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\exnumber{4482}
\heading{Precession in a three sites system} 
\auname{Shimon Elkabetz}


{\bf The problem:}

\Dn

A system is characterized by an energy spectrum $E_{n}$.
The system is set up in a state $\psi$.
Given the survival probability $p_n=\left|\left\langle E{n}|\psi\right\rangle \right|^{2}$
 which describes the initial state of the system.

$P\left(t\right)$ is the probablity find the system in the initial state after a time t.

 \Dn

$\overline{P}=\displaystyle \lim_{t\rightarrow\infty}\frac{1}{t}\intop_{0}^{t}P\left(t\right)$

\Dn

$\left(0\right)$ Express $\overline{P}$ in terms of the survival probablity $p{}_{n}$.

A particle of charge $q$ is in a three site system. 

The standard basis is represented by the states $\left|x=1\right\rangle$ 
 ,$\left|x=0\right\rangle$ ,$\left|x=-1\right\rangle$.
 \Dn
 
The Hamiltonian of the system is:$ \left(\begin{array}{ccc}
0 & c & 0\\
c & 0 & c\\
0 & c & 0
\end{array}\right)$
 
\Dn
An external electric field f is inserted along the x-axis.   
The initial state of the particle is given by : $\left|x=1\right\rangle$.

\Dn


(1) Express the Hamiltonian in terms of the generators of rotations $S_{x},S_{y},S_{z}$
    for a spin 1 particle.

(2) what is the fundamental frequency $\omega$ of the periodic motion obtained by the system?

(3) Explicitly represent $R\left(t\right)=\left\langle x\right\rangle _{t}$
 

(4) The probability P(t) can be expressed or written as a Fourier sum of the cosine sine functions, Write explicitly the expression of the Fourier sum of $P(t)$ of the non zero terms. There is no need to calculate the $A_{n} \& B_{n}$.

(5) Use the results obtained in clause $\left(0\right)$ in order to get an explicit expression of $\overline{P}$.   

(6) What is the minimal value $\overline{P}$ can obtain? For what electric field does this occur?
\Dn

In clause (3) it is helpful to use the known results for precession ( Rabi's Formula ).

In clause (5) it is recommended to use the known polarization states of spin 1.


\Dn\Dn


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

\Dn
(0) The initial state:

\Dn
\begin{center}
$\psi(0)=\displaystyle \sum_{n}|E_{n}><E_{n}|\psi>$ 
\end{center}

\Dn
The general state:
\begin{center}
$\psi(t)=\displaystyle\sum_{n} e^{-iE_{n}t}|E_{n}><E_{n}|\psi>$
\end{center}
 

So:

\begin{center}
 $P(t)=|<\psi(0)|\psi(t)>|^{2} =|\displaystyle \sum_{n} <E_{n}|\psi>^{*}<E_{n}|\displaystyle \sum_{n}<E_{n}|\psi> e^{-iE_{n}t}|E_{n}>|^{2}=$
\Dn 
 $|\displaystyle\sum_{n}<E_{n}|\psi>^{*}<E_{n}|\psi><E_{n}|E_{n}>e^{-iE_{n}t}|^{2}=|\displaystyle\sum_{n}|<E_{n}|\psi>|^{2}<E_{n}|E_{n}>e^{-iE_{n}t}|^{2}$
 \Dn
$ =\left|\displaystyle\sum_{n} p_{n}e^{-iE_{n}t}\right|^{2}$

\end{center}

\begin{center}
$ \overline{P}=\displaystyle \lim_{t\to\infty}\frac{1}{t}\int_{0}^{t}P(t)dt=\displaystyle \lim_{t\to\infty}\frac{1}{t}\int_{0}^{t}\left|\displaystyle \sum_{n} p_{n}e^{-iE_{n}t}\right|^{2}dt=\displaystyle \lim_{t\to\infty}\frac{1}{t}\int_{0}^{t} \displaystyle \sum_{n,m} p_{n}p_{m}e^{-i(E_{n}-E_{m})t}dt$
\end{center}
\Dn
For sum $n\neq m$ the Sine terms cancel each other

\begin{center}
 $=\displaystyle \lim_{t\to\infty}\frac{1}{t}\int_{0}^{t} \displaystyle \sum_{n} p_{n}^2 +\sum_{n\neq m}2cos{((E_{n}-E_{m})t)}dt$
 \end{center} 

\begin{center}
$\bar{P}=\displaystyle \lim_{t\to\infty}\frac{1}{t}[(\displaystyle\sum_{n} p_{n}^2)t + \frac{2sin((E_{n}-E_{m})t)}{E_{n}-E_{m}}]$
\end{center}

\Dn
\begin{center}
 $\bar{P}=\displaystyle\sum_{n} p_{n}^2$

 \end{center} 
\Dn
%%%%%%%%%%%%%%%
(1)The Hamiltonian is:
\Dn
\begin{center}
$
H=H_{0}+V=\left(\begin{array}{ccc}
0 & c & 0\\
c & 0 & c\\
0 & c & 0
\end{array}\right)-qfx$
\end{center}
 

The intial state is $ \left|x=1\right\rangle$.
\Dn

The matrix representation for the position operator x in the standard basis is:
\Dn
\begin{center}
$\
x=\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & -1
\end{array}\right)$
\end{center}
  and for the generators of rotations: 
  \Dn
  
 \begin{center}
  $S_{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 1\\
0 & 1 & 0
\end{array}\right)$ ,$ S_{z}=\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & -1
\end{array}\right)$
 
 \end{center}

It easy to see that :

\begin{center}
$H=\sqrt{2}cS_{x}-qfS_{z}$
\end{center}
 

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

(2)
The evolution operator is:\begin{center}
$U\left(t\right)=e^{-it\hat{H}}=e^{-i(\overrightarrow{\Omega t})\overrightarrow{s}}$
\end{center}

\Dn
\begin{center}
$H=\overrightarrow{\Omega}\cdot\overrightarrow{S}=\Omega_{x}S_{x}+\Omega_{y}S_{y}+\Omega_{z}S_{z}$
\end{center}

 \Dn

\begin{center}
$\overrightarrow{\Omega}=(\sqrt{2}c,0,-qf)$
\end{center}
 
\begin{center}
$\vert\overrightarrow{\Omega}\vert=\sqrt{(\sqrt{2}c)^2+(qf)^2}$

\end{center}


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

(3)\begin{center}
$\left\langle x\right\rangle =\left\langle \left(\begin{array}{ccc}
1\\
 & 0\\
 &  & -1
\end{array}\right)\right\rangle$
\end{center}
\Dn
\begin{center}
 $\langle x \rangle=<\psi(t)|x|\psi(t)>=P(x=1)*1+P(x=0)*0+P(x=-1)*-1$

\end{center}
\Dn
 
 now its look like spin $\frac{1}{2}$ problem.
 
 \Dn
 if we take a new $S_x$ dim=2 with changing the 'c' to 2'c' as the energy gain, we preserve the eigenvalues.
  The evolution in time operator stay the same.
  
so
\Dn

\begin{center}
$\theta=\arctan(\frac{\sqrt{2}c}{qf})$

\Dn
$U(t)|\uparrow>=R(\overrightarrow{\theta}(t))= cos(\frac{\Omega t}{2})-isin(\frac{\Omega t}{2})[cos(\theta)\sigma_z+ \sin(\theta)\sigma_x]$
\Dn

$|\psi(t)>=U(t)|\uparrow>=[cos(\frac{\Omega t}{2})-isin(\frac{\Omega t}{2})cos(\theta)]|\uparrow> - i[\sin(\theta) sin(\frac{\Omega t}{2}])|\downarrow>$
\Dn

$\langle R(t)\rangle=\langle x\rangle=<\psi(t)|x|\psi(t)>=<\psi(t)|\lbrace[cos(\frac{\Omega t}{2})-isin(\frac{\Omega t}{2})cos(\theta)]|\uparrow> + i[\sin(\theta)sin(\frac{\Omega t}{2}])|\downarrow>\rbrace$
\Dn

$=cos(\frac{\Omega t}{2})^2+sin(\frac{\Omega t}{2})^2cos(\theta)^2-\sin(\theta)^2 sin(\frac{\Omega t}{2})^2$
\Dn
$=1-2\sin(\theta)^2 sin(\frac{\Omega t}{2})^2$

