Quantum Mechanics and Atomic Physics 1
Quantum Mechanics and Atomic Physics 1 PHYS 3220
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eeWnuw EA E 13 z r MZKQE Wmmam mm vxl 9W M awm WWW Mwa w my Fwn39 4 M m RMw gm mam g wavMa Aaniyn39m r I 1 neg Rm A e W a Amw W m v 47 a M an e a 39w w R Fl Zab NOTE r we Ls cu es xcn e am quotmmwwum up 1m Rtr0 0 gtwmmaw W Kira 9 MM km WW Wm M H WM W yW no 11qu ammuuuw 5 GM mtkonouw 9 w W gm W in 1 NW 1 ulgt 0 in 9 7 7 39 I m 2mm gt M 44m IltW cf c cquot 1 Wm Chin c 539quot lei MI 2 c t h I g m OWW g A Am Mml n A Am 4377 Au mm w V W 7 WW W wmve 44mm 1 s77 W W Wmf W 1 Mn Wm I DI WAM 443 A w AWgt 03 pm mum Ni ii mm mam 17gt ZdMW Y mgggi Ami NW A k mu m JM1 MAN W has ltmwgt 4mm Z cwwmw 7 A g ja W Tm s the Nu m MWW 4 mmlxxmw d quot quotc d1 My Adv a A An S LSW A 2 J23 Mn M Wm Ways km a n Um gm H amen Em SM Mn ffmwgm gt CE 0 E Akigt O LEK 3 l 9 J W WM MAvxgt a W W QWMR CM Jw W 0 A m W15 7 NM m r 7 MM MtW Eastfr 5W W 4 MW mmW39mx AI 1 641 5 WW T mu wgmwmg W i M u M A 5 A WWWquot gt Am AWN i A e 67W ltm ngt mm mm gt A Aim Smdany Jam Vii1 14 FM MM Miro ma 0 Wmwpum4amp xmwavma AiJ MM MM 39 WIN Ml HAW14AM memwaa MMJMW W4 m 15 Wmn1m Lani TNT we 2le mlo WM Mex 4 in rm 1 WW glam W ma 044W iv A 5 d WWV X wl WWm mywmg kwm R yigx9 ykslmmmw MAL 39 ELM WMAWM JJ MV W Ja BD 39 WMMWMM JMWQAWW ARM Mug VALme Ma WMM s a anmru Hrutovm was fuzllj lmulolrl any Wm M m w n 49 m 411 1mWyw4wm tx39vhpa CE W dam 1mm 1 1 l I Lm amp2ltvlmgtu 73x wx Wain ltm1Czwgt kc o O D u o o 0 c o x 7 7 L m ltV 1C1vgt 12 l cgt o a WWW Lx L o MW auf wanmwwmgM V A 4WD 1 max MW 4 2 am 39 4mm W4 Mm W ww Tn ilxz WWW gt LLEJM Spin 12 Recall that in the Hatom solution we showed that the fact that the wavefunction I r is singlevalued requires that the angular momentum quantum nbr be integer 0 l 2 However operator algebra allowed solutions 0 12 1 32 2 Experiment shows that the electron possesses an intrinsic angular momentum called spin with K 12 By convention we use the letter s instead of for the spin angular momentum quantum number s 12 The existence of spin is not derivable from non relativistic QM It is not a form of orbital angular momentum it cannot be derived from L E X The electron is a point particle with radius r 0 Electrons protons neutrons and quarks all possess spin s 12 Electrons and quarks are elementary point particles as far as we can tell and have no internal structure However protons and neutrons are made of 3 quarks each The 3 halfspins of the quarks add to produce a total spin of 12 for the composite particle in a sense T Nx makes a single T Photons have spin 1 mesons have spin 0 the deltaparticle has spin 32 The graviton has spin 2 Gravitons have not been detected experimentally so this last statement is a theoretical prediction Spin and Magnetic Moment We can detect and measure spin experimentally because the spin of a charged particle is always associated with a magnetic moment Classically H a magnetic moment is de ned as a vector p associated with a loop of current The direction of p is perpendicular to the plane of the current loop i righthandrule and the magnitude is u iA inrz The connection between orbital angular momentum not spin and magnetic moment can be seen in the following classical model Consider a particle with mass m q In charge q in circular orbit of radius r speed v period T 211r qv qv t qu v 3 1 1A 21 r T 4262008 Dubson Phys3220 langularmomentumlLpr mvr so vrLm and u 2L m So for a classical system the magnetic moment is proportional to the orbital angular momentum it orbital The same relation holds in a quantum m system In a magnetic eld B the energy of a magnetic moment is given by E FLB uz B assuming 13 B2 In QM LZ hm Writing electron mass as n1e to avoid confusion with the magnetic quantum number m and q 46 we have uz im where m Z The quantity HE E 26h is called the Bohr me me 2 magneton The possible energies of the magnetic moment in 13 B2 is given by Em HZBHBBm For spin angular momentum it is found experimentally that the associated magnetic moment is twice as big as for the orbital case Fl i g spin We use S m instead of L when referring to spin angular momentum This can be written uz hm 2pBm The energy ofaspin inafield is Espin 2uB Bm m il2 a fact which has been veri ed experimentally The existence of spin s 12 and the strange factor of 2 in the gyromagnetic ratio ratio of it to g was first deduced from spectrographic evidence by Goudsmit and Uhlenbeck in 1925 Another even more direct way to experimentally determine spin is with a StemGerlach device next page 4262008 Dubson Phys3220 This page from QM notes of Prof Roger Tobin Physics Dept Tufts U SternGerlach Experiment W Gerlach amp 0 Stem Z Physik 9 349252 1922 a a A A BB F VuB uVB Fzuz De ection of atoms in zdirection is proportional to zcomponent of magnetic moment uz which in turn is proportional to L2 The fact that there are two beams is proof that K s 12 The two beams correspond to m l2 and m 712 If 1 then there would be three beams