0
$\begingroup$

This question is related to How spin-orbit coupling makes spin-forbidden reactions possible? - Matter Modeling Stack Exchange

Even more importantly, [H,S2]=0 fails for multi-electron systems even under the no-pair approximation, and in this case the violation should be orders of magnitudes larger than the one found in the hydrogen atom. This is because we can now construct scalar states of different multiplicities even without creating electron-positron pairs. Now the scalar states are at best only a few eVs (frequently just a fraction of an eV) apart, so their mixings are sufficiently large to have observable consequences even without the aid of high-precision spectrometry, like spin-forbidden reactions, intersystem crossing and phosphorescence.

I would like to understand how "construct scalar states of different multiplicities" would let spin-forbidden reactions happen.

From a comment from Nike Dattani

you can make the electrons paired (leading to a spin multiplicity of 1) or you can have them unpaired (leading to different multiplicities).

if I use two electrons to form singlet and triplet (if they are "scalar states"), for a non-relativistic Hamiltonian, they still commute with the Hamiltonian. Where is the role of letting spin-forbidden reactions happen?

Does it imply one starts from some scalar relativistic Hamiltonian, obtains eigenstate $|I\rangle$ and $|J\rangle$ with different multiplicities as "scalar states", adds spin-orbit coupling into the Hamiltonian, and sees the $$\langle I | \hat{H}_{\mathrm{SO}} | J \rangle \neq 0 \tag{1} $$, then why different multiplicity will let (1) non vanish? Eq. (12) in "Combining spin-adapted open-shell TD-DFT with spin–orbit coupling"

$\endgroup$
1
  • $\begingroup$ For any users curious about why this question has a net score of 0: I was curious myself, so I had to do a bit of digging. It turns out that the OP Bottisham may have annoyed several users here, which may have had a negative effect on people's desire to upvote. $\endgroup$ Jan 26, 2023 at 17:38

1 Answer 1

3
$\begingroup$

No, the presence of scalar states with different multiplicities is only a necessary condition for spin-forbidden reaction to take place (because otherwise there are no spin-forbidden reactions), but not a sufficient condition. Indeed, one may easily come up with a two-orbital two-electron system with just 3 singlet states and 1 triplet state, and set all the SOC matrix elements to zero; now you have a system with scalar states with different multiplicities, but no spin-forbidden reactivity at all.

However, in reality, any system with at least one electron has an infinite number of eigenstates (and if you consider pair production effects, then even the vacuum has an infinite number of eigenstates). It then becomes exceedingly unlikely that all of the infinite number of possible SOC matrix elements are zero. And as long as there are scalar states with different multiplicities, an infinite number of SOC matrix elements will couple them, so that if even one of the matrix elements are zero, spin-forbidden reactions will be possible. Mathematically proving that spin-forbidden reactions must be always (or almost always) possible when there is more than one electron, is a different matter and does not seem as trivial as I thought, but I strongly believe that this is possible.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .