Why is the change in total spin ∆S = 0 during an absorption of a photon, in contrast to the ∆L = ±1, where the photon transfers its angular momentum?

I am aware that spin-orbit coupling breaks this, I want to know why its a rule in the first place.

  • $\begingroup$ Photons have one unit of angular momentum only ($S= 1$), and on absorption or emission the angular momentum of the atom or molecule+photon is always conserved. For example singlet to triplet transfers occur but only if angular momentum from elsewhere (other than spin) is used, as in your example of spin-orbit coupling. $\endgroup$
    – porphyrin
    Mar 9 at 9:17
  • $\begingroup$ Thanks @porphyrin Do photons also have orbital angular momentum %l = 1%? If they do, why does transferring this to the molecule not change their orbital angular momentum to 0? If they don't, how can absorption of a photon cause a change the orbital angular momentum of the molecule? $\endgroup$ Mar 9 at 11:53
  • $\begingroup$ yes photons have angular momentum, quantum numbers $S=1, m_s=\pm 1$ and yes it changes the angular momentum of the molecule so that the total in the molecule plus photon is the same as in the excited molecule after the photon is destroyed by being absorbed. The angular momentum of the atom/molecule is changed by $\pm 1$ unit from whatever it was before. You see this clearly in a vibrational-rotational spectrum such as from HCl, where only $J$ changes by $\pm 1$ are observed and $J'=0 \to J''=0$ is not observed. $\endgroup$
    – porphyrin
    Mar 9 at 14:41
  • $\begingroup$ Sorry, I'm a little confused. You're saying photons have $S = 1$ (spin angular momentum), the change in spin orbital angular momentum is usually 0 during the absorption of a photon. Do they have orbital $L = 1$ as well as $S$ in order to get $\Delta L = 0$? If they can freely re-allocate angular momentum between $L$ and $S$ to get the general change in $J$, then why is $\Delta S$ required to be 0? $\endgroup$ May 6 at 2:12
  • $\begingroup$ I'm not sure I understand your question, but it is quite simple in outline. Photons have 1 unit of angular momentum, however, the electron spin is not affected by this but in a dipole transition the atom/molecule orbital angular momentum has therefore to change by 1 unit, i.e. orbitals $s \to p,\; p \to d$ etc for a dipole transition. Similarly in molecules but now you need to look at the point group to determine which symmetries are involved. $\endgroup$
    – porphyrin
    May 6 at 14:45


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