corresponding to m 71 0 l The separation ofthe beams is a direct measure of Hz which provides proof that uz 2 B m The extra factor of 2 in the expression for the magnetic moment of the electron is often called the quotg factorquot and the magnetic moment is often written as uz g B m As mentioned before this cannot be deduced from nonrelativistic QM it is known from experiment and is inserted quotby handquot into the theory However a relativistic version of QM due to Dirac 1928 the quotDirac Equationquot predicts the existence of spin s 12 and furthermore the theory predicts the value g 2 A later better version of relativistic QM called Quantum Electrodynamics QED predicts that g is a little larger than 2 The g 4262008 Dubson Phys3220 factor has been carefully measured with fantastic precision and the latest experiments give g 20023193043718i76 in the last two places Computing g in QED requires computation of a infinite series of terms that involve progressively more messy integrals that can only be solved with approximate numerical methods The computed value of g is not known quite as precisely as experiment nevertheless the agreement is good to about 12 places QED is one of our most wellverified theories Spin Math Recall that the angular momentum commutation relations L2LZ0 LiLjihLk ijkcyclic were derived from the definition of the orbital angular momentum operator 1 r X The spin operator S does not exist in Euclidean space it doesn39t have a position or momentum vector associated with it so we cannot derive its commutation relations in a similar way Instead we boldly postulate that the same commutation relations hold for spin angular momentum SZSZ 0 Si 8 ihSk From these we derive just abefore that 2 2 3 2 S Isms h sslsmsgt if Isms s1nces12 SZ smsgt hms s ms i hlsms sincems ss l2l2 Notation since s 12 always we can drop this quantum number and specify the eigenstates of L2 LZ by giving only the m5 quantum number There are various ways to writethis Isms lms These states exist in a 2D subset of the full Hilbert Space called spin space Since these two states are eigenstates of a hermitian operator they form a complete orthonormal set 4262008 Dubson Phys3220 S5 within their part of Hilbert space and any arbitrary state in spin space can always be a written as all bli b J Gri iths notation is X ax bx Matrixnota onm p NotethatltTTgt qt 1 T 0 If we were working in the full Hilbert Space of say the Hatom problem then our basis states would be In 6 n1Z ms Spin is another degree of freedom so that the full speci cation of a basis state requires 4 quantum numbers More on the connection between spin and space parts of the state later Note on language throughout this section I will use the symbol SZ and SK etc to refer to both the observable quotthe measured value of SZ is h 2quot and its associated operator quotthe eigenvalue of SZ is h2 quot The matrix form of S2 and SZ in the mZ basis can be worked out element by element Recall that for any operator A Amn n S 2 S 2 T s2y hz my 0 etc T l 0 l 0 sz 2h2 s 1 4 0 l 2 0 l Operator equations can be written in matrix form for instance 1 0 1 1 T in 3 2 2 2 0 1 0 2 0 We are going ask what happens when we make measurements of SZ as well as SK and l T Eh i 0 etc S 2 Sy using a StemGerlach apparatus Will need to know What are the matrices for the operators SK and Sy These are derived from the raising and lowering operators 4262008 Dubson Phys3220 s SXiSy sx s s s SX iSy 3 3y s s To get the matrix forms of 8 S we need a result from the homework ms h ss1 mm1 lsams1 51115 h ss1 mm 1 For the case s 12 the square root factors are always 1 or 0 For instance s 12 s n1s 1 m 12 gives s s 1 mm 1 3 i L 1 Consequently 2 2 2 S i th S T 0 and SilT fill Sill 01eadingto T S T 0 T S L h etc and 0 1 0 0 8 h S h Not1ce that S S are not herm1t1an 0 0 1 0 Using sx s s and sy as s yields h 0 1 h 0 i S Sy These are herm1t1an of course a h a 0 1 0 i 1 0 Often written S 36 where 6X 6y 5 52 are called the Pauli spin matrices a Now let s make some measurements on the state alT bli b Normalization 1 3 a2 lbl2 1 a Suppose we measure SZ on a system in some state b Postulate 2 says that the possible results of this measurement are one of the SZ eigenvalues h 2 or h 2 4262008 Dubson Phys3220 Postulate 3 says the probability of nding say h 2 is lt0 In of this measurement which found h 2 the initial state collapses to 2 Z Prob nd h2 x gt b2 Postulate 4 says that as a result But suppose we measure Sx Which we can do by rotating the SG apparatus What will we nd Answer one of the eigenvalues of Sx which we show below are the same as the eigenvalues of SZ h 2 or h 2 Not surprising since there is nothing special about the zaXis What is the probability that we nd say Sx h 2 To answer this we need to know the eigenstates of the Sx operator Let s call these so far unknown eigenstates ITO and Isl1 Grif ths calls them x0 and How do we nd these We must solve the eigenvalue equation S x 0 h2 a a 7t h2 a 7 wh1ch can be rewritten 0 In linear h2 