Matrix Elements of Many-Electron Wavefunctions. noninteger principal quantum number. solutions to Schröd. Eq. outside sphere of radius r

Transcription

1 30 - Matx Eements of Many-Eecton Wavefunctons Last tme: ν = R En, f ( ν, ) g ( ν, ) need both f and g to satsfy bounday condton fo E < 0 as ν = n µ πµ s phase shft of f ( ν, ) nonntege pncpa quantum numbe soutons to Schöd. Eq. outsde sphee of adus 0 nfnte set of ntege-spaced ν-vaues that satsfy bounday condton Wave emeges fom coe wth ν-ndependent phase. Coe tansfoms wave wth coect 0 mtng behavo nto one that exts magnay sphee of adus 0, whch contans the coe egon, wth πµ phase shft. Coe samped by set of dffeent s. Today: next few ectues Wavefunctons and Enegy States of many-eecton atoms. obtas confguatons L - S states. eectons ae Femons must be antsymmeted : KEY PROLEM 3. Sate detemnants ae antsymmetc wt a e, ej pemutatons. omaaton SO. Matx Eements of one - e Opeatos: e.g. H = a ( ) s C. Matx Eements of two - e Opeatos: e.g. H (a vey bad petubaton ) eff k k 4. H n tems of εn, F, G, ζn paametes 3 obta enegy Sate-Condon j spnobt e = > j e j Page 3-9 s an exampe of what we w be abe to do. * Intepetabe tends: Peodc Tabe * tomc enegy eves: mysteous code no atom-to-atom eatonshps evdent wthout magc decode ng.

2 Many-eecton H H= h() e + + a ( ) s = 4 34 > j= j H (0) sum of hydogenc Z - e tems: ε n unsheded obta eneges n How do we set up matx epesentaton of ths H? H (0) defnes bass set (compete, othonoma, ) H h H as sum, E as sum, as poduct spata pat Eectonc Confguaton: st of obta occupances e.g. C s s p sx e spn pat 30 - the φ ( ) s coud be hydogenc o sheded-coe Rydbeg-ke obtas. not suffcent to specfy state of system 3 sevea LS, tems ase fom ths confguaton: e.g. p D, P, S L S s H () R = ( ) φ s (0) = () = ( ) () sm = = spn-obta [ () + ( )] ()() = + h h φ φ h( ) φ( ) φ( ) h( ) φ( ) φ( ) we know that L, L, S, S commute wth h so we can use these to bock dagonae H. () + ote that athough does not commute wth e j, ths s not a pobem fo s and s because h() and j do not nvove spn. destoys but not L! j a constant wth espect to h(). e j

3 How do we get egenstates of L, L, S, S? 30-3 ethe: I. Method of M L, M S boxes dvanced Inoganc Whch LS tems exst, not the specfc nea combnatons of spn-obta poducts that coespond to these tems. II. ngua momentum coupng technques 3-j addes pus othogonaty pojecton opeatos We w etun to ths pobem and appoach t both ways. One goous symmety must be mposed: Pau Excuson Pncpe: eectons ae Femons and theefoe any acceptabe wavefuncton must be antsymmetc wth espect to pemutaton of Y pa of e e.g., = u ( ) u ( ) P obtas, = u ( ) u ( ), e ae ndstngushabe, [ HP, ]= 0 eectons + oson ( ntege spn) Femon ( ntege spn) j a 's must beong to + o egenvaue of P (note that P = I) j j P ± =, ±, = [, ±, ]=± ± + bosons femons

4 geneae to 3 e? 3! combnatons needed! Hobe 30-4 s have! tems (each a poduct of spn-obtas) matx eements have (!) addtve tems! TRICK! Sate Detemnants u u () u () ( ) u ( ) e = u ( ) u ( ) u ( ) u ( ) obta ow s abe fo e coumn s abe fo spn-obta you show that 3 3 Sate detemnant gves 6 addtve poduct tems Detemnants *! tems n expanson of detemnant * detemnant changes sgn upon pemutaton of Y two ows [e s] o coumns [spn-obtas] * detemnant s eo f any two ows o coumns ae dentca. * detemnant may be unquey specfed by man dagona MUST SPECIFY I DVCE STDRD ORDER I WHICH THE SPI-ORITLS RE TO E LISTED LOG MI DIGOL e.g. s α, s β, p α, p β, p0 α, p0 β, p - α, p - β, [o fo p, suppess p n notaton: αβ0α M L =, M S = ] eed a fancy notaton to demonstate how Sate detemnants ae to be manpuated n evauatng matx eements. Ths notaton s to be fogotten as soon as t has seved ts mmedate pupose hee.

