photons - In QED, why is the $e^- + e^+leftrightarrowgamma$ process forbidden on-shell?

QED has a vertex that couples a single photon to two fermions. This vertex describes the annihilation of an electron-positron pair into a photon. Why is this process forbidden for all three particles being on shell?

Answer

(I henceforth assume $c= \hbar=1$.) It is forbidden by the four-momentum conservation law. Put yourself in the centre of mass reference frame of the couple of massive particles (electron and positron). There $P_{e\overline{e}} = (2E,\vec{0})$ with $E\geq m_e>0$. Just because four momentum is conserved, this four-momentum must be the same as the one of the photon: $P_\nu=(k,\vec{k})$. So $k=2E$ and $\vec{k}=\vec{0}$. Since, for the photon $k= |\vec{k}|$, the two conditions cannot hold simultaneously and four-momentum conservation is violated. At least one of the three particle must be virtual.