On the annihilation of a neutron with an antineutron In my introductory to elementary particle physics course, I am asked to consider the process
\begin{equation}
n+\bar{n}\to\pi^++\pi^0+\pi^-
\end{equation}
and to determine whether it is possible or not. In particular, I have been introduced to quantum baryonic, lepton and quantum charge numbers.
Based on these, I see no reason why the process could not be possible. However, a quick Google research seems to indicate the reverse. What am I missing? Here is the Feynman diagram I draw for this process:
$n+\bar{n}\to\pi^++\pi^0+\pi^-$ process" />
 A: Of course the reaction can go, since its twin, $p+\bar{p}$ can, and does go, section 6.1 of this review.
P, C, J, and isospin go, but since you did not mention I, I won't cover it--it is not a problem.

*

*On the left-hand-side, $n+\bar{n}$ you have

Either ortho-neutronium, $^3S_1$, hence C= - and P= - ;
or else para-neutronium,  $^1S_0$, hence C= + and P= - . Hence,

*

*On the right-hand side,
$$C|\pi^+\pi^-\rangle= (-)^L |\pi^+\pi^-\rangle \\
C|\pi^0\rangle = +|\pi^0\rangle,
$$
where L is the orbital angular momentum between the two charged pions; so a total of C= - for L=1 (think of a ρ !); or a C= + for L=0 (think of a σ). Observe how the total angular momenta match between left and right.
So ortho- goes to $\rho^0\pi^0$ and para- goes to $\sigma \pi^0$, in terms of placeholders for quantum numbers. There is no pretense of real dynamics, just as the baryonia on the l.h.s. are mere notional carriers of quantum numbers, not necessarily real resonant states.

However, in both cases, the total parity must be negative, and one unit of orbital angular momentum contributes a - sign to the three - signs of the three pions.
So, it appears you need a further orbital angular momentum of 1 between the ρ and the  π in the first case, which does not change the total angular momentum; the para case is fine as is and goes without a hitch.
