What is a "Higgs mode" in superconductors and why is it called that way?

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I understand spontaneous symmetry breaking and the Higgs mechanism pretty well in the context of high energy physics (Relativistic QFT). Although I do also understand these phenomena in condensed matter physics (Non-Relativistic QFT), I might ignore some of the details.

For the sake of simplicity, I will talk about an abelian U(1) (global or gauge) symmetry.  My understanding is the following:

RQFT

A complex Klein-Gordon field (charged or uncharged under a gauge interaction depending on whether we are talking about a SSB or a Higgs mechanism) takes a constant VEV, so the vacuum is full of this field. The fluctuations around the VEV are described by another complex field (which only differs from the first one by the the VEV). In the RQFT case, the latter complex field may be discomposed in a component in the direction of the U(1) symmetry ("the bottom of the Mexican hat")  and another radial component, that is, a goldstone boson (a massless propagating particle in the case of a SSB and a longitudinal mode of the gauge boson in the Higgs mechanism) and a massive independent particle (a sigma meson in a linear sigma-model -- where there is SSB -- and the Higgs particle in the case of the Higgs mechanism). These two degrees of freedom -- the massless goldstone (or the longitudinal polarization of the gauge boson) and the radial massive particle -- are in correspondence with the two degrees of freedom associated to the original complex Klein-Gordon field (the would-be particle and antiparticle).

NRQFT

The situation is different in non-relativistic QFT (condensed matter). Here, a complex Schrödinger field represents a single degree of freedom, as the Lagrangian only contains a time derivative. This is the Schrödinger field connected with the atoms of a superfluid  (SSB) or with the Cooper pairs of a superconductor (Higgs mechanism).  In theses cases, as far as I understand, there is a massless excitation (Goldstone boson) so-called phonon in the former case and an additional longitudinal polarization of the photon in the superconductor case. But there are no radial massive excitations (what would-be a sigma meson or a Higgs boson, respectively) in any case. The reason is clear: the Schrödinger field leads to a single degree of freedom instead of the two degrees of freedom due to the a complex Klein-Gordon field.

My question is the following. If, as I stated, there is no possibility of a Higgs-like excitation in a superconductor just because of the conservation of the number of degrees of freedom, what is a "Higgs mode" in superconductors and why is it called this way?

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The answer may be found in http://arxiv.org/abs/cond-mat/0109409v1

answered Sep 4, 2015 by (875 points)
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Your counting of degrees of freedom for non-relativistic systems is wrong. In a superconductor, the order parameter is a complex scalar. Of course it has a magnitude and a phase. The phase fluctuations give rise to Goldstone mode and the magnitude fluctuations are massive.

answered Sep 4, 2015 by (550 points)

I said that it is a complex field. The point is how it evolves in time. Does it obey a first order (Schrodinger-like?) or a second order (Klein-Gordon) equation? A scalar complex field that follows a Schrodinger-like (e.g, Gross–Pitaevskii) equation corresponds to a single degree of freedom (note that the number operator, the integral of the square of the radial component, and the phase component are canonical conjugated pairs). Dirac equation is first order, but it mixes different components. Each component follow a Klein-Gordon equation. That's why A Dirac field describes 2 dof despite it's first order. Anyway, you said "scalar", and a scalar field doesn't obey Dirac equation.

So either there is an emergent approximated Lorentz invariant equation in the condensed matter system or the amplitude (or radial or Higgs) mode is not independent of the other degrees of freedom of the system.

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