# 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 $\vec r \times \vec p$. – my2cts 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? – Pregunto Oct 9 '18 at 20:39

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.