# Bloch Hamiltonian of a Hexagonal Lattice

As explained here (link in the first comment. couldn't post more than 2 links here!), the Bloch Hamiltonian of a lattice is obtained as $$h_k = \sum_je^{i\mathbf{k.R_j}}h_j$$ where $h_j$ is the hopping matrix, and $j$ runs over all lattice vectors.

For a square lattice, again as said in the linked lectures, for $h_j$ and then $h_k$ we have:

My question is about the next example in the lectures which is a hexagonal lattice. According the formula above for $h_j$, with $j$ running over all lattice vectors, the hexagonal shouldn't be any different than the square lattice above. But this is what the lecture gives for $h_k$:

Which shows $h_j$ has an extra $e^{i\mathbf{k.(a_1-a_2)}}$ term! Where does this term come from? Shouldn't we just have $a_1,a_2,-a_1,-a_2$ translation vectors, as in the square lattice?

If you consider only $a_1, a_2, -a_1, -a_2$, you miss 2 neighbours. You also need to consider $a_1 - a_2$ and its opposite, which gives you the top-most and down-most vertices on your last diagram, respectively (starting from the central vertex).
• Thanks. But my confusion was that, isn't the unit cell that we should cover by the $R_j$ vectors the highlighted rhomboid part of the whole hexagon, rather its whole vertices and structure? (Because we can choose the unit cell to be that rhomboid) Commented Oct 20, 2017 at 6:47