From the well-known AZ Tenfold Classification Table, a 1D system with square-positive particle-hole symmetry belong to class D and hence is characterized by a $Z_2$ topological invariant. I suppose that this means given a band Hamiltonian $H(k)$ and a unitary operator $C$, whose $C^2=1$ and anticommutes with $H(k)$, the topological index defined by $\frac{1}{\pi}\int\langle u_k|i\partial_k|u_k\rangle\mathrm dk \mod 2$ would be quantized for each band.

I can see how this is the case in the specific model. If we consider a two-band system, and let the particle-hole operator be $\sigma_z$, then the condition $\{H,\sigma_z\}=0$ is equivalent to $H$ lying in the $\sigma_x$-$\sigma_y$ plane. In this specific case I can show that the quantity defined by $\frac{1}{\pi}\int\langle u_k|i\partial_k|u_k\rangle\mathrm dk \mod 2$ must be either $0$ or $1$, by connecting $|u_k\rangle$ to the original $|u_0\rangle$ through $|u_k\rangle = \exp\left[\frac{i}{2}\Delta\theta(k)\sigma_z\right]|u_0\rangle$. But I don't see how this result follows from the general fact that a particle-hole operator $C$ exists.

  • $\begingroup$ You need C to be antiunitary, and so \sigma_z is not an example of a possible C operator. $\endgroup$ – Terry Loring Jun 16 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.