\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{E451: Operating rotation matrices on spin states (2002A2)}
\auname{Adam Reichenthal}


{\bf The problem:}

\Dn

Find the state vector ${\Psi}_m$ in the standard base for a spin polarized in the XY plane at angle ${\varphi}= 60^\circ.$

Relate to the following cases:
\Dn

(1) Spin 1/2.

(2) Spin 1 with circular polarization.

(3) Spin 1 with linear polarization.
\Dn

You must express the elements of the state vector in your final answer with ${e^{i{\pi}/integer} }$, ${\sqrt{2}}$ etc.

In order to prevent losing all points because of an algebraic mistake, please write first the analytic formula and only after calculate it to receive your final answer.

\Dn\Dn

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

(1) 
We can receive any spin 1/2 polarization state by combining a rotation round the Y axis and a rotation round the Z axis. The result is:

\[
|\vec{e}_{\theta,\varphi}>=R(\varphi)R(\theta)|\uparrow>=e^{-i \varphi S_z}e^{-i \theta S_y}|\uparrow> \rightarrow \left(\amatrix{e^{-i \varphi/2}cos(\theta/2) \cr e^{i \varphi/2}sin(\theta/2)}\right)
\]

The rotation action that has been done on the spin 1/2 up is a rotation by $\theta=\pi/2$ round the Y axis and after that a rotation by $ \varphi = \pi/3 $ round the Z axis. Therefore:

\[
|\Psi_m>=e^{-i(\pi/6)}cos(\pi/4)| \uparrow> + e^{i(\pi/6)}sin(\pi/4)| \downarrow>
\]

\Dn

Analytic result:
\[
 \Psi_m  = 
\left(\amatrix{ e^{-i \frac{\pi}{6}}cos(\frac{\pi}{4}) \cr e^{i \frac{\pi}{6}}sin(\frac{\pi}{4}) }\right)
=
\frac{1}{\sqrt{2}} \left(\amatrix{ e^{-i\frac{\pi}{6}} \cr e^{i\frac{\pi}{6}} }\right)
\]

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


(2) 
We will use a similar technique as at (1) once again but this time using dim=3 rotation matrices to be able to take care of spin 1 and using the linear basis.

\Dn

We can rotate the state  $ | \Uparrow > $ in order to get other circularly polarized states:
\[
|\vec{e}_{\theta,\varphi}>=U(\varphi\vec{e}_z)U(\theta\vec{e}_y)|\Uparrow>=e^{-i \varphi S_z}e^{-i \theta S_y}|\Uparrow> \rightarrow 
\left(\amatrix{\frac{1}{2} (1+cos\theta) e^{-i \varphi} \cr \frac{1}{\sqrt{2}} sin \theta \cr \frac{1}{2} (1-cos\theta) e^{i \varphi})}\right)
\]

The rotation action that has been done on the spin 1 up is a rotation by $\theta=\pi/2$ round the Y axis and after that a rotation by $ \varphi = \pi/3 $ round the Z axis. Therefore:

\[
|\Psi_m>=
\frac{1}{2}(1+cos(\pi/2)) e^{-i \pi/3}| \Uparrow> +
\frac{1}{\sqrt{2}} sin(\pi/2) | \Updownarrow>+
\frac{1}{2} (1-cos(\pi/2)) e^{i \pi/3}| \Downarrow>
\]

\Dn

Analytic result:
\[
\Psi_m=
\frac{1}{2} \left (\amatrix { e^{-i\frac{\pi}{3}} \cr  \sqrt{2} \cr  e^{i\frac{\pi}{3}}}\right)
\]

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


(3) 
Using the same technique as in (2) this time with a rotation matrix that rotates the state $|\Updownarrow>$ in order to get other linearly polarized states:

\[
|\vec{e}_{\theta,\varphi}>=
U(\varphi\vec{e}_z)U(\theta\vec{e}_y)|\Updownarrow>=
e^{-i \varphi S_z}e^{-i \theta S_y}| \Updownarrow> \rightarrow
\left(\amatrix{-\frac{1}{\sqrt{2}} sin\theta e^{-i \varphi} \cr cos\theta \cr \frac{1}{\sqrt{2}}sin\theta e^{i \varphi}}\right)
\]

We receive:
\[
|\Psi_m>=
-\frac{1}{\sqrt{2}} sin(\pi/2) e^{-i \pi/3} |\Uparrow> +
cos(\pi/2)|\Updownarrow> +
\frac{1}{\sqrt{2}}sin(\pi/2) e^{i \pi/3}|\Downarrow>
\]

Analytic result:
\[
\Psi_m=
\frac{1}{\sqrt{2}} \left (\amatrix { -e^{-i\frac{\pi}{3}} \cr 0 \cr  e^{i\frac{\pi}{3}}}\right)
\]


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