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\heading{E442: Determining polarization state of spin $\frac{1}{2}$} 
\auname{Lihor Komanski}


{\bf The problem:}

\Dn

Find the orientation of the polarized state 
\[|\psi\rangle = \left(\eexp{-i \frac{\pi}{4}} \cos \frac{\pi}{8}\right)
|\uparrow\rangle + \left(\sin \frac{\pi}{8} \right) |\downarrow\rangle\]
using the two following methods:

\Dn

(1) The first method is getting the polarized state by using the rotation matrices (around $y$ axis and then around $z$ axis) on the up $|\uparrow\rangle$ state.


(2) The second method is by calculating the polarization vector (i.e. the expectation values of $\sigma_{i}$)


\Dn\Dn


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{\bf The solution:}

\Dn

(1)first we find the polarization for a general state by rotating an up $|\uparrow\rangle$ state first by $\theta$ around $y$ axis and then $\phi$ around $z$ axis. It is given by:

\[
|\vec{e_{\theta,\phi}}\rangle = R_{z}(\phi)R_{y}(\theta)|\uparrow\rangle = \eexp{-i\phi S_{z}} \eexp{-i\theta S_{y}} |\uparrow\rangle=
 \left( \begin{array}{c} \eexp{-i\phi/2} \cos\left(\frac{\theta}{2}\right) \\
\eexp{i\phi/2} \sin\left(\frac{\theta}{2}\right)\end{array} \right)
\]
Where
\[
R_{y}(\theta)=\eexp{-i \theta \cdot \frac{1}{2} \sigma_{y}} = \cos \frac{\theta}{2} -i \sigma_y \sin \frac{\theta}{2} =
\left( \begin{array}{cccc} \cos \frac{\theta}{2} & -\sin \frac{\theta}{2} \\ \sin  \frac{\theta}{2} & \cos \frac{\theta}{2} \end{array} \right)
\]
\[
R_{z}(\phi)=\eexp{-i \phi \cdot \frac{1}{2} \sigma_{z}} = \cos \frac{\phi}{2} -i \sigma_{z} \sin \frac{\phi}{2} =
\left( \begin{array}{cccc} \eexp{-i \frac{\phi}{2}} & 0 \\ 0 &  \eexp{i \frac{\phi}{2}} \end{array} \right)
\]
Thanks to gauge freedom, we can write the general state as:
\[
|\vec{e_{\theta,\phi}}\rangle = \left( \begin{array}{c} \eexp{-i\phi/2} \cos\left(\frac{\theta}{2}\right) \\ \eexp{i\phi/2} \sin\left(\frac{\theta}{2}\right) \end{array} \right) \cdot \eexp{i \alpha}
\]
So the state can be written with $\alpha= -\frac{\phi}{2}= -\frac{\pi}{8} $ 
\[
|\vec{e_{\theta,\phi}}\rangle =\left( \begin{array}{c} \eexp{-i\phi} \cos\left(\frac{\theta}{2}\right) \\
\sin\left(\frac{\theta}{2}\right) \end{array} \right)
\]
Comparing $|\psi\rangle$ and $|\vec{e_{\theta,\phi}}\rangle$
gives $\theta = \frac{\pi}{4} \text{ and } \phi = \frac{\pi}{4} $





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

(2) The polarization vector is defined as:
\[ \vec{M} = \left( \langle\sigma_{x}\rangle ,  \langle\sigma_{y}\rangle , \langle\sigma_{z}\rangle \right) \]
Where
\[ \langle\sigma_{i}\rangle = \langle\Psi\vert\sigma_{i}\vert\Psi\rangle \]
For $\vert\psi\rangle$ state:
\[ 
\vert\psi\rangle= \left( \begin{array}{c} \eexp{-i \frac{\pi}{4}} \cos\left(\frac{\pi}{8}\right) \\
 \sin\left(\frac{\pi}{8}\right)\end{array} \right)
\]
\[\langle\psi\vert\sigma_{x}\vert\psi\rangle= \cos\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}\right)=\frac{1}{2} \]
\[\langle\psi\vert\sigma_{y}\vert\psi\rangle= \sin\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}\right)=\frac{1}{2} \]
\[\langle\psi\vert\sigma_{z}\vert\psi\rangle=\vert\psi_{1}\vert^{2} -\vert\psi_{2}\vert^{2}
=\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \]

Hence the polarization vector will be:
\[ \vec{M} = \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{\sqrt{2}} \right) \] 
Therefore the spacial orientation will be: $\theta = \frac{\pi}{4} \text{ and } \phi = \frac{\pi}{4} $


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