Tadpole diagram in QED

Question

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 $e barpsi gamma^mu psi A_mu$, 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 $barpsi$ 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.)

Noiralef · Accepted 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.

Jan Polane · Answer

Because of momentum conservation at the vertex the photon has four momentum zero. So there cannot be a photon at all.

Tadpole diagram in QED

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