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Choose one free, massless complex scalar field with unit charge magnitude and bosonic statistics, minimally coupled to $A$. To fix the sign, use the conventional gauge covariant derivative $D_{\mu }={\partial }_{\mu }-iq{A}_{\mu }$ and take $q=-1$, so $D_{\mu }={\partial }_{\mu }+i{A}_{\mu }$ and $\phi \mapsto e^{-i\alpha }\phi$. Its antiparticle has charge $+1$. The matter data areThere is no scalar self-interaction. For real fields before formal continuation, its action is $S_{\phi }=\int (D_{\mu }\phi {)}^{\ast }(D_{\mu }\phi )d^{4}x=\int \bar{\phi }(-D^2)\phi d^{4}x$, by integration by parts. The Gaussian functional integral gives the inverse functional determinant and thus ${\Gamma }_{\phi }=+{\mathrm{Tr}}^{\prime }\log (-D^2)$, up to normalization.

The key operator identity isOn the formal complex gauge slice, $-D^2=-M_A={\mathcal{O}}_A$. With matching regulators and omitted zero modes, this gives ghost-scalar determinant cancellation in a complex quadratic gauge, leaving the Maxwell action as the background functional:where gauge-fixing terms and field-independent constants are understood separately. For a fixed background the complex scalar field and Faddeev-Popov ghost field integrals are Gaussian, so this determinant cancellation is exact within the formal construction. It is a cancellation of closed bosonic matter loops against ghost loops in that background functional. It does not prove that physical scalar quantum electrodynamics has no radiative corrections: the printed real Euclidean gauge slice contains only $A=0$ and misses every nonzero-curvature orbit. On that literal slice the equality is trivial; a nontrivial version requires the additional complex-contour interpretation.