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Lecture 17 Outline Test 1 review o Solvay talk Lorentz 0 Chapters 1 and 2 partial overreview from httphscicasoueduSolvayCongre55719277 Brusselsjpg Ensembles of identical wavefunctions 0 Statistical interpretation of WW t2as prob density i many identical systems before measurement 9 l 2m Mill Hill H HMHH ii many independent measurements r P v WW o The expectation value is the average of repeated measurements on an ensemble of systems 93gt g 9 new dx Co ordinate space sandwiches 93gt Ildex IJ2 dx 1 dx Twiksy bits Ehrenfest s p mdltxgtdt The rule classical gt quantum mechanical Operators replace all p s With th x A 8 ltQxpgt Qa ih xp dx While the T 1922771 While for the SH0 V mw2x2gt Heisenberg s uncertainty principle 0 de Broglie Wavelength is related to the momentum h 27Th p A A 39 o For a given 11 spread uncertainty in wavelength implies a momentum spread uncertainty h 0 Classical world example Timeindependent Schr dinger eqn 0 From the timedependent Schredinger eqn m xn h23211xt 3t 2m 3x2 0 using separable solutions x t xgpt VIJU t o x t x eiEthstati0nary states since dltQxpgtdt 0 Combination Fried Rice 0 The timeindependent Schrodinger eqn has an in nite number of solns 1x 2x 3x 0 each With a corresponding E1 E2 E3 1 17t 1 6 iE1th7 1 27t 205 6 iE2th7 o The general soln x t to the timedependent Schrodinger eqn is a linear combination of in nite set of separable solns Ilnr 15 from timeindependent eqn was t 2 2mm e Z Enth n21 now the 4 cases In nite square well Simple Harmonic Oscillator 2 methods Free particle Finite square well bound and scattering states The in nite square well 0 Since Vx O we have an effective SHO like equation d2 nx ZmEn ki nx Where kn h dx2 00D Well Solutions n27r27i 2 PnCU 15 I 71 6 1Enth sin 6 1 2 ma2 t a a 0 General soln to time dependent Schrodinger equation 00 2 TNT n27r2h x t Z on sin 6 1 2m t n21 a a 0 Expand initial wavefunction on nxllx 0 dx g sin Q4930 dx 0 on tells you the amount of w in x t o cn2 probability of measuring En7 2021 lCn2En SHO solutions Hermite Polys 1 2 1 14 mHn e 2 En n 0 x W 9 Hermite polys are Hn 1101 H12 H24 2 2 H5 32g5 160g3 120g 0 Method 1 was the analyticFrobenius method we me am feven a0 262 and fodd G16 363 a 2ltn jgt a 2 j1j29 SHO Analytic 0 Method 2 The ladder Operators came from 32313 iii q 23913 mwi a 50 1 A 1 a E i xZEmw 0 Starting with ground state solnz WW mw 14 w 962 d E lh h 6 25 an 0 2 w awnwn wn awn wn1 1 009 1 W H hwampiamp i 5 thus mam E i wampi o 3 hmeamp amp and 13 z39 hmw2amp amp Scattering States Free particle stationary states are propagating waves ZmE x t Aeikxt and k i Either x t Aexpiz39lcC moving to the left right But we can build a wavepaoket 1 00 2 Wm E kelltkx tgtdk 1 00 0 2 7TOO 1 70 Bound states x t Z onwncr eXp iEnth e zkxdx Finite D Well bound states 0 Bound states V0 lt E lt O k xZmEh 0 since VLL symmetry even or odd solns about a O K Dcoscx V0 lt a lt a Fez m W x D sincx Fez m Vxgta V93ltO V0ltxlta Vxgta V93ltO 0 D7 F Via boundary cond xxa and d xdxxa 0 Introducing z 1m zo axZmVOh we have tanz zOz2 1 or cotz zOz2 1 FEW scattering 0 Now looking at scattering states E gt O 0 so we have is and H Wh A617 Bea m V a lt a T ECU N CsinOw D oos r V a lt a lt a Fell V a gt a 0 Apply boundary cond xxia and d xdxxia 0 Transmission probabilities T F2A2 0 Re ection probabilities R B2A2 Lecture 23 Outline Guts of Hermitian 0 Quiz Number 6 o Determinate States Section 