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\begin{document} 

\heading{Bell's inequality and mermin's EPR} 
\auname{Dotan Davidovich}

{\bf Bell's inequality}\\\\

Two particles are emitted in opposite directions , in each side there is a device which measure the particle spin in the angles $ \alpha $ or $\gamma $ with respect to some arbitrary axis , the second device on the other side measure angles$ \beta$ and $\gamma$ (same $\gamma$ ). The possible result for  $\alpha$ $ \beta$ $\gamma$  are respectively     \[\tag{1} a=\pm1$ , $ b=\pm1$ , $ c=\pm1 \]  we assume casuality , so after emission the particles do not effect each other and so does the measuring devices.
\\

From (1) we can get the following condition : \[\tag{2} a(b-c)=\pm(1-bc) \] 
If we repeat this experiment many times, each time has its own set of hidden variables denoted by j.  We now can write: 
\[\tag{3}  a_jb_j-a_jc_j=\pm(1-b_jc_j)  \]
Now averaging over all j , we get the following inequalities :
\[\tag{4} \langle ab\rangle  - \langle ac \rangle +\langle bc\rangle \leq 1 \;\; , \;\; \langle ab\rangle  - \langle ac \rangle -\langle bc\rangle \geq -1\]
Which are the original bell inequalities.\\
For a spin 1/2 singlet state Quantum mechanics predicts $ \langle ab\rangle=cos(\theta_{ab})  $
When $\theta$ is the angle between the detectors angles $\alpha$ and $\beta$ so we get: \\ \\
\[ \tag{5} \vert cos(\theta_{ab}) - cos(\theta_{ac})\vert + cos(\theta_{bc})  \leq 1 \] 
\\ This inequality is most strongly violated when $  \theta_{ab}=60^{\circ} $ $  \theta_{ac}=120^{\circ} $ $  \theta_{bc}=60^{\circ} $       namely a 60 degrees separation between 3 angles , then (5) becomes :\[\tag{6} \vert \dfrac{1}{2} + \dfrac{1}{2} \vert +\dfrac{1}{2} \leq 1 \rightarrow  \dfrac{3}{2}\leq 1 \] 
The CHSH inequality is a generalization of bells , it takes the second angle of the second observer different so we have 4 possible angles and we get: \[\tag{7} \langle ab\rangle  + \langle ad \rangle + \langle bc \rangle - \langle cd \rangle \leq 2\] 
If we go back to only 3 angles then c and d have maximum correlation so in every measurement they are opposite, so we have :
\[\tag{8} d=-c \;\; and \;\;    \langle cd \rangle =-1 \]
Inserting into (7) we get bell's original inequality (4).







\newpage
{\bf Detector efficiency loophole}\\\\

When measuring the correlation between a and b , real detector not always take a measurement, they have an efficiency limit $\eta$ which is $ 0 \leq \eta \leq 1 $ 
This makes the inequality (7) depend  on the detectors efficiency .
If particle detection depends only on the efficiency of the detector $\eta$  the correlation function $ \langle ab\rangle$ becomes: \[\tag{9} \langle \tilde{ab}\rangle = \dfrac{\eta}{2-\eta}\langle ab\rangle\] \\
Inserting into (7) we get:\\ \[\tag{10} \dfrac{\eta}{2-\eta}\vert \langle ab\rangle  + \langle cb \rangle + \langle ad \rangle - \langle cd \rangle \vert\leq 2 $  $\rightarrow  $   $ \vert \langle ab\rangle  + \langle cb \rangle + \langle ad \rangle - \langle cd \rangle \vert\leq \dfrac{4}{\eta} -2 \]\\
And for the most violating case $ \theta =45^\circ $ we get:\;\;
$ 2\sqrt{2} \leq \dfrac{4}{\eta} -2  $\\\\
 This will be violated only if  $ \dfrac{4}{\eta} -2 \leq 2\sqrt{2} $   $\;\; so\;\;$   $ \eta \geq \dfrac{4}{2+2\sqrt{2}}\simeq 0.828 $\\ \\
 \\ 
{\bf Mermin EPR}
\\ \\
In analogy to the EPR experiment we now consider a 3 particle system. 3 particles are emitted in 3 different directions at each one there is a detector , set to one of two possible setting ,each set can give the answer +1 or -1 \\

 

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\includegraphics[scale=0.5]{mermin.eps} 

