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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?

Qmechanic
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Aaron
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2 Answers2

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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.

Darkseid
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Yes, with care. While $\Sigma$ is a loop correction, $$G_0^{-1} = p^2 - m_B^2 = \Gamma^{(2)}_{\text{tree}} $$ is the tree-level one containing the bare mass.

So, we have indeed. $$ \begin{split} \Gamma^{(2)}_{\text{loop-corrected}} & = G_c^{-1} \\ &= G_0^{-1} - \Sigma \\ &= p^2 - m_B^2 - \Sigma. \end{split} $$

I think this is also what Adam said. But this is not the end of the story. To extract observable, we need field-strength renormalization. So there must be additional correction from the renormalization.

user112002
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