I'm already familiar with the "Mexican hat" story of Anderson-Higgs mechanism in Landau-Ginzberg theory. However, I have never seen anyone talking about Goldstone and Higgs mechanism in the context of BCS theory of superfluidity and superconductivity. I need the answer to the following questions:

  1. In a BCS superfluid, what does the Goldstone mode creation operator look like, in terms of BCS quasi-paritcle operators $\alpha_k=u_k a_k-v_k a^\dagger_{-k}$ ($a_k, a^\dagger_k$ are fermion annihilation/creation operators of momentum $k$)?

  2. In BCS superconductivity, what is the operator representation of the gauge boson $A^\mu(k)$ and Higgs boson $H(k)$?

Or to put it another way, say if I want to create a Goldstone boson (with momentum $k$) in superfluid BCS ground state $|\Phi_{BCS}\rangle$, what does the resulting excited state look like?

  • 1
    $\begingroup$ What are you calling $a_k$ and $a_k^\dagger$ here, the bare fermion creation and annihilation operators? Please edit this information into your question to make it intelligible. $\endgroup$ Nov 24, 2016 at 17:05
  • $\begingroup$ @MarkMitchison Yes. Anyway I need the operator repesentation of the Goldstone and Higgs. $\endgroup$
    – Lagrenge
    Nov 24, 2016 at 17:26
  • $\begingroup$ Goldstone and Higgs modes can be expanded in normal modes, and thus will get the usual form of creation/destruction operators. Classically, the Anderson-Higgs mode follows a massive wave equation, as can be seen from Ginzburg-Landau formalism, which you can quantise if you want. They representation from the quasi-particle is harder to give, since they are clearly dressed states. A Higgs mode is intrinsically a bosons mode dressed by the fermionic quasi-particles. You get it either by integrating out the fermionic modes in a path-integral approach in the mean-field approximation. $\endgroup$
    – FraSchelle
    Nov 25, 2016 at 11:46
  • $\begingroup$ One more time, be careful with your nomenclature. A superfluid is a quantum condensate of neutral bosonic particles (as He-4), a BCS superfluid is a quantum condensate of neutral fermionic particles (as He-3) (in a sense, they are not spin-neutral, but remains charge-neutral), a BCS superconductor (or a superconductor in short) is a quantum condensate of charged fermionic particles (as electrons). This nomenclature is not accepted by everyone, so I suggest you to define what you mean each time you use them and want to compare these different mechanism. $\endgroup$
    – FraSchelle
    Nov 25, 2016 at 11:50

2 Answers 2


$\def\kk{{\bf k}} \def\rr{{\bf r}} \def\ii{{\rm i}} \def\ee{{\rm e}} \def\qq{{\bf q}}$ I can't give a complete answer but I will attempt to address question 1). (To be absolutely clear, that means I will talk only about a BCS-type superfluid which is electrically neutral and therefore supports a Goldstone mode. In the charged case, the Goldstone mode is lifted to the plasma frequency by the Anderson-Higgs mechanism.) The original BCS theory actually does not contain a Goldstone mode. This is because one assumes a static condensate described by the order parameter $$\Delta(\rr) = \frac{1}{V}\sum_{\kk,\qq} \ee^{\ii\qq\cdot\rr} \langle a_{\kk+\qq/2,\uparrow}a_{-\kk+\qq/2,\downarrow}\rangle.$$ Here I'm considering a homogeneous $s$-wave superfluid with periodic boundary conditions in a volume $V$. In the ground state, one has a spatially constant order parameter $\Delta(\rr) = {\rm const.}$, which means that Cooper pairs condense into states with zero centre-of-mass momentum $\qq =0$.

The typical BCS mean-field treatment predicts an elementary excitation spectrum consisting of gapped quasiparticles, which are created by breaking apart condensed pairs. However, in an uncharged superfluid, the low-energy excitations actually correspond to gapless collective oscillations of the Cooper-pair condensate, i.e. the Goldstone mode. This leads to a time- and space-varying order parameter $\Delta(\rr,t)$ describing a macroscopic number of Cooper pairs carrying non-zero centre-of-mass momentum $\qq\neq 0$. In other words, the Goldstone mode excitations correspond to the entire condensate being coherently displaced slightly in momentum space. But since in BCS theory the condensate is a classical variable (the mean field), there is no operator describing the dynamics of $\Delta(\rr,t)$. Nevertheless it is possible to compute the spectrum of its excitations using a dynamical extension of BCS theory where the mean-field is time-dependent. In the end, this procedure turns out to be equivalent to the random-phase approximation. A quite comprehensive study along these lines was done by Combescot et al.

