$2PI$ contribution to the $2PI$ effective action

Question

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 !

Answer 1

Okay, I figured it out and for do that in the futur nobody stumbles across the same problem, here is the answer:

The point is that $\Sigma[G]$ consists of $2PI$-graphs in $G$, that means the edges (the propagators) in the graphs are the full propagator $G$ instead of the bare propagator $C.$ The Full propgator fullfills the Dyson equation:

$$G = C + C \Sigma[G]G,$$

this is equivalent to the equation

$$G^{-1}= C^{-1} + \Sigma[G],$$

which is the on shell case of the equation in the question.

Now the single tadpol does occur in $\Sigma[G].$ Now plugging in a single tadpole into the propagator of the tadpole yields the double tadpole.

So basically: If one would calculate $\Sigma[G]$ in terms of $C,$ one would also find $2PR$ diagrams but in $G$ they are $2PI$