= Solution
Two local gapped Hamiltonians are <topologically equivalent> when a continuous path of local Hamiltonians joins them without closing the bulk gap. The <Bogoliubov--de Gennes Hamiltonian> has energies
$$
E_\pm(k)=\pm\sqrt{(2t\cos k+\mu)^2+4\Delta^2\sin^2k}.
$$
For $t,\Delta>0$, this gap can close only at
$$
\mu=-2t\quad(k=0),
\qquad
\mu=2t\quad(k=\pi).
$$
The whole region $-2t<\mu<2t$ is connected and gapped, so its parameters can be continuously deformed to $t=\Delta,\mu=0$. Therefore
$$
\boxed{H\simeq H_0\quad\text{for }-2t<\mu<2t}.
$$
Writing
$$
H_{\rm BdG}(k)=d_y(k)\sigma_y+d_z(k)\sigma_z,
\qquad
d_y=-2\Delta\sin k,\quad
d_z=-2t\cos k-\mu,
$$
shows the topological distinction. As $k$ crosses the Brillouin zone, $(d_y,d_z)$ traces an ellipse. It encloses the origin once when $|\mu|<2t$, giving nonzero <winding number of a one-dimensional Bogoliubov--de Gennes Hamiltonian>. For $\mu<-2t$ it does not enclose the origin and has winding zero. Changing this integer requires the ellipse to pass through the origin, exactly the bulk gap closing.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-342-kitaev-ellipse-winding.png]
{title=Kitaev-chain ellipse and its winding around the origin}
{description=For chemical potential zero the Bogoliubov--de Gennes vector traces an ellipse enclosing the origin once. For chemical potential minus 3t the translated ellipse misses the origin and has winding number zero.}
Back to article page