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While repeating this experiment many times we notice that some setting of all 3 detectors , yield a certain result : for settings  $ a_1b_2c_2$ , $a_2b_1c_2$ , $a_2b_2c_1 $  we always get an even number of -1 in the detectors. so : \[\tag{11}a_1b_2c_2=a_2b_1c_2=a_2b_2c_1 =1\]
Using this result with the LHVT we can predict what will be the result for an $a_1b_1c_1$ setting : taking into account that $a_2^2=b_2^2=c_2^2=1 $ we can write : \[\tag{12} a_1b_2c_2 \cdot a_2b_1c_2 \cdot a_2b_2c_1 = a_1b_1c_1=1\]
We can now see that the outcome will be, so an even number of detector will measure -1 in this setting too.\\\\
This experiment can be modelled by a quantum mechanical state :\[\tag{13}\vert\psi\rangle=\dfrac{1}{\sqrt{2}}(\vert\uparrow\uparrow\uparrow\rangle - \vert\downarrow\downarrow\downarrow\rangle)\]
And the detectors as:  $ a_1=\sigma_x^a$\;\; $a_2=\sigma_y^a$ \;,\; $b_1=\sigma_x^b$ \;\;$b_2=\sigma_y^b$ \;,\; $c_1=\sigma_x^c$ \;\; $c_2=\sigma_y^c $\;.\\\\
What would be the possible outcomes for\; $ \sigma_x^a \sigma_y^b \sigma_y^c$ \;in this state ? (corresponding to a measurement of  \; $ a_1b_2c_2$ ).
We remember that: \[\tag{14} \vert\uparrow\rangle=\dfrac{1}{\sqrt{2}}(\vert x \rangle + \vert \bar{x} \rangle) = \dfrac{1}{\sqrt{2}}(\vert y \rangle + \vert \bar{y} \rangle) $  ,  $ \vert\downarrow\rangle=\dfrac{1}{\sqrt{2}}(\vert x \rangle - \vert \bar{x} \rangle) = \dfrac{-i}{\sqrt{2}}(\vert y \rangle - \vert \bar{y} \rangle) \]\\
Then : \\ \[\tag{15} \vert\psi\rangle=\dfrac{1}{4} \left[(\vert x \rangle + \vert \bar{x} \rangle)\otimes(\vert y \rangle + \vert \bar{y} \rangle)    \otimes(\vert y \rangle + \vert \bar{y} \rangle)     +     (\vert x \rangle + \vert \bar{x} \rangle)\otimes(\vert y \rangle - \vert \bar{y} \rangle)    \otimes(\vert y \rangle - \vert \bar{y} \rangle)  \right] \Rightarrow \] 
\[\vert\psi\rangle=\frac{1}{2}(\vert \bar{x} y \bar{y} \rangle + \vert \bar{x} \bar{y} y \rangle + \vert x \bar{y} \bar{y} \rangle + \vert x y y \rangle\] \\
Now we see that for  measurement setting of \;$ \sigma_x^a \sigma_y^b \sigma_y^c $ \; we will always get an even number of -1 as assumed in the experiment above.  
the same can be shown for : $ \sigma_y^a \sigma_x^b \sigma_y^c$  \;\; and  \;\; $ \sigma_y^a \sigma_y^b \sigma_x^c$ .\\

Now we can look on a different measurement : \[\tag{16}A\vert\psi\rangle=(\sigma_x^a \sigma_y^b \sigma_y^c)(\sigma_y^a \sigma_x^b \sigma_y^c)(\sigma_y^a \sigma_y^b \sigma_x^c)\vert\psi\rangle =\vert\psi\rangle\] 
Under the assumptions of LHVT and using the fact that $(\sigma_y^i)^2=1  $  we get: \\
\[\tag{17}(\sigma_x^a \sigma_y^b \sigma_y^c)(\sigma_y^a \sigma_x^b \sigma_y^c)(\sigma_y^a \sigma_y^b \sigma_x^c)=\sigma_x^a \sigma_x^b \sigma_x^c \Rightarrow \sigma_x^a \sigma_x^b \sigma_x^c\vert\psi\rangle=\vert\psi\rangle\]\\

Meaning that a measurement of which will yield the same result as before .\\


What does QM predicts ?\\\\
If we look at $\; \vert\psi\rangle \;$ for $\; \sigma_x^a \sigma_x^b \sigma_x^c \;$ we get:\\
\[\tag{18} \vert\psi\rangle=\frac{1}{2}(\vert \bar{x} x x \rangle + \vert x \bar{x} x \rangle + \vert x x \bar{x} \rangle + \vert \bar{x} \bar{x} \bar{x} \rangle\   \]\\ 
And finally : 
\[\tag{19} \sigma_x^a \sigma_x^b \sigma_x^c\vert\psi\rangle=-\vert\psi\rangle\]

Therefore QM predicts that only an odd number of -1 result will always be measured , in opposite to the LHVT which predicts an even number of -1 will always be measured.\\


The same result can be obtained from the operator's point of view. First,
in order that the measurement (16) will have a meaning all operators must commute. using  :
\[\tag{20}  \sigma_x^a\sigma_y^a=-\sigma_y^a\sigma_x^a  \;\; and \;\;  (\sigma_y^i)^2=1  \]
The commutations relations are :\\
\[\tag{21}[   \sigma_x^a \sigma_y^b \sigma_y^c   , \sigma_y^a \sigma_x^b \sigma_y^c ] = \sigma_y^a \sigma_x^b \sigma_y^c \sigma_y^a \sigma_x^b \sigma_y^c - \sigma_y^a \sigma_x^b \sigma_y^c  \sigma_x^a \sigma_y^b \sigma_y^c =\sigma_y^a\sigma_x^a\sigma_x^b\sigma_y^b  - \sigma_y^a\sigma_x^a\sigma_x^b\sigma_y^b =0\]\\

The same can be shown for the other commutation relations.\\

What will be the result of this measurement( $ A\vert\psi\rangle $ )?\\\\
Using  the commutativity of the a,b,c noted operators and (20) we get :\\
\[\tag{22}(\sigma_x^a \sigma_y^b \sigma_y^c)(\sigma_y^a \sigma_x^b \sigma_y^c)(\sigma_y^a \sigma_y^b \sigma_x^c)= \sigma_x^a\sigma_y^b\sigma_x^b\sigma_y^b\sigma_x^c = -\sigma_x^a\sigma_y^b\sigma_y^b\sigma_x^b\sigma_x^c=-\sigma_x^a\sigma_x^b\sigma_x^c \]\\

Therefore :
 \[\tag{23} \sigma_x^a \sigma_x^b \sigma_x^c\vert\psi\rangle=-\vert\psi\rangle\]
 Which is the same contradiction as before.




\end{document} 
