 Solution

Choose the covariant derivative $D_{\mu }={\partial }_{\mu }+i{\mu }^{ϵ/2}e{A}_{\mu }$, so the interaction term is $-{\mu }^{ϵ/2}e\bar{\psi }{\gamma }^{\mu }{A}_{\mu }\psi$. The gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha$ and $\psi \mapsto e^{-i{\mu }^{ϵ/2}e\alpha }\psi$. The derivative of this local phase cancels the added vector-potential term in $D_{\mu }\psi$, making $i\bar{\psi }{\gamma }^{\mu }{D}_{\mu }\psi$ gauge-invariant.

For renormalized fields and parameters, a complete local QED counterterm parametrization isWith ${\psi }_0=\sqrt{Z_2}\psi$, $A_0=\sqrt{Z_3}A$, the definitions are $\delta Z_i=Z_i-1$, $\delta m_L=Z_2{m}_0-m$, and $e_0={\mu }^{ϵ/2}e{Z}_1/(Z_2\sqrt{Z_3})$. In particular $\delta m_L$ is the coefficient in the Lagrangian density, not just $m_0-m$.

The Ward identity requires $Z_1=Z_2$ in a gauge-preserving renormalization scheme. The kinetic and vertex counterterms then combine into $i\delta Z_2\bar{\psi }/D\psi$, so gauge invariance is explicit and $e_0={\mu }^{ϵ/2}e/\sqrt{Z_3}$. A photon mass term is not an allowed counterterm in unbroken QED. For perturbative inversion one separately adds gauge fixing, ${\mathcal{L}}_{\mathrm{gf}}=-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$, taking $\xi =1$ in Feynman gauge. That quantization term is not itself part of the gauge-invariant density. The Abelian Faddeev-Popov determinant is field independent for this linear gauge and introduces no interacting ghost contribution.