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\heading{E4440: Identification of Rotation}
\auname{Uri Argaman and Yair Margalit}


{\bf The problem:}

\Dn

Assume a particle with spin 1/2. The following rotation matrices are defined (in the standard basis):

\[
\mathcal{R_A}
= \left( \amatrix { 0 & 1 \cr 1 & 0}\right)
\]
\Dn
\Dn
\[
\mathcal{R_B}
=\frac{1}{\sqrt{2}}\left( \amatrix { 1 & -1 \cr 1 & 1}\right)
\]
\Dn
\Dn
\[
\mathcal{R_C}
=\frac{1}{\sqrt{2}}\left( \amatrix { 1 & 1 \cr 1 & -1}\right)
\]
\Dn

(1) Find the angle of rotation ${\Phi}$ and the axis of rotation
${\vec n}$ for all of the matrices.

(2) Write a rotation matrix 4x4 that operates $R_C$ on 2 spins.

\Dn\Dn


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



(1) A general rotation of spin $\frac{1}{2}$ is given by:
\[
R(\Phi)=cos(\frac{\Phi}{2})I-isin(\frac{\Phi}{2})\sigma_n
\]
\[
\sigma_n=\vec n \cdot \vec \sigma
\]
(A)For $R_A$: We can see that $R_A=\sigma_x$. So using gauge freedom,
we will demand:
\[
R(\Phi)=[cos(\frac{\Phi}{2})I-isin(\frac{\Phi}{2})\sigma_n]e^{i\alpha}=\sigma_x
\]

If we use $\Phi=\pi$ and $\vec {n}=(1,0,0)$ we get:
\[
R(\pi)=-ie^{i\alpha}\sigma_x
\]

So for $\alpha=\frac{\pi}{2}$ we get the desired result.

\Dn

(B)Similarly for $R_B$; We can see that:
\[
R_B=\frac{1}{\sqrt2}(I-i\sigma_y)
\]
We will choose $\Phi=\frac{\pi}{2}$ and $\vec n=(0,1,0)$:
\[
R(\frac{\pi}{2})=\frac{1}{\sqrt2}[I-i\left(\amatrix{ 0 & -i\cr i &
0}\right)]=R_B
\]

(C)For $R_C$, we see that:
\[
R_C=\frac{1}{\sqrt2}(\sigma_x+\sigma_z)
\]
So we will choose $\Phi=\pi$ and $\vec n=\frac{1}{\sqrt2}(1,0,1)$.
Using gauge freedom we will demand:

\[
R(\pi)=-ie^{i\alpha}\frac{1}{\sqrt2}(\sigma_x+\sigma_z)=-\frac{i}{\sqrt2}e^{i\alpha}\left(\amatrix{ 1 & 1\cr 1 &
-1}\right)
\]
And if we take $\alpha=\frac{\pi}{2}$ we get the desired result.

\Dn
\Dn

(2)
The 2 dimensional standard basis is: $|\uparrow>=\begin{pmatrix} 1 \\ 0\end{pmatrix}$ and $|\downarrow>=\begin{pmatrix} 0 \\ 1\end{pmatrix}$. So we will take:
\[
|\uparrow\uparrow>=|\uparrow>\otimes|\uparrow>=\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \qquad |\uparrow\downarrow>=|\uparrow>\otimes|\downarrow>=\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}
\]
\[
|\downarrow\uparrow>=|\downarrow>\otimes|\uparrow>=\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} \qquad
|\downarrow\downarrow>=|\downarrow>\otimes|\downarrow>=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\]
And the matrix that rotates the two spins together: 
\[
R_C^{(4)}=I \otimes R_C^{(2)} + R_C^{(2)} \otimes I=\frac{1}{\sqrt2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\otimes\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}+\frac{1}{\sqrt2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\otimes\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=
\]
\[
=\frac{1}{\sqrt2}\begin{pmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \end{pmatrix}+\frac{1}{\sqrt2}\begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \end{pmatrix}=\frac{1}{\sqrt2}\begin{pmatrix} 2 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & -2 \end{pmatrix}
\]
We can easily see that the first matrix (before summation) only rotates the second spin, and the second matrix only rotates the first spin.
\end{document}
