Calculating the degeneracy of three electrons coupled quantum state where Hamilton is c*(S1*S2+S2*S3+S3*S1).c=Const.The answer in non coupling representation is 8.
while I choose(H,S,Sz)as CSCO (S=S1+S2+S3,Sz=Sz1+Sz2+Sz3 )
S=1/2 or 3/2 ,corresponding degeneracy equal to 2*S+1 by 4or2，total degeneracy equal to 6,weird.
Then I rectify the CSCO to ( H,S,Sz,S12 ) (S12=S1+S2 ) and find the extra degeneracy of S12 while S=1/2,bravo ! Has everything been corrected?
Degeneracy in S12 mean difference in coupling, meaning that we have to discriminate 3rd with group of 1st&2nd electron.
Does this violate the Identity or symmetry？
Using 2*(S1*S2 + S2*S3 + S3*S1 ) = S_tot^2 - S1^2 - S2^2 - S3^2, which is also true quantum mechanically, you find a spin-3/2 multiplet (4 degenerate states) plus 2 degenerate spin-1/2 multiplets (also 4 degenerate states). Which adds to 8 states, as required.
The permutation group acts by the identity representation on each of the four states of the spin-3/2 multiplet, but by two 2-dimensional irreducible representations on the four states of the two spin-1/2 multiplets.
Using 2*(S1*S2 + S2*S3 + S3*S1 ) = S_tot^2 - S1^2 - S2^2 - S3^2, which is also true quantum mechanically, you find a spin-3/2 multiplet (4 degenerate states) plus 2 degenerate spin-1/2 multiplets (also 4 degenerate states). Which adds to 8 states, as required.
The permutation group acts by the identity representation on each of the four states of the spin-3/2 multiplet, but by two 2-dimensional irreducible representations on the four states of the two spin-1/2 multiplets.
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