In many cases I see articles that introduce a material that has zero gap in the absence of spin-orbit interaction and then the spin-orbit opens a gap and author study the topological phase.
My question is, is it possible to have a material that has large band gap (1 eV) (even without spin-orbit) and by considering the spin-orbit become topological insulator?
The spin-orbit interaction in certain materials can lead to inversion of the bands, i.e., some of the valence bands cross the gap to become conduction bands and vice versa. A topological band theory requirement states that at the boundaries of this material, there must be gapless edge states. In 3D TIs, the edge states are in both the kx and ky momentum directions, so they form "cones" in Fermi space at the outer surfaces of the material, and hence they're called surface states. Such states of matter have been known prior to the discovery of TIs, such as quantum Hall and quantum spin-Hall insulators.
Look at the formalism introduced by Fu and Kane (doi:10.1103/PhysRevB.76.045302), which is utilized by Zhang et al., (doi:10.1038/nphys1270, Fig 2, 3) to predict Bi2Se3, Bi2Te3 and Sb2Te3 as 3D TI materials. All three materials already possess a bandgap (albeit small, ~0.2-0.3 eV) and are "semiconductors" from conventional band theory ignoring spin-orbit coupling (SOC). Addition of SOC to the band structure calculations leads to inversion of the bands at the Γ point in the momentum space. As a requirement of a topological invariant (more info in doi:10.1103/RevModPhys.82.3045, Section II), conducting gapless states are introduced in such "gapped" systems at their edges/boundaries. This happens as the topological invariant changes as it goes through the boundary between the strong SOC material and another "non-topological material", e.g. vacuum or free-space.
On the other hand, Bismuth and Antimony are semi-metals, i.e., materials without a bandgap. But when mixed in a certain ratio, the SOC in the material opens up a bandgap and introduced conducting edge or surface states (doi:10.1103/RevModPhys.82.3045, Fig 8). Pure Bismuth, when grown in thin layers on a substrate, goes through a semi-metal to semiconducting transition, and is expected to be a 2D TI due to its strong SOC (doi: 10.1103/PhysRevLett.109.166805, 10.1103/PhysRevLett.113.206802).
So to answer your question: if there is a material system, in which strong SOC exists and leads to inversion of bands, and this band-inversion gap is ~1 eV, then you can have gapless states at the boundaries of the material and an insulating phase in the bulk. However, for reasons that I personally am unaware of, most TI material to date tend to have smaller insulating bandgaps in the bulk. Moreover, they are not truly insulating in the bulk due to material defects and dopants.
The spin-orbit interaction in certain materials can lead to inversion of the bands, i.e., some of the valence bands cross the gap to become conduction bands and vice versa. A topological band theory requirement states that at the boundaries of this material, there must be gapless edge states. In 3D TIs, the edge states are in both the kx and ky momentum directions, so they form "cones" in Fermi space at the outer surfaces of the material, and hence they're called surface states. Such states of matter have been known prior to the discovery of TIs, such as quantum Hall and quantum spin-Hall insulators.
Look at the formalism introduced by Fu and Kane (doi:10.1103/PhysRevB.76.045302), which is utilized by Zhang et al., (doi:10.1038/nphys1270, Fig 2, 3) to predict Bi2Se3, Bi2Te3 and Sb2Te3 as 3D TI materials. All three materials already possess a bandgap (albeit small, ~0.2-0.3 eV) and are "semiconductors" from conventional band theory ignoring spin-orbit coupling (SOC). Addition of SOC to the band structure calculations leads to inversion of the bands at the Γ point in the momentum space. As a requirement of a topological invariant (more info in doi:10.1103/RevModPhys.82.3045, Section II), conducting gapless states are introduced in such "gapped" systems at their edges/boundaries. This happens as the topological invariant changes as it goes through the boundary between the strong SOC material and another "non-topological material", e.g. vacuum or free-space.
On the other hand, Bismuth and Antimony are semi-metals, i.e., materials without a bandgap. But when mixed in a certain ratio, the SOC in the material opens up a bandgap and introduced conducting edge or surface states (doi:10.1103/RevModPhys.82.3045, Fig 8). Pure Bismuth, when grown in thin layers on a substrate, goes through a semi-metal to semiconducting transition, and is expected to be a 2D TI due to its strong SOC (doi: 10.1103/PhysRevLett.109.166805, 10.1103/PhysRevLett.113.206802).
So to answer your question: if there is a material system, in which strong SOC exists and leads to inversion of bands, and this band-inversion gap is ~1 eV, then you can have gapless states at the boundaries of the material and an insulating phase in the bulk. However, for reasons that I personally am unaware of, most TI material to date tend to have smaller insulating bandgaps in the bulk. Moreover, they are not truly insulating in the bulk due to material defects and dopants.
Dear Tanuj and Elman,
Thanks for your answers. It seems that Bi2Se3, Bi2Te3 and Sb2Te3 are good example for semiconductors that becomes TI by considering the spin-orbit interaction.
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