# Physically, why is the ground state wavefunction of Neon be spherically symmetric?

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.
• 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. – jacob1729 Oct 21 '20 at 9:54
• @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). – jacob1729 Oct 21 '20 at 16:55
• 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. – SuperCiocia Oct 21 '20 at 21:50