solid_state_physics/tutorial/1_02s.tex

   1 \pdfoutput=0
   2 \documentclass[a4paper,11pt]{article}
   3 \usepackage[activate]{pdfcprot}
   4 \usepackage{verbatim}
   5 \usepackage{a4}
   6 \usepackage{a4wide}
   7 \usepackage[german]{babel}
   8 \usepackage[latin1]{inputenc}
   9 \usepackage[T1]{fontenc}
  10 \usepackage{amsmath}
  11 \usepackage{ae}
  12 \usepackage{aecompl}
  13 \usepackage[dvips]{graphicx}
  14 \graphicspath{{./img/}}
  15 \usepackage{color}
  16 \usepackage{pstricks}
  17 \usepackage{pst-node}
  18 \usepackage{rotating}
  19
  20 \setlength{\headheight}{0mm} \setlength{\headsep}{0mm}
  21 \setlength{\topskip}{-10mm} \setlength{\textwidth}{17cm}
  22 \setlength{\oddsidemargin}{-10mm}
  23 \setlength{\evensidemargin}{-10mm} \setlength{\topmargin}{-1cm}
  24 \setlength{\textheight}{26cm} \setlength{\headsep}{0cm}
  25
  26 \renewcommand{\labelenumi}{(\alph{enumi})}
  27
  28 \begin{document}
  29
  30 % header
  31 \begin{center}
  32  {\LARGE {\bf Materials Physics I}\\}
  33  \vspace{8pt}
  34  Prof. B. Stritzker\\
  35  WS 2007/08\\
  36  \vspace{8pt}
  37  {\Large\bf Tutorial 2 - proposed solutions}
  38 \end{center}
  39
  40 \section{Phonons 1}
  41 \begin{enumerate}
  42  \item \begin{itemize}
  43         \item $r_i=r_{i0}+u_i$\\
  44               $\rho=r_2-r_1=r_{20}+u_2-r_{10}-u_1=(r_{20}-r_{10})+(u_2-u_1)
  45                    =\rho_0+\sigma$
  46         \item $\Phi-\Phi_0=\frac{D}{2}(\rho-\rho_0)^2
  47                           =\frac{D}{2}(\rho^2+\rho_0^2-2\rho_0\rho)$\\
  48               $\rho^2=\rho_0^2+\sigma^2+2\rho_0\sigma$
  49               $\Rightarrow$ $\rho=\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}$\\
  50               $\Rightarrow$ $\Phi-\Phi_0=\frac{D}{2}
  51                              [2\rho_0^2+\sigma^2+2\rho_0\sigma-
  52                               2\rho_0\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}]$
  53        \end{itemize}
  54  \item $\sigma \parallel \rho_0$:
  55        \begin{enumerate}
  56         \item \begin{flushleft}
  57                \includegraphics[height=6cm]{elongation_p01.eps}
  58                \includegraphics[height=6cm]{elongation_p02.eps}
  59                \includegraphics[height=6cm]{elongation_p03.eps}
  60               \end{flushleft}
  61         \item $\sigma = \sigma_{\parallel}$:\\
  62               $\rho_0 \sigma_{\parallel} = |\rho_0| |\sigma_{\parallel}|$\\
  63               $\Phi-\Phi_0=\frac{D}{2}\left(2\rho_0^2+\sigma_{\parallel}^2+
  64                            2\rho_0\sigma_{\parallel}-
  65                            2\rho_0\sqrt{(\rho_0+\sigma_{\parallel})^2}\right)
  66                           =\frac{D}{2}\sigma_{\parallel}^2$
  67        \end{enumerate}
  68  \item $\sigma \perp \sigma_0$:
  69        \begin{enumerate}
  70         \item \begin{flushleft}
  71                \includegraphics[height=5.3cm]{elongation_n01.eps}
  72                \includegraphics[height=5.3cm]{elongation_n02.eps}
  73                \includegraphics[height=5.3cm]{elongation_n03.eps}
  74               \end{flushleft}
  75         \item $\sigma=\sigma_{\perp}$:\\
  76               $\sigma_{\perp} \rho_0 = 0$\\
  77               $\Phi-\Phi_0=\frac{D}{2}\left[2\rho_0^2+\sigma_{\perp}^2-
  78                            2\rho_0\sqrt{\rho_0^2+\sigma_{\perp}^2}\right]$
  79
  80         \item $\sigma_{\perp} = \alpha \rho_0$, $\alpha \ll 1$\\
  81               $\sqrt{\rho_0^2+\sigma_{\perp}^2}=
  82                \sqrt{\rho_0^2+\alpha^2\rho_0^2}=
  83                \rho_0\sqrt{1+\alpha^2}\stackrel{Taylor}{=}
  84                \rho_0(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)$\\
  85               $\Rightarrow \Phi-\Phi_0=
  86                \frac{D}{2}\left[\rho_0^2\left(2+\alpha^2-
  87                2(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)\right)\right]=
  88                \frac{D}{2}\left[\rho_0^2(\frac{\alpha^4}{4}+\ldots)\right]$\\
  89               $\Rightarrow \Phi-\Phi_0\stackrel{\alpha\ll 1}{=}
  90                \frac{D}{2}\rho_0^2\frac{\alpha^4}{4}=
  91                \frac{D}{2}\sigma_{\perp}^2\frac{\alpha^2}{4}$
  92         \item $\sigma_{\parallel}$, $\sigma_{\perp} \ll \rho_0$\\
  93               $\Rightarrow$ potential contribution of $\sigma_{\perp}$
  94               compared to contribution of $\sigma_{\parallel}$
  95               negligible small.
