This is just a conceptual question I guess. The annihilation of a particle with a finite mass and its anti-particle cannot lead to the emission of only one photon, and this is due to the conservation of energy and linear momentum.
However, how could this be shown in a mathematical way? Could this be done, perhaps, with a consideration of the four-vector momentum of the two particles?
Answer
As dmckee says in a comment, the proof is ridiculously simple. Suppose we work in the centre of momentum frame so the total momentum is zero. The particle comes in with some momentum $p$ and the antiparticle comes in with the opposite momentum $-p$, and the two annihilate.
Suppose the annihilation produced a single photon. The momentum of a photon is:
$$ p = \frac{h}{\lambda} $$
but the problem is that the momentum of a photon is always $h/\lambda$. Unlike a massive particle a photon has no rest frame i.e. no frame in which it's momentum is zero. So creation of a single photon means the momentum would change from zero to $h/\lambda$ and momentum wouldn't be conserved.
For momentum to be conserved we have to create a minimum of two photons moving in opposite directions i.e. with momenta $h/\lambda$ and $-h/\lambda$.
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