\documentclass[11pt,fleqn]{article}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% Page setup
\topmargin -1.5cm \oddsidemargin -0.04cm \evensidemargin -0.04cm
\textwidth 16.59cm \textheight 24cm \setlength{\parindent}{0cm}
\setlength{\parskip}{0cm}


% Fonts
\usepackage{latexsym}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bm}


% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}

% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}

% Mass symbol
\newcommand{\mass}{\mathsf{m}}
\newcommand{\Mass}{\mathsf{M}}

% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}}
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}}
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}

% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}

% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}

% heading
\newcommand{\heading}[1]{\begin{center} \Large {#1} \end{center}}
\newcommand{\auname}[1]{\begin{center} \bf Submitted by: #1 \end{center}}


\newcommand{\nl}{\vspace*{3mm}}


\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\heading{Ex471: Quantum Mechanics II} 

\auname{T. Arie Weiss}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Given a spherical shell with radius R and a particle with mass M and
charge q.in this question we have to assume that the particle can be
excited from ground state to first energy level but not beyond so
the state space is four dimensional.

\nl

 we prepare the particle as "concentrated" as possible in the
"north pole" of the shell. \nl


1. write the wave function $\psi(\theta,\phi)$ of the particle \nl

 this question demands
"thinking", the answer is simple and do not demand "algebra". the
state you wrote is not stationary. the particle oscillate between
the north and the south pole. \nl

 2. what is the period of the
oscillations?

\nl

we define the of the system in the
following:$P(t)=<\psi(t)\arrowvert\hat{z}\arrowvert\psi(t)>$

\nl

3. find the polarization P(t) as function of time \nl

given

\[Y_{00}=\frac1{\sqrt{4\pi}}, ~~~~  Y_{10}=\sqrt{\frac{3}{4\pi}}cos\theta, ~~~~ Y_{1,\pm1}\propto sin(\theta)\]

\nl

\nl answer:


\nl


1. We want to find a solution that gives us the best probability to
find the particle in the north pole of the shell. we are only able
to use the linear combination of $Y_{00}, Y_{1-1}, Y_{10}, Y_{11}$.
we can see that the function $\psi=1+cos\theta$ gives us that
probability. after normalization and using $Y_{lm}$ we get:
\[\psi=\frac{\sqrt{3}}{2}Y_{00}+\frac12Y_{10}\]

2. the energy level of each state will be $E_l=\frac{l(l+1)}{2mR^2}$
so the energy levels of the given problem are:
\[E_0=0\]
\[E_1=\frac1{mR^2}\]
we will write the time depended wave function:

\[\psi=\frac{\sqrt{3}}{2}Y_{00}+\frac12e^{-iE_1t}Y_{10}\]
from here we can see easily that the frequency is
$\omega=E_1=\frac1{mR^2}$ and the period is
\[T=2\pi mR^2\]

3. and now to make a sandwich...

\[P(t)=<\psi(t)\arrowvert\hat{z}\arrowvert\psi(t)>\]
\[z=R\sqrt{\frac{4\pi}{3}}Y_{10}\]
\[P(t)=R\sqrt\frac{4\pi}{3}\int_{-1}^1Y_{10}[\frac{\sqrt3}{2}Y_{00}+\frac12e^{iE_1t}Y_{10}][\frac{\sqrt3}{2}Y_{00}+\frac12e^{-iE_1t}Y_{10}]\]
after doing all the math the solution is:
\[P(t)=\frac12R\cos(E_1t)\]


\Dn
\Dn
\Dn
\Dn
\begin{verbatim}
Remarks:


(3) The integration should be written in a more simple way,
    involving only two Y_{lm}'s.

\end{verbatim}






\end{document}
