[Physics] Tight-binding model of graphene – filling in a few steps

graphenesolid-state-physicstight-binding

I'm trying to figure out the tight-binding model for graphene after not using any of my solid-state knowledge for a few years. I am getting lost in notation and I was hoping for some help filling in the gaps.

I am following the derivation here. I'm not entirely sure where to start since I have so many questions, but I suppose let's start with this:

They give the tight-binding wavefunction of graphene as

$\frac{1}{\sqrt{N}}\sum_{j}\exp(i\vec{k}\cdot\vec{R_j})\left[c^A\phi(\vec{r}-\vec{R_j})+c^B\phi(\vec{r}+\vec{\delta_3}-\vec{R_j})\right]$

where I've chosen to center the model on the A atoms and picked a vector to give me the B atoms.

The authors then give the hamiltonian to be diagonalized as

$-t\sum_{<ij>}|\phi^A_j><\phi^B_i|+h.c.$

where does this form come from? Can someone shed some light on how they arrived there? And what exactly are the eigenfunctions here? I believe from the way the derivation goes after this, the eigenfunctions are the sums over the A or B atoms in the wavefunction, but it's not clear why.

Finally, in all the derivations I can find, the hamiltonian ends up with zeros on the diagonal. Why are diagonal elements zero? Is there no energy contribution from the atoms themselves, but rather only from the hopping terms?

Best Answer

This is my favorite graphene reference. It's more concerned with the high magnetic field behavior of graphene (quantum Hall regime), but it's introduction is still very well done.

A few points:

-The wavefunction is the sum of two Bloch functions (one for each sublattice of graphene). The tight binding approximation assumes that the electronic wavefunction is well approximated by a sum of atomic orbitals, which are the states being summed over (the unbound pi orbital of each carbon atom). The overall wavefunction is described in a sublattice basis. So each coefficient describes the "weight" of the electronic state on a given sublattice.

-The tight binding Hamiltonian only considers the influence of neighboring atomic sites. Because of the structure of graphene, each carbon atom on sublattice A only has nearest neighbors on sublattice B. (If you were considering next-nearest neighbors, you would take into account interactions within the same sublattice as well). Thus, to capture this geometrical effect, you Hamiltonian has some term that looks like $|\phi_j^A><\phi^B_i|$ to describe the nearest neighbor coupling with strength $t$ between atomic orbitals on neighboring sites $i,j$.

-There is generally a diagonal term. You're exactly right, though, that this can be ignored within the nearest neighbor approximation. There is an on-site energy term which represents the energy of the atomic orbital at each lattice site. Because this wavefunction is written in a sublattice basis, the diagonal terms of the matrix represent the on-site energy of each distinct sublattice. Of course for graphene this is identical (since each sublattice is chemically identical). Thus, any contribution of the form $\epsilon \begin{pmatrix} 1&0\\0&1\end{pmatrix}$ can be ignored. If you did a similar treatment of a material like hexagonal boron nitride (sublattice of boron, sublattice of nitrogen) you would never be able to shift off the diagonal contributions because it would look like $\begin{pmatrix}\epsilon_A & 0\\0 & \epsilon_B\end{pmatrix}$ with distinct on-site energy terms.

-One caveat - you can only subtract off a constant on the diagonal. If you were to take into account next-nearest neighbor (nnn) hopping, you would find that there's also a diagonal contribution since each atom of one sublattice only has next-nearest neighbors on the same sublattice. In general, this corrective term will be proportional to some constant $t'$ and it will depend on the momentum $k$. So your new matrix would look like: \begin{equation} H = \begin{pmatrix} t'f(k) & t g(k) \\ t g^*(k) & t' f(k) \end{pmatrix}. \end{equation} where $f$ and $g$ are function of crystal momentum $k$ related to phase factors you pick up in your wavefunction.

It's tempting to think I could also shift off this diagonal term, but because I'm interested in the energy dispersion with respect to momentum $k$, I can't shift off an entire function. I could shift off this function evaluated at one point $k_0$, but I will still have nonzero diagonal terms at different points in the Brillouin zone.