# Why do the diagrams in $\Gamma[\Phi]$ differ from those in $\Phi\Gamma^{\rm int}_{\Phi}[\Phi]$ only by numerical prefactors?

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?

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.
• 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? – amilton moreira Apr 27 '19 at 18:39