# 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.

• 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.
• 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