0 b b h2 7 b algebra this last equation is called the characteristic equation x Alx where 7 are the unknown eigenvalues In matrix form this is This system of linear equations only has a solution if 7 h 2 7 h 2 Det h 2 7 0 So M hzf 0 2 7tih2 h2 L As expected the eigenvalues of Sx are the same as those of SZ or Sy Now we can plug in each eigenvalue and solve for the eigenstates 22 an m 2 26 an m 2 Sowehavenltx and ltx11 4262008 Dubson Phys3220 1 Now back to our question Suppose the system in the state I ll 0 and we measure Sx What is the probability that we nd say Sx h 2 Postulate 3 gives the em 03 a Question for the student Suppose the initial state is an arbitrary state b and we recipe for the answer 2 2 2 L J l2 Prob nd SX h2 Tm Tu gt measure Sx What are the probabilities that we nd Sx 752 and h2 Let s review the strangeness of Quantum Mechanics 1 Suppose an electron is in the Sx h2 eigenstate ITO Ifwe ask What is the value of Sx Then there is a definite answer h 2 But if we ask What is the value of SZ then this is no answer The system does not possess a value of 2 If we measure SZ then the act of measurement will produce a de nite result and will force the state of the system to collapse into an eigenstate of 82 but that very act of measurement will destroy the de niteness of the value of Sx The system can be in an eigenstate of either Sx or 2 but not both 4262008 Dubson Phys3220 Quantum PHYS 3220 concept questions Operators A wavefunction 1px has been expressed as a sum of energy eigenfunctions unx s 1110 261 unoc Compared to the original 1px the set of numbers 010203 contains A more information B less information C the same information Dcannot be determineddepends 11708 Consider a complex vector V lVgt e VJ2 Where V1 and V2 are complex numbers they are the components of V If we want the inner product of V with itself ltVVgt to be positive for nonzero V what should ltABgt be A A1 B1 A2 B2 B A1 B1 A2 B2 QIAl Bl A2 3239 D More than one of these options E NONEof these makes sense 11708 If fx and gx are wave functions and c is a constant then ltcfggt A cltfggt B cltfggt C Cltflggt DcltFggt E None of these A vector can be written as a column of its components likewise a vector in Hilbert space a wave function can be written as an in nite column of its components in a basis of the uns EA The dot product oftwo vectors A and B is A c I ALAJ 111 3 A B EAiBi A i1 1 5 The inner product u w wEcM and qgt2dlp is fax urcp A 3 B Eleni id c Elalzldnlz DElal2ldnl2 E something else Do you plan to attend today s Tutorial on functions as vectors A Yes at 3 pm B Yes at 4 pm C Perhaps more likely at 3 D Perhaps more likely at 4 E No can t comenot planning on it 11708 Do the set of all normalized wave functions form a vector space A Yes B No A wavefunction 1px has been expressed as a sum of energy eigenfunctions ux s 1110 26 mo Viewing 1px as a vector in Hilbert space what role do the cn s and un s play A un s are basis vectors cnis are norms of vectors un s are components cn s are norms of vectors OED un s are basis vectors cn s are components un s are components cn s are basis vectors E Something elsel don t understand U A wavefunction 1px has been expressed as a sum of energy eigenfunctions ux s iv ch in Viewing 1pgt as a vector in Hilbert space what role do the cn s and ugt s play A un s are basis vectors cnrs are norms of vectors un s are components cn s are norms of vectors OED un s are basis vectors cn s are components un s are components cn s are basis vectors E Something elsel don t understand U 11708 If a wave flinction fX is squareintegrable ffx2dx lt oo does that mean that fX is always normalizable That is can we always find a number A such that fAfx2dx1 m AYes B No True A or False B The operator i ie multiplying by the constant i 1 is a hermitian operator The momentum operator 3 is hermitean l x meaning f 13ggt lg Is 32 hermitean A Yes B No 11708 True A or False B If fx is a wave function then 91quot M idx 139 dx True Always A False Always B or True Sometimes C dgx dx fxgxlhm f dx gltxgt hmns f dx for hmns Given that Q is a Hermitian operator what can you say about ltQ2gt ie Ileleygt 7 A It will be real ifand only if 11 is a stationary state eigenstate of H B It will be real ifand only if lIJis an eigenstate of Q C It must be real and ltQgt2 D It must be real but cannot ltQgt2 E It must be real and may or may not ltQgt 11708 Suppose f1gt and f2gt are eigenvectors of operator Q with eigenvalues q1 and q2 respectively s afygtbf2gt an eigenvector of Q A Yes always B No never C Only if ab Only if qiqz E Not suresomething else7 U In the simple harmonic oscillator the eigenvalues of H are E hmn12 and a measurement ofenergy will always observe one of these values What can we say about ltHgt A It must always be one of the En B It will never be one ofthe En C It is sometimes one of the E but only for a stationary state D It is sometimes one of the En not necessarily for a stationary state E Something else Postulate 3 A measurement of