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Decay of a pi0 meson?

  1. Oct 12, 2013 #1
    About the pi0 meson:

    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
  2. jcsd
  3. Oct 12, 2013 #2


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    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.

    It is the strong interaction. It is not useful to view it as a force in low-energetic processes like bound hadrons.
    Quarks don't "touch" each other at all. Their wavefunctions might overlap, but that is true all the time in hadrons.

    I deleted your identical thread in the homework section and moved this thread to particle physics.
    Last edited by a moderator: May 6, 2017
  4. Oct 12, 2013 #3
    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.

    Thanks for the reply-
    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?
  5. Oct 12, 2013 #4


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    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.

    The lifetime is an inherently statistical property. If we watch a single meson, it could decay immediately or it could take many lifetimes for it to decay. The individual events are truely random and it is only by observing many events that we would see an exponential distribution emerge, parameterized by the actual lifetime value.

    In an explicit calculation, the probability to decay does depend on the overlap of the wavefunctions that mfb referred to. In fact, it depends on the probability that we find the antiparticle at the same location as the particle, so the idea that the particles "touch" to annihilate is not really horrible. The reason it's not encouraged to think in terms of such classical ideas is that the classical, Bohr-type picture of the particles revolving around one another is not accurate and would give the wrong answers for the energy, etc., of a bound state. QM only gives the probability to find a particle at a particular point, as encoded in the wavefunction; it does not describe the particle as having some specific trajectory between measurments.
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