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

#### Aleolomorfo

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?

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Staff Emeritus
Look at the Clebsch-Gordan coefficients.

#### Aleolomorfo

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.

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

#### clem

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.

• vanhees71

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