solid_state_physics/tutorial/1_02s.tex

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  29 \begin{document}
  30
  31 % header
  32 \begin{center}
  33  {\LARGE {\bf Materials Physics I}\\}
  34  \vspace{8pt}
  35  Prof. B. Stritzker\\
  36  WS 2007/08\\
  37  \vspace{8pt}
  38  {\Large\bf Tutorial 2 - proposed solutions}
  39 \end{center}
  40
  41 \section{Phonons 1}
  42 \begin{enumerate}
  43  \item \begin{itemize}
  44         \item $r_i=r_{i0}+u_i$\\
  45               $\rho=r_2-r_1=r_{20}+u_2-r_{10}-u_1=(r_{20}-r_{10})+(u_2-u_1)
  46                    =\rho_0+\sigma$
  47         \item $\Phi-\Phi_0=\frac{D}{2}(\rho-\rho_0)^2
  48                           =\frac{D}{2}(\rho^2+\rho_0^2-2\rho_0\rho)$\\
  49               $\rho^2=\rho_0^2+\sigma^2+2\rho_0\sigma$
  50               $\Rightarrow$ $\rho=\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}$\\
  51               $\Rightarrow$ $\Phi-\Phi_0=\frac{D}{2}
  52                              [2\rho_0^2+\sigma^2+2\rho_0\sigma-
  53                               2\rho_0\sqrt{\rho_0^2+\sigma^2+2\rho_0\sigma}]$
  54        \end{itemize}
  55  \item $\sigma \parallel \rho_0$:
  56        \begin{enumerate}
  57         \item \begin{flushleft}
  58                \includegraphics[height=6cm]{elongation_p01.eps}
  59                \includegraphics[height=6cm]{elongation_p02.eps}
  60                \includegraphics[height=6cm]{elongation_p03.eps}
  61               \end{flushleft}
  62         \item $\sigma = \sigma_{\parallel}$:\\
  63               $\rho_0 \sigma_{\parallel} = |\rho_0| |\sigma_{\parallel}|$\\
  64               $\Phi-\Phi_0=\frac{D}{2}\left(2\rho_0^2+\sigma_{\parallel}^2+
  65                            2\rho_0\sigma_{\parallel}-
  66                            2\rho_0\sqrt{(\rho_0+\sigma_{\parallel})^2}\right)
  67                           =\frac{D}{2}\sigma_{\parallel}^2$
  68        \end{enumerate}
  69  \item $\sigma \perp \sigma_0$:
  70        \begin{enumerate}
  71         \item \begin{flushleft}
  72                \includegraphics[height=5.3cm]{elongation_n01.eps}
  73                \includegraphics[height=5.3cm]{elongation_n02.eps}
  74                \includegraphics[height=5.3cm]{elongation_n03.eps}
  75               \end{flushleft}
  76         \item $\sigma=\sigma_{\perp}$:\\
  77               $\sigma_{\perp} \rho_0 = 0$\\
  78               $\Phi-\Phi_0=\frac{D}{2}\left[2\rho_0^2+\sigma_{\perp}^2-
  79                            2\rho_0\sqrt{\rho_0^2+\sigma_{\perp}^2}\right]$
  80
  81         \item $\sigma_{\perp} = \alpha \rho_0$, $\alpha \ll 1$\\
  82               $\sqrt{\rho_0^2+\sigma_{\perp}^2}=
  83                \sqrt{\rho_0^2+\alpha^2\rho_0^2}=
  84                \rho_0\sqrt{1+\alpha^2}\stackrel{Taylor}{=}
  85                \rho_0(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)$\\
  86               $\Rightarrow \Phi-\Phi_0=
  87                \frac{D}{2}\left[\rho_0^2\left(2+\alpha^2-
  88                2(1+\frac{\alpha^2}{2}-\frac{\alpha^4}{8}+\ldots)\right)\right]=
  89                \frac{D}{2}\left[\rho_0^2(\frac{\alpha^4}{4}+\ldots)\right]$\\
  90               $\Rightarrow \Phi-\Phi_0\stackrel{\alpha\ll 1}{=}
  91                \frac{D}{2}\rho_0^2\frac{\alpha^4}{4}=
  92                \frac{D}{2}\sigma_{\perp}^2\frac{\alpha^2}{4}$
  93         \item $\sigma_{\parallel}$, $\sigma_{\perp} \ll \rho_0$\\
  94               $\Rightarrow$ potential contribution of $\sigma_{\perp}$
  95               compared to contribution of $\sigma_{\parallel}$
  96               negligible small.
