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by: Dr. Carissa Rowe


Dr. Carissa Rowe
GPA 3.97


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This 105 page Class Notes was uploaded by Dr. Carissa Rowe on Tuesday October 20, 2015. The Class Notes belongs to PHYS610A at San Diego State University taught by M.Bromley in Fall. Since its upload, it has received 34 views. For similar materials see /class/225319/phys610a-san-diego-state-university in Physics 2 at San Diego State University.




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Date Created: 10/20/15
Lecture 17 taking it LZ o Spherical ng Harmonics section 75 o L2LxLyLZLi section 72737475 a touch on Rotation ops section 72737475 Separation of Spherical Coords v2w w Ed using separable solns Mr 6 gb R7 Y6 gb we found 27717 2 am 1 i Singal L a2Y 0 Y sin686 86 sin26 82gb constants F7 CQ gb 66 1 66 1 0 Time indep SE Tried further separation Y6 gb 6ltIgtgb sinQ 8 1 821 e 86 O 89 s1n6 1s1n2 6 1 82 86 So introduced separation constant m2 m2 0 Two angular equations azimuthal 0 Another separation constant m2 m2 0 a means two angular eqns for 6 next slide and gb 16 2ltIgt2 182gb This gb eqn has solns Aexpiz mgb Boundary condition on aziumuthal gb 27139 gb ezmqb l ZW zmqb 27mm ezmqb 8 thus exp27m39m 1 so we have m 07 i1 i2 The coef cient A we absorb over into the polar soln Two angular equations polar 8 89 s1n6 s1n6g 66 ls1n2 6 m2 9 0 Has general soln the associated Legendre functions 96 APEmCOS 6 de ned in terms of the th Legendre polynomials lml lml d l x2 T P a lt gt dx eltgt to evaluate the Legendre polys use Rodrigues formula 13693 ya 1 eg P0 1 P1 13 P2 3x2 1P3 5593 313 1335 associated Legendre functions Legendre polynomial FAQS has degree 6 With parity eveness or oddness depending 011 6 assoc Legendre funks Pema 1 x2 ixyml Pew d Note for gt 6 then Pgna 0 so 26 1 possible m For 5 0 1381 For 5 1 PP cos67 P11 sin6 1 13205 1325 53932 1 1 332P2a Sam1 512 132293 1 5132 i 2 13293 31 5132 P20 3cos26 17 P213sin6cos6P223sin26 l 2 polar plots 0 assoc Legendre funks where 7 P mCOS 6 look like Spherical Harmonics 0 Finally we need to normalise these angular functions 0 W 9 9b RUM9 9b RT Y97 9b Rr9lt1gt ib2 d3r W03 6 gb2 r2 sin6 drdegb R2r2 drY6gb2sin6d6dgb 0 require both radial and angular integrals normalised 00 27139 7139 R2r2 dr 1 and Y67 gb2sin6 degb 1 0 0 0 o normalised angular functions are spherical harmonics We Q5 WM 1 1 MEWS 9 o 6 azimuthal quantum number m magnetic quantum Back to the Classical Picture 0 Polar plot of le6 gb for high K m NOW Lac ypz Zpya Ly pr 517192 L2 xpy ypxa Lam Ly z39hLZ LEW L2 239th L2 L30 ihLy 0 ie all incompatible observables eg aanLy 2 Angular Momentum eigenfunctions a separation of variables gives wn7g7me7 6 gb E n 6 mg a spherical harmonics YEWW gb are the eigenfunctions of MWWMWMW 0 But squared total L2 Li L L is h2 8 8 1 8 L2 W 39 6 YW 71 1 YW E sin6 39 81 39 sin68gb E E 0 ie simultaneous eigenstates of both L2wgt 04W While Lzlwgt lwgt L2 mgth2 1 mgt and L26mgthm mgt 0 50 1 l 3 5 andm l l Raising Lowering Operators Note that i slipped in factors 6 g If we don t work in position space can show this Following Shankar 125 Assume angular basis states L2O gt O