# Factorization of a closed shell Slater determinant into spatial and spin part

I'm studying the HF method, so I have a non-relativistic time-independent Hamiltonian.

Since in the Hamiltonian, there is no spin, we have:

$$\left[\hat{H}, \hat{S}_z \right] = 0 \quad \text{and} \quad \left[\hat{H}, \hat{S}^2 \right] = 0$$

This means that it should be possible to factorize the total wave function as:

$$\Psi(1,2, \dots)=\underbrace{\Phi(1,2,\dots)}_{\text{Spatial part}}\underbrace{\chi(1,2,\dots)}_{\text{Spin part}}$$

From Szabo, Modern Quantum Chemistry, chapter 2.5.2, I read that closed shell Slater determinants are eigenfunctions of $$\hat{S}_z$$ and $$\hat{S}^2$$. So a closed shell Slater determinant should be factorizable into spatial and spin parts as above. Is this correct? I know the question sounds obvious, but I want to make sure I understood correctly.

• Kindly check this link You will see that the determinant can be factored out as spatial and spin components. Aug 9, 2023 at 10:32
• I remember that such factorization can be done only for one- and two-electron systems. It is generally not possible for a system with >2 electrons. Aug 11, 2023 at 2:06
• @jxzou I'm now studying "Spin Eigenfunction, construction and use" by Ruben Pauncz. I will understand the topic in detail and then maybe leave the answer here. I want to clarify this because I think it's of central relevance. In fact, I think this is strictly related to spin contamination. Aug 11, 2023 at 9:59
• I gave my +1 long ago, but are you able write an answer now? If so, please do not do it in a comment like you did last time. Dec 19, 2023 at 20:19
• @NikeDattani Unfortunately to this, I'm not able to answer yet :(. I'm very sorry ... This is more complicated than expected. I can tell you that you have to look for Configuration State Functions (CSFs). But I still need to study the topic in detail. Dec 21, 2023 at 10:11