3
$\begingroup$

In gamma decay, there is a transition between two states where energy is released by the emission of a gamma ray. We know that light is composed of photons, gamma rays are in fact photons emitted. It is common to analyze radiation fields by their multipole character. Hence, we can classify the transitions between excited states (or to ground state) by their electromagnetic mutlipolarity. Next, we restrict our multipole order L by the conservation of angular momentum during gamma decay. Since angular momentum is quantized, we say that a multipole transition of order L transfers an angular momentum of L$\hbar$ per photon. From this we say that the change in the spin (total angular momentum) of the states must be equal to the angular momentum of the photon. Such that the angular momentum of the photon is restricted by $$ | I_i -I_f| \le L \le I_i + I_f$$ Which is the first selection rule of gamma decay. The other can be obtained by conservation of parity. But what I don't understand is the fact that it seems like the intrinsic spin of the photon which equals 1 is just been neglected in these selection rules? Because they also say that E0 transitions cannot occur is this because of the fact that photons carry spin 1? Am I wrong or do I misunderstand the concept of the gamma ray carrying angular momentum?

$\endgroup$
1
$\begingroup$

Everything you say seems 100% correct.

But what I don't understand is the fact that it seems like the intrinsic spin of the photon which equals 1 is just been neglected in these selection rules?

It hasn't been neglected. It's part of the angular momentum described by your notation $L$. Your $L$ is made up of the sum of two angular momenta: the photon's spin, and the photon's orbital angular momentum.

So in fact, the "angular momentum" is the total angular momentum of the photon.

(This answer was originally a comment by Ben Crowell.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.