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?

  • $\begingroup$ 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. $\endgroup$ – jezzo Mar 10 '20 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.