From a548ec888564c4aca659f187c41f3f995842a7f6 Mon Sep 17 00:00:00 2001 From: hackbard Date: Wed, 23 Apr 2008 11:45:34 +0200 Subject: [PATCH] fix + proposed solutions initial checkin --- solid_state_physics/tutorial/2_01.tex | 2 +- solid_state_physics/tutorial/2_01s.tex | 116 +++++++++++++++++++++++++ 2 files changed, 117 insertions(+), 1 deletion(-) create mode 100644 solid_state_physics/tutorial/2_01s.tex diff --git a/solid_state_physics/tutorial/2_01.tex b/solid_state_physics/tutorial/2_01.tex index 7ccbbaf..cddf85b 100644 --- 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 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 $I = (\textrm{charge}) \cdot (\textrm{loops per time}) + \stackrel{1/T=\omega_L/2\pi}{=} + (Ze)(\frac{1}{2\pi}\frac{-e}{2m}B)$\\ + $\mu=IA=I2\pi<\rho^2>=-\frac{Ze^2B}{4m}<\rho^2>$\\ + $== \Rightarrow =3=3$\\ + $<\rho^2>=+=\frac{2}{3}$\\ + $\mu=-\frac{Ze^2B}{6m}$ + \item $\chi=\frac{\mu_0N\mu}{B}=-\frac{\mu_0NZe^2}{6m}$ + \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} -- 2.39.2