How can ΔJ = 0 for an electric dipole transition?

[lawlieto]

Consider a multi-electron atom. (In our course we deal with alkalis mostly so that we have energy levels which are similar to the hydrogenic ones with quantum defect. I don't know if that is relevant here)

Edit: l = orbital angular momentum of a single electron, L = total orbital angular momentum of the electrons, J = L+S with S = total spin

For an electric dipole transition in hydrogen Δl = ±1. Since we only consider one electron jumping at a time in multi electron atoms, for that single electron it's still true that Δl = ±1. However, one of the selection rules for multi electron atoms is that ΔJ=0,±1. Since ΔS=0, ΔL=0,±1. (if L=0 initially, then the 0 change is not possible). How is it possible for total angular momentum change to be 0 though? If only one electron transitions at a time and one photon is emitted, that photon carries away ±1 unit of angular momentum, but if the overall angular momentum change of the atom is 0, how is angular momentum conserved?

For instance, if an electron jumps from 4s->3p there is already +1 change in ΔL. This single electron cannot go 4s->3s so that ΔL=0, since that is forbidden for an electric dipole transition. So for a e.g. 4s->3p transition, do the other electrons rearrange themselves somehow? I was thinking the other electrons must have some role in this, as this 0 change in L is not possible for a single electron atom.

 

[mjc123]

J is not necessarily L + S. J can take values L + S, L + S - 1, ... L - S (as far as those values are positive). Thus ΔJ = 0 does not imply that ΔL = 0.
A single electron is 4s gives a term ²S_1/2. In 3p, it gives ²P_1/2 and ²P_3/2, both of which are accessible from ²S_1/2. The transition ²S_1/2 → ²P_1/2 has ΔS = 0 and ΔJ = 0, but ΔL = 1.

 

[DrClaude]

To complete @mjc123 good answer, the angular momentum of the photon "goes into" a reorientation of the relative orientation of L and S. This is why L = 0 ↔ L = 0 and J = 0 ↔ J = 0 transitions are forbidden: if L = 0 or J = 0, then there is no relative orientation to begin with, so the angular moment of the photon must go into changing L or J.

 

[lawlieto]

J is not necessarily L + S. J can take values L + S, L + S - 1, ... L - S (as far as those values are positive). Thus ΔJ = 0 does not imply that ΔL = 0.
A single electron is 4s gives a term ²S_1/2. In 3p, it gives ²P_1/2 and ²P_3/2, both of which are accessible from ²S_1/2. The transition ²S_1/2 → ²P_1/2 has ΔS = 0 and ΔJ = 0, but ΔL = 1.

That makes a lot of sense! Thank you.

To complete @mjc123 good answer, the angular momentum of the photon "goes into" a reorientation of the relative orientation of L and S. This is why L = 0 ↔ L = 0 and J = 0 ↔ J = 0 transitions are forbidden: if L = 0 or J = 0, then there is no relative orientation to begin with, so the angular moment of the photon must go into changing L or J.

Thanks for the clarification!