\section{Phonons 2}
\begin{enumerate}
-\item Derive the dispersion relation for a linear chain with two different
- alternating types of atoms.
-\item Discuss the two solutions for $\omega^2$.
+\item \begin{itemize}
+ \item Convention:\\
+ Atom type 1: $M_1$, $u_s$ (elongation of atom $s$ of type 1)\\
+ Atom type 2: $M_2$, $v_s$ (elongation of atom $s$ of type 2)\\
+ Lattice constant: $a$, Spring constant: $C$
+ \item Equations of motion:\\
+ $M_1\ddot{u}_s=C(v_s+v_{s-1}-2u_s)$\\
+ $M_2\ddot{v}_s=C(u_{s+1}+u_s-2v_s)$
+ \item Ansatz:\\
+ $u_s=u\exp{i(ska-\omega t)}$\\
+ $v_s=v\exp{i(ska-\omega t)}$
+ \item Solution of the equation system:\\
+ $-\omega^2M_1u=Cv[1+\exp(-ika)]-2Cu$\\
+ $-\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
+ Non trivial solution only if determinant of coefficients
+ $u$ and $v$ is zero.\\
+ $\Rightarrow
+ \left|
+ \begin{array}{cc}
+ 2C-M_1\omega^2 & -C[1+\exp(-ika)]\\
+ -C[1+\exp(ika)] & 2C-M_2\omega^2
+ \end{array}
+ \right|=0$\\
+ $\Rightarrow
+ M_1M_2\omega^4-2C(M_1+M_2)\omega^2+2C^2(1-\cos(ka))=0$
+ \end{itemize}
+\item \[
+ \omega^2=C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)\pm
+ C\sqrt{\left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2-
+ \frac{2(1-\cos(ka))}{M_1M_2}}
+ \]
+ \begin{itemize}
+ \item $ka\ll 1$:\\
+ $\rightarrow \cos(ka)\approx 1-\frac{1}{2}k^2a^2$\\
+ Optical branch: $\omega^2\approx
+ 2C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)$\\
+ Acoustic branch: $\omega^2\approx
+ \frac{C/2}{M_1+M_2}k^2a^2$\\
+ \item $k=0$:\\
+ Optical branch: $u/v = - M_2/M_1$ (out of phase)\\
+ \item $k=\pm \pi/a$:\\
+ $\rightarrow \omega^2=2C/M_2,2C/M_1$
+ \end{itemize}
\end{enumerate}
\end{document}