I am learning QFT on my own via Mark Srednicki's book, and I have a bit of trouble following the author in chapter 9, where the path integral for an interacting theory is introduced, as well as Feynman diagrams. The theory considered is as follows: $$ \mathcal{L} = -\frac{1}{2} Z_\phi \partial^\mu\phi\partial_\mu\phi - \frac{1}{2} Z_m m^2 \phi^2 + \frac{1}{6} Z_g g \phi^3 + Y\phi.\tag{9.1} $$ If we want to use the free theory results and Feynman propagators, then it is useful to identify the following Lagrangian: $$ \mathcal{L} = \mathcal{L_0} + \mathcal{L_1} $$ where $$ \begin{align} \mathcal{L}_0 &= -\frac{1}{2}\partial^\mu\phi\partial_\mu\phi - \frac{1}{2}m^2\phi^2,\tag{9.8} \\\\ \mathcal{L}_1 &= \frac{1}{6}Z_g g \phi^3 - \frac{1}{2}(Z_\phi - 1)\partial^\mu\phi\partial_\mu\phi - \frac{1}{2}(Z_m - 1)m^2\phi^2 + Y\phi\\ &= \frac{1}{6}Z_g g \phi^3 + \mathcal{L}_\text{ct}. \tag{9.9}\end{align} $$ With this, the author proposes to calculate the following path integral: $$ Z[J] = \int [\mathcal{D}\phi] \space e^{i\int d^4 x \space (\mathcal{L}_0 + \mathcal{L}_1 + J\phi)}\tag{9.5} $$ and says that the counterterms (i.e. $\mathcal{L}_\text{ct}$) can be ignored ("for now"). After some works which I think I understand, the author comes to the following expression: $$ Z_1[J] \propto \sum_{V = 0}^\infty \frac{1}{V!}\left[\frac{i}{6} Z_g g \int \text d^4 x \space \left(-i\frac{\delta}{\delta J(x)}\right)^3 \right]^V \times \sum_{P = 0}^\infty\frac{1}{P!}\left[\frac{i}{2}\int \text d^4 x \text d^4x' \space J(x)\Delta(x-x')J(x') \right]^P.\tag{9.11} $$ If I understand correctly, this expression (roughly) corresponds to the starting path integral but where the counterterms have been "ignored" (for some reason I don't know yet but I think this is the point of the author for the rest of the chapter), hence the name "$Z_1$" and not "$Z$". The author then introduces Feynman diagrams as a way of representing the different ways of expressing $Z_1[J]$. Feynman diagrams (and the way they arise from the latter formula) aren't super-clear to me yet, and I hope that will come clearer with more practice, but I can't get anywhere from what's follow; after a few paragraphs, the author says this:
Were it not for the counterterms in $\mathcal{L}_1$ , we would have $Z[J] = Z_1 [J]$. Let us see what we would get if this was, in fact, the case. In particular, let us compute the vacuum expectation value of the field $\phi(x)$, which is given by $$ \langle 0 | \phi(x) | 0 \rangle = \cdots = \frac{\delta}{\delta J(x)} W_1(J) \space |_{J=0}. \tag{9.17} $$ This expression is then the sum of all diagrams that have a single source, with the source removed: $$ \langle 0 | \phi(x) | 0 \rangle = \frac{i g}{2} \int \text d^4 x \space \frac{1}{i} \Delta(x - y) \frac{1}{i} \Delta(y - y) + \mathcal{O}(g^3) \tag{9.18} $$ where we assumed that $Z_g = 1 + \mathcal{O}(g^2) \approx 1$.
The expression for $W_1(J)$ is given by the sum of all (non-vacuum) connected diagrams,
$$
W_1(J) = -i\sum_{I \neq \{0\}} C_I.\tag{9.16}
$$
From the author description of what is a Feynman Diagram, I can recognize that $(9.18)$ has an associated diagram of the following form: ─◯ (i.e. a propagator connected to a "self-connected" propagator (I don't know what else to call it at the moment, I think it's called a bubble but the book hasn't yet introduced a nomenclature for it at this stage)).
