High spin for $Δ^{++}$ resulting from angular momentum of pion in relation to proton

When a pion (Spin 0) collides with a proton (Spin 1/2) the result can be a Δ++ (Spin 3/2). When I tried to look up how the resulting Spin is possible, I came about an answer I do not understand.

It says that the Δ++ particle is not created through a head-on collision between the pion and the proton (which would indeed follow the regular pattern of 0 + 1/2). Rather it is created when the pion has an angular momentum of one h-bar (i.e. Spin 1) in relation to the proton (see picture; thus explaining the Spin 3/2 of the resulting Δ++). I do not understand the statement that the pion has an angular momentum of one h-bar in relation to the proton.

Can anybody help out?

Picture from Yoram Kirsh, Fundamentals of Physics B - Tel Aviv, 1998

• Just apply the formula $\vec r \times \vec p$. Oct 9 '18 at 20:19
• @my2cts Thanks, but why how does that formula produce "1" as a result? (assuming I have to add it to the 1/2 of the proton) And why does that not apply when the collision is head on? Oct 9 '18 at 20:39

This explanation belongs to the era before quarks, but it is mathematically correct.

At the level of resonances and particles one is in the quantum mechanical regime. This means that the modeling of any "measurements" or "resonance creations" has to follow the operator formalism , in this case the angular momentum operator. This has an operator algebra and group representations.

The diagram shows that in scattering, the angular momentum between the pion and the proton is quantized , and the values depend on $$l$$ the angular momentum quantum number. The probability to generate a $$Δ$$ depends on the pion - proton distance to be such that "spin" $$l=1$$ . Otherwise no resonance is produced.