# Spin-Adapted Configuration and Slaters determinants for Open- and Closed-shell systems

Some time ago, I asked a question about spin-adapted Configuration State Functions. I read that, when we ignore the electronic repulsion in a atom or a molecule, we can write the ground state using a simple Slater determinant (SD) constructed with spin-orbitals for closed-shell systems.

However, if we have an open-shell system, in general we will need to take linear combinations of SDs. the exceptions are the cases when the magnetic quantum numbers have maximum or minimum values: $$M_{L} = \pm L,\ M_{S} = \pm S$$. I studied the answer I received and I read more about it, and now I think I understand why this happens, but I'm not entirely sure and I would like to know if my explanation is correct:

The wavefunction $$\phi_{0}$$ can be constructed using single one-electron wavefunctions. This wavefunction has to be an eigenfunction of the Hamiltonian operator $$H$$ (without the electronic repulsion). However, when we ignore relativistic effects, the Hamiltonian commutes with the square of the total spin of the system $$S^{2}$$ and its component in the z direction $$S_{z}$$. So, $$\phi_{0}$$ also has to be a eigenfunction of $$S^{2}$$ and $$S_{z}$$. As is shown in the Szabo and Ostlund book, "Modern Quantum Chemistry", any single SD will be a eigenfunction of $$S_{z}$$ with the eigenvalue $$M_{S}$$. If the determinant is a closed-shell determinant, that is, if all spatial orbitals are doubly occupied, the SD will also be a eigenfunction of $$S^{2}$$ with eigenvalue 0 (singlet state). So, for closed-shell systems, a single SD is an eigenfunction of $$H, S^{2}$$ and $$S_{z}$$, and can be used for describe the state of the system.

Second these same authors, a single open-shell determinant is not an eigenfunction of $$S^{2}$$, except when all the single electrons have parallels spins. In this case, the wavefunction can also be taken as a SD. However, I don't understand how this is related to what was written above, that we have to have $$M_{S} = \pm S$$ and $$M_{L} = \pm L$$ in an open-shell case to describe it using only one determinant. If this is not the case, we can construct wavefunctions that are eigenfunctions of $$S^{2}$$ taken linear combinations of SDs.

My last question: It is true for restricted determinants, but what happen if we consider unrestricted determinants?

• I'd be careful about phrasing: "This wavefunction has to be an eigenfunction of the Hamiltonian operator 𝐻 (without the electronic repulsion)". The Hamiltonian of a many-electron system without the electronic repulsion is not the Hamiltonian of the many-electron system. – jezzo Mar 10 at 18:03