Is there a way to calculate the photoelectric effect in QED via a Feynman diagram? The photoelectric effect is the historic origin of the quantum particle description of light. From it we learn that when light is shone onto a metal single photons interact with single electrons in the metal which are ejected if the absorbed energy is larger than the binding energy of the metal. 
The (free) process is:
$$e^-+\gamma\rightarrow e^-.$$
However, this process violates conservation of energy (all final states are real particles). Of course the reason the process occurs is because the electron is initially bound (not free), and some energy goes into releasing the electron from the metal potential.
The question is, is there anyway to do this calculation in QED, somehow incorporating the binding energy in the calculation?
 A: There is a  publication that calculates the photoelectric effect using Feynman diagrams.( Here is a pdf )
Here are two screen captures. 

this one reduced so as to capture the Feynman diagram.

Anybody interested should go to the link above.
A: The problem is that photoelectricity is a messy process involving multiple interactions.
The photon transfers energy to the metal by exciting an electron within it, however the excited electron rattles around in the metal lattice transferring energy to anything it collides with. The quantum efficiency of photoelectron ejection from a metal surface is in the order of $10^{-5}$ to $10^{-6}$, so in all but 0.001% to 0.0001% of cases the electron energy is transferred to lattice vibrations (heat!) and no photoelectron is produced. In a very small number of cases enough energy is transferred to another electron to eject it from the metal.
I don't know enough about QED to speak definitively, but I'd be very surprised if anything this complicated could be usefully described using QED.
