In $\phi^4$ theory with the $\lambda \phi^4 / 4!$ interaction term gives rise to “tadpole” diagrams like this:
http://chaos.stw-bonn.de/users/mu/uploads/2015-11-21/phi4.png
If I have a standard QED interaction with $e \bar\psi \gamma^\mu \psi A_\mu$, can I also have tadpole diagrams like this one?
http://chaos.stw-bonn.de/users/mu/uploads/2015-11-21/qed.png
In the $\phi^4$ case the propagator connects to the same vertex twice and therefore $\phi^2$ is used up. In the QED case I am not sure if the $\bar\psi$ and $\psi$ can connect to the same vertex and propagators. I have not seen any of these one-photon fermion loops in my QFT class. Since it is possible in $\phi^4$, I was wondering if the same is possible in QED.
(The diagrams are made with tikz-feynman.)
Best Answer
First of all, these diagrams are no tadpole diagrams. A tadpole diagram is a diagram with exactly one external leg.
Nevertheless, the QED diagram exists, of course. When you calculate it, you need to "connect" the electron propagator $S_F = \frac{i (\gamma \cdot p + m)}{p^2 - m^2}$ corresponding to the loop from both sides with the $\gamma^\mu$ from the vertex. That gives $( S_F )^A_B\, ( \gamma^\mu )^B_A = tr(S_F \gamma^\mu)$ (where $A$ and $B$ are fermion indices).
When you actually write down the complete expression for the tadpole
you will see that it gives zero, because there is an integral over $p$ and the integrand is an odd function of $p$.
Therefore also all diagrams including this tadpole (like yours) are zero.