= Solution
After integrating out the shell, let a chosen operator's coefficient be $\kappa+\Delta\kappa$ and write the two-derivative quadratic term as $\tfrac12(1+\Delta Z)(\partial\phi)^2$. Here $\Delta\kappa$ is the shell-induced change before rescaling, and $\Delta Z$ is the shell correction to the <wavefunction renormalization>. Work in an operator basis in which that quadratic term has this form; the other generated operators remain in the <Wilsonian effective action>.
Under $x'=bx$, the volume element changes by $d^4x=b^{-4}d^4x'$ and each derivative by $\partial_x=b\partial_{x'}$. <Canonical field normalization> is restored by defining
$$
\phi'(x')=b^{-1}\sqrt{1+\Delta Z}\,\phi(x'/b),\qquad
\phi(x)=\frac b{\sqrt{1+\Delta Z}}\phi'(bx).
$$
Indeed, the volume factor, two derivatives and two field factors cancel in the <kinetic term>. A term with $n$ fields and $m$ derivatives consequently obtains the factor $b^{-4}b^m b^n(1+\Delta Z)^{-n/2}$. Hence
$$
\boxed{\kappa'=\frac{\kappa+\Delta\kappa}{(1+\Delta Z)^{n/2}}\,b^{m+n-4}
=\frac{\kappa+\Delta\kappa}{(1+\Delta Z)^{n/2}}\left(\frac{\Lambda'}\Lambda\right)^{m+n-4}.}
$$
This is <Wilsonian rescaling of a scalar coupling>. The exponent is minus the coupling's engineering <mass dimension>, $[\kappa]=4-n-m$. The rescaling also returns the low-momentum cutoff to $\Lambda$, because $p'=p/b$.
Back to article page
