%<PLAINTEXT>
\question
{Kouki Yonaga}                  % Enter your name
{12/7/11}                        % Enter mm/dd/yy
{yona@cmpt.phys.tohoku.ac.jp}   % Enter e-mail address
{3-11}                          % Enter the number and question.
%question
{Consider a $2(N - 1)$ linear chain of atoms in which two different atoms A and B with masses $M_{A}$ and $M_{B}$ are connected along a chain (wall-A-B...-B-wall).
 Obtain and plot the phonon dispersion for this configuration.}
%answer
{The equations of motion for the displacement of the $j-$th atoms are given by 
\begin{equation}
  \begin{array}{c}
   M_{A}\ddot{x_{j}} = -K(x_{j} - y_{j-1}) - K(x_{j} - y_{j}) \\
   M_{B}\ddot{y_{j}} = -K(y_{j} - x_{j}) - K(y_{j} - x_{j+1}). \\
  \end{array}
 \label{Q3-11eq:equation}
\end{equation} 
 where $x_{j}$, $y_{j}$ and $K$ are, respectively, the displacements of two different atoms in $j-$th unit cell and force constatnt.
 When we consider the Bloch therem, the $x_{j}$, $y_{j}$ are written as 
\begin{equation}
  \begin{array}{c}
   x_{j} = A\exp(i2kja - i\omega t) \\
   y_{j} = B\exp(i2kja - i\omega t). \\
  \end{array}
 \label{Q3-11eq:solutions}
\end{equation}
 where $A$, $B$ represent amplitude, and $k$ and $a$ are, respectively, the wave number and lattice constant. For the boundary conditions that the end atoms are fixed, ${\rm Re}(x_{N})$ and ${\rm Re}(x_{0})$ are zero.
 When we consider $t = 0$, ${\rm Re}(x_{0})$ is written as
\begin{equation}
 {\rm Re}(A) = 0.
 \label{Q3-11eq:A}
\end{equation}
 From Eq.\,(\ref{Q3-11eq:A}), $A$ is a purely imaginary number and satisfies
\begin{equation}
 A = iR,
 \label{Q3-11eq:imaginary}
\end{equation}
where $R$ is real number. From Re$(x_{N}) = 0$, the Re$(x_{N})$ is written as
\begin{equation}
 {\rm Re}(x_{N}) = -iR\sin(kNa) = 0.
 \label{Q3-11eq:kna}
\end{equation} 
The wave numbers which satisfy Eq.\,(\ref{Q3-11eq:kna}) are given by
\begin{equation}
 k = \frac{m\pi}{Na}, \ \ \ (m = 1, 2, ..., N -1).
 \label{Q3-11eq:k}
\end{equation}

Since length of the unit cell is $2a$, the first Brillouin zone is defined by a region in the $k-$space;
\begin{equation}
 -\frac{\pi}{2a} < k < \frac{\pi}{2a}.
 \label{Q3-11eq:B.Z.}
\end{equation}
Using Eq.\,(\ref{Q3-11eq:equation}), Eq.\,(\ref{Q3-11eq:solutions}) can be written as
\begin{eqnarray}
 -\omega^{2}M_{A}A & = & -2KA + K(1 + e^{-i2ka})B  \nonumber \\
 -\omega^{2}M_{B}B & = & K(1 + e^{i2ka})A -2KB .
 \label{Q3-11eq:omega}
\end{eqnarray}
From Eq.\,(\ref{Q3-11eq:omega}), the simultaneous equations of motion can be expressed by 
\begin{equation}
 D(k)\mbox{\boldmath $u$} = 0,
 \label{Q3-11eq:matrix}
\end{equation}
where $D(k)$ and $\mbox{\boldmath $u$}$ are defined, respectively, by
\begin{equation}
  D(k) = 
    \left[
   \begin{array}{cc}
    {\displaystyle -\frac{2K}{M_{A}}} + \omega^2 & {\displaystyle\frac{K}{M_{B}}}(1 + e^{i2ka}) \\
      & \\
    {\displaystyle \frac{K}{M_{A}}}(1 + e^{-i2ka}) & -{\displaystyle\frac{2K}{M_{B}}} + \omega^{2} \\
   \end{array}\right],
 \label{Q3-11eq:dynamical_matrix}
\end{equation}
and
\begin{eqnarray}
 \mbox{\boldmath $u$} = 
   \left[
  \begin{array}{c}
      A \\
      B \\
  \end{array}\right].
 \label{Q3-11eq:vector}
 \end{eqnarray}

