= Solution
Take the <Minkowski metric> with signature $(+---)$ and a coupling $e\ne0$. For a <charged scalar field> use the <gauge covariant derivative>
$$
D_\mu\phi=(\partial_\mu-iea_\mu)\phi,
\qquad \phi'=e^{i\alpha}\phi,
\qquad a'_\mu=a_\mu+e^{-1}\partial_\mu\alpha.
$$
A <scalar electrodynamics> Lagrangian is
$$
\boxed{\mathcal L=-\frac14f_{\mu\nu}f^{\mu\nu}
+(D_\mu\phi)^*D^\mu\phi-V(|\phi|^2).}
$$
Any real potential bounded below and depending only on the modulus gives <gauge invariance>. Indeed,
$$
D'_\mu\phi'=e^{i\alpha}D_\mu\phi,
\qquad f'_{\mu\nu}=f_{\mu\nu},\qquad |\phi'|^2=|\phi|^2.
$$
The derivative of the phase cancels the shifted <gauge field> in the first identity; commuting partial derivatives proves the second. These identities verify invariance of every term, including the interaction hidden in the kinetic term.
For an unbroken example choose $V=m_\phi^2|\phi|^2+\lambda|\phi|^4$ with $m_\phi^2>0$ and $\lambda>0$. The minimum is at zero. There is no vector <mass> term in the quadratic expansion and no <Higgs mechanism>. Pure electromagnetic theory has a neutral massless <spin>-one <photon> with two physical transverse polarizations, equivalently helicities $+1$ and $-1$; longitudinal and time-component polarizations are gauge redundancies. In the unbroken scalar theory, the same massless <photon> is accompanied by a <spin>-zero charged particle and its oppositely charged <antiparticle>, both of <mass> $m_\phi$. A complex scalar has two real physical <degrees of freedom>, rather than two unrelated charged species.
For a Higgs example choose
$$
V=\lambda\left(|\phi|^2-\frac{v^2}{2}\right)^2,\qquad \lambda>0,\quad v>0.
$$
The vacuum modulus is nonzero. Around one vacuum representative, <unitary gauge> removes the phase and writes $\phi=(v+h)/\sqrt2$. Then
$$
(D_\mu\phi)^*D^\mu\phi
=\frac12\partial_\mu h\partial^\mu h+\frac12e^2(v+h)^2a_\mu a^\mu,
\qquad V=\lambda\left(vh+\frac{h^2}{2}\right)^2.
$$
Thus the quadratic spectrum is
$$
\boxed{m_A^2=e^2v^2,\qquad m_h^2=2\lambda v^2.}
$$
The <gauge boson> is a massive <spin>-one particle with three polarizations, and $h$ is a neutral massive <spin>-zero <Higgs boson>. The scalar phase supplies the longitudinal vector polarization; it is not an extra physical massless <Goldstone boson>. The degree count is unchanged: two massless-vector polarizations plus two scalar degrees become three massive-vector polarizations plus one radial scalar degree. The underlying <gauge invariance> remains a redundancy of the description.
For the two-charge theory, use
$$
D_\mu\phi=(\partial_\mu-iea_\mu)\phi,
\qquad D_\mu\psi=(\partial_\mu-2iea_\mu)\psi,
$$
and retain the same <gauge transformation> of $a_\mu$. Both derivatives transform with the phase of their own field. A manifestly stable <two-charge scalar gauge potential> is, for positive $m_\phi^2,M^2,\lambda_\phi,\lambda_\psi$, $g\ge0$ and a nonzero complex constant $b$,
$$
V(\phi,\psi)=m_\phi^2|\phi|^2+M^2|\psi-b\phi^2|^2
+\lambda_\phi|\phi|^4+\lambda_\psi|\psi|^4+g|\phi|^2|\psi|^2.
$$
Since $\psi-b\phi^2$ has charge $2e$, its modulus is <gauge-invariant>. Expanding its square exhibits the direct coupling
$$
-M^2\bigl(b\psi^*\phi^2+b^*\psi\phi^{*2}\bigr)
+M^2|b|^2|\phi|^4.
$$
In particular the charge in $\psi^*\phi^2$ is $-2e+2e=0$, which uses the stated charge ratio. The potential is real, bounded below and has the zero-field minimum, while the kinetic terms have the standard positive signs. A complete example is therefore
$$
\boxed{\mathcal L_2=-\frac14f_{\mu\nu}f^{\mu\nu}
+(D_\mu\phi)^*D^\mu\phi+(D_\mu\psi)^*D^\mu\psi-V(\phi,\psi).}
$$
There is genuine direct interaction even with $g=0$ because $b\ne0$. In four spacetime dimensions $b$ has <mass dimension> $-1$, but the expanded potential contains only quadratic, cubic and quartic field monomials; the cubic coefficient $M^2b$ has dimension one. Thus the example is also power-counting renormalizable.
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