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