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Why is the decay of a neutral $\rho$ meson into two neutral pions forbidden? (Other modes of decay are possible though.) Is it something with conservation of isospin symmetry or something else? Please explain in a bit more detail.

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As far as I understand, due to conservation of angular momentum, the resulting system of neutral pions would need to have angular momentum 1, therefore, the identical neutral pions would be in an anti-symmetric state, which does not seem possible as they are bosons. Note that a neutral rho meson can decay into two neutral pions and a $\gamma$, although this decay is suppressed.

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If we look at isospin, $\rho = |1,0\rangle$ and $\pi^0= |1,0\rangle$.

Since SU(2) isospin is a really good symmetry in strong interactions, it must be conserved. Looking at the isospin of the final state:

\begin{equation} |1,0\rangle \otimes |1,0\rangle = \sqrt{\frac{2}{3}} |2,0\rangle + 0 |1,0\rangle - \sqrt{\frac{1}{3}} |0,0\rangle \end{equation}

That is, there is no $|1,0\rangle$ component in the final state, and therefore the process is not allowed by SU(2) isospin symmetry.

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$$ \rho^0 \longrightarrow \pi^0 \pi^0 $$ The parity af $\rho^0$ and $\pi^{0}$ are $$ \begin{aligned} & P_{\rho^0}=(-1)(+1)(-1)^0=-1 \\ & P_{\pi^0}=(-1)(+1)(-1)^0=-1 \end{aligned} $$ Due to conservation of parity in strong interactions, we require $$ P_{\rho^0}=P_{\pi^0} \cdot P_{\pi^0} \cdot(-1)^2 $$ which means the orbital angular momentuen of the final state is $L=1$. Due to conservation of total angular momentum, the total angular momentum of the final state must be equal to the total angular momentum of the initial $\rho_0$ $$ J=J_{\rho^0}=1 $$ Since $\pi^{0}$ are bosons so we require the total wavefunction to be symmetric. $L=1$ means the spacial part is anti-symmetric. so the spin part has to be anti-symmetric, which is impossible for two spin-0 sion.

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  • $\begingroup$ Although your answer put in more effort and is clear, it is actually an expanded version of an earlier, accepted, answer. Please do not revive old questions with duplicates of answers. $\endgroup$ May 19, 2023 at 5:15
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    $\begingroup$ @naturallyInconsistent There doesn’t seem to be an “accepted” answer. I didn’t understand the most upvoted answer myself at first as it’s too brief. I thought new learners like me would benefit from a more detailed answer like this one. Is there a better way to do so? $\endgroup$
    – L L
    May 19, 2023 at 11:27
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Clebsch-Gordan isospin rule suggest decay of I=1,Iz=0 particle into two I=1,Iz=0 products has zero coefficients of Clebsch-Gordan [1x1].

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    $\begingroup$ This doesn't look like a very clear explanation. You should expand on the details to make your answer somewhat understandable. $\endgroup$ Jun 13, 2014 at 13:10
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The charge conjugation symmetry is conserved in the strong interaction. In the above decay mode, neutral rho meson decay into two neutral Pion, neutral rho meson has -1 and neutral Pion has +1 for C such that decay violates C symmetry. Therefore it is forbidden.

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  • $\begingroup$ This answer is completely wrong, please vote it down so people do not get confused $\endgroup$
    – Juanjo
    Nov 3, 2018 at 15:31

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