Factorization into disconnected Feynman diagrams I am studying "Quantum theory of many particle systems" by Fetter and Walecka, and just had my first encounter with Feynman diagrams. They are introduced to represent the terms in the first order terms in the perturbative expansion of the numerator of the interacting Green's function, given as

Here all coordinates are four vectors of position and time. I understand how to obtain this expression via Wicks theorem. The book explains that all terms may be represented via the following set of Feynman diagrams, labelled correspondently with the terms in equation (9.1),

I understand how these diagrams follow. However the book then makes the point that since terms A and B contain disconnected diagrams, they may be factored out of the perturbative expansion, and thus equation (9.1) may be just as well represented as,

At this point I am kind of lost. I assume that this should be interpreted as just distributive multiplication of diagrams, in which case we have suddenly obtained 15 different terms? But equation (9.1) and fig 9.1 contained just 6 terms. I realise figure 9.2 also includes the zeroth order term but where do all the extra terms come from? It seems to me like we are adding the non-connected diagrams to all terms, but this is not what I see in the original expansion. Can anyone explain what I am missing here?
 A: *

*The partition function
$$ Z[J,\bar{J}]~=~e^{\frac{i}{\hbar}W_c[J,\bar{J}]} \tag{A} $$
consists of all diagrams and is the exponential of the connected diagrams, cf. the linked cluster theorem.


*In particular, all vacuum bubbles
$$ Z[0,0]~\stackrel{(A)}{=}~e^{\frac{i}{\hbar}W_c[0,0]} \tag{B}$$
are the exponential of the connected vacuum bubbles.


*The numerator of the 2-point function is
$$\begin{align} \left.  \frac{\delta^2 Z[J,\bar{J}]}{\delta J_k\delta \bar{J}_{\ell}} \right|_{J=0=\bar{J}}
~\stackrel{(A)}{=}~&\left.  \frac{\delta^2 e^{\frac{i}{\hbar}W_c[J,\bar{J}]}}{\delta J_k\delta \bar{J}_{\ell}} \right|_{J=0=\bar{J}}\cr
~=~&\frac{i}{\hbar} \left.\frac{\delta^2 W_c[J,\bar{J}]}{\delta J_k\delta \bar{J}_{\ell}}\right|_{J=0=\bar{J}} 
e^{\frac{i}{\hbar}W_c[0,0]},\end{align} \tag{C} $$
i.e. it factorizes into the connected 2-point function times the vacuum bubbles, cf. OP's title question. In eq. (C) we have assumed that the 1-pt functions vanish. (In OP's case this follows because a fermion line cannot end/start.)


*In particular, the 2-point function
$$\begin{align}
\langle \phi^k \phi^{\ell}\rangle_{J=0=\bar{J}} ~:=~&\frac{1}{Z[0,0]}\left.  \frac{\delta^2 Z[J,\bar{J}]}{\delta J_k\delta \bar{J}_{\ell}} \right|_{J=0=\bar{J}}\cr
~\stackrel{(B)+(C)}{=}&\frac{i}{\hbar} \left.\frac{\delta^2 W_c[J,\bar{J}]}{\delta J_k\delta \bar{J}_{\ell}}\right|_{J=0=\bar{J}}\cr ~=:~& \langle \phi^k \phi^{\ell}\rangle^c_{J=0=\bar{J}}.\end{align} \tag{D} $$
becomes the connected 2-point function.
