Posts Plotly: Tight-binding Model for Graphene

Plotly: Tight-binding Model for Graphene

Honeycomb lattice of graphene

The unit cell of graphene’s lattice consists of two different types of sites, which we will refer to as $A$ and $B$ sites (see Figure 1).

Figure 1. Honeycomb lattice of graphene where different colors are used to denote the two sublattices. The basis vectors of the unit cell are shown with black arrows. This figure is generated by TikZ/LaTeX.

With the basis vectors, the cell can be defined by the cell vector

\[\begin{equation} \boldsymbol{R}_n = j\cdot\vec{a}_1 + k\cdot\vec{a}_2 \end{equation}\]

Below we will used $(j, k)$ to denote the cell index.

Nearest-neighbour Tighting Binding Model for Graphene

The Hamiltonian of the graphene lattice

\[\begin{equation} {\hat H} = -\frac{\hbar^2}{2m}\nabla^2 + U(\boldsymbol{r}) \end{equation}\]

where $U(\boldsymbol{r})$ is the periodic potential of the crystal. The Schrödinger equation follows

\[\begin{equation} \left[ -\frac{\hbar^2}{2m}\nabla^2 + U(\boldsymbol{r}) \right] \psi_{\boldsymbol{k}}(\boldsymbol{r}) = E(\boldsymbol{k}) \psi_{\boldsymbol{k}}(\boldsymbol{r}) \end{equation}\]

where $\psi_{\boldsymbol{k}}(\boldsymbol{r})$ is the wavefunction of the electron, and $E(\boldsymbol{r})$ is the corresponding energy.

Assuming that $w_\alpha(\boldsymbol{r} - \boldsymbol{R}_n)$ are orthonormal wavefunctions located at site $\alpha$ of the cell $\boldsymbol{R}_n$, i.e.

\[\begin{equation} \int_V\, w^*_\alpha(\boldsymbol{r} - \boldsymbol{R}_n) w_\beta(\boldsymbol{r} - \boldsymbol{R}_m) \, \mathrm{d}\boldsymbol{r} = \delta_{\alpha\beta} \delta_{mn} \end{equation}\]

According to the Bloch’s theorem, the wavefunction of the electron in a periodic potential can be written as linear combination of the Bloch sum, i.e.

\[\begin{equation} \psi_{\boldsymbol{k}}(\boldsymbol{r}) = \frac{1}{\sqrt{N}}\sum_\alpha C_\alpha(\boldsymbol{k}) \sum_n e^{i\boldsymbol{k}\cdot\boldsymbol{R}_n} w_\alpha(\boldsymbol{r} - \boldsymbol{R}_n) \end{equation}\]

By expanding the Hamiltonian in the Bloch sum basis, we have

\[\begin{equation} H(\boldsymbol{k}) = \begin{bmatrix} H_{11} & H_{12} \\ H_{21} & H_{22} \end{bmatrix} \end{equation}\]

where $H_{\alpha\beta}$ is given by

\[\begin{align} H_{\alpha\beta} &= \frac{1}{N} \sum_{nm} \int_V\, e^{i\boldsymbol{k}\cdot(\boldsymbol{R}_m - \boldsymbol{R}_n)}\, w^*_\alpha(\boldsymbol{r} - \boldsymbol{R}_n) \, {\cal H}(\boldsymbol{r}) \, w_\beta(\boldsymbol{r} - \boldsymbol{R}_m) \, \mathrm{d}\boldsymbol{r} \nonumber \\[6pt] &= \sum_n \int_V\, e^{i\boldsymbol{k}\cdot\boldsymbol{R}_n}\, w^*_\alpha(\boldsymbol{r} - \boldsymbol{R}_n) \, {\cal H}(\boldsymbol{r}) \, w_\beta(\boldsymbol{r} - \boldsymbol{R}_0) \, \mathrm{d}\boldsymbol{r} \cr \end{align}\]

where we used the fact that the integrand only depends on the distance between the site $\boldsymbol{R}_n$ and $\boldsymbol{R}_m$.

In the nearest-neighbours tight-binding model, we only considter those terms with $ | n | \le 1 $.

