Is the interaction $\gamma p \to \pi^+n$ allowed? I'm doing an undergraduate lecture course in particle physics and I'm still getting to grips with the basics of interactions.
One of the example interactions I've been given in an exercise is $\gamma p \to \pi^+n$, and the question implies that this is allowed. However, I don't see how angular momentum is conserved. If I'm looking up the right numbers...
$$
J(\gamma)=1 \\
J(p)=J(n)=1/2 \\
J(\pi) = 0
$$
which means that I have $J=3/2$ on the LHS and $J=1/2$ on the RHS. From other similar (but not identical) questions, I have seen mention of the relative angular momentum between the two final-state particles, but I cannot find a good reference on how to deduce this...!
 A: I agree with most of the discussion in @annav answer. But the key point is missing.
The J value that you quote are related to the modulus of the spin angular momentum of the particles ($J^2 = j(j+1) \hbar^2$). Therefore you can't add them without care.
The resulting angular momentum must not match the sum of its components but must fulfill the triangular inequality:
 If $J=S_1+S_2$ for vectors  one has 
$|S_1-S_2|\leq J \leq|S_1+S_2| $
for lengths
Hence the combination of a spin 1 and a spin 1/2 can give either 1/2 or 3/2. 
This fixed the paradox, whatever is the pion spin 0 or 1.
A: In quantum mechanics there is angular momentum between two particles when they are bound it is called the L quantum number, and it corresponds to the macroscopic angular momentum concept.Angular momentum exists also between two scattering particles, the usual classical rxp definition.
Angular momentum conservation is a strict law both classcically and quantum mechanically.
In particle interactions we are in the quantum mechanical regime where operators will give us the state of angular momentum , and they impose a quantization on the angular momenta available for the interaction.
Because of conservation of angular momentum, researchers were forced to give an intrinsic angular momentum to elementary particles, called spin, so that the interactions will still conserve angular momentum. That is how we know the neutrino is spin1/2, for examle, so that the decay data of the muon do not violate angular momentum conservation.
In the specific reaction, the target proton has spin 1/2, and the incoming photon spin 1. Depending on the wavefunction  of the scattering, there is a probability that an l=1 could scatter, and then the conservation would give  5/2 on the left!. The scattered products can balance any incoming quantum numbers by going out with a relative angular momentum of the pion and neutron of 1 or 2 or .... It is all in the wavefunctions.
There is no problem. Conservation of angular momentum adds both intrinsic (spin) and relative angular momenta in the interactions by construction of spin. The spin definition is part of the successful fit to the data by the standard model for particle physics .
