# Feynman diagram for pair production in matter

I'm aware that electron pair-production from a single photon requires the presence of matter — say some large nucleus — able to absorb momentum, as in the process \begin{align} N \gamma \rightarrow N e^- e^+ \end{align} where $N$ represents the nucleus. My question is, how would one represent this process with a Feynman diagram? I'm having trouble constructing one with valid vertices that makes sense. Would the neutron interact with the photon? With the electron-positron pair? Both? Any help would be appreciated.

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The spectator is sometimes drawn off to one side with a dashed line (or dashed double line) connecting to a circle that contains the pair production. However, that is not a diagram you can calculate. If you need to calculate the diagram go with Ross' suggestion. – dmckee Mar 5 '14 at 22:29

You have a photon that comes in and splits into $e^-e^+$. One of the electrons escapes, the other has a short segment, where it sheds another photon. The second electron now escapes and the photon is absorbed by the nucleus. The center electron segment and the second photon can be off the mass shell.

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So then there is indeed a vertex with a single photon on one end and and an electron positron-pair on the other. How can this vertex simultaneously satisfy mass-energy conservation and momentum conservation? That's the whole point of having the nucleus present, I thought – mikefallopian Mar 5 '14 at 22:42
It satisfies energy and momentum conservation, but the electron that leaves and interacts with the nucleus need not (and will not) be on the mass shell. That gets rectified by the interaction with the nucleus. – Ross Millikan Mar 5 '14 at 22:44
The internal line (from pair production to interaction with the spectator) is off-shell; it doesn't have to have the right mass. – dmckee Mar 5 '14 at 22:45
Can someone explain this concept (or link me to a good discussion) of being "on/off the mass shell" to me? I'm only just learning particle physics and QFT and I've never heard of it prior to now. I imagine it has something to do with finite-time violations of mass-energy conservation via the uncertainty principle? – mikefallopian Mar 5 '14 at 22:53
A particle in a Feynmann diagram that does not go off to infinity need not satisfy $E^2=m^2+p^2$ As it lives a short time, it is covered by the uncertainty principle. – Ross Millikan Mar 5 '14 at 23:02