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I heard that antiquarks are just like antimatter. Does that mean that they cancel quarks out? If so, does that mean that there are more quarks than antiquarks?

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    $\begingroup$ Yes, there are more quarks than antiquarks. Usually people phrase this as "there are more baryons than antibaryons", but a baryon is just 3 quarks. $\endgroup$
    – knzhou
    Dec 11, 2018 at 21:43
  • $\begingroup$ @knzhou Since virtually all baryonic matter in the Universe is composed of protons (uud) and, in some cases, neutrons (udd), wouldn't it be more accurate to say that antiquarks are extremely rare? $\endgroup$ Dec 12, 2018 at 0:52

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It is indeed true, that antiquarks are antimatter and annihilate with quarks. As knzhou pointed out in his comment, there are more baryons (3 quarks) than antibaryons (3 antiquarks).

There is also another kind of hadronic (made of quarks) matter, mesons, consisting of one quark and one antiquark each. Those, however, are not stable.

Addendum

As Chappo pointed out in his comment, quarks don't exist in isolation. Thus, we cannot observe these processes with typical scattering experiments. However, from the decay products of mesons, we can reconstruct the decay process using conservation laws.

For example, $\pi^0$ (a superposition of $u\bar u$ and $d\bar d$) can decay into two photons, which can be interpreted as the quark-antiquark pair annihilating.

It is also possible for quarks and antiquarks of different flavour to interact in a similar process. Due to conservation laws, however, these processes involve the weak interaction, so they happen at a much lower rate. Instead of photons, which mediate the electromagnetic force, gauge bosons of the weak interaction are produced in this case.

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    $\begingroup$ This probably needs a bit more detail, as it's too simplistic. How do quarks and antiquarks "annihilate" if they don't exist in isolation (due to colour confinement)? Can an up quark annihilate with a down antiquark, or do they have to be the same flavour? How can a kaon exist if it's a quark and antiquark? $\endgroup$ Dec 12, 2018 at 0:25

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