Transitions in diatomic molecules

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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!
 

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Dr. Courtney
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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
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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).
 
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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?
 
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Dr. Courtney
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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?

Yes, single photon angular momentum changes of 1 are allowed. Sorry for the awkward, ambiguous grammar.
 
  • #6
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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!
 
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DrClaude
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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
https://jila.colorado.edu/~ajsh/astr5110_13/notes/selec.pdf
 

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