# I Pion+/- decay - what *Exactly* is happening?

1. Aug 19, 2017

### ChrisVer

OK people, I will agree with you and say that the neutral pion is only the $u\bar{u}-d\bar{d}$ state... In that manner the decay $\pi \rightarrow \pi^0 e \nu$ can't happen since there is no meson with the $d\bar{d}$ (or $u\bar{u}$) alone quark content...
Also what's the meson that has the state: $u\bar{d} \pm u\bar{s} \pm u \bar{b} \pm c \bar{s}\pm c \bar{d} \pm c \bar{b}$ (that would come from the W hadronic decay addition of Feynman Diagrams)?

2. Aug 19, 2017

### vanhees71

3. Aug 19, 2017

### ChrisVer

because I couldn't produce a neutral pion according to the definition I am getting: I either end up with processes resulting to $u\bar{u}$ or to $d\bar{d}$ (so projections of the neutral pion, as well as the sigma mesons)... I originally said that the decay was possible (like a semileptonic meson decay), since those were actually neutral pions but I was wrong.

4. Aug 19, 2017

### vanhees71

But it's not wrong, as you can see from the particle data booklet (the branching ratio is only about $10^{-8}$, but it's a measured decay of the charged pions). I still don't understand, why you come to the conclusion that this decay cannot exist, only because the pion is the given superposition?

5. Aug 19, 2017

### Staff: Mentor

They have a neutral pion component.
There is no such thing, and a W cannot decay to a single meson anyway.

We had a great question about partial widths a while ago, where you can see the effect of the superpositions of quark states directly in the branching fractions.

6. Aug 21, 2017

### ChrisVer

Because that decay ends up with $u\bar{u}$ or $d\bar{d}$ and not in a superposition of the two... and that is not a neutral pion (as I was repeatedly told in this discussion)...
On the other hand, if we are to "add" every possible outcome (and so say we can have both the $u\bar{u}$ and $d\bar{d}$ in a combination, al), then I would ask on this counter-example (a possible W decay to a meson):
https://arxiv.org/pdf/1001.3317.pdf (Fig2)
$D^0 \rightarrow \rho^+ \pi^-$ : $c \bar{u} \rightarrow W (d\bar{u}) \rightarrow (u \bar{d}) (d \bar{u})$
my question would then be why do we get a $\rho^+$ (with the $u\bar{d}$ decay of a W ) and not a $\alpha u\bar{d} + \beta u \bar{s}$ (forgetting the bottoms due to energy). Instead we call the 1st a rho, and the second a possible Kaon (2 different outcomes)... Is it related to a flavour symmetry or something?

7. Aug 21, 2017

### Orodruin

Staff Emeritus
This is self contradictory. If it can end up in uubar and ddbar, clearly it can end up in a linear combination of those states. It is a question of what the amplitude is.

What makes you think it doesn't decay to both linear combinations? The only difference is that the s quark is heavy enough for the mass eigenstates to be essentially equivalent to the flavour states, which does not happen in the pion system.