Assume that we have a QFT with one scalar field $\phi$ with mass $m$ and the Lagrangian $$\begin{aligned} \mathcal{L}_{\mathrm{EFT}, \mathrm{off}}=& \frac{1}{2}\left(\partial_{\mu} \phi\right)^{2}-\frac{1}{2} m^{2} \phi^{2} \\ &-\frac{C_{4}}{4 !} \phi^{4}-\frac{C_{6}}{6 ! M^{2}} \phi^{6}-\frac{\tilde{C}_{6}}{4 ! M^{2}} \phi^{3} \square \phi-\frac{\hat{C}_{6}}{2 M^{2}}(\square \phi)^{2} \end{aligned}.$$ The propagator for $\phi$ in momentum space will then be something like $$\frac{i}{p^2-m^2 + i\epsilon}.$$

The Feynman rule for this propagator is usually represented by a straight line.

In some lecture notes (that I'm unfortunately not allowed to share here) we consider all $1PI$ diagrams at tree-level which contribute to the two-point function, i.e. only one diagram, a straight line. The amplitude of this diagram is written down as $$\mathcal{M}_2 = i (p^2-m^2).$$


I don't understand why the propagator and amplitude don't coincide. I mean, just looking at the units these two things don't seem to be related, but we still use the same description in terms of Feynman diagrams, which seems weird. Is there a connection? How can I see it?


1 Answer 1


The main point is that the 2-pt functions for the generator $W_c[J]$ of connected diagrams and the generator $\Gamma[\phi_{\rm cl}]$ of 1PI diagrams are each other's inverse (up to factors of $i$), cf. e.g. this & this related Phys.SE posts.

In particular note that for a 1PI diagram the external legs are stripped/amputated. In this process, the free propagator/2pt-function then turns into its own inverse.


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