In class we were briefly discussing, that one way to see if the edge of a TI carries a state is to consider the scattering of a lead that is attached to this edge. In fact the argument was more general: consider the surface of a solid with a lead attached, that is cut off at some point. A state living on the surfaces running into the lead is connected to a state coming from the lead via the reflection matrix. $$\Psi_{in} = R(\theta) \Psi_{out}$$ If there is a state living on the surface then scattering on the end of the lead means acquiring a phase of -1 (the end is a hard wall). Therefore $$\Psi_{in} = - \Psi_{out}$$ combining the statements a condition that a (topological) state lives on the surface (edge) is then $$det(R(\theta)+1) = 0$$ Now the argument continued that the scattering matrix can be viewed as a phase $e^{i\theta}$. Here $\theta$ is depending on the momentum along the surfaces, so that $\theta = \theta(k)$. Continuity now demands that $$\theta(k+2\pi) = \theta(k)+n \cdot 2\pi$$ The main point now is that if $n \neq 0$ the phase winds at least once while moving through the Brillouin zone and that guarantees that there is a state at the given energy living on the surface. Can someone explain to me why a phase winding looking like the figure below does exclude any state from the surface?