observable can only result in one of the eigenvalues of O lfwe measure the momentum ofa free particle in 1D What outcomes are possible A Any real value is possible B Any positive value is possible C Any value is possible including complex values D Only integer values are possible E For a given particle there is only ONE possible value or perhaps 2 ipo To think about What ifwe measure x instead of p 11708 WHITEBOARDS Come up with two different normalized states Lux for a particle in a harmonic oscillator potential such thatthe probability of measuring E0 ground state energy is exactly 13 For each state you come up with What is ltHgt To think about ifyou have time 0 lfl give you the value ofltHgt is your Lux uniquely determined 0 How does your state evolve with time Luxt 0 Given only that ProbE013 What is the range of all possible ltHgt s Suppose we take the particle in the state you came up with measure H and happen to get E0 the energy of the ground state Sketch Luxquot2 immediately after measurement What happens as time goes by lfyou have time resketch if 0 you had been given a particle measured H and happened to get E the energy of the first excited state 0 you had been given a particle measured X and happened to get x0 to high precision Observable A Alp mp normalized eigenstates 1M Ipz eigenvalues a1 a2 Observable B B b no alized eigenstates 1 2 eigenvalues b1 b2 The eigenstates are related b 2 3 3 2 1 76 2 76 Observable A is measured and the value a1 is found What is the system39s state immediately after measurement A 1V1 B 1P2 CV l i l Csz Ci amp 52 HOH ZCFO D at E 1 11708 Observable A Alp mp normalized eigenstates 1M Ipz eigenvalues a1 a2 Observable B B b no alized eigenstates 1 2 eigenvalues b1 b2 The eigenstates are related b 2 3 3 2 1 76 2 76 Immediately after the measurement of A the observable B is measured What is the probability that b1 will be found A0 B1 C 05 132413 E413 Observable A Alp mp normalized eigenstates 1M Ipz eigenvalues a1 a2 Observable B B b no alized eigenstates 1 2 eigenvalues b1 b2 The eigenstates are related b 2 3 3 2 1 76 2 76 Ifthe grad student failed to measure B but instead measured A for a second time What is the probability that the second will yield a1 A0 B1 C 05 132413 E413 A system is in a state which is a linear combination of the n1 and n2 energy eigenstates 1 1 1POM e quot1l1x e 2391l1 x E 1 IE 2 What is the probability that a measurement of energy will yield energy E Aexp i2wlt B 12 012 Dl4 E Something else 11708 You measure the energy of a particle in a simple harmonic oscillator and nd E1 ie the rst excited energy 32 hm Which graph best represents Lpxquot2 immediately after the measurement To think about how does the plot evolve in time A B C D Or E None ofthese is remotely correct If we change the potential Vx do the eigenvectors ofx change ie gX0X How about the eigenvectors of p ie fDO X A gX0 will change and so will fDO B gxowill change butfpowill NOT C gxowill NOT change butfpowill D Neither gX0 norfpowill change E It depends Suppose 1IJxt is known to be an energy eigenstate state n 1Px 2 uxexp iEth Can that energy eigenstate be written as 1m t fdpclgtprfx wherefp x mJexpip AYes B No C Maybe 11708 Do the set of deltafunctions 6XXo for all values of X0 form a complete set That is can any function fX in the Hilbert Space be written as a linear combination of the delta function like so fx fFx05x x0dx0 AYes B No If you answer Yes you should be able to construct the function FxO An isolated system evolves with time according to the TDSE with V Vx The wave function 1P 1IJxt depends on time Does the expectation value ofthe energy ltFlgt depend on time A Yes always B No never C Sometimes depending on initial conditions A system described by PE VX is in state 11 xt when a measure of the energy is made The probability that the measured energy will be the nth eigenvalue En is c expiE39 quot h Does the probability of finding the energy E when the system is in state WXt depend on the time t of the measurement A Yes B No 2 KM 1Pxt2 11708 Can the wave function 1I39xt describing an arbitrary physical state always be written in the form iEnt 1I x t u xe 4 where LlnX and EI1 are solutions of Hunx Enunx AYes B No A system described by PE Vx is in state 11 xt when a measurement ofthe energy is made Does the probability of finding the energy En depend on the time t of the measurement A Yes B No C Depends on lIJXO D Depends on Vx Given two quantum states labeled fgt and ggt which relation below must be true A ltffgt ltgggt 2 ltfggt2 B ltffgt ltgggt s C ltffgt ltgggt ltfggt2 D None of the above is guaranteed it depends on the states fand g 11708 In general given Hermitian operators A and B and a state 1 and with the usual notation ltAgtltwlAlwgt What can you say about ltw ltAgt B wgt A ltABgt B ltBAgt C ltBgtltAgt D MORE than one of these is correct E NONE of these is in general correct The proof of the generalized uncertainty principle involves inner products like ltltAgt1P 1Pgt Does this equal Hint