEPENDENT PERTURBATIONS RADIATION THEORY 39 eld is Bm mPO mn Bn impo39mn 8 29 8d The Einstein Transition Probabilities The coef cients Bm and BMquotIL are known as the Einstein transition probability eoe icients for induced emission and absorption respectively Since a system in an excited state can emit radiation even in the absence of an electro magnetic eld the completion of the theory of radiation requires the calculation of the transition probability coef cient Amm for spon taneous emission The direct quantum mechanical calculation of this quantity is a problem of great dif culty but its value has been determined by Einstein1 by a consideration of the equilibrium between two states of different energy If the number of systems in the state with energy Em is N m and the number in the state with energy En is N n then the Boltzmann distribution law states that at equilibrium amp kT an E1quot e 3 CT N 11 39 n e kT where T is the absolute temperature and k is the Boltzmann constant Expressed in terms of the transition probability coefficients discussed above the number of systems making the transition from39m to n in unit time is39 leAm m Bm gtnPanl Similarly the number of systems making the reverse transition is NanHmP I mn Since at equilibrium these two numbers are equal we39have after elim N m inating the ratio F by means of equation 830 39 Bn wnPO mn c 7 Ina n gm quot1P an hymn Ant me W or pvmn h 831 Bm ne kT Bn am Using relation 829 the energy density may therefore be written as Am b Bm rn P047131 hymn quot I 832 e W l 1 A Einstein Physik z 18 121 1017 THE EINSTEIN TRANSITION PROBABILITIES 115 The energy density in a radiation eld in equilibrium with a black body at a temperature T may be calculated by the methods of quantum statistical mechanics Chapter XV and as mentioned in Chapter I is given by Planckfs radiation distribution law as 87rhu3I 1 9051111 63 n M mg 6167 1 Comparing 832 and 833 we see that h 3 Am m 87F mn Bm m C or 32w3v3 2 Amsn 363h lRmnI 834 To the degree of approximation used above the coef cient for spon taneous emission depends only on the matrix element for the electric dipole moment between the two states If the variation of the eld over the molecule is not neglected there will be additional terms in the expression for A the rst two additional terms being those corre sponding to magnetic dipole and electric quadripolo radiation In eluding these terms gives2 m n 3211313quot 39 Ayn m ch mierlni2 2 raw 5Imlerrln2 835 We may estimate the relative orders of magnitude of these terms as follows Disregarding the constant term we have lmlerlnl2 N can2 r9 65 X 10 36 cgs 2 h 2 Km 71 87 x10 41cgs 2mc 2 T367911 merr39nl2 68 X 10 43cgs x 5000 A 6 2 27w p L 27m VP We thus see that the probability of transition due to magnetic dipole or electric quadripole radiation will be negligible in comparison to the probability of transition due to electric dipole radiation The higher terms in 835 will therefore be of importance only in those cases in which 2E U Condon and G H Shortley The Theory of Atomic Spectra p 96 Cam bridge University Press 1935 1 16 TIMEDEPENDENT PERTURBATIONS RADIATION THEORY the electric dipole matrix element mlcrln vanishes because of the symmetry properties of the states m and n The actual intensity of radiation of frequency vmn due to sponta neous emission willquotof course be vmn NmhvmnAmm 836 8e Selection Rules for the Hydrogen Atom According to the results derived above the only transitions of importance in the hydro gen atom will be between those states a and b for which alerlb eialxlb jalylb kalzlb is different from zero The eigenfunctions for the hydrogen atom may be written as ybn z m frP39 quot cos 0 elm so that we must investigate the values of such integrals as fa z martin dr fa z mt cos 9 l39z39m39 dr This integral may be written as the product of integrals in 739 0 and p The integrals in 739 will be non vanishing so that we may concentrate our attention on the integrals over the angular coordinates For the 2 component of the electric dipole moment we must therefore investigate the integral I fle39 cos 0 PPquotI sin 0 d0 fe mwm dltp 837 From the recursion formulas for the associated Legendre polynomials equation 485 we have 1 lm 1 21 1 Z m cos OPP Pli L39 838 H The integral 837 is therefore nonvanishing only if m m39 I l39 l1 839 These relations are the selection rules for the emission of light polarized in the z direction Rather than calculate the matrix components for a and39y separately it is more convenient to calculate those for the coma binations a z39y rsin 0 8quot and x 13y 7 sin 6 6quot For the combination as iy we have the integrals fle39 sin a 1913quotquot sin 9 d0 jam H s dgo 840 SELECTION RULES FOR THE HARMONIC OSCILLATOR 117 The integral over 0 is non vanishing only if m m39 1 From the recursion formula 486 1 sin 3 Ptquot Plath1 Plaza 841 quotn1 we see that the integral over 6 is nonvanishing only if 2 239 i 1 Similarly for the combination a in we have the selection rules m m 1 Z l t l The selection rules for the hydrogen atom may therefore be written as A1 i1 Am 0 i1 842 The selection rules for the different types of polarization are of course signi cant only when there is a unique 2 direction due for example to the presence of a uniform magnetic eld This subject will be dis cussed more fully in the following chapter where the Zeeman effect is considered It is apparent from the derivation of the above selection rules that they are not limited to the hydrogen atom but are valid for any central eld problem where the angular pertion of the wave function is identical with that of the hydrogen atom 8f Selection Rules for the Harmonic Oscillator Let us new con sider the system consisting of a particle moving along the 1 axis with a harmonic motion Let the charge on the particle be 8 and let the charge at the position of equilibrium be we The instantaneous value of the electric dipole moment will then be em The wave functions for the system will be the eigenfunctions of the harmonic oscillator sec tion 5e A transition between the states n and am will be possible only if the integral efll nxll m d3 is different from zero From the recursion formula equation 4 99 we see immediately that the integral 843 will be zero unless m a h 1 This is the selection rule for the harmonic oscillator only the funda mental frequency v can be emitted or absorbed by this system We shall later require the exact values of the integrals in 843 The wave unetions are m 1 5an Nae 2 1135 845 a it 1 H 5 ii NW t 2 Where 118 TIME DEPEN DENT PERTURBATIONS RADIATION THEORY so that e F j niexln 1 Nanl ffe 2HRSEH1 d3 8 46 V oz By the use of the recursion formula 844 this reduces to nlexln 1 2Vth1 fc52f1nsn 1Hndx Nn1 2 Nn1 3ltn1 N fwxne down N 847 n 11 Introducing the explicit expressions for the normalizing factors we Obtain n1 elexln 1 3 20 848 Similarly quot nleznln 1 NRNRHI faquot naillime dx 3 N n1 n 2v NR 8 5 v 8 49 8g Polarizability Rayleigh and Raman Scattering If a is the polarizability of an atomic system then an electric eld E inducesa dipole moment Re 013 in the system Let us consider classically a system in which an electron of charge e is bound elastically to an equilibrium position at which there is a charge 6 If the system is subjected to an alternating electric eld of strength E Re cos wt the classical equation