# Strong Decay and Parity Conservation?

The following decay is possible according to the PDG and according to my notes it is a strong decay:

$$\omega(1420) \to \rho^0 + \pi^0$$

The JPC values are:

$\omega(1420)$ 1--

$\rho$ 1--

$\pi$ 0-+

So, all three particles have, for themselves, a parity of -1.

The combined parity on the right side should then be (-1)*(-1)=1. But the left side has a parity of -1. This violates parity, but parity should not be violated in a strong decay.

1) What's going on and where is the error in my argument?

2) How can I calculate the orbital angular momentum the two decay products have in relation to each other?

• Welcome to Physics.SE, Nx1990. I've replaced your unicode greek letters with LaTeX alike markup for MathJax to render as it allow the use of superscripts. – dmckee Jan 30 '12 at 21:42
• Question for the student: how does the parity of a state depend on it's angular momentum quantum number? – dmckee Jan 30 '12 at 21:44
• Regarding to your question: I thought parity is an intrinsic property of a particle, and does not depend on the angular momentum. However, I seem to be wrong. There seems to be an additional factor of (-1)^L. Since the omega is a vectormeson, it has spin 1. Because J=1 for the omega, L must be 0. The pion has J=0 and S=0, so L=0. The rho has J=1 and S=0, so L=1. Is that correct? – Nx1990 Jan 30 '12 at 21:51
• Now, if that is correct, the rho gets an additional factor of (-1)^1, so the parity of the rho is +1, and parity is conserved again. The relative angular momentum seems to be 1 then?! – Nx1990 Jan 30 '12 at 21:52
• Feel free to write it up as an answer: self-answers are allowed and encouraged. Then the votes will tell you if you're right. – dmckee Jan 30 '12 at 23:59