# Decay of a pi0 meson?

## Main Question or Discussion Point

I have a vague understanding of the idea of quantum super-position giving the linear combination of uubar and ddbar, even though it is very hard to picture - if possible I would like to avoid going off onto a tangent about this...

My question is: am I right in saying that the decay of pi0 is through EM?

Is it caused by annihilation of the 'quark/antiquark?' Is this a too simplistic way of looking at it?

Also: In general- why doesn't the strong force prevent annihilation of a quark-antiquark pair if it becomes repulsive at short distances? Or, is it naive of me to be thinking that the quark and antiquark actually need to "touch" to annihilate?

Any responses wold be welcome...

Thank you

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mfb
Mentor
The electromagnetic interaction is dominating, the decay via a Z is possible but rare. You can see this in the http://pdg8.lbl.gov/rpp2013v2/pdgLive/Particle.action?node=S009 [Broken] - most decays have at least one photon, and neutrinos are extremely rare.
Is it caused by annihilation of the 'quark/antiquark?
Right.

In general- why doesn't the strong force prevent annihilation of a quark-antiquark pair if it becomes repulsive at short distances?
It is the strong interaction. It is not useful to view it as a force in low-energetic processes like bound hadrons.
Or, is it naive of me to be thinking that the quark and antiquark actually need to "touch" to annihilate?
Quarks don't "touch" each other at all. Their wavefunctions might overlap, but that is true all the time in hadrons.

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Thank you for sorting out the thread for me- I am unfamiliar with how to do things like that. I also don't know how I missed that there was a particle physics thread.

Your response has cleared things up for me for the most part.
Obviously all mesons are unstable, and if I remember correctly the lifetime of pi0 is quite short. Why doesn't the qq annihilation happen immediately though?

fzero
This is a feature of quantum mechanics. QM assigns every event a given probability that the event happens, but does not directly give the time at which the event will occur. Even the quoted lifetime is not the exact time that would take for a specific meson to decay. Rather, if we are given a sample of $N_0$ mesons at rest, the lifetime is the time over which we can expect a bit over half of the mesons to decay Specifically, if we measure $N_0$ mesons at time $t$, then at time $t=\tau$, the lifetime, we should measure $N_0/e$ mesons, where $e$ is the base of the natural log.