of motion is 032 m git239 kr eE0 one at 850 Where r is the displacement of the electron from the origin and k is the force constant of the forces binding the electron to the equilibrium position The steadystate solution of this equation is 8 6E cos a Eocoswt 39 a r 2 2 851 tic men 22qu k where wo J The dipole moment of the system is R er N N 3N NOTATION q3 q1q2q3q3N Use 3N 3N 3N q p E q1P1q2P2q3NP3N zqipi i1 dP3N dPldP2dP3N Holonomic system described by a minimum set of independent quantities NUMBER DEGREES OF FREEDOM minimum number of independent coordinates required to de ne mechanical state Holonomic constraints fq1 qk 0 each such equation reduces the number of degrees of eedom by one Example two particles tied together gt x a y b 0 z c 0 Nonholonomic constraints given by inequalities harder example friction Statistical mechanics only has holonomic constraints HOW COORDINATES EVOLVE IN TIME Newton s Equations of Motion Cartesian coordinate in aVBxi i12N conservative systems in SM Vtime independent gt F VV Two other use il formulations Lagrange and Hamilton Lagrange Equations of Motion Take Newton ax V Vx1x2 Z3N I de ne kinetic energy 1 2 2 2 T 2mkxk yk Zk k1 a 3x mixi De ne Lagrangian L T V Then from Newton s equations of motion a a 8T 51 95 219 dt ml dt dt axi SO 1 3L a L LAGRANGE 3N equations dt Bit ax EQS OfMOTION L Lx1Z3NJ39c1z393N i e positions and velocities Important property Lagrange s equations have same form in any coordinate system YOU PROVE IN PROBLEM SET Advantages 1 easy to generalize to nonconservative non holonomic 2 scalar equations not vector as Newton 3 natural eld equations 4 easy to quantize Hamiltonian equations of motion Note that Lagrange is 3N 2nd order in time Want to replace them by 6N lst order in time More convenient for SM Use new coordinate system General Momentum p De nes pi 1i M Example Cartesian coordinates 3L px 39 mixi I 8x so pxi is ordinary linear momentum If qi is angle 17 is angular momentum We want the equations of motion in terms of qN pN not qN q39N Recall the Legendre transformation We have LqN q39N t and want to go to some other function say H HqN pN t De ne in anticipation W QM 5 M34 23 ch WL r MP 7 32 Mb Wm H quot va W12 quotHauevlmz M Ma team A WM CE 2 a 3 7 V7 3 it it a g m V V651 4L 75592310 I k a M 4 Em M 0 7amp9 1amp6 M41 m lgkta39 412 mu fogzt m g 76 Boerk7 5 7 3am in c pigmacaw mgf w A wcw hat Ft tVJk is o posnldi QdZFd t 7L MW 512 wrheAmdz h a 32M 674 jun Md by 2395 3 v f 58 050W M m m M dxdad t wt in 1 r d Cow46539 aw w W396 Wong J bray Quavane c a art 49 Wm M m M Wi km 5mg 1 of 3 034a CayWtquot Ma 5 may 075 quot5 My W Com a A mi k th FCMI F i ln Kcaw a Ohr gie lmo Cam Mrwufcol A aw Emil fdxz xz I I I 77m mei d f5 w Macy 2 M bmm h ume H g aw r f kwa JR dlfiw 1W qrod J 17f lmgm 1quot Z Lgquot Viiquot be 74a 7a m s3 A Mm Q3 71 M m 3 0 3 gi 4 coma a 77 W 6 5c WE WW 1703 Q i5 4 79 f u egg debt x 7amp0 39We anzr Q mng 5 7A A Wm m 3W 1m Wm M mm 3f 4 3 0 3 egW L mmhc g g l I quot Z M f4 Wkly 0 wry 03 uxquot M m6w c7 Cidgt491quot a DWI5quot quotit X E ugt n3 lx 190191 7J3 r1 0am0 a 711 mow Mac x 39 quotQ 4 7 Llt chn of f faUI39vCTJWm i4 W wwm fw zg ych m 772 494 41 cf 22446 M W 79 Q u 1 A WEE l7j elm 1 jxgx 3 Arr 1 f Vfquotp 13 quotg cfj h Ala acadu 5 i Vi Vf if 35765 Rang kg 5 kn 312 1 3a WG jigMac 2sz if ML 1iji397 7 W WM GLHW 5lm H i271 L V Vy LLVzy 76173531 V MWJ 74 WAS 6 a 5 V v WWW 7 li 7iquot 39 6 939 f0 6 if f VNquotW WV f t 177 7 77 6 Jt 1M a 2 H 11 o 2m 39 I M 32 7h M hers m W a Mr PM mg W down 3 7M1 mtchz39feoxc 3 Man 5 W 52 K f Z g 6 79mm 04 mwa Qfowya M2 quotkW74 3 714 W malfuw 47gt 7L Avg14 quot491quot 3 f fdxiadf 3 ti fdw daw VV 1 Vj V fu Ed th 7399 W n 32 fawis Ar dz dxdeigamp In 0 wquot M WW M WWW 4 4 WWW W M Awwaam W Y gtLg M e 06 A 7 15 7 whats MasMm Manea 4 awkwin 744m 7L emu 39 t 39 Y 7 A M 1ng 4x434 0 V1 n 9 WP fat quotWmA skunk A 5114 Q judlto pain H jg 1 m fdt 4 75 71L 39W a 2 mawa 9 maria 12 General Concepts with two arbitrary functions 1 and 22 both complex and only chosen so that the integrals exist With 9 the operator 73 and a real this yields h an Elixv M h 611 8021 7qu l alx E tka ax Jd3xl alx5a5 n or dexux 3av ffxx iauv 6x 6x Reordering leads on to fxxiwwr z 20 teacu wI 8x or by partial integration on the lefthand side to lfxu v Zeldgxuquot 0 which of course yields the result of a Eq 75 again Problem 8 Derivatives of an operator Let fpx be an integer function of the operators phxk Then the general relations a f gg fvpk and 3f EE ack 82 with the abbreviation 139 L9 Elm9f shall be derived from the commutation rules Solution The commutation rules are Pepi 2 0 xkix 0i pk xl 2 5hr 83 From these we construct 81 and 82 in four consecutive steps 1 Let fp then we have fax 0 and if812 z 5 Hence Eqs 81 and 82 become ppm 2 0 and pbxk 6 Le they are satisfied according to 83 In the same way for fx 6 f Mn 6 af pk 0 they may be proved to hold Problem 9 Time rate of an expectation value 13 2 Let 81 and 82 hold For two functions f and g Then they hold as well for any linear combination elf r 629 with complex numbers c1 and 62 in consequence of their linearity 3 Withfand g they hold for the productfg For 81 this is easily checked by direct computation 6 5 8f gnaw3cquot an gpitrprg 7 fartfar9fprgprfal MPG with the two central terms cancelling An analogous computation is easily performed to prove 82 4 It then follows that the relations hold for any linear combination of products consisting of an arbitrary number of pk s and xk s ie for any integer function in these variables as had to be proved Problem 9 Time rate of an expectation value Let A be the expectation value of an operator not explicitly dependent upon time in a time dependent state w How does A change with time What follows for in and pk Solution The expectation value ltAgtlt Al gtldtWitlAWitl 91 has the time rate gltAgtldrtwAixwAwt 92 The time derivatives of the two wave functions 1 and up satisfy the Schrodinger equations h it a1 ar Trimmer 93 with the hamiltonian operator H being hermitian Harri Putting 93 into 92 we find 039 i Eon gldrttmwAtr r AHM or in Hilbert space notation d 3 on ltH 1Angt ltwAHwgt 94 LECTURE 139 1 o M May cit9 23 57 1 Inna MMN now M NH 13391 ffw 3 AM i 39 257 M 74 m ame nm femo 26 I General Concepts Problem 1 Law of probability conservation If the normalization relation Howe 1 11 is interpreted in the sense of probability theory so that d3xtfltli is the probability of nding the particle under consideration in the volume eiement d3x then there must be a conservation law This is to be derived How may it be interpreted classically Solution The conservation law sought must have the form of an equation of continuity 39 divs 550 12 d with guymime pow 13 the probability density and s the probability current density As p is a biiinear form of gt and its complex conjugate Eq 12 can be constructed only by a combination of the two Schrodinger equations it 811 it Mrquot H H 14 g 391 i a quotI i a with the hamiltonian h2 v2 W 15 2m the same for both equations Thus we nd it 80 H H I 8 According to 12 it ought to be possible to write the left hand side in the form of a divergence Indeed we have 2 t General Concepts ma nktaxi Leg 2 h g WHIIIIiH Iz r fgitllwztthJIVZWH W Wrdivtl WtfHIIIWIquot 2m so that we may identify si V th 16 2m Classical interpretations may be arrived at as follows If the quantities