We usually read the case for a hydrogen atom where we find that the potential energy depends only on the variable $r$ and the energy eigenvalues are dependent on principal quantum number only. What will be the dependence of the energy eigenvalues for the case of a multi-electron system?
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2$\begingroup$ As written, this is too general to meaningfully answer. Multi-electron systems includes Helium atoms, condensed matter systems, plasmas, etc. Which is to say, most of the universe. Do you mean specifically the case of multi-electron atoms? $\endgroup$– By SymmetryCommented Sep 1, 2021 at 13:51
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This depends on the level of approximation that you are considering. For example, in the hydrogen atom, if you consider relativistic approximation, you will find that the energy eigenvalues depend on both $n$ and $l$.
The best way to describe this for a multi-electron atom is through the following :
If you solve the base Schroedinger equation with no approximations ( corrections ), you'll find that the energy eigenvalues depend only on $n$. However, you then remember that the different electrons would interact with each other. Hence, you consider the electron-electron interaction. If you go a little further, you shall see that electrostatic repulsion between the electrons also plays a role in determining the energy eigenvalues. Finally, you also need to consider spin-orbit interaction which can be done in $2$ ways - $LS$ coupling or $JJ$ coupling. This would give rise to the fine structure that we see in the spectrum of different elements. This is usually a good place to stop.
If you proceed through $LS$ coupling, the energies would depend on $n, L,S$, where $L,S$ is the Total orbital/spin angular momentum. See Vector model of the atom to see, how to add angular momentum.
Similarly, if you take the $JJ$ coupling route, the energies would depend on $n, L,S,J$, where $J$ is the total angular momentum.
If you include Zeeman effect or hyperfine splitting, the energies would also depend on $m_j$.
So, you can see, energy eigenvalues depend on many quantum numbers, and not just the principal quantum number. It depends on how many corrections you consider into your calculations. Remember, more corrections is equivalent to a more accurate result, but also a more complicated expression.
answered Sep 1, 2021 at 13:56
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1$\begingroup$ It always amazes me (when I think about it) that the hydrogen-atom-model still pretty much works with 100 electrons around. $\endgroup$ Commented Sep 1, 2021 at 14:04
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$\begingroup$ @JonCuster yeah, all it needs is a couple of correction factors, and that's all. Apart from just studying it, or using it, it feels good thinking how fascinating the entire thing is. $\endgroup$ Commented Sep 1, 2021 at 21:47