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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?

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

1 Answer 1

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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 $$A_2~=~D_2^{-1}.\tag{3}$$

Eq. (2) can be argued in at least 2 ways:

  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].

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  • $\begingroup$ 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 $E_1 = E_2$? $\endgroup$
    – CBBAM
    Feb 29 at 4:56
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Feb 29 at 7:00
  • $\begingroup$ 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? $\endgroup$
    – CBBAM
    Feb 29 at 15:55

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