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.


1 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.

  • $\begingroup$ For example, people used to think the proton was an elementary particle. After they found out it wasn't, they didn't want to have to go back and change every reference to the "spin" of the proton. $\endgroup$
    – user4552
    Commented Nov 26, 2017 at 23:43

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