- #1

Aleolomorfo

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- TL;DR 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?

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?