# Molecular electronic state symmetry

I am trying to work out the ground state symmetry of electronic states. The paper states the GS symmetry is $^6A_1$ for the right hand side electron configuration and $^4B_1$ for the LHS in the D$_{2\mathrm{d}}$ point group. I understand that the symmetry is determined from direct products of the molecular orbital symmetries of the singly occupied orbitals. However I cannot get to the two symmetries listed. For the left electronic configuration it is $b_1 \times a_1 \times b_2 \times e \times e = [a_1+a_2+b_1+b_2]$ so I'm unsure of how to get $A_1$. Likewise the product e x e for the right electronic configuration produces the same results and again I'm unsure of how to arrive at $B_1$.

Each of the terms you generate has a different energy, and may have different spin multiplicity. The spin multiplicity $(2S+1)$ is given by the left superscript (6 and 4) . The quartet (right) it seems has the lowest energy with spin of a triplet (sym e) x doublet ($b_2) \to$ triplet + quartet. To do this add spin vectors triplet $S_T=1/2+1/2$ , doublet $S_D= 1/2$ and make series $S_T+S_D \to |S_T-S_D|$ in unit steps, (Clebsh-Gordon series) this gives total spin 3/2 & 1/2 or quartet($\uparrow\uparrow\uparrow$) + doublet($\uparrow\uparrow\downarrow$). The symmetry has next to be associated with the spin multiplicity and this is complicated. It is necessary to find the character ($\chi$) under, say, quartet multiplicity for each symmetry operation $R$ for each symmetry species $A_1,A_2 ..$ etc. $R^2$ means square the operation. In this case this is
$$\chi(quartet) = (1/6)\left([\chi(R)]^3-3\chi(R)\chi(R^2) +2\chi(R^3)\right)$$
The result has to be reduced into symmetric $\chi^+$ and antisymmetric components $\chi^-$ as $\chi^+=\left( [\chi(R)]^2+\chi(R^2)\right)/2$ and as $\chi^-=\left( [\chi(R)]^2-\chi(R^2)\right)/2$ and the set of characters produced reduced to find the symmetry species.