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Is it true that both linear momentum and energy cannot be conserved if the production of an electron-positron pair via photon annihilation were to occur in free space?

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I guess that depends on what you mean by free space. Yes that is true that a gamma ray in a perfect vacuum can not decay by pair production since it can not conserve both energy and momentum. It will undergo pair production however if near a nucleus. A nucleus will enable energy and momentum to be conserved.

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  • $\begingroup$ Thank you. You have confirmed that I, indeed was correct! $\endgroup$ – Douglas D. Beatenhead May 30 '17 at 22:26
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There are two ways that I could interpret this question. If you're interested in the interaction

$$ \gamma \to e^+e^- $$

taking place in free space, that's forbidden. The reason is that the electron-positron pair always has a rest frame where the momentum is zero, but the photon has no such rest frame.

If, on the other hand, you're asking about

$$ \gamma \gamma \to e^+ e^- $$

that is allowed. The matrix element is exactly the same as for the more familiar process $e^+e^-\to\gamma\gamma$, at least in the limit where time-reversal is a good symmetry of quantum electrodynamics. However the phase space for the $\gamma\gamma$ process is very different and it has never been observed with free photons. Much more common is an interaction between a hard (MeV+ energy) photon and the virtual photons that make up the electric field around atomic nuclei.

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  • $\begingroup$ I find your answer interesting. You're correct, but I used math and a few simple diagram to solve the problem. Thank you. $\endgroup$ – Douglas D. Beatenhead May 31 '17 at 19:01

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