Su–Schrieffer–Heeger model

Example

The Su–Schrieffer–Heeger (SSH) model is a tight-binding chain with two sites $A$ and $B$ in each unit cell and alternating hopping amplitudes. Let $v, w \ge 0$ be the hopping inside a cell and between neighbouring cells. On $\ell^2(\Z; \C^2)$, with $\psi(n) = (\psi_A(n), \psi_B(n))$, the Hamiltonian is $$(H\psi)_A(n) = v\,\psi_B(n) + w\,\psi_B(n-1), \qquad (H\psi)_B(n) = v\,\psi_A(n) + w\,\psi_A(n+1).$$ Its Bloch Hamiltonian is $$H(k) = \begin{pmatrix} 0 & h(k) \\ \overline{h(k)} & 0 \end{pmatrix}, \qquad h(k) = v + w e^{-ik}, \quad k \in [-\pi, \pi],$$ so the band structure is $E_\pm(k) = \pm\abs{v + w e^{-ik}}$.

Check of the gap: $\abs{v + we^{-ik}}$ ranges over $[\abs{v - w}, v + w]$, so the bands are separated by a gap around $0$ exactly when $v \ne w$.

Check of the topology: $H(k)$ anticommutes with $\sigma_z = \operatorname{diag}(1, -1)$ (chiral symmetry), so when $v \ne w$ the loop $k \mapsto h(k)$ avoids $0$ and has a winding number around $0$. The circle $v + w e^{-ik}$ has centre $v$ and radius $w$, so the winding number has absolute value $1$ if $w > v$ and is $0$ if $w < v$.

Check of the edge states: take the half-infinite chain $n \ge 1$ with $v = 0$ and $w > 0$. The site $A$ of the first cell has no partner, since its only bond would go to cell $0$. So $\psi = \delta_{A,1}$ satisfies $H\psi = 0$, a zero-energy state at the end, as predicted by the nonzero winding number.