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# QUANTUMMECHANICS PHYS610A

SDSU

GPA 3.75

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This 17 page Class Notes was uploaded by Citlalli Sauer 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 16 working the angles end of 1 D SHO Via Dirac section 74 skip Chapter 8 on Path lntegralsdon t read skip Chapter 9 on Uncertainty bitsread 3 D Systems section 102 might be back to 10 skip Chapter 11 on Symmetriesmight also be back Spherical geometry ng Harmonics section 7 5 L2LxLyLZLi Rotation ops section 72737475 End of SHO 0 1 D problems work in position or momentum space 0 SHO ladder operators along with energy eigenstates A 1 Fz39anmwi a i V 2hmw A h A A A m A A 9 2mwaa and 192 wa a 0 Given the ground state soln and its energy mad 2 1 T000 14 W 9 and E0 Em End of 1D o the SH0 has time dependent superposition solns Wt 220 CneiEMRIEM o applet time http WWWphysuriedu yoonqhornainhtrnl 0 nally want to emphasise bound vs scattering energies 3D timedep Schr dinger eqn 0 Easy to generalise timedependent 1 D SE to 3 D ih11rtgt HIM tgtgt 0 just modify the Classical Hamiltonian Operator 1 110672172 193 19 19 Vyz a a a o Wherein replace pm gt zh pg gt zh8 y p2 gt zh 3 E2 39h V2IJ W 2 3t 2m c 39 7 22 i o Laplaman V 82x82y822 CCR s o canonical commutation relations 3D timeindep Schrodinger eqn Probability of nding particle Ilr If within a volume elernent d3r dx dy dz is Ilr t2d3r with Ilrt2d3r 1 Given potential Vx y z t Vx y z time indep we again have a complete set of stationary states as Ilnr t nreiEnth E and w are eigenvalues functions of timeindep SE E2 V w ET 2m and Ilr t Z Cn nreiEnth is tirne dep SE soln 3D timeindep Schriidinger eqn 0 in nite 3 D square well separation of variables Spherical Co ordinates Time indep SE v2 w E16 Atoms etc intuitive to work in spherical not cartesian o where for example the potential is only V0 z7 cos6 937 cosgbsin6 y7 singbsin6 rm cos6E tangby 7 9 Applying chain rule 2a222 3x 3x 37 3x 36 3x 3gb 0 gets us to the Laplacian in Spherical coordinates V2 7 22 1 2 sin6g 1 3 2 7 2 37 37 7 2 sin 6 36 36 7 2 sin2 6 32gb Separation of spherical variables 0 Time indep SE V2 Vw Em Where 0 Look for separable solns 03 6 gb RrY6 gb Y 8 3R R 8 Y R 323 2 Y 2 V R 7 2 37 74 3rgtr2 11636 81n636gtr2 sin26 32gb V2RY Mr E RY 0 Ig Y Wm 2 V7 E 0 o splits into time indep SE into radial and angular terms 1 1 a 633 1 323 O Y sineae S 66 Sin26 82 Angular equation Fr G396 gb O 0 So introduce an arbitrary separation constant 1774 EM 1 and G6 gb 1 0 Consider the angular equation rst radial later 1 1 8 dY 1 821 7 19 W 1 8 81 821 2 sin 6 s1n6gt 1Y sin 6 0 Try further separation of variables Y6 gb 6ltIgtgb sin d 39 1821 1 2 9 36s1n636gt 811 16 Pygb O 0 So introduce another separation constant m2 Two angular equations azimuthal 0 Another separation constant m2 means two angular eqns sin 6 g 39 86 9 36 sin6 gt 1sin26m2 1 821 m2 c1gt 82 The second eqn has solns gb Aexpiimgb Boundary condition on aziurnuthal gb 27 gb emu562mm ezmqb thus exp2m39m 1 so we have m 0 i1 i2 The coef cient A we absorb over into the polar soln Two angular equations polar sin 62 86 sinQZ i EM 1 sin2 6 m2 9 O Has general solutions the associated Legendre functions 96 AP mCOS 6 de ned in terms of the E th Legendre polynomials lml PM 1 x2 2 a to evaluate the Legendre polys use Rodrigues formula 1 d g P 2 H 93 2dx x 6gP01P1P232 17P3 associated Legendre functions Legendre polynomial Pgx has degree E with parity eveness or oddness depending on E assoc Legendre funks P mx 1 92 yml Pgl 1 Note for gt K then P mx 0 so 25 1 possible m For E 0 PC 1 For E 1 P10 cos Pl1 sin6 13201 P2U 3x2 1 d P21x 1 P x 3xx1 92 P2295 1 932 if 13293 31 932 P20 300s26 1 1321 3sin600s6 1322 3sin26 Spherical Harmonics 0 Finally we need to normalise these angular functions 0 W7 9 b RUM97 b RUM97 b RT99 I 2 d3r W0 6 gb2 r2 sin6 drd dgb R2r2 drY6gb2 sin6d6dgb 0 require both radial and angular integrals normalised oo 27r 7r R2r2 dr 1 and Y6 gb2 sin6 d dgb 1 0 0 0 o normalised angular functions are spherical harmonics m imq m Yl 47F 6 Pl C086 o E azimuthal quantum number m magnetic quantum Angular Momentum eigenstates 0 Back to Classical E 77 X 13 Where Lac ypz zpy Ly sz mpg L2 my ypx Law Ly ith Lw L2 239th L2 L96 ihLy 0 ie all incompatible observables eg aLmaLy Z 0 But squared total L2 Li L L2 is L27 O 0 ie simultaneous eigenstates of both L2lwgtalwgt While Lzlwgt lwgt L2 mgth2 1 mgt and LZ mgthm mgt 0 30 113andm E1 1 757 757 Angular Momentum eigenfunctions 0 separation of variables gave T njgameOquot 6 gb E n E mg 0 spherical harmonics ngew gb are the eigenfunctions of 412 a a 1 a L2YW 39 6 YW W E 1 YW sin6 666111 86gt sinea l Mwammww 0 ie good quantum numbers Hn E Wggt Enn E Wggt o Shankar 122 operator formalism rotate in a 3 plane URgb 15 11m N eiqboLZh 0 hNZ o in position space e 088 p cb 7MP b 0 Raising Lowering Operators Following Shankar 125 Assume angular basis states L2oa gt 04045 and Lzoa gt oa gt De ne raising lowering operators Li L96 i iLy they commute as L2 Li ihLi and L2 Li O L2Llo gt LL2 hLgta gt Lm hLgta gt hLloz gt similarly L2La gt LL2oa gt oaLoa gt but there are bounds on 04 52 Z 0 since lta IL2 Lilam a lLi Lilam 2 o Loa maxgt O and Loa mmgt O

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