3
$\begingroup$

So I am currently taking a class on advanced quantum field theory, and I came across something that I don't understand. I have been thinking about it for quite some time and I cannot get my head around it.

So the $1PI$ effective action is given by:

$$\Gamma[\Phi]= J \Phi- W[J[\Phi]]$$

Plugging this into an exponential yields:

$$e^{-\Gamma[\Phi]} = e^{-J \Phi} e^{W[J[\Phi]]} = LHS.$$

Plugging in the definiftion, of $W$ yiels:

$$LHS = e^{-J \Phi} \int [d \phi] e^{-S[\phi]+ J[\Phi] \phi}.$$

Now, we can do a Feynman expansion of the action and write:

$$S[\phi]= S[\Phi]+ \frac{1}{2} \psi S''[\Phi] \psi + S'[\Phi] \psi + \sum_{n \geq3} \frac{S^{(q)}}{q!} \psi^q$$

plugging in yields

$$LHS = e^{-J \Phi} \int [d \psi] e^{-S[\Phi]- \frac{1}{2} \psi S''[\Phi] \psi - S'[\Phi] \psi - \sum_{n \geq3} \frac{S^{(q)}}{q!} \psi^q -J[\Phi](\Phi +\psi) }$$

The $J \Phi$ terms in the exponent cancel and we can take out the action $S[\phi],$ we get:

$$LHS = e^{-S[\phi]} \int [d \psi] e^{- \frac{1}{2} \psi S''[\Phi] \psi - S'[\Phi] \psi - \sum_{n \geq3} \frac{S^{(q)}}{q!} \psi^q -J[\Phi]\psi} $$

Now the second part of this product is again a path integral. Since we Taylorexapnded around $\Phi,$ we should find

$$\langle \Phi +\psi \rangle= \Phi \Leftrightarrow \langle \psi \rangle = 0,$$

This additional restriction of the path integral is enforced by the source term $J[\Phi] \psi$

We can therefore say that the second factor is a sum of diagrams, with propagator $S''[\Phi]$ and vertices $S'[\Phi] \psi$ and $\frac{S^{(q)}[\Phi]}{q!} \psi^q.$

Now taking the $\mathrm{ln}$ on both sides, yields

$$\Gamma[\Phi]= S[\Phi] + \mathrm{ln} \left( \int [d \psi] e^{- \frac{1}{2} \psi S''[\Phi] \psi - S'[\Phi] \psi - \sum_{n \geq3} \frac{S^{(q)}}{q!} \psi^q -J[\Phi]\psi}\right).$$

Now, my professor called the second term "$\Gamma^{1PI}[\Phi]$" and therefore writes

$$\Gamma[\Phi] = S[\Phi] + \Gamma^{1PI}[\Phi].$$

Now what comes next seems to be very important but is nowhere proven. He claims, that the $\Gamma^{1PI}$ is a sum over $1PI$ graphs.

My question is: How can this be seen from the representation that he chose. Of course, one finds

$$G^{-1}=\frac{\delta^2 \Gamma [\Phi]}{\delta \Phi \delta \Phi}= S''[\Phi] +\frac{\delta^2 \Gamma^{1PI} [\Phi]}{\delta \Phi \delta \Phi} = S[\Phi] - \Sigma,$$

if we define $\Sigma_{a,b}= -\frac{\delta^2 \Gamma[\Phi]}{\delta \Phi_a \delta \Phi_b}.$

And coming from another approach to QFT, one finds via the Dyson equation

$$G = G_0 + G_0 \Sigma G \Rightarrow G^{-1} = G_0^{-1} - \Sigma,$$

Where $G_0$ is the bare propagator, $G$ the dressed propagator and $\Sigma$ the $1PI$-function.

So it appears to be correct to say that $\Gamma^{1PI}$ governs the $1PI$ (self energy contribution), i.e. is a sum over $1PI$ graphs. But I cannot see how it follows from the definition that we originally found (via the path integral over $\psi.$) In other words:

How can we see that the additional term indeed is a sum over the $1PI$ contributions.

I hope, I could make my question clear. If not, please ask

$\endgroup$

1 Answer 1

3
$\begingroup$
  1. The effective action $\Gamma[\phi_{\rm cl}]$ is the generating functional of 1PI diagrams, cf. e.g. this Phys.SE post.

  2. The effective action $\Gamma[\phi_{\rm cl}] = S[\phi_{\rm cl}] +{\cal O}(\hbar)$ is the original classical action up to quantum corrections, cf. e.g. this Phys.SE post, where calculations very similar to OP's are performed.

  3. Points 1 & 2 yield that OP's sought-for difference $\Gamma^{1PI}[\phi_{\rm cl}]:=\Gamma[\phi_{\rm cl}] - S[\phi_{\rm cl}]$ is the generating functional of 1PI loop-diagrams, cf. the $\hbar$/loop-expansion.

$\endgroup$
2
  • $\begingroup$ Hey Qmechanic, thanks for the answer. 1. It is very useful and yet another argument, why this is true, but how do you see that $S[\Phi] $ consists of merely non-loop diagrams? 2. Furthermore, in the lecture notes my professor argued that it also just follows from the fact that $\langle \psi \rangle = 0.$ Any idea, how that follows from there? $\endgroup$
    – jabru
    Jul 2 at 16:31
  • $\begingroup$ 1. I updated the answer. $\endgroup$
    – Qmechanic
    Jul 2 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy