I'm somewhat confused by this decay. Associated vertex seem to be related to QCD residual terms contributing to the nuclear force, therefore they should manifest conservation of Isospin and Parity.

However, given the last one, the pseudoescalar and vector nature of the particles involved would imply the identity: $$ (-1)=(-1)^{L+1} $$ being $L$ the orbital angular momenta of the three $\pi$ system in the final state. Since angular momenta conservation is sacred, for it raises from Lorentz invariance itself, one would expect to have: $$ L=1 $$ which seems to contradict Parity conservation. Is something wrong in this argument? Or is it that this decay is of a different nature than that I'm assuming? and in case the last one is the answer, how could one describe such vertex from those in the Standard Model?

P.D. Bear with me, this is my first question here. Thanks in advance.


1 Answer 1


Your mistake is coming from your treatment of the orbital angular momentum in the case of a 3 body-decay. You have to take into account the orbital angular momentum between 2 pions $L_1$ and the orbital angular momentum $L_2$ between the third pion and the barycenter of the first 2 pions. The conservation of the total angular momentum imposes that $$\vec{1} = \vec{L_1} + \vec{L_2}$$ Restricting $L_1$ and $L_2$ to the lowest values, it's possible with $L_1=0$ and $L_2=1$, or $L_1=1$ and $L_2=0$, or $L_1=1$ and $L_2=1$ (the latter being possible since the addition of $\vec{1}+\vec{1} = \vec{2}, \vec{1}$, or $\vec{0}$ ). The only way to conserve parity is to choose $L_1= L_2=1$ since the parity of a 3 body system is: $$ \eta_{\omega} = (\eta_\pi)^3 \times (-1)^{L_1} \times (-1)^{L_2}$$ This decay is very similar to the one of the $\omega(1420)$ wiich occurs via the $\rho$ resonance: $$\omega(1420) \to \rho + \pi$$ $$\rho \to \pi + \pi$$ the $\rho$ having a spin 1. The $\rho$ is equivalent to the 2 pions system with $L_1=1$. The orbital angular momentum between the $\rho$ and $\pi$ being $L_2=1$.


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