# Why do closed shells have a spherical symmetric charge density?

When repeating multi-electron atoms, I often read that closed shells have a symmetric charge density. It was justified by the following addition theorem for spherical harmonics everytime:

$\sum\limits_{m=-l}^{l} Y^{*}_{lm}(\theta,\phi)Y_{lm}(\theta,\phi)=\frac{2l+1}{4\pi}$

which is the sum of the squares of the magnitude of the angular part of the wave functions. Since the radial parts aren't functions of the angles, the whole probability density should be sphericaly symmetric.

But is the probability density of the whole multi-electron wave function really just the sum of the magnitude of the filled orbitals? I mean the wave function for two electrons in two different states (1 and 2) are:

$\Psi(r_1,r_2) \propto \psi_1(r_1) \psi_2(r_2) \pm \psi_2(r_1) \psi_1(r_2)$

so $|\Psi|^2$ is not just a sum over the magnitude of the single states.

• Are you asking if the whole multi-electron wave function is actually spherical harmonics? Or are you asking if the spherical harmonics are in fact a basis set? Sep 22, 2016 at 21:52
• I don't understand, that spherical harmonics form a basis, too, but this is another question I should post. I don't see that the angular momentum of one of the electrons commutes with the haniltonian, because of the electron-electron interaction terms. Sep 23, 2016 at 7:10

Ok i think i found a solution. The whole wave function $\Psi$ of $n$ electrons is either symmetric or anti-symmetric, depending on the spin state. The symmetrisation or antisymmetrisation can be done by considering all permutations of the electrons, like the state i found in my question.
To calculate the electron density of the $i$-th electron, $|\Psi|^2$ has to be integrated over the space of all electrons but the $i$-th. Neglecting spin, $|\Psi|^2$ will lead to products of:
$(Y_{l,-l}(j)Y_{l,-l+1}(k)...)(Y^*_{l,-l}(j')Y^*_{l,-l+1}(k')...)$, with particle $j$ in the state $L=l$ and $m=-l$ and so on.
Because of the orthonormality of spherical harmonics only the products with $j=j'$, $k=k'$ and so on, will be nonzero. Then automaticly also $i=i'$. So the $i$-th electron density will be proportional to the sum:
$\sum\limits_{m=-l}^{l} Y^{*}_{lm}(i)Y_{lm}(i)$