\question
{Yuki Tatsumi}
{02/05/12}
{tatsumi@flex.phys.tohoku.ac.jp}
{2-12}
%question
{Plot the density of states for the parabolic energy band, $E = a \left( {k_x}^2 + {k_y}^2 \right)$.}
%answer
{We now consider the parabolic energy band in two dimension as follows:

\begin{equation}
E = a \left( {k_x}^2 + {k_y}^2 \right) = ak^2 ,
\label{eq:Q2-12-1}
\end{equation}

where $k$ is $ \sqrt{{k_x}^2 + {k_y}^2} $. If the length of $x$ and $y$ axis in the system is $L_x$ and $L_y$, the area which a microscopic state occupies in the reciprocal space is ${\displaystyle \frac{2\pi}{L_x} \cdot \frac{2\pi}{L_y} }$. When we consider the isotropic two dimensional plane in $k$ space, the number of states $dN$ with the wave number from $k$ to $k+dk$ is written as:

\begin{equation}
dN = 2 \cdot \frac{2{\pi}kdk}{\displaystyle \frac{2\pi}{L_x} \cdot \frac{2\pi}{L_y}} = \frac{Sk}{\pi}dk ,
\label{eq:Q2-12-2}
\end{equation}

where the factor $2$ is the spin degeneracy and ${S = L_x \cdot L_y }$ is the area of the system. Using Eq.\,(\ref{eq:Q2-12-2}), we obtain

\begin{equation}
\frac{dN}{dk} = \frac{Sk}{\pi} .
\label{eq:Q2-12-3}
\end{equation}

Using Eq.\,(\ref{eq:Q2-12-1}), we obtain

\begin{equation}
\frac{dE}{dk} = 2ka .
\label{eq:Q2-12-4}
\end{equation}

Using Eqs.\,(\ref{eq:Q2-12-3}) and (\ref{eq:Q2-12-4}), we can obtain the density of states for the parabolic energy band $E = a \left( {k_x}^2 + {k_y}^2 \right)$ as follows:

\begin{equation}
D(E) = \frac{dN}{dE} = \frac{\displaystyle \frac{dN}{dk}}{\displaystyle \frac{dE}{dk}} = \frac{S}{2{\pi}a} .
\label{eq:Q2-12-5}
\end{equation}

In Fig.\,(\ref{fig:Q2-12-1}), the density of states for the parabolic energy band is plotted as a function of the energy $E$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=8cm,clip]{Q2-12-1.eps}\\
\begin{minipage}{12cm}
\caption{Density of states (DOS) for the parabolic energy band $E = a \left( {k_x}^2 + {k_y}^2 \right)$. The DOS is constant for $E>0$.
\label{fig:Q2-12-1}
}
\end{minipage}
\end{center}
\end{figure}

}
