# In QED, why is the $e^- + e^+\leftrightarrow\gamma$ 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?

• – Qmechanic Oct 22 '14 at 12:36

(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.

• Can one also use Furry's theorem here to show that the amplitude is zero? – Your Majesty Oct 22 '14 at 10:54
• No, Love Learning, one can't. Furry's theorem is only about the expectation of the product of currents - and one needs both creation of a current (the electron and the positron Dirac operator) as well as a photon. The "dynamical part" of the amplitude behind this forbidden process doesn't really vanish; it is purely the kinematical part - one that only depends on the momenta - that vanishes for reasons that Valter described. – Luboš Motl Oct 22 '14 at 11:07