# Wilson effective action: why can additional terms be neglected?

I'm currently studying the Wilson effective action and have a question regarding its application for scalar fields. We consider a scalar field theory and defined the relevant path integral up to a momentum scale M: $$W = \int [\mathcal{D}\phi]^M~ \exp \left( - \int d^dx ~ \frac{1}{2} (\partial_\mu\phi)^2+ \frac{1}{2}m^2\phi^2 + \frac{\lambda}{4!}\phi^4 \right)$$ (in Euclidean spacetime), with $$[\mathcal{D}\phi]^M = \prod_{|k| $$\phi(x) = \int_{|k|

Then the scalar field is split into a "fast" and "slow" field: $$\phi(k) = \Phi(k) + \varphi(k)$$ with $$\Phi(k):~ 0 $$\varphi(k):~ \mu

Plugging in $$\Phi+\varphi$$ in the path integral $$W$$, one obtains the Wilson effective action as $$\exp{(-S_\mathrm{eff}[\Phi])} = \int [\mathcal{D}\varphi]_\mu^M \exp{(-S[\Phi+\varphi])}.$$

Then goal is to build a "quantum theory" for $$\varphi$$ to check for possible divergences which could be a problem in integrating the fast field $$\varphi$$ out. To get an action that describes $$\varphi$$, one simplifies the expression $$S[\Phi+\varphi]$$:

$$S[\Phi+\varphi] = S[\Phi] + \int d^dx ~\varphi(x) \frac{\delta}{\delta\Phi(x)}S[\Phi ] + S[\varphi] + \int d^dx~ \left( \frac{\lambda}{4}\Phi^2(x)\varphi^2(x)+\frac{\lambda}{6}\Phi(x)\varphi^3(x) \right)$$

As further steps it is mentioned that the second term vanishes since one can neglect terms linear in $$\varphi$$ (they seem to be only small contributions). But why are they small?

• Which derivation? Which page? Sep 6, 2022 at 14:25
• Hello Qmechanic, this is an example from a lecture I took, I did not find it in any book. Sep 6, 2022 at 14:26
• Symmetries are key in the construction of an effective action. Are there any additional symmetries of the fields given in your problem? Something like $\mathbb{Z}_2$, i.e. $\phi \mapsto -\phi$? Sep 6, 2022 at 14:36
• Thank you for your answer! We have not been given any additional symmetries, but I did not consider this aspect. Sep 6, 2022 at 15:08

Moreover, a $$\mathbb{Z}_2$$-symmetry would rule out terms $$\varphi^n\Phi^m$$ with $$n+m$$ odd, cf. above comment by scaphys.
• $\mathbb{Z}_2$ symmetry wouldn't rule it out because it would transform both $\phi$ and $\Phi$ at the same time. Sep 6, 2022 at 22:03