lo gt and Lzlo gt la gt De ne raising lowering operators Li Lx i iLy they commute L2 Li ihLi and L2 Li 0 so we can do the same approach as Li Lx i iLy L2Lloz gt m hLO gt Lm hLlolt gt hLlolt gt similarly L2LOz gt LL2oz gt ozLoz gt Top Bottom Rungs a As per the SH0 Dirac treatment a but there are bounds on 04 2 2 0 since WW Lime MILE Lime 2 0 Loz maxgt 0 and Loz mmgt 0 L2 as Generators of Rotation we got by Classical to quantum a gt 1 and p gt 13 but there is another point of View this bit is not assessable as seen in advanced course Shankar 122126 rotation formalism eg a y plane N I 2 0 2 e z quLzh UiRlt oIEgt1 hm so working in position space 2 i7i8 n gtoo EN O P7 gb 0 H a under rotation we have H invariant 0 ie good quantum numbers Hn 6 mg Enn 6 mg Lecture 19 Outline Hbomb o everything you want to know about the Hatorn Section 1317 132 133 o Periodic table warm fuzzy version Section 134 tie me Coulomb down sport When H is rotationally invariant V7 can separate solns Mr 6 gb anw o 712 d2u 712 KM 1 V E 2mdr2 r2m 7 2 J H For Coulomb use p HT and pm me227T60h243 then d2 u 1 Ln m u M p p2 Where up pg1 e p vp from p gt 0 and p gt oo again we re searchin for a power series Froebenius soln WP 0ij 22p recursion termination With up pg1 e p vp and vp 20 cjpj the power series solution led us to C C 9 j1j2 2 5 So for large j ie large p cc 7 consider 2339 2 c m c c 1 H1179 j1j j o if this was exact then cj 0 and 00 oo 2739 vltpgt 2cm 2 co 2 739 p so um j0 j0 39 a which is bad so the series must terminate at jmax 22p Allowed 801115 a the series terminates at some jmax so ijax1 0 18 2jmax 1 pm 0 o and nally we de ne n jmax 6 1 thus pm 2n 2j 1 pm 2j 1 n Cy j1j2 2 6939 j1j2 2 Ci E h2lt2 712 77182 l 2 m e2 21 n 2m 2m 27reoh2 pm 2712 47reo n2 0 ie the Bohr formula The Bohr radius a0 is de ned V 2mE 77182 1 1 h 47reoh2 0 note that we also have p HT If n mm L Lon 39 Radial togetherness With up pg1 e p vp and vp 20 ijj serles termmates at jmax n 6 1 1e ija l 0 so vp is a polynomial of degree jmax and p raon Ground state hydrogen has n 1 then 6 0 and m e2 2 E 136 V 1 2712 1mm 8 CO 1 e rao 6 R Y0 6 e do W1007 Cb 107 0 9b a0 V4 7Ta3 a coef cient 0 determined from f R2r2dr 1 Coulomb radial solutions First excited states Series terminates at k n 6 1 and p raon First excited state hydrogen has n 2 and energy E2 El 1364 eV 0 Want coef cients of vp 0 31p 32p2 60 gt Cl CQ gt 020010 61 gt 01000 gt 310 0 thus vp 301 p or vp 0 CO r 39r 2a CO 7 2CL 1 0 0 R20 2amp0 2a 8 R21 4amp8 r e o Twovalues 0m00r 1 m0i1 o 4 degenerate states with the same binding energy Hydrogen States a L pa is an associated Laguerre polynomial o Lqa exddaqexaq is the qth Laguerre polynomial vltpgtLieallt2pgt where L pltxgtlt 1gtP Lqltasgt o L0lL1 xlL2x2 4x2 and L8lL xlLgx2 4x2 o For arbitrary 71 possible 6 012n l a while for each 6 possible m 07 i1 i2 i6 WW7 6 b 3 Warnae gtlt WW TLCLQ 3D Hydrogen Pictures 0 The radial wavefunctions in all their gory R10 2amp0 2 eXp 7 a0 R20 R21 R30 R31 R32 3 a 2 T 35 1 a Oexp r2a0 y w explt r2aogt 3 20 7a 72 02 7 1 gg eXp 7 3a0 80 1 1 7quot 7quot 27 5 5gg KM 773 4 81m eXp Ir3a 0 a 3 D structures see wwwdaugerxtom for maCOSX Just Accidentally Degenerate For arbitrary n possible 6 012n 