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magnesium12
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Homework Statement
What terms are generated by the configuration (e1g)2(e2u)2 in D6h symmetry?
Homework Equations
configuration (χd)2 gives terms 1(symmetric product) + 3(antisymmetric product) where χ=symmetry of an orbital and d = degenerate
(e1g)x(e1g) = A1g + [A2g] + E2g
(e2u)x(e2u) = A1g + [A2g] + E2g
where brackets [ ] indicate antisymmetric product
s=singlet, d=doublet, t=triplet, q=quartet, qu=quintet
sxs = s
sxd = d
sxt = t
sxq = q
dxd = s + t
dxt = d+q
dxq = t+qu
txt = s+t+qu
txq = d+q+sextet
qxq = s+t+qu +septet
The Attempt at a Solution
So first I found the terms generated by e1gxe1g and e2uxe2u seperately
(e1g)2x(e1g)2 = 1A1g + 3A2g + 1E2g
(e2u)2x(e2u)2 = 1A1g + 3A2g + 1E2g
Then find direct product of [(e1g)x(e1g)]x[(e2u)x(e2u)]
1A1g 1A1g
3A2g x 3A2g =
1E2g 1E2g
1A1gx1A1g + 1A1gx3A2g + 1A1gx 1E2g
3A2gx1A1g + 3A2gx3A2g + 3A2gx 1E2g =
1E2gx1A1g + 1E2gx3A2g + 1E2gx 1E2g
1A1g + 3A1g + 2A2g + 4A2g + 2E2g
1A1g + 3A1g + 5A1g + 2A2g + 4A2g + 3E2g
1A1g + 1A2g + 1E2g + 2E2g + 3E2g
= 31A1g + 23A1g + 5A1g +1A2g +22A2g +24A2g + 1E2g + 22E2g + 23E2g
The correct answer is
31A1g + 3A1g + 5A1g +1A2g +23A2g + 31E2g + 23E2g
I assume I'm messing up in when doing this direct product: [(e1g)x(e1g)]x[(e2u)x(e2u)].
but I'm not sure exactly what I'm doing wrong. Any help is really appreciated!