Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking into the decay of a $\Phi$-meson decaying into $K^+$, $K^-$. My problem is, the $\Phi$-meson has a total angular momentum of 1 and the two Kaons have a total angular momentum of 0. On the first glance this seems to be a violation of the angular momentum conservation law.

I tried to explain it to myself using the quark picture of the system. So the $\Phi (s \bar{s})$ goes under the production of a $u\bar{u}$ pair into $K^+(u \bar{s})$ and $K^- (\bar{u}s)$. Using Clebsch Gordon coefficients i noted down the possible combinations and that works out so far.

My question is, how can the 'effective' total angular momentum be explained as it seems to me, that the conservation law is not conserved, which can not be. Do i miss something or have i got something wrong?

share|cite|improve this question
up vote 4 down vote accepted

The $\Phi$ meson does indeed have angular momentum of $1$ and so the Kaons must have a total angular momentum of $1$. Since the Kaons are both spin zero that means that they must come out in a state with non-zero orbital angular momentum state, i.e., an excited state. In general any orbital angular momentum is allowed, however I believe that the general trend is that the higher the necessary orbital angular momentum, then more the process is suppressed.

share|cite|improve this answer
I agree, that as a general trend: the higher orbital angular momentum is needed for a decay, the process should be more suppressed. But referring to the pdg summary tables the decay into $K^+$,$K^-$ is the dominant decay channel with a branching ratio of 48% - which seems to be a result of a so called OZI rule. – user46328 May 14 '14 at 8:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.