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Are there Quark-Antiquark Particles?

  1. Jul 20, 2015 #1
    There are exotic atoms such as the protonium (proton+antiproton) and positronium (electron+positron); I was wondering if quark-antiquark particles could appear even if they only exist for a fraction of a second.
     
  2. jcsd
  3. Jul 20, 2015 #2
    Yes, they are called mesons.
     
  4. Jul 20, 2015 #3

    ohwilleke

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    Not only are there quark-antiquark particles called mesons. There exist a particular subset of mesons called quarkonium, which are composed of a quark of a particular flavor and an anti-quark of the same flavor. For example, charmonium is a meson made of a charm quark and an anticharm quark.
     
  5. Jul 20, 2015 #4
    Why is it important if the quark and anti-quark have the same flavor?
     
  6. Jul 20, 2015 #5

    ohwilleke

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    Matter-antimatter annihilation can only take place if the quarks are the same flavor (for reasons that are obvious from the rules of Feynman diagrams). So, once you know that rule, the fact that you can have a meson with quarks and antiquarks of the same flavor is much more remarkable than the fact that you can have a meson with a quark of one flavor and an antiquark of another flavor in it. A meson with a charm quark and an antidown quark shouldn't be able to annihilate, while a meson with a charm quark and an anticharm quark should be able to annihilate, yet they can briefly form a composite object anyway.
     
  7. Jul 21, 2015 #6

    ChrisVer

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    Quarkonia are just a subset of mesons.
    I don't find anything important about it.

    This is wrong. A meson with a charm and an antidown quark would be the [itex]D^+[/itex] charmed meson.
    The [itex]c,\bar{d}[/itex] can still go through a s-channel (with Ws) into other particles, and you can have [itex]D^+ \rightarrow l^+ \nu_l[/itex]. They have a small Branching fraction, but they are not "forbidden". The BR of [itex]l=e, \tau[/itex] haven't been observed and pdg gives upper bounds, but [itex]l=\mu[/itex] has been measured.

    A better example for proving that wrong are the Kaon decays. eg [itex]K^+ \rightarrow \mu^+ \nu_\mu[/itex] (63.55%) and K+ is a [itex]u\bar{s}[/itex].

    What you say would be true if there was no possible transition between generations (no CKM matrix). It would be true however to say that J/ψ (cc*) or Y (bb*) do have much shorter lifetime. But it's difficult to make use of it, because they have so many different decay modes (either the mesons or the quarkonia).
     
    Last edited: Jul 21, 2015
  8. Jul 21, 2015 #7

    mfb

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    Neutral mesons of any kind can decay to two or more photons. It is a rare process because it cannot happen at tree level and needs both the electromagnetic and weak interaction, but it is possible.

    ##B^0 \to \pi^0 \pi^0 \to 4 \gamma## has been observed.

    Charm+antidown has charge, so the decay products need at least one charged particle obviously.
     
  9. Jul 22, 2015 #8

    ohwilleke

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    When I say that a meson is not able to annihilate, I do not mean that it is not able to decay. What I mean instead when I say that I meson is not able to annihilate is that it cannot decay directly to photons, leaving behind no quarks, directly from its current state without intermediate W boson interactions.

    For example, in a case like a B0 meson decay to photons, you have a b quark and an anti-d quark leaving a neutral charge. For this to decay to photons, you must first have the b quark transition by emitting to W- boson to, for example, a charm quark, and then have the charm quark emit a W+ boson while it transforms to a d quark, and then have the d quark and the anti-d quark annihilate, and also have the W+ and W- bosons either produce equal and opposite decay products and annihilate, or directly annihilate each other (if the timing of the sequential interaction permitted, which it probably wouldn't given the short mean lifetime of a W boson and the relatively slow speed of weak force interactions).

    I would describe that sequence of events a B0 meson decaying into a D+ meson and a W- boson, followed by the decay of the D+ meson into a pion (or given that a pion isn't really d-antid, a pair of pions) and a W- boson, followed by the annihilation of the pions and the annihilation of the W+ and W- boson products with each other.

    Collapsing the original meson to the final decay process oversimplifies what is going on which is something much more complex than annihilation.

    In contrast, quarkonium can go directly from meson to a pair of photons without decaying into something else as an intermediate stage of the process. which is fairly called annihilation.

    This matters because it isn't all that amazing that something can exist even though after being transformed through multiple future steps of non-instantaneous subsequent reactions, the end decay product can be photons. In contrast, it is amazing that a composite object made up of particles that can instantly annihilate into photons in a single step as is the case in quarkonium.
     
    Last edited: Jul 22, 2015
  10. Jul 22, 2015 #9

    ChrisVer

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    The thing with those mesons is also that they don't decay into photons... they'll preferably (I guess) go through the gluon exchange (hadronize). So I interpreted the "annihilation" as an s-channel diagram, where the W could exist.
    For example for the [itex]J/\psi (1S)[/itex] the e/m decay to [itex]3\gamma[/itex] (2 are C-forbidden), is like [itex]\Gamma_{3\gamma}=0.001 \%[/itex]
     
  11. Jul 24, 2015 #10

    mfb

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    That restriction sounds a bit arbitrary.
    That does not work, that process is forbidden by energy conservation. The W boson in the overall decay is not real.
     
  12. Jul 28, 2015 #11

    ohwilleke

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    Even if the D+ and W- step was virtual, the pion step would not be.
     
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