 Solution

Let the bare fields obey the wave-function renormalization definitions ${\psi }_0=Z_2^{1/2}\psi$, $A_{0\mu }=Z_3^{1/2}{A}_{\mu }$, and put $m_0=Z_{m}m$. Define the vertex renormalization factor by $e_0={\mu }^{ϵ/2}e{Z}_1{Z}_2^{-1}{Z}_3^{-1/2}$. Substitution into the bare QED Lagrangian density givesEquivalently, write $Z_i=1+{\delta }_i$ and ${\delta }_m=Z_2{Z}_m-1$. The counterterm Lagrangian isFor the subsequent Feynman gauge calculation include $-(\partial \cdot A{)}^2/2$ in the gauge-fixed action. The Ward identity for QED renormalization constants implies $Z_1=Z_2$ when the regularization preserves gauge invariance; this is a consequence, rather than an extra independent definition of the renormalization constants.