# Why does particle-hole symmetry in 1D lead to a $Z_2$ topological invariant?

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.

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