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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.

enter image description here

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

enter image description here

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

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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?

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    $\begingroup$ x, z, w, y are the 4 vertices... $\endgroup$ Mar 12 at 15:01
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    $\begingroup$ In the expansion you have written down, you have kept only the terms or order $\lambda^0$. This reproduces the free theory; $n$-point correlators are obtained from $Z[J]$ by differentiating $n$-times with respect to $J$ and then setting $J=0$. If you want the interaction effect you have to look for terms of order $\lambda$. The loops arise since in the interaction you have terms like $\phi(x)^3$. $\endgroup$ Mar 12 at 16:12
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    $\begingroup$ I've done the calculations. If I am right you should come with four tadpoles and a "double propagator" (two non-intersecting lines). $\endgroup$ Mar 12 at 16:49
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    $\begingroup$ Ah, I've done the calculations for the $\Delta_F (x-y) \Delta_F(z-\xi)$ term appearing in the expansion of the exponential of the functional differential operators, so not the right one... But yes given the form of the loop you seek, it should be something like you said. $\endgroup$ Mar 12 at 17:17
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    $\begingroup$ You can't do what you've done because the square of an integral is not the double integral of the square of the integrand: $\left[\int dx f(x) \right]^2=\int dx f(x) \int dy f(y) \neq \int d^{2x} f^2(x)$. $\endgroup$ Mar 12 at 17:42
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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.

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