\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}}


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



\heading{E452: Operating with rotation matrices on spin states (2003A3)} 
\auname{Yuri Khodorkovsky}


{\bf The problem:}

\Dn

Consider a spin $\frac{1}{2}$ state: $|\psi\rangle = \left(\eexp{-i \frac{\pi}{4}} \cos \frac{\pi}{8}\right)
|\uparrow\rangle + \left(\sin \frac{\pi}{8} \right) |\downarrow\rangle$ .


\Dn

(1) What is the spacial orientation of the state represented by $|\psi\rangle$ ?

(2) Write the rotation matrices needed to rotate a spin up $|\uparrow\rangle $ state to the
$|\psi\rangle$ state.

\Dn

Now consider a spin $1$ up state (circular $z$ polarization)

\Dn


(3) Which state do we get after rotating it as in (2)?

(4) What is the probability to measure a linear $z$ polarization after the rotation done in (3)?


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
The general spin $\frac{1}{2}$ polarization state, in the direction $\theta , \varphi$ is given by:

\[
\left( \begin{array}{c} \eexp{-i\varphi/2} \cos\left(\frac{\theta}{2}\right) \\
\eexp{i\varphi/2} \sin\left(\frac{\theta}{2}\right)\end{array} \right)
\]

After multiplying it by the (immaterial) phase $\eexp{-i\frac{\varphi}{2}}$ , we get:

\[
\left( \begin{array}{c} \eexp{-i\varphi} \cos\left(\frac{\theta}{2}\right) \\
\sin\left(\frac{\theta}{2}\right)\end{array} \right)
\]

Now it is evident, that in our state $|\psi\rangle$:  $ \theta = \varphi = \frac{\pi}{4}$

%%%%%%%%%%%%%%%
\Dn

(2)
The rotation written explicitly is:

\[
|\psi\rangle = \eexp{-i\varphi/2}R_{z}(\varphi)R_{y}(\theta)|\uparrow\rangle = \eexp{-i\pi/8}\eexp{-i\varphi S_{z}} \eexp{-i\theta S_{y}} |\uparrow\rangle
\]

The two rotation operators are represented by:

\[
\eexp{-i \frac{\pi}{4} \cdot \frac{1}{2} \sigma_{y}} = \cos \frac{\pi}{8} -i \sigma_y \sin \frac{\pi}{8} = 
\left( \begin{array}{cccc} \cos \pi/8 & -\sin \pi/8 \\ \sin \pi/8 & \cos \pi/8 \end{array} \right)
\]
\[
\eexp{-i \frac{\pi}{4} \cdot \frac{1}{2} \sigma_{z}} = \cos \frac{\pi}{8} -i \sigma_z \sin \frac{\pi}{8} =
\left( \begin{array}{cccc} \eexp{-i\pi/8} & 0 \\ 0 & \eexp{i\pi/8} \end{array} \right)
\]
%%%%%%%%%%%%%%
\Dn

(3)
Given the state:  $|\Uparrow \rangle \longmapsto \left( \begin{array}{c} 1\\0\\0 \end{array} \right)$ , we will rotate it to get some other state $|\varphi\rangle$ :

\[ |\varphi\rangle = \eexp{-i \frac{\pi}{4} S_z} \eexp{-i \frac{\pi}{4} S_y} |\Uparrow \rangle \]

Using $\eexp{-i\vec \Phi \cdot \vec S} = 1 - i\sin(\Phi)S_n - (1-\cos\Phi)S_n^2$ we obtain:

\[
|\varphi\rangle = 
\left( \begin{array}{ccc} \eexp{-i\pi/4} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \eexp{i\pi/4}\end{array} \right)
\left( \begin{array}{ccc} \frac{1}{2} \left(1+\cos\frac{\pi}{4} \right) & -\frac{1}{\sqrt{2}} \sin\frac{\pi}{4} & \frac{1}{2} \left(1-\cos\frac{\pi}{4}\right) \\ \frac{1}{\sqrt{2}} \sin\frac{\pi}{4} & \cos\frac{\pi}{4} & -\frac{1}{\sqrt{2}} \sin\frac{\pi}{4} \\ \frac{1}{2} \left(1-\cos\frac{\pi}{4}\right) & \frac{1}{\sqrt{2}} \sin\frac{\pi}{4} & \frac{1}{2} \left(1+\cos\frac{\pi}{4}\right) \end{array} \right)
\left( \begin{array}{c} 1\\0\\0 \end{array} \right)
\]

Obviously, we need to use only the first column of the $R_y$ matrix.
After a little algebra we get:

\[
|\varphi\rangle = \left( \begin{array}{c} \frac{\sqrt{2}+1}{4} (1-i) \\ \frac{1}{2} \\ \frac{\sqrt{2}-1}{4} (1+i)\end{array} \right)
\]


%%%%%%%%%%%%%
\Dn

(4)
First, we check and see that this ket is normalized.
Then, we can find the probability:

\[|\langle\Updownarrow|\varphi\rangle|^2 =  \frac{1}{4}\]






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document} 
