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Consider the question whether a photon can decay, in the free space, into an electron and a positron - a pair production event. Such an event I don't think is possible, because both energy and momentum cannot be conserved (please correct me if I am wrong). When trying to show this I worked totally in the center of mass frame of the decay products. We then get:

Conservation of energy: $$E_\gamma=E_{e^-}+E_{e^+}$$ and conservation of momentum: $${E_\gamma \over c}=0$$ I would then have concluded that since in the center of mass frame we have $E_{e^-}=E_{e^+}$ that $E_{e^-}=0$ and that this is not possible.

I, however, think that this analysis is wrong since it totally ignores the potential energy between the positron and the electron due to their opposite charges. If this was included we could indeed have $E_{e^-}=0$ and then a photon could 'decay' into a positron and an electron in empty space.

So which analysis is right? Also how would I incorporate potential energy into this situation?

Sofia
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Quantum spaghettification
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2 Answers2

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Your analyses is not wrong, the photon is a stable particle. The most immediate argument is that linear momentum cannot be conserved. In the center-of-mass frame of the produced pair, the total linear momentum is zero, while the photon should have a linear momentum since there is no frame in which the photon is at rest.

Sofia
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Sorry but your question has a little error. The photon cannot decay into a pair in empty space. It has to be a quantized spectrum of energy levels present for this process to happen. This quantized energy level configuration can be provided with the presence of an atom (where the threshold energy for the process is $2mc^{2}$) or an electron (where the threshold energy for the process is $4mc^{2}$). In the case of the presence of an atom the recoil energy of the atom is very little due to his large mass compared with the $e-p$ but in the case of the presence of an electron his recoil momentum cannot be neglected and in this case we speak of an "triplet production" because we have a $e-p$ pair and a recoil electron.

Tomas Libutti
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