Using isospin notation $$ \Delta^+=\left|\frac 3 2,\frac 1 2\right\rangle=\frac{1}{\!\sqrt{3}}\bigg(|duu\rangle+|udu\rangle+|uud\rangle\!\bigg) $$
It is known all of the $\Delta$ baryons with mass near $1232 \,\operatorname{MeV}$ quickly decay via the strong force into a nucleon (proton or neutron) and a pion of appropriate charge.
My question is... is it just experimental evidence or rather a theoretical necessity?
Going ahead with isospin formalism:
\begin{aligned} \Delta^+=\left|\frac 3 2,\frac 1 2\right\rangle\longrightarrow & \frac{1}{\!\sqrt{3}}\left|\frac{1}{2}, -\frac{1}{2}\right\rangle|1,1\rangle+\sqrt{\frac{2}{3}}\left|\frac{1}{2}, \frac{1}{2}\right\rangle|1,0\rangle\\&\frac{1}{\sqrt 3}|n\rangle|\pi^+\rangle+\sqrt{\frac 2 3}|p\rangle|\pi^0\rangle \end{aligned}
This reminds me a lot of a change of basis for the state $|\Delta^+\rangle$ from the coupled basis to the uncoupled basis, if we suppose to apply the algebra of addition of angular momenta. Is it just a coincidence? Why $1/2$ and $1$ though? Aren't there other ways of obtaining $3/2$?
Is this decay predictable using just the expression for $\Delta^+$ in terms of quarks?