p and s are both multiplied by m the mass of the particle we obtain mass density pm and momentum density 9 P m 3 N in y pmmm gms yrswhat 17 and the equation of continuity may be interpreted as the law of mass conservation In the same ways multiplication by the particle charge e yields charge density pa and electric current density j p ep jxes 18 and 12 becomes the law of charge conservation It is remarkable that the conservation laws of both mass and charge are essentially identical This derives from the fact that one particle by its convection current causes both The expression for the total momentum of the Schrodinger eld derived from 16 and 17 a P J 43X 2J diriWVtilllVtwh i may by partial integration in the second term be reduced to pJd3xw Vw its corresponding to its explanation as the expectation value of the momen tum operator hi V in the quantum state 1 cf Problem 3 2 ID p Problem 2 Variational principle of Schrodinger To replace the Schrodinger equation 52 v2uvnvuan 21 2m by a variational principle for the energy Problem 2 Variational principle of Schrodinger 3 Solution Since the constraint ld3x we 1 t 22 holds for any solution it of the differential equation 21 the energy will be found by multiplying 21 with w and integrating over the whole space hz Ezj d3Xl WV2 VU I 23 2m A partial integration in the rst term yields according to Green s law d3xtll k39i72tll df VtIj a exv v 24 Now the normalization integral 22 exists only if at large distances r the solution if vanishes at least as tilocr 3gt0 Under this condition however the surface integral in 24 vanishes when taken over an in nitely remote sphere so that 23 may be written EJd3x VWtht tii VHNI 25 2m This equation is completely symmetrical in the functions 11 and It as is the normalization 22 so that it might equally well have been derived from the complex conjugate of Eq 21 2 V2wVrltZ Etll 21 2m it would not be dif cult to Show that 21 and 21 are the Euler equations of the variational problem to extremize the integral 25 with the constraint 22 We shall however make no use of the apparatus of variational theory and prefer a direct proof instead Let tilI be a solution of 21 that belongs to its eigenvalue 131 It will give the integral 25 the value E A Let us then replace if by a neigh bouring function gift 5w with latl being small but arbitrary except for 22 still to hold for A6 as well as for tilAI ld3xll i5wll i5t l 1 and therefore fd3xltp25l 5 fdgx5 15t 0 26 4 General Concepts Setting 811445111 into the energy integral 25 the energy becomes 8116i511 with 52 SE lagx EEWI39VW VII139 WitH WWWquot 35 Jd3x V6 V5 V5w5 2 2m Here the rst order changes stand in the rst and the second order changes in the second line By partial integration in the sense opposite to the one above we fall back in the first line on 561 V2 11 and 5119 Valli1 where 21 and 24 may be used to eliminate the derivatives E g we then have 2 J43 Verve View 51 mistm m so that with the help of Eq 26 the rst line of 2 may finally be reduced to secondorder contributions only 2 ugliest WMZHV Emaelz 28 Since no linear contribution in 3gb or MI remains E a clearly is a maxi mum or a minimum for W 0 ie for i113 being a solution of the Schro dinger equation Whether we get a maximum or a minimum will be decided by the sign of 28 To get some insight into this last question we make use of the set ti5 of solutions to 21 to form a complete orthogonal system of func tions ld3xtlZti39 5 29 We then expand 541 with respect to this system 511 chdxv 210 Eq 28 then renders 2 8 2 cjfcv 33 2h thJIV bVV EA IJIV if V 9 m 2 226364 Weill2 VztfvHVEiltfv a v 2m or using 21 and 29 SEAZlculWEH EA 211 Problem 3 Classical mechanics for 8748 averages S If i51 is the ground state we have E 2 E1 for all states p so that the sum 211 is positive The variational principle therefore makes 8 a mini mum No such general rule can be established for excited states where the sum 211 consists of positive and negative terms Problem 3 Classical mechanics for space averages To show that Newton s fundamental equation of classical dynamics d quot F 3391 I t l with p the momentum of and F the force acting upon the particle still holds for the space averages expectation values of the corresponding operators inquantum mechanics Solution If the force F derives from a potential F VV and momentum is replaced by the operator ft8V then the two space aver ages in Eq 31 are de ned by p fd3xd vtlu 32 F J39 d3x 9519 VNJ 33 Our task then is to prove that 31 is valid for the integrals 32 and 33 if if and dz satisfy the Schrodinger equations i1 an it V2 v 1 8 2m if w a aw h 34 V2 n V at 1 6 2m d1 d1 We start our proof with the time derivative of Eq 32 h it P 7 33xttfr Vii ltwl P VWI llW l S W where the surface contribution of the game integration in the last term vanishes and has been omitted Replacing W and u according to 34 we may proceed to A Wt d om Wye W eWi M 4 1amp0 P ld3x VV fF as was to be proved I Agt x448 Macao ME maze Claw 1 0 3 5 g 5 543 for FVx Viagt0 Ffa xOixJX ff Pro em 4 Classical laws for angular motion 34 pix Wow FIX Kw To show that the classical relation between angular momentum L 2 r x p and torque T r x F where p stands for linear momentum and F for force dL 5 u w Eng am still holds for the space averages in quantum mechanics Solution As in the preceding problem we start by constructing the space averages h 3 7dewaxm mm and T fd3xtixrxVVtf 43 The wave functions 10 and w are again supposed to satisfy the Schr dinger equations 34 Problem 4 Classical laws for angular motion 7 We begin the proof of Eq 41 by differentiating L Eq 42 h L dax W Trx V lllr x In the second term we use the identity WWI VWMWVW to the rst term of which we apply the general vector rule jd3xrfo0 44 with f in ll Thus we arrive at h L J 3er tzV 1 tJV I z where we replace the time derivatives V and g5 according to 34 g JAM xV2 Vz1V2ip fd3x erwwwww 45 Now in the rst integral the bracket WWW 72 W WWquot W0 is the gradient of the scalar function f ng t V111 so that according to 44 this integral vanishes In the second integral the bracket is equal toVWx up We then use the identity VV I i1 VWWt Ft WVV and for f Virgil again the vector rule 44 Then the integral becomes nahy 3313er VVwWA z fe3xrxtp vm ie it becomes identical with the torque average 43 aSwas to be proved 5y NW3 9 C m gf sz effdef 39rf Agra gm axezdatox 03 2 72 apr 2 me j 42 E w CPU 77lt tamam mam m 42444 6 if 5 WM lt5 mfg Mfma Qha 34m WV H Rh S VgJAW J an 1 37 9 3quotquot V V 4 X x g v vv r Mcf xd wzmrgfwb jAEA39 lt1 at fat y wait quot 54 x m mugya J8 gm M W WM Jjagww Gnd 24an ux 94f LUWM 77 lt2 fgt f f VP 47 1 Wm 9 at Sklrg57Lo In 32 a 7 3 M 5 EE39ffE 76043 M a1 mww 5 W m m k5 pg MR 413 ltf39gt fa r fvm VOW3 m ms WW vmw wv M ltfjgt f a VWf WM 3793 ltfy zz af xW lt5gt W 3 NM 3 r 56 a fee garw 9 Z mjf ts r xx w z WM m g 31 aimwe 93mm miW I ttkg If 0 FIgt quot 2 dz VIA W W 3F W S 1499 wkm tfnt k 3 w 34quot Wm e Jac t m mug3 r mwgm wgmmmac Q31 t5 W1 cueuw 76quot x aquot 9630 a w cowhfym 8 General Concepts Problem 5 Energy conservation law If the energy content of a Schrodinger wave field is described by the space integral 25 of problem 2 the law of energy conservation should be of the form aW td39 S 0 51 1 a V t with W the energy density and S the energy ux vector This law shall by derived by constructing the ux vector 5 Solution We