Violation of spin conservation in pion annihilation

[Trollfaz]

$\pi^\pm$ are mutual particle-antiparticle pairs, while $\pi^0$ is it's own antiparticle. All has a spin of 0.
In any annihilation reaction of a particle $x$, the equation is
$$x+\bar{x}\to \gamma+\gamma$$
Photons have no charge but a spin 1. I can see charges are conserved but spin is not since the total spin before the reaction is 0 while after the reaction is 2. Is there a solution to this or am I just wrong?
Sorry the LaTeX here behaves differently from the one on my computer.

 

[renormalize]

Is there a solution to this or am I just wrong?

What's conserved in these reactions is the total angular momentum. For spin-0 pions in the center-of-mass frame, the reaction $\pi+\bar{\pi}\rightarrow\gamma+\gamma$ results in two spin-1 photons travelling in opposite directions with their spins antiparallel, so the total angular-momentum is zero both before and after the reaction.

 

[Trollfaz]

So p is your symbol for pion? The general symbol for pion is $\pi$ p represents proton

 

[renormalize]

So p is your symbol for pion? The general symbol for pion is $\pi$ p represents proton

Yes, and I've edited post #2 to reflect that notation. But same conservation reasoning also applies to protons and antiprotons: two spin-$\frac{1}{2}$ particles annihilate to two spin-1 photons with antiparallel spins.

 

[Vanadium 50]

In any annihilation reaction of a particle , the equation is

False.

total spin before the reaction is 0 while after the reaction is 2.

False.