Now, my issue is that I don't understand how the whole is linked. I don't really understand how does the author comes from $(9.17)$ to $(9.18)$ and, since $(9.17)$ deals with diagrams (because of $W_1$), from $(9.17)$ to Feynman diagrams. I can only roughly understand how one draw a diagram from $(9.18)$ because Srednicki explicitly stated that "a line segment stands for a propagator $-i\Delta(x-y)$", so one could indeed recognize that $(9.18)$ must associate a diagram with two propagators, one being a line and the other begin a loop.
My question is then: how does the author derive $(9.18)$ from $(9.17)$ and how can this derivation be interpreted in terms of Feynman diagrams, given that $(9.17)$ implies them?
Edit -- more details
In fact, I think understand how $(9.18)$ can be obtained from the first equality in $(9.17)$, so I'll do the derivation. $(9.17)$'s first equality gives $$ \langle 0 | \phi(x) | 0 \rangle = \frac{1}{i} \frac{\delta}{\delta J(x)} Z_1[J] \Bigg|_{J=0}.\tag{9.17} $$ Now, explicitly, $$ \begin{align} \langle 0 | \phi(x) | 0 \rangle &= \frac{1}{i}\frac{\delta}{\delta J(x)} \left\{ \exp\left[\frac{i}{6} Z_g g \int \text d^4 y \space \left(-i\frac{\delta}{\delta J(y)}\right)^3\right] Z_0[J] \right\} \Bigg|_{J = 0} \\ &= \frac{1}{i}\left[\frac{i g}{6}\int\text d^4 y \space \left(-i\frac{\delta}{\delta J(y)}\right)^3 \frac{\delta}{\delta J(x)} Z_0[J]\right] \Bigg|_{J = 0} \times \exp(0) \end{align} $$ where we considered $Z_g \approx 1$ and where $$ Z_0[J] = \exp\left[\frac{i}{2} \int\text d^4 z\space\text d^4 z'\space J(z) \Delta(z - z') J(z')\right].\tag{9.7} $$ Earlier in the chapter, the author mentions Wick's theorem. Wick rule says that the functional derivative of $Z_0$ w.r.t. $J$ is given by the sum of all the ways there are to contract the derivative by pairs, and then by evaluate to $J=0$ this will produce null terms as well as propagators. Each contraction between two points $x$ and $y$ gives a factor $$ \frac{\delta}{\delta J(x)}\frac{\delta}{\delta J(y)} Z_0[J] \Bigg|_{J=0} = \frac{1}{i}\Delta(x - y). $$ In our case, we have four functional derivatives and the Wick rule indicate that we have to sum over each ways there are to contract these four derivatives by pairs. The only contractions that give non-zero factors (when evaluate at $J=0$) and/or do not cancel each other out are in our case
- one contraction between $\delta/\delta J(x)$ and one of the three $\delta/\delta J(y)$, this gives a numerical factor of $3$, because we have three choices in the derivatives, and a propagator $-i \Delta(x - y)$ ;
- the contraction between the two remaining $\delta/\delta J(y)$ which gives a propagator $-i\Delta(y - y)$.
So, in the first order in $g$, one finally has $$ \begin{align} \langle 0 | \phi(x) | 0 \rangle &= \frac{i g}{6} \int\text d^4 y\space\frac{3}{i}\Delta(x - y)\frac{1}{i}\Delta(y - y) + \mathcal{O}(g^3) \\ &= \frac{1}{2} i g \int\text d^4 y\space\frac{1}{i}\Delta(x - y)\frac{1}{i}\Delta(y - y) + \mathcal{O}(g^3) \end{align} $$ and this is Srednicki's result. Now, where I lose the author a bit is with the $W_1$ and the diagrams. The author gives that $$ W_1(J) = \frac{1}{i} \sum_{I \neq \{0\}} C_I\tag{9.16} $$ and the same calculation starting instead from $W_1$ without going through $Z_1$ as I did giving the same result is not that clear for me. How do we start only from $\langle 0 | \phi(x) | 0 \rangle$ (or $W_1$ but I think this is the same deal) to get to the diagrammatic result?