have found in 25 that E jd3x W 52 with WiVIJ39V it V213 53 2m where the rst term is the kinetic the second the potentiai energy density According to Eq 51 we need the derivative h2 2 3VIZIquotWIV V VWWWI W 54 Since Vt W VMWW WVW WWW VMVWFtWZW we can reshape the kinetic energy part of 54 and write and 52 it it wv ttwtw ww WWW m 2m 2m th1 IIVtii 55 In the last terms use of the Schrodinger equations 34 permits us to replace space derivatives and potential by time derivatives The resulting terms F ii t ttto 1 I exactly cancel so that Eq 55 indeed is of the form 51 to be proved with s iwthwo 56 2m the energy flux vector Problem 6 Hermitian conjugate 9 Problem 6 Hermitian conjugate The hermitian conjugate 5 of an operator 2 is defined by 5362 tit a at ifquot 2quot 39 61a or in Hilbert space notation mila Wm 61 b with if and p any two functions normalized according to lll 1 alto 1 551 This de nition shall be translated into a matrix relation What follows for the eigenvalues of a hermitian operator de ned by Q 9 Solution The matrix of an operator is de ned with respect to a com plete set of orthonormal functions u 14 u 5 63 The arbitrary but normalized functions it and go then may be expanded it Esau 925pm 64 v 1 Putting 64 into 61 we get 2 Zatbimuvlun Z zatbauvlmun u v p v and since this is supposed to hold for any pair viz a it must hold for each term separately maize 919 65 We now use the set 21 for matrix de nition writing the righthand side of 65 39 or at u min The lefthand side may be reshaped as follows Qu u dr 2a a Ha mom 7 ungnVV 2 23 Hence it follows from 65 for the matrix elements of Q and 2 that till 91quot 66 ie the elements of the hermitian conjugate or adjoint matrix are obtained by transposing Univ and taking the complex conjugate of the elements of Q It may be noted that from 66 we get immediater 2 Q 26 OneBody Problems without Spin OneDimensional Problems Problem 16 Forcefree case Basic solutions The onedimensional wave equation shall be solved in the case V0 and the physical signi cance of the solutions shall be discussed Solution The wave equation 52 aw h an a 161 2m 5x2 i at i permits factorization illxtuxettl 162 because by putting 162 in 161 one arrives at 22 quot h V i 39 m 163 2m a i Q where he stands as abbreviation for the separation parameter Splitting 163 into two separate equations we obtain gimg ie gte quot A 164 and aajw bv if With real to the wave function is periodic in and MP independent of time stationary state with a positive the constant W M 3 8 z k2 166 becomes positive too and the solutions of 163 are as well periodic in space It is an essential feature of quantum mechanics that time dependence is of the eomplex form 164 the real functions sinwt and 05 on are not solutions of the differential equation 164 This behaviour so different from classical physics is a consequence of the Schrodinger equation being of the rst order in time MM 3 The physical meaning of the parameter on may be further interpreted by considering the operator on the lefthand side of16 1 to be the hamilt 3 onjanampgsistin only of Linekinetic energy 0 erator in our case It layera follows mam kinetic energy of a particle and must ence39 f w e positive real Our solution therefore is an eigenstate of the hamiltonian c 4 ty a Vv In the following we shall write i again for the onedimensional part of the Xe wave function satisfying Eo AA and u for its space part w Problem 16 Forcevlree case Basic solutions 27 Since k2 is a positive constant the complete solution of 165 or u k2 u0 167 nxAe Bequot quot 1688 so that the onedimensional wave function mxlgftl 5814 ithmwtl B ikxagtt If 1 Vpnl 1 t Rk 1 consists oi th running in opposite directions both with phase velocity ooh wk The physical signi cance of the space part 168 a of the wave function becomes clear when we derive density palWI 169 and ux h dill 6gbquot S zwf 5393 1610 S from 168 b We find amp 2 2 at 25ch a 23 plAl 18 a as x awful hk amt j 9 W 5lAl3 Bl3l em 4 The two waves of amplitudes A and 3 apparently correspond to two opposite currents Awhose intensity is given by their respective normalization constants and is proportional to k The density shows interference of the two tooherentl waves causing a space periodicity As long as no special reason like boundary conditions is given to achieve coherence it will be reasonable to take either of the two waves putting 80 and obtaining sgt0 or A0 giving slt0 The result then corresponds to the linear motion of paWMlmrdirectmn Admitting both signs of k we may therefore summarize the nal result as follows W WAS Ci IllxtCe quotxquot quot V Ehco k2 1611 paveVOCi h hk p Cl2 s lCl2 m 28 OneBody Problems without Spin OnoDimensional Problems Elimination of 1 yields frz k2 E 1612 2m so that pzhk 1613 is the momentum and U E 1614 m the classical velocity of the particle The latter is by no means identical with the phase velocity of the wave 271211 to Hg 1 U U ph k p 2 it is however identical with the group velocity 17 UgrchwJL 7211 m M 7lt ILAAJLS L CAM4k M NB The fundamental differential equation 161 may be interpreted as a Mahdi di hsion equation with an imaginary constant of diffusion D 621 a h W a Dr39m Whereas factorization plays rather an important role in quantum theory but not in diffusion problems the typical sourcetype solutions of real diffusion theory 3 m 1x o 1 I v lIxt I7 ldi k 2quot t are of no importance in quantum theory Time reversal in 161 changes 11 in 11quot Problem 17 Forcefree case Wave packet A wave packet shall be constructed and its development as a function of time be investigated Solution We start from the special solution of the wave equation found in 1611 WtkxtCke39 quot 171 with h w z k2 172 2m O A if With JO ake S m nyf rnh 5 991 A So adeum Problem 1 Forcefree case Wave packet 29 and Ck an arbitrary amplitude Constant Here It still is a free parameter so that the complete solution of the wave equation is obtained as any convergent integral of 171 over k Mm 1 dkrlxtk xr 173 Eq 173 describes the most general form of an onedimensional wave packet The amplitude function C k must in order to make the integral converge tend to zero at least as 11 with klwe Every suitable choice of 02 yields a Special solution Let us now construct the wave packet so that for an initial time t0 the probability of nding the particle it describes differs appre ciably from zero only inside a small region around x0 and that the particle moves with momentum pozfrkg This can be accomplished ifwechoose 3 7 M quot5 0 Wmiwa 2 t M a Rm Wm llX10Aexplfkoxl 174 4 because then the density x2 PX 0 lll Itx ll2 124112 exp as localizes the particle within lxlSa and the flux 1610 becomes f1 x2 h sx0 szolAl2 exp a l x pg k0 so that the particle velocity is v0 ko and pozmnahko the mo mentum of the packet Since the wave function represents one particle the normalization condition CO I a xo 1 CQ holds i e 1 til2 The expression 174 may be decomposed into plane waves using 173 and 171 CO wx0 f dkCke quot i 176 1412 175 anal W 1 A quot153 6quot Q E 51 tit3quot lit a quot2 I u 2 2w 7 VS 5 a f2k fa if ecu a Lot CA18 N 11 gag9R wot Wm CosL 2 mi Wff y 3197 xi l o 3ampXo 3 A e M e x 996354 74quot 53 L I20gt07L 31 ag2 cr3ugt36 a t 35 WW6 5750 W m 2 m QQ g W2z of 35 21 rmwm wd 2quot W me alwg 50 os f0 We i o v 5 Mk6 iboM f 30 OneBody Problems without Spin OneDimensional Problems