1 While for each 6 possible m 07 i1 i2 i6 Thus the total degeneracy for energy level En is n l dn 226 1 n2 60 But degenerate wavefunctions are all nicely orthogonal W mwnlglml 7 2 Sin 6 6nn6gg6mm Since spherical harmonics are orthogonal and since eigenfunctions of H With distinct En Hydrogen Spectrum Quantum Electrodynamics interaction of matter EM enlightens us with photons E by H E7 3 SM Consider the hydrogen atom and level transitions Ei Ef 136 Vlt gt gt Where E7 by hcA The Rydberg constant is 3 wa m e2 2 R 1097gtlt107 1 y 47Tch3 47reo m Bohr Radius and other expects a Rydberg Ry is a natural atomic energy scale a Bohr radius a0 is the natural length scale H The periodic table Coulornbic many body Hamiltonian xed nucleus requires anti symmetric electron wavefunction h2v2 1 gt212 r e2 jzl 2m 7 47T 0 2 47T 0 jkrj rk 71739 ie With ibr1r2 rZ gtlt XltSl52 sZ One electron states 1ng have n2 degeneracy n8 shells have 716 0 mg 0 thus 2 electrons np shells have 716 1 mg i1 0 thus 6 electrons ll up shells according to the Pauli exclusion principle Hund s rules give con guration 2S1LJ eg Carbon 3P0 1522522192 has possible L 2 S 01 and J 3 2 10 Lecture 25 Outline end of RT Hydrogen ne structure 173 Hydrogen hyper ne structure 151 Zeeman effect 145 time dep PT 181182 CharPOOR Finite nuclear rnass u Gravitation H FS relativistic corr H gives E11 oz FS Spin orbit H o i 1 Lamb shift quantized electric eld effect E11 oz HFS rnagnetic int of e and p dipoles E11 Hydrogen ne structure The Bohr energy levels E2 WE are of order 042mc2 Where the ne structure constant 04 is given by e2 1 d 47reohc 137036 Possible corrections to the Bohr energy levels AE 5 x 104E9 dirnensionless mM mM GmM gives E11 10 41E9 T 4mc2 SL gives E11 a4mc2 87T 0 m2 c2 r3 5mc2 m4 2 Mame Fine Structure Relativistic 0 really need QED for this but include semiclassically 0 Via the electrons KE H 1922771 HIT 627 2 2 4 me p p T 2 mc2 V62102 772204 77202 m 1 110 8771302 H p 44m2 E21 HO 2 T 8772362 8772362 2m 277262 7 o where used H 0 p22m which simpli es 1 1 1 E1 E0 2 2E0 2 4 T 2mC2 n me 74 n m e 742 n m 77162064 3 1 2 4n2 n3 0 Note we still used the n mgt states with H T Fine Structure SpinOrbit o Coulomb interaction only for e at rest 0 according to the e the proton is moving gt B eld o interacts with electrons magnetic moment ue 2 62 e L J2 L2 32 2m2c2r3 4m2c2r3 0 good quantum numbers n j mj With j E i H o M 39B 1 1 Eslo L2 iltigt 52 jj1 e1 gt 4m2c2r3 7 3 M 4 77162064 E OR 3 1 4 n3 1 0 Note ifn2canbuildjfrom Oor1 Total Fine Structure 0 Combining E n E and Eight j 0 we get the rst order total ne structure splitting 2 4 2 1 3 E1 E1 E1 mCO O fs T 80 2mg n 4n 0 ie the same effect upon both states with j E i Hyper ne Structure SpinSpin 0 One small dipole bumps into a larger dipole and says your being down always seems to make me feel up 0 electron and proton have instrinsic magnetic moments gee e gpe S g S gtgt Me 2me 6 me 6 up 2mp p 0 consider electron s ue interacts with proton s B eld o For ground state 713 1 O the perturbation is just 1 1 11 839 82 32 32 839 2239 e Plsm 22gt o homework singlet OO has E 136eV 3E2A4 triplet s 1i1 0 llgt and E 136eV 1h2A4 22 3 2 M09 6 Eif 56 51 p e 0 ie gives 21 cm microwaves