Can Exact Joint Probability Distributions Predict Violations in CHSH Tests?

In summary, the model predicts a CHSH violation in some actual experiments. It would take a lot of repetitions to detect the violation even in a near perfect experiment.
  • #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)##
EPR-sim-1.png
 

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  • #2
Mentz114 said:
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##

What is [itex]\theta_0[/itex]?
 
  • #3
stevendaryl said:
What is [itex]\theta_0[/itex]?

Misunderstood your question. It is the inital polarization of the pair.

There's more info in the attached.
 

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  • #4
To be honest, I do not really see your point. Each photon from an entangled photon pair when viewed on its own is unpolarized on average, so you cannot cherry pick one pair of polarizations, but you have to average such that each photon viewed on its own becomes unpolarized in the ensemble average. How do you achieve that in your model?
 
  • #5
Cthugha said:
To be honest, I do not really see your point. Each photon from an entangled photon pair when viewed on its own is unpolarized on average, so you cannot cherry pick one pair of polarizations, but you have to average such that each photon viewed on its own becomes unpolarized in the ensemble average
Thanks for the reply. If you can't see the point of my post I'm sorry and it is no doubt because of my poor communication skills.
How do you achieve that in your model?
The model has achieved what it was meant to.

[Edit]
In an attempt to make my point explicitly I would like to add this.

We have here a non-local model of an idealized EPR/Aspect experiment and the joint distribution of outcomes (from photon detectors) for the setup . This distribution appears to predict the outcomes of some actual experiments including violations of the CHSH and Clauser-Holt/Eberhard inequalities. That is all. Nothing is proved or disproved. If the probabilities are a fair description then so are deductions from them.

My other hope is that someone will check the crucial calculation.
 
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  • #6
stevendaryl said:
What is [itex]\theta_0[/itex]?
These settings ##\theta_0=0.03\pi,\ a=0.02\pi,\ a'= 3\pi/2,\ b=-0.10\pi,\ b'=\pi## give ##B=2.177##. A simulation of 50 runs of 25000 photons each gives a mean B of 2.174 with standard deviation 0.017.

So this non-local model has this CHSH violation ( at least). But it would take a very large number of repetitions to detect it it even in a near perfect experiment. It also tells me that it is possible to produce a set of data which displays this violation. I can provide any number of them.

There could be a quantum model with interference that can push the probabilities round a lot more. Something to think on.
 
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  • #7
Mentz114 said:
Misunderstood your question. It is the inital polarization of the pair.

There's more info in the attached.

Okay. Let me summarize what I think your model is:
  1. Each photon has a polarization at each moment
