Selection rules for atomic transitions

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 4K views
copernicus1
Messages
98
Reaction score
0
Hi, I'm a little confused by the selection rules for atomic transitions. In pretty much any standard QM text, they derive the fact that an electron can't transition in an atom unless [itex]\Delta\ell=\pm1[/itex], i.e. unless the orbital angular momentum changes by one unit. Then the books go on to say that this can be interpreted as a consequence of the fact that the photon carries one unit of spin.

This seems strange to me, since the Schrödinger equation doesn't predict spin, and as far as I can tell, spin is not even put in by hand in these calculations. How do the transition rules somehow predict that the angular momentum must change by one unit when the whole formalism doesn't incorporate spin angular momentum?

It seems like someone working out this effect without any knowledge of spin would have to come up with some other interpretation.

Thanks!
 
Physics news on Phys.org
My understanding was that it's down to the parity of the original and new electronic wavefunctions. The electric dipole operator has odd parity (swap all the positions and it points in the opposite direction), and the transition probability is proportional to <new state|electric dipole operator|old state>, so for this integral to be nonzero the new and old states must have different parity values, i.e. <+|-|-> or <-|-|+>, otherwise the two wave-functions will cancel. States with the same l have the same parity, so transitions must involve the electron either gaining or losing one quanta of orbital angular momentum or the integral is zero and the transition is "forbidden".
 
Then the books go on to say that this can be interpreted as a consequence of the fact that the photon carries one unit of spin.

This seems strange to me, since the Schrödinger equation doesn't predict spin, and as far as I can tell, spin is not even put in by hand in these calculations. How do the transition rules somehow predict that the angular momentum must change by one unit when the whole formalism doesn't incorporate spin angular momentum?

It seems like someone working out this effect without any knowledge of spin would have to come up with some other interpretation.

The selection rule is just a rule of thumb; the radiation at frequency corresponding to "forbidden transition" has certain non-zero presence in the spectrum of the gas of atoms. The corresponding lines are just too faint when compared to regular ones.

You are right, there is never one correct interpretation. That's why the word "interpretation" - it is often very subjective view of what it all means and there are usually more than one.

Spins and photons indeed do not figure in the basic Schr. equation, but the selection rule still follow from the latter. This can be viewed as a purely mathematical fact, that certain eigenvibrations of Schr. equation are hard/impossible to excite by external light wave. This happens in simpler models too, for example a set of coupled harmonic oscillators can have eigen-modes which are hard to excite by external light wave.
 
Schrödinger eqn does not predict the nature of intrinsic spin,but it does predict orbital angular momentum.The change in total angular momentum is ±1 which can be derived from other means like invoking group theory.The result can be interpreted if we say photon has spin 1.By the way it is possible to prove using maxwell eqn that photon does have spin 1 character.