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I expect that in many-body problems of electrons, spin should cause same-spin-electrons to repel more strongly than opposite spin electrons because the Pauli exclusion principle is the observation that a slater determinant with repeated columns vanish identically. HF often capture the physics of many-body problems, and therefore we expect that there should be some repulsion due to spin in theories that HF works well. I am trying to pinpoint how this repulsion (1) enters mathematically in the exact solution (as it does not appear in the Hamiltonian) and (2) how the repulsion is expressed in $|\psi|^2$. The hamiltonian of an atom is an example that I have been using to understand this. The simplest is helium:

$$ H = -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{r_{12}}. $$ As it is a spin-independent theory, any solution of the time-independent Schrödinger equation cannot depend on spin through the Schrödinger equation. I expect that any dependence of spin must enter by additional restrictions of $\psi$. Wikipedia claims in the third paragraph of this section that if $i,j = \uparrow,\downarrow$, a solution is a matrix of functions $\psi_{ij} : \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{C}$.

This raises a couple of questions:

  1. Does Wikipedia demand anything more than that the spatial part be antisymmetric? I suppose so, otherwise there would be no reason to label the different spin configurations individually. Wikipedia claims that the $\psi_{ij}$ are restricted by the pauli exclusion principle. How could that possibly provide a condition $\psi_{ij}$ here?

  2. How can I see that these restrictions show up as a repulsion in $|\psi_{ii}|^2$?

It would be helpful if you could be as mathematically precise as possible.

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  • $\begingroup$ For the interaction between magnetic moments you have a Hamiltonian proportional to $\vec\mu_1\dot\vec\mu_2=\tfrac{1}{2}(\vec\mu_1+\vec\mu_2)^2-\tfrac{1}{2}(\vec\mu_1^2+\vec\mu_2^2)$. $\endgroup$
    – Daniel
    Commented Dec 16, 2019 at 3:24
  • $\begingroup$ Pauli's exclusion principle means that fermionic states do not appear in identical copies (as bosonic states do), and is not related with electrodynamics; is much more universal. $\endgroup$
    – Daniel
    Commented Dec 16, 2019 at 3:35

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Some information about the physical system is written in the Hamiltonian, and some information is enforced in the types of solutions that you allow. For instance, in the case of SCF and other wavefunction methods, only wavefunctions that meet the Pauli exclusion principle are allowed to be placed in the expansion. Trying to put more than one electron in a specific spin-orbit is not allowed in the trial wavefunctions.

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The total wavefunction for a system of identical fermions must always be antisymmetric under particle exchange. For the particular case of two particles, there is the nice feature that the wavefunction can be factored into two parts, $\Psi(r_1, \sigma_1; r_2, \sigma_2) = \chi_{\sigma_1 \sigma_2} \psi(r_1, r_2)$. Since $\Psi$ must be antisymmetric there are two possibilities: $\psi$ is antisymmetric under $r_1 \leftrightarrow r_2$ (and $\chi$ is symmetric), or $\psi$ is symmetric (and it is $\chi$ that is antisymmetric). When the spatial wavefunction $\psi$ is symmetric it is sometimes called the bonding orbital, whereas the antisymmetric case is called the antibonding orbital.

If $\psi(r_1,r_2)$ is antisymmetric, it must necessarily be zero at $r_1 = r_2$. This is a manifestation of the Pauli exclusion principle. As a consequence the fermions are more tightly confined, and so have a higher kinetic energy, and this state thus has a higher energy than the symmetric solution. Note that this case corresponds to the fermions having the same spin, as $\chi_{\sigma_1,\sigma_2}$ will be symmetric. So a state with same-spin electrons will (in general) have a higher energy than a state with mixed-spin electrons.

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  • $\begingroup$ Thanks. What is the form of your $\chi_{\sigma_1\sigma_2}$? $\endgroup$
    – Mikkel Rev
    Commented Dec 16, 2019 at 17:35
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    $\begingroup$ There are 4 possibilities for $\chi_{\sigma_1 \sigma_2}$, namely: $\chi_{\uparrow \uparrow}$, $(\chi_{\uparrow \downarrow} + \chi_{\downarrow \uparrow})/\sqrt{2}$ and $\chi_{\downarrow \downarrow}$ which are the triplet states (they are all symmetric under exchange of spin indices). The remaining state is the singlet state $(\chi_{\uparrow \downarrow} - \chi_{\downarrow \uparrow})/\sqrt{2}$, which is symmetric. $\endgroup$
    – Clara Díaz Sanchez
    Commented Dec 16, 2019 at 18:02
  • $\begingroup$ Thank you. So when I quoted Wikipedia's $\psi_{\uparrow \downarrow}$ and $\psi_{\downarrow \uparrow}$, how would I translate this notation into the one that has been used in the present answer? And how is it possible to work out which of $\psi_{\uparrow \downarrow}$ and $\psi_{\downarrow \uparrow}$ are symmetric and antisymmetric, such that I can apply the information given in the present answer? $\endgroup$
    – Mikkel Rev
    Commented Dec 16, 2019 at 19:32
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    $\begingroup$ Looking at the Wikipedia article, I think that the second order spinor with 4 components $\psi_{ij}(r_1,r_2)$ (where $i$ and $j$ are $\uparrow$ or $\downarrow$) is simply my $\Psi(r_1,\sigma_1;r_2,\sigma_2)$. The $\chi_{ij}$ I gave in my previous reply are what the article calls $\sigma^i_k$. $\endgroup$
    – Clara Díaz Sanchez
    Commented Dec 17, 2019 at 0:33
  • $\begingroup$ Thank you. The question is then -- Wikipedia indexes using two indices for the $\phi$-orbitals: $\phi_i^k$. It seems almost certain that there is a particular symmetric solution associated with each spin of the triplet spins, otherwise $\phi_i^k$ would commute with $\sum_i$. That means the wikipedia authors believe there is an additional criterion for each spin projection. And what is that criterion? $\endgroup$
    – Mikkel Rev
    Commented Dec 17, 2019 at 15:27

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