Transitions in diatomic molecules

[kelly0303]

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!

 

[Dr. Courtney]

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.

 

[TeethWhitener]

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).

 

[kelly0303]

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]

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.

 

[kelly0303]

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]

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