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That's probably a plain question but I wonder: Which mesons can exist? The only limitation I'm aware of is the charge (antiparticle+particle). Ok, finally also the color (anticolor+color) but I assume this is not really reducing the possibilities.

Can you please shine a bit light on me?

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You can classify mesons via two questions:

  1. How do they transform under P and C?

    $P$ is the eigenvalue under parity transformations and $C$ is the eigenvalue under charge conjugation. They can take on the values $\pm 1$, so the possibilities are $$PC = \big\{(++), \, (+-),\,(-+),\, (--)\big\}$$

  2. What is their quark structure?

    Here I'm talking about constituent quarks, that is no quark-antiquark pairs that may appear inside a meson.

If you answer those two questions, you can find an associated name in tables like this one, which is a good summary of all the different mesons.

Additionally, you can ask:

  1. What is their total angular momentum?

    As you can see in the review I linked above, the lowest possibilities for the total angular momentum $J$ are $$ J^{PC} = 0^{++},\, 1^{+-},\, 0^{-+},\, 1^{--}$$ They are the ground state, whereas higher-$J$ represent excited states.

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  • $\begingroup$ Thank you! I fear my question is even more simple: Is every single quark-antiquark-combination possible to form a meson? So e.g. (u,d_), (u,s_), (u,c_), u(t_), (u,b_), (d, s_), (d, c_),.. Seizing this assumption would lead to about 15 combinations resp. including their antiparticles then 30. n over k with n=6 and k=2? I know the rules above would tell that but is there a very general point of view to allow or deny a combination/meson? $\endgroup$ – Ben Nov 16 '17 at 10:03
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    $\begingroup$ Yes, since “meson” just means “quark+antiquark”, you can have every combination of six flavors. And since e.g. $u\bar u$ is also possible, there’s $6^2=36$ combinations! These 36 combinations relate to the second question in my answer. Then you can also have 4 different $PC$ structures, which gives $4\times 36=144$ possibilities for the ground state. Then there are who-knows-how-many excited states... $\endgroup$ – Stephan Nov 16 '17 at 10:57

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