about the $v=w$ case. By symmetric we mean that $v > w$ case is exactly the same as $w > v$ case. But this is not quite right. The information obtained from the ‘eigenvalues’ is not complete! We need to also look at the eigenvectors to get a complete information about these cases. Interestingly, the topological aspect of this problem is hidden in the eigenvectors.

Since the band Hamiltonian is a two dimensional Hermitian matrix it can be written in a more insightful notation as, $$\begin{aligned}
H(k) &= &\vec{h}(k)\cdot\vec{\sigma}\end{aligned}$$ using Pauli’s matrices, $\{\sigma_i\}$, as the basis for a two dimensional Hermitian matrix [@sakurai]. By comparing terms we get, $$\begin{aligned}
h_x(k) & = & v+w\cos{k} \nonumber \\
h_y(k) & = & w\sin{k} \nonumber \\
h_z(k) & = & 0\end{aligned}$$ As we saw earlier, the eigenstates are parametrised by $\phi=\tan^{-1} (\frac{w\sin{k}}{v+w\cos{k}})=\tan^{-1}{\frac{h_y(k)}{h_x(k)}}$. Therefore, in the $h_x-h_y$ space, the direction of the vector $\vec{h}(k)$ denotes an eigenstate and the magnitude of the vector will give its eigenvalue.

A plot of the trajectory of $\vec{h}(k)$ over the first Brillouin zone {$-\pi \rightarrow \pi$} gives the full set of eigenstates. If we plot the trajectories for the two seemingly equivalent cases: $v<w$ and $v>w$ we observe, 