322 o Eigenfunctions of Hermitian Operators Section 33 o Generalised Statistical Interpretation Section 34 Determinate States 0 Observables are represented by herrnitian ops QT o Deterrninate States are Eigenfunctions of 0 These are states Where every Q measurement q 0 eg energy of stationary states En 03 I Q Q2gt 0146 Q2 1 gt Q Q 1 Q Q 1 gt 0 0 Where q Q Note Q is Herrnitian so is q o The only vector that has zero inner product is O OOgt KC 110 0gt SO Q P 1 11 o Time independent Schrodinger equation Hun nwn Eigenfunctions Discrete Spectra For normalisable eigenfunctions of hermitian operators De ne spectrum degenerate QM mac and 0 lt f fltxgtldx lt oo Two theorems a Eigenvalues are real b Eigenfunctions belonging to distinct eigenvalues are orthonormal Proof of a Qfx mas ltflcgtfgt lt flfgt qltffgt qltffgt Thus since fa y o q q Eigenfunctions Discrete Spectra o b Eigenfunctions belonging to distinct eigenvalues are orthonormal Qf qf93 and QQW Q g93 0 Since Q is Hermitian ltf 9gt ltlefgt q ltf9gt qltf9gt 0 We know that q q are real distinct thus fggt O 0 note if not distinct can construct orthogonal set Completeness Axiom Eigenfunctions of an observable operator1 are complete any function2 can be expressed as a linear combination of them Footnotes 1 read Hermitian 2 function that is in the Hilbert space Generalised Statistical Interpretation 0 If you measure an observable Qx p on a particle of state x If you are certain to get one of the eigenvalues of the hermitian operator Qx ih Qf qf93 and QQW Q g93 o If Q has a discrete spectrum the probability of observing the particular eigenvalue qnassociated With orthonormalised eigenvector is ICnI2 ltfn 1 gt2 0 Upon measurement the wavefunction collapses to the corresponding eigenstate 0 That s the generalised statistical interpretation along With Schrodinger eqn time evolution QM foundation GSI take 2 part I Observable operator has complete set of eigenfunctions W937 t Zn Cnfn93 Given orthonormal eigenfunctions use Fourier trick an ltnt 1 gt fzltxgtwltxtgtdx So on is how much of fn is in 11 as before For a normalised 11 total probability Zn cn2 1 1 IJIJgt Cmfm95gt Cnfn95gtgt chcwmlm 2032071 Z 6n2 GSI take 2 part II o The expectation value of Q is sum of the possible outcomes times the probability of that outcome ltQgt 011m ltltZ emfmltxgtgt 62 Z ammo ZQnCZLCnltfmlfngt ZQnICn2 0 Where we used an qnfn 0 ie 11 collapses to an eigenfunction of that operator Lecture 37 81 32 S ymmetry o Finish addition of angular momentum Section 443 0 Symmetry of Fermions and Bosons Section 511 0 Application to atoms He Section 521 Triplet VS Singlet states I 0 ie are triplet and singlet states eigenvectors of S2 with S2s m h2ss 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 SZIS mgt Sp28622SpSels mgt h28818 mgt SHSem 2 iT Ti 819mm Z12 Ti ill Consider the triplet l 0gt 2 Ti H SPS 10gt i m Ti I10gt while SP210gt h210gt and S 210gt h210gt thus s2l 0gt 3 3 2110gt 2h2l 0gt Consider the singlet 0 0gt E at m SPS O 0gt i553 ti 3 it l 0gt while SPVIO 0gt EEQIO 0gt and Se2O 0gt EWIO 0gt thus 820 0gt 3 3 2 30 0gt 00 0gt Triplet vs Singlet states III S2s m Sp2S 22SpS s m h2ss1s m Triplet state S21 0 2h21 0 Singlet state S20 0 00 0 Triplet state S21 1 2h21 1 Triplet state S21 1 2h21 1 All are simultaneous eigenstates of S2 and 82 In general with individual 51 and 52the total spin can be 5 51525152 1sl 52 The tricky bit is to construct a combo eigenstate sms Isms Z 05555ij W4er