Orbital angular momentum values of $\Delta^{++}$ decay

Question

Considering the following decay: $$\Delta^{++}\rightarrow n\space + \pi^{+}$$

We know that for $\Delta^{++}$, $J^{P}=\frac{3}{2}^{+} \rightarrow$ the only possible value for the orbital angular momentum is $L=0$ (Correct me if I'm wrong here) What would that imply for the $L$ values on the R.H.S. ? Considering the fact that the spin of the neutron is $\frac{1}{2}$ and that the spin of the pion is $0$, in order to conserve $J$, would the only possible value of $L$ be $-1$ ? (where I assumed that $J=\mid L-S \mid$)

Answer 1

The $\Delta$ that we start with is a particle on its own (its internal quark structure is irrelevant) so there is no orbital angular momentum so $L=0$, as you say.

$J$ is conserved ($L$ and $S$ separately are not) so the total angular momentum on the two decay particles must be the same as that of the parent, $J={3 \over 2}$.

The neutron and the pion are two separate particles so they can have non-zero orbital angular momentum $L$ between them. The total $J$ is made up of the vector sum of the $L$ and the $1 \over 2$ spin of the neutron. So conservation of angular momentum allows $L=1$ or $L=2$, as you can make $3 \over 2$ from $1+{1 \over 2}$ or $2-{1 \over 2}$

But then look at the parity. For the $\Delta$ it is +. For the $n\pi^-$ system it is the product of the intrinsic parities (+1 and -1 respectively) and a factor $(-1)^L$ from the space part of the wave function. So conservation of parity (this is a strong decay) requires that $L$ is odd.

That rules out the $L=2$ possibility, showing that $L=1$