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

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 $$2PI$$

• Posting images of math is very strongly discouraged on Physics SE. Please use Mathjax instead. Jul 18, 2022 at 23:36
• More on 2PI effective action. Jul 19, 2022 at 6:20

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 $$2PI$$