\documentclass[11pt,fleqn]{article}


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


% 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}
\usepackage{graphicx}


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

\makeatletter
\def\fps@figure{htbp} % place figure in this order: 'here' 'top' 'bottom' ...
\makeatother

\begin{document} 

\exnumber{4484}
\heading{Spin $\frac{1}{2}$ in a rotating magnetic field} 
\auname{Diana Moskvitin and Semion Dekhtyaryuk}


{\bf The problem:}

\Dn
The Hamiltonian of spin $\frac{1}{2}$ in a magnetic field is  $H = h(t) \cdot S$. \\
Consider a magnetic field with magnitude  $|h(t)| = \Omega_0$, which rotates in the XY-plane with angular velocity $\omega$. \\
After time $t$ the angle between X-axis and magnetic field is $\varphi = \omega \cdot t$.\\
The Hamiltonian of the “stationary system” (the lab) is $\tilde{H} = H - \omega \cdot S_z$.
\begin{enumerate}
\item
Find the angle $\theta_0$ in which one should prepare the spin in order for it to follow the magnetic field with a constant angle ($\theta_0$ is respected to Z-axis) 

\begin{flushleft}
From now on, the spin is prepared (t=0) in X-axis direction 
\end{flushleft}

\item
Letting the system to evolve, what is the maximal deviation angle of the spin from the X-axis

\begin{flushleft}
From now on, we let the magnetic field to propagate for half a cycle ($\varphi = \pi$)
\end{flushleft}

\item 
What is the final direction of the spin in the stationary system in the limit $\omega \mapsto 0$ ?
\item
What is the final direction of the spin in the stationary system in the limit $\omega \mapsto \infty$ ?
\item
For which $\omega$ values the deviation of the spin from X-axis would be 0?
\end{enumerate}

\textbf{Hint:} this question does not require algebraic effort, only geometrical and physical understanding.

\Dn\Dn

\newpage
{\bf The solution:}
\Dn
\begin{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%			1st Part	     %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item 
In order for the spin to follow the magnetic field at a constant angle, it should be prepared in the direction of the magnetic field. By switching to a stationary system we get an effective magnetic field, so one must prepare the spin in the direction of the effective magnetic field.\\

Magnetic field rotates in XY-plane:
\begin{eqnarray}
\vec{\Omega} &=& \Omega_0(\cos{\varphi},\sin{\varphi},0)\\
\Vec{S} &=& (S_x,S_y,S_z)
\end{eqnarray}


Hamiltonian of the stationary system would be: 
\begin{eqnarray}
\tilde{H} &=& \vec{\Omega} \cdot \Vec{S} - \omega \cdot S_{z} = \frac{\Omega_0}{2} \cos{\varphi} \begin{bmatrix}
0 & 1 \\
1 & 0 
\end{bmatrix} 
+ 
\frac{\Omega_0}{2} \sin{\varphi}
\begin{bmatrix}
0 & -i \\
i & 0 
\end{bmatrix} 
- 
\frac{\omega}{2}  
\begin{bmatrix}
1 & 0 \\
0 & -1 
\end{bmatrix} =\\
&=& 
\frac{1}{2}  
\begin{bmatrix}
-\omega & \Omega_0 e^{-i\varphi} \\
\Omega_0 e^{i\varphi} & \omega 
\end{bmatrix}
\end{eqnarray}
Without the loss of generality one could assume that the effective magnetic field is displaced on XZ-plane, hence $\varphi = 0$:
\begin{eqnarray}
\tilde{H} &=& \frac{1}{2}  
\begin{bmatrix}
-\omega & \Omega_0  \\
\Omega_0  & \omega 
\end{bmatrix} = \frac{1}{2} \left(\Omega_0\cdot\sigma_x - \omega\cdot\sigma_z\right) = \Tilde{\Omega}\cdot\Vec{S}\\
\tilde{\Omega} &=& (\Omega_0,0,-\omega) 
\rightarrow\tilde{\theta}_0 = \arctan{\left(\cfrac{\omega}{\Omega_0}\right)}
\end{eqnarray}
So the angle $\theta_0$ would be:
\begin{eqnarray}
\theta_0 &=& \frac{\pi}{2} + \tilde{\theta}_0 = \frac{\pi}{2} + \arctan{\left(\cfrac{\omega}{\Omega_0}\right)}
\end{eqnarray}
\begin{figure}
    \centering
    \includegraphics[width=0.4\textwidth]{1.JPG}
    \caption{stationary system}
    \label{fig:my_label}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%			2nd Part	     %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newpage
\item Consider the effective magnetic field on XZ-plane with $\tilde{\theta}_0$ angle between the field and X-axis, by preparing the spin on X-axis, the spin would do a precession motion around the effective magnetic field, as a result the spin will be at the maximum angle deviation  $\Delta\theta_{Max} = 2 \tilde{\theta}_0$ 

\begin{figure}
    \centering
    \includegraphics[width=0.4\textwidth]{2.JPG}
    \caption{Spin precession in the stationary system}
    \label{fig:my_label}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%			3rd Part	     %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item 
For the $\omega \mapsto 0$ limit, the change is adiabatic, hence the system will conserve its energy level, as a result the spin would follow the magnetic field, (continuing its percession motion) facing $\varphi = \pi$ respectfully to XY-plane.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%			4th Part	     %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item
For the $\omega \mapsto \infty$ limit, the change is sudden, so the state of the system before the change is conserved, hence the spin would stay in its initial state in $\varphi =0 $ respectfully to XY-plane.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%			5th Part	     %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item 
Magnetic field rotates to $-\hat{x}$, finishing half of its period. Full period would be:
\begin{eqnarray}
\omega \cdot T_{m} &=& 2\pi \rightarrow T_{m} = \frac{2\pi}{\omega}
\end{eqnarray}
For each half a period of the magnetic field the spin must finish a full period, by using this constrain on the spin's precession, and the fact that the spin precession frequency is $\Omega$ we get:
\begin{eqnarray}
T_{Spin} &=& \frac{\pi}{\omega}\\
\Omega &=& \sqrt{{\Omega_0}^2+{\omega}^2}\\
\Omega \cdot T_{Spin} &=& 2 \pi \cdot n\\
\omega_{n} &=& \frac{\Omega_0}{\sqrt{4 n^2 - 1}} , \ \ \ n = 1, 2, 3, ...
\end{eqnarray}
\end{enumerate}
\end{document} 