Why does electron-positron annihilation conserve parity? I think I'm missing something quite basic here but consider the process:
$$ e^- + e^+ \rightarrow 2\gamma$$
Fermions have opposite parity to antifermions so the parity quantum number before the process is $P=-1 \times (-1)^L$ where $L$ is the relative orbital angular momentum, which should vanish in the zero momentum frame where the collision is head on. So we start with $P=-1$. The photons on the other hand have the same parity as each other so have parity $P=+1$ (after again noting that they must be heading in opposite directions to each other in the ZMF so have $L=0$).
I think my reasoning about the angular momentum must be wrong, but I don't see why. (Additionally, the same argument says that if this is $L:0\rightarrow 0$ then charge congugation $C$ is not a conserved quantity either.)
 A: The electron-positron collision state in question has the same quantum numbers as   positronium, which  may exist in two states. The first of these is the short-lived singlet state $^1S_0$, so $L=S=J=0$, hence $P=-$ and $C=(-)^{L+S}=+$, and this is your parapositronium two-photon mode (must have even number of photons, from C), with symmetric parity-odd wavefunction,
$$
\propto (|\hat \epsilon _1\rangle  |\hat \epsilon _2\rangle - |\hat \epsilon _2\rangle |\hat \epsilon _1\rangle)(e^{ik(\hat {\mathbf k} \cdot ({\mathbf r}_1- {\mathbf r}_2)-2t)}- e^{ik(\hat {\mathbf k} \cdot ({\mathbf r}_2- {\mathbf r}_1)-2t)}  )  , 
$$
for respective photon polarizations $\hat \epsilon_1, \hat\epsilon_2$. 
Should remind you of the neutral pion (a pseudoscalar), and decays just like it into an odd-parity couple of photons.
The other, not sought here, long-lived  bound state is orthopositronium, triplet and longer lived, $^3S_1$, so $P=-, ~ C=-$. Hence, it must decay to 3γ, unless it were a virtual γ in production through collision of electron-positron. 
(So, does this remind you of the  ρ or the J/ψ ?) 
