The discussion centers on transitions in diatomic molecules as analyzed through molecular spectroscopy. It confirms that changes in total angular momentum (J) can only be -1, 0, or +1, corresponding to the P, Q, and R branches, respectively, due to the single photon carrying a spin of 1. Multiphoton events, which require larger electromagnetic (EM) fields, are necessary for changes greater than one. The selection rules dictate that transitions must adhere to the conditions of Δl = ±1 and allow for Δm = 0, despite the photon’s polarization components.
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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 said: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.
kelly0303 said: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?
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!Dr. Courtney said:Yes, single photon angular momentum changes of 1 are allowed. Sorry for the awkward, ambiguous grammar.
Indeed, the selection rule is ##\Delta l = \pm 1##, due to the spin of the photon.kelly0303 said: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##.
No. The quantization axis is not the same as the direction of propagation of the photon, so ##\Delta m =0## is allowed.kelly0303 said: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?
Things are a bit more complicated when you go beyond the dipole approximation. Seekelly0303 said:Is this true for all orders in an EM interactions (i.e. E1, E2, M1 and higher)? Thank you!