%% This LaTeX-file was created by <jl> Tue Mar 11 11:57:02 1997
%% LyX 0.10 (C) 1995 1996 by Matthias Ettrich and the LyX Team

%% Don't edit this file unless you are sure what you are doing.
\documentclass[12pt,letterpaper,oneside,onecolumn]{article}
\usepackage[]{fontenc}
\usepackage[dvips]{epsfig}

%%
%% BEGIN The lyx specific LaTeX commands.
%%

\makeatletter
\def\LyX{L\kern-.1667em\lower.25em\hbox{Y}\kern-.125emX\spacefactor1000}%
\newcommand{\lyxtitle}[1] {\thispagestyle{empty}
\global\@topnum\z@
\section*{\LARGE \centering \sffamily \bfseries \protect#1 }
}
\newcommand{\lyxline}[1]{
{#1 \vspace{1ex} \hrule width \columnwidth \vspace{1ex}}
}
\newenvironment{lyxbibliography}
{
\begin{thebibliography}{99}}
{\end{thebibliography}}
\def\lxq{"}
\newenvironment{lyxcode}
{\list{}{
\rightmargin\leftmargin
\raggedright
\itemsep 0pt
\parsep 0pt
\ttfamily
}%
\item[]
}
{\endlist}
\newcommand{\lyxlabel}[1]{#1 \hfill}
\newenvironment{lyxlist}[1]
{\begin{list}{}
{\settowidth{\labelwidth}{#1}
\setlength{\leftmargin}{\labelwidth}
\addtolength{\leftmargin}{\labelsep}
\renewcommand{\makelabel}{\lyxlabel}}}
{\end{list}}
\newcommand{\lyxletterstyle}{
\setlength\parskip{0.7em}
\setlength\parindent{0pt}
}
\newcommand{\lyxaddress}[1]{
\par {\raggedright #1 
\vspace{1.4em}
\noindent\par}
}
\newcommand{\lyxrightaddress}[1]{
\par {\raggedleft \begin{tabular}{l}\ignorespaces
#1
\end{tabular}
\vspace{1.4em}
\par}
}
\newcommand{\lyxformula}[1]{
\begin{eqnarray*}
#1
\end{eqnarray*}
}
\newcommand{\lyxnumberedformula}[1]{
\begin{eqnarray}
#1
\end{eqnarray}
}
\makeatother

%%
%% END The lyx specific LaTeX commands.
%%

\pagestyle{plain}
\setcounter{secnumdepth}{3}
\setcounter{tocdepth}{3}
\begin{document}

An example of the use of TDPT is a spin 1/2 system in an oscillating
magnetic field.

This can be:

\begin{itemize}

\item electron bound in an atom 

\item nucleon bound in nucleus

\item nucleus (NMR)

\item free electron

\end{itemize}

states in absence of magnetic field is eigenstate of \( H_{0} \), \( E=E_{0}=\hbar \omega _{0} \).

\begin{itemize}

\item spin states are \( |+> \), \( |-> \)

\item \( H_{0}=E_{0}|+><+|+E_{0}|-><-| \)

\item Then the the hamiltonian with a z magnetic field will be: \( H_{0}=(E_{0}+(K=g_{e}\frac{|\mu _{e}|}{2}B_{0}))|+><+|+(E_{0}-K)|-><-| \). Splitting
of degeneracy occurs. Let \( |g\mu |=\hbar \omega _{12} \)= magnitude of split in degeneracy

\item A rotating magnetic field perpendicular to the z direction will give
\( H'(t)=C(cos(\omega t)-isin(\omega t))|+><-|+C(cos(\omega t)+isin(\omega t))|-><+| \) where \( C=\frac{|g\mu |}{2}B_{1} \)

\end{itemize}

In the interaction picture, \( i\hbar (b_{-}'(t)|->+b_{+}'(t)|+>=Ce^{-i(\omega -\omega _{12})t}|+><-|+Ce^{i(\omega -\omega _{12})t}|-><+| \)

Now apply the results of TDPT. Initially \( b_{-}=1 \) so \( b_{1}(t)=-2\iota C\frac{1}{\hbar (\omega -\omega _{12})}e^{-i(\omega -\omega _{12})t/2}sin((\omega -\omega _{12})t/2) \).

This corresponds to a probability of transition of:

\( \frac{4C^{2}}{\hbar ^{2}(\omega -\omega _{12})^{2}}sin^{2}(\frac{\omega -\omega _{12}}{2}t) \)

near resonance, this breaks down. Away from resonance this goes as
\( sin^{2} \). Near resonance and with short times, you get: \( C^{2}t^{2}\frac{1}{\hbar ^{2}} \).

The exact solution is solvable. Go into rotated frame. \( b_{i}(t)=c_{i}(t)e^{\mp (\omega -\omega _{12})t/2} \), then you
get: \( i\hbar c_{i}'=\mp (\delta =\frac{\omega -\omega _{12}}{2})c_{i}+C\epsilon _{ij}c_{j} \). eigenvalues of diagonalization are: \( \lambda =\pm \hbar \sqrt{\frac{C^{2}}{\hbar ^{2}}+\delta ^{2}} \). Eigenvectors can be
written as \( |\lambda _{+}>=cos(\theta )|+>+sin(\theta )|-> \) and similarly for \( |\lambda _{-}> \). where \( \theta =\frac{1}{2}atan(\frac{-C}{\hbar \delta }) \). 

Time dependence in nonrotating frame is: \( sin(\theta )cos(\theta )(e^{-i\lambda t/\hbar }-e^{i\lambda t/\hbar })e^{-i(\omega _{0}+\omega /2)t}|+> \) \( +(sin^{2}(\theta )e^{-i\lambda t/\hbar }-cos^{2}(\theta )e^{i\lambda t/\hbar })e^{-i(\omega _{0}-\omega /2)t}|-> \).

The probability of changing state is then: \( P_{1}=\frac{\frac{C^{2}}{\hbar ^{2}}}{\frac{C^{2}}{\hbar ^{2}}+(\frac{w-w_{12}}{2})^{2}}sin^{2}(t\sqrt{\frac{C^{2}}{\hbar ^{2}}+(\frac{w-w_{12}}{2})^{2}}) \). At resonance, the probability
is 1. Around the peak, you have lorentzian behavior.

\end{document}
