# (Topological) Superconductor is an Insulator

1. Superconductor may be viewed as an Insulator, if the Superconductor is formed by the BCS pairing where the excitations around the Fermi surface are gaped everywhere. However, superconductor also superconducts. So how can a superconductor be both an Insulator, as well as a superconductor?

2. Topological Superconductor may be viewed as an Insulator, if the Superconductor is formed by the BCS pairing where the excitations around the Fermi surface are gaped everywhere. However, Topological superconductor also superconducts. So how can a Topological superconductor be both an Insulator, as well as a superconductor?

3. The classifications of Topological Superconductor requires the assumption that the phases of matter to be gapped in the energy spectrum between ground state and 1st excited states, such as the work of Kitaev and Ryu et all. If Topological superconductor also superconducts, it should be gapless, should the classifications fail when Topological Superconductor becomes gapless and superconduct?

• First and second questions are the same aren't they ? The only requirement for the classification is the presence of a gap, hence it works for superconductvity as well. In fact superconductivity enlarges the classification by adding a few extra groups (the one you can reach by requiring particle-hole symmetry), as found by Altland and Zirnbauer in 96. A gapless state would not superconduct, because the gapless state would be influenced by impurities. All states (superconductor, topological insulator and topological superconductors), have edge states which conduct, though of different origins. – FraSchelle Jan 8 '17 at 10:28
• @FraSchelle, thanks However, for superconductor, at DC limit of conductivity $\sigma(\omega)= \infty$ as $\omega \to 0$ . So other than the gap, there is a density at $E=0$ that looks like to be gapless? – user32229 Jan 8 '17 at 21:02
• A superconductor is gapped. There is no excitation (i.e. no state) below the gap. Still the DC conductivity is theoretically infinite (in fact it's not because a current generates a magnetic field which destroys the superconductivity if too strong ; it's called the critical current of a bulk superconductor). To reconcile all the ideas, the supercurrent has to be on the edge, see the book by London on Superfluids. This edge state are not of the same origin than the bulk-boundary correspondence in topological systems. – FraSchelle Jan 9 '17 at 12:30

1. Answer to questions 1 and 2:

The energy gap that is most commonly referred to in a superconductor is the gap in the dispersion of Bogoliubov quasiparticles. These quasiparticles do not contribute to superconductivity. The supercurrent, which is composed of Cooper pairs, is responsible for superconductivity. For any non-zero temperature below the critical temperature ($T_{\rm c}$), the conduction electrons in a metal (which now superconducts) contribute to the formation of two types of quasiparticles:

1. Cooper pairs (bound state of two electrons), or
2. Bogoliubov quasiparticles (superposition of electron and hole)

It is the Cooper pairs that flow without any resistance. Despite the existence of a non-zero density of (non-superconducting) Bogoliubov quasiparticles at finite temperatures, the reason the resistance remains zero below $T_{\rm c}$ is that the Cooper pair condensate "shorts out" the Bogoliubov quasiparticles. In other words, in terms of a circuit diagram, you have a finite resistor (Bogoliubov quasiparticles) in parallel with a zero resistor (Cooper pair condensate).

Ironically, the energy gap helps in maintaining superconductivity. Here's why: below $T_{\rm c}$, when we break a Cooper pair, by supplying an energy equal to the energy gap, we get a pair of Bogoliubov quasiparticles (not electrons). If the gap were zero (everywhere in the Brillouin zone) we would not have Cooper pairs, thus no supercurrent, and thus no superconductivity!