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In this video https://www.youtube.com/watch?v=tq_y1qOmUBE&t=783s it's mentioned that the structor of atoms of multiple particles can be approximated using the Schrodinger Equation of the Hydrogen Atom. The number that can fit in each shell is the number of configurations for the corresponding energy level for the hydrogen atom multiplied by 2.

I know that for the case of the multi electron atom $$V_n=-\frac{e^2}{4\pi\varepsilon_0r_n}$$ and $$V_{mn}=\frac{e^2}{4\pi\varepsilon_0r_{mn}}$$

Let's say that instead we have a bound state, for which $$V_n=f(r_n)$$ and $$V_{mn}=-f(r_{mn})$$ with $$V(r)\not\propto-\frac{1}{r}$$

and each particle in this bound state is an identical fermion.

Can we approximate the structor of this system using the Time Independent Schrodinger equation for the case of a single particle with $V=f(r)$?

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The interactions between electrons in a multielectron atom break the spherical symmetry and they make the calculation a many body problem rather than a two body problem like a hydrogen atom. The Schrodinger equation can still be solved in principle, but it becomes a formidable numerical calculation.

The approximation you mention is based on averaging the electron-electron interactions to calculate an average potential that restores the spherical symmetry. That is, the interaction between any two electrons is not spherically symmetric, but since the electrons can be anywhere in the atom we can average out all the instantaneous interactions to produce a spherically symmetric average interaction. Do this for all the electrons and we end up with a mean field potential that is the same for all the electrons. This reduces the calculation to a two body problem and we can (numerically) calculate the eigenstates for the system.

This procedure doesn't depend on the exact form of the pair potentials. It requires only that the mean field is a good approximation. There may be pathological potentials where the pairwise correlations become too strong to be usefully averaged, but for any potential reasonably similar to an inverse square I would expect the approximation to hold.

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