I am trying to understand the difference between the 2-point vertex function and the self energy. In many presentations, they are described in ways that seem nearly equivalent, yet as I work through the details I run into seemingly nonsensical statements.

To start with definitions, we know that the 2-point vertex $\Gamma^{(2)}$ is characterized by $\Gamma^{(2)} = G_c^{-1}$, the inverse of the 2-point connected Green's function. The self-energy $\Sigma$ is the sum of all 1-particle irreducible (1PI) diagrams of the two-point function after amputating the external legs.

It is also said that $\Gamma^{(n)}$, the $n$-point vertex, is said to be the sum of all amputated 1PI diagrams with $n$ external legs. This statement is not a definition, but a derived consequence. However, this statement seems strange to me in the case $n=2$, in which case it would be equivalent to saying $\Gamma^{(2)} = \Sigma$.

We know that $G_c^{-1} = G_0^{-1} - \Sigma$, where $G_0$ is the free propagator. If $\Gamma^{(2)} = \Sigma$, then we would arrive at the nonsensical statement $2\Gamma^{(2)} = G_0^{-1}$. This seems to imply that $\Gamma^{(2)}$ should not be thought of as the sum of all the 1PI diagrams for the amputated 2-point correlator.

Am I missing something here, or are textbooks being imprecise in the statement that $\Gamma^{(n)}$ is the sum of all amputated 1PI diagrams with $n$ external legs?

  • $\begingroup$ Related: physics.stackexchange.com/q/193419/2451 $\endgroup$
    – Qmechanic
    Commented Oct 3, 2017 at 18:07
  • $\begingroup$ One should not forget that the effective action (when obtained perturbatively) contains both the bare action and the diagrams. The diagrams are then 1PI due to the construction of $\Gamma$, but there is also a contribution (usually for small $n\leq 4$) from the bare action. $\endgroup$
    – Adam
    Commented Jul 1, 2018 at 8:19

1 Answer 1


It really looks like there is missing $n > 2$ for that definition. Clearly 1-PI diagrams for n-point functions with $n > 2$ are what the vertex function is, because none of them can be amputated if stacked together. On other hand, with 2-point function one can always stack 1-PI diagrams by connecting them in such way that all but one of them can be amputated.

  • $\begingroup$ Sorry to comment on an older post but I am having the same problem as the original poster where I don't see what's special about $\Gamma^{(2)}$. What do you mean by "stacking" 1-PI diagrams? $\endgroup$
    – CBBAM
    Commented Mar 29 at 6:13

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