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I understand that the term "vector meson" denotes a meson particle (composite particle with a quark and an antiquark) which transforms as a vector under the Lorentz group. Since both the quark and the antiquark are Lorentz spinors, we have the tensor product

$$ \frac 12\otimes\frac 12 = 1\oplus 0\tag{1} $$

so from a group theoretic viewpoint, it makes sense to me that a meson is either a scalar or a vector.

But what's the exact difference between the $\pi$ and $\rho$ mesons? How does a scalar $\bar ud$ differ from a vector $\bar ud$? I suspect it has to do with the Clebsch-Gordan decomposition in $(1)$, so the scalar $\bar ud$'s spin state must be a superposition $\frac{1}{\sqrt{2}} \left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right)$

Is that correct so far?

If yes, then is it possible that a quark in a $\rho$ meson "flips its sign" such that the vector meson becomes a scalar meson? This would be some process like $\rho\to\pi+X$ with $X$ some other particle that must be there because of energy conservation, for example a photon.

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    $\begingroup$ $X$ is another pion. $\mathcal{B}(\rho \to \pi \pi)$ is basically 100% $\endgroup$ – dukwon Mar 16 '17 at 21:31
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I suspect it has to do with the Clebsch-Gordan decomposition in $(1)$, so the scalar $\bar ud$'s spin state must be a superposition $\frac{1}{\sqrt{2}} \left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right)$

Yes, I would agree with this (though there are multiple isospin scalar states that determined the charge of the $\pi$ meson)

Is it possible that a quark in a $\rho$ meson "flips its sign" such that the vector meson becomes a scalar meson?

Yes, you could say that one of the quarks in the meson can 'flip' it's spin when it reacts with some other particle, causing the vector meson to become a scalar meson. It is also possible for the $\rho$ meson to decay into two pions, since the $\rho$ mesons are charged and this charged has to be conserved, decays like $\rho^\pm \to \pi^\pm + \pi^0$ or $\rho^0 \to \pi^+ + \pi^-$ are possible.

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  • $\begingroup$ My apologies for my earlier messy answer, I was confused between isospin and spin states for a moment. $\endgroup$ – JgL Mar 16 '17 at 22:37

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