Glyillyzgt 50le2 A Yes B No Suppose the state function 11 is known toAbe the eigenstate 1P1 of operator A with eigenvalue al What is the standard deViation A A 2 0A ltIrg A A In A Zero always B Nonizero always C Zero or nonizero depending on the details of the eigenfunction 1P1 11708 Suppose two observables commute A B 0 Is voB zero or nonzero A Zero B Nonizero C Zero or nonizero depending on details of the state function 11 used to compute O39AO39E Consider a Hamiltonian such as A2 H p V02 m What is the value of d d A 15gt lt PIH Pgt dt dt A Zero always B Nonizero always C Zero or nonizero depending on 11 Consider a Hamiltonian such as A2 H p V02 2m What is the value of d d E 1P dtlt dtlt A Zero always B Nonezero always C Zero or nonezero depending on 11 9 What is the value of d d A d A 11111P l11111J 1 dtlt gt dtlt gtdtltgt where the operator is defined as 111 I for any wave function 1P A Zero always B Nonezero always C Zero or nonezero depending on 11 11708 If 13163 0 then 197 Qt ltw 1 an 0 any 11 can you see Why If then does it follow that for any 11 O 0 A Yes B No If you have a single physical system with an unknown wave function 1P can you determine E lt11J FIIIJgt experimentally A Yes B No 11708 If you have a system initially with some state function 1P and then you make a measurement ofthe energy and find energy E how long will it take afterthe energy measurement for the expectation value ofthe position to change significantly A Forever ltxgt is a constant B hE C neither of these Complex number 2 a ib z 9 Z Z What is the value of A a2 b2 B a Cb D1 E 0 The wave function lIJxt below is a solution to the timeindependent Schrodinger equation for an infinite square well that goes from 0 to 1 How many energy eigenstates of the system have non fx zero amplitude A1 B 2 C 3 or more D Not enough info 11708 3D Consider a particle in 3D Is there a state where the result of position in the ydirection and momentum in the zdirection can both be predicted with 100 accuracy A Yes every state B Yes at least one state but not all C No there is no such state D Yes but only for free particles E Yes but only for a spherically symmetric potential notjust free particles Is the 3D wave function 2 I n 7 Innyny Inn HZ Xyz 7 sm 7 s1 7 s1 17 a a a a A h2 92 an eigenfunction of HX 772 2m 9x A Yes B No 11708 For the particle in a 3D box is the state nx ny nz 1 0 1 allowed A Yes B No The ground state energy ofthe particle in 2 2 a3D box is 121212fm2 ma 35 What is the energy ofthe 1st excited state A4e BSe C6 D8 E9e Considerthree functions fx gy and hz f x is a function ofx only gy is a function ofy only and hz is a function ofz only They obey the equation fx gy hz C constant What can you say aboutf g and h A f g and h must all be constants B One of f g and h must be a constant The other two can be functions of their respective variables C Two of f g and h must be constants The remaining function can be a function of its variable 11708 In the 3D infinite square well what is the degeneracy of the energy corresponding to the state nx ny nz 123 A1 B3 C4 D6 E9 Separation of variables has gotten us to 2 c ltIgt dqgt2 Is there anything we can say about the sign of the constant c in that equation A c must be 20 B c must be 50 C c can be or 7 but it cannot be 0 D Can t decide without knowing more what s the potential what are the boundary conditions for our particular problem JWMWle Jjwcp e 11 WW gxigrpvwwg m av dam WWW Wm39mmg MWT39 i m WM mpum 0 M ltu3gt39 an Kuww i m p amp m3 a 13gt M on Um gm v m v c a 43m waAvyM 30 339 my w bilgkrt km 4d kn M H 39A x mm 4194 muMr k H u g wotW wbv uagkvou 1 1w Wu Wm M M x m m M m M Wu 4qu M4 MN v w ramig x39K 11mm Wu IUIWgt Fgk quotyefv5 Mumugt I e 0 um I Ix ltpWgt A eraw Wm t we E vu Ln mm ramaw vepvewAJuAMWw w12FI way arr r rcM zwt wtwmew uw w w opevmov 39 Rm Sm mnwan k ww lt31 WWI L 3915 r Wm WM WM 910 a m H 4W w a My mm W N L39x quHJLuTAAMW 9L WM gmw W Lgt R s LN UKH 77 a lu V43 lt44 ml 7 NH lta dgHJLLMLD W lt31kgt LMW why kra s 0wa E quotgt WM V r a w h 1 W wao M W si MM Wmquot PM vexquot Wire J y Wuhan W 4 w LMW7L7 XVa 2 wd wg v24 3m WMMW AH RK v A W WW m WWWWW law A mm am 33 S 1mm IMWMMW WWW m 4141qu 5 y i WM R Lin ma mmw Nip amp gmwwr Ln 3 an A WW4 n WWJWM39 saw WWW f U16V A39 Twpy A L mns umrp wup I lmy mam HA Q j TW L va a4 MTWMM w Wm 1 WW wakwdc 94 3 Army km a m A Rquot IT Lug v 1amp39 5 Rhomum 1 T u quotLu srlku Hk3 us run Wm hm n W kinMann M Musth 4m A4Wm nv 7 lik m gtlt3Hgt Hwy offgm a n V 4 My EH A Hm Envgt 1 Wm 2 H m u WNW mum 3 M i 2 cum g M gt mum 5 Z Innl gt11 V39qummwmmquot m gag Ame WW m mi u wa bi x absiwr ngt arrLE 29 5938 an r9 5 92w FS Laxgzz 9an 3975 u f3 3 3 x E m wfv n mv u w my 53 32 1 vw ZZv 5 H TSVES t m 23 38 F M 3w in p432 95 s i 15 f mt 5 3 5 gr 3 s aif i 35 infi T5 3 H M13 mi a 33 1 uX me 3 w w w w E W 291st i n 37 MT l H Fi SEE e 4 DUE xii 3amp3 was mg n Ari u x ArthxEv WW T min 6 5v J39JN MMWM mum 3 AM 4 van6K MJAMJ m mumv 4 wk Wm Wu J 5quotng w Ax w 5147 L muw m M Ni Mn M X n