# Feynman diagram for photon absorption

What is the simplest Feynman diagram for photon absorption by an atom?

Is it described by an incoming photon and bound atomic electron interacting at one vertex with an outgoing virtual electron which then, at a second vertex, emits a virtual photon and a real bound excited electron. The virtual photon is finally absorbed by the nucleus of the absorbing atom at the 3rd vertex.

What are the dimensionless factors involved in the cross-section for the process. If my 3 vertex diagram is correct then I guess the fine structure constant cubed will be in the mix but are there any other numeric constants?

Some background to the question: I'm trying to determine the cross-sections for photon absorption of Hydrogen Balmer lines, specifically I'm trying to link these cross-sections to the Einstein absorption/emission rate coefficients. I've published this work, which is still being developed, on my website. If I assume a 3 vertex diagram and hence a $4\pi \alpha^3$ dimensionless constant then I obtain a good estimate for the thickness of the Sun's photosphere.

Thanks

• I recall this subject coming up in the chat room recently, though I've searched for it without any success. I'm sure the conclusion was that QFT isn't good at describing bound states and there is no simple Feynman diagram to describe the excitation of an electron in an atom. However I don't recall the details so I'm unwilling to commit to an answer. – John Rennie Jul 8 '16 at 15:39
• Feynman diagrams are just a notation for the terms of the perturbative expansion of an amplitude $A$. So you have to ask yourself two questions: what is the unperturbed Hamiltonian $H_0$ you start from? What is the perturbation $H_1$ you add to $H_0$, which will become the variable of your perturbative series? – user154997 Oct 11 '17 at 14:41

I think it is just a one vertex diagram with $A$ at the initial state and $A^*$ at the final state, where $A$ and $A^*$ denote the atom in the ground and an excited state respectively. If you look at the QM calculation of such an absorption, you will see what should be ascribed to the vertex. Say, it may be a dipole interaction of a bound electron, but with conservation of the total momentum of the "photon + Atom" system. It is not a QFT problem, but that of QM.
The full atomic wave function $\Psi({\bf{r}}_{\rm{e}},{\bf{r}}_{\rm{N}})\propto {\rm{e}}^{ {\rm{i}}{\bf{P}}_{\rm{A}}{{{\bf{R}}_{\rm{A}}}}}\psi({\bf{r}})$ takes already into account the binding potential exactly - no vertices are necessary for describing it as you are not going to "build" it perturbatively.