&=& H_{ion,kin} + H_{ion-ion} + H_{el,kin} + H_{el-el} + H_{ion-el} \nonumber
\end{eqnarray}
which accounts for the ionic and electronic subsystem as well as the coupling between these two.
+Obviously, without any approximation this problem is not solvable.
+{\bf Born-Oppenheimer} or {\bf adiabatic approximation:}
Lighter valence electrons move much faster than the nuclei and thus follow the ionic motion adiabatically.
For the electrons the nuclei appear fixed in position.
On the other way round the electrons appear blurred to the nuclei adding an extra term to an effective potential.
-This is called the Born-Oppenheimer or adiabatic approximation basically switching off electron-phonon interactions.
+This approach is basically switching off the interaction of electrons and lattice vibrations.
+{\bf Independent electron approximation:}
Having separated the ionic and electronic degrees of freedom the Hamiltonian still involves all electronic coordinates which results in a many-particle wave function as a solution of the Schr"odinger equation depending on the positions of all electrons.
-By completely neglecting the electron-electron interaction the Hamiltonian can be written as a sum of single particle Hamiltonians
+By completely neglecting the electron-electron interaction the idealized Hamiltonian can be written as a sum over single particle Hamiltonians
\[
H = \sum_i - \frac{\hbar^2}{2m} \nabla_i^2 + v_{ext}({\bf r}_i)
\]
where $v_{ext}$ is the combination of ion-electron and a constant ion-ion interaction.
-This is called the independent electron approximation.
-Thus, it is sufficient to consider a single electron located in an effective time-independent potential due to the static ions and all other electrons.
+Using these approximations it is sufficient to consider a single electron located in an effective time-independent potential constituted by the static ions and all other electrons.
Since most materials condense into almost perfect periodic arrays the periodicity should also hold for the potential style.
Within this tutorial even the periodic potential is simplified.
-Consider a single particle (mass $m$) enclosed in a box (side length $L=V^{1/3}$) where the potential is constant ($V_0$) inside the box and infinite at the surface.
+Consider a single particle (mass $m$) enclosed in a box (side length $L=\mathcal{V}^{1/3}$) where the potential is constant ($V_0$) inside the box and infinite at the surface.
\begin{enumerate}
\item Write down the Schr"odinger equation and boundary conditions
showing some values allowed for $k_x$ and $k_y$.
\end{enumerate}
+\section{Reciprocal lattice}
+
+Show that the volume $V_{rec}$ of the unit cell of the reciprocal lattice is given by $V_{rec} = (2\pi)^3/V_{real}$, where $V_{real}$ is the volume of the unit cell in real space.
+{\bf Hint:} Use the relation $a \times (b \times c) = b(ac)-c(ab)$ and $a(b \times c)=b(c \times a)$.
+
\end{document}
--- /dev/null
+\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}
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