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

$$\pi^0 \to \gamma+\gamma$$


$$C|\pi^0\rangle = 1$$

$$C|\gamma\rangle = -1$$

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

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

$$P|\gamma\rangle = -1$$

the parity of the system of two photons is given by $(-1)^lP|\gamma\rangle P|\gamma\rangle$ 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?

  • $\begingroup$ 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. $\endgroup$ Sep 3 '18 at 5:09

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.


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