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I am trying to understand how to conserve angular momentum in a elementary decay/reaction.

Consider the elementary reaction:

$$ K^{-}(J = 0) +p(J = 1/2) ~\to~ \Omega^{-}(J = 3/2) + K^{+}(J = 0) + K^{0}(J=0) $$ We can see that net angular momentum is increased by one unit. So as per angular momentum conservation law this reaction should not be allowed.But, I know that this is allowed with explanation given that extra one unit angular momentum being taken care by orbital angular momentum. If that is the case than this conservation law for angular momentum will be redundant as we can argue for any change in angular momentum being taken care of by change in orbital angular momentum.

If any change in angular momentum is allowed how much can be adjusted for orbital angular momentum?

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  • $\begingroup$ Not every hole can be patched by orbital angular momentum... show your work. $\endgroup$ – Cosmas Zachos Nov 13 '20 at 3:53
  • $\begingroup$ "If any change in angular momentum is allowed" Angular momentum is quantized at this level of interactions. $\endgroup$ – anna v Nov 13 '20 at 6:29
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I think what's going on here is two things: (1) the forbidden processes, unlike the one in your example, involve either a single particle in the initial state or a single particle in the final state; and (2) there is a hidden assumption that we're dealing with pointlike particles and interactions that happen at a pointlike vertex.

A simple example is a particle of spin 0 decaying to a spin 0 plus a spin 1. We can get into a frame in which the initial particle is at rest at the origin. Then because the particles are pointlike and decay occurs at a pointlike vertex, each of the final particles has zero orbital angular momentum in the final state. This process is therefore forbidden.

There is then the question of why the above argument still holds for an example like photon emission by an excited spin-0 state of hydrogen, decaying to the ground state, which is indeed forbidden. (It goes only by two-photon emission.) The hydrogen atom is not pointlike, it's composite. I'm tempted to say that it still works because angular momentum has to be conserved at that vertex, which doesn't include the proton, but that may be wrong. We do, after all, have to have a proton present for this decay to happen -- a free electron can't gamma decay.

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  • $\begingroup$ This doesn't seem right. Consider a theory with a massive, heavy vector particle coupled to pairs of scalars. Everything is pointlike but the vector can certainly decay. $\endgroup$ – knzhou Nov 13 '20 at 3:50
  • $\begingroup$ @knzhou can you give and example? the only point particles are the particles in the SM table, and there the only scalar is the Higgs. $\endgroup$ – anna v Nov 13 '20 at 6:21

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