This is a Fourier integral whose inversion is 00 Cic 1 I dx x0equot quot oe A x2 E J dxexp Eg Nico kn This integral may be evaluated using the well known formula 00 j dzequotz 1E 177 co The result can A expwa2kk02 178 TE can easily be understood in terms of Heisenberg s uncertainty rule In the initial state the coordinate uncertainty of the particle is according to Eq 174 of the order of sza as 18 shows to this wave func tion there contributes a spectrum of wave numbers it or of momenta phk around iltk0 of a width Ak1a or Apfza Therefore independently of the choice of a there holds the relation AxApzh 179 which is in fact Heisenberg s principle of uncertainty Having determined Ck from the initial state at the time 20 we are now prepared to evaluate the full integral 173 at any time viz the integral 00 4419 1224312 Mintquot fdkexp 2 a k k0 ikx 12mk 71 The exponent is a quadratic form in k so that again reduction to the complete error integral 177 can be performed The result is a 5 55 xz Ztazkox H Zn kgaz 5 W1quot EliXJ 2 Am Aexp m if 1 i 2 2a31iw ma2 ma2 9 SPmct FM my 53760 1710 Ni apV Problem 1 Forcefree ease Wave packet 31 A good understanding of this rather cumbersome expression can be obtained by again discussing density p and ux 5 but now at any time The former becomes t rkot 2 W m o 2 DXslll xytl2 Wexp w W z39 2 Noll aid 2 ma quot18 As a function of x the density giant still is a bellshaped curve whose X 1711 k 67252fo maxrrnum however has now shifted from x20 to x m 0t The WY Wot the wave group represented by 1710 therefore prop agates with a velocity 9079 group velocity particle velocity Hugh The denominator of the exponent in 1711 shows that at the same 4 a time the wave packet has broadened from its initial width at at t0 to an s 1 m 2 1 t quot39 W I quotWquot 2 a 81ltma21 mat at tt This effect can easily be explained from the spectral function 178 The wave number spectrum having the width Alc1a the veloc h f itres of the partial waves cover a region of Width AvEAkE so that the packet broadens by Axt Av 5t as derived above The ux is obtained from 1710 with the help of the relation t 39 wik walk v r 8x 0 fzt lz ma straightforward calculation yields by comparison with 1111 fitx 4 it sou peeve 93 In 1 ma2 1 1712 1 it follows that we by no means have spoo for all times as we had for t0 This again is a consequence of the nite width of the velocity pig spectrum At the acket maximum x 0 t E I39MZ leads to t E r 3 5 9V0 3 m g JXo S g f limpet 3 agged aoeessM g a f lz ihznu aer 23 5 9 WM nadir4mm in 13 z 1 V n 2 5 is 1th Mod ewe Ac 1 ml z a z a o N b h Wilma ampea le QIdflo k39li T WD quot 393 thm 3 qu 22 General Concepts It should be noted that from the normalization fa 19113 1 14 there follows la 3klfk11 148 This can be shown by setting 141 into 147 for 41 1 d3 WP lt2 n3931extcPkIdak39e tquotfkgtftks If here the integration over coordinate space is first performed2 we almost immediately arrive at the expression 148 Problem 15 Momentum space Periodic and aperiodic wave functions To deduce the probability interpretation of momentum space wave functions in the continuous spectrum by starting from periodic wave functions Mr in ordinary space and investigating the limiting process for infinitely large periodicity cube Solution Let L be the period in each of the three space directions x 522 Then the Fourier series h whenctzcze 39w gt m WM 151 k 2m includes terms only with components 27 ki In ni0 i1 i2 152 of each vector k This means that in 3 space for large L a volume element le includes L 3 d3k 153 27 states of different it s The normalization of series 151 can still be chosen by suitable choice of the coef cients 631118 square integral over the periodicity cube is l 53 illle L 3 2 2c cquot e tuquot w39 j d3x crux wry 13 I k 2 Cf however the remark at the end of the following problem m roo ng delZQm 53 Angular Momentum of a OneParticle System 83 FIGURE 57 spherical polar coordinates three components has a de nite value In quantum mechanics when angular momentum is conserved only its magnitude and one of its components are speci able We could now attempt to nd the eigenvalues and common eigen functions of L and L by using the forms for these operators in Cartesian coordinates However we would nd that the partial differential equations obtained would not be separable For this reason we carry out a trans formation to spherical polar coordinates Figure 57 illustrates these coordinates The coordinate r is the distance from the origin to the point x y z The angle 0 is the angle the vector r makes with the positive 2 axis The angle that the projection of r in the xy plane makes with the positive x axis is denoted p Mathematics texts often interchange 0 and 1 Most physics texts use the designations of Fig 57 A little trigonometry gives x 1 sin 0 cos 9 579 y r sin 0 sin p i 580 z r cos 0 581 r2 x3 y2 z2 582 cos 6 583 z tan q yx 584 To transform the angularmomentum operators to spherical polar coordinates we must transform aax aay and 382 into these Coordinates This transformation may be skimmed if desired Begin reading again after Eq 5111 To perform this transformation we use the chain rule Suppose we have a function of r 0 and 9 fr 3 qa If We carry out a change of independent variables by substituting r rx v 2 0 9xy 2 w w z 585 84 Angular Momentum into 1 we transform it into a function of x y and z rx y z 0X y z 90 y 2 gx y z 586 For example suppose that fr 6 p 3r cos 9 2 tan 9 587 Using 582 583 and 584 we have 39 gx y z 32 Zy xquot 588 The chain rule tells us how the partial derivatives of gx y z are related to those of f r 0 p In fact dyanwinA3plawaaxawwalawqalaw m d e l9wrlquotamp LLLLLLL am a a o sf 9 31 333v 39 69 T3 2 lt3 Y a c x 5 I o convert these equations to operator equations we delete f and g However it would not do to wnte for example c 6 6r if 3 9 71 a 545 a 6 It a X 8 gt6 or the rst term on the right side of the operator equation corresponding to 589 because this would imply that aar was to operate on GEM whereas according to 589 it should operate only on 12 For the operator equation corresponding to 589 we thus write 6 er a a9 a a a a 5 57 32 as 567 592 with similar equations for my and aaz The task now is m evaluate the partial derivatives such as Br3x Taking the partial derivative of 582 with respect to x at constant y and 2 we have 2r 2x 2rsin ones 9 593 x 2 s39n 0 39 5 94 ax m 1 cos 91 Differentiating 582 with respect to y and with respect to 2 we nd 6r 39 53 sm 0 sm qn 595 1 cos a 596 a any 53 Angular Momentum of a OneParticle System 85 Suppose we had di erentiated 579 with respect to r to get ax 5 sm 0 cos q 597 Then since 3r 1 5 a axar 593 we conclude from 597 that ar 1 5 H sin 8 cos tp 7 599 which disagrees with 594 In fact 599 is wrong To obtain 597 we differentiated 579 with respect to r keeping 0 and q constant Hence 597 is more fully written as sin ocos p 5100 r om Moreover we can only use an equation like 598 when the variables kept constant are the same on each side of the equation Thus 3r 1 5 axa r 5101 6r 1 33 ax6r 5102 By not paying proper attention to what variables were being kept constant during the partial differentiation we obtained the erroneous equation 599 Thermodynamics offers rich opportunities for committing such errors From 583 we nd 5 7 5103 39 cos 0 cos p quotquot r 5104 Also 30 cos 0 sin 9 a0 sin 