each $k$. Now our task is simply to diagonalise this two dimensional matrix that will give us the bulk dispersion relation in terms of the energy eigenvalues: $$\begin{aligned}
& E(k)=\pm \sqrt{v^2+w^2+2vw\cos(k)}\end{aligned}$$ Likewise, the eigenstates (eigenvectors) look like, $$\begin{aligned}
 \ket{\pm(k)} = \begin{pmatrix}
\pm e^{-i\phi(k)} \\
1
\end{pmatrix} \nonumber \\
\phi(k)=\tan^{-1} \left(\frac{w\sin{k}}{v+w\cos{k}}\right) \end{aligned}$$

Note that our Hamiltonian has two parameters: $v$ and $w$. It is clear that different choices of these parameters will lead to different dispersion relations. Let us try to see what these band structures look like and what can one conclude from the dispersion. Note that here we have considered the Fermi energy to be the zero energy level – this is the origin in our energy scale relative to which all other energies are measured here.

The plots in FIG. \[sshdispersion\] suggest that when there is staggering, $v\ne w$, in the Hamiltonian, the dispersion is gapped and hence a staggered case would be an insulator. Only in the case of $v=w$ the gap closes and there are states available at arbitrary low energies above the Fermi level, therefore $v=w$ case is a metal.

Beyond energy-band description
------------------------------

From the Bulk Hamiltonian, it seems like the problem is symmetric 