  96        \end{enumerate}
  97  \item \begin{itemize}
  98         \item As long as the displacements and thus the elongation is small
  99               compared to the equilibrium state the change in the potential
 100               due to the perpendicular elongation is negligible small.
 101         \item Regarding a possible existence of perpendicular elongation
 102               the model of the linear chain is unproblematic.
 103         \item In a real crystal couplings in other directions exist.
 104               These can only be neglected if they are small compared to the
 105               coupling of the considered direction.
 106        \end{itemize}
 107 \end{enumerate}
 108
 109 \section{Phonons 2}
 110 \begin{enumerate}
 111 \item \begin{itemize}
 112        \item Convention:\\
 113              Atom type 1: $M_1$, $u_s$ (elongation of atom $s$ of type 1)\\
 114              Atom type 2: $M_2$, $v_s$ (elongation of atom $s$ of type 2)\\
 115              Lattice constant: $a$, Spring constant: $C$
 116        \item Equations of motion:\\
 117              $M_1\ddot{u}_s=C(v_s+v_{s-1}-2u_s)$\\
 118              $M_2\ddot{v}_s=C(u_{s+1}+u_s-2v_s)$
 119        \item Ansatz:\\
 120              $u_s=u\exp(i(ska-\omega t))$\\
 121              $v_s=v\exp(i(ska-\omega t))$
 122        \item Solution of the equation system:\\
 123              $-\omega^2M_1u\exp(i(ska-\omega t))=
 124              C\exp(-i\omega t)[v\exp(iska)+v\exp(i(s-1)ka)-2u\exp(iska)]$\\
 125              $\Rightarrow -\omega^2M_1u=Cv(1+\exp(-ika))-2Cu$\\
 126              $-\omega^2M_2v\exp(i(ska-\omega t))=
 127              C\exp(-i\omega t)[u\exp(i(s+1)ka)+u\exp(iska)-2v\exp(iska)]$\\
 128              $\Rightarrow -\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
 129              Non trivial solution only if determinant of coefficients
 130              $u$ and $v$ is zero.\\
 131              $\Rightarrow
 132               \left|
 133               \begin{array}{cc}
 134               2C-M_1\omega^2 & -C[1+\exp(-ika)]\\
 135               -C[1+\exp(ika)] & 2C-M_2\omega^2
 136               \end{array}
 137               \right|=0$\\
 138              $\Rightarrow
 139               4C^2+M_1M_2\omega^4-2C\omega^2(M_2+M_1)-
 140               \underbrace{C^2(1+\exp(ika))(1+\exp(-ika))}_{
 141               C^2(\underbrace{1+1+\exp(ika)+\exp(-ika)}_{
 142                               2+2\cos(ka)=2(1+\cos(ka))})}$\\
 143              $\Rightarrow
 144               M_1M_2\omega^4-2C(M_1+M_2)\omega^2+2C^2(1-\cos(ka))=0$
 145       \end{itemize}
 146 \item \[
 147       \omega^2=C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)\pm
 148                C\sqrt{\left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2-
 149                       \frac{2(1-\cos(ka))}{M_1M_2}}
 150       \]
 151       \begin{itemize}
 152        \item $ka\ll 1$:\\
 153              $\rightarrow \cos(ka)\approx 1-\frac{1}{2}k^2a^2$\\
 154              Optical branch: $\omega^2\approx
 155                               2C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)$\\
 156              Acoustic branch: $\omega^2\approx
 157                                \frac{C/2}{M_1+M_2}k^2a^2$\\
 158        \item $k=0$:\\
 159              Optical branch: $u/v = - M_2/M_1$ (out of phase)\\
 160        \item $k=\pm \pi/a$:\\
 161              $\rightarrow \omega^2=2C/M_2,2C/M_1$
 162       \end{itemize}
 163 \end{enumerate}
 164
 165 \end{document}