- #1

Mentz114

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I've spent some time setting up my simulation of a photonic EPR experiment and came up with an exact joint probability distribution for the outcomes ##P_{xy}(\alpha,\beta)##.

To calculate the CHSH number we need the probability of a coincidence for each of the possible settings which becomes a correlation by doubling it and subtracting 1.This gets ##C_{ab}=2(P_{11}(a,b)+P_{00}(a,b))-1## and so on for ##ab'##, ##b'a## and ##a'b'##. The value we want is ##B=C_{ab}-C_{ab'}+C_{a'b}+C_{a'b'}##.

Using ##a=0.03412,\ a' = 3\pi/2, b=0, b'=\pi## and ##\theta_0=0.03412## gves ##B=2.0147##

This can be easily checked (please !). The value obtained here is not a statistic but an expectation and so indicates an actual violation inherent in the probabilities.

The simulation (as always) gives the same number but with statistical fluctuations.

I can provide Maxima scripts and the simulator if anyone wants them.

The probabiities are :

[tex]

\begin{align*}

P_{11}(\alpha,\beta,A) &= {\cos\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\,{\cos\left( \frac{\theta_A-\theta_B}{2}\right) }^{2},\

P_{10}(\alpha,\beta,A) ={ \cos\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\,{\sin\left( \frac{\theta_A-\theta_B}{2}\right) }^{2}\\

P_{01}(\alpha,\beta,A) &= {\sin\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\,{\cos\left( \frac{\theta_0-\theta_B}{2}\right) }^{2},\

P_{00}(\alpha,\beta,A) = {\sin\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\,{\sin\left( \frac{\theta_0-\theta_B}{2}\right) }^{2}

\end{align*}

[/tex]

[tex]

\begin{align*}

P_{11}(\alpha,\beta,B) &={\cos\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}\,{\cos\left( \frac{\theta_A-\theta_B}{2}\right) }^{2},\

P_{10}(\alpha,\beta,B) ={\sin\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}\,{\cos\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\\

P_{01}(\alpha,\beta,B) &= {\cos\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}\,{\sin\left( \frac{\theta_B-\theta_A}{2}\right) }^{2},\

P_{00}(\alpha,\beta,B) = {\sin\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}\,{\sin\left( \frac{\theta_0-\theta_A}{2}\right) }^{2}

\end{align*}

[/tex]

Summing over ##S## gives ##P_{xy}(\alpha,\beta)= \tfrac{1}{2}\left(P_{xy}(\alpha,\beta, A)+ P_{xy}(\alpha,\beta, B)\right)##

To calculate the CHSH number we need the probability of a coincidence for each of the possible settings which becomes a correlation by doubling it and subtracting 1.This gets ##C_{ab}=2(P_{11}(a,b)+P_{00}(a,b))-1## and so on for ##ab'##, ##b'a## and ##a'b'##. The value we want is ##B=C_{ab}-C_{ab'}+C_{a'b}+C_{a'b'}##.

Using ##a=0.03412,\ a' = 3\pi/2, b=0, b'=\pi## and ##\theta_0=0.03412## gves ##B=2.0147##

This can be easily checked (please !). The value obtained here is not a statistic but an expectation and so indicates an actual violation inherent in the probabilities.

The simulation (as always) gives the same number but with statistical fluctuations.

I can provide Maxima scripts and the simulator if anyone wants them.

The probabiities are :

[tex]

\begin{align*}

P_{11}(\alpha,\beta,A) &= {\cos\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\,{\cos\left( \frac{\theta_A-\theta_B}{2}\right) }^{2},\

P_{10}(\alpha,\beta,A) ={ \cos\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\,{\sin\left( \frac{\theta_A-\theta_B}{2}\right) }^{2}\\

P_{01}(\alpha,\beta,A) &= {\sin\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\,{\cos\left( \frac{\theta_0-\theta_B}{2}\right) }^{2},\

P_{00}(\alpha,\beta,A) = {\sin\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\,{\sin\left( \frac{\theta_0-\theta_B}{2}\right) }^{2}

\end{align*}

[/tex]

[tex]

\begin{align*}

P_{11}(\alpha,\beta,B) &={\cos\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}\,{\cos\left( \frac{\theta_A-\theta_B}{2}\right) }^{2},\

P_{10}(\alpha,\beta,B) ={\sin\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}\,{\cos\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\\

P_{01}(\alpha,\beta,B) &= {\cos\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}\,{\sin\left( \frac{\theta_B-\theta_A}{2}\right) }^{2},\

P_{00}(\alpha,\beta,B) = {\sin\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}\,{\sin\left( \frac{\theta_0-\theta_A}{2}\right) }^{2}

\end{align*}

[/tex]

Summing over ##S## gives ##P_{xy}(\alpha,\beta)= \tfrac{1}{2}\left(P_{xy}(\alpha,\beta, A)+ P_{xy}(\alpha,\beta, B)\right)##

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