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 $$ [\mathcal{D}\phi]^M = \prod_{|k|<M}d\phi_k $$ $$ \phi(x) = \int_{|k|<M} \frac{d^dk}{(2\pi)^d} \mathrm{e}^{ikx} \phi(k). $$

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

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?

  • $\begingroup$ Which derivation? Which page? $\endgroup$
    – Qmechanic
    Sep 6, 2022 at 14:25
  • $\begingroup$ Hello Qmechanic, this is an example from a lecture I took, I did not find it in any book. $\endgroup$
    – physicsCat
    Sep 6, 2022 at 14:26
  • 1
    $\begingroup$ 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. $\phi \mapsto -\phi$? $\endgroup$
    – scaphys
    Sep 6, 2022 at 14:36
  • $\begingroup$ Thank you for your answer! We have not been given any additional symmetries, but I did not consider this aspect. $\endgroup$
    – physicsCat
    Sep 6, 2022 at 15:08

1 Answer 1


Since each term in the action should satisfy momentum conservation, one argument is that a mixed term of heavy and light modes has little or no kinematic phase volume [1].

Moreover, a $\mathbb{Z}_2$-symmetry would rule out terms $\varphi^n\Phi^m$ with $n+m$ odd, cf. above comment by scaphys.


  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; p. 396 between eqs. (12.5) & (12.6).
  • $\begingroup$ Thank you for your fast answer, also for the reference in Peskin and Schröder, I totally overlooked this one. $\endgroup$
    – physicsCat
    Sep 6, 2022 at 15:09
  • $\begingroup$ $\mathbb{Z}_2$ symmetry wouldn't rule it out because it would transform both $\phi$ and $\Phi$ at the same time. $\endgroup$ Sep 6, 2022 at 22:03
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Sep 7, 2022 at 0:49

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