 Solution

The change of field and parameters separates the Lagrangian density in Euclidean space intoThe two-leg counterterm interaction vertex is $-(\delta Z{p}^2+\delta m^2)$, with the overall momentum conservation delta function understood. With $D_0(p)=1/(p^2+m^2)$, the explicit first-order contribution to the propagator isThe minus sign follows either from expanding $e^{-S_{\mathrm{ct}}}$ or from expanding $(p^2+m^2+\delta Z{p}^2+\delta m^2{)}^{-1}$.

For $k=2$, the loop correction is $-{\tilde{\lambda }}_4{T}_D(m)D_0(p{)}^2/2$. Its ultraviolet pole from c isThere is no external $p^2$ dependence in the amputated tadpole diagram, so no one-loop wave-function renormalization is required. The minimal subtraction scheme therefore givesOnly the pole is subtracted: including $1-{\gamma }_E+\log 4\pi$ would impose a different finite subtraction prescription. The defined $\delta m^2$ is the coefficient in $Z{m}_0^2=m^2+\delta m^2$, not automatically the difference $m_0^2-m^2$ beyond this order.