all61th PMS my ltw391vi gtltnuvgt tJaw PM 1 3276 ij fwwa A QV NLK bayuanu A A Mm mm 4 me Wm PM Jk A Muhmww A VAMM C M M ou rW 0 Wm 1390 Pmlwgt Wm 3an mammww ummwrynmy wwuiMM QIVWH WM MW m Lwm m waxm l Z a WAquot 7gt w Em1numgtltn mwgt Aim M 1 MJWWM inn1W 1w ma M1Qi4i md n plum wi W wwm We MAW W X Wm M MW nb 13 f Mxmmwm A z Pwn mm EM Cnuan Gum MW to Zuw Z npnamvmw M Mu mnmmmg Quantum PHYS 3220 concept questions 92508 Clicker Intro Do you have an iCIicker Set your frequency to CB and vote A Yes B No Have you looked at the web lecture notes for this class before now A Yes B No 92508 ICLICKER frequency is CB Have you done the assigned reading for today A Yes Grif ths only B Yes Web notes only C Yes both text and notes D Not really but I will soon E Nope Intro to Quantum Mechanics In Classical Mechanics can this equation be derived A Yes B No 92508 Can this equation be derived s dL T dt net A Yes B No ICLICKER frequency is CB Have you done the assigned reading for today A Yes Grif ths only B Yes Web notes only C Yes both text and notes D Not really but I will soon E Nope Postulate 3 says wltxgt2dx Probpartice is between x and xdx What conclusion can you draw A Twx2dx must be exactly 1 B Twx2dx is nite but needn t 1 00 C l1Jxl2 must be nite at all x D More than one of these E Onemore are true but do not follow 92508 Is this wave function normalized This wave function is pure real PRe PA In A Yes B No l l 01 0 o3 04 05 06 07 x How would you physically interpret the wave function in the sketch A This doesn t look very physical B QM doesn t let you interpret wave functions like this C It s a large particle D a small particle E a particle located at a de nite spot x0 2 WI After measureme r x0 X Statistics and Probability 92508 You flip an ordinary coin in the air and get 3 heads in 3 tosses On the 4th toss the probability of heads is A greater than 50 B less than 50 C equal to 50 0 O O O O O O 0 O O O U O 0 O 0 O O O O O O 0 O O O O O O O O O O O l O O O O 0 O O O O O O O O O O O O O 0 0 O O O O O O O O O O 0 O O O 0 0 O O O O O O O O O I O O O 0 0 O O 0 0 O O HHHHHHHHHHH Plinko A marble is released from the same starting point each time Classical physics says identical systems with exactly the same initial conditions always lead to the same final result in a deterministic and repeatable way Is the distribution of final outcomes for the Plinko game played 300 times in this example in conflict with our theories of classical systems 10 A Yes B No The probability density lwl2 is plotted for a normalized wave function wx What is the probability that a position measurement will result in a measured value between 2 and 5 A 23 AT B 03 r oo4 2 D05 ml E 06 92508 The probability density 1p2 is plotted for a nonnormalized wavefunction 1px What is the probability that a position measurement will result in a measured value between 3 and 5 A23 B49 Do you plan on attending Tutorial today 4 PM basement Tutorial bay A Yes I ll be there B Maybe C Nocan t come N independent trials are made of a quantity X The possible results form a discrete spectrum X1 X2 H XM M possible distinct results Out of N trials n of the trials produce result x If you add up all the results of all N trials What is the sum of the results A 2X1 D N B Enlxl E Nle C EN 92508 N 342 trials 6 different possible results in each trial What is the best estimate ofthe probability that a token picked from the bag will be an 8 A zero D g 342 342 342 1 c 7 6 For a large number N of independent measurements of a random variable x which statement is true A ltX2gt 2 ltxgt2 always B ltx2gt 2 ltxgt2 0r ltx2gt lt ltxgt2 depending on the probability distribution A ball is released from rest You take many pictures as it falls to xH pictures are equally spaced in time What is ltxgt the average distance from the origin in randomly selected pictures 92508 AH2 B larger than H2 but C larger than H D smaller than H2 E 3927 xo less than H XH Waves A traveling wave is described by Y1xt 4 sin2x t All the numbers are in the appropriate SI mks units To 1 digit accuracy the wavelength A is most nearly A 1m B 2m C 3m D 4m E Considerany more than 4m Two traveling waves 1 and 2 are described by the equations Y1xt 8 sin4x 2t Y2xt 2 sinx 2t All the numbers are in the appropriate SI mks units Which wave has the higher speed A Wave 1 B Wave 2 C Both waves have the same speed 92508 Have you ever studied the classical Wave Equation A Yes B No C Not sure Let y1xt and y2xt both be solutions of the same wave equation that is zy Lc92y xz v2 tz where ican be 1 or 2 and v is a constant Is the function ysumxt ay1xt by2xt still a solution ofthe wave equation with a b constants A Yes always B No never C Sometimes depending on y1 and y2 Two impulse waves are approaching each other as shown Which picture correctly shows the total wave when the two waves are passing through each i C other i or E None of these is i D M remotely correct i 92508 A twoslit interference pattern is viewed on a screen The position of a particular minimum