\question
{Yuki Tatsumi}
{01/05/12}
{tatsumi@flex.phys.tohoku.ac.jp}
{2-13}
%question
{By expanding $f(k)$ of Eq. (2.28) near the K point, show that the Hamiltonian matrix is written by Eq. (2.35).}
%answer
{
We use the definition of the real space unit vectors ${\bf a}_{1}$ and ${\bf a}_{2}$  and the reciplocal space unit vectors ${\bf b}_{1}$ and ${\bf b}_{2}$, shown in Fig.\,(\ref{fig:Q2-13-1}). This definition is the same as the textbook.

\begin{figure}[h]
\begin{center}
\includegraphics[width=15cm,clip]{Q2-13-1.eps}\\
\begin{minipage}{10cm}
\caption{Definition of the real space unit vectors ${\bf a}_{1}$ and ${\bf a}_{2}$,  and the reciplocal space unit vectors ${\bf b}_{1}$ and ${\bf b}_{2}$. The gray zone of the reciprocal space is the first Brillouin zone.
\label{fig:Q2-13-1}
}
\end{minipage}
\end{center}
\end{figure}


Using the definition, the sum of the phase factors $f(k)$ (Eq. (2.28) in the textbook) is given by


\begin{equation}
f(k) = e^{ik_{x}a/\sqrt{3}} + 2e^{-ik_{x}a/2\sqrt{3}}\cos{\left(\frac{k_{y}}{2}a\right)} .
\label{eq:Q2-13-1}
\end{equation}

By expanding $f(k)$ around the K point $(0 , -4\pi/3a)$, 

\begin{eqnarray}
f\left(k_{x} = 0 , k_{y} = -\frac{4\pi}{3a}\right)
& = &
1 + 2{\cos\left(-\frac{2}{3}\pi\right)}
\nonumber \\
& = &
1 + 2\cdot\left(-\frac{1}{2}\right)
\nonumber \\
& = &0 ,
\label{eq:Q2-13-2}
\end{eqnarray}

\begin{eqnarray}
\frac{{\partial}f}{{\partial}k_{x}}{\left(k_{x} = 0 , k_{y} = -\frac{4\pi}{3a}\right)}
& = &
\left.\frac{ia}{\sqrt{3}}\cdot{e^{ik_{x}a/\sqrt{3}}} - 2\cdot\frac{ia}{2\sqrt{3}}\cdot{e^{-ik_{x}a/2\sqrt{3}}}\cos{\left(-\frac{2}{3}\pi\right)}\right|_{k_{x} = 0 , k_{y} = -\frac{4\pi}{3a}}
\nonumber \\
& = &
\frac{ia}{\sqrt{3}} - \frac{ia}{\sqrt{3}}\cdot\left(-\frac{1}{2}\right)
\nonumber \\
& = &
\frac{\sqrt{3}}{2}ai ,
\label{eq:Q2-13-3}
\end{eqnarray}

\begin{eqnarray}
\frac{{\partial}f}{{\partial}k_{y}}{\left(k_{x} = 0 , k_{y} = -\frac{4\pi}{3a}\right)}
& = &
\left.2{\cdot}{e^{-ik_{x}a/2\sqrt{3}}}{\cdot}\frac{a}{2}\left\{-\sin{\left(-\frac{2}{3}\pi\right)}\right\}\right|_{k_{x} = 0 , k_{y} = -\frac{4\pi}{3a}}
\nonumber \\
& = &
2\cdot\frac{a}{2}\cdot\frac{\sqrt{3}}{2}
\nonumber \\
& = &
\frac{\sqrt{3}}{2}a ,
\label{eq:Q2-13-4}
\end{eqnarray}

we can write $f(k)$ as follows:

\begin{eqnarray}
f(k) & = & 
\frac{\sqrt{3}}{2}aik_{x}+\frac{\sqrt{3}}{2}a\left(k_{y}+\frac{4\pi}{3a}\right)
\nonumber \\
& = & 
\frac{\sqrt{3}}{2}a\left(i{k'}_{x}+i{k'}_{y}\right) ,
\label{eq:Q2-13-5}
\end{eqnarray}

where $k_{x} = {k'}_{x}$ , $k_{y}+\frac{4\pi}{3a} = {k'}_{y}$. Using $f(k)$, the Hamiltonian matrix ($\varepsilon_{2p} = 0$) is described by

