# Tight binding on sawtooth 1D lattice

I am reading the paper "Bose condensation in flat bands" (arXiv).

The authors consider a tight-binding model on the one-dimensional "sawtooth" lattice, comprised of two sites A and B in the unit cell $a$. They arrive at a Hamiltonian (their equation 2), something like this

$H = \pmatrix{ 2 \cos (k) && 1+ e^{i k a} \\ 1+ e^{-i k a} && 0}$

I am not understanding how there is an asymmetry on the diagonal components. My naive thinking is site A and site B should produce the same matrix element.

Where is my mistake?

Note that the authors define $\vec{b}_k=[b_{B,k},b_{A,k}]^T$ and write the Hamiltonian $$H_\text{kin}=\sum_k \vec{b}_k ^\dagger\left[\begin{array}{cc} 2t\cos(ka) & t'(1+e^{ika})\\ t'(1+e^{-ika}) & 0 \end{array}\right]\vec{b}_k.$$ The $1,1$ element therefore corresponds to hopping between $B$ sites, which, by Fig. 1(d), is just the tight-binding chain. Similar cannot be said for the $A$ sites, hence the $2,2$ element being zero.
• @Nigel1 No, the point of the sawtooth lattice is that there is no hopping between $A$ sites (otherwise it would be a triangular lattice). FYI, lines in the figure indicate hopping terms. Oct 10, 2015 at 12:06