is marked This spot on the screen is further from the lower slit than from the top slit How much further 6 7 gt 39 slit 1 4 gt shtz fringes wavelength A screen A 2A B 15K C 3A D O5gt E None of these Two radio antennae are emitting isotropic radio signals at the same frequency f in phase The two antennae are located a distance 105A apart Acf A technician with a radio tuned to that frequency fwalks away from the antennae along a line through the antennae positions as shown b h n i 3 1 It I l s I 1quot v 1 gt s I s39 a quot 39 10 quot k As the technician walks she notes the tone from the radio is A very loud all the time B alternates loud and quiet as she walks C very quiet all the time 2D quiet at rst and then loud all the time 1O Starting with the assumed solution 1IJxt A expikx wt how can one obtain a factor of 00 perhaps with other factors Use the operator 9 9 52 A 7 B 7 C 7 x 9t 9x2 92 D 7 E 77 62 7t 92508 Starting with the assumed solution 1IJxt A expikx wt how can one obtain a factor of k2 perhaps with other factors Use the operator 9 9 52 A7 B7 c7 x m 5 Do you know the de Broglie relations A Yes B No Two particles 1 and 2 are described by plane wave of the form expikx wt Particle 1 has a smaller wavelength than particle 2 A1 lt A2 Which particle has larger momentum A particle l B particle 2 C They have the same momentum D It is impossible to answer based on the info given 92508 Consider the eigenvalue equation d 7 x C39 x dx2f f How many of the following give an eigenfunction and corresponding eigenvalue fx sinkx C k2 FX expx C 1 I FX expi k X C k2 IV FX x3 C 6 A1 B2 C3 Dall4 ENone d w I i 8 dz fiwlp xt1Pxtdx f yxl x 1 dx 7 AYes no problemo BThere s something not right about this Which expression below would be the QM equation for ltKEgt 2 2 A quoth k 111xz111xzdx 2m w hz 92 B iw yhy xt1Pxtdx w hz 92 c LEW xtylpxt dx D None of these E More than one 92508 After assuming a product form solution 111xt wx t the TDSE becomes 2 2 1 a h 1 c9 i II v E p at 2m 1 x If the potential energy function V in the Schrodinger Equation is a function of time as well as x V Vxt would separation of variables still work that is would there still be solutions to the SE of the form 11th 1PX39 t7 A Yes always B No never C Depends on the functional dependence of Vonxandt l1J1x t and l1J2x t are two solutions of the timedependent SE ls lysumx t a1P1x t b11J1x t also a solution ofthe TDSE A Yes B No C Depends on l 1x t and l 2x t Do you know what the momentum operator is A Yes B No 92508 Do you plan to attend today s Tutorial on relating classical to Quantum an qualitative sketching of wave functions A Yes at 3 pm B Yes at 4 pm C Perhaps more likely at 3 D Perhaps more likely at 4 E No can t comenot planning on it Given 1Pnx t as one of the eigenstates of Pity Emlyn what is the expectation value ofthe Hamiltoniansquared H2gtf PnHH1Pndx A En B En2 C zero D En2 En E Something elseit really depends 1P1 and 1P2 are two energy eigenstates of the Hamiltonian operator They are nondegenerate meaning they have different eigenvalues E1 and E2 FIIIJ1 E111J1 and Pity2 EZIIJ2 and EM E2 Is 11 1P1 115 also an energy eigenstate A Yes always B No never C Possibly yes depends 92508 Given ux A sink x B cosk x the boundary condition uO 0 implies what AAO BBO Ck0 Dkn139rnl23 E None of these Given ux A sink X the boundary condition ua 0 implies what AA O B k O Ckn139rnl23 D None of these An electron and a neutron have the same speed Which particle has the shorter wavelength A The electron B The neutron C They have the same wavelength 92508 How does the energy E1 of the ground state n1 of an infinite square well of width a compare with the energy of the ground state of a well with a larger width The larger well has A lower energy B higher energy C the same energy D Need more information How does the energy E of the n 3 state of an infinite square well of width a compare with the energy of the n3 state of a well with a larger width The larger well has gt A lower energy B higher energy C the same energy 17 In an infinite square well the lowest two stationary states are 1Xand M206 ttime t0 the state of a particle in this square well is 111m 0MIX 2a Is this particle in a stationary state A Yes l is a stationary state B No l is not a stationary state at any time C No l is a stationary state only when t D Not enough information 92508 In an infinite square well the lowest two stationary states are 1X and MAX At time t0 the state of alparticle in this square well is wxx07uxuzx What is the wave function at time t WOW At t0 could the for an electron in an in nite square Well of Width a 0ltxlta be 1mm AsinZUDc Where A is a suitable normalization constant Assume it is zero outxide the region 0ltxlta A Sure B No it s notphysical The energy eigenstates un form an orthonormal set meaning f uxu xdx am What is f Mx20nunxdx 7 A Er B czcm C cm D cquot E None of these 92508 Given a particle in a box size a with 1110 t 0 2Jisimmca JisiqucMJ a 3 3 What is the probability of measuring E1 