I have an issue understanding a point that my professor made in the lecture. We started with the following derivation:
Consider an even Theory, i.e. the action takes the form:
$$S[\phi] = \phi \cdot C^{-1} \cdot \phi + V[\phi], $$
where $V[\phi]$ is symmetric in $\phi$
Then the free energy in the presence of a bilocal source is given by:
$$W[K] = \mathrm{ln}\left(\int d \phi e^{-S[\phi]+ \frac{1}{2} \phi K \phi} \right).$$
We can know take the Legendre transform with respect to $G = \frac{\delta W}{\delta K,}$ and find the $2PI$ action
Now, we find that given that the propagator fulfills the Dyson equationand the covariance is $C^1-K$ that
Now defining $\Sigma = - \frac{\delta \Gamma^{2PI}}{\delta G},$
We can find an explicit expression for the effective action:
Now my question is, why is $\Gamma^{2PI}$ a sum over connected $2PI$ graphs? If I understand the notes correctly, then we can even go further and find the $\Sigma$ is a sum over $2PI$- Graphs. Can anyone explain why this is? I could nowhere find a rigorous proof for this.
I actually need this as well for an exercise
In order to be very specific, consider the $\lambda \phi^4$-theory. The self energy for the on-shell solution ($K[G] = 0$) should contain the following diagram, I think.
However, this is not $2PI$
thanks in advance !