9 55 r 7 T 5105 From 584 we have 2 sin tp 31 s rsin 0 5106 31 cos w 3y r sin a 5107 a 5 5108 86 Angular Momentum Substituting 594 5104 and 5106 into 592 we nd a 0 a 39 a smocoswar 5 i2i 1109 ax r 60 rsin 069 Similarly a 6 60595111973 cos a 6 sm 05m pa r4 5110 a a sin 9 a 6 COSBEH r ET 5111 At long last we are ready to express the angularmomentum components in spherical polar coordinates Substituting 580 581 5110 and 5111 into 563 we have 6 sin 9 6 L zhr sin 03m 90cos 06 T 6 cos asian 8 cosltp a rcososm051nqa5 r ao mag a a L zhsin 99 cot 9cos 995 5112 Also L i cos 3 cot 95in 3 5113 1 39 I L i a 5114 2 23 By squaring each of Lx L and Lquot and then adding their squares we can construct L Eq 566 The result is Problem 510 32 a 1 an 2 a L 893 cot 0 30 sina39g39 312 5115 OccaSionany L2 is written in the form 1 6 a 1 an 2 2 L h sin 6 as 5 9 39 ring 3 avg 5116 Using the de nition of the product of operators we readily verify the equivalence of 5116 and 5115 Although the angularmomentum operators depend on all three Cartesian coordinates x y and 2 they involve only the two spherical polar coordinates 0 and 92 We now nd the common eigenfunctions of La and L which we denote by Y Since these operators involve 0 and 9 only Ywill be a func tion of these two coordinates Y Y 0 9 Of course since the operators 92 714 F 4V z z gz 5 I e lt15 j ow Lay21w L I 5 MM CHAPTER 7 SPHERICALLY SYMMETRIC SYSTEMS 7 1 The Schrodinger equation for spherically symmetric potentials The potential energy of a particle which moves in a central spherically syn metric field of force depends only upon the distance s between the particle and the center of force The Schrodinger equation for the energy states of such a system is therefore W 33112 was o quotz11 Spherical polar coordinates Fig 71 x w rsin icoso y rsin dsin at r cos 9 72 are appropriate to the symmetry of the problem since the potential func tion Vr is independent of the angular variables 8 and o The Schrodinger equation 7 1 expressed in these coordinates is 35 11 3 as 1 6 r26r 6r 72 sinaaa Smear sin266755 1s more o M The energy states of the system are 2 determined by those solutions of this equation which are continuous have continuous derivatives in r 8 and qt and are for bound states quadrat ically integrable Solutions of Eq 7 3 can be con structed by the method of sepa ration of variables To apply this method we attempt to nd a solu tion of the form is RCTY9 74 in WhiCh RC7 is independent 0f the FIG 71 Spherical polar coordi angles and Y0 o is independent nate system 188 I 7 11 V sonnonmenn EQUATION srnnnrcan SYMMETRY 189 X9 of r Substituting Eq 74 into Eq 7 3 and rearranging we obtain 1 9amp5 33 E mews Rdr dr 1 1 a ar 1 3 Mlism 5561118597 sin2 3133 7M5 In this equation the lefthand member depends by hypothesis only upon the variable r while the right hand member is independent of 7 Conse quently the equation can be satisfied identically only if each member is a constant C The energy is determined by the equation for the radial wave function 127 1 a T2 are 2mm m VW R 0 7 6 583 d7 7i whereas the angular part of the solution Y9 it satis es 1 a or 1 a2Y m 55 81139 l aha 9 53 E nation 74 is inde endent of the ener E and of the potential energy VOL Therefore the angular dependence of the wave functions is determined solel b th 1quot a s 39 39 solutions of E 39 icgar ess of the s ecia We shall rst give attention to the solutions of the angular equation and return to the radial equation in the discussion of speci c examples Equation 7 7 for the functions Y can be separated again by the substitution YCG lt1 P04 78 We obtain 2 7 115sin 6sin 9 051112 a 2 gig 7712 7 9 in which the separation constant is written as m2 The second of Eqs 9 is 1qu 2 which has the solutions 39 e ei m 711 e e fir 190 SPHERICALLY SYMMETRIC SYSTEMS case 7 By the substitution u cos 6 the rst of Eqs 7 9 is reduced to d we lt M hl idpi 1 2 P Oi which is the differential equation de ning the associated Legendre fanc tt ons 1 It is possible to solve this equation by the method of series that is by a procedure similar to that used for the harmonic oscillator Section 59 This is done in detail in mathematical works devoted to the subject of spherical harmonics and it is found that bounded differentiable solutions of Eq 7 12 exist if and only if the constant C is 7 12 0 ll 1 7 13 where l is a non negative integer and 171 has one of the integer values l ml l l 1 2 These are therefore the only admissible values of C and m if xix is to be the wave function for a physical system Note that the functions de ned by Eq 7 4 1 are therefore single valued functions of o In other subjects such as electrostatics the functions Y0 it must represent physical quantities known in advance to be single valued so that the requirement that m be an integer follows immediately from Eq 7 11 Wench it however is interpreted physically through the product 3V4 which is independent of m r 39 real 39 hence the condition of single valuedness of Qprlied directly The requirement that m be an inte er 39 ther from the conditio f boundedness of it anm m 7 12 are irregular at the poles u 13 72 Spherical harmonics We shall now construct the functions Y0 e by a method due to Kramers quot which is more direct than the method of series We have already remarked that the angular functions Y are independent of E and Vr therefore no generality is lost in this 1 E T Whittaker and G N Watson A Course of Modern Analysis 4th ed Cambridge Cambridge University Press 192 Section 155 ff 2W Pauli Die Allgemet nen Prinzipt cn der Wellenmechcnik Ann Arbor J W Edwards Publisher 1947 p 126 3 H A Kramers Quantum Mechanics Amsterdam NorthHolland Publishing 00 1957 45 p 168 H C Brinkman Applications of Spinor Invariants in Atomic Physics Amsterdam NorthHolland Publishing Co 1956 Cf also R Courant and I Hilbert Methods of Mathematicei Physics New York Inter science Publishers Inc 1953 Vol I Appendix to Chapter VII 7 2 SPHERICAL nanmomcs 191 part of the problem if Eq 7 1 is replaced by Lsplace s equation var 0 Ha Solutions of this equation are called harmonic functions A single valued harmonic function which is continuous in a neighborhood of the origin can be approximated arbitrarily well by a polynomiai in x y 2 If such a polynomial is to have the form of Eq 7 4 it must be homogeneous in these variables ie it must have the form its 2 plt11l apqrxpyqzh 7 15 in which the numbers a must be chosen so that Eq 03944 is satis ed The integer l is of course the degree of the polynomial The number of terms in the sum 245 is 96 1l 2 For a given value of p the index q can have the l p 1 values 0 l 2 11 while 1 can have any of the l 1 values 0 1 2 2 thus the total number of combinations Hows tow1gt1sctnc2i is the number of linearly independent polynomials of degree I Now the Laplacian of the functiOn 7 15 is a polynomial of degree 5 2 hence the requirement that Laplace s equation be satis ed imposes am 1l conditions on the coefficients up There remain therefore iil1l2 ol 1l 211 linearly independent harmonic polynomials of degree 2 Explicitly these may be taken to be I 2 0 1 l a 1 3 y a la 2 my ye 22 x2 y2 222 02 yz l m 3 2032 322L913 3212 3102 3232 yy2 322 222 3x2 222 3y2 xyz and so on 746 It is evident that each term in the polynomial 7 15 is proportional to 7 Hence it can be written lab 2 rlYl0gt db in which Yzw gt is a