How about measuring E2 A ProbE1 13 ProbE2 23 B PE1 Sqrt13 PE2 Sqrt23 C PE1 l3 PE2 O D PE1 Sqrtl3 PE2 O E None of these is correct MM 0 J isiuma JZSiu3JDC a 3 3 Assuming your system is isolated does the probability of measuring E1 depend on the time t of the measurement A Yes B No Mathematica haslnumerically solved the TISE with Vx5mw2x2 starting from u01 with an assumed energy E What should be our next try AMake E a little bigger B Make E a little smaller C Make uO a little bigger D Make uO a little smaller E None of thesemore than onesomething else 92508 Mathematica haslnumerically solved the TISE with Vx5mw2x2 starting from u01 with an assumed energy E What should be our next try AMake E a little bigger B Make E a little smaller 0 Make w a little bigger D Make w a little smaller E None ofthesemore than onesomething else 2 What is the behaviour of ux Axe 1 as x goes to infinity Au blows up B u goes to a nonzero constant C u goes to 0 but ux is not normalizable D u goes to 0 and ux is normalizable E Can t decide Without knowing A and a 20 In QMwhat is 13 fx where f is the momentum operator A O for all functions fx B 71k fx C dfdx D ih df dx E None of thesesomething else 92508 xPi Is pxih Is ipcx cxip A Yes and yes B Yes and no C No and yes D No and no A A 1 A 1 aa7 7H 7 ha 2 What is leg 21 F77 m A eif w fzm wvwlk F 70 1 1wa a MM Gym wry lt F Hgt ml fd EWV39F39MA VHqu 5911 WWW W F P39 MM gum 2 S Tf ax 1thYV r as r V 7 7 7 A z up iris 7 WM Mi meu I L n ys Flmgksv115 TMM 5075 Hut 2 39 ha W aux r 7W x AN 4 lt mffkiue Wu 1 FOO Jm w 7quot 7 NM 3 Yumt m a gm 0 QMMAME 57 X7 r tiki V IL L at fa ghxw e wi dkfii Ekwi 71 a Ygxti ei w ifmgyzg nyiv ipVhrdkdn vamg ft 35 quPtk Na swung p2 l j l 39 1 fxk 1 mi m fa 1mg e f A EM 47 lquotLx e Vquot q a LW W k JJK f L 150 JP W U 4F 10 JrLWllllr39 Nan F mg m MW Wm W Yaw mm a W mam Ann Lwo m 7 m aw m 4quot FW WWle 1 v w W w at Jruu fww okm 13v woe wquot Mimi do 7 39N KKMM J1 W 4 an wanw hm Jay mbw M Lm WM Nut uWW zyh 0 n3 m MAXIM A mxmwwm AW wuviWWA L A 7 FW tumJim 2121 MJ 1211 ng VM ghk quotaquot 39m w quotF 194 anw w A Mm WJW w YLr r if w w NAmg A y 3 fuNrLOLnxias Fmwa W old43 m To0 gm robin m yewwh 7 1113 UM M39ruflmw M W W L me am Mu in rm L W 5 Postulwti 36 a MMW o m 1mm Yup MAJ a WAIKW 01k dwummkh 1 Va W cm 4 AdjAW WinUL on ma T Wow I M wmxms 1quot m MJ i mew 1 31 x J n gm n JAE 4 7 w AhWyR EMA 5w Jvmmwx w m L wmuh drums Jw M WW W amw x mm yrs T Ql ATMmm v mew 341 3 64 6 rwm ilm 4 mi P M 4 K1 1 mm Mix Hgiwmzxa a 7339 7 SW1 AW JAM AW ind m JAM WlAUA 7 lt1 in sin LH2 W WyeMug m Make JW AWampW 1 Wmt X cwc w w g c e wm ACAAR c e ig fi 9 J37 n 1gt Jwv a WW J MM M Wm E WAR EvobdeLIty 1 MW 7 L 1 f 6 IWLWH mil a mu F 15 9a 2Lva All ra wryxnv n chjn Vin Hie1 quot gt bath i ahaquot u wwm L MAD AU an an o aw WMVW yg my my 42 EN r C 12 l w L Mm 39 wx y 1 maxum W 444411 of WW WW 77 g 71 7 SM Mm mg 1 um KMWM Waiki C1 m 4qu Gz wheaHl w mamLu Jig Avmm Mt W WWW M Wu New SE V JAM MMV MM mJLuu oi a 52gt Ami 6 ltrlc3gt cm NW damn aw mm 93 4Www 1 ma a ixh gt2Qgt 1 17 a ZACM 96 w z i a 3 gm gag 5 Limp m i 1 CW lt9quot WWW wmghleuvcmge APSE 777 F23 Notice that in 1D problems like the 1D in nite well or the 1D SHO we only needed one number n to uniquely specify an eigenstate This state label is called a quantum number or qnumber and it is always in a ltol correspondence with the eigenvalues of an observable operator But in 3D problems we need 3 quantum numbers nx ny nZ to fully specify a state or equivalently n ny nZ n nx nZ etc where n n n n Just specifying n or just nx is insufficient since the eigenstates of H or x are degenerate In cases with degeneracy more than one quantum number is required to specify a state and the other quantum numbers are associated with other operators that must commute with the first Iftwo operators commute examples HHX 0 flX Hy 0 then there exists a set of orthonormal simultaneous eigenstates of both operators We proved this for the case of operators with nondegenerate states but it is also true when there are degeneracies We will show below that when operators do not commute it is impossible to find simultaneous eigenstates Claim If N quantum numbers example nx ny nZ are required to uniquely specify a state then there must exit N commuting operators example x fly HZ whose simultaneous eigenstates are nondegenerate and whose N eigenvalues provide the quantum numbers that uniquely label the state Such a set of operators is called a complete set of commuting operators CSCO We will give a proof later when we talk about matrix formulation of QM F z NW N w JAM ADM VWWAIWWQ w WMM miuxww ye ow W39 W WM w TMWu n Tmmm rm NW LWWMM Wm W mm a A A ltagt ltL lt gt 1mm jamsa 37b 19gt qglww lgaa a MW fawn Q23 W lair yltltnaggt w Uvsm 5 New u vwk H gt 1 A A A w unuvgta lt wu gt A Erma AMUWW 1 wwmaam ltmamp1gt 0 MM WQMWM LW Md Mmmxwwa m lt61 Ww if WMM qu Io 7 C dogtltqgtwagtqabmf 1W Wmvmm5iaaqw rm hasywm 1nQMmwamgagtltagtM WWWQJWXLWNQ MMWW iw mewMW 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