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\question
{Kouki Yonaga}                  % Enter your name
{12/7/11}                        % Enter mm/dd/yy
{yona@cmpt.phys.tohoku.ac.jp}   % Enter e-mail address
{2-10}                          % Enter the number and question.
%question
{Obtain Eqs.(2.33) and (2.34). Evaluate the value of the Fermi velocity.}
%answer
{
Using the phase factor $f(k)$ in Eq.\,(2.28);
\begin{equation}
    f(k) = 1+4\cos\left(\frac{\sqrt{3}k_{x}a}{2}\right)\cos\left(\frac{k_{y}a}{2}\right)+4\cos^{2}\left(\frac{k_{y}a}{2}\right),
 \label{Q2-10eq:phase}
\end{equation}
The eletronic energy $E^{\pm}(k_{x},k_{y})$ of graphene is given by Eq.\,(2.32);
 \begin{eqnarray}
    E^{\pm}(k_{x},k_{y}) & = & \pm t \sqrt{\left|f(k)\right|^{2}} \nonumber \\
                        & = & \pm t \sqrt{1+4\cos\left(\frac{\sqrt{3}k_{x}a}{2}\right)\cos{\left(\frac{k_{y}a}{2}\right)}+4\cos^{2}\left(\frac{k_{y}a}{2}\right)},
   \label{Q2-10eq:energy}
 \end{eqnarray}
where $t$ is the nearest neghbor transfer integeral which is also called $-\gamma_{0}\ (t = -\gamma_{0})$. $\gamma_{0}$ is given a positive value. We can evaluate the Fermi velocity as follows.
At first, we calculate the lowest order term in  $f(k)$ around the K points. Because the K point has a three-fold rotational symmetry in the $k$-space, the first order term in $f(k)$ becomes zero.
For example, when we consider a K point;  
\begin{equation}
    K = (0,-\frac{4\pi}{3a}),
\end{equation}
the lowest order term in $f(k)$ is written as
\begin{eqnarray}
 \frac{\partial f(k)}{\partial k_{x}}\Big|_{k_{x}=0}&=&0 \nonumber \\
 \frac{\partial f(k)}{\partial k_{y}}\Big|_{k_{y}=\frac{4\pi}{3a}}&=&0.
\end{eqnarray}

Thus the $f(k)$ is written as

\begin{eqnarray}
 f(k)& = &
  f(k)|_{k_{x}=0, k_{y}=-\frac{4\pi}{3a}} + \frac{\partial^{2} f(k)}{\partial k_{x}^{2}}\Big|_{k_{x}=0} + \frac{\partial^{2} f(k)}{\partial k_{y}^{2}}
                         \Big|_{k_{y}=-\frac{4\pi}{3a}} + 2\frac{\partial^{2} f(k)}{\partial k_{x}\partial k_{y}}\Big|_{k_{x}=0, k_{y}=-\frac{4\pi}{3a}} \nonumber \\
       & = &\frac{3a^{2}k^{2}}{4},
 \label{Q2-10eq:phase_K}
\end{eqnarray}
where $k^{2}$ is defined by $k^{2} = k_{x}^{2} + k_{y}^{2}$. If we use the difinition of the Fermi velocity;
\begin{equation}
 v_{\rm F} = \frac{1}{\hbar} \frac{\partial E}{\partial k}\Big|_{E=E_{\rm F}},
\end{equation}




The $E^{\pm}(k)$ can expressed by
\begin{equation}
 E^{\pm}(k) = \pm \hbar v_{\rm F}\mid k \mid.
\end{equation}
Using Eqs.\,(\ref{Q2-10eq:energy}) and (\ref{Q2-10eq:phase_K}),  we got Fermi velocity as follows;
\begin{equation}
 v_{\rm F} = \sqrt{3}\left(\frac{\gamma_{0}a}2\hbar\right),
\end{equation}



}