  97        \end{enumerate}
  98  \item \begin{itemize}
  99         \item As long as the displacements and thus the elongation is small
 100               compared to the equilibrium state the change in the potential
 101               due to the perpendicular elongation is negligible small.
 102         \item Regarding a possible existence of perpendicular elongation
 103               the model of the linear chain is unproblematic.
 104         \item In a real crystal couplings in other directions exist.
 105               These can only be neglected if they are small compared to the
 106               coupling of the considered direction.
 107        \end{itemize}
 108 \end{enumerate}
 109
 110 \section{Phonons 2}
 111 \begin{enumerate}
 112 \item \begin{itemize}
 113        \item Convention:\\
 114              Atom type 1: $M_1$, $u_s$ (elongation of atom $s$ of type 1)\\
 115              Atom type 2: $M_2$, $v_s$ (elongation of atom $s$ of type 2)\\
 116              Lattice constant: $a$, Spring constant: $C$
 117        \item Equations of motion:\\
 118              $M_1\ddot{u}_s=C(v_s+v_{s-1}-2u_s)$\\
 119              $M_2\ddot{v}_s=C(u_{s+1}+u_s-2v_s)$
 120        \item Ansatz:\\
 121              $u_s=u\exp(i(ska-\omega t))$\\
 122              $v_s=v\exp(i(ska-\omega t))$
 123        \item Solution of the equation system:\\
 124              $-\omega^2M_1u\exp(i(ska-\omega t))=
 125              C\exp(-i\omega t)[v\exp(iska)+v\exp(i(s-1)ka)-2u\exp(iska)]$\\
 126              $\Rightarrow -\omega^2M_1u=Cv(1+\exp(-ika))-2Cu$\\
 127              $-\omega^2M_2v\exp(i(ska-\omega t))=
 128              C\exp(-i\omega t)[u\exp(i(s+1)ka)+u\exp(iska)-2v\exp(iska)]$\\
 129              $\Rightarrow -\omega^2M_2v=Cu[\exp(ika)+1]-2Cv$\\
 130              Non trivial solution only if determinant of coefficients
 131              $u$ and $v$ is zero.\\
 132              $\Rightarrow
 133               \left|
 134               \begin{array}{cc}
 135               2C-M_1\omega^2 & -C[1+\exp(-ika)]\\
 136               -C[1+\exp(ika)] & 2C-M_2\omega^2
 137               \end{array}
 138               \right|=0$\\
 139              $\Rightarrow
 140               4C^2+M_1M_2\omega^4-2C\omega^2(M_2+M_1)-
 141               \underbrace{C^2(1+\exp(ika))(1+\exp(-ika))}_{
 142               C^2(\underbrace{1+1+\exp(ika)+\exp(-ika)}_{
 143                               2+2\cos(ka)=2(1+\cos(ka))})}$\\
 144              $\Rightarrow
 145               M_1M_2\omega^4-2C(M_1+M_2)\omega^2+2C^2(1-\cos(ka))=0$
 146       \end{itemize}
 147 \newpage
 148 \item \begin{eqnarray}
 149       \omega^2&=&C\left(\frac{2C(M_1+M_2)}{2M_1M_2}\right)\pm
 150                  \sqrt{\frac{4C^2(M_1+M_2)^2}{4M_1^2M_2^2}-
 151                        \frac{2C^2(1-cos(ka))}{M_1M_2}} \nonumber \\
 152               &=&C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)\pm
 153                  \sqrt{C^2\frac{(M_1+M_2)^2}{M_1^2M_2^2}-
 154                        \frac{1}{M_1M_2}2C^2(1-cos(ka))} \nonumber \\
 155               &=&C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)
 156                  \stackrel{{\color{red}+}}{{\color{blue}-}}
 157                  C\sqrt{\left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2-
 158                         \frac{2(1-\cos(ka))}{M_1M_2}} \nonumber
 159       \end{eqnarray}
 160       \begin{figure}[!h]
 161
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 212       \end{figure}
 213       \begin{itemize}
 214        \item $ka\ll 1$:\\
 215              $\rightarrow \cos(ka)\approx 1-\frac{1}{2}k^2a^2$ (Taylor)\\
 216              Optical branch: $\omega^2\approx
 217                               2C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)$\\
 218              Acoustic branch: $\omega^2\approx
 219                                \frac{C/2}{M_1+M_2}k^2a^2$\\
 220        \item $k=0$:\\
 221              $\rightarrow u/v = - M_2/M_1$ (out of phase)\\
 222        \item $k=\pi/a$\\
 223              $\rightarrow \omega^2=2C/M_2,2C/M_1$
 224       \end{itemize}
 225 \end{enumerate}
 226
 227 \end{document}