# Amputated connected 2-point function is inverse to connected 2-point function

Let $$D_n$$ denote the $$n$$-point correlation function consisting of only connected diagrams. We may decompose this as an integral of two products. The first factor consists of a product over the $$n$$ external legs and their corrections, each denoted $$E^{(n)}_i$$, so this factor becomes $$\prod_{i=1}^n E^{(n)}_i$$. The second factor is the amputated diagrams which I will denote as $$A_n$$. In other words, $$A_n$$ are the diagrams that remain once the external legs and their corrections have been removed. Thus $$D_n = \int \Big(\prod_{i=1} E^{(n)}_i\Big) A_n$$ where the integral is taken over all internal points.

With the above setup, I have read that the amputated 2-point correlation function is the inverse of the 2-point correlation function, i.e. $$\int A_2 D_2 = \delta$$ but I can not see where this follows from.

Is there anyway to see why the amputated 2-point correlation function is the inverse to the 2-point (connected) correlation function without using effective actions?

• Which lectures? Which page? Feb 29 at 4:46
• @Qmechanic These are lectures that have been shared with me so unfortunately they're not available online. Feb 29 at 4:57

TL;DR: Use the relations $$D_2~=~\int E_1A_2E_2\tag{1}$$ and $$E_1~=~D_2~=~E_2\tag{2}$$ to conclude OP's sought-for relation $$A_2~=~D_2^{-1}.\tag{3}$$
1. Either we assume that the propagator $$D_2$$ is a matrix of all field species in the theory, and hence unique.
2. More commonly, we assume [or can prove e.g. via conservation laws] that the propagator $$D_2$$ is diagonal in the field species [so that there in principle are eqs. (1)-(3) for each field species].
• For the first definition of $D_2$ you used, why can we omit the integrals and shouldn't we have $D_2 = E_1A_2E_2$? How do we know that $E_1 = E_2$? Feb 29 at 4:56
• Thank you! I've been trying to better understand Eq. (2). Wouldn't the external legs $E_1$ or $E_2$ need to start at an external point and end at an internal point? If so, how could they be equal to $D_2$ which contains both external points? Feb 29 at 15:55