# 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?

• Why would the process violate energy conservation? You just said yourself where the energy goes. Can QED analyze an electron that is bound weakly to a very large crystal lattice? Why not? arxiv.org/abs/0910.1809. In scribd.com/doc/148001167/… Lamb and Scully argue that one does not have to quantize the radiation field, at all, to explain the photoelectric effect, and they give a reference to a paper of theirs in which they do. This sounds a bit like a case of insufficient effort. Did you even google this? – CuriousOne Aug 28 '14 at 18:39
• Consider the simpler case of the photo-ionization of a hydrogen atom - what is known in astrophysics as a "bound-free transition". It looks just like bremsstrahlung in reverse, except that the electron is bound to start with. How this is handled in QED I couldn't say, but it certainly can be. – akrasia Aug 29 '14 at 22:46
• @CuriousOne, before asking I did find 0910.1809. As the authors state, "the photo electric effect in the early experiments is produced by weak, non-coherent radiation of high frequency... whereas [the subject of the present paper] the radiation is weak, of high frequency, and coherent." Firstly, I'm interested in the photon description (not coherent radiation). Secondly, they use non-relativisitc QED, not fully covariant QED. And lastly, what they do looks highly rigorous and mathematical, whereas you can see from the title I'm looking for a Feynman diagram type explanation. – Jase Uknow Aug 29 '14 at 23:04
• @CuriousOne, the second paper you mention is off topic. I'm not asking how to solve the problem, I'm asking how to solve the problem in QED. – Jase Uknow Aug 29 '14 at 23:06
• @akrasia, this is exactly the kind of thing that I was thinking. And I agree the model would need to be something simple like an atom. Actually, I was thinking of something like Mott scattering where the $otimes$ is inserted to handle elastic scattering. But in the photoelectric effect, is is inelastic, so not sure how to proceed. – Jase Uknow Aug 29 '14 at 23:13

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.

• I have found again the quotes, the original link got broken. It would be good if people who downvote leave a comment. I would have contacted him/her with the correction :( – anna v Feb 4 '18 at 5:38

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.

• Well, how about we change from "metal" to "semiconductor" and just use QED to calculate the initial energy of the electron at ejection? – Carl Witthoft Aug 28 '14 at 11:27
• @Carl, I don't mind if metal or semiconductor. If the calculation is easier with a semiconductor because of the problems with the metal mentioned by John, then please provide that calculation as an answer. I'm still not sure how to "use QED" in this case. – Jase Uknow Aug 28 '14 at 12:40
• The photoelectric effect can be made to have a quantum efficiency of nearly one these days. Your digital camera wouldn't work otherwise, neither would solar panels. And please check the quantum efficiency of real world photomultiplier tubes. – CuriousOne Aug 28 '14 at 18:28
• @CuriousOne: the question asks light is shone onto a metal and the quantum efficiency for photoelectron ejection from a metal is in the range I gave. Neither CCDs nor solar panels eject electrons, they work by exciting electrons between bands. This happens in a metal too, but that's just the first step in a process leading to ejection of an electron from the metal surface. – John Rennie Aug 28 '14 at 19:14
• @JohnRennie: If you insist, PMTs do emit electrons and they have quantum efficiencies near one, just as well. It does not seem important to me to make a distinction between the photoelectric effect and the inner photoelectric effect within the question of the OP. – CuriousOne Aug 28 '14 at 19:18