Topological insulator vs. topological superconductors in any dimension My question today is simple. What is the difference between a topological insulator and a topological superconductor? How that difference depends on the dimensionality of space(time)? What is the difference in terms of the so-called edge states?
PS: I only know that a topological insulator is some material that is conductor on the edge but insulator in the bulk...Is that definition right in ANY dimension?What about topological superconductors?
 A: For the sake of simplicity, let us limit this discussion to non-interacting fermionic phases: e.g. band insulators with electron quasiparticles and fully gapped superconductors with Bogoliubov quasiparticle excitations. Classification of interacting and/or non-fermionic phases is still work in progress. Before outlining basic differences between topological insulators (TIs) and topological superconductors (TSCs) in words, it is worth mentioning that classification of non-interacting fermionic topological phases, with different symmetries and dimensionality, is elegantly summarized in the following “periodic table”

where $\Theta$ and $\Xi$ represent time-reversal symmetry (TRS) and particle-hole symmetry (PHS) respectively.
Under the above assumptions, we can represent both TIs and TSCs by at least four- and two-band models (with a gap) respectively. The TI has twice the number of bands due to Kramers degeneracy, which is a result of TRS. That’s the first difference: TIs must have TRS, but TSCs may or may not possess TRS.
Another difference is that a TI is protected by conservation of ${\rm U}(1)$ (particle conservation) symmetry, whereas a TSC (or any superconductor) necessarily has to break ${\rm U}(1)$ symmetry. In other words, a superconductor must possess PHS, which is a manifestation of ${\rm U}(1)$ symmetry breaking; whereas an insulator may or may not possess PHS (depending on the model/material parameters of the insulator) while still preserving ${\rm U}(1)$ symmetry. You might want to check out:

Why does a superconductor obey particle-hole symmetry?

for subtleties associated with PHS in a superconductor; but we’ll just take it for granted here.
Here are some examples, of different topological phases, classified in terms of their boundary states (in any dimension), chosen in order to illustrate some of the above differences:


*

*The experimentally discovered TIs, i.e. Quantum Spin Hall Effect (QSHE), such as HgTe/CdTe quantum wells, and 3D band TIs, such as Bi$_{2}$Se$_{3}$, belonging to columns $d = 2$ and $d = 3$ respectively, of class ${\rm AI\,I}$, both have $\mathbb{Z}_{2}$ classification, which indicate the number of topological protected Kramers pairs and Dirac cones respectively: 0 or 1. Both TIs (obviously) satisfy TRS.

*The experimentally studied superconducting nanowires, belonging to the $d=1$ column of class ${\rm D}$, also has $\mathbb{Z}_{2}$ classification, which indicates the existence or absence of Majorana zero modes (MZMs) at the edges of the nanowire. The 2D analog of the nanowire, i.e. the $p+{\rm i}p$-superconductor, has a $\mathbb{Z}$ classification, which indicates the number of well-separated vortices, and consequently the number of unpaired MZMs, in the system. Note that class ${\rm D}$ superconductors break TRS.

*Class ${\rm DI\,I\,I}$ superconductors satisfy TRS. The 1D version is like the class ${\rm D}$ nanowire; except each end of the wire has two species (Kramers pairs) of MZMs which do not annihilate each other. For the 2D case, we have a similar situation; except the two Kramers pair MZMs are counter-propagating edge modes (like the QSHE). Hence classification for both 1D and 2D is $\mathbb{Z}_{2}$. For 3D, however, we have a surface “Majorana cone,” instead of a Dirac cone, like in a 3D TI. The only difference in the case of the 3D TRS superconductor is that every Majorana cone is protected. Hence, depending on the material parameters, the surface can have any number of Majorana cones. Hence the classification is $\mathbb{Z}$.


In order to see this pattern emerge, in a rigorous fashion (besides just looking at examples), you need to take the so-called K-theory approach; for details you can look at:

Alexei Kitaev. “Periodic table for topological insulators and superconductors.” In ADVANCES IN THEORETICAL PHYSICS: Landau Memorial Conference, vol. 1134, no. 1, pp. 22-30. AIP Publishing, 2009. (arXiv)

As alluded to earlier, generalization of this classification procedure is still work in progress. One example is the study and classification of the so-called “Symmetry Protected Topological” (SPT) phases, which relaxes the non-interacting and/or fermionic constraints. At least one of the appeals of this classification approach to topological phases is the possibility of theoretically and/or experimentally discovering unexpected and mysterious new phases of matter.
A: *

*The topological insulator has the charge $U(1)$ symmetry, while the topological superconductor does not. In fact, this difference is general, and applies to all insulators and superconductors. In condensed matter physics, insulators and superconductors are both gapped fermion systems. Depending on whether the charge $U(1)$ symmetry is presented or not, they are further classified into these two categories.

*The difference between insulators and superconductors is just a matter of symmetry class, and is independent on the dimensionality. As you can see in the periodic table classification, the symmetry class (rows) and the dimensionality (columns) are two independent variables of free fermion SPT phases.

*The edge state of topological insulators are gapless Dirac fermions, while the edge state of topological superconductors are gapless Majorana fermions.
