Graphene's Tight Binding Hamiltonian

Question

Graphene has two atoms in its primitive unit cell. This makes it intuitive to see that the tight binding Hamiltonian can be constructed as a $ 2 \times 2 $ matrix $H$ acting on a spinor $S$ that consists of the wavefunction from an atom in sublattice A and B.

$H_{monolayer}=\gamma \cdot \begin{pmatrix} 0 & k_x-ik_y \\ k_x+ik_y & 0 \end{pmatrix}$

$S_{monolayer}=\begin{pmatrix} |\psi_A\rangle\\ |\psi_B\rangle \end{pmatrix}$

Bilayer Graphene has four atoms in a primitive unit cell and its tight binding Hamiltonian is a 4x4 matrix whose matrix elements represent the hopping between said lattice sites (depending on how it is stacked and what hopping parameters you wish to involve in the calculation). An example might be as follows:

$H_{bilayer}=\begin{pmatrix} 0 & 0 & 0 & v(k_x-ik_y)\\ 0 & 0 & v(k_x+ik_y) & 0\\ 0 & v(k_x-ik_y) & 0 & \gamma'\\ v(k_x+ik_y) & 0 & \gamma' & 0 \end{pmatrix}$

$S_{bilayer}=\begin{pmatrix} |\psi_{A1}\rangle\\ |\psi_{B2}\rangle\\ |\psi_{A2}\rangle\\ |\psi_{B1}\rangle \end{pmatrix}$

Where the basis is chosen is in an arbitrary order (1 and 2 indices refer to layer number).

How does one write this in a "two-component basis" and what does that mean? Also, what is this hamiltonian acting on in this case? The bilayer hamiltation in this basis (which I do not know what it represents) is written as follows:

$H'_{bilayer}=-\dfrac{\hbar^2}{2m}\begin{pmatrix} 0 & (k_x-ik_y)^2 \\ (k_x+ik_y)^2 & 0 \end{pmatrix}$

Answer 1

The Hilbert space of the bilayer system, which is spanned by the four-component basis, can be divided into the low-energy subspace and the high-energy subspace. The low-energy subspace is spanned by $|\psi_{A1}\rangle$ and $|\psi_{B2}\rangle$, while the high-energy subspace is spanned by $|\psi_{A2}\rangle$ and $|\psi_{B1}\rangle$. It is the interlayer coupling $\gamma'$ that makes the four basis different. $|\psi_{A2}\rangle$ and $|\psi_{B1}\rangle$ are coupled to each other, forming the bounding and anti-bounding states: $|\psi_{A2}\rangle\mp|\psi_{B1}\rangle$ (of the energy $\mp\gamma'$), and hence pushed away from the Fermi surface (or the zero-energy level). While $|\psi_{A1}\rangle$ and $|\psi_{B2}\rangle$ are not coupled, and remain zero energy in the $k\to0$ (long-wave-length) limit. We wish to consider the effective Hamiltonian for the low-energy electrons near the Fermi level (because transport and many other physical properties are dominated by them). These low-energy electron states are mainly made up of $|\psi_{A1}\rangle$ and $|\psi_{B2}\rangle$, so we seek to reduce the 4-by-4 Hamiltonian to its upper 2-by-2 block. This can be done in terms of a 2nd-order perturbation. The resulting 2-by-2 Hamiltonian is acting in the low-energy subspace spanned by the basis $|\psi_{A1}\rangle$ and $|\psi_{B2}\rangle$.

Here are the details of how it works. To simplify our notation, we introduce the Pauli matrices $\sigma_x$, $\sigma_y$ and $\sigma_z$. The 4-by-4 bilayer Hamiltonian can be written as $$H_\text{bilayer}=\left(\begin{matrix}0&v\vec{k}\cdot\vec{\sigma}\\ v\vec{k}\cdot\vec{\sigma}&\gamma'\sigma_x\end{matrix}\right),$$ where $\vec{k}\cdot\vec{\sigma}=k_x\sigma_x+k_y\sigma_y$. Around the momentum $k\to 0$, $v \vec{k}\cdot\vec{\sigma}$ term can be treated as the perturbation. Explicitly, $H_\text{bilayer}=H_0+H_1$, with $H_0$ being the 0th-order Hamiltonian and $H_1$ being the 1st-order perturbation, $$H_0=\left(\begin{matrix}0&0\\ 0&\gamma'\sigma_x\end{matrix}\right),H_1=\left(\begin{matrix}0&v\vec{k}\cdot\vec{\sigma}\\ v\vec{k}\cdot\vec{\sigma}&0\end{matrix}\right).$$ Now we want to consider the perturbative correction brought to the upper 2-by-2 block of $H_0$ by $H_1$. According to the 2nd-order perturbation formalism, the correction (in the upper block) reads $$H_2'=-(v\vec{k}\cdot\vec{\sigma})(\gamma'\sigma_x)^{-1}(v\vec{k}\cdot\vec{\sigma})=-\frac{v^2}{\gamma'}((k_x^2-k_y^2)\sigma_x+2k_xk_y\sigma_y).$$ The effective Hamiltonian in the upper block should be given by $H_0'+H_1'+H_2'+\cdots$, but $H_0'=H_1'=0$, so we eventually have $$H_\text{bilayer}'\simeq H_2'=-\frac{v^2}{\gamma'}\left(\begin{matrix}0&(k_x-ik_y)^2\\(k_x+ik_y)^2&0\end{matrix}\right),$$ as expected (with the effective mass $m$ properly defined to make $\hbar^2/(2m)=v^2/\gamma'$).