Let's have an atomic carbon with the following electron configuration:

$$ 1s^2 2s^2 2p^2 $$

One of it's levels is ${}^1S_0$, which is corresponding with the following state:

$$ \begin{align} \left| S=0, M_S=0 \right> &= \frac{1}{\sqrt{2}}\Big( \alpha(1)\beta(2) - \alpha(2)\beta(1) \Big)\\ \left| L=0, M_L=0 \right> &= \frac{1}{\sqrt{3}}\Big( \left| p^+(1)p^-(2) \right> - \left| p^0(1)p^0(2) \right> + \left| p^-(1)p^+(2) \right> \Big)\\ \left| S=0, M_S=0, L=0, M_L=0 \right> &= \frac{1}{\sqrt{6}} \Big( \left| p^+(1)\alpha(1)p^-(2)\beta(2) \right>- \left| p^+(1)\beta(1)p^-(2)\alpha(2) \right> \\&- \left| p^0(1)\alpha(1)p^0(2)\beta(2) \right> + \left| p^0(1)\beta(1)p^0(2)\alpha(2) \right> \\&+ \left| p^-(1)\alpha(1)p^+(2)\beta(2) \right> - \left| p^-(1)\beta(1)p^+(2)\alpha(2) \right> \Big) \end{align} $$

Now I know, that ${}^1S_0$ state is supposed to be totally symmetric, i.e. $A_g$ in the $D_{2h}$ point group. But is it possible to derive it "rigorously" from the wavefunction?

  • 1
    $\begingroup$ Too late for an answer right now, but the idea is you need to perform all the symmetry operations of the D2h point group on the wavefunction. For example $\sigma_{xy}$ reflects points (x,y,z) into new points (x,y,-z), so if you replace all instances of z in the wf by -z, and you find that the wf remains unchanged, that corresponds to a character of +1. If the sign of the wf is flipped, that’s a -1. You can ignore the spin part because symmetry operations don’t act on them. $\endgroup$ Commented Jan 15, 2019 at 1:49
  • $\begingroup$ Only the spatial part is important here so why not use the symmetry of the spherical harmonics with the appropriate L and ml values ? The right-hand most columns in most point group tables give these functions. $\endgroup$
    – porphyrin
    Commented Jan 15, 2019 at 9:28
  • $\begingroup$ @porphyrin I understand, what you're saying, but I don't see the connection somehow... Should I "redraw" every microstate wavefunction as p-orbitals? $\endgroup$
    – Eenoku
    Commented Jan 15, 2019 at 13:57
  • $\begingroup$ No I don't think so: if, for example, L = 0 there is no orbital angular momentum and the wavefunction must be totally symmetric, so $A_g$. $\endgroup$
    – porphyrin
    Commented Jan 15, 2019 at 14:59


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