A full quantum-mechanical treatment of the Goldstone mode can of course be performed by going beyond mean-field theory. However, usually this is done in the path integral formalism, where there are no operators whatsoever. In this case, the Goldstone mode is an excitation of a Hubbard-Stratanovich field, which is introduced in the Cooper pair channel and used to integrate out the bare fermion field. Quantum fluctuations of the pair condensate are described as Gaussian (or higher-order) fluctuations around the saddle-point describing the BCS ground state. A good original reference for this formalism is Engelbrecht et al. and references therein (unfortunately behind a paywall), although there are many more modern treatments that can also be found by Googling. The topic of fluctuations in BCS-type neutral superfluids is currently very active because of experiments on the BCS-BEC crossover in ultracold atomic gases.

  • $\begingroup$ Does this mean that, in the original BCS mean-field treatment, only the ground state $|\Phi_{BCS}\rangle$ gives a good approximation of the true ground state of original (number-conserving) Hamiltonian; the excited states cannot be obtained by $\alpha^\dagger_k|\Phi_{BCS}\rangle$? (The BCS prediction of excitation spectrum shouldn't be taken seriously at all as it doesn't contain the Goldstone?) $\endgroup$
    – Lagrenge
    Nov 24, 2016 at 20:35
  • 2
    $\begingroup$ @Lagrenge In the uncharged superfluid, yes: the BCS quasiparticle spectrum is completely wrong. But in the typical condensed-matter context (charged superconductor) the Anderson-Higgs mechanism lifts the Goldstone mode way up to the plasma frequency, so that the gapped BCS quasiparticle spectrum is still correct at low energies. Even in the uncharged case, the mean-field equations do give qualitatively correct equations of state because the thermodynamics of the superfluid phase transition is determined by the fraction of condensed pairs, and not by the collective mode excitations. $\endgroup$ Nov 24, 2016 at 20:47
  • $\begingroup$ @MarkMitchison I'm surprised by your previous comment. Your answer should make it clear that a BCS superconductor does not have Goldstone mode, only Anderson-Higgs mechanism. In ultracold gases, the so-called BCS-BEC crossover is misleading, since atoms are always neutral, whereas a BCS condensate is necessarily charged. I know this is the usual nomenclature in this community, but it is completely misleading regarding the quantum field theory nomenclature. Your previous comment make this point clear, not your answer. It is unfortunate that this point is not clear in your answer. $\endgroup$
    – FraSchelle
    Nov 25, 2016 at 11:38
  • $\begingroup$ @FraSchelle My answer addresses only question 1) about the uncharged case, precisely because I am not an expert on the Higgs mechanism. I thought it would be clear therefore that I consider only a neutral system (note the italicised "uncharged" and the fact that the OP already appears to be quite aware of the nonexistence of Goldstone modes in the charged system). However, to avoid any confusion I have now explicitly stated that I am only talking about neutral systems. Please let me know if this is clear. Perhaps you'd also like to add an answer discussing the Higgs mechanism? $\endgroup$ Nov 25, 2016 at 14:09
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    $\begingroup$ Well the famous argument explaining superfluidity (zero-resistance flow) due to Landau is predicated on a gapless excitation spectrum (see e.g. this impromptu Google result, Sec. 2.1). So in other words, you cannot explain superfluidity in the neutral system without invoking the collective mode. Obviously electrical resistivity/superconductivity doesn't arise in a neutral system, which doesn't respond to electric fields at all :) $\endgroup$ Sep 24, 2020 at 9:53

In a superfluid Bose-Einstein condensate, the Goldstone excitations are oscillating perturbations of the order parameter $\psi(\mathbf{r},t)$ above its equilibrium value $\psi_0$. In the long-wavelength limit, they are just sound waves in a superfluid, described by a linearized Gross-Pitaevskii equation.

In a BCS superconductor, the Goldstone modes are oscillations of a Cooper pair condensate. Because of the Coulomb interactions, these excitations are very similar to usual plasma oscillations (bulk plasmons) in a metal, as was first described by P.W. Anderson and G. Rickayzen. In 3D, these modes acquire a gapped dispersion due to long-range character of Coulomb interaction. In contrast, in 2D superconductors the Goldstone modes, as well as usual 2D plasmons, remain gapless.

  • $\begingroup$ Is this the reason why fluctuations kill SC in 2D? Maybe you can help me answering this question physics.stackexchange.com/q/338943 $\endgroup$
    – JWDiddy
    Jun 19, 2017 at 21:01
  • $\begingroup$ @JWDiddy I don't know much about this subject, but I think Goldstone modes don't destroy superconductivity unless they violate the Landau criterium. In 2D you need to look at vortices and related BKT transition (see, e.g., journals.aps.org/prl/abstract/10.1103/PhysRevLett.47.1542). In 1D superconducting ordering can be absent theoretically, in infinite system, but can still be present in finite system. $\endgroup$ Jun 20, 2017 at 12:31

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