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## Homework Statement

What terms are generated by the configuration (e

_{1g})

^{2}(e

_{2u})

^{2}in D

_{6h}symmetry?

## Homework Equations

configuration (χ

_{d})

^{2}gives terms

^{1}(symmetric product) +

^{3}(antisymmetric product) where χ=symmetry of an orbital and d = degenerate

(e

_{1g})x(e

_{1g}) = A

_{1g}+ [A

_{2g}] + E

_{2g}

(e

_{2u})x(e

_{2u}) = A

_{1g}+ [A

_{2g}] + E

_{2g}

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

(e

_{1g})

^{2}x(e

_{1g})

^{2}=

^{1}A

_{1g}+

^{3}A

_{2g}+

^{1}E

_{2g}

(e

_{2u})

^{2}x(e

_{2u})

^{2}=

^{1}A

_{1g}+

^{3}A

_{2g}+

^{1}E

_{2g}

Then find direct product of [(e

_{1g})x(e

_{1g})]x[(e

_{2u})x(e

_{2u})]

^{1}A

_{1g}

^{1}A

_{1g}

^{3}A

_{2g}x

^{3}A

_{2g}=

^{1}E

_{2g}

^{1}E

_{2g}

^{1}A

_{1g}x

^{1}A

_{1g}+

^{1}A

_{1g}x

^{3}A

_{2g}+

^{1}A

_{1g}x

^{1}E

_{2g}

^{3}A

_{2g}x

^{1}A

_{1g}+

^{3}A

_{2g}x

^{3}A

_{2g}+

^{3}A

_{2g}x

^{1}E

_{2g}=

^{1}E

_{2g}x

^{1}A

_{1g}+

^{1}E

_{2g}x

^{3}A

_{2g}+

^{1}E

_{2g}x

^{1}E

_{2g}

^{1}A

_{1g}+

^{3}A

_{1g}+

^{2}A

_{2g}+

^{4}A

_{2g}+

^{2}E

_{2g}

^{1}A

_{1g}+

^{3}A

_{1g}+

^{5}A

_{1g}+

^{2}A

_{2g}+

^{4}A

_{2g}+

^{3}E

_{2g}

^{1}A

_{1g}+

^{1}A

_{2g}+

^{1}E

_{2g}+

^{2}E

_{2g}+

^{3}E

_{2g}

= 3

^{1}A

_{1g}+ 2

^{3}A

_{1g}+

^{5}A

_{1g}+

^{1}A

_{2g}+2

^{2}A

_{2g}+2

^{4}A

_{2g}+

^{1}E

_{2g}+ 2

^{2}E

_{2g}+ 2

^{3}E

_{2g}

The correct answer is

3

^{1}A

_{1g}+

^{3}A

_{1g}+

^{5}A

_{1g}+

^{1}A

_{2g}+2

^{3}A

_{2g}+ 3

^{1}E

_{2g}+ 2

^{3}E

_{2g}

I assume I'm messing up in when doing this direct product: [(e

_{1g})x(e

_{1g})]x[(e

_{2u})x(e

_{2u})].

but I'm not sure exactly what I'm doing wrong. Any help is really appreciated!