How spin-orbit coupling makes spin-forbidden reactions possible?

Question

My question is related to the Wikipedia page of spin-forbidden reactions https://en.wikipedia.org/wiki/Spin-forbidden_reactions

When a reaction converts a metal from a singlet to triplet state (or vice versa):

1. The energy of the two spin states must be nearly equal, as dictated by temperature,
2. A mechanism is required to change spin states.

Strong spin-orbital coupling can satisfy the 2nd condition.

But, in https://en.wikipedia.org/wiki/Spin%E2%80%93orbit_interaction

It can be shown that the five operators $H_0$, $J^2$, $L^2$, $S^2$, and $J_z$ all commute with each other and with $\Delta H$.

The above quotation would imply under the spin-orbit coupling operator in the spin-orbit interaction Wikipedia page, $S^2$ is a good quantum number, and the expectation value of $S^2$ will not change, though $S_z$ could change.

Though in Sakurai's Advanced Quantum Mechanics, the author mentioned the Dirac equation for the hydrogen atom, the Hamiltonian commutes with $K, J^2$, and $J_z$, no longer commutes with $L^2$, I assume it won't commute with $S^2$. That leads to my question(s):

Does the spin-forbidden reaction Wikipedia page implies another way of introducing spin-orbit interaction, make $[H, S^2]\neq 0$? Or means $S_z$ is not a good quantum number; Or when people talk about this issue, people imply the full Dirac equation or anything possibly equivalent to it somehow (exact 2-component method)? Or anything else?

Answer 1

TL;DR: The statement $[H,S^2]=0$ indeed holds for no-pair (i.e. without considering the production of electron-positron pairs) calculations of hydrogen-like ions, but not for general atoms and molecules. It also fails when the creation of electron-positron pairs is considered.

A hydrogen-like ion has only one electron, so all scalar states of a hydrogen-like ion must be doublet. The (no-pair) spin-orbit interaction, which mixes different scalar states but does not create new electrons, thus invariably leads to pure doublet states, so $H$ commutes with $S^2$ because the latter is a constant operator within the Hilbert space studied.

The situation changes when the full Dirac equation is solved (EDIT: when not only the full Dirac equation is solved, but also you take into account the Dirac sea), where you'll have additional states with e.g. two electrons and one positron, which can potentially constitute a quartet state. These quartet states may then mix with the one-electron doublet states, giving states that are not eigenstates of $S^2$. But of course, these quartet states are ~1 MeV above the ground state, so their mixings into the ground state are negligible, except possibly for ultra high-precision spectroscopies. (EDIT: when you solve the full Dirac equation for a hydrogenic ion but assume that all the negative energy states are empty, you should still get $[H,S^2]=0$, for the same reason as above: the use of the Dirac equation does not change the fact that the electron spin is exactly 1/2, and since there is only one electron, all the Dirac eigenstates must be pure doublets.)

Even more importantly, $[H,S^2]=0$ 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.

Answer 2

Inspired by wzkchem5's answer, I tried to figure out some formulae on my own (I still think Dirac sea $\neq$ QFT, but the latter issue will be beyond the scope of this question)

1. There is one mistake in my question, for the exact solution of the Dirac equation of the hydrogen atom (without any pair production), $[H, S^2]= 0$. Since one may construct $S^2 := \Sigma^2 = \begin{bmatrix} \sigma^2 &0\\0&\sigma^2\end{bmatrix} = 3 \mathbf{I}_{4\times4} $, which would commute with the Hamiltonian. Here $ \Sigma_p:= \begin{bmatrix} \sigma_p &0\\0& \sigma_p \end{bmatrix} $, $p=x,y,z$. The notation of $S^2$ may not be standard here, since it is commonly denoted to 2-component Pauli matrices directly.

2. Concerning many-electron systems, the $\mathbf{S}^2$ may be constructed as $ \left( \sum_i \mathbf{S}_i \right)^2 $ and the spin-orbit term may be written as $ \mathbf{\Sigma}_i \cdot \mathbf{L}_i = \mathbf{S}_i \cdot \mathbf{L}_i$, $i=1,2,\cdots N_e$ (https://arxiv.org/pdf/1306.0479.pdf)

Here is the thing,
$$ \mathbf{S}^2 = \sum_i \mathbf{S}_i^2 + 2 \sum_{i \neq j} \mathbf{S}_i \cdot \mathbf{S}_j \tag{1}$$

The first term at the right-hand side of the above equation will commute with the spin-orbit coupling, not the second. Take a two-electron system for example,

$$ \sum_{pq} 2 [S_1^p S_2^p, S_1^q L_1^q + S_2^q L_2^q]\ \,\,\,\,\,\, p,q, =x,y,z\tag{2} $$ $$ = 2 \sum_{pq} [S_1^p,S_1^q] S_2^p L_1^q + [S_2^p,S_2^q] S_1^p L_2^q \tag{3} $$ $$ = 2i \sum_{pq} \varepsilon_{pqr} S_1^r S_2^p L_1^q + \varepsilon_{pqr} S_2^r S_1^p L_2^q \tag{4} $$ $$ = 2i \sum_{pq} \varepsilon_{pqr} S_1^r S_2^p L_1^q + \varepsilon_{pqr} S_1^p S_2^r L_2^q \tag{5} $$ $$ = 2i \sum_{pq} \varepsilon_{pqr} S_1^r S_2^p L_1^q + \varepsilon_{rqp} S_1^r S_2^p L_2^q \tag{6} $$ $$ = 2i \sum_{pq} \varepsilon_{pqr} S_1^r S_2^p L_1^q - \varepsilon_{pqr} S_1^r S_2^p L_2^q \tag{7} $$ $$ = 2i \sum_{pq} \varepsilon_{pqr} S_1^r S_2^p \left( L_1^q - L_2^q \right)\tag{8} $$ which is not zero in general. Hopefully that's right.