Electric charge of the Higgs mode in superconductivity I have a question about the Higgs mode in superconductivity. In this doc, it is said, page 12, that the Higgs mode has no electric charge. But it couples nonlinearly with the photon (in the phenomenological Ginzburg-Landau theory there is a term $\propto e^2 A^2 h$ with $A$ the vector potential and $h$ the Higgs field). I just don't understand how a field can be coupled to $A$ without having a charge. I hope to be clear.
 A: Indeed, it should be evident from your Abelian Higgs mechanism that your σ, or Higgs h, etc... mode is "dark"... a U(1) singlet, so it does not respond to gauge transformations, even before you coupled its larger doublet to photons via a charge. The thicket of symbols you have been thrown into obscures the basic symmetry structure that governs the mechanism.
Recall the ungauged, globally U(1) symmetric scalar doublet (I'm being cavalier with normalizations), of action
$$
\partial_\mu \phi^* \partial^\mu \phi -\lambda (\phi^* \phi - v^2/2)^2 .
$$
Its U(1) symmetry transformations act as $\delta \phi= i \epsilon\phi$, with $\epsilon$ the real, dimensionless infinitesimal parameter of the transformation. Thus,  the conserved Noether current is $J_\mu\propto i(\phi^*\partial_\mu \phi- \phi\partial_\mu \phi^*). $
Now, as you learned, SSBreaking is more transparent in the polar representation of the scalar field,
$$
\phi (x)  \equiv {h(x)+v \over \sqrt{2}} e^{i\theta(x)/v} ~~\leadsto \\
\delta h=0, ~~ \delta \theta =\epsilon v , ~~ J_\mu = -(v+h)^2\partial_\mu \theta /v, \\
{\cal L} =\frac{1}{2} \bigl( \partial_\mu h \partial ^\mu h +(1+h/v)^2 \partial_\mu \theta \partial ^\mu \theta\bigr ) -{\lambda \over 4} (h^2+2vh)^2 ,
$$
where the massless Goldstone field θ has dimensions of momentum ― like φ, h, and the order parameter v. The dross mode, h, has mass $\sqrt {2\lambda} ~ v$, as you can check, and is a singlet (invariant) under the U(1). Significantly, the potential does not depend on  θ!
Now, gauge this model with minimal coupling, $\partial_\mu \phi\mapsto (\partial_\mu -ieA_\mu)  \phi$, to obtain an irrelevant Maxwell term, the same Higgs potential, the same dark higgs kinetic terms; and, significantly, a few extra terms, in the goldston kinetic term,
$$
\frac{1}{2} (v+h)^2 (\partial_\mu \theta/v -e A_\mu  )  ^2.
$$
Through a gauge transformation, you may now completely absorb the goldston into the photon, to reduce this term to
$$
\frac{e^2}{2} (v+h)^2   A_\mu^2,
$$
which includes a mass term, m=ev, for the photon (inverse penetration length for the magnetic field); and a trilinear coupling $e^2v hA_\mu A^\mu$ over and above the evident quadrilinear coupling γγhh.

*

*Note there is no trace or remnant of θ in the system anymore: it has left the problem! Its degree of freedom is the extra longitudinal polarization mode of the massive γ,  that distinguishes it from its massless former self. Because of this mass, a magnetic field attenuates (dies off exponentially) as it enters the superconducting medium.

So, indeed, the higgs does couple to the photon with strength $e^2$, so it decouples for vanishing e.  Your reference by Shimano & Tsuji calls this "nonlinear coupling" to the extent it requires superstrong EM fields to express itself. It is of O($e^2$) in the amplitude, and not of O(e), as expected in the conventional weak field EM couplings―see below.
But, in this topsy-turvy medium, where goldstons pop into and out of the medium,  the higgs is still characterized as "dark". It certainly does not couple to EM minimally, O(e),  as you saw from its unaffected kinetic term, so you cannot slam a photon onto it and observe a deflection, because there is no such linear EM γhh coupling. That is what the admittedly imprecise rubric "dark" indicates.
