I've deduced following postulates from studying my chem books.
1) Slater determinants are eigenfunctions of an unperturbed atomic Hamiltonian, which contains kinetic and central potential energy parts of each electrons only, since spin orbitals constituting the determinants are originated from one-electron Hamiltonian eigenfunctions.
Here, atomic Hamiltonian = (kinetic part of each electrons) + (central potentials between atomic nucleus and electrons) + (interelectronic potentials)
Spin orbit interaction and the other effects are neglected.
2) Slater determinants or their linear combinations are eigenfunctions of total spin angular momentum and total orbital angular momentum operator($S^2$ and $L^2$) simultaneously.
In addition to them, I would refer to information from Quantum Chemistry, 6th Ed., written by I. N. Levine.
(p. 312) Since $S^2$ and $L^2$ commute with the atomic Hamiltonian and with the exchange operator, the zeroth order functions should be eigenfunctions of $S^2$ and $L^2$. (The zeroth order functions Levine mentioned above is indicating the single or linear combinations of Slater determinants in a same configuration)
Well, I can accept the fact that the atomic Hamiltonian, $S^2$ and $L^2$ commute.
However if the single or linear combinations of Slater determinants in a same configuration are the zeroth order functions (eigenfunctions of the unperturbed atomic Hamiltonian), how the fact that $S^2$ and $L^2$ commute with the atomic Hamiltonian makes the Slater determinants be the eigenfunctions of the atomic Hamiltonian??
For example, a Slater determinant corresponding to the one of the Helium-first excited states, $|1s\alpha~2s\beta|$, is an eigenfunction of the unperturbed atomic Hamiltonian(this determinant is a zeroth order function), $S^2$ and $L^2$, but not of the atomic Hamiltonian.