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.

Answer 2

Light mesons, wich are the entities envolved in this decay, have all one thing in common: their orbital angular momentum is 0. They are, however, grouped in the pseudo-scalar mesons (with spin s=0 for the pions, $\pi$, for example) and the vector-mesons (with s=1 for the rho, $\rho$, and omega, $\omega$, for example).

The parity of a meson state is the product of the parity of its constituents by the parity of it's orbital wave function like this: $P(q)P(q*)(-1)^l$ where q* is the anti-quark and the term $(-1)^l$ is the parity of the orbital wave function. Check Modern Particle Physics (Mark Thompson) on page 229!

Since this is a strong force mediated decay, the parity must be conserved from the initial state to the final state. We can consider the center of mass referential of the two resulting products and refer to the orbital angular momentum of the final state there. Thus, the following $l$ refers to the angular momentum of the composite final state $\pi+\rho$. Because $P(q)=1, P(q*)=-1$:

$P(\omega) = P(\rho)P(\pi)(-1)^l$

$P(q)P(q*)(-1)^l = P(q)P(q*)P(q)P(q*)(-1)^l$

$1\times(-1)\times(-1)^0 = 1\times(-1)\times1\times(-1)\times(-1)^l$

$-1 = (-1)^l$

Thus, in order to conserve parity the orbital wave function of the final state must present an odd value for $l$ (like $l= 1, 3$ etc).Furthermore, this must be also be consistent with the conservation of angular momentum:

The initial state ($J_i$) has $J_i=1$ ( since $J=s+l=1+0$). The final state ($J_f$) has the sum $J_f= s_1+s_2+l$ where $s_1$ and $s_2$ are the spins of the rho and pi mesons. Acording to the rules of addition of angular momenta we first add two of them: $s_1+s_2 = 1+0 = 1$ and then we add the third giving $J_f=1+l$. We can then conclude that $l$ must either be 0, 1 or 2 since this is the only way we can get $J_f$ to equal $J_i$. (to add angular momentum: $|l_1-l_2|,|l_1-l_2|+1,|l_1+l_2|+2$ ,.... stopping only when we have he value $|l_1+l_2|$). Thus we can clearly see that in this case, to respect both parity and angular momentum conservation the final state $l$ must really be $l=1$.