Figure 1: Vector lattices and Brillouin zone of honeycomb lattice
Honeycomb lattice (or hexagonal lattice) is realized by graphene. The lattice consists of two carbon atoms (hereafter we call them $A$ and $B$ sites) per unit cell. The lattice vectors can be written as
All of above mentioned vectors are illustrated in Figure 1. On the right hand side of the figure first Brillouin zone of graphene is presented, which is the hexagon with two vertices at Dirac points $K$ and $K^\prime$. The reasons for choosing such BZ zone and for the name of those two points will be explained elsewhere.
where $a_{\vec{R}}$ and $b_{\vec{R}}$ are annihilation operators on sites $A$ and $B$ at position $\vec{R}$ respectively. To compute the energy band, one needs to perform Fourier transformation on the site operators as follows
As we expected, the Fourier-transformed Hamiltonian is off-diagonal, indicating there is no tunelling between $A$ and $B$ sublattices. It is straightforward to find eigenvalues of the above Hamiltonian and hence, find the analytical formula of energy band.
where $f(k_x, k_y) = 2 \cos (\sqrt{3} k_y a) + 4 \cos(\frac{\sqrt{3}}{2} k_y a) \cos(\frac{3}{2} k_x a) $. A visualization of graphene's energy band can be found in Figure 2.
Energy band of graphene
Honeycomb lattices
Honeycomb lattice (or hexagonal lattice) is realized by graphene. The lattice consists of two carbon atoms (hereafter we call them $A$ and $B$ sites) per unit cell. The lattice vectors can be written as
\[ \vec{a}_1 = \frac{a}{2} (3, \sqrt{3}), \quad \vec{a}_2 = \frac{a}{2} (3, -\sqrt{3}) ,\]
where $a$ denotes carbon-carbon distance. The corresponding reciprocal lattice vectors read
\[ \vec{b}_1 = \frac{2 \pi}{3a} (1, \sqrt{3}), \quad \vec{b}_2 = \frac{2 \pi}{3a} (1, - \sqrt{3}) .\]
In actual calculations, it is convenient to name these nearest-neighbor vectors
\[ \vec{\delta}_1 = a(1,0), \quad \vec{\delta}_2 = \frac{a}{2} (-1, \sqrt{3}), \quad \vec{\delta}_3 = \frac{a}{2} (-1, -\sqrt{3}). \]
All of above mentioned vectors are illustrated in Figure 1. On the right hand side of the figure first Brillouin zone of graphene is presented, which is the hexagon with two vertices at Dirac points $K$ and $K^\prime$. The reasons for choosing such BZ zone and for the name of those two points will be explained elsewhere.
Tight-binding Hamiltonian
The tight-binding Hamiltonian of graphene reads
\[ H = \sum_{\vec{R}, j} a^{\dagger}_{\vec{R}} b_{ \vec{R} + \vec{\delta}_j } + \text{h.c.}, \]
where $a_{\vec{R}}$ and $b_{\vec{R}}$ are annihilation operators on sites $A$ and $B$ at position $\vec{R}$ respectively. To compute the energy band, one needs to perform Fourier transformation on the site operators as follows
\[ \begin{pmatrix} a_{\vec{R}} \\ b_{\vec{R}} \end{pmatrix}^T= \frac{1}{\sqrt{N_{\text{unit}}}} \sum_{\vec{k}} e^{-i \vec{k} \cdot \vec{R}} \begin{pmatrix} a_{\vec{k}} \\ b_{\vec{k}} \end{pmatrix}^T .\]
The transformed Hamiltonian is thus represented in a compact form
\[ H = \sum_{\vec{k}} c^\dagger_{\vec{k}} h(\vec{k}) c_{\vec{k}},\]
where $c_{\vec{k}} = \begin{pmatrix} a_{\vec{k}}, b_{\vec{k}} \end{pmatrix}^T$ and
\[ h(\vec{k}) \begin{pmatrix} 0 & \underset{j = 1, 2, 3}{\sum} e^{i \vec{k} \vec{\delta}_j} \\\ \underset{j = 1, 2, 3}{\sum} e^{-i \vec{k} \vec{\delta}_j} & 0 \end{pmatrix}. \]
As we expected, the Fourier-transformed Hamiltonian is off-diagonal, indicating there is no tunelling between $A$ and $B$ sublattices. It is straightforward to find eigenvalues of the above Hamiltonian and hence, find the analytical formula of energy band.
\[ E_{\pm } (k_x, k_y) = \pm t \sqrt{3 + f(k_x, k_y)}, \]
where $f(k_x, k_y) = 2 \cos (\sqrt{3} k_y a) + 4 \cos(\frac{\sqrt{3}}{2} k_y a) \cos(\frac{3}{2} k_x a) $. A visualization of graphene's energy band can be found in Figure 2.
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