In $\phi^4$ theory with the $\lambda \phi^4 / 4!$ interaction term gives rise to “tadpole” diagrams like this:


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?


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.)


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 QED tadpole diagram 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.

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    $\begingroup$ I see the point with the odd integrand now. Perhaps the authors and professors have asked the same thing themselves and after finding the solution discarded all those diagrams directly. That could explain why I did not find anything while reading Peskin & Schroeder's “Intro to QFT”. $\endgroup$ – Martin Ueding Nov 22 '15 at 8:56

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