= Solution
Choose the <covariant derivative> $D_\mu=\partial_\mu+i\mu^{\epsilon/2}eA_\mu$, so the <interaction term> is $-\mu^{\epsilon/2}e\bar\psi\gamma^\mu A_\mu\psi$. The <gauge transformation> is $A_\mu\mapsto A_\mu+\partial_\mu\alpha$ and $\psi\mapsto e^{-i\mu^{\epsilon/2}e\alpha}\psi$. 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 is
$$
\boxed{\begin{aligned}
\mathcal L={}&i\bar\psi\not\partial\psi-m\bar\psi\psi-\frac14F_{\mu\nu}F^{\mu\nu}
-\mu^{\epsilon/2}e\bar\psi\gamma^\mu A_\mu\psi+\mathcal L_{\rm ct},\\
\mathcal L_{\rm ct}={}&i\delta Z_2\bar\psi\not\partial\psi-\delta m_L\bar\psi\psi
-\frac{\delta Z_3}{4}F_{\mu\nu}F^{\mu\nu}
-\mu^{\epsilon/2}e\delta Z_1\bar\psi\gamma^\mu A_\mu\psi.
\end{aligned}}
$$
With $\psi_0=\sqrt{Z_2}\psi$, $A_0=\sqrt{Z_3}A$, the definitions are $\delta Z_i=Z_i-1$, $\delta m_L=Z_2m_0-m$, and $e_0=\mu^{\epsilon/2}eZ_1/(Z_2\sqrt{Z_3})$. 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 $i\delta Z_2\bar\psi\not D\psi$, so <gauge invariance> is explicit and $e_0=\mu^{\epsilon/2}e/\sqrt{Z_3}$. A <photon> <mass> term is not an allowed <counterterm> in unbroken <QED>. For perturbative inversion one separately adds <gauge fixing>, $\mathcal L_{\rm gf}=-(\partial_\mu A^\mu)^2/(2\xi)$, 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.
Back to article page
