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Mesons are bosons, therefore their wavefunction must be symmetric under particle exchange. Overall, the meson wave function ($\text{WF}$) has the following contributions:

$$\text{WF} = \lvert \text{flavor}\rangle \lvert \text{spin}\rangle \lvert \text{radial}\rangle \lvert\text{color}\rangle.$$

Mesons are a color singlet, their color wavefunction looks like: $$\frac{1}{\sqrt{3}}(\lvert r\bar{r}\rangle + \lvert b\bar{b}\rangle + \lvert g\bar{g}\rangle)$$ which is symmetric.

For the pseudoscalar mesons the spin part is antisymmetric because the spins are a spin singlet.

The radial part is symmetric for the pseudoscalar/vector mesons, because the angular momentum $\ell$ is zero.

When it comes to the flavor part of the wavefunction for the pseudoscalar mesons, it is symmetric as well: for example $$\pi^+: \frac{1}{\sqrt{2}}(\lvert u\bar{d}\rangle + \lvert \bar{d}u\rangle).$$

This gives us an overall antisymmetric wavefunction because of the antisymmetric spins. But the wavefunction has to be symmetric! The same goes for the vector mesons: there we just have an antisymmetric flavor part times symmetric spin. So it is again overall antisymmetric.

What's wrong with this reasoning?

Qmechanic
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MmeTautou
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    A wavefunction for several particles is antisymmetric under exchange of any two identical fermions. The spin of the total system (e.g. whether a composite particle is a boson or fermion) is not important. – diracula Mar 12 '17 at 12:32
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    More importantly, mesons do not contain identical particles. The exchange you are discussing is I believe a parity operation (which acts to swap the particles), so the resulting sign is the parity of the meson. – diracula Mar 12 '17 at 12:34
  • Thank you, I now understand. So there is no special requirement concerning the symmetry of the meson wave function? On the other hand, the reason for introducing color was that the baryon decuplet wave functions had to be antisymmetric. The Delta resonances do not always have the same quark content. So why is antisymmetry necessary for baryons? – MmeTautou Mar 12 '17 at 12:49
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    There cannot be, as observed mesons have different parities. – diracula Mar 12 '17 at 12:52
  • I'm not sure whether you have read the second part of my previous reply as I have modified it shortly before you answered. Nevertheless, I still keep thinking why we even bother then to write down the flavor wave function of the pion in a symmetric way and not antisymmetric as for the vector meson. The pion has a symmetric flavor part while the rho mesons have the same quark content but are described by an antisymmetric flavor part. Why is that? – MmeTautou Mar 12 '17 at 13:20
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    The $\Delta^{++}$ baryon has quark content $|uuu\rangle$, so contains three identical fermions, hence arguments from exchange symmetry can be made, leading to the necessity of introducing colour. This is distinct from arguments involving parity, I believe, and parity does not act to swap particles as it did in a meson. To the rest of your comments, maybe this question will be helpful. (I do not think I can give a good answer.) – diracula Mar 12 '17 at 13:38
  • I have another fundamental problem of comprehension: the pion wave function has to be antisymmetric because pions have a negative parity? Is that correct? Because protons have a positive parity but are also required to have a fully antisymmetric wave function. How does that go together that protons have a positive parity and a antisymmetric wave function? Is the correct explanation that in SU(3) you assume the u,d and s quarks to be identical in terms of the strong interaction so that you have a system of "identical" fermions that need an overall antisymmetric wave function? – MmeTautou Mar 28 '17 at 20:19
  • As I said above it is only in mesons that a parity operation corresponds to particle exchange; this is not the case in baryons. (But note a $\pi^+$ does not even contain two identical quarks, so swapping particles there has nothing to do with symmetry under exchange of identical fermions.) The antisymmetry you refer to corresponds to particle exchange I believe, and this is in general completely distinct from parity. Additionally, you may find it helpful to look at the question I linked in my previous comment, where a conventional choice of parity for baryons is discussed. – diracula Mar 29 '17 at 20:36
  • Yes, that was a huge misunderstanding that I mixed up parity with the anti-symmetry for exchanging particles. So the meson parity you mentioned in your second post is just about the mirror image. But why do you construct the wave function (for pions for example or protons) in such a way, that they are antisymmetric for particle exchange? We did this very much in detail in the lectures and used the proton-wave function to compute for example the magnetic momentum. But since protons have both u- and d-quarks, why is this necessary? The same argument should apply to the mesons I think. – MmeTautou Mar 30 '17 at 05:58
  • I now came to the conclusion that you could probably treat a meson similiar to an electron-positron-system where you can impose antisymmetry of the product-wave-function under particle exchange if you regard them as two different charge states of the same particle. Do you think this can be applied to the meson as well? If I assume that antiquarks are just different states of the same particles and can be treated as equivalent in terms of the same mass? – MmeTautou Mar 30 '17 at 15:47
  • I think it would be best if you started a new question with your specific query, rather than extending this discussion in the comments section. Additionally I'm not sure I can answer your questions satisfactorily, but you might get a nice answer on a new question. – diracula Mar 31 '17 at 14:54

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No, it has to be anti-symmetric because you are exchanging two fermions. It doesn't matter that they form a boson together. Two bosons are symmetric under exchange but then you would have 4 constituents, 2 each.

Alex Roberts
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