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

Question

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.

SuperCiocia · Answer

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.

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

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