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I don't understand what is meant by "gapless superconductors". As far as I know, the superconducting gap is necessairy for both zero resistance and the Meissner effect, the two hallmarks of superconductivity. So, what does it mean to have gapless superconductivity and how does it work?

Thanks in advance, Greetings

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    $\begingroup$ There is a related unanswered question What are gapless superconductors?. In a comment, Alexander gives a reference that might be helpful. $\endgroup$
    – Anyon
    Apr 24, 2021 at 16:46
  • $\begingroup$ Consider also that you can have zero resistance in an electric circuit which has a loss resistor in parallel with a zero resistance resistor. So having a gapless system (which is lossy) that still has superconducting electrons does not imply nonzero resistance even in principle. $\endgroup$
    – KF Gauss
    Jun 2, 2021 at 17:57

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Superconductivity is characterized by a charged order parameter such as $\langle\psi_\alpha \psi_\beta\rangle \sim \epsilon_{\alpha\beta}\Phi$, where $\psi_\alpha$ is a spin 1/2 fermion of charge $e$, and $\Phi$ carries charge $2e$. If $\psi_\alpha$ is coupled to a $U(1)$ gauge field we find a Higgs (Anderson-Higgs-Meissner) effect, and the photon acquires a "mass".

The existence of a gap is a separate question. We look at the fermion propagator $G_{\alpha\beta}(\omega,k)=\langle\psi_\alpha\psi^\dagger_\beta\rangle_{\omega,k}$, and ask if there are any gapless excitations, that is, if $G(\omega,k)$ has poles $\omega=\epsilon_k$ with $\epsilon_k\to 0$ for some $k$. In (s-wave) BCS theory the absence of gapless modes is an automatic consequence of the presence of an order parameter, but that does not have to be the case.

A trivial counter-example is an order parameter with higher spin/orbital angula momentum. This occurs in $^3He$ ($p$-wave, though neutral) and high $T_c$ compounds (believed to be $d$-wave). In these cases the order parameter has nodal points or surfaces.

What is usually meant by gapless superconductivity is a more complicated situation, $s$-wave superconductivity in the presence of impurities (for example magnetic impurities, or ferro-magnetic order). In this case there is a range of temperatures where the order parameter persists, but the gap vanishes. The standard references are chapter 8 of de Gennes, Superconductivity of Metals and Alloys, as well as chapter 21 of Abrikosov, Fundamentals of the Theory of Metals.

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  • $\begingroup$ Thanks for your answer. There is still something which I don't really understand about it. So if the zero resistance can be explained only with the coherence effect, then I don't understand the following situation: If one applies an electric field to the superconductor in order to induce a supercurrent, the field also acts on the normal electrons which are still present in a sc. So why isn't there resistance due to the normal electrons? (If we consider a gapless situation) $\endgroup$
    – Motionx
    Jun 15, 2021 at 8:13
  • $\begingroup$ @Motionx Zero resistance is a more complex phenomenon than the Meissner effect, which is why the Meissner effect is usually taken to be the defining feature. If there is an order parameter, $\Phi$ than we have a supercurrent $j_s\sim\nabla\phi$, where $\phi$ is the phase. In both gapped and non-gapped superconductors there will in general (at $T\neq 0$) be a normal, dissipative, current as well. $\endgroup$
    – Thomas
    Jun 16, 2021 at 18:32
  • $\begingroup$ So why does the supercurrent not decay? The situation is easiest in a ring geometry, because decay is prevented by quantization of flux/circulation. In a straight wire we usually appeal to the Landau criterion. Here, indeed, a gapless superconductor is complicated because superficially the Landau criterion would seem to give a zero critical velocity/current. $\endgroup$
    – Thomas
    Jun 16, 2021 at 18:35
  • $\begingroup$ Abrikosov mumbles something about the Landau criterion not being applicable. I'm not sure I understand his reasoning. $\endgroup$
    – Thomas
    Jun 16, 2021 at 18:37

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