Suppose we have a diatomic molecule in a $^1\Sigma$ state, and it transitions to an excited $^1\Pi$ state. Note that the total spin of the electrons remains unchanged ($\Delta S=0$) as we can assume spin-orbit coupling is negligible and hence the dipole operator involved in the transition cannot act on the spin. The electrons have gained some orbital angular momentum ($\Delta\Lambda=1$) which is fine, this comes from the incoming photon, which has 1 unit of angular momentum. Then if we look at the rotational fine structure, we see that a $\Delta J=0$ transition is allowed, since we've already changed angular momentum elsewhere. A "Q-branch" in my fine structure is allowed, angular momentum is conserved, everyone is happy.

My question is what about the "P" and "R" branches, which are where $\Delta J=\pm1$? These are usually introduced in spectroscopy courses first, before the Q-branch, because you consider a simple transition from one rotational state to another, ignoring possible electronic changes. You argue that because angular momentum must be conserved, when the molecule absorbed the photon it has to take its angular momentum, hence $\Delta J=\pm1$.

In my reading, I've noticed when people talk about electronic spectroscopy, where the orbital angular momentum of electrons could change (e.g. $\Sigma\rightarrow\Pi$), they say that one might expect P, Q and R (i.e. $\Delta J=0$ and $\pm1$) fine structure. I just don't see how the P and R branches are allowed though, if we've already used up the angular momentum to change the orbital.

  • $\begingroup$ It is essentially because in a $\Pi$ state there are two angular momenta, end over end and along the internuclear axis (in a diatomic) and it is involved. Best to look for a full explanation as in book by M.J. Hollas Chapter 7, p257 4th edition of 'Modern Spectroscopy' publ. Wiley $\endgroup$
    – porphyrin
    Commented Apr 21, 2022 at 17:59


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