For a may electron atom, a closed subshell structure implies $$L=S=0$$ and therefore also, $J=0.$ Therefore, the ground state wavefunction of such an atom is spherically symmetric because the rotation operator does not alter the state. I understand this mathematically.

But physically, why should this be true? If we consider the Neon, it has the ground state electronic configuration $1s^22s^22p^6$ i.e. the outermost orbitals are $p_x,p_y$ and $p_z$ none of which are spherically symmetric.


Say there are $6$ observers who measure one electron each at the same time. They'll each find a different wavefunction, whose angular part will be $Y_{1,1}, Y_{1,0},$ or $Y_{1,-1}$. They can each compute the probability density of their electron, so $|Y_{1,1}|^2, |Y_{1,0}|^2,$ or $|Y_{1,-1}|^2$.

They then decide to sum their results to get the total probability density of finding an electron: $$ 2|Y_{1,1}|^2 + 2|Y_{1,0}|^2 + 2|Y_{1,-1}|^2 = 1, $$ which is spherically symmetric.

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  • $\begingroup$ Does this explain why the wavefunction is spherically symmetric? I think this only explains the electron density (which could have $e^{im\phi}$ like factors in it still) being spherical. $\endgroup$ – jacob1729 Oct 21 '20 at 9:54
  • $\begingroup$ @jacob1729 See SuperCiocia's answer to the question linked above by Qmechanic. Also see my Wikipedia link, and Unsöld's theorem. $\endgroup$ – PM 2Ring Oct 21 '20 at 13:03
  • $\begingroup$ @PM2Ring no neither of those address my concern. The point is that the state has $J=0$ and so the wavefunction $\psi(\theta,\phi,r)$ is symmetric, not just the mod square. Unsold's theorem only addresses the electron density (which is of course spherical). $\endgroup$ – jacob1729 Oct 21 '20 at 16:55
  • $\begingroup$ In my other answer I argue that with a closed shell each electron has a partner with opposite angular monentum. So it doesn’t matter that the orbitals on their own are not spherically symmetric, each orbital has 2 electron that “rotate the opposite way” and hence cancel each other out. $\endgroup$ – SuperCiocia Oct 21 '20 at 21:48
  • $\begingroup$ If you want to show that the state vector itself has zero angular momentum, you note that for a filled shell $|\Psi\rangle$ Is a product of all $m_\ell$. Applying the ladder operators will inevitable raise or lower one of the external ones which gives you 0 automatically. And the state for which $J_z|\Psi\rangle = J_+|\Psi\rangle = J_-|\Psi\rangle = 0$ is the zero angular momentum state. $\endgroup$ – SuperCiocia Oct 21 '20 at 21:50

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