# Parity conservation in electromagnetic decays

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

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

and:

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

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

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?

• 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. Sep 3 '18 at 5:09