573lec33

573lec33 - 5.73 Lecture#33 33 1 L-S Terms via L2 S2 and...

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33 - 1 5.73 Lecture #33 updated September 19, L-S Terms via L 2 , S 2 and Projection LAST TIME: * method of M L , M S boxes. [For 3 L states, cross out boxes starting from both (M L =L,M S =1) and (M L =L,M S =0).] complete (2L + 1)(2S + 1) dimension for each L-S term [# of boxes] *n l 2 pattern *( n l ) 2 n l * method of ladders plus orthogonality TODAY: L 2 , S 2 method to obtain |LM L SM S , especially for M L ,M S boxes in which the where method of ladders plus orthogonality is most inconvenient : M L = 0, M S = 0 other, strong spin-orbit basis sets Modern calculations use projection operators: designed to project away all unwanted parts of ψ yet preserve normalization. * L 2 L + L only for M L = 0 block. Every L–S term is represented in this most evil block. * set up and diagonalize S 2 — easy — by forming ± linear combinations (singlet and triplet) αβ βα αβ + βα * transform L 2 to singlet, triplet basis (block diagonalization), then diagonalize L 2 by knowing (from crossing out boxes method) eigenvalues: L(L + 1)
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33 - 2 5.73 Lecture #33 updated September 19, Look at the M L = 0, M S = 0 block of f 2 and construct all L – S basis states. All extant L-S terms of f 2 are present once in the M L = M S = 0 block. Never try to get to this block by ladders and orthogonality! Cute trick that works especially well in M L = 0 and M S = 0 blocks because many otherwise awful terms vanish. diagonal but vanishes in M L = 0 nondiagonal ψα β ψβ α β α β α β 1 2 3 4 5 6 7 33 22 11 00 =− = Do d 2 in lecture L L LL L L L L L L 2 2 1 2 1 2 2 =+ + () + [] = + + +− −+ + − zz z , , h hh same as 2 So for M L = 0 block only, can replace L 2 by L + L (or L L + ) and, for M S = 0 only, replace S 2 by S + S .
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33 - 3 5.73 Lecture #33 updated September 19, ψα β ψ ψβ α ψψψ 1 2 112 2 113 2 2 212 2 224 3 2 334 2 31 3 5 4 2 434 2 4 33 6 6 6 6 22 6 1 6 1 0 =− = + = + = + = + = + =+ + = + = SL 6 1 61 0 1 1 10 22 12 1 1 10 22 12 00 0 1 2 1 2 2 4 246 5 2 55 6 2 5 3 5 7 6 2 6 5 6 2 6467 7 2 7 2 7 5 67 ψψψψ ++ = + = + + = + = + = = + all easy require a bit more work now we know, for 2e , S 2 can only have 2 h 2 and 0 h 2 eigenvalues (triplet and singlet) diagonalize S 2 by inspection ψψ 1 12 1 2 34 2 3 5 63 5 6 47 ts s () = −− // t s : αβ βα + This also has αβ − βα form ignore factors of h 2 SS LL L L 2 21 2 6 23 6 12 6 3 3 12 6 2 2 3 2 2 ββ −= −+− [] + +− + / / etc. For f 2 :
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33 - 4 5.73 Lecture #33 updated September 19, Confirm that these functions diagonalize S 2 and give correct diagonal elements. 0 0
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This note was uploaded on 11/28/2011 for the course CHEM 5.74 taught by Professor Robertfield during the Spring '04 term at MIT.

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573lec33 - 5.73 Lecture#33 33 1 L-S Terms via L2 S2 and...

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