+\pdfoutput=0
+\documentclass[a4paper,11pt]{article}
+\usepackage[activate]{pdfcprot}
+\usepackage{verbatim}
+\usepackage{a4}
+\usepackage{a4wide}
+\usepackage[german]{babel}
+\usepackage[latin1]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{amsmath}
+\usepackage{ae}
+\usepackage{aecompl}
+\usepackage[dvips]{graphicx}
+\graphicspath{{./img/}}
+\usepackage{color}
+\usepackage{pstricks}
+\usepackage{pst-node}
+\usepackage{rotating}
+
+\setlength{\headheight}{0mm} \setlength{\headsep}{0mm}
+\setlength{\topskip}{-10mm} \setlength{\textwidth}{17cm}
+\setlength{\oddsidemargin}{-10mm}
+\setlength{\evensidemargin}{-10mm} \setlength{\topmargin}{-1cm}
+\setlength{\textheight}{26cm} \setlength{\headsep}{0cm}
+
+\renewcommand{\labelenumi}{(\alph{enumi})}
+
+\begin{document}
+
+% header
+\begin{center}
+ {\LARGE {\bf Materials Physics I}\\}
+ \vspace{8pt}
+ Prof. B. Stritzker\\
+ WS 2007/08\\
+ \vspace{8pt}
+ {\Large\bf Tutorial 1 - proposed solutions}
+\end{center}
+
+\section{Free electron in a box}
+\begin{enumerate}
+ \item
+
+ \begin{itemize}
+ \item Schr"odinger equation:\\
+ \[
+ - \frac{\hbar^2}{2m} \nabla^2 \Psi({\bf r}) + V({\bf r}) \Psi({\bf r})
+ = E \Psi({\bf r}) \textrm{ , }
+ V({\bf r}) = 0 \textrm{ for } {\bf r} \in [0,L]
+ \]
+ \item Boundary conditions: $\Psi({\bf r}) = \Psi(x,y,z)$\\
+ \[
+ \Psi(0,y,z) = \Psi(L,y,z) = 0 \qquad
+ \Psi(x,0,z) = \Psi(x,L,z) = 0 \qquad
+ \Psi(x,y,0) = \Psi(x,y,L) = 0
+ \]
+ \end{itemize}
+
+
+ \item
+ \begin{itemize}
+ \item Product ansatz: $\Psi({\bf r})=F_x(x)F_y(y)F_z(z)$, with\\
+ $F_x(x)=0$ for $x=0,L$\\
+ $F_y(y)=0$ for $y=0,L$\\
+ $F_z(z)=0$ for $z=0,L$.
+ \item Schr"odinger equation:\\
+ Use: $\nabla^2=\frac{\partial^2}{\partial x^2} +
+ \frac{\partial^2}{\partial y^2} +
+ \frac{\partial^2}{\partial z^2} \Rightarrow$\\
+ \[
+ - \frac{\hbar^2}{2m} \Big[
+ F_y(y) F_z(z) \frac{d^2}{dx^2} F_x(x) +
+ F_x(x) F_z(z) \frac{d^2}{dy^2} F_y(y) +
+ F_x(x) F_y(y) \frac{d^2}{dz^2} F_z(z)
+ \Big] =
+ E F_x(x) F_y(y) F_z(z)
+ \]
+ \item Schr"odinger equation fullfilled if:
+ \[
+ - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} F_x(x) = E_x F_x(x), \quad
+ - \frac{\hbar^2}{2m} \frac{d^2}{dy^2} F_y(y) = E_y F_y(y),\quad
+ - \frac{\hbar^2}{2m} \frac{d^2}{dz^2} F_z(z) = E_x F_z(z).
+ \]
+ \[
+ \Rightarrow \Big[E_x + E_y + E_z\Big] F_x(x) F_y(y) F_z(z) =
+ E F_x(x)F_y(y)F_z(z) \textrm{, } \quad E = E_x + E_y + E_z
+ \]
+ Three eigenvalue problems of the same character.
+ Sufficient to examine only one!
+ \item Solution of the Schr"odinger equation:\\
+ \begin{itemize}
+ \item Ansatz: $F_x = A_x \exp(ik_xx) + B_x \exp(-ik_xx)$
+ \item Boundary conditions:\\
+ $F_x(0)=0 \Rightarrow B_x=-A_x$, \quad
+ let ${A}_x = \frac{1}{2i}\tilde{A}_x$
+ $\Rightarrow$ $F_x(x) = \tilde{A}_x \sin(k_xx)$\\
+ $F_x(L)=0 \Rightarrow k_x L = n_x \pi$ or rather
+ $k_x=n_x \frac{\pi}{L}$, \quad $n_x=0,\pm1,\pm2,\ldots$
+ \item Forbidden values for $n_x$:\\
+ $n_x \ne 0$: otherwise wave function zero for all $x$\\
+ $n_x > 0$: wave functions for $+n_x$ and $-n_x$
+ not linearly independent (same quantum sate)
+ \end{itemize}
+ \[
+ \Rightarrow
+ \Psi_{n_x n_y n_z} = A \sin(\frac{n_x \pi}{L}x)
+ \sin(\frac{n_y \pi}{L}y)
+ \sin(\frac{n_z \pi}{L}z),
+ \quad A=\tilde{A}_x\tilde{A}_y\tilde{A}_z
+ \]
+ \item Energy\\
+ \[
+ E_{n_x n_y n_z} = \frac{\hbar^2 \pi^2}{2m L^2}(n_x^2+n_y^2+n_z^2)
+ \]
+ \end{itemize}
+
+ \item Ground-state:
+ \[
+ \Psi_{111} = A \sin(\frac{\pi}{L}x) \sin(\frac{\pi}{L}x)
+ \sin(\frac{\pi}{L}x) \qquad
+ E_{111} = \frac{\hbar^2 \pi^2}{2m L^2} (1+1+1)
+ = \frac{3 \hbar^2 \pi^2}{2m L^2}
+ \]
+ \item $n_x,n_y,n_z=1,2,3\ldots$\\
+ Allowed $k_{x,y,z}$ values located in positive octant only.
+ \begin{center}
+ \includegraphics[width=10cm]{feg_kvals.eps}
+ \end{center}
+
+\end{enumerate}
+
+\section{Reciprocal lattice}
+
+Convention:
+\begin{itemize}
+ \item basis of unit cell in real space: $a_1,a_2,a_3$
+ \item basis of unit cell in reciprocal space: $b_1,b_2,b_3$
+\end{itemize}
+Prove:
+\[
+V_{real}=a_1(a_2 \times a_3)
+\]
+\[
+V_{rec}=b_1 ( b_2 \times b_3)
+ =\frac{(2\pi)^3}{(a_1(a_2 \times a_3))^3} (a_2 \times a_3) [
+ (a_3 \times a_1) \times (a_1 \times a_2) ]
+\]
+\[
+\textrm{hint 1: }
+(a_3 \times a_1) \times (a_1 \times a_2) =
+a_1((a_3 \times a_1)a_2) - a_2((a_3 \times a_1)a_1) =
+a_1((a_3 \times a_1)a_2)
+\]
+\[
+\Rightarrow V_{rec}= \frac{(2\pi)^3}{(a_1(a_2 \times a_3))^3}
+(a_2 \times a_3) (a_1(a_3 \times a_1) a_2)
+\]
+
+\end{document}