\begin{eqnarray}
H & = &
\left(
\begin{array}{cc}
\varepsilon_{2p} & tf(k) \\
tf(k)^\ast & \varepsilon_{2p} \\
\end {array}
\right)
\nonumber \\
 & = & 
-\frac{\sqrt{3}}{2}at
\left(
\begin{array}{cc}
0 & -i{k'}_{x}-{k'}_{y} \\
i{k'}_{x}-{k'}_{y} & 0 \\
\end {array}
\right)
\nonumber \\
& = & 
\frac{\sqrt{3}}{2}a\gamma_{0}
\left[
\left(
\begin{array}{cc}
0 & -i \\
i & 0 \\
\end {array}
\right)
{k'}_{x}
-
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end {array}
\right)
{k'}_{y}
\right]
\nonumber \\
& = & 
{\hbar}v_{\rm F}\left[\sigma_{y}{k'}_{x} - \sigma_{x}{k'}_{y}\right] ,
\label{eq:Q2-13-6}
\end{eqnarray}

where the Fermi velocity $v_{\rm F} = \frac{\sqrt{3}}{2{\hbar}}a\gamma_{0}$, transfer integral $t = -\gamma_{0}$ and the $\sigma_{x}$ and $\sigma_{y}$ are the Pauli matrices. To obtain the Hamiltonian of plain formula, we change of the coordinates. It must be that the eigenvalues are invariant when the coordinates are changed. Then we change of the coordinate ${k'}_{x}$ and ${k'}_{y}$ by rotating by $\pi/2$. The $\pi/2$ rotation matrix is given by

\begin{equation}
C_{\frac{\pi}{2}} = 
\left(
\begin{array}{cc}
\cos{\frac{\pi}{2}} & -\sin{\frac{\pi}{2}} \\
\sin{\frac{\pi}{2}} & \cos{\frac{\pi}{2}} \\
\end {array}
\right)
 = 
\left(
\begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end {array}
\right).
\label{eq:Q2-13-7}
\end{equation}

Using Eq.\,(\ref{eq:Q2-13-7}), ${k''}_{x}$ and ${k''}_{y}$ which rotate $\pi/2$ from ${k'}_{x}$ and ${k'}_{y}$ are given by

\begin{eqnarray}
\left(
\begin{array}{cc}
{k''}_{x} \\
{k''}_{y} \\
\end {array}
\right)
& = &
\left(
\begin{array}{cc}
\cos{\frac{\pi}{2}} & -\sin{\frac{\pi}{2}} \\
\sin{\frac{\pi}{2}} & \cos{\frac{\pi}{2}} \\
\end {array}
\right)
\left(
\begin{array}{cc}
{k'}_{x} \\
{k'}_{y} \\
\end {array}
\right)
\nonumber \\
& = &
\left(
\begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end {array}
\right)
\left(
\begin{array}{cc}
{k'}_{x} \\
{k'}_{y} \\
\end {array}
\right)
\nonumber \\
& = &
\left(
\begin{array}{cc}
-{k'}_{y} \\
{k'}_{x} \\
\end {array}
\right) .
\label{eq:Q2-13-8}
\end{eqnarray}

Using Eq.\,(\ref{eq:Q2-13-8}), we can obtain

\begin{equation}
{k'}_{x} = {k''}_{y} , {k'}_{y} = -{k''}_{x} .
\label{eq:Q2-13-9}
\end{equation}

Therefore we can exchange ${k'}_{x}$ and ${k'}_{y}$ for ${k''}_{x}$ and ${k''}_{y}$. Hamiltonian matrix in Eq.\,(\ref{eq:Q2-13-6}) is described by

\begin{eqnarray}
H & = &
{\hbar}v_{\rm F}\left[\sigma_{y}{k'}_{x} - \sigma_{x}{k'}_{y}\right]
\nonumber \\
& = &
{\hbar}v_{\rm F}\left[\sigma_{x}{k''}_{x} + \sigma_{y}{k''}_{y}\right]
\nonumber \\
& = &
{\hbar}v_{\rm F}({\sigma}{\cdot}{\kappa}) ,
\label{eq:Q2-13-10}
\end{eqnarray}

where $\kappa = -i\nabla$. This is a form of Dirac Hamiltonian. In general, the part of Pauli matrices in the Dirac Hamiltonian have arbitrary property if they satisfy any conditions. These descriptions have the same eigenvalues respectively. Thus we can obtain the same eigenvalues from Eq.\,(\ref{eq:Q2-13-6}) and Eq.\,(\ref{eq:Q2-13-10}) and we can alter Eq.\,(\ref{eq:Q2-13-6}) to the massless Dirac Hamiltonian in Eq.\,(\ref{eq:Q2-13-10}). 
}