ie V AEh 1420 MHZ Stark effect clari cation Effect of Bohr energy levels in weak uniform Eat field For ground state E1100 68100z100gt E O 6823mgz2 mggt E O For excited states E21ng invoke matrix elements A 68200z210gt 3 8a0 ie perturbation has no 1st order effect on Mg i1 but H513 eigenstates 200gt i 210gt split Zeeman effects Bohr energy levels change in uniform Bat eld ignoring the proton s coupling to the Bad eld HZ u M8 B xt 2 2SBext 6 m Effect depends on strength compared to internal elds Weak elds total J L S conserved due to spin orbit HQO good quantum numbers n j my 6 E weak B t39ltL ie splits mj states EEL21 136eV1 0424 i LLBBext 0 Strong BM in z dir good q numbers n E mg m8 6 Egstmng B xtn mgmsLzZSZn mgm8gt MBBextmgl 2ms Simultaneous Diagonalisation remember matrices if 9 A 7 0 cannot nd a set of common eigenvectors that simultaneously diags both Degeneracy needs a basis that is diagonal in all of H rstly H 0 and simultaneously H Stark effect for n 2 needed H513 and H 0 eigenstates are 211gt 21 1 200gt i 210gt Good quantum numbers means that the basis is close to diagonal Wrt H eg intermediate Zeeman effect weak lt Bm lt strong is complicated need Clebsch Gordon s etc TimeDep Schriidinger Eqn Recall so far tirne indep H0 ie ihtgt Howe solns Wt Zn Cntln0gt Zn Cn0e iE3mn0gt which came from ih tgt Howe 0gt d A0 anOf 0 0 0gt H gt Mt dt Enema m gt o By linear independence defn iff Ci O n 39EO 0 dealt Cnt solns Cnt Cn0e Enth 0 now we use those solns to solve Ht H0 H t 0 ie if system starts in i0gt we want to know the Prob of nding f0gt due to H t Time dependent Perturbation Theory 0 for H t 7 0 consider solns tgt Zn dnteiE3thn0gt o where dnt contains evolution due to H t 0gt H0 Hot Wt Z 03 E00n Hootn H tdngt e iEgthn0gt 0 now project Wrt nal states f0eiE th ddn Z 0 0 0 Z Z mmfolnw ltf0H tln0gt Ef E thd gt ZltfOIH lttgtn0gteWntdn ltfOIH lttgtIz390gte W 0 Where we assumed that dn0 6W Timedep PT contd 0 given initially only in a pure eigenstate i0gt then 39 t M cm ltf0H lttgtIz390gte Wwtdt 0 example of 1 D SHO Lecture 26 Outline end 0 road a Zeeman effect 145 and NMR untangled o Entanglement EPR Bell s Theorem the Cat Zeno o if you want intro to time dep RT 181182 Magnetic Fields and Atoms Zeeman effect weak Zeeman strong Paschen Back NMR now including nucleus Entanglement Production Entanglement measurement QM measurement EPR Einstein Podolsky Rosen pa Bell s inequality Bell consequences the Cat and Zeno Timedependent Perturbation Theory 0 for H t y 0 consider solns wtgt Zn dnteiE3thn0gt o Where dnt contains evolution due to H t d A 0gt H0 M was ddn A gt o Z in Eodn Hodn H tdn e iEnthn0gt E0 0 now prOJect Wrt nal states ltfOeZ fth ddn i 0 0 0 Z mw na dnltf0H tln0gt e if EM 239 iw0 z w0 dt Zltf0H39tn0gte rmtdn ltf0IH lttgtI20gte 0 Where we then assume that dn0 6m TimeDep PT 1D SHO example a Prob to measure particle in nal state is Pft dft2 i t W aft h ltf0H lttgtIz390gte Wdt TimeDep PT 1D SHO contd Lecture 10 evolution of the species 0 Schrodinger Eqn Soln end of Section 43 0 Quantum State time evolution 0 Free Particle Section 51 o if time Wavepacket propagation Section 51 Schriidinger Eqn General soln 405 HWW when H is Hermitian eigensolutions must