Quark compositions in $\pi^+$ to $\pi^0$ pion decay

Question

Pions can undergo a rare beta-like decay into leptons:

Pion beta decay (with probability of about $10^{−8}$) into a neutral pion plus an electron and electron antineutrino (or for positive pions, a neutral pion, positron, and electron neutrino).

• Why is the quark composition of the neutral pion is so different to the charged pion after pion decay?

$$\pi^{+}(\overline u,d) \to \pi^{0} \left(\frac {\overline u,u-\overline d, d}{\sqrt 2} \right)+e^{+}+ \nu_e$$

$$\pi^{-}(\overline d,u) \to \pi^{0} \left(\frac {\overline u,u-\overline d, d}{\sqrt 2} \right)+e^{-}+\overline \nu_e$$

• Why is the quark composition of the neutral pion is so different with neutral hadron (like neutron)?

Answer 1

Because your formulae for the pion wave functions are written in a confusing way. The quark composition of the positively charged pion is $$ |\pi^+\rangle = |u\bar d\rangle $$ while the neutral pion is $$ |\pi^0\rangle = \frac{|u\bar u\rangle - |d\bar d\rangle}{\sqrt{2}}. $$ These two composite formulae are completely analogous: these pions are mesons, i.e. quark-antiquark bound states, and they just differ in which quarks from the double $(u,d)$ are used. The neutral pion is a superposition of two pieces: superpositions of these types are omnipresent everywhere in quantum theory. The charged pions only contain one term because there is only one quark-antiquark combination involving $u,d$ quarks and antiquarks whose total charge is $\pm 1$. To get $Q=0$, one may either combine $u\bar u$ or $d\bar d$ and the relative complex coefficient between these two terms is a priori arbitrary and determines whether the superposition is a mass eigenstate and if it is, whether it's lighter or heavier.

Of course that the detailed formulae for the two pions must be different in some respects, otherwise they would be an identical particle.

During the decay, all the conservation laws and other laws of physics are satisfied.

Answer 2

I think I know where your misconception lies-- you appear to think that the individual quarks are the real thing, so you wonder why the $\pi^{+}$, made up of ''pure'' states of $|{\bar u}\rangle$ and $|d\rangle$, could decay into a ''mixed'' state.

The problem here is that the physical thing is not the quarks, themselves, but the quark field. And this field is spanned by four basis states (so long as you don't allow higher generations of quarks): $|{\bar u}\rangle$, $|u\rangle$, $|{\bar d}\rangle$, and $|d\rangle$. Since we know that free states cannot carry color charge, and we are interested in the lightest possible strongly interacting particles, we are restricted to build states out of quark/antiquark pairs. It turns out that the three low energy eigenstates to the hamiltonian are $|u{\bar d}\rangle$, $|d{\bar u}\rangle$ and $\frac{1}{\sqrt{2}}\left(|u{\bar u}\rangle - |d{\bar d}\rangle\right)$, but any other linear combination of quark/antiquark states would be, in the abstract, an eqally good value for the quark field.