Conservation of angular momentum in a collision using quantum mechanics In an undergraduate course in particle physics we are learning about angular momentum conservation.
In particular, at some point we analyzed the colision between a pion ($J=0, p=-1$) and a deuteron ($J=1, p=+1$) to produce two neutrons ($J = 1/2, p=+1$). We assumed first of all that the relative angular momentum of the deuteron and the pion was zero.
Thus the equation of conservation of angular momentum is:
$$ 0 + 1 = \frac{1}{2} + \frac{1}{2} + L $$
Where $L$ is the relative angular momentum of the system of neutrons and the sum is performed as a "sum of quantum vectors" (i.e. by Clebsch-Gordan method). Without getting too much into the details, we deduce that the final state of the system of neutrons must have $L = 1$ and $S = 1$ (just because the total wave function must be totally anti-symmetric). 
On another problem involving a gamma decay of a nucleus from an excited state we did not include a term for the relative angular momentum between the nucleus and the photon.
My questions are:


*

*How can I decide if there should or should not be term corresponding to the relative angular momentum? In particular why isn't it there on the gamma decay problem?

*In the problem about neutrons, if I think about it in terms of classical physics, I imagine that the deuteron and the pion collide radially (so no angular momentum as seen from the collision). But in that case the outgoing neutrons should go out in such a way that the position and the momentum are parallel, so again, there should be no angular momentum, but there is... What am I doing wrong here?

*Finally, when talking about conservation of angular momentum in this kind of context, shouldn't we specify an origin for the angular momentum? Why is it that it is never mentioned from where is angular momentum measured?
 A: 
How can I decide if there should or should not be term corresponding to the relative angular momentum? In particular why isn't it there on the gamma decay problem?

Such a term should always exist, but the orbital angular momentum might not change. This may be the case in the decay you refer to, but you haven't given any information about it.

In the problem about neutrons, if I think about it in terms of classical physics, I imagine that the deuteron and the pion collide radially (so no angular momentum as seen from the collision). But in that case the outgoing neutrons should go out in such a way that the position and the momentum are parallel, so again, there should be no angular momentum, but there is... What am I doing wrong here?

The deuteron and the pion both have finite sizes, and the range of the interaction between them is nonzero, so for all these reasons a purely radial collision isn't necessary.

Finally, when talking about conservation of angular momentum in this kind of context, shouldn't we specify an origin for the angular momentum? Why is it that it is never mentioned from where is angular momentum measured?

This is the same as in classical physics. If it's conserved for one choice of axis, then it's conserved for others as well.
A: One has to clarify what quantum mechanics means. It means that the appropriate solutions of the QM differential equations define the interactions, definite wavefunctions $ψ$  which will give probability distributions for the system under study by using $ψ^*ψ$ . That is what can be measured, not an orbit, but an orbital. 
In this framework, there are bound states. One cannot say that the electron in an orbital around a nucleus has an angular momentum corresponding to the classical definition. The angular momentum is contained in the energy level quantum numbers that describe the electron's orbital. Thus for particles coming from bound states  it is the quantum numbers that tell you how much angular momentum is carried by a particle.
For scattering problems , elementary particles are point particles and their interactions are described by Feynman diagrams at vertex points, in a sense it is only "head on" collisions that can happen. Feynman diagrams can be used for the pion deuteron problem.
For decays, the only angular momentum is the one of the quantum energy level from which the decay happens. That is the only angular momentum available for a quantum mechanically bound state, there is no "orbit", only orbitals. This is the quantum mechanical frame. Also one has to remember that the spin quantum number defining particles was invented so that angular momentum conservation would hold at the quantum level too.
This complicated system has been validated by a plethora of experimental evidence, so one should avoid using classical mechanics intuition at the quantum mechanical level. This link may help.
