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\heading{E453: Operating Rotation Matrix on Spins} 
\auname{Reut Balaly}


{\bf The problem:}

\Dn

1) Write the standard representation of spin 1 with linear polarization in the $\hat{z}$ direction. \\
2) Calculate the standard representation of spin 1 with circular polarization in the $\hat{x}$ direction. \\
One measeures whether the spin is linear polarized in the $\hat{y}$ direction. \\ 
3) What is the probability of getting a positive result for the 1st preparation? \\
4) What is the probability of getting a positive result for the 2nd preparation?  

\Dn


\Dn\Dn


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{\bf The solution:}

\Dn

(1) The representation of spin one with linear polarization in the $\hat{z}$ direction is: \\
\begin{center}
$\left|\bar{e_{z}}\right\rangle = \left|\Updownarrow\right\rangle\rightarrow \left(  \amatrix {{0} \cr  {1} \cr {0} } \right)$
\end{center}
\\ \\
(2) If we rotate the circular polarization state (of the standard basis) by 90 degrees round the y axis we get: \\
\begin{center}
$\left|\vec{e_{x}}\right\rangle =R(\frac{\pi}{2} \bar{e_{y}} )\left|\Uparrow\right\rangle \rightarrow \left(  \amatrix { {\frac{1}{2} & \frac{-1}{\sqrt{2}} & \frac{1}{2} } \cr { \frac{1}{\sqrt{2}} & 0 & \frac{-1}{\sqrt{2} }} \cr {\frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2}} } \right) \cdot\left(  \amatrix {{1} \cr  {0} \cr {0} } \right) = 
\left(  \amatrix {{\frac{1}{2}} \cr  {\frac{1}{\sqrt{2}}} \cr {\frac{1}{2}} } \right)$
\end{center}
\\ \\
(3) The probability of getting a positive result from the 1st preparation is:
\begin{center}
$ \left|\left\langle \bar{e_{y}} | \bar{e_{z}} \right\rangle \right|^{2}=0$
\end{center}
because of the 90 degree orthogonality of the linear basis. \\ \\
(4) due to the freedem of choice for the axis system we can say that  $ \left|\left\langle \bar{e_{y}} | \vec{e_{x}} \right\rangle \right|^{2}= \left|\left\langle \bar{e_{z}} | \vec{e_{x}} \right\rangle \right|^{2}$, and the probability of getting a positive result for the second preparation is: \begin{center}
$ \left|\left\langle \bar{e_{z}} | \vec{e_{x}} \right\rangle \right|^{2}=\left|\left(  \amatrix{0} \cr  {1} \cr {0 } \right)\cdot\left(  \amatrix {{\frac{1}{2}} \cr  {\frac{1}{\sqrt{2}}} \cr {\frac{1}{2}} } \right)\right|^{2}=\frac{1}{2}$
\end{center}
 \\\\
We can confirm the results by direct calculations.


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