# 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. Commented Jun 16, 2021 at 16:49

$$H(k)C = C H(-k)$$
Therefore when the Hamiltonian is spectrally gaped around zero and you call $$P_-(k)$$ and $$P_+(k)$$ the projectors over the negative and positive eigenvalue you get that:
$$P_+(k) C = CP_-(-k)$$.
This implies that the integral quantity you wrote when computed using the frame $$u_j (k)$$ of $$P_-(k)$$ is identical to the one computed using the simmetrical frame $$C u_j(k)$$ of $$P_+(k)$$. However the sum of those two quantities must be an even number, so if one of them is even the other must even as well and same happens when one of them is odd.