Strong Decay and Parity Conservation?

Question

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?

Answer 1

1) 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.

Now, if that is correct, the rho gets an additional factor of (-1)^1, so the parity of the rho is really +1, and parity is conserved again: (-1) = (+1)*(-1).

2) From the arguments of 1), the relative angular momentum seems to be L_rho - L_pion = 1 - 0 = 1.