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

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)?

## 2 Answers

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.

• why baryon beta decay does not works in this way
– Neo
Commented Sep 23, 2012 at 9:29
• Which way? The fundamental processes behind beta decays are always the same. The wave functions of the decaying particles and final particles are different because they're different particles. But it's always the case that the beta decay transforms a $d$ quark to $u e^+ \nu$ or some rearrangement from the left to right hand side and/or some replacement of particles by antiparticles. Both quarks in the charged pion may get transformed by this elementary beta process at the level of quarks and leptons which is why the superposition is possible on the right hand side. Commented Sep 23, 2012 at 10:05
• Why is the quark composition of the neutral pion is so different with neutral hadron
– Neo
Commented Sep 23, 2012 at 10:10
• It is not "so different". The neutral pion is a bound state of a light quark and its antiparticle; neutron is a $udd$ bound state. They are different composite particles which means that they use different combinations of the elementary particles but all of the light hadrons use $u,d$ quarks and antiquarks. Commented Sep 23, 2012 at 15:33
• Dear FrankH, because the final state is a superposition, there are actually two diagrams that contribute to the decay amplitude. In one of them, for $\pi^+$ decay, $\bar u$ is just running horizontally from the initial to final state while $d$ converts to a $u$ by emitting a virtual $W$ which later decays to $e^+$ and $\nu$ by another vertex. In the other diagram, producing the second term, $d$ runs from the initial to final state while $\bar u$ becomes $\bar d$ by emitting a virtual $W$ which also splits to $e^+\nu$ on the right side. Commented Sep 23, 2012 at 17:02

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.