ms 5 m gt5jmjgt Hydrogen Atom One particle state includes spinor n mg msgt Whose total orbitalspin angular momentum range is j s s l s H h2v2 82 47reo 7 Consider a H atom electron quantum numbers n mgsmsz 1 for electron only j 6 orj lg 5 electron proton J 61 or J 6 or J 1 the J 6 has two distinct combos with i IF important When H includes eg H80 oc SL spin orbit interaction gt hyper ne structure H Z oc L S Bem magnetic eld Zeeman effect Helium atom and symmetry a Two electrons E wr1r2xslsg o For the He atom xed nucleus Z 2 With 2 electrons 712 1 Z82 712 1 Z82 H V2 V2 2m 1 4713960 7 1 2m 2 4713960 7 2 1 e2 47T 0 r1L r2 soln ibr1r2 war1wbr2 wn1 1m31rlwn2 2me2r2 But QM says identical particles are indistinguishable singlet wr1r2 Almanac I392Wbril antisymmetry 11172gt 111271gt lies in spin spatial 0k triplet ibr1 r2 Awarlwbr2 Waltr2r1l Helium atom and symmetry 0 anti symmetry has real effects consider states 2SHL lworbgt8 X lepingtA X 8 He3S lworbgtA lepingtS X 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 08 Outline 1D SHO quiz 2 escaping the in nite well section 22 simple harmonic oscillator section 23 rstly Via power series section 232 httpwwwfalstadC0mqm1d Simple Harmonic Oscillator the most pervasive ubiquitous around consider small fluctuations about some potential ininiina Va Va0 d V 2 dx SE 0 at 02 effective potential Va ka mw2x2 at n1inin1a Classical a mass oscillates on a spring Hooke s law 8V d2a k Fat Kam with w 81 dt2 m has a solution 513t A sinwt B coswt Power Series Method 2 d2 2h m it mw2az2 Was Emgr a rewrite With Change of length and energy scales 2E 5 51 and KE dw 5 0 looking out towards a gt 00 ie g gtgt K CHM d5 0 so B 0 to not have explosion at a gt 007 making me f erg2 381 soln mg g Ae 522 Beg2 just try and try again a we have our rescaled SE digg 2 Kw a so working with Mg f 8 522 We want Hg d d 2 2 1 d2w d2f df 2 d g Te 254N52 U e 52 a which makes the Schredinger equation look like d2f df 2 2 d g 25d g 2 1fe 5 2 2 Kf e 5 2 d2f df d g2 25d K 1f0 Power series hunt d2f df d e 25 a hunt for a power series solution ale Frobenius method K Uf0 f aoa1 a2 2 Zajgj j0 d 00 39 d Jg a1 mg 3 236ng j0 d2 00 39 d g 2a2 2 x 3a3 ZU1gtU 2 25 7 j0 Z J1j 2mm 27 K 1ajgj 0 Power series formula Polynomials cant be written as eg 513 512 1 ie uniqueness of 57 linear independence means that Vj a 2339 1 K a 39 2 I 39 9 3 HM 2 7 o A recursion formula that is the Schrodinger equation Starting With a0 and a1 you get the rest 0 Split the complete solution f g feven fodd fevenlt gt a0 l 252 l and foddlt gt Q16 353 l Truncate Power series a 2j1 K 92j1j2 Where to go With this Consider for large j W 1 1 39 2 f5 0 5730 fg2 73085 E 32 E 3 Which results in badness 115 f e52 2 08522 2 aj2 yaj and aj so we must truncate the power series gt quantises K E 1 an20 gt K2n1 gt Enn hw Meanwhile the other odd or even series an 0 Power series coef cients feven CLQCL2 2 and fodd a1 a3 3 o in terms of the principle quantum number n a Ma 7 j1j2 7 o For n 0 one term a0 with a2 0 and a1 0 fo a0 thus mg fog 8 82 a0 852 0 For n 1 one term a1 with a3 0 and a2 0 W5 f1 e 522 Q15 8 82 a For n 2 two terms with a2 a0 ie a4 0 we 135 52 a0 age 882 aou 252 8 82 Hermite Polys 0 nally we also need to normalised and also given a g xmwhas the Hermite polynomials are Hn mw 1 We 5 WH 22 H01 H12 H24 2 2 H5 3255 16053 1205 0 some useful relations are Hn1 2 Hn 2an1 Hn 1quote 2 dig n 8 52 SHO Wavefunctions SHO nonbounded solns Lecture 38 