Thanks a lot! One question, if I use the full Dirac equation for the hydrogen atom, is there any additional state, as two electrons and one positron? The exact solution is known (under the fixed nuclear approximation), e.g., en.wikipedia.org/wiki/…, I am stupid enough that did not to find this kind of state (2e- 1e+). — Bottisham Oct 5 at 19:33

@Bottisham This seems to be the exact solution under the assumption that there is no Dirac sea, i.e. the infinite number of negative energy states are not filled by electrons. In this case there is no electron-position pair to talk about, but also the wavefunction does not exactly reflect reality, which is mentioned by the wiki page you linked (i.e. the micro-Hartree errors). In reality the Dirac sea is filled, which gives rise to states with electron-positron pairs. AFAIK such electron-positron pairs have precluded an exact Dirac equation solution in the presence of the Dirac sea. — wzkchem5 Oct 5 at 19:50

+1. @Bottisham if you found the answer helpful then I also recommend upvoting it! This whole platform works based on voting :) — Nike Dattani Oct 5 at 20:02

@NikeDattani Thank you for reminding me. I need to ask a few additional questions. — Bottisham Oct 5 at 20:11

#1. In wikipedia page has a section "Negative-energy solutions", seems to be the "Dirac sea" (please let me know if it's not); #2 "In reality the Dirac sea is filled", I think the sea picture has been replaced by QFT en.wikipedia.org/wiki/Dirac_sea "Dirac sea theory has been displaced by quantum field theory, though they are mathematically compatible. " #3 is "when the full Dirac equation is solved "related to "rise to states with electron-positron pairs"? I think the energy contribution of electron-position pair is beyond the Dirac equation, since all solutions in the wikipedia page — Bottisham Oct 5 at 20:19

Would you provide a derivation (possibly in reference) for $[H,S^2] \neq 0$ for many-electron systems, under the no-pair approximation? Whether it comes from the Breit or Coulomb interaction? (sorry that my previous comment may have ignored you editing, and I cannot edit it now) — Bottisham Oct 5 at 20:24

@Bottisham If you're going to ask so many further questions, it would be better if you post them on the main site so that everyone can see them. Comments are considered "temporary" and are not to be used for further questions. In fact your question got "flagged" for having too many follow-up questions! — Nike Dattani Oct 6 at 3:13

@NikeDattani Thanks for reminding me. If this is the convention/rule of this site, I can post them to another question(s). Though I initially expect to have discussions within one question, perhaps it should be the other way around. — Bottisham Oct 6 at 4:11

@Bottisham #1: Yes, but I think the negative-energy solutions can only be called a Dirac sea when they are filled. If this is wrong then I'll rephrase my answer; in the answer I mean a completely filled Dirac sea. #2: Yes, so it may be slightly more rigorous to say "in reality, the QFT is operative" instead of "the Dirac sea is filled", but obviously this does not change the main idea. #3: Maybe - one may as well argue that if you don't consider the Dirac sea as filled, you are not solving the Dirac equation in the right way. Anyway this point has been made clear with my last edit. — wzkchem5 Oct 6 at 8:49

@Bottisham My answer already contains a derivation of that fact. The mere fact that $H$ mixes different $S^2$ eigenstates suffices to prove that $[H,S^2]\neq 0$. If you prefer formulas over text, I may translate this into formulas, i.e. explicitly write out why $H$ mixes different $S^2$ eigenstates and what the matrix elements look like. — wzkchem5 Oct 6 at 8:52

@Bottisham As Nike suggested, the site is structured in a way where any given post should focus on just one question. Some basic clarifying questions can be done in comments, but longer discussions are generally reserved for chat. Having questions in separate posts makes it easier for future users to find these questions and to evaluate how well an answer addresses that question. When questions and answers are continually updated/expanded, it makes it harder to evaluate the quality as a whole. — Tyberius ♦ Oct 6 at 14:10

As somehow related to the moderator's comment in my own answer, would you possibly further explain how "construct scalar states of different multiplicities" work, to let $S^2$ not commute with $H$? Possibly in formulas. Would you please also explain the "scalar state"? It seems to be a special terminology, cannot find much on google combining relativistic quantum, etc — Bottisham yesterday

@Bottisham "construct scalar states of different multiplicities" was written in the paragraph about multi-electron systems. When there's more than one electron, you can make the electrons paired (leading to a spin multiplicity of 1) or you can have them unpaired (leading to different multiplicities). This is how you create states with different multiplicites. If you have only one electron you usually can't do that: it's basically always a doublet. — Nike Dattani 12 mins ago

@NikeDattani I can have paired and unpaired electrons, then, so what? For a non-relativistic Hamiltonian, in a two-electron system, I can have singlet and triplet states. It does not imply making spin-forbidden reaction happen. — Bottisham 8 mins ago

As for your question about the "scalar state" terminology, I thought it was possible to infer what was meant based on what's in the answer already, but if wzkchem5 wants to clarify that in their answer it would be okay. We just don't want too many follow-up questions in the comment section, for many reasons. One of them is because comments are temporary so they can be deleted at any time with no notice, and you can't get them back. This would mean that all the time that wzkchem5 spent (including careful typesetting in MathJax) would... — Nike Dattani 8 mins ago

...be in some sense wasted. Recently a candidate in a Stack Exchange election said that they would not tolerate comment threads that involve more than 10 comments. As for your question to me one minute ago, I would highly recommend to ask it in a new question on the main site. I just made an attempt at answering the question in your comment about constructing states of different multiplicities; and now you're asking a completely different question about what that has to do with making spin-forbidden reactions happen. — Nike Dattani 3 mins ago

I can't open a chat room. Then, please explain scalar state in the answer. I may create another question. But, the link said "mercilessly sending comments to chat once there's more than 10 comments under a given post. ", I am fine with that. Since, I don't have enough privilege to create a chat room. — Bottisham 3 mins ago