 Solution

Use the wave-function renormalization coefficient and anomalous dimension definition stipulated in this part, keeping their signs and normalization. Put $A=1/[6(32{\pi }^2{)}^2]$. To the required order,At fixed bare ${\lambda }_4$, the leading change of the dimensionless coupling is $\mu d{g}_4/d\mu =-2ϵg_4+O(g_4^2)$, because ${\lambda }_4$ has mass dimension $2ϵ$. Therefore the chain rule, applied before setting $ϵ=0$, givesThusThe vanishing engineering contribution $-2ϵg_4$ multiplies a pole and leaves a finite answer. Dropping it at the start would incorrectly give zero. The numerical answer is conditional on the supplied two-loop coefficient. A convention defining the field anomalous dimension as ${\gamma }_{\varphi }=\frac{1}{2}\mu d\log Z/d\mu$ instead gives ${\gamma }_{\varphi }=-\gamma /2$ for the same $Z$; substituting that convention silently would answer a different question.