Understanding how to derive Feynman Rules for the 2-point correlation function ($\phi^3$ theory)

Question

This is a follow-up question (first one here). I am aimed to work out $V(\phi) = -\lambda \frac{\phi^3}{3!}$ theory in detail to understand how Feynman rules can be derived from the functional integral

\begin{equation*} Z[J] := \int d[\phi] e^{iS[\phi] + i\int d^d x J(x) \phi(x)} \tag{1} \end{equation*}

I performed a completely analogous computation to that of Jeanbaptiste Roux, up to second order in perturbation theory, and obtained

\begin{align*} &\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right)\times \exp\left(i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0} \\ &=1+\frac{i}{2}\times 0+\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y)) \\ &+ \frac{(i)^2}{4}\int d^d x \int d^d y \int d^d t \int d^d \xi\,(iJ(x))\Delta_F(x-y)(iJ(y)) (iJ(t))\Delta_F(t-\xi)(iJ(\xi)) + \mathcal{O}(\Delta_F^3) \end{align*}

So it seems to me that the perturbative expansion is determined by the number of propagators!

So, up to first order, we encounter one propagator and $2$ external legs $iJ$

Up to second order we encounter two propagators and $4$ external legs $iJ$

I've just learned that the source $J$ is represented by a vertex with one outgoing line i.e.

With this information, I was trying to understand why, for instance, the 2-point correlation function $\langle \phi(x_1) \phi(x_2) \rangle$ has the following contributions up to second order

Why should we draw a bubble like shown above? If I am not mistaken, we should have two external legs due to having two propagators and $4$ vertices due to having $4$ sources. However, I only see two vertices here... what am I missing?

PS: Please note this is not a homework question. I am studying Osborn notes, section 2.2. Interacting Scalar Field Theories, and I want to understand how he constructed the Feynman rules (page 23) via working out the simplest example I could find: $\phi^3$ theory

EDIT 0

Let me go slowly here. I will only focus on second order terms. As stated in the comments, we should get $\propto \int \mathrm{d}^4 z \int \mathrm{d}^4 w J(x) \Delta_F(x-z) \Delta_F^2(z-w) \Delta_F(w-y) J(y)$

This is what I have done so far

\begin{align*} &\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right) \exp\left(i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0} \\ &=\left[\cdots +\frac{(i)^2}{4}\int d^{2d} z \int d^{2d} w \left(\frac{\delta}{\delta \phi(z)}\right)^2\Delta_F^2(z-w) \left(\frac{\delta}{\delta \phi(w)}\right)^2+\cdots \right]\left.e^{i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \\ &=\left[1+\frac{(i)^2}{4}\int d^{2d} z\int d^{2d} w \left(\frac{\delta}{\delta\phi(z)}\right)^2\Delta_F^2(z-w)\left(\frac{\delta}{\delta\phi(w)}\right)\left( -\frac{\lambda}{2}i\phi^2(w)+iJ(w)\right) +\cdots\right] \\ &\times \left.e^{i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \end{align*}

Is this OK so far?

Answer 1

So, like it can be seen in the comments, I will perform the calculations for the second term of the expansion of the exponential containing the functional derivatives. I will denote $\phi(x)$ by $\phi_x$ and $\frac{\delta}{\delta \phi(x)}$ by $\delta_{\phi_x}$. \begin{align*} &-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\delta_{\phi_x}\delta_{\phi_y}\delta_{\phi_z}\delta_{\phi_w}\left.e^{i \int d^d \xi(-\frac{\lambda}{3!}\phi^3+J\phi)}\right|_{\phi=0} \\ &=-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\delta_{\phi_x}\delta_{\phi_y}\delta_{\phi_z}\left[i\left( -\frac{\lambda}{2}\phi^2_w+J_w \right) \right] \\ &\hphantom{==}\times\left.e^{i \int d^d \xi(-\frac{\lambda}{3!}\phi^3+J\phi)}\right|_{\phi=0} \\ &=-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\delta_{\phi_x}\delta_{\phi_y}\left[\vphantom{\frac{1}{2}}i\left( -\lambda \phi_w \delta_{w,z} \right) \right. \\ &\hphantom{==}\left. - \left(-\frac{\lambda}{2} \phi^2_w +J_w \right)\left( -\frac{\lambda}{2} \phi^2_z +J_z \right) \right]\left.e^{i \int d^d \xi(-\frac{\lambda}{3!}\phi^3+J\phi)}\right|_{\phi=0} \\ &=-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\delta_{\phi_x}\left[\vphantom{\frac{1}{2}} -i\lambda \delta_{w,y}\delta_{w,z}-\left( -\lambda \phi_w \delta_{w,y} \right) \right. \\ &\hphantom{==}\left.\times\left( -\frac{\lambda}{2}\phi^2_z+J_z \right)-\left( -\frac{\lambda}{2}\phi^2_w+J_w \right) \left( -\lambda \phi_z \delta_{zy} \right)+\left[-i\lambda \phi_w \delta_{w,z}-\left( -\frac{\lambda}{2}\phi^2_w+J_w\right)\right.\right. \\ &\hphantom{==}\left.\left.\times \left(-\frac{\lambda}{2}\phi^2_z+J_z\right)\right]\left( -i\frac{\lambda}{2}\phi^2_y+iJ_y\right)\right]\left.e^{i \int d^d \xi(-\frac{\lambda}{3!}\phi^3+J\phi)}\right|_{\phi=0} \end{align*} From here we just have to see that only the terms having zero $\phi$ in them after the ultimate derivation will survive to the limit $\phi \rightarrow 0$. So using one last time $(uv)'=u'v+uv'$ we arrive at the conclusion that the result ought to be: \begin{align*} &-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\left[\lambda \delta_{w,x}\delta_{w,y}J_z \right. \\ &\hphantom{==}+\lambda\delta_{z,y}\delta_{z,x}J_w+\lambda\delta_{w,x}\delta_{w,z}J_y+\lambda \delta_{w,y}\delta_{w,z}J_x+J_x J_y J_z J_w] \end{align*} We see that the integrand is composed of four tadpoles and one double propagator (two non-intersecting lines). Equating the terms that are equal to each other one finds: \begin{align*} &-\lambda \int d^d z \int d^d w \Delta_F (0) \Delta_F (z-w) \\ &-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w) J_x J_y J_z J_w \end{align*} I didn't check my calculations, hope there is no sign problem.