The diagonal matrix elements

Apparently, only $n=0$ contributes to the matrix elements $H_{11}$ and $H_{22}$, therefore

\[\begin{align} H_{11} = H_{22} &= \int_V\, w^*_1(\boldsymbol{r} - \boldsymbol{R}_0) \, {\cal H}(\boldsymbol{r}) \, w_1(\boldsymbol{r} - \boldsymbol{R}_0) \, \mathrm{d}\boldsymbol{r} \nonumber \\[6pt] &= \int_V\, w^*_2(\boldsymbol{r} - \boldsymbol{R}_0) \, {\cal H}(\boldsymbol{r}) \, w_2(\boldsymbol{r} - \boldsymbol{R}_0) \, \mathrm{d}\boldsymbol{r} \nonumber \\[6pt] &= \varepsilon_0 \end{align}\]

The off-diagonal matrix elements

Let’s first inspect the $H_{12}$ term. One can see from Figure 1 that the nearest-neighbours of the $A$ site in cell $(j, k)$ located in the cell $(j, k$), $(j-1, k)$ and $(j, k-1)$, respectively. As a result, the $H_{12}$ term can be readily written.

\[\begin{align} H_{12} &= \int_V\, w^*_1(\boldsymbol{r} - \boldsymbol{R}_0) \, {\cal H}(\boldsymbol{r}) \, w_2(\boldsymbol{r} - \boldsymbol{R}_0) \, \mathrm{d}\boldsymbol{r} \nonumber\\[6pt] &+ e^{-i\boldsymbol{k}\cdot\boldsymbol{a}_1}\, \int_V\, w^*_1(\boldsymbol{r} - \boldsymbol{R}_{(-1, 0)}) \, {\cal H}(\boldsymbol{r}) \, w_2(\boldsymbol{r} - \boldsymbol{R}_0) \, \mathrm{d}\boldsymbol{r} \nonumber \\[6pt] &+ e^{-i\boldsymbol{k}\cdot\boldsymbol{a}_2}\, \int_V\, w^*_1(\boldsymbol{r} - \boldsymbol{R}_{(0, -1)}) \, {\cal H}(\boldsymbol{r}) \, w_2(\boldsymbol{r} - \boldsymbol{R}_0) \, \mathrm{d}\boldsymbol{r} \end{align}\]

By further assuming that the integral of the last three terms are the same, i.e.

\[\begin{equation} t = \int_V\, w^*_1(\boldsymbol{r} - \boldsymbol{R}_0) \, {\cal H}(\boldsymbol{r}) \, w_2(\boldsymbol{r} - \boldsymbol{R}_0) \, \mathrm{d}\boldsymbol{r} \end{equation}\]

we have

\[\begin{equation} H_{12} = t\left[ 1 + e^{-i\boldsymbol{k}\boldsymbol{a}_1} + e^{-i\boldsymbol{k}\boldsymbol{a}_2} \right] \end{equation}\]

Similarly, we have

\[\begin{equation} H_{21} = t\left[ 1 + e^{i\boldsymbol{k}\boldsymbol{a}_1} + e^{i\boldsymbol{k}\boldsymbol{a}_2} \right] \end{equation}\]

The complete matrix

Putting together all the matrix element, we have

\[\begin{equation} H(\boldsymbol{k}) = \begin{bmatrix} \varepsilon_0 & t\left[ 1 + e^{-i\boldsymbol{k}\boldsymbol{a}_1} + e^{-i\boldsymbol{k}\boldsymbol{a}_2} \right] \\ t\left[ 1 + e^{i\boldsymbol{k}\boldsymbol{a}_1} + e^{i\boldsymbol{k}\boldsymbol{a}_2} \right] & \varepsilon_0 \end{bmatrix} \end{equation}\]

The eigenvalues $E(\boldsymbol{k})$ can then be easily obtained

\[\begin{align} E(\boldsymbol{k}) &= \varepsilon_0 \pm t\sqrt{ 1 + e^{-i\boldsymbol{k}\boldsymbol{a}_1} + e^{ i\boldsymbol{k}\boldsymbol{a}_1} + e^{-i\boldsymbol{k}\boldsymbol{a}_2} + e^{ i\boldsymbol{k}\boldsymbol{a}_2} + (e^{-i\boldsymbol{k}\boldsymbol{a}_1} + e^{-i\boldsymbol{k}\boldsymbol{a}_2}) (e^{ i\boldsymbol{k}\boldsymbol{a}_1} + e^{ i\boldsymbol{k}\boldsymbol{a}_2}) } \nonumber \\[6pt] &= \varepsilon_0 \pm t\sqrt{ 3 + 2\cos(\boldsymbol{k}\boldsymbol{a}_1) + 2\cos(\boldsymbol{k}\boldsymbol{a}_2) + 2\cos(\boldsymbol{k}(\boldsymbol{a}_1-\boldsymbol{a}_2)) } \end{align}\]

One of the definitions for the hexagonal lattice is

\[\begin{align} \vec{a}_1 &= {a\over2}(3, \sqrt{3}) \\[6pt] \vec{a}_2 &= {a\over2}(3,-\sqrt{3}) \end{align}\]

where $a$ is the bond-length of the graphene. With this definition, the energy band of graphene follows.

\[\begin{equation} E_(k_x, k_y) = \pm t \sqrt{ 3 + 2 \cos(\sqrt{3}k_y a) + 4 \cos({3\over2}k_x a)\cos({\sqrt{3}\over2}k_y a) } \end{equation}\]

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