Suppose $W$ is the generator of connected Feynman diagrams in $\Phi^4$ theory. We define $$\Gamma[\Phi]=W[j]-W_jJ,\tag{13.37}$$

where $$W_jJ=\int{dxW_j(x)j(x)}\tag{13.38}$$ and $$ \Phi\equiv\frac{\delta W[j]}{\delta j(x)}.\tag{13.39}$$

Now we define
$$\Gamma^{int}[\Phi] \equiv \Gamma[\Phi]-\frac{1}{2} \Phi iG_0^{-1}\Phi, \tag{13.51}$$ where $G_0$ is the bare propagator.

The claim is that the diagrams in $\Gamma[\Phi]$differ from those in $\Phi\Gamma^{int}_{\Phi}[\Phi]$ only by numerical prefactors, where $$\Gamma^{int}_{\Phi}[\Phi]=\frac{\delta \Gamma^{int}[\Phi]}{\delta \Phi(x)}\tag{13.41}$$

This is done in Kleinert's Chapter 13: Notes on formal perturbation theory.

Why is this claim true?


1 Answer 1


Kleinert is observing below eq. (13.64) that the Euler vector field$^1$

$$ V~:=~\int \!d^Dx~\Phi(x)\frac{\delta}{\delta \Phi(x)}$$ counts$^2$ the number of $\Phi$-powers in each term of the effective action $\Gamma[\Phi]$. So e.g. $V[\Phi^n]~=~n\Phi^n$, and so forth.


$^1$Note that Kleinert is using deWitt's condensed notation, cf. e.g. eq. (13.38).

$^2$ In the heat of the argument, Kleinert overlooks the quadratic free part $\Gamma_0=\Gamma[\Phi]-\Gamma^{\rm int}[\Phi]$, but that is anyway trivial to account for.

  • $\begingroup$ what you mean is if we expand $\Phi\Gamma^{int}_{\Phi}[\Phi]$ and $\Gamma[\Phi]$ in powers of $\Phi$ and then compare them we will arrive in the result? $\endgroup$ Commented Apr 27, 2019 at 18:39
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Apr 28, 2019 at 10:31

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