A massless complex bosonic <charged scalar field> with $D_\mu=\partial_\mu+iA_\mu$ contributes $[\det(-D^2)]^{-1}$ to a Gaussian background-field integral. A <Faddeev-Popov ghost field> pair contributes $\det\mathcal O_A$. On the formal <complex quadratic Abelian gauge condition>, $\mathcal O_A=-D^2$, so the two determinants cancel and their effective-action contributions sum to zero. Use matching regulators, boundary conditions and removed zero modes. This is a formal cancellation of closed loops; the <reality obstruction for a complex Euclidean gauge condition> precludes deducing that physical <scalar quantum electrodynamics> has no quantum effects.

Ghost-scalar determinant cancellation in a complex quadratic gauge

For this <charged scalar field>, $\sqrt{-g}=r^3\sin\theta\cos\theta$. On $e^{-i\omega t}$, the <gauge-covariant derivative of a charged scalar field> supplies the shifted frequency
$$
\omega+qA_t=\omega-\frac{qQ}{r^2}.
$$
The wave equation becomes
$$
\frac1{r^3}\frac{d}{dr}\left(r^3f\frac{dR}{dr}\right)+\frac{(\omega-qQ/r^2)^2}{f}R
+\frac{R}{r^2\Theta\sin\theta\cos\theta}\frac{d}{d\theta}\left(\sin\theta\cos\theta\frac{d\Theta}{d\theta}\right)=0.
$$
<Separation of variables> with constant $\lambda$ gives
$$
\boxed{\frac1{\sin\theta\cos\theta}\frac{d}{d\theta}\left(\sin\theta\cos\theta\frac{d\Theta}{d\theta}\right)+\lambda\Theta=0}
$$
and
$$
\boxed{\frac1{r^3}\frac{d}{dr}\left(r^3f\frac{dR}{dr}\right)+\left[\frac{(\omega-qQ/r^2)^2}{f}-\frac{\lambda}{r^2}\right]R=0}.
$$
The angular equation is the axisymmetric <spherical harmonic> equation on the <three-sphere>; regular solutions have $\lambda=\ell(\ell+2)$ with $\ell=0,2,4,\ldots$, the even values selected by the absence of $\phi$ and $\psi$ dependence.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 311 / 1 / b / iv / Solution

A charged scalar field is a complex <scalar field> transforming nontrivially under a gauge symmetry. For an electromagnetic potential $A_a$, its gauge-covariant derivative is $\mathcal D_a=\nabla_a-iqA_a$.

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 Incoming links: Charged scalar field