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Any examples?

$$? \rightarrow \pi^0 \pi^0$$

If such a process exist, could there be nonzero total orbital angular momentum in the final states of the two neutral pions? But then how to understand the interaction between the two neutral pions that make them rotate with each other to contribute the non-zero L?

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1 Answer 1

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The decay of the $K^0_s$ meson.

The PDG lists a branching fraction around 32%.


But then how to understand the interaction between the two neutral pions that make them rotate with each other to contribute the non-zero L?

They don't "rotate" they come out of the interaction with different "orbital" angular momentum. The term "orbital is a little unfortunate here, it really just means angular momentum of motion (AKA $\vec{r} \times \vec{v}$) as opposed to intrinsic angular momentum (AKA spin).

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  • $\begingroup$ The fractions come from isospin rules, and actually the clebsh gordan coefficient for pi0pi0 is 0. $\endgroup$
    – anna v
    Commented Nov 7, 2012 at 6:50
  • $\begingroup$ for rho to pi0pi0 $\endgroup$
    – anna v
    Commented Nov 7, 2012 at 10:13
  • $\begingroup$ Thanks @dmckee, But I agree with anna. $\rho^0 \rightarrow \pi^0\pi^0$ is forbidden, due to several reasons: 1). isospin rules as stated by anna, 2) angular momentum conservation 3) charge conjugation, the initial state is -1, but the right hand side is 1...could we treat $\pi^0\pi^0$ as a free system as opposed to be a bound system? $\endgroup$
    – Osiris Xu
    Commented Nov 8, 2012 at 4:55
  • $\begingroup$ Hmmm...yes. Will Edit. $K^0_s$ certainly does. $\endgroup$ Commented Nov 8, 2012 at 4:59

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