As an alternative “way” to obteined the electronic state of a molecule, the direct product of the irreducible representations of the single occupied molecular orbitals has to be performed. In this escenario, when two MO's are degenerate they are assigned a single irreducible representation for both orbitals, but for the performance of the direct product, this irreducible representation needs to be multiplied twice (once for each orbital), why is that? and why only the irreducible representations of the single occupied molecular orbital participate in the direct product? Thanks for your time
Hi Elizabeth! In principle, you obtain the irreducible representation of your system by multiplying the representations of all electrons - not just the unpaired ones. But you can simplify that calculation because the products for fully occupied orbitals will just give you total symmetry. (It's actually quite easy to show that in non-degenerate point groups). For degenerate orbitals, things get a little more complicated, but the principle is the same: full occupation = total symmetry; anything below full occupation = some other than total symmetry.
So if you see an irreducible representation being calculated for any system, all fully occupied orbitals are typically neglected because they just give you total symmetry anyway. Only the orbitals with partial occupation will add something other than total symmetry to the final product.
Hi Elizabeth! In principle, you obtain the irreducible representation of your system by multiplying the representations of all electrons - not just the unpaired ones. But you can simplify that calculation because the products for fully occupied orbitals will just give you total symmetry. (It's actually quite easy to show that in non-degenerate point groups). For degenerate orbitals, things get a little more complicated, but the principle is the same: full occupation = total symmetry; anything below full occupation = some other than total symmetry.
So if you see an irreducible representation being calculated for any system, all fully occupied orbitals are typically neglected because they just give you total symmetry anyway. Only the orbitals with partial occupation will add something other than total symmetry to the final product.
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