\end{center}\Dn 
 

%%%%%%%%%%%%%

(4)The probability to find the system in the initial state after time t is:

\begin{center}
$P\left(t\right)=\left|\left\langle 1|\psi\left(t\right)\right\rangle \right|^{2}=\displaystyle\sum_{n,m}p_{n}^{*}p_{m}e^{i(E_{n}-E_{m})t}=\left\{ E_{n}-E_{m}= \tilde {\Omega}\right\} $
\end{center}
 

we know that $E_n$ get only values from $\lbrace 0,\Omega,-\Omega\rbrace$ and $\tilde{\Omega}$ get $\lbrace 0,\Omega,-\Omega,2\Omega,-2\Omega\rbrace$:
\Dn

For n=m $\Rightarrow A_{0}$

\Dn
For n$\neq m\Rightarrow$ we have for $e^{i(E_{n}-E_{m})t}$ 4 options(without duplications):
\Dn

\begin{center}
$cos(\Omega t)+isin(\Omega t)$, $cos(2\Omega t)+isin(2\Omega t)$,$ cos(-\Omega t)+isin(-\Omega t)$,$ cos(-2\Omega t)+isin(-2\Omega t)$
\end{center}

The sine function cancelled and we stay only with:

\begin{center}
$A_{0}+A_1cos(\Omega t)+A_2cos(2\Omega t)$
\end{center}
 


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

(5)Explicit expression of $\overline{P}$ is:
\Dn

Any rotation can be given by a combination of rotations around z axis and a rotation around the y axis.
we find the state $|\uparrow>$ after rotations is circularly polarized.
the circularly polarized state is the time average of the initial state rotation.
\Dn


\Dn

for $\theta=\arctan(\frac{\sqrt{2c}}{\epsilon})$

\Dn
\begin{center}
$U(\varphi\overrightarrow{n_z})U(\theta \overrightarrow{n_y})\left(\begin{array}{c} 1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}
\frac{1}{2}\left(1+cos\theta\right)\\
\frac{1}{\sqrt{2}}sin\theta\\
\frac{1}{2}\left(1-cos\theta\right)
\end{array}\right)=\left(\begin{array}{c}
\psi_{1}\\
\psi_{2}\\
\psi_{3}
\end{array}\right)$
\end{center}
 

\begin{center}
$
P_{i}=\left|\left\langle \psi_{i}|\psi\right\rangle \right|^{2}$
\end{center}
 

\begin{center}
$
\bar{P}=\displaystyle\sum_{n} p_{n}^{2}=P^{2}\left(x=1\right)+P^{2}(x=0)+P^{2}(x=-1)=\left(\frac{1}{2}\left(1+cos\theta\right)\right)^{4}+\left(\frac{1}{\sqrt{2}}sin\theta\right)^{4}+\left(\frac{1}{2}\left(1-cos\theta\right)\right)^{4}$
\end{center}

%%%%%%%%%%%%%
\Dn
(6)The minimal value $\overline{P}$ can obtain is:

\begin{center}
$\bar{P}=\frac{1}{16}\left[\left(1+cos\theta\right)^{4}+4sin^{4}\theta+\left(1-cos\theta\right)^{4}\right]$ 
\end{center}

\begin{center}
$0=\frac{\partial P}{\partial\theta}=\frac{1}{4}\left[-sin\theta\left(1+cos\theta\right)^{3}+4cos\theta sin^{3}\theta+sin\theta\left(1-cos\theta\right)^{3}\right]$
\end{center}

\begin{center}
$0=-\left(1+cos\theta\right)^{3}+4cos\theta sin^{2}\theta+\left(1-cos\theta\right)^{3}$
\end{center}
 

\begin{center}
$0=-\left(1+cos^{3}\theta+3cos\theta+3cos^{2}\theta\right)+4cos\theta sin^{2}\theta+\left(1-cos^{3}\theta-3cos\theta+3cos^{2}\theta\right) $
\end{center}
 

\begin{center}
$
0=2cos\theta sin^{2}\theta-4cos^{3}\theta-6cos\theta$
\end{center}

It can be true only when:
\begin{center}
$
sin\theta=0,cos\theta=0$
\end{center}
 

we get maximum for $sin\theta=0$
and minimum for $cos\theta=0$
 

\begin{center}
$\bar{P}=\frac{1}{16}\left[\left(1\right)^{4}+4+\left(1\right)^{4}\right]=\frac{6}{16}=\frac{3}{8}$ 
\end{center}

The required electric field is:

\begin{center}
$0=cos\theta=\frac{qf_{min}}{\sqrt{2c^{2}+q^{2}f^{2}}}$
\end{center}
 


\begin{center}
$f_{min}=0$
\end{center}
 


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