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 2 pattern * (n ) 2 n ′ ′ * 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 ψ α β ψ β α ψ α β ψ β α ψ α β ψ β α ψ α β 1 2 3 4 5 6 7 3 3 3 3 2 2 2 2 1 1 1 1 0 0 = = = = = = = Do d 2 in lecture L L L L L L L L L L L L L L L L L L L L L L L L L L 2 2 2 2 2 2 1 2 1 2 2 = + + ( ) = + + [ ] ( ) [ ] = = + = + + + + + + + + + + z z z z z z z , , ( ) 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, ψ α β ψ ψ ψ ψ ψ ψ ψ β α ψ ψ ψ ψ ψ ψ ψ α β ψ ψ ψ ψ ψ ψ ψ ψ β α ψ ψ ψ ψ 1 2 1 1 2 2 1 1 3 2 2 2 1 2 2 2 2 4 3 2 3 3 4 2 3 1 3 5 4 2 4 3 4 2 4 3 3 6 6 3 3 6 6 2 2 6 16 10 2 2 = = + = + = = + = + = = + = + + = = + = S L S L S L S L 6 16 10 1 1 10 22 12 1 1 10 22 12 0 0 0 12 12 24 2 4 6 5 2 5 5 6 2 5 3 5 7 6 2 6 5 6 2 6 4 6 7 7 2 7 2 7 5 6 7 ψ ψ ψ ψ α β ψ ψ ψ ψ ψ ψ ψ ψ β α ψ ψ ψ ψ ψ ψ ψ ψ α β ψ ψ ψ ψ ψ + + = = + = + + = = + = + = = = + S L S L S L all easy require a bit more work now we know, for 2e , S 2 can only have 2 2 and 0 2 eigenvalues (triplet and singlet) diagonalize S 2 by inspection ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ 1 1 2 1 2 1 1 2 1 2 2 1 2 3 4 2 1 2 3 4 3 1 2 5 6 3 1 2 5 6 4 7 2 2 2 2 2 2 t s t s t s s = + ( ) = ( ) = + ( ) = ( ) = + ( ) = ( ) = / / / / / / t s : : αβ βα αβ βα + This also has αβ − βα form ignore factors of 2 S S S S L L L L 2 2 1 2 1 2 1 2
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