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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?

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    $\begingroup$ 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. $\endgroup$
    – diracula
    Mar 12, 2017 at 12:32
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    $\begingroup$ 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. $\endgroup$
    – diracula
    Mar 12, 2017 at 12:34
  • $\begingroup$ 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? $\endgroup$
    – MmeTautou
    Mar 12, 2017 at 12:49
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    $\begingroup$ There cannot be, as observed mesons have different parities. $\endgroup$
    – diracula
    Mar 12, 2017 at 12:52
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    $\begingroup$ 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.) $\endgroup$
    – diracula
    Mar 12, 2017 at 13:38

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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.

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