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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?

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    $\begingroup$ 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. $\endgroup$ – dmckee --- ex-moderator kitten Jan 30 '12 at 21:42
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    $\begingroup$ Question for the student: how does the parity of a state depend on it's angular momentum quantum number? $\endgroup$ – dmckee --- ex-moderator kitten Jan 30 '12 at 21:44
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    $\begingroup$ 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? $\endgroup$ – Nx1990 Jan 30 '12 at 21:51
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    $\begingroup$ 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?! $\endgroup$ – Nx1990 Jan 30 '12 at 21:52
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    $\begingroup$ 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. $\endgroup$ – dmckee --- ex-moderator kitten Jan 30 '12 at 23:59
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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.

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First lets clarify one thing: Light mesons, wich are the entities envolved in this decay, have all one thing in common: their orbital angular momentum is l=0. They are, however, grouped in the pseudo-scalar mesons (with s=0 for the pions for exemple) and the vector-mesons (with s=1 for the rho and 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)xP(q*)x(-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!

Thus, since this is a strong decay, the parity must be conserved from the initial state to the final state:

P(omega)x(-1)^l = P(rho)xP(pion)x(-1)^l

(here we can consider the center of mass referential of the two resulting products and refer to the orbital angular momentum there without loss of generality because we arealy know tha l=0) Now, since P(q)=1 and P(q*)=-1:

P(q)xP(q*)x(-1)^l = P(q)xP(q*)xP(q)xP(q*)x(-1)^l

1x(-1)x(-1)^0 = 1x(-1)x1x(-1)x(-1)^l

-1 = 1*(-1)^l

What can be concluded in order to conserver parity is the following: the orbital wave function of the final state must present an odd value for l (like l= 1, 3 etc) meaning that s waves (l=0), d waves (l=2) etc are excluded and only p waves (l=1), f waves (l=3) etc can be observed.

Furthermore this must be consistent with the conservation of angular momentum:

The initial state (Ji) has Ji=1 ( since J=s+l=1+0) The final state (Jf) has the following sum : Jf= s1+s2+l where s1 and s2 are the spins of the rho and pi mesons. Acording to the rules of addition of angular momenta we first add two of them: s1+s2 = 1+0 = 1 and then we add the third giving Jf=1+l. We can then conclude that l must either be 0, 1 or 2 since this is the only way we can get Jf to equal Ji. (remember that to add angular momentum: |l1-l2|,|l1-l2|+1,|l1+l2|+2,.... stopping only when we have he value |l1+l2| and so, the values 1 and 2 will spam the sbspace with J=1 as well!). Thus we can clearly see that in this case, to respect both parity and angular momentum ocnservation the final state l must really be l=1.

Hope this gives a clear sight on what is happening in this deacay!

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