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Jul 29, 2023 at 3:35 history edited Qmechanic CC BY-SA 4.0
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Jan 15, 2020 at 17:00 comment added Lautron @ZeroTheHero So, using your example, the electron transition from state l=1 ml=1 to state l=1 ml=0 would mean that a photon with an angular momentum -1 (spin down) was absorbed. In this case we make a vector sum of the electron orbital angular momentum +1 (ml =1) with the photon angular momentum -1 (spin down) and both vectors cancel each other on the z direction, and in the end of the day we have a l=1 ml=0 electron, with no angular momentum on the z direction, but still with the same magnitude of angular momentum (because the transition didn't change the quantum number l). Am I right?
Jan 8, 2019 at 14:25 comment added lawliet But 3p->2p is forbidden for electric dipole transitions, Δ_lowercase_L= 0 is not allowed (change in angular momentum of a single electron), yet ΔL=0 (total orbital angular momentum of a multielectron system) is still somehow possible for an electric dipole transition. It puzzles me because we only consider one electron jump at a time for such transitions. I think you're misunderstanding what I'm asking. (and change in spin is also 0, adding to its weirdness)
Jan 7, 2019 at 15:15 comment added ZeroTheHero Well of course for $3s\to 2p$ $\Delta L$ is NOT zero since $p$ states have $L=1$ and while $s$ states have $L=0$. You need to look at some transitions of the $3p\to 2p$ type or something like this. See pdfs.semanticscholar.org/7abe/… for a nice discussion of applicability of vector model.
Jan 7, 2019 at 15:03 comment added lawliet Is your lowercase L the total orbital angular momentum? By real example, I mean that how do the electrons rearrange themselves in a multi-electron atom so that ΔL=0? The vector model is about arrows only, I wanted something like an electron dropping from 3s->2p level and in that case how does ΔL end up being zero. (e.g. what rearrangements happen with other electrons?)
Jan 6, 2019 at 14:16 comment added ZeroTheHero what do you mean by “real example”? The identification of an atom and an energy level? why won’t $\vert \ell=1,m=1\rangle \to \vert \ell=1,m=0\rangle$ do? Remember that angular momentum is a vector so it’s perfectly possible to vectorially add the angular momentum of the photon to that of the state to get a state with final value of angular moment: add to vectors with relative angle $60^o$ to form an equilateral triangle.
Jan 6, 2019 at 9:17 comment added lawliet @zerothehero thanks, but the sources mentioned there still only talk about rules and how it's possible, there is no mention of a real example unfortunately.
Jan 5, 2019 at 22:40 comment added ZeroTheHero related to physics.stackexchange.com/q/445901
Jan 5, 2019 at 21:35 review First posts
Jan 5, 2019 at 22:40
Jan 5, 2019 at 21:33 history asked lawliet CC BY-SA 4.0