I have this generator $$ Z(J,\lambda)=\int D\varphi e^{ i\int \!d^4x\,\{ \frac{1}{2}[(\partial\varphi)^2-m^2\varphi^2 ]-\frac{\lambda}{4!} \varphi^4+J\varphi \} } .\tag{11} $$ Which I first expand in $J$ to get $$\begin{align} Z(J,\lambda)&=\int D\varphi e^{ i\int \!d^4x\,\{ \frac{1}{2}[(\partial\varphi)^2-m^2\varphi^2 ]-\frac{\lambda}{4!} \varphi^4\} } \left[1+\int d^4w\, iJ\varphi+ \frac{1}{2!}\left( \int d^4w\, iJ\varphi \right)^{\!2}+\dots\right] \\ &= \sum_{s=0}\frac{1}{s!}\int d^4x_1\dots d^4x_s J(x_1)\dots J(x_s) \int D\varphi e^{ i\int \!d^4x\,\{ \frac{1}{2}[(\partial\varphi)^2-m^2\varphi^2 ]-\frac{\lambda}{4!} \varphi^4\} } \end{align}\tag{13} $$ where $J$ and $\varphi$ are functions of the $w$ which I have specified as distinct $x_k$ in the second line. In the usual way, the Green's functions are $$G(x_1,\dots,x_s)= \frac{1}{Z(0,0)}\int D\varphi e^{ i\int \!d^4x\,\{ \frac{1}{2}[(\partial\varphi)^2-m^2\varphi^2 ]-\frac{\lambda}{4!} \varphi^4\} }\varphi(x_1)\dots\varphi(x_s) $$ so that I may write concisely $$ Z(J,\lambda)= Z(0,0)\sum_{s=0}\frac{1}{s!}\int d^4x_1\dots d^4x_s J(x_1)\dots J(x_s) \,G(x_1,\dots,x_s).$$ Now I can begin to form my question: I want to examine the $\mathcal{O}(\lambda)$ term in the 4-point Green's function $G(x_1,x_2,x_3,x_4)\equiv G^{(4)}$. First, I expand in $\lambda$ as $$G^{(4)}=\frac{1}{Z(0,0)}\int D\varphi e^{ i\int \!d^4x\,\{ \frac{1}{2}[(\partial\varphi)^2-m^2\varphi^2 ] }\varphi(x_1)\varphi(x_2)\varphi(x_3)\varphi(x_4) \left[ 1-\frac{i\lambda}{4!}\int d^4w \,\varphi^4 +\dots\right]\tag{15}$$ so that $$G^{(4)}_{\mathcal{O}(\lambda)}=\frac{1}{Z(0,0)}\frac{-i\lambda}{4!}\int d^4w\int D\varphi e^{ i\int \!d^4x\,\{ \frac{1}{2}[(\partial\varphi)^2-m^2\varphi^2 ] }\varphi(x_1)\varphi(x_2)\varphi(x_3)\varphi(x_4) \varphi^4(w) .\tag{16}$$ Now I will simplify this integral by Wick contraction and that brings me finally to my exact question: The source text I am using, Zee, says this is equal to $$G^{(4)}_{\mathcal{O}(\lambda)}=-i\lambda\int d^4w D(x_1-w)D(x_2-w)D(x_3-w)D(x_4-w).\tag{17} $$ I see how the integral over $D\varphi$ will produce $Z(0,0)$, and I see how all the contractions with the four identical $\varphi(w)$ will produce $4!$ for the term with $D(x_1-w)D(x_2-w)D(x_3-w)D(x_4-w) $, but when I do the Wick contraction, I get very many more Wick permutations such as, for example, terms with $D(x_1-x_2)$. Evidently, these other terms correspond to disconnected diagrams, and Zee is only talking about this connected diagram below. Here is my question: Why should these other Wick contracted terms not show up in $G^{(4)}_{\mathcal{O}(\lambda)}$? Is Zee just ignoring them because they don't describe the given diagram, or does the Green's function notation mean "only the connected parts."

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    $\begingroup$ Ah, you are correct and I wasn't paying attention! In this case I agree with you, Zee should also include disconnected diagrams in addition to the one you have pointed out. See the discussion around equation (6) in that same section in Zee: you really want to have $Z(0,\lambda)$ in the denominator if you want the disconnected diagrams to cancel out. See also equation (10) and the paragraph below it mentioning that there are disconnected pieces in $G^{(4)}$. Perhaps Zee is just implicitly not showing the disconnected parts of your expression. $\endgroup$ Commented Sep 23, 2020 at 21:10
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    $\begingroup$ @Seth Whitsitt: That seems like an answer. $\endgroup$
    – Qmechanic
    Commented Sep 23, 2020 at 21:36
  • $\begingroup$ @SethWhitsitt Could you do a full answer? Since each term in the sum of Wick permutations would have the same $Z(0,\lambda)$ in the bottom, I don't see how it could cancel only the disconnected parts. Regarding the part around Eq's (6,10), that's exactly what my question is getting at but it seems like $G$ has connected and disconnected parts without regard for the separation between $Z$ and $W$. I'm not so curious as to the principle of the thing, which I do understand in theory, as to the exact method by which the connected parts can be cancelled from $G^{(4)}$, even with $\lambda\neq0$. $\endgroup$ Commented Sep 23, 2020 at 22:42
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    $\begingroup$ I think even my previous comment contained some errors in regards to how the connected correlator looks, but I think I essentially agree that Zee is omitting disconnected terms and you are correct. I will write an answer when I get time to clarify how the connected case looks. $\endgroup$ Commented Sep 25, 2020 at 18:01


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