# Electronic Spec - Terms generated by config of (e1g)2(e2u)2

## 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

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!

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DrDu
You messed some things up: This is probably a typo:
"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" and should read
"So first I found the terms generated by e1gxe1g and e2uxe2u seperately
(e1g)x(e1g) = 1A1g + 3A2g + 1E2g
(e2u)x(e2u) = 1A1g + 3A2g + 1E2g"

Then a singlet times a triplet does not give a quartett:
1A1gx3A2g is not 4A2g but 3A2g.
Similarly 1A1gx 1E2g is not 2E2g but 1E2g. There are more of this kind.
It would be helpful if you could write down for each term in the sum how it is reduced out.
E.g. 1A1gx1A1g=1A1g, ...