# Is there a topological insulator without Spin Orbit Coupling (SOC)?

• A
There are some famous materials is determined as TI induced by SOC, like graphene and so on. But from some formula, for instance, Kane-Fu formula, they just need parities to get Z2 number. So I wonder if there is a known TI with weak soc.

• atyy

king vitamin
Gold Member
For non-interacting electrons, the answer for when you can get a topological insulator is given by looking at the general classification scheme, which has been worked out, and seeing what symmetry class your Hamiltonian is in, see https://arxiv.org/abs/0803.2786 (especially Table I). The Kane-Mele state is in the d=2 AII class, and the Fu-Kane state is d=3 AII.

The key point in applying the classification scheme to these particular models is that they have time-reversal (TR) symmetry implemented by $T = i \sigma^y K$ where $K$ is complex conjugation and $\sigma^y$ is the spin matrix. If the spin does not appear explicitly in the Hamiltonian but you still have TR symmetry, then you would have invariance under $T = K$ and you end up in the AI class, and there are no topological insulators. (Also, without TR symmetry, you cannot get a TI in three dimensions, but you can in two: the Haldane model.)

But more abstractly, the existence of a possible TI state "just" boils down to how it transforms under symmetries as outlined in the linked paper. So if you have time reversal symmetry which acts on your Hamiltonian as $T = UK$ where $U^2 = -1$, you can get a TI whether the matrix $U$ is acting on spin or some other degree of freedom.

Finally, if you're interested in topological insulators outside of the Kane-Mele/Fu-Kane Z2 classifications, the answer is emphatically yes! As shown in the linked paper, there are topological superconductors with full SU(2) spin symmetry.

I'm afraid I'm not familiar enough with TIs to discuss the specific microscopic models which have been proposed or realized experimentally, but the above generalities may be helpful.