# How is total angular momentum conserved in electric dipole transitions for $ΔJ = 0$ in a multi-electron atom?

One of the selections rules is that

ΔJ=0,±1

for an electric dipole transition in a multi-electron atom, where J is the total angular momentum. Since

ΔS=0

(as spin is not affected), the change in total orbital angular momentum is also 0 or ±1. Could someone give me a physical example how a zero change is possible? As far as I can understand, the change is ±1 for a single-electron atom and cannot be 0. The photon emitted carries the ±1 angular momentum away and hence angular momentum is conserved. We only consider 1 electron transitioning at a time, so I don't understand how the overall angular momentum change can be 0 then, if it can't be 0 for a single electron atom. I've been trying to look this up but the furthest I got was "this is just possible because of how angular momentum combines".

Could someone describe a real example to me how this works in practice?

• related to physics.stackexchange.com/q/445901 Jan 5, 2019 at 22:40
• @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 6, 2019 at 9:17
• 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 14:16
• 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 7, 2019 at 15:03
• 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 15:15

A good example when $$\Delta J = 0$$ is a Rubidium D1 line transition:
$$5 ^2 \text{S}_{1/2} \rightarrow 5 ^2 \text{P}_{1/2}$$
at 795 nm. In that case: $$\Delta S = 0$$ and the only available electron changes its angular momentum $$l$$ by 1 (from 0). In that case, you have two available resultant $$J=\{1/2, 3/2\}$$. Both of them relate to different "orientations" of $$l$$ with respect to electron spin $$s$$. By the way, the transition to state $$J=3/2$$ is a well-known D2 transition at 780 nm.
Regarding $$\Delta L = 0$$, I provided an example with an explanation in that post: https://physics.stackexchange.com/a/750497/358408 .