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.


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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.