We know that the hamiltonian SSH model in the presence of on-site potential(V) can be written on the basis of the Pauli matrix.
$$h(k)=V\sigma_0+h_x\sigma_x+h_y\sigma_y,$$
and the term V breaks the chiral symmetry by shifting the zero energy topological edge state.
So, my question is: Does the identical matrix affect topology?
In the past, I came across the same problem. Here is what I figured out so far, but I would appreciate if someone could give a better answer.
In short, the $V$ term in your Hamiltonian can be gotten rid of by going into a suitable rotating frame. Consider the following unitary $$U(t) = \exp(-i V \sigma_0 t/\hbar)$$ transforming the Hamiltonian leads to \begin{eqnarray} h'(k) &=& U^{\dagger}(t) h(k) U(t) - i\hbar U^{\dagger}(t) \frac{\partial}{\partial t} U(t)\\ &=& h(k) - \hbar V\sigma_0 = h_x \sigma_x + h_y \sigma_y \end{eqnarray} which is chiral symmetric, i.e. $$\sigma_z h'(k) \sigma_z = - h'(k)$$ So, choosing $U(t)\sigma_z$ as your unitary operator for the chiral symmetry, this should mean that also $h(k)$ is chiral symmetric.
The point I am not sure about is whether or not one is "allowed" to choose a time-dependent unitary operation for the chiral symmetry operation.
