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.

  • $\begingroup$ Kindly check this link You will see that the determinant can be factored out as spatial and spin components. $\endgroup$ Aug 9, 2023 at 10:32
  • $\begingroup$ 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. $\endgroup$
    – jxzou
    Aug 11, 2023 at 2:06
  • $\begingroup$ @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. $\endgroup$
    – Al1010
    Aug 11, 2023 at 9:59
  • $\begingroup$ 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. $\endgroup$ Dec 19, 2023 at 20:19
  • $\begingroup$ @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. $\endgroup$
    – Al1010
    Dec 21, 2023 at 10:11


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