exist HEgt EEgt Wt Z EgtltE 0gt iEth ZaE0 iEtF IEgt Eg for practice spin particle in magnetic eld 0 1 1 H BOSZ B0h Eigenstates are the same as those 0f 82 with eigenvalues E l39yB0h and E 7B0h 2 Quantum State timeevolution 0 our spin particle enters magnetic eld at t O o and it takes a length of time t 0 to pass through 0gtalEjgtbEZ gta 19 2 Z a Determinate Stationary States 0 If initial state is eigenstate of Hamiltonian Ogt c then we calculate the state evolution simply as Wt Z EgtltE 0gt iEth 6 iEtRIEgt Etgt o BUT if we do any other measurement 9 of this state POOH ltM tgt2 ltMEtgt2 Z39EthltwlEgt2 ltMEgt2 POW7t 0 0 ie the probability of measuring M is constant Engineering Quantum States If initial state is eigenstate of 32 Ogt but we place the magnetic eld along y direction 1 O i H yBOSy B0h 2 239 So the state evolution Via expansion in eigenveotors of H cld1 xiz39xi After some time we then do measurements of 32 Wt CIEJgt dlEy gt Z Schrodinger vs Heisenberg 0 Given a time independent Hermitian Hamiltonian H Wt Z EgtltEl 0gt iEth Ut 0gt 0 ie time evolution is a unitary transformation U05 Z ltEle z Eth E e thh 0 state evolves tgt 6 thh 0gt in Hilbert space check this by inserting into the Normalisation lt t tgt lt 0UltUtl 0gt lt 0 0gt 1 o This is Schrodinger picture of time evolution 0 Heisenberg Picture transform to basis that rotates at same rate ie W constant time evolution in operators Free particle o Schrodinger equation for free particle V 0 WW HIW 5 mlwtgt 0 We already know eigensolns stationary modes ie A2 Wt lee W HEgt f mEgt EEgt o noting that an eigenstate 13pgt ppgt is also an eigenstate 132W MW p2lpgt we can trial A2 2 p lpgtElpgt gt p E pgt0 gt pi2mE 2m 2m 0 each eigenvalue E has two orthogonal eigenstates so IE9 E gt alpgt IP gt 0 ie probability particle measured moving left or right The free particle in mspace Note that that proof didn t rely on a particular space but for now choose position space p iii d 252 fid dw Ll 2 2m dx2 d 2 k w we ve already found solns to which add time bits Aeikx Be ikx went Be ikx iEth x t Aeikx t Be ikwt Ew thus Waves have momentum p his cred to de Broglie so called planewaves have wavenumber k V ZmEh Stationary Propagation free particle stationary states are propagating waves wkcvyt Aeikx t Be ikxt consider x vt 0 every point moves along with same velocity v in x dirn waveform isnt changing Either 1M1 t AeWCH is wave moving to the right or 1M1 t Be WC l is wave moving to the left Worst is that ff wzwkdx A2 ff dx A2oo A free particle cannot exist in a stationary state or such a free particle does not have a de nite energy Lecture 20 Outline Spin 0 Spin Section 141142143 o Spin Dynamics and SternGeriach Section 144145 0 intro to Angular momentum coupling Section 15 Hermitian Commuting operators Angular momentum operators commute L2 L2 0 L2 H LZ7 H 0 when H rotationally invariant simultaneously diagonalise with common eigenbasis with 6 mg Enn 6 mg with 2n gm Wu ln am with 6 mg hmgn 6 mg ie we can simultaneously measure those observables ie remember eigenstate collapse Separation of variables single particle wn7g7me7 6 gb BUT the algebraic theory allows half integer 6 and me Spin angular momentum o Particles have