Phy5670/Phonon in Graphene

Structure of Graphene

Carbon atoms in graphene are constructed on a honeycomb lattice, which is shown in Fig. 1. The circle and solid point together combine into a two-point basis in the Bravais lattice. The carbon-carbon distance is $a~1.42{\mathrm {\AA} }$. The primitive vectors are

${\vec {a_{1}}}={\frac {\sqrt {3}}{2}}a{\hat {x}}+{\frac {3}{2}}a{\hat {y}}$

${\vec {a_{2}}}=-{\frac {\sqrt {3}}{2}}a{\hat {x}}+{\frac {3}{2}}a{\hat {y}}$

Then the reciprocal lattice parameters can by generated by using ${\vec {b_{i}}}\cdot {\vec {a_{j}}}=2\pi \delta _{ij},(i,j=1,2)$

${\vec {b_{1}}}={\frac {2\pi }{a}}({\frac {\sqrt {3}}{3}}{\hat {x}}+{\frac {1}{3}}{\hat {y}})$

${\vec {b_{2}}}={\frac {2\pi }{a}}(-{\frac {\sqrt {3}}{2}}a{\hat {x}}+{\frac {3}{2}}a{\hat {y}})$

General phonon dispersion relations

Before we move on to solving phonon in graphene, we discuss how to solve phonon problem for general case. Suppose we have particle system, which is constructed by unit cells. For each unit cell, there are N atoms. Let's use ${\vec {u_{i}}}=(x_{i},y_{i},z_{i})$ to represent the displacement of the $i^{t}h$ particle coordinate. Thus, the equations of motion for these N atoms in each unit are:

$M_{i}{\vec {{\ddot {u}}_{i}}}=\sum _{j}K^{(ij)}({\vec {u_{j}}}-{\vec {u_{i}}})$ $\qquad (i=1,2,...N)$

where,$M_{i}$is the mass of each particle, and $K^{(ij)}$ is the $3*3$ force constant tensor between the $i^{th}$ and the $j^{th}$ particles.

By Fourier transform, we have

${\vec {u_{i}}}={\frac {1}{\sqrt {N_{\Omega }}}}\sum _{\vec {k^{'}}}\exp {-i({\vec {k^{'}}}\cdot {\vec {R_{i}}}-\omega t)}{\vec {u_{\vec {k^{'}}}^{(i)}}}$

and, in turn,

${\vec {u_{\vec {k}}^{(i)}}}={\frac {1}{\sqrt {N_{\Omega }}}}\sum _{\vec {k^{'}}}\exp {i({\vec {k}}\cdot {\vec {R_{i}}}-\omega t)}{\vec {u_{i}}}$

where, $N_{\Omega }$ is the number of the wave vectors ${\vec {k_{i}}}$ in the first Brillouin zone, and ${\vec {R_{i}}}$ is the original position of the $i^{t}h$ particle.

We assume that all ${\vec {u_{i}}}$ have the same eigenfreguency $\omega$, which means:

${\ddot {\vec {u_{i}}}}=-\omega ^{2}{\vec {u_{i}}}$

$(\sum _{j}K^{(ij)}-M_{i}\omega ^{2})\sum _{\vec {k^{'}}}\exp {-i{\vec {k^{'}}}\cdot {\vec {R_{i}}}}{\vec {u}}_{\vec {k^{'}}}^{(i)}=\sum _{j}K^{(ij)}\sum _{\vec {k^{'}}}\exp {-i{\vec {-i{\vec {k^{'}}}\cdot {\vec {R_{j}}}}}{\vec {u_{\vec {k^{'}}}^{(j)}}}}$

Multiplying $\exp {i{\vec {k}}\cdot {\vec {R}}_{i}}$ and taking the summation of Failed to parse (unknown function "\math"): {\displaystyle \vec{R}_i<\math>, we obtain: <math>\sum_{\vec{R_i}}\exp{i(\vec{k}-\vec{k^'})\cdot\vec{R_i}}=N_\Omega\delta_{\vec{k},\vec{k^'}}<\math> Thus, <math>(\sum_{j}K^{(ij)}-M_i\omega^2(\vec{k})I)\vec{u}_\vec{k}^{(i)}-\sum_{j}K^{(ij)}\sum_{\vec{k^'}}\exp{i\vec{-i\vec{k}\cdot\Delta\vec{R_{ij}}}\vec{u_{\vec{k}}^{(j)}}}} where, I is a $3\times 3$ unit matrix, and $\Delta {\vec {R_{ij}}}={\vec {R_{i}}}-{\vec {R_{j}}}$ is the relative coordinate of the $i^{th}$ particle and the $j^{th}$ particle.