I am learning about parity of nucleus and (elementary) particles. For example, proton has a intrinsic parity of $+1$. However, a more general term for parity of proton is $(-1)^l$ where $l$ is orbital quantum number. So proton parity is $+1$ only when $l=0$ which implies that proton is in ground state (in some nucleus I guess, not sure "ground state" is applicable for free particles?).
My question is can free particle, like proton have orbital quantum number? (eg. proton that has just been released in beta+ decay so he flies away as free particle)
I ask this because there is this nuclear reaction:
$$
\rm p^+ + {^{7}_{3}Li} → {^{8}_{4}Be} → {^{4}_{2}He}+ {^{4}_{2}He}+ 17.2\,MeV
$$
If I understood this reaction correctly, we have free flying proton that hits nucleus of Li and then Be is created. This Be is unstable so it decays to two He nuclei. Given the law of conservation of parity (that applies for EM and strong interaction) we must have equal parity on far left and far right side.
Parity for proton is +1 while for $\rm{^{7}_{3}Li}$ is -1 which gives total parity -1 on the left side (they are multiplied). On the right side we have two He nuclei which both have +1 so we have total parity of +1 on the right side.
So this reaction will not happen (because we have odd parity on the left and even on the right) unless proton has $l=1$ which will make it parity to $(-1)^l = (-1)^1 = -1$. Then on the left side we will have parity of $(-1)(-1) = +1$ which will make this process possible. In my notes from the class it says "this reaction will not happen if proton is in ground state, meaning it has $l=0$, because then parity is not conserved". I am wondering how can proton be in the ground state, or how can it have orbital angular momentum, if it is a free particle like in this process?
Best Answer
Recall that the value and sign of $\mathbf{r} \times \mathbf{p}$ angular momentum depends on the point around which you chose to measure it.
A free particle can (indeed, does) have angular momentum around any and all points relative which it is moving and has a non-zero impact parameters.
But, frankly, that's not a very interesting statement.