extrinsic orbital angular momentum L o empirically intrinsic spin angular momentum with S2s m8 h2ss ls m8 and SZ5 m8 hmss m8 1 3 8 0757175w and ms s sls ls a consider electrons with xed spin 5 o with two eigenstates Sgt and Sgt ms S2Sgt 3h28gt and szsgt Z s2sgt h2sgt and szsgt hsgt E Co gt 1 mS Derivation of spin matrices o Deriving the matrices S2 and 82 o from spin up and spin down in the spinor notation 1 Sgt 7 O Pauli spin matrices Deriving the matrices for sac and Sy as per WI9112 we have Sac7 Sy 27182 and can use ladder Operators Si 8x i 23983 to show that Sx E hax Sy E am and 82 E haz Where 1 0 0y 10 2390 z39 10 0 0392 0 1 NOW an electron is described by spinor in Hilbert Space 1 0 was 24 Z 1549372473 a 2 b 2 are probabilities 0f 82 measurement of ilh 2 Classical Charge in a B eld Classically magnetic dipole moment i yg Where spinning angular momentum g gyromagnetic ratio y In a magnetic eld E torque i x energy E so we have a Hamiltonian H y Spin in a magnetic eld Originally spectroscopy suggested particles have spin So we have a classical Hamiltonian H y Imagine spin half particle in g BQIAC we have 1 1 0 H B0h 2 1 Eigenstates Egt are those of 82 with eigenvalues E yB0h and E YB07L Sidenotes QED tells us that yquantum m 2 x yclassical electron s dipole moment is g 2 x 100115965219 Larmor precession o with initial state Xgtt 0 aSgt bSgt and o the TDSE ie ihxgt HXgt o with a time independent magnetic eld has 301118 a ei yBotZ a S iEth b S iEth Xgt gt gte o with a 008062 and b sina2 real ltSzgt ltXtISzXtgt Sac ltXtSxXtgt hsinozcoshBot lt81 ltXtSyXtgt hsinasinltvBotgt o with Larmor frequency of rotation w yBO Classical SternGerlach and precession 0 Heavy particle eg Silver with classical spin B eld Ba y z x 0 B0 z Where B0 gtgt Z I spmiup y W X SPi tlidown lnhomogenous Force F VuB y Sx 0 82 a eg more usually F2 1123 HCOS 1 B eld creates Larmor precesssion B2 w B0 dominates a meaning rapid a y rotation w yBO smears Fx 8x oc cosyB0t averages Fx m 0 Thus classical Force F2 y SZ in Z direction Quantum SternGerlach 0 With random initial state Xt 0 aSgt bSgt o Ignore rapidly rotating hsinm COSyB0t Addition of angular momentum 0 Consider ground state H proton electron 6 0 a Each can have spin up or spin down TT Ti H H c Total angular momentum S E Sp Se 0 All four states are 82 E 85 S eigenstates Szlxpxegt 55 S pregt Emplxpxa Emelxpxa Wm meXpXegt 39 SO m mp 77 1TT7 0 Ti H 1ii Claim two spin particles tarry total 5 0 or s 1 there are 5 l triplet states in s m notation l l ETT and l 0 E mi H and l 1 Eli 0 AND one s 0 singlet state 0 0 E i H Triplet VS Singlet states I 0 ie are triplet and singlet states eigenvectors of S2 with S2s m hss 1s m 82 SP sens Se SP S 2 2SPS rstconsider SpS Ti 85 85 SE 8 5 SEN lt2 lt2 lt2 lt2 lt2 lt2 lt2 ltlt lt2 lt lt2 2 E2 2 z392 2 m 1pm m NID NID NID NID NID 1 O PHDquot Triplet VS Singlet states II SQIS mgt Sp2 S 22SpSels mgt h8818 mgt SHSem 2 iT Ti 819mm Z12 Ti ill Consider the triplet l 0gt 2 Ti H SPS 10gt i m Ti I10gt while SP21Ogt71210gt and S 210gt h210gt thus 5210gt 3 3 2110gt 2h210gt Consider the singlet 0 0gt E at m SPS O 0gt gi ti 3 it l 0gt while SP2O 0gt h20 0gt and S 20 0gt gl o 0gt thus 820 0gt 3 3 2 30 0gt 00 0gt