Particle physics: Why is J^P called spin parity if J is the total angular momentum?

Question

Here is the question I am working on: "The Ξ- has spin parity=½+. It decays through the weak interaction into a Λ0 and a π- meson. If the spin parity of the Λ0 particle is 1/2+ and the spin parity of the π- particle is 0- what are the allowed relative orbital angular momenta for the Λ0 + π- system?"

I am not looking for an exact answer to this question, but rather an explanation.

If J gives the total angular momentum as the sum of the orbital and spin angular momenta (J=L+S), why is J^p termed the 'spin' parity? In the above question the spin of the resultant particles is given as 1/2 and 0 respectively, but this is labelled J, although it is spin.

What I understand from reading is that the final total angular momentum is the sum of L+S, and given J and S, L can be solved for, but it seems that it will always be zero. Unless J is not the spin. If so how can you determine the spin just from the information given above?

Also is the final total angular momentum J just the sum of the angular momenta of the products (i.e. 1/2+0)?

Any help in understanding this would be appreciated.

Answer 1

"Spin parity" isn't a thing. It's saying the xi baryon has spin $\frac{1}{2}$ and positive parity; they're separate properties whose names tend to be run together for some reason.

As for why we use the word spin even though some of the angular momentum may be orbital: it allows you to imagine the $\Xi^-$ as an elementary particle which has the same amount of angular momentum as a spin-$\frac{1}{2}$ fundamental particle, namely $\frac{\hbar}{2}$. When you're treating the $\Xi^-$ as an elementary particle, you don't care whether part of its angular momentum comes from the orbital angular momentum of even more fundamental constituents. In a sense, all of quantum field theory is based on the idea that we can ignore small-scale structure in this way, under appropriate conditions.