The SSH model is a well-known topological model in one dimension, whose topological invariant is the Zak phase. The Zak phase is defined as:

$$ \gamma = \int_{-\pi}^{\pi} dk\ A(k) $$ where $A(k)$ is the Berry connection given by: $$ A(k) = i \langle u(k)|\frac{d}{dk}|u(k)\rangle $$ and $u(k)$ are the periodic part of Bloch wavefunctions of the SSH model. The Zak phase is quantized to $n \pi$, where $n = 0,1$, by inversion symmetry. When the Zak phase takes on the value of $\pi$, edge states appear at the ends of the SSH model chain, and they are topologically protected. However, if a symmetry-broken term is introduced, such as staggered onsite energies, the Zak phase is no longer quantized, but its numerical value can still approaches $\pi$, if the term is tiny.

If a SSH model with tiny symmetry-broken term has the numerical value of the Zak phase close to $\pi$, it will host edge states at the ends of a finite chain. Although the edge states are not degenerate, could they be considered approximately protected by the Zak phase since its value is close to $\pi$?


1 Answer 1


It's hard to tell that you might have a non quantized value of the Zak phase because your integral is a closed loop with respect A. If the function is analytic in the closed integration path, then the result will be always zero. Otherwise, will always be quantized to values of pi.

  • $\begingroup$ Zak phase is quantized only when certain symmetry is present (For SSH model, it's the chiral symmetry, e.g. Hamiltonian contains $\sigma_x$,$\sigma_y$ but no $\sigma_z$ where $\sigma_i$ is the Pauli matrix.). When symmetry is broken, the Zak phase is not quantized. $\endgroup$ Commented Dec 8, 2023 at 7:37

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