Slater determinant as an unperturbed atomic wave function

Question

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.

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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.

Answer 1

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 S2 and L2 commute with the atomic Hamiltonian makes the Slater determinants be the eigenfunctions of the atomic Hamiltonian??

Actually, you go it right, but let us spell this out step-by-step. First, forget about angular momentum operators, since you do not need them to understand that a single Slater determinant is indeed not an eigenfunction of the atomic Hamiltonian (as you called it). That is it, you are right. A Slater determinant is an eigenfunction of the unperturbed Hamiltonian which describes a system of independent electrons, but not of the exact one.

Nevertheless, a Slater determinant can be used as a trial wave function in a variational procedure. That is the whole point of the Hartree-Fock method: a Slater determinant $\Phi$ is not an eigenfunction of the atomic Hamiltonian, but we could evaluate it energy $\left\langle \Phi \mid H \mid \Phi \right\rangle$ using Slater rules, and consequnetly, we could minimize it to find an upper bound to the ground state energy. The resulting Slater determinant obtained by minimizing the energy is indeed only an approximation to the ground state wave function.

Note also that the exact wave function (which is an eigenfunction of atomic Hamiltonian) can be expressed as a linear combination of Slater determinants for the various possible electronic configurations, not just a single Slater determinant. And this is the theoretical basis for configuration interaction (CI) method.

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Now back to all this business with angular momentum operators. A single Slater determinant is not necessarily an eigenfunction of neither $\hat{L}^2$ nor $\hat{S}^2$. However, as you mentioned, the Hamiltonian indeed commutes with all these operators (in non-relativistic approximation), and thus, the exact wave function is also an eigenfunction of $\hat{L}^2$ and $\hat{S}^2$ (as well as $\hat{L}_z$ and $\hat{S}_z$). To be more precise, the exact wave function can be chosen to be a simultaneous eigenfunction of all these commuting operators $\hat{H}, \hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$. And since it is really desirable, we require the same from out trial wave function: we would like it to be an eigenfunction of these angular momentum operators.

So we want to construct our trial wave function being an eigenfunction of angular momentum operators $\hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$, but a single Slater determinant would not necessarily do it, so we form what is called a spin-adapted configuration state function (CSF) - a linear combination of Slater determinants which is an eigenfunction of angular momentum operators.

Note, however, that this is not always possible. For instance, for the case of unrestricted determinants using a linear combination would not help in making the trial wave function an eigenfunction of $\hat{S}^{2}$.