To obtain the eigenvalues $\omega^{2}(k)$ for dynamical matrix with $\mbox{\boldmath $u$} \neq 0$, we solve a seculr equation

\begin{equation}
 \begin{array}{|cc|}
     {\displaystyle -\frac{2K}{M_{A}}} + \omega^2 & {\displaystyle\frac{K}{M_{B}}}(1 + e^{i2ka}) \\
     & \\
    {\displaystyle\frac{K}{M_{A}}}(1 + e^{-i2ka}) & {\displaystyle-\frac{2K}{M_{B}}} + \omega^{2}
    \end{array}
     = 0.
 \label{Q3-11eq:det}
\end{equation}

From Eq.\,(\ref{Q3-11eq:det}), we get $\omega^{2}_{\pm}$ as follows;
\begin{equation}
 {\displaystyle\frac{\omega^{2}_{\pm}}{\omega_{0}^{2}}} = 1\pm\sqrt{1-\frac{4M_{A}M_{B}\sin^{2}(ka)}{(M_{A}+M_{B})^{2}}},
 \label{Q3-11eq:omega^2}
\end{equation}  
where $\omega_{0}$ is difined by $\omega_{0} = \sqrt{K\left({\displaystyle\frac{1}{M_{A}}+\frac{1}{M_{B}}}\right)}$. When we denote $M_{B}$ with a positive number $l$
\begin{equation}
 M_{B} = lM_{A},
 \label{Q3-11eq:M}
\end{equation}
we get the $\omega_{\pm}$;
\begin{eqnarray}
 {\displaystyle\frac{\omega_{+}}{\omega_{0}}} & = & \sqrt{1+\sqrt{1-\displaystyle\frac{4l\sin^{2}(ka)}{(1 + l)^{2}}}} \nonumber \\
 {\displaystyle\frac{\omega_{-}}{\omega_{0}}} & = & \sqrt{1-\sqrt{1-\displaystyle\frac{4l\sin^{2}(ka)}{(1 + l)^{2}}}}. 
\end{eqnarray}

In Fig.\,(\ref{Q3-11fig:1}), ${\displaystyle\frac{\omega_{+}}{\omega_{0}}}$ and ${\displaystyle\frac{\omega_{-}}{\omega_{0}}}$ are plotted within the first Brillouin zone.
 Here we consider the case that ${\displaystyle\frac{4l}{(l+1)^{2}}} = 0.9$.
\begin{figure}[htbp]
 \begin{minipage}{0.5\hsize}
  \begin{center}
   \includegraphics[width=70mm]{Q3-11.eps}
  \end{center}
  \caption{Energy Dispersion. }
  \label{Q3-11fig:1}
 \end{minipage}
 \begin{minipage}{0.5\hsize}
  \begin{center}
   \includegraphics[width=75mm]{Q3-11-2.eps}
  \end{center}
  \caption{$\Delta\omega/2$ vs $l$.}
  \label{Q3-11fig:2}
 \end{minipage} 
\end{figure}

The $\omega_{+}$ and $\omega_{-}$ are called, respectively, optical mode and acoustic mode. When we define an energy gap $\Delta\omega$ at $k = {\displaystyle\pm\frac{\pi}{a}}$, $\Delta\omega$ is expressed by
\begin{eqnarray}
 \Delta\omega &=& \frac{\omega_{+}^{2}}{\omega_{0}^{2}} - \frac{\omega_{-}^{2}}{\omega_{0}^{2}}, \nonumber \\
              &=& 2\sqrt{1 - \frac{4l\sin(ka)^{2}}{(1 + l)^{2}}} \nonumber \\
              &=& 2\frac{\mid 1-l\mid}{(1+l)^{2}}.
\end{eqnarray}
$\Delta\omega$ is plotted as a function of $l$ for $l > 1$. Fig.\,(\ref{Q3-11fig:2}), here we consider $\sin ka = 1$.
Fig.\,(\ref{Q3-11fig:2}) shows that, when the atoms in unit cell is equal\,($l = 1$), the gap becomes zero. 
In the limit of $l = \infty$, $\omega_{+}$ and $\omega_{-}$ goes, respectively, $2\omega_{0}$ and $0$
}





