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\heading{E443: Identification Hadamard rotation of spin 1/2} 
\auname{Maya Zeevi}

{\bf The problem:}

\Dn
Given a particle of spin 1/2 let's define the next operation (in the standard basis): \\
$ R=\frac{1}{\sqrt{2}} \left(  \amatrix  {1 & 1 \cr 1 & -1 } \right) $ \\
(1) Operating R on the up state. What will be the polarization direction? \\
(2) Operating R on the down sate. What will be the polarization direction? \\
(3) What is the rotation angle represented by R? \\ \\
It is possible to realize R at the lab by operating magnetic field for 1 sec. At this time the Hamiltonian is of the form of $H=h\cdot\sigma+c$. \\
(4) Decide the direction and the size of the magnetic field h. \\
(5) Decide the value of the constant c. \\

\Dn\Dn

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{\bf The solution:}

\Dn

(1) By operating R on the up state, we get: 
\begin{center}
$R\left|\uparrow\right\rangle=\frac{1}{\sqrt{2}}(\left|\uparrow\right\rangle + \left|\downarrow\right\rangle$
\end{center}
(2) By operating R on the down state, we get: 
\begin{center}
$R\left|\downarrow\right\rangle=\frac{1}{\sqrt{2}}(\left|\uparrow\right\rangle - \left|\downarrow\right\rangle$
\end{center}
(3)We can see the R matrix as the rotation matrix multiplying by a phase factor: \\
   Thus we will compare them.\\
   
$\frac{1}{\sqrt{2}} \left(  \amatrix  {1 & 1 \cr 1 & -1 } \right)=\frac{1}{\sqrt{2}}\left(\sigma_{x} + \sigma_{z}\right)=\frac{1}{\sqrt{2}}\left(1,0,1\right)\cdot\left(\sigma_{x},\sigma_{y},\sigma_{z}\right)=  \vec{n}\cdot\vec{\sigma} $\\

$ \vec{n}\cdot\vec{\sigma}=e^{i\alpha}\left[cos\left(\frac{\Phi}{2}\right)\hat{1}-i sin\left(\frac{\Phi}{2}\right)\cdot\sigma_{n}\right]$\\


One can see from the equation \\

$ cos \left(\frac{\Phi}{2}\right)=0 \hspace{0.5cm} \Rightarrow \Phi=\pi $ \\

$ e^{i\alpha}\cdot\left(-i sin\left(\frac{\Phi}{2}\right)\right)=1 \hspace{0.5cm} \Rightarrow \alpha=\frac{\pi}{2}$\\




(4)+(5) After 1 sec the evolution operator is equal to R.\\
        Thus we will compare them.\\
 

$\hat{U(t)}=e^{-i\hat{H}t}=e^{-i(h\cdot\sigma+c)t}$ \\
$U(t=1sec)=e^{-i\hat{h}\cdot\sigma}\cdot e^{-i c}=R=e^{i \alpha}\cdot e^{\frac{-i \hat{\Phi}\sigma}{2}}$\\
 

Derive from the comparison: \\
$c=- \alpha=-\frac{\pi}{2}$\\
$\hat{h}=|h|\cdot\hat{n}$ \\
 $|\hat{h}|=\frac{\pi}{2}, \hat{n}=\frac{1}{\sqrt{2}}(1,0,1)$\\
 



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