I Forbidden decay ##\rho^0\rightarrow \pi^0\pi^0##

Summary
Reasons behind the forbidden decay ##\rho^0\rightarrow \pi^0\pi^0## and considerations about the wave function of two bosons system
Hello everybody!

I have a question regarding the forbidden decay ##\rho^0 \rightarrow \pi^0\pi^0##, but it is a general doubt.

My book states that one of the reasons why the decay is forbidden is Bose-Einstein statistics, the final state of two equal pions must be in an antisymmetric state.
My reasoning is the following. The wave function of two pions is:
$$\psi_{\pi^0\pi^0} = \psi_{space}\psi_{isospin}$$
I neglect the spin part since pions are spin-0 particles.
##\rho## is a spin-1 particle, so the orbital momentum of the two pions system is ##l=1##; analogously for isospin ##I=1##.

$$(exchange)\psi_{\pi^0\pi^0} =(exchange) \psi_{space}\psi_{isospin} = (-1)^l(-1)^I = (+1)$$

My result is not in agreement with the book. However, if I do not consider the isospin wave function everything is ok.
My question is if I should consider the isospin wavefunction. If not, why?
 

Vanadium 50

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Look at the Clebsch-Gordan coefficients.
 
Look at the Clebsch-Gordan coefficients.
$$|\rho^0> = |1,0> = \frac{1}{\sqrt{2}}|1,+1>|1,-1> + 0 |1,0>|1,0> - \frac{1}{\sqrt{2}}|1,-1>|1,+1> = $$ $$ =2\frac{1}{\sqrt{2}}|1,+1>|1,-1> + 0 |1,0>|1,0> = \frac{1}{\sqrt{2}}|\pi^+\pi^-> + 0 |\pi^0\pi^0>$$

From the Clebsch-Gordan coefficients I see that the decay in ##\pi^0\pi^0## is forbidden, but I do not see the link with my way of looking at the problem.
 

Vanadium 50

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I'm not sure what to tell you. The amplitude is zero. Might it be zero for some other reason too? I guess.
 

clem

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The rho has Ispin 1. pi0-pi0 cannot be in an I=1 state.
The decay is Ispin forbidden, but that is not absolute.
The spin-statistic argument (without Ispin) is absolute.
 

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