On $\Delta^{+}$ particle decay

isospin-symmetryparticle-physicspionsstrong-force

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?

Best Answer

My question is... is it just experimental evidence or rather a theoretical necessity?

Both. It is an experimental fact, which confirmed/motivated the theory of strong interactions, early on. Strong means fast decay, short lifetimes. Theory went hand in hand with experiment back then, the mid 1950s. The theory quickly appreciated Heisenberg's spin-like treatment of charge-irrelevance (isotopy) was a conserved property of the strong interactions, and it helped organize all experimental data, which then confirmed further logical-connection predictions of it, like the one you are discussing.

Is it just a coincidence? Why $1/2$ and $1$ though? Aren't there other ways of obtaining $3/2$?

Not really, in pragmatic terms. While you could imagine, in a different world, adding isospin 1/2 with isospin 2 to also get isospin 3/2, there are no isospin 2 mesons below the mass of the Δ to serve...

Is this decay predictable using just the expression for $\Delta^+$ in terms of quarks?

Yes and no. People have used this decay in conjuring up and cleaning up the quark model in the first place. (The complete symmetry of the uuu fermion quarks of the $\Delta^{++}$ led to a spin-statistics paradox which ushered in the hypothesis of color.) Now, any half-decent quark model incorporates these facts and , in reverse, "postdicts" them!

The specific decay you have is the standard illustration/first-exercise of isospin applications in virtually every introductory textbook on the subject. It is supposed to remind you of Clebsching, because it is formally identical to that problem: theory is application of mathematical rules of logic. That is the whole point of isospin: you handle it exactly like spin! In particular, squaring the respective channel amps gives you double the rate for the proton versus neutron mode.

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