1
$\begingroup$

When a $\pi^0$ decays into photons, the only possible number of photons to which it can decay is 2$\times n$, with $n$ a natural number. This is because in electromagnetic decays charge conjugation is conserved, let's assume that it decays into two photons

$$\pi^0 -> \gamma+\gamma$$

and:

C|$\pi^0$> = 1

C|$\gamma$> = -1

in this decay, parity also needs to be conserved, and

P|$\pi^0$> = -1

P|$\gamma$> = -1

the parity of the system of two photons is given by $(-1)^l$P|$\gamma$>P|$\gamma$> where $l$ is the orbital angular momentum of the system. Does this mean that if parity is conserved, the orbital angular momentum of the resulting system must be $l=1$ (or 3, 5...) or am I deducing something wrong?

$\endgroup$
  • $\begingroup$ You are doing fine. In real life, you may confirm the l=1 directly by inspection of the explicit amplitude you find in QFT texts, calculated by J Steinberger in the late 40s. $\endgroup$ – Cosmas Zachos Sep 3 '18 at 5:09
1
$\begingroup$

You are deducing it right. Both C and P should be conserved in the decay, so to obtain a negative parity, since you have two identical paricles in the final state, you need the proper value of angular momentum l.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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