fix + proposed solutions initial checkin

diff --git a/solid_state_physics/tutorial/2_01.tex b/solid_state_physics/tutorial/2_01.tex
index 7ccbbafbde24babfd58a84bf8cbcd98224064ffb..cddf85b0739c828b74d058088fab7396a0e83cca 100644 (file)
--- a/solid_state_physics/tutorial/2_01.tex
+++ b/solid_state_physics/tutorial/2_01.tex
@@ -104,7 +104,7 @@ atom or ion.
         \item Calculate the magnetic suscebtibility in a state $\phi$.
              What term is responsible for the diamagnetic contribution?
              {\bf Hint:} The magnetic suscebtibility is defined as
-             $\chi=-\frac{1}{V}\frac{\partial^2 E}{\partial B^2}$.
+             $\chi=-\frac{1}{V}\mu_0\frac{\partial^2 E}{\partial B^2}$.
        \item Assuming a spherically symmetric charge distribution the equality
              $<\phi|x^2|\phi>=<\phi|y^2|\phi>=\frac{1}{3}<\phi|r^2|\phi>$
              is valid. Rewrite the diamagnetic part of the suscebtibility

diff --git a/solid_state_physics/tutorial/2_01s.tex b/solid_state_physics/tutorial/2_01s.tex
new file mode 100644 (file)
index 0000000..a892853
--- /dev/null
+++ b/solid_state_physics/tutorial/2_01s.tex
@@ -0,0 +1,116 @@
+\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})}
+\renewcommand{\labelenumii}{\arabic{enumii})}
+\renewcommand{\labelenumiii}{\roman{enumiii})}
+
+\begin{document}
+
+% header
+\begin{center}
+ {\LARGE {\bf Materials Physics II}\\}
+ \vspace{8pt}
+ Prof. B. Stritzker\\
+ SS 2008\\
+ \vspace{8pt}
+ {\Large\bf Tutorial 1 - proposed solutions}
+\end{center}
+
+\section{Diamagnetism}
+\begin{itemize}
+ \item Magnetic field ${\bf B}$
+ \item Magnetization ${\bf M}$
+ \item Suscebtibility $\chi=\frac{\mu_0 {\bf M}}{{\bf B}}$
+\end{itemize}
+
+\begin{enumerate}
+ \item {\bf Classical approach:}
+       \begin{enumerate}
+        \item Maxwell: $\oint_{\partial A} E \, ds
+                       = -\frac{d}{dt}(\int_A B \, dA)
+                       \stackrel{B(r)=B}{=}-\frac{d}{dt}(BA)$\\
+             $-\frac{d(BA)}{dt}=-\pi r^2 \dot{B}=U_{ind}$\\
+             $U_{ind}=\oint_{\partial A} E \, ds 
+              \stackrel{E(s)=E}{\Rightarrow}
+              E=-\frac{\pi r^2}{2\pi r}\dot{B}$\\
+              $\dot{v}=a=\frac{e}{m}E=-\frac{e}{2m}r\dot{B}
+              \Rightarrow v=-\frac{e}{2m}rB$\\
+             $\omega_L=\frac{v}{r}=-\frac{e}{2m}B$
+        \item $I = (\textrm{charge}) \cdot (\textrm{loops per time})
+              \stackrel{1/T=\omega_L/2\pi}{=}
+              (Ze)(\frac{1}{2\pi}\frac{-e}{2m}B)$\\
+             $\mu=IA=I2\pi<\rho^2>=-\frac{Ze^2B}{4m}<\rho^2>$\\
+             $<x^2>=<y^2>=<z^2> \Rightarrow <r^2>=3<x^2>=3<y^2>$\\
+             $<\rho^2>=<x^2>+<y^2>=\frac{2}{3}<r^2>$\\
+             $\mu=-\frac{Ze^2B}{6m}$
+        \item $\chi=\frac{\mu_0N\mu}{B}=-\frac{\mu_0NZe^2}{6m}<r^2>$
+       \end{enumerate}
+ \item {\bf Quantum mechanical theory:}
+       \begin{itemize}
+        \item vector potential ${\bf A}$
+        \item ${\bf B}=\nabla\times{\bf A}$
+        \item $
+              H_{kin}=\frac{1}{2m}(-i\hbar\nabla_{r}-e{\bf A})^2
+              =H_{kin}^0 + H_{kin}'
+             $
+       \end{itemize}
+       \begin{enumerate}
+        \item \begin{eqnarray}
+             H_{kin}&=&\frac{1}{2m}(-\hbar^2\nabla_{r}^2+e^2{\bf A}^2
+                                    +i\hbar \nabla_{r}e{\bf A}
+                                    +e{\bf A}i\hbar \nabla_{r})\nonumber\\
+             H_{kin}^0&=&\frac{-\hbar^2}{2m}\nabla_r^2\nonumber\\
+             H_{kin}'&=&\frac{i\hbar e}{2m}(\nabla_r{\bf A}+{\bf A}\nabla_r)+
+                        \frac{e^2{\bf A}^2}{2m}\nonumber
+             \end{eqnarray}
+             Terms in $H_{kin}'$ can be treated as small perturbation.
+        \item ${\bf A}=\left(-\frac{1}{2}yB,\frac{1}{2}xB,0\right)$, since:
+             $\nabla_r\times{\bf A}=\left(0,0,\frac{1}{2}B+\frac{1}{2}B\right)=
+              \left(0,0,B\right)$\\
+             Note: $\nabla_r{\bf A}=0$
+        \item $
+             H_{kin}'=\frac{i\hbar e}{2m}\left(
+              x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}
+                                         \right)+\frac{e^2B^2}{8m}(x^2+y^2)
+             $\\
+             $
+              L_z=x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}
+              \Rightarrow  H_{kin}'=\frac{i\hbar e}{2m}L_z+
+              \frac{e^2B^2}{8m}(x^2+y^2)
+             $
+        \item $\chi=-\frac{1}{V}\mu_0\frac{\partial^2 E}{\partial B^2}
+              \Rightarrow$
+             only second term contributes to $\chi$!
+             $\chi=-\frac{1}{V}\mu_0\frac{e^2}{4m}<\phi|(x^2+y^2)|\phi>$
+       \item $<\phi|x^2|\phi>=<\phi|y^2|\phi>=\frac{1}{3}<\phi|r^2|\phi>$\\
+             $\Rightarrow \chi=-\frac{1}{V}\mu_0\frac{e^2}{6m}
+              <\phi|r^2|\phi>$\\
+             Consider all $Z$ electrons and all atoms per volume:\\
+              $\chi=-\frac{\mu_0NZe^2}{6m}<\phi|r^2|\phi>$
+       \end{enumerate}
+\end{enumerate}
+
+\end{document}