= Solution
{c}
Let $a=\lambda/6$. Within terms already suppressed by $\Lambda^{-2}$, the leading <equation of motion> gives $\Box\phi=-m^2\phi-a\phi^3$. Equality here means equivalence under a perturbative <field redefinition>, together with <integration by parts>, rather than a pointwise identity inside arbitrary <off shell> <correlation functions>. Hence
$$
\begin{aligned}
\phi^3\Box\phi&\doteq-m^2\phi^4-\frac\lambda6\phi^6,\\
(\Box\phi)^2&\doteq m^4\phi^2+\frac{m^2\lambda}{3}\phi^4+\frac{\lambda^2}{36}\phi^6.
\end{aligned}
$$
For the second relation, factor the difference as $(\Box\phi-m^2\phi-a\phi^3)(\Box\phi+m^2\phi+a\phi^3)$, which is explicitly a <redundant operator>. If
$$
\Delta\mathcal L=\frac1{\Lambda^2}\left[c_1\phi^6+c_2\phi^3\Box\phi+c_3(\Box\phi)^2\right],
$$
the reduced correction is
$$
\Delta\mathcal L\doteq\frac1{\Lambda^2}\left[
c_3m^4\phi^2+\left(-c_2m^2+\frac{c_3m^2\lambda}{3}\right)\phi^4
+\left(c_1-\frac\lambda6c_2+\frac{\lambda^2}{36}c_3\right)\phi^6\right].
$$
The first two coefficients shift the already present <mass> and quartic <coupling constant (physics)>. After these shifts,
$$
\boxed{\text{one genuinely new dimension-six interaction remains: }\quad
\frac{c_{6,\rm red}}{\Lambda^2}\phi^6,\qquad
c_{6,\rm red}=c_1-\frac\lambda6c_2+\frac{\lambda^2}{36}c_3.}
$$
This is the <equation-of-motion reduction of scalar operators> through $O(\Lambda^{-2})$; terms of $O(\Lambda^{-4})$ are outside that truncation. The lower-dimensional shifts should not be silently dropped when $m\ne0$.
Back to article page