Outline Still Perturbed o time independent perturbation theory 0 nondegenerate Perturbation theory Chapter 611 0 First and second order theory Chapter 612613 o degenerate Perturbation theory Chapter 62 o brie y Hydrogen ne structure Chapter 63 0 very brie y Zeernan Stark effects Chapter 64 Nondegenerate perturbation theory 0 Summary of formulae for timeindependent potentials E s E E W ltw2IH Iw2gt ngt W W m Z lt E39If 13 W mat n But how to know how accurate the solution really is o Mosttimes compare with 2nd order energy correction 0 HI 0 2 EnmE2E lEfl E ZM ma E2 E31 Rerun Problem 61 run through 64 and then 62 Obvious problem if we have degenerative energy levels Degenerate perturbation theory Obvious problem if we have degenerative energy levels H0 3gt E0 3gt 7 H0 2gt E0W8gt rescue degenerative perturbation theory Section 62 H0 0gt H0O 3gt lm E0O 3gt lm following as per non degenerate theory can Show etc Waa Wab 05 1 oz E Wba Wbb 5 5 Where Wm W H W30 ie an additional eigenproblem 1 1 2 2 Ei 5 Wm Wbb i W Wbb 4Wab In effect diagonalisation of the degenerate part of H 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 energy levels 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 Zeeman and Stark effects How do the Bohr energy levels change in uniform Bat and Eat elds Depends on eld strength compared to internal elds 6 HZ u M8 B xt 2SBext weak elds total J L S is conserved with good quantum numbers n E j mj due to spin orbit strong elds the good quantum numbers are 713 Mg m8 As for weak Stark e ect eg H g eEz can calculate E11 8 100z 100gt E O and E oc e282 quadratic Lecture 28 Outline End of Chapter 3 a Quiz time a Will cover the following in terms of 13 and H Determinate States Section 322 Eigenfunctions of Hermitian Operators Section 332 Generalised Statistical Interpretation Section 34 Matrix Elements Section 36 GenStatInterp measurement Every real number A is an eigenfunction of a To compute the CA we compute with g x 65 A co ltgi111gt OO 00 g x 1115 tda 111M 25 the probability of getting a result in the range d is P ICWI26M lt9A 1 gt2dA I 1 A fl26M wavefunction collapses to a range which depends on measuring device precision the average of measurements is Eigenfunctions 13 Spectra Example 32 Find eigenvalues eigenfunctions of 13 ie the eigenfunctions are again not square integrable BUT they are not in Hilbert space for any complex p but can again de ne a kind of Dirac orthonormality OO 00 ltfp lfpgt gp 9pd A2 e p lmhda A227Th 529 p 529 29 39 haVing 01108811 A lx 27th the eigenfunctions satisfy ltfp lfpgt 6p p and are complete GenStatInterp p measurement a Complete write any square integrable function as 11152 ff cpfpadp 1 ff cpefpxhdp o Where the expansion coefficients are a function 1 C29 mm 00 mmdx MW 0 Thus the probability of getting a result in the range dp is P Cpl2dp lt1gtptl2dp 0 Again range depends on measuring device precision o Eigenfunctions of 13 have wavelength A 27Thp proof of de Broglie formula but no particle has exact p o momentum space CM19 2 position space 1115 25 Generalised Uncertainty Principle 0 Last lecture de ned determinate state as 0393 ltQ lt61gt2gt ltQ Q 1 Q Q I gt 0 0 But for indeterminate states and two observables 0324 I W1 ltAgt 1 lfl ltAgt 1 gt 0 I 3 ltBgt 1 l3 ltBgt 1 gt a can show using Schwartz inequality and more that 2 inmi o remembering 51913 271 thus ago 2 712 0 ie incompatible observables cannot have a complete set of common eigenfunctions