# Transitions in diatomic molecules

• I
Hello! I started reading some stuff about molecular spectroscopy and I see that the changes in the total angular momentum, J, can be -1, 0 or +1 (corresponding to the P, Q and R branches). Is this because the photon carries a spin of 1? Also is this the most general case (assuming only EM interactions), or we can have changes greater that 1 in the J values (but probably very weak)? Thank you!

## Answers and Replies

Dr. Courtney
Gold Member
2020 Award
Changes in total angular momentum of greater than one (or zero) require multiphoton events. Multiphoton events require much larger EM fields and usually result from a tuned and focused laser excitation.

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TeethWhitener
Gold Member
In addition, the Q branch (##\Delta J =0##) is forbidden for diatomics with electronic angular momentum of zero for the same reason (angular momentum conservation).

Changes in total angular momentum of greater than one (or zero) require multiphoton events. Multiphoton events require much larger EM fields and usually result from a tuned and focused laser excitation.
Thank you for your reply. So for a not intense EM field, a change of 1 is the best you can get, right? What do you mean by "or zero"? Can't you, in general, easily have a change of 1, without violating any conservation?

Dr. Courtney
Gold Member
2020 Award
Thank you for your reply. So for a not intense EM field, a change of 1 is the best you can get, right? What do you mean by "or zero"? Can't you, in general, easily have a change of 1, without violating any conservation?

Yes, single photon angular momentum changes of 1 are allowed. Sorry for the awkward, ambiguous grammar.

Yes, single photon angular momentum changes of 1 are allowed. Sorry for the awkward, ambiguous grammar.
Thank you! One more question, can we have transition for which ##\Delta l = 0##? Given that the photon has a spin of 1, and we conserve parity in EM interaction, I assume that we can only have ##\Delta l = \pm 1##. Is this true for all orders in an EM interactions (i.e. E1, E2, M1 and higher)? Also I assume we can have ##\Delta m = \pm 1##, but can we have ##\Delta m =0##? Given that the photon has as polarization components just ##\pm 1## but not 0, I assume that one can't have ##\Delta m =0##. Is that true? Thank you!

DrClaude
Mentor
Thank you! One more question, can we have transition for which ##\Delta l = 0##? Given that the photon has a spin of 1, and we conserve parity in EM interaction, I assume that we can only have ##\Delta l = \pm 1##.
Indeed, the selection rule is ##\Delta l = \pm 1##, due to the spin of the photon.

Also I assume we can have ##\Delta m = \pm 1##, but can we have ##\Delta m =0##? Given that the photon has as polarization components just ##\pm 1## but not 0, I assume that one can't have ##\Delta m =0##. Is that true?
No. The quantization axis is not the same as the direction of propagation of the photon, so ##\Delta m =0## is allowed.

Is this true for all orders in an EM interactions (i.e. E1, E2, M1 and higher)? Thank you!
Things are a bit more complicated when you go beyond the dipole approximation. See