0
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Why can the same not be said for A sites? Is hopping between A sites not also a tight-binding chain? $\endgroup$
    – Nigel1
    Oct 10, 2015 at 12:01
  • $\begingroup$ @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. $\endgroup$
    – sxwzd
    Oct 10, 2015 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.