Triplet vs Singlet states III S2s m Sp28622SpS s m h2ssl5 m Triplet state S2l 0 2h2l 0 Singlet state S20 0 00 0 Triplet state S2l l 2h2l l Triplet state S2l l 2h2l 1 All of these are eigenstates of S2 but also eigenstates of 827 ie Szsmsgt hmp m smsgt Lecture 15 Dirac s SHO o SHO discussion section 7 3 o SHO Via Dirac section 7 4 Hermite polynomials W5 1196 52 NH 6 52 0 so the power series de nes inner polynomial behaviour 0 which we demand are only physical Hilbert space solns 0 ie terminate series eg tail solns n 001 001 as 2ltn jgt 32 f E O49051hw j1j2aj or gt main SHO dot points 1 Energy is quantised E n hw the classical oscillator has continous spectra macroscopic particle E gtgt ha Bohr s correspondence principle n E 1027 c c 2 n 1 2 uniforrn energy level spac1ng square well E n E useful creation and annihilation of how quanta solid state phonons E M oscillator eld photons 3 non zero point energy cannot have x 019 0 best we can do is do With AgrAp 732 WOIHW0gt W0T 0gt WellW0 gt O 4 solutions alternate odd and even but Why main SHO dot points contd 5 Wavefunotions extend beyond Classical turning points 6 Probability of measuring Classical particle varies as P oc l 1 Cl Mac w xgx2 ie Classical particle fast in middle slows at turns Quantum probability density Px r2 tends as n gt 00 to Classical PC a Analytic Dirac Method Worked SHO out in position basis ie can also do the same in momentum space next HW A2 p 1 2A2 EEE 2m2mwxgtgt gt ever clever Dirac worked in energy basis ie no w i p2 mm Egt EEgt H IE only assuming that we know GCCR 32313 th iii s0 factorise Operators by de ning Griz as if they were numbers U2 v2 v z u v i3mwi and amp i13mw93 1 1 V 2hmw V 2hmw Ze GCCR gmwa Egt EEgt 0 So working with ampi W 233 mwi A A 1 A A A A a 2hmw 2p mwx 2p mwx 1 A2 A 2 AA AA 2hmw p mum zmwxp 1993 1 A2 A 2 i A A 2mm p mm 2h567p aa and H hwampamp aa and H hwampamp o and thus the commutator a our 1 Ladder operators 0 re writing the Schr dinger eqn with SHO HIEgt Mach i Egt EEgt o Prove the rst of the following on the board HampEgt EhwamplEgt and HOE IE E hwamp Egt Bottom of the Ladder but we want H E but eigenfunctions O mgt E 13gt EhwampEgt and Egtgt E hwamp Egt We cell ampi the ladder operators since given is a eigensoln nds a new one diEgt Each solution diEgt with energy E i how but what happens for E S O demand ampE0gt O A A H 1 A 1 O aaE0gt E0gt thus HE0gt EhwE0gt which then gives us En n hw Algebraic Normalisation Pqeedfact0rs amp lngtzcnllin1gt2u1dampll ngtdnll n1gt aga ltheEi nik ngerequamknr f l EJEQgt 1 1 hwampiamp i hwn A A 1 1 ma i gtIEngt n gtIEngt ampamp Engt amp ampEngt n given En llEn 1gt En1En1gt 1 and ldn2 ldn2ltEn1lEn1gt Enamp ampEngt n 1 ME W 1En1gtand aEngt IEHgt thus lEngt ampnE0gt 1 W Coolish stuff 0 Can also de ne number Operator N amp i amp 0 now Check out amp i amp ampamp i anwzi nLi anwzi 2hmw 1 A meA 2hmw mex h 9 amp amp i mwi i mwi 2hmw 1 A 2 A 2hmw 2zp Z mwhp A h A A A h A A 9 2mwaa and 192 ZW a Matrix elements 0 Consider the matrix elements of i in EmampEngt n 1ltEmEn1gt Emamp Engt xWEmlEn D Using Matrix elements 0 calculate Emi 3Engt in Energy basis Zeroeth wavefuncion 0 use E1gt ampE0gt O to nd E0gt o by nally jumping into x land 93E0gt