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\heading{E485: Interwoven states of 2 spins} 
\auname{Ilan Shemesh}


{\bf The problem:}

\Dn
There are 2 particles of spin $\frac{1}{2}$. The spin states of the system are being describe in the standart base:\\

$ |\uparrow\uparrow\rangle,|\uparrow\downarrow\rangle,|\downarrow\uparrow\rangle,|\downarrow\downarrow\rangle$
\\
\\
The system has being prepered in the state: $|\psi \rangle \propto |\uparrow\uparrow\rangle + |\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle$

Notice: The system has being prepered in a symertic state compere to the pormutation, so the tw particles has the same polarization state.\\
\\
(1) Find the expration of the operator $S_x$, which describe the spin of one of the particles as a matric in the standart base, which has defined abouve. \\ 
\\
(2) Find the polarization vector $\vec{M}=(\langle S_x \rangle, \langle S_y \rangle, \langle S_z \rangle)$ which describe the polarization state of each particle.\\
\\
(3) Write the polarization direction $\vec{n}$ and polarization rate $p=\frac{|\vec{M}|}{M_{max}}$ of the state of one particle. \\
\\
(4) What is the state vector in the base: $up / down$ , which describes single spin with full polarization in $\vec{n}$ direction.\\
\\
(5) What will be the system state if the 2 particles are being prepered at a state of full polarization, in $\vec{n}$ direction.\\
\\
     
\Dn

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{\bf The solution:}

\Dn
(1) The standard base is:
\Dn
$\left|\uparrow\uparrow\right\rangle=\left(\begin{array}{c} 1\\0\\0\\0\end{array}\right),  \left|\uparrow\downarrow\right\rangle=\left(\begin{array}{c} 0\\1\\0\\0\end{array}\right),  \left|\downarrow\uparrow\right\rangle=\left(\begin{array}{c} 0\\0\\1\\0\end{array}\right),  \left|\downarrow\downarrow\right\rangle=\left(\begin{array}{c} 0\\0\\0\\1\end{array}\right), $
\Dn

The system is defined by: $\Psi=|S_a,S_b\rangle$. \\ \\
In order to write the  operator $S_x$, which describe the spin of one of the particles, we need to write it, in the standart base witen abouve. \\ \\ 
For the first particls: 
$S_x^1=S_x\otimes\hat{I} = S_{x}=\frac{1}{2}\left(\amatrix{0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1
\cr 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0}\right); \quad $

\Dn

And, for the second particle:
$S_x^2= \hat{I}\otimes S_x = \frac{1}{2}\left(\amatrix{0 & 1 & 0 & 0 \cr 1 & 0 & 0 & 0
\cr 0 & 0 & 0 & 1 \cr 0 & 0 & 1 & 0}\right). \quad $\\ \\

\Dn
(2) We are being asked to find $\vec{M}=(\langle S_x \rangle, \langle S_y \rangle, \langle S_z \rangle)$ for each particle. \\

For each one of the particles $j=1,2$ we define: \\
\\
$\langle S_i^{j} \rangle = \langle \Psi | \hat{S_i^{j}} | \Psi \rangle,$  where $i=x,y,z$ and $j=1,2$ (the particle index)\\
\\ 
For the \textbf{First} particle ($j=1$): let's find the operators $\hat{S_i^{1}}$ : \\

When we define:
\[S_{x}^1=\frac{1}{2}\sigma_x\otimes\hat{I}\ , \quad
S_{y}^1=\frac{1}{2}\sigma_y\otimes\hat{I}\ , \quad
S_{z}^1=\frac{1}{2}\sigma_z\otimes\hat{I}\]

we obtain the matrices $S_{x}^1, S_{y}^1, S_{z}^1$ in the standart basis,

\[S_{x}^1=\frac{1}{2}\left(\amatrix{0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1
\cr 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0}\right), \quad
S_{y}^1=\frac{1}{2}\left(\amatrix{0 & 0 & -\imath & 0 \cr 0 & 0
& 0 & -\imath \cr \imath & 0 & 0 & 0 \cr 0 & \imath & 0 &
0}\right), \quad
S_{z}^1=\frac{1}{2}\left(\amatrix{1 & 0 & 0 &
0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & -1 & 0 \cr 0 & 0 & 0 &
-1}\right)\]

