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:

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

  • $\begingroup$ The low energy Hamiltonian of bilayer graphene (describing the two bands closest to the chemical potential) is given by a $2 \times 2$ matrix. Depending on the type of stacking the electron will exist, in this approximation, only on two types of sites. Hence the two spinor components correspond to sublattices corresponding to these type of sites. Please let me know if that helps. Otherwise I can give an explicit example in the form of a formal answer. $\endgroup$
    – NanoPhys
    Commented Aug 11, 2013 at 1:44
  • $\begingroup$ The answer below and this comment clears things up greatly - Ill accept an official answer soon. Are there are good books or papers on this? Conceptually, one thing I can't see right away, is why are the electrons predominately on these two lattice sites? $\endgroup$
    – Fire
    Commented Aug 11, 2013 at 6:05
  • $\begingroup$ @Fire It is the low-energy electrons that dominate these two sites. The other two sites host high-energy electrons, which we do not care much when considering the low temperature physics of the system. The reason that the other two sites are of higher energy is because they are stacked right above each other such that the interlayer coupling between them is strong. The strong coupling locks their electrons together, as if to form the valence bound, such that effectively the electron degrees of freedom on the high-energy sites are quenched (or localized, no longer itinerant). $\endgroup$ Commented Aug 11, 2013 at 7:35
  • $\begingroup$ I see. Can you always write the bilayer as a 2x2 matrix, or is it only when it is AB stacked and low energy approx? $\endgroup$
    – Fire
    Commented Aug 11, 2013 at 15:19

1 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 $\hbar^2/(2m)=v^2/\gamma'$).

  • $\begingroup$ What if there be a potential difference between top and bottom layers of bilayer graphene? Then where should we add the potential? $\endgroup$
    – Roy
    Commented Sep 5, 2014 at 16:15
  • 1
    $\begingroup$ @Hesam Then you should add the potential term on the diagonal using $\sigma^z$ matrices. In that case, the degeneracy will be completely removed, and the problem becomes a non-degenerated perturbation. $\endgroup$ Commented Sep 10, 2014 at 4:31
  • $\begingroup$ yes, but if we do so the band structure is not exactly similar to the literature. $\endgroup$
    – Roy
    Commented Sep 12, 2014 at 13:38

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