5 30-5 = ( ) # of bnay pemutatons p ( )! u ( ) u ( )! dffeent s s OE pescpton fo eaangng the obtas fom the ntay specfed ode s poduct of sevea P j s o, moe usefu fo povng theoems, a poduct of factos P whch te whethe the -th eecton s to be eft n the -th spn-obta o tansfeed to some unspecfed spn-obta [ u() u ( ) ]= Pu (). omaaton Vefy that (! ) s coect nomaaton facto, = [ ] [ ] = (! ) ( ) u ( ) u ( ) u ( ) u ( ). ow eaange nto poducts of one - e oveap ntegas,, = = (! ) ( ) Pu () P u (). u ae othonoma u() u(j) has no meanng because ba and ket must be assocated wth same e The ony noneo ega tems n ae those whee ECH P = P othewse thee w be T LEST ms- matched ba - kets u ( k) u ( k) u ( ) u ( ) j j = 0 = 0 (Hee the eecton names match n each ba-ket, but the spn-obta quantum numbes do not match.)

6 30-6 Thus t s necessay that =, p= p, ( ) = + and = (! ) u ( ) u ( ) u ( ) u ( ) = = each tem n sum ove gves +, but thee ae possbtes fo P, possbtes fo P! possbtes fo sum ove = (! ) = Thus the assumed (!) nomaaton facto s coect.. Matx eements of one-eecton opeatos F f e.g. L= = ( ) F ( ) p ( )! a ( ) a ( ) ( ) p ( )! b ( ) b ( ) = ( ),,,, ( )! a ( ) f b( ) = ( ) ( ) [ P P ]! a ( ) b( ) Pa () f Pb() ( ) ( ) [ ] [ Pa ( ) P b ( ) ] Poduct of obta matx eement factos n each tem of sum. Of these, ae obta oveap ntegas and ony one nvoves the one-e opeato.

7 30-7 SELECTIO RULE ΨF Ψ = 0 f and dffe by moe than one spn - obta (at east one of the obta oveap ntegas woud be eo) two cases eman:. dffe by one spn-obta = u () a ( k) u ( ) = u () b ( k) u ( ) k k use u to denote common spn-obtas use a k, b k 0 to denote unque spn-obtas the msmatched obtas ae n the same poston fo ths choce, a P factos of each must be dentca to a factos of addtona equement: must bng msmatched obtas nto -th poston so that they match up wth the f( ) opeato to gve Y OTHER RRGEMET GIVES ( ) a () f b () k k ak( ) bk( ) u( ) f u( ) = 0 ( ) = 0 0 ( )! ways of aangng the e n the othe matched obtas and thee ae dentca tems (n whch the e s n the pveged ocaton) n the sum ove ( ) F = (! )! ak fbk

8 30-8 If the ode of spn-obtas n o must be aanged away fom the standad ode n ode to match the postons of a k and b k, then we get an addtona facto of () p whee p s the numbe of bnay pemutatons F = ( ).e. = 5 7 = 3 5 = 5 3 F = 7F3 p a k f b k fo dffeence of one spn-obta. = Dffe by eo spn-obtas F! Pa( ) f Pa( ), = ( ) ( ) a othe factos ae = [ ( ) ]! dentca tems fom sum ove agan! F = a f ( ) a * omaaton * e Opeato F Exampes of f 3 : = 3αα α L = h( 3+ ) LS comes out amost the same as nave expectaton WITHOUT need fo antsymmetaton! ( ) 3 = h + J 3αα α= L 3αα α+ S 3αα α = h αα α + 0 3αα α next tme G(,j)