Our wave function is given: $\Psi=\frac {1}{\sqrt{3}}\left( \amatrix { 1 \cr 1 \cr 1 \cr
0}\right)$

so:

\[\langle S_{x}^1 \rangle=\frac{1}{6}(\begin{array}{cccc}1& 1& 1& 0 \end{array})
\left(\amatrix{0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr 1 & 0 & 0 & 0
\cr 0 & 1 & 0 & 0}\right)\left( \amatrix { 1 \cr 1 \cr 1 \cr
0}\right)=\frac {1}{3}\]

\[\langle S_{y}^1 \rangle=\frac{1}{6}(\begin{array}{cccc}1& 1& 1& 0 \end{array})
\left(\amatrix{0 & 0 & -\imath & 0 \cr 0 & 0 & 0 & -\imath \cr
\imath & 0 & 0 & 0 \cr 0 & \imath & 0 & 0}\right)\left( \amatrix {
1 \cr 1 \cr 1 \cr 0}\right)=0\]

\[\langle S_{z}^1 \rangle=\frac{1}{6}(\begin{array}{cccc}1& 1& 1& 0 \end{array})
\left(\amatrix{1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & -1 & 0
\cr 0 & 0 & 0 & -1}\right)\left( \amatrix { 1 \cr 1 \cr 1 \cr
0}\right)=\frac{1}{6}\]

\[\vec{M^1}=\frac{1}{6}( 2,0 ,1 )\] \\ 


For the \textbf{Second} particle ($j=2$): let's find the operators $\hat{S_i^{2}}$ : \\

When we define:
\[S_{x}^2=\hat{I} \otimes \frac{1}{2} \sigma_x\ , \quad
S_{y}^2=\hat{I} \otimes \frac{1}{2}\sigma_y\ , \quad
S_{z}^2=\hat{I} \otimes \frac{1}{2}\sigma_z\]

we obtain the matrices $S_{x}^2, S_{y}^2, S_{z}^2$ in the standart basis,

\[S_{x}^2=\frac{1}{2}\left(\amatrix{0 & 1 & 0 & 0 \cr 1 & 0 & 0 & 0
\cr 0 & 0 & 0 & 1 \cr 0 & 0 & 1 & 0}\right), \quad
S_{y}^2=\frac{1}{2}\left(\amatrix{0 & -\imath & 0 & 0 \cr \imath & 0
& 0 & 0 \cr 0 & 0 & 0 & -\imath \cr 0 & 0 & \imath &0}\right), \quad
S_{z}^2=\frac{1}{2}\left(\amatrix{1 & 0 & 0 &
0 \cr 0 & -1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & -1}\right)\]

so:

\[\langle S_{x}^2 \rangle=\frac{1}{6}(\begin{array}{cccc}1& 1& 1& 0 \end{array})
\left(\amatrix{0 & 1 & 0 & 0 \cr 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & 1 \cr 0 & 0 & 1 & 0}\right)
\left( \amatrix { 1 \cr 1 \cr 1 \cr 0}\right)=\frac{1}{3} \]

\[\langle S_{y}^2 \rangle=\frac{1}{6}(\begin{array}{cccc}1& 1& 1& 0 \end{array})
\left(\amatrix{0 & -\imath & 0 & 0 \cr \imath & 0
& 0 & 0 \cr 0 & 0 & 0 & -\imath \cr 0 & 0 & \imath &0}\right)
\left( \amatrix {1 \cr 1 \cr 1 \cr 0}\right)=0\]

\[\langle S_{z}^2 \rangle=\frac{1}{6}(\begin{array}{cccc}1& 1& 1& 0 \end{array})
\left(\amatrix{1 & 0 & 0 &0 \cr 0 & -1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & -1}\right)
\left( \amatrix { 1 \cr 1 \cr 1 \cr 0}\right)=\frac{1}{2}\]

\[\vec{M^2}=\frac{1}{6}( 2, 0,3 )\] \\ 

\Dn

(3) Now we need to find the polarization direction $\vec{n}$ and polarization rate $p=\frac{|\vec{M}|}{M_{max}}$.\\
\\
If a spin is completely polorized so, it point to a spesifice direction. in case of spine half the maximum Amplitude of this vector is $1/2$ , as a result of stern-gerlach experience $\Rightarrow M_{max}=1/2$, (for spin one we get: $M_{max}=1$) 
\\

So, for the \textbf{First} spin we get:
\[ |\vec{M^1}|=\frac{\sqrt5}{6},\quad \Rightarrow \quad \vec{n^1}=\frac{\vec{M}}{|{\vec{M}}|}= \frac{1}{\sqrt 5}  (2,0,1); \quad p^1=\frac{|\vec{M}|}{M_{max}} =\frac{\sqrt 5} {3}\] 

And, for the \textbf{Second} spin we get:
\[ |\vec{M^2}|=\frac{\sqrt {13}}{6}, \quad \Rightarrow \quad \vec{n^2}=\frac{\vec{M}}{|{\vec{M}}|}= \frac{1}{\sqrt {13}}(2,0,3); \quad p^2=\frac{|\vec{M}|}{M_{max}} =\frac{\sqrt {13}} {3}\] 

(4) What is the state vector in the base: $up / down$? \\ 
The base $up / down$ is:
\[ | \uparrow \rangle \mapsto \left( \amatrix { 1 \cr  0}\right) \quad
 | \downarrow \rangle \mapsto \left( \amatrix { 0 \cr  1}\right)\]

we will take spin $up$ which is $\left( \amatrix { 1 \cr  0}\right)$ and spin it around y ases in $\theta$:
\[
\left(\amatrix{\cos{\frac{\theta}{2}} & -\sin{\frac{\theta}{2}}  \cr \sin{\frac{\theta}{2}} & \cos{\frac{\theta}{2}} }\right) \left( \amatrix { 1 \cr  0}\right) = \left( \amatrix { \cos{\frac{\theta}{2}} \cr  \sin{\frac{\theta}{2}}}\right)
\]

