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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
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Nuclearforce
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  • Related: http://physics.stackexchange.com/q/13513/2451 , http://physics.stackexchange.com/q/22916/2451 , http://physics.stackexchange.com/q/12488/2451 – Qmechanic Oct 22 '14 at 12:36

1 Answers1

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(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.

Valter Moretti
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  • Can one also use Furry's theorem here to show that the amplitude is zero? – Physics_maths 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