Why are hydrogenic levels used in writing electronic configuration? I recently started taking a course in Atomic and Molecular Physics. We learned about Hartree Fock approach to solving the many-electron atom problem. If I understand correctly, the electron orbitals that we refer to as 1s, 2s, 2p etc. are eigenstates of the hydrogenic Hamiltonian. When electron-electron repulsion is included these states are no longer the solutions to the problem, and the total wave function is an anti-symmetric wave function in N spatial/spin co-ordinates.
Then why do we use 1s, 2s, 2px etc. when writing the electron configuration of many electron atoms?
I also don't understand how can we use that description to write the spectroscopic term; I learned about LS and JJ coupling where the calculations start with taking the L values of the unpaired electrons in the outermost shell. But how can we justify that these electrons would have that L value. For e.g. How do we know that the outermost electron of sodium ([Ne] 3s1) actually has l=0, if it does not really reside in a hydrogenic evergy level 3s1?
 A: The issue here is that although the atomic single-electron orbitals have the same quantum numbers as the hydrogenic orbitals (n,l,$m_l$,$m_s$), they are not hydrogenic orbitals because they result from the self-consistent (central) HF potential.  This is an approximation, but quite a good one for valence orbitals (the closed shells generate central potentials in HF approximation).  This permits the use of hydrogenic names (1s, 2p, etc), but they are not hydrogenic orbitals.  As with many approximations in physics, the justification is that it works for many cases of interest.
A: You've got two separate questions there. The first question is why we use the hydrogenic orbitals as our basis set in HF-SCF calculations.
We can use absolutely any functions we want in an HF-SCF calculation and the calculation will always give us the best approximation possible using the basis we provide. But the more closely our basis function resemble the atomic orbitals the faster the calculation will converge and the more accurate the final result will be.
The hydrogenic and the atomic orbitals are both solutions for a spherically symmetric potential so their angular part will be spherical harmonics in both cases. The only difference is going to be the radial dependence. So the two sets of functions are going to be fairly similar, and that makes the hydrogenic orbitals an obvious choice for a basis. There isn't any more fundamental reason, it's just that of all the possible functions we could choose the hydrogenic orbitals are an obvious choice and experience tells us that in practice they work well.
The second question is how we know that the outermost electron in sodium has $\ell=0$.
And this is because, as I mentioned above, both the hydrogenic and SCF potentials have spherical symmetry. The angular parts of the wavefunction are dictated by the symmetry so we can be confident that the $3s$ atomic orbital in sodium has zero angular momentum just like the $3s$ orbital in hydrogen.