For the \textbf{First} case: 
\[ \hat{n_1} =\frac{1}{\sqrt 5}  (2,0,1)  \quad  \text{Therefore:} \]

\[ \cos{\theta}= \frac {1}{\sqrt 5}, \quad \sin{\theta}= \frac{2}{\sqrt 5} \Rightarrow \quad \]
 \[\cos{\frac{\theta}{2}}= \pm\sqrt{\frac{1+\cos{\theta}}{2}}= \pm\sqrt{\frac{1+\frac {1}{\sqrt 5}}{2}}  , \quad 
 \[\sin{\frac{\theta}{2}}= \pm \sqrt{\frac{ 1- \cos{\theta}}{2}} = \pm \sqrt{\frac{ 1- \frac {1}{\sqrt 5}}{2}} \] 
 \[ \text{while} \quad \theta_1 =  \arctan (2) \]
 Hence: 
\[ \hat{\psi_1}= \left( \amatrix { \cos{\frac{\theta}{2}} \cr  \sin{\frac{\theta}{2}}}\right) = \left( \amatrix {  \sqrt{\frac{1+\frac {1}{\sqrt 5}}{2}} \cr  \sqrt{\frac{1-\frac {1}{\sqrt 5}}{2}}}\right)\]

For the \textbf{Second} case: 
\[ \hat{n_2} = \frac{1}{\sqrt {13}}(2,0,3) \quad \text{Therefore:} \]

\[ \cos{\theta}= \frac {3}{\sqrt {13}}, \quad \text{and} \quad \sin{\theta}= \frac{2}{\sqrt{13}} \Rightarrow \quad \]
\[ \cos{\frac{\theta}{2}}= \pm\sqrt{\frac{1+\cos{\theta}}{2}}= \pm\sqrt{\frac{1+\frac {3}{\sqrt {13}}}{2}} \quad \sin{\frac{\theta}{2}}= \pm \sqrt{\frac{ 1- \cos{\theta}}{2}} = \pm \sqrt{\frac{ 1-  \frac {3}{\sqrt {13}}}{2}} \]
\[ \text{while} \quad \theta_2 =  \arctan (\frac{2}{3})\]
Hence: 
\[ \hat{\psi_2}= \left( \amatrix { \cos{\frac{\theta}{2}} \cr  \sin{\frac{\theta}{2}}}\right) = \left( \amatrix { \sqrt{\frac{ 1+  \frac {3}{\sqrt {13}}}{2}} \cr  \sqrt{\frac{ 1-  \frac {3}{\sqrt {13}}}{2}}}\right)\]

\Dn
(5) What will be the system state if the 2 particles are being prepered at: (a) $ \hat{n_1} $, (b) $\hat{n_2}$ ?
\Dn
Here there are 2 particles. The wave function will be tensor product between them:
\[ | \Psi \rangle = \left( \amatrix { \cos{\frac{\theta}{2}} \cr  \sin{\frac{\theta}{2}}}\right) \otimes \left( \amatrix { \cos{\frac{\theta}{2}} \cr  \sin{\frac{\theta}{2}}}\right) = \left( \amatrix { \cos^2{\frac{\theta}{2}} \cr \frac{1}{2}\sin{\theta} \cr \frac{1}{2}\sin{\theta} \cr \sin^2{\frac{\theta}{2}}}\right)\]

So, For \textbf{(a)} we will get: $ | \Psi_a \rangle = \left( \amatrix { \cos^2{\frac{\theta}{2}} \cr \frac{1}{2}\sin{\theta} \cr \frac{1}{2}\sin{\theta} \cr \sin^2{\frac{\theta}{2}}}\right) = \left( \amatrix { \frac{1}{2}\left(1+\frac {1}{\sqrt 5}\right) \cr \frac {1}{\sqrt 5} \cr \frac {1}{\sqrt 5} \cr \frac{1}{2}\left(1-\frac {1}{\sqrt 5}\right)}\right)$ 

\Dn
And, For \textbf{(b)} we will get: $ | \Psi_b \rangle = \left( \amatrix { \frac{1}{2}\left(1+\frac {3}{\sqrt {13}}\right) \cr \frac {1}{\sqrt {13}} \cr \frac {1}{\sqrt {13}} \cr \frac{1}{2}\left(1-\frac {3}{\sqrt {13}}\right)}\right)$ 

\Dn
\Dn
\Dn
\Dn
\begin{verbatim}
Remarks:


All the results are the same for both particles.
The results for the "first" particle are correct
and the results for the "second" particle are wrong.

\end{verbatim}

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