# How does Delta baryon decay conserve angular momentum?

I'm a chemist so bear with me:

I understand the Delta baryons $\Delta^{+}$ and $\Delta^{0}$ to be in some sense spin (and isospin) quartet states of the proton and neutron. These can decay straight to a proton or neutron through emission of a pion, which is spin-0.

I've been puzzling about the mechanism by which this decay conserves spin. I note via Wikipedia that pions decay into either a lepton pair or a photon pair, for which it's easy to show spin is conserved, however pion production seems to induce a loss of $J=1$.

This seems to suggest momenta of:

$\Delta^{+} (J=\frac{3}{2}) \rightarrow{} p^{+} (J=\frac{1}{2}) + \pi^{0} (J=0)$

Am I missing a boson? Is my understanding of spin flawed? I assume that if this decay exists, similar things turn up everywhere else.

The missing $J=1$ angular momentum is carried by the orbital angular momentum known as $\vec L$ – the relative orbital angular momentum of the proton and the pion.
The $L\gt 0$ wave functions vanish near $r=0$ as $r^L$ (this is familiar from quantum chemistry, think of the hydrogen atom) which is why the probabilities of similar decays would be negligible in chemistry etc. However, in nuclear physics, the baryons are large enough so that such wave functions, despite their vanishing at the origin, yield significant probabilities.