If a particle decay has quark annihilation will it always emit a photon?

Question

From what I can gather the decay $\pi^0 \rightarrow \gamma \gamma $ emits two photons because $\pi^0 $ has quark content $(u\bar u-d\bar d)/\sqrt{2} $, and the $u$ annihilates with $\bar u$, and $d$ annihilates with $\bar d$. So each annihilation has it's own photon. If this is true does that mean that $\pi^0 \rightarrow \gamma \gamma \gamma $ is not allowed because there aren't three annihilations?

Answer 1

One reason that a $\pi_0$ can’t decay into one photon is that a massive particle decaying into a single massless particle violates the conservation of momentum. This is obvious if you think about it in the rest frame of the pion. Another reason is that this would violate charge-conjugation symmetry. The C-parity of a neutral pion is $+1$, while the C-parity of $n$ photons is $(-1)^n$.

It can decay into two photons because this is compatible with momentum conservation. In the pion’s rest frame, the two photons can come out in opposite directions, with zero total momentum. But there don’t have to be two kinds of quarks inside for it to do this. For example, charmonium ($c\bar{c}$) decays into two photons, as does positronium on the leptonic side.

It cannot decay into three, five, etc. photons because this is forbidden by charge-conjugation symmetry.

It can decay into four, six, etc, photons, but the probability of each extra photon is reduced by a factor of $\alpha\approx1/137$, the fine structure constant. (There can be other numerical factors also.)

A neutral pion does not necessarily decay to just photons. For example, among other possibilities are: one photon, one electron, and one positron; no photons, two electrons, and two positrons; just one electron and one positron.