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by: Citlalli Sauer


Citlalli Sauer
GPA 3.75


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Class Notes
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This 11 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 22 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 22 Outline Method Madness 0 Spin and 2 particle Symmetry Section 151 2 particle Symmetry Section 101 102 151 Variational Principle and example Section 161 He calc Section 161 into Perturbation Theory timeindep non degen 171 Hyper ne Interaction 0 Consider the following 6 0 Hamiltonian ltHgt WHO thlw En Alt8msSpS 8msgt a Two particles have relative orbital angular momentum 0 also have coupled spin W n mggt X smsgt Where S2s m8 h2ss ls m8 0 Triplet l l Triplet l Triplet ll 0 Singlet 0 0 using SPSE s 01 are and SZsmsgt hmssmsgt igt H 15 gt E an H 3 Ti ii S2 SP2 S 2 the energies different H hyper ne splitting 21cm line Wavefunction Symmetry Direct product combines different vector spaces eg W172gt lt12Wgt E lt1 lt2 1gt W2gt eg particle l is at 5131 and particle 2 is at 5132 But QM says identical particles are indistinguishable Bosons wgt5 w1w2gt5 w1w2gt w2w1gt Fermions WM w1w2gtA w1w2gt w2w1gt ensures antisynnnetry w1w2gtA w2w1gtA ie Pauli principle wwgtA wwgt wwgt 0 ICUlCU2gtS X lwn ni w2m2gtA IW1W2gtA 1m3gt Helium atom and symmetry 0 anti symmetry has real effects consider states 2SHL lworbgt8 X lepingtA X 8 He3S lworbgtA lepingtS X Variational Principle 0 Variational principle gives an upper bound on E98 0 pick any normalised trial function Wt then Egs S 0 The proof since there exists eigenfunctions um of H M chwn suchthat Hwngt Enwn thus 1 we ECCnltwmlwngt Z W ltHgt WHIW ZCRCnWmIHWM ZCRCnEnWmeM a so Zn cn2En 2 E98 Zn cn2 E98 as reqd 1D Harmonic oscillator dumb guess a To nd an upper bound on E98 hw Choose 2 trial functlon wt to be a Gaussmn 4513 A8 bx The parameter b is arbitrary While normalisation gives 00 2 2b i 1 A2 8 2 da A2 2 so A next calculate T V h 2 mw2az2 2m oo x2 ltTgt h2lAl2 e bw2d2e b day h 2b 2m 00 das2 2m mw2A2 0 2 ma V 51328 2 da 2 OO 86 2 mwQ r r a 2 mwQ 0 So 3 7er and thus H g m W 0 0 wt With b 73 2 gives a hw minima Helium ground state guess 1 a Want an upper bound on E98 79005 8V considering o the Cou101nbicquot Hamiltonian H H1 H2 V66 712 2 2 2 1 H V V e 47T 0 7 1 7 2 r1 r2 ignoring the V86 interaction has hydrogenic soln 8 41171 100r1 100r2 e 2 7quot2 0 0 o wlOOr energy HQ Z2EL m 4 x 136 544 8V if non interacting H1 H2 8E1 109 8V ie wt is eigenfunction of most of H wt 8E1 Vee bt a but this is for two non interacting electrons next add V66 Helium ground state guess 1 contd a Want an upper bound on E98 79005 8V given trial Triage 2Tl l397a2OJO lt lr1ir2lgt 82 8 2 4r1r2a0 V d3 d3 lt 66gt 47T 0 7mg r1 r2 r1 r2 such a running Coulomb integral is a pain but do able 5 e2 5 Vee E m 34 V lt gt 4a 47TEO 2 1 e so our trial function gives quite a good upper bound 5 Et ltH1H2l egt 8V but we can do better with a bit of thought Helium ground state guess 2 guess 1 wtr1 r2 e2rlr2 0 gave E m 75 8V 7TCLO guess 2 consider shielding of nucleus ie Z lt Z 2 trial wtr1 r2 Lie Zt7 17 2 0 exible Zt rewriting 7139CLO h2 82 Z75 Z75 H v2 v2 1 2gt47T 0 7 1 7 2 82 Z15 Z15 1 47T 0 7 1 7 2 r1 r2 where ltHgt 2sz1 2Zt m ltV gt a can show 5 while Veg ZtE1 s0 ltHgt 2Zt2 4ZtZt 2 gngl 223 217243 d 27 0 when Zt E m 169 so m 775eV Helium ground state research a guess con guration interaction wt Ci wmlimi wn l m ma NdZm Energy 6V 0 990 7834239169 1 1656 7892709569 2 2286 789883375O 3 2916 7900341868 4 3546 7900879312 5 4176 7901116327 6 4806 790123651O 7 5436 7901303773 8 6066 7901344293 9 6696 7901370133 10 7326 7901387377 11 7956 790139932O 12 8586 7901407849 00 7901439039 Exact 790143939O NIST ASD 31 790051506O Accuracy of Variational Method 0 Given a Hamiltonian ie potential Your trial wavefunction should try to include short range and long range behaviour and nodes so now that you ve done some examples assume i lib 1 15 0 2 lEogt 1i0E1gtgt ie that your trial contained 10 pollution ltE0HE0gt ltE1HE1gt EO 001131 E wt 1 101 so Em s 099E0 001E1 0 ie rst order error in Mt gives second in Et 0 that s a really really nice property


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