wow ampE0gt gm ipmwa E0gt O W mwx 0x x and x a dm 0 Which has solution In 0 g 92 some constant 2 And 3po hwampamp 5 Em thus EO gm General solution 0 Given the ground state solution and its energy mw mu 1 0 h 14 KM g 92 and E0 gm 0 we can now generate the set of stationary states was Anltagtwoltxgt while 12 n gym 0 Although it really is easier to write the wavefunctions as mw14 1 was g WHn e 22 Analytic Method Examples 0 The ladder operators 2 23913 mwzi A 1 ai xZEmw 0 Given the ground state solution and its energy mad 2 1 T000 14 W 9 and E0 Em 105 A1 50 005 A1 d mu 14 x2 A1 y4 9 lt as x2hmw 7T5 o remember also that Hamp E hwamp Lecture 9 Outline observables SE o QM postulates Section 41 and 42 o Incompatible observables Uncertainty principle 0 eg Gaussians and Stern Gerlach o Schrodinger Eqn Section 43 Generalised Uncertainty eg Gaussian 0 2 2 lt9Agt o non zero for non commuting Operaters eg 32313 iii 0 Last time Gaussian z x 7r02 ie x a2202 wwwmwwif wwwwwwxfwwwwmmxa 0 Recall K defn 0f eigenfunctions kx A617 mmawwfmmwwwwm1mwwwmwx I l oo e szh e x al22U2 7 2 4 ipah P2U22h2 7T 0 SO Generalised Uncertainty eg Gaussian 0 Gaussian plots of both x and gbp 19W 0 The variance 0 92 93gt2 022 a2 a2 022 0 By inspection gbp has p 0 thus 0p h a i i E x5 V50 2 0 Thus a Gaussian is a minimum uncertainty state 0960p Noncommuting operators Discrete 0 2 2 mm o for matrices if 9 A 0 cannot nd a set of common eigenvectors that simultaneously diagonalises both 0 The S G Sm Sy 32 measurements are the archetype Sm MID O 1 1 1 O O 1 0 ie all are Hermitian measure spin in 9 y z dirs 0 But they do not commute With each other lack of Simultaneous Eigenfunctions c all are Hermitian with eigenvalues h 2 and h 2 Eigenbasis expansion Diagonalisation 0 ie all are Hermitian measure spin in as y z dirs choice of phase 0 ie Ivan s innocent question Determinate States 0 if wigt ie eigenfunction of Q 0 These are states Where every 9 measurement wi 032 Q Wi2gt lt l9 wi2 gt Q W l9 wi gt 0 0 Where wz Note 9 is Herrnitian so is Q w o The only vector that has zero inner product is O OOgt I WW 0gt SO W cow 0 Time independent Schrodinger eqn HM En ngt 0 eg energy of stationary states nH ngt Schrodinger Eqn 4 The state vector evolves according to Schrodinger Eqn ihl lt gt HIW o Hamiltonian of a Simple Harmonic Oscillator potential 2 1 A2 I9 A p A Hclassical l 5771092972 gt H l 5771002972 0 Can choose Whether to use or pgt as basis A E2 ol2 1 A p2 h2mw2 ol2 H mw2x2 or HE 2m dx2 2 2m 2 dp2 0 depends on which is easier to solve With a wavefunction 0 We later on solve the SH0 problem With a 3rd basis Schr dinger Eqn General soln mlttgtgt 19W HIEgt EEgt 0 when H is Hermitian eigensolutions must exist 0 H eigenproblein is timeindependent Schredinger Eqn Wt Z EgtltEl tgt ZaEtEgt New rewrite ih tgt 13mm 0gt A d 0gt H Wt Z 05 EaEtgt Engt o By linear independence defn iff al 0 dt aEt s01ns aEtaEOeiEth W ZaE0gt iEt Egt Z EgtltElw0gte iEt Lecture 12 Outline Feeling Trapped o Homework 5 comments c Mid term review a still stuck in a box Section 52 o if time stuck in a nite well problem 526 Homework 5 time evolution a Physical picture of S G state evolution


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