spherical harmonic of order l The separation constant 717 196 SPHERICALLY SYMMETRIC SYSTEMS CHAR 7 The integral in this expression can be evaluated most easily by sub stituting the expression 7 28 for one of the Piquot and 7 31 for the other thus 1 m m 1 m 1 1 Pl 10lsz l W W 1 z 111 1 1 X 1012 1l2 1Id 7 40 The function 12 11 has a zero of order I at each of the points u 1 whence if 7410 is integrated by parts I m times the integrated part will vanish each time The result is 1 m l 1 P1 lPTl2 dIJ 222092 1 21 X 1012 10 2 1ld 1 2212ll211 2ldl 741 The integral in this equation is Problem 7 6 I 2 z 211 2 nu MP2 lt2l1 whence 1 2 l 1P zquot12d r 1 E i Zip 742 and Eq 739 becomes 2 l m 1 7 43 m2 11 2quot 2z1z m1 The orthonormal functions Y2quot are therefore rm 4 HM2141 amp cos a 7 44 in which the phase factor quot has been chosen to agree with that most commonly used in the literaturem The function Y1 is related to Y r 1E U COndon and G H Shortley The Theory of Atomic Spectra Cam bridge Cambridge University Press 1953 Chapter 111 Section 4 J Blatt and V F Weisskopf Theoretical Nuclear Physics New York John Wiley and Sons 1952 Appendix A Section 2 7 2 SPHERICAL HARMONICS 197 through the identity 7 32 YT mY zquot 7 45 The functions Pi can be computed from the Legendre polynomials PP id l2 1l 7 46 39 2m dy by means of the relation m m d39quot P 5111 0 dIL m P1 In this way one can construct the table 1 0 0 V47r Y cos 0 Y e sin 9 o 5 i 2 1 15 39e 748 Y2 Tea Traces 0 1 Y2 8 sm cos Y e2 sin2 9 The spherical harmonics YIquot0 4 are a complete orthonormal set of functions ie Y l39y 51139 6mm and any function f0 qs which is continuous and has continuous rst and second derivatives can be expanded in the form an 1 m 4 Z 2 mm 4 7 50 10 mix l The coef cients ffquot are given by I Yr 1 lt7 51 and the value of f0 45 at the pole 0 0 is Problem 7 11 m gt imam a i 214 11 2 7 52 10 l0 In the particular case that f0 4 is a spherical harmonic Y we have f 39 YI I Y which is zero unless l 1 Consequently Eq 7 52 198 SPHERICALLY SYMMETRIC SYSTEMS lCHAP 7 FlG 72 Coordinate systems for the proof of the addition theorem becomes cf Problem 7 10 Mo 45 214 1 Y9 Y2 2144 P cos a in 7 53 This relation can be used to prove an important formula called the addition theorem for spherical harmonics which establishes the relation between spherical harmonics referred to two differently oriented systems of axes Let the point P on the unit sphere be de ned by the coordinates 9 and with respect to the axes c y z and by 9 and with respect to rc y 2 Fig 7 2 The rectangular coordinates of P in the two coordinate systems are related linearly by means of the table of direction cosines of the angles between the primed and unprimed axes Consequently a homogeneous polynomial of degree l in 9 112 becomes after transformation to the primed coordinates a homogeneous polynomial of degree I in x y 2 Also the Laplace equation 2 62 62 6 vii 63 W 52 2 0 7 54 is invariant to this change of variables Problem 7 12 ie it becomes 2 I 2 H W W o 7 55 2 V 1p 6x 2 6y 2 62 2 1 G Joos Theoretical Physics New York G E Stechert and Co 1934 Chapter I 7 2 SPHERICAL HARMONICS 199 It follows that a spherical harmonic of degree l in r y 2 is also a spherical harmonic of degree 1 although one of different form in x y 2 The meaning of these considerations is clear The physical system is spherically symmetric and the direction chosen for the coordinate axes is of no signi cance for the mathematical description of the system The angular dependence of the wave functions has been shown to be given by the functions Yquot6 5 which are a complete set of 2t 1 harmonics of degree I Consequently it must be possible to express the spherical harmonics of the same degree Yquot9 35 with respect to any other system of axes as linear combinations of the YIquot6 4 This invariance to rotation of the coordinate system characterizes the symmetry It is the fundamental property upon which the entire theory of spherical harmonics depends In particular the function P cos 6 is a spherical harmonic of degree l and therefore 1 P cos 939 2 am rm a 7 56 I m in which the coef cients am are a me i P cos 639 P cos at We 7 57 The last equation follows from the fact that the Legendre polynomial is a real function Substituting 6 45 for 6 in Eq 7 53 we obtain Pi cos 639 ms 45 214 11 Y39zquota elm 7 58 Now at 6 O the angles 6 and 4 are equal to the angles a and B which are the polar coordinates of the z axis with respect to the 3 y z system Fig 7 2 Hence 47139 m 39 am 21 Yi a ti 7 59 and l P cos 039 2 We New 4a 7 60 mI l The relation between 8 and the angles 6 45 is determined from the geometry of Fig 7 2 cos 0 cos a cos 6 sin a sin 9 cos d 3 7 61 1 The two polynomials have different coef cients agar 200 SPHERICALLY SYMMETRIC SYSTEMS loan 7 The angles 6 and q5 are determined by the direction of the unit vector i pointing from the origin toward P and it is often convenient to use the notation Y39zquot6 45 Y39zquotf 762 Thus if r is a vector in the direction of the z axis we have cos 6 f f and Eq 7 60 becomes A A 47quot l a A m A P10 r m m Y7quot rm r 7463 Equation 7 60 or Eq 7 63 is the addition theorem 73 Degeneracy angular momentum The energies of the stationary states of a spherically symmetric system are those values of E for which the radial wave equation 7 6 has solutions which are admissible as wave functions This equation is equivalent to d2 2m ll 1 d T 7L31E Vr u 0 7 64 where TR u Except for a change in the range of the independent vari able 1 Z 0 this equation is of the same form as the onedimensional Schrodinger equation 5 3 provided the function 2 ll 1 57 Tz 7 65 V V is regarded as an equivalent one dimensional potential energy function The solutions of Eq 7 64 are similar in character to those of Eq 5 3 in that there is in general a partly continuous and partly discrete spec trum of allowed values of E It is clear that the energy eigenvalues de pend in general upon the quantum number l but that they are inde pendent of m Therefore each of the functions RYfquotm l l is a solution of the Schrodinger equation 7 1 for the same value of E and since these functions are linearly independent the states of energy E are 2l 1fold degenerate The reason for this degeneracy is the rotational symmetry of the system In order to see this more clearly let Mr 6 4 be an eigenfunction be longing to the energy E ie H l T1 9 E MT 6 39 waM i 201 awf39m Slnce the angle 4 can be measured with respect to any direction per pendicular to the z axis the function iIT 0 QB 5 must also be an eigen function for the same value of E ie 39 quot l w 7 3 DEGENERACY ANGULAR MOMENTUM l l H I r1 0 4 e Ell 0 0 4 e Where e is an arbitrary angle Now if e is small we have Mr 0 e Mr 0 as e 1W o 0 hence introducing the operator D 6643 Eq 7 67 becomes Hw 6 eHDw 9 4 Ear 6 05gt was 0 03 By Eq 7 66 this reduces to HD DENr 0 45 0 7 68 This equation is true for all eigenfunctions of the Hamiltonian and smce these are by hypothesis a complete set of states of the system we have H Dl 0 769 ie the operator D 66 commutes with the Hamiltonian cf Section 639 9 The relation 7 69 which can be verified by writing out the Schro dinger equation in the form 7 3 Problem 7 18 states in quantum mechanical terms that H describes a system invariant torotation about the zaxis According to the general theory of Chapter 6 the operators H and D must have simultaneous eigenfunctions The functions YIquot contain the angle 4 only in the factor 6quot whence Demquot imamquot The eigenvalues of D are purely imaginary and hence D is not Hermitian The operator 6 L 6 D HIE 7 70 ela 6 however satis es the eigenvalue equation Lze M min 7 71 Saw 2 fd 2ltJ d ltLwgt 1Lz 0 d 0 1 z J my where the second integral is obtained from the rst by integration by parts The physical interpretation of the operator Lz follows immediately from the rules 6 83 Thus we have i 202 SPHERICALLY SYMMETRIC SYSTEMS CHAR 7 and is Hermitian Explicitly if it is any singlevalued function of we have whence h a h i 772 Lzx yax quot620y ypz The operator Lz therefore corresponds to the z component of the angular momentum of the particle The classical angular momentum vector is de ned to be L r X p The corresponding vector operator Lop 1013 X Pop has the components L Eq 7 72 and L yp 21 L 217 cpz 7 74 Note that 25 pg 0 etc so that the order of the factors in the terms of Eq 7 72 and Eq 7 74 can be changed if desired Lrgtltp pigtlt 139 7 75 The negative sign in this equation results of course from the de nition of the vector product Equation 7 69 which is equivalent to H L 0 7 76 expresses the fact that the zcomponent of L is a constant of the motion for any spherically symmetric system Hence the law of conservation of angular momentum holds in quantum as well as in classibal mechanics as it must since it is a direct consequence of the geometrical symmetry of the system Furthermore since 17 L1 11 Ly 0 777 In k ld d 7 EthL j to D 123264 a Jnml mm 7 E m 09w NHd Goa duneu Ar39m fwn 39 I 7 3 DEGENERACY ANGULAR MOMENTUM 203 or in general H L 39 0 7 78 an eigenfunction of H can be simultaneously an eigenfunction of L2 or L or L2 or of any linear combination of these operators The function RYIquot is now seen to be a simultaneous eigenfunction of H and Lz the 2t 1 fold degenerac has been resolved b means of the commuting The rotational degeneracy of the states does not occur of coursm a system which is not spherically symmetric If forces are introduced which destroy the spherical symmetry then the energy will depend in general upon m as well as l Such forces are represented by additional terms in the Hamiltonian which do not commute with L so that Eq 7 78 is 39 no longer true The introduction of such forces removes the degeneracy We shall show in Section 105 that for a charged particle the degeneracy is completely removed by the application of an external magnetic eld which de nes a special direction in space 39 The functions RY are eigenfunctions of the z component of angular momentum and represent states which are quantized with respect to the I zact39s We have seen that any direction in space can be chosen as the axis of quantization since every component of L commutes with H However the components of L are not commuting operators hence it is usually im possible to form a simultaneous eigenfunction of two different components of L It is a matter of algebra to show Problem 7 19 that the operators L3 Ly Lz satisfy the commutation rules L1 Ly ihLz L L ihLz L2 L3 ihLy 7 79 or in vector form Problem 7 20 L X L ihL 7 80 The equations 7 79 are the fundamental relations among the com ponents of any angular momentum vector They ex ress 39 cise form WWW ferent directions are not commutable operational We shall show in Chapter 9 that a complete characterization of the angularmomentum operators can be obtained from the relations 7 79 Some of the conse quences of these relations which result from the de nitions 7 72 7 74 V and the commutation rules 6 83 are listed in Problems 7 21 and 7 22 1 Cf eg L Page Introduction to Theoretical Physics 2nd ed New York D Van Nostrand Co Inc 1935 p 101 G Joos Theoretical Physics New York G E Stechert and Co 1934 p 132 W a a Wm M in 5 74quot 2 10IZ mm 1 Male au dd 2 76M 204 gtSPHERICALLY SYMMETRIC srsrsms can 7 The square of the angularmomentum operator L is L2 LL L L Lf 7 81 The operator L2 commutes with every component of L ie u m rv W L2Lz L2Li1iL2L1 0 7 82 for it follows from Eqs 7 81 and 7 79 that for example 1L2in L3 Li L3L1 L3 L4 LUle Lzl L111 LtlLU LzlLZr L2 Lu I4le ih LL LzLu LzLy LyLz 0 Furthermore because of the relations 7 76 and 7 77 L2 commutes with the Hamiltonian for a spherically symmetric system H L2 0 783 The operators H L and L2 are commuting operators and the energy states of our problem can therefore be written as simultaneous eigenfunc tions of these operators of Section 6 8 We have already seen that the functions RYquot0 45 are eigenfunctions of H and L2 and it will now be shown that Lzmo Id 1 2Y zquot0 4b 784 ie the spherical harmonic of degree l is an eigenfunction of the square of the total angular momentum belonging to the eigenvalue 1 1h2 The spherical harmonics are functions of the angles 0 and gt and the operator L has been de ned in terms of the rectangular coordinates x y z A change of variables is therefore required which can be carried out by straightforward substitution Eqs 7 2 Problem 7 25 Alternatively using vector methods we can proceed directly from the de nition L r x v 7 85 to arrive at L2 h2rgtlt VrX w h2rVX rX w h2r rvzw w v v W rvvp 7 86 Now if A is any vector and 5 any scalar function we have by the de nition of the gradient operator E 31 Avr A rVA rar rV Tar 74 THE THREEDIMENSIONAL HARMONIC OSCILLATOR 205 and Vr3 Application of these rules in Eq 7 86 yields 6 6 L251 h2r2V2 39 71239 6 3r r r g V gt 222 2 a hrViI hl 276T arO r V V 39Tf 222 26 26 hrV ha rr Tar 0139 l 6 WLZHII V2b1 T2 135 6r Now if V231 is expressed in terms of spherical polar coordinates the pre ceding equation becomes 2213 i 192 L l h isino aa s Mao sin26 a 2 7 87 ie the operator L2h2r2 is just the angular part of the Laplacian operator The relation 7 84 is therefore equivalent to Eq 7 7 in which 0 ll 1 Eq 7 18 The results of this section are summarized in the statement that the spherical harmonics YZ 0 45 are simultaneous eigenfunctions of the operators L2 and L2 belonging to the eigenvalues mh and ll 1h2 respectively Since these functions are a complete set with respect to functions of 6 and 45 L and L2 are a complete set of commuting operators for this class of functions The Hamiltonian H representing a spherically symmetric system commutes with L and L2 and these three operators form a complete set with respect to the quantum states of the system 7 4 The three dimensional harmonic oscillator A particle attracted toward a xed point by a force proportional to the distance from the point has the potential energy W Her2 new y2 22 which is spherically symmetric The Schrodinger equation for this system is h2 2 2 v ikwm 1 yf 0339 9 Ziox Xf i v 41 mamwk w f m m 27w WW Q A c B I 39 X gz afx ixl gh quot szyX gX 2le Wk M Ma flec 5 7 O C I if 75M ZZHrby Mfr 2 7quot Xf f ffg a7c SLOW AX395 97x Xf A spac M lg at 15431 A m 2 771 ho lle WI ymef Co mm f ll 2 3 K M r 2 ZXZZ7ZZZ PIC74 9344 M omLanw Er 214 92ihia m 5 ampY 6X 0 07 ax 939 i z w 1x 3 L hm w x 1 47 d3 5 MM 321 2 ugh M g ma53 253 r a 1L 9 quotr it w 172 Mak W 4 Game 379 A 2 J 7 714 m H 1 F2 irra2zizV 107 W AM t WM agomTfM L s CY f aw A 2 ve vey 1ewzm vo f 71 3 Meaw we 95m 751 HJLZ 0 39 zzh 441 L h 7 Maiz 7 LL mfp 1134sz 1 mg n c 99 oys39m n 44a cowh 1 W may 4311 emu re gv uv we her J 120 mmw m Z Zak A ritm fum CF a4 m39 n L2