Does the connected Green's function's decomposition into 1PIs have connected contributions, or can it be written exclusively using 1PIs?

Question

1. While reading this article by Abbot on the background field method, in Fig 5. on page 38 (page 6 in the pdf file), we can see the relation between connected contributions to the two point function and the 1PI contributions

   

   Of course, we can recursively insert this expression for connected diagrams to obtain a chain

   C-1PI-C-1PI-C-1PI-C-1PI-C-1PI-C-1PI-C-1PI-C-1PI-C-1PI-C

   which can be extended as much as we want.

2. Nonetheless, while reading this notes of Matthew Schwartz, in eq. 8 on page 3 we can see

   

   where only 1PI-s are considered.

Are both this expressions equivalent? if so why?

Answer 1

As Chris Drost mentions in his answer, the difference between the figures is caused by whether the various$^1$ definitions include or exclude the bare/free propagator $b$ [cf. e.g. the last sentence in the first paragraph on page 3 of the second reference].

1. The first figure (fig. 5) $$ C~=~CAC \tag{5}$$ states that the 1PI 2-pt fct (amputated, incl. bare/free contribution) $$ A~\equiv~ b^{-1}-a~=~C^{-1}$$ is the inverse of the connected 2-pt fct $C$ (non-amputated, incl. bare/free contribution) [which is the full propagator].

   Note that $a$ here is defined as minus the 1PI 2-pt fct (amputated, excl. bare/free contribution), i.e. the self-energy $\Sigma$ (up to conventional factors of $i$).

2. The second figure (eq. 8) states a formula for the connected 2-pt fct (non-amputated, excl. bare/free contribution) $$\begin{align}c~\equiv~& C-b~\stackrel{(5)}{=}~\frac{1}{b^{-1}-a}-b\cr ~=~&~b\frac{1}{1-ab}-b~=~b\frac{ab}{1-ab}\cr ~=~&b\sum_{n=1}^{\infty}(ab)^n.\end{align} \tag{8}$$

--

$^1$ Both authors implicitly assumes that there are no tadpoles so that the self-energy is 1PI, cf. my Phys.SE answer here.

Answer 2

They are subtly different. First: the first document seems to be happy to assume that something could be in the C part that is not in the 1PI; the second document seems to be trying to reduce everything that happens to the 1PI. Second: both documents seem a little sketchy about whether they include the trivial contribution (-->--).

You can definitely do recursive diagrams, though. To do the second diagram sum as a finite recursive diagram, you can simply write:

(-->--[ C ]-->--)  =  (-->--[ 1PI ]-->--)  +  (-->--[ 1PI ]-->--[ C ]-->--) 

and then we find that indeed C is all-order in 1PI:

(-->--[ C ]-->--)  =  (-->--[ 1PI ]-->--) + (-->--[ 1PI ]-->--[ 1PI ]-->--) + ...

But often the recursive diagram leads to a finite form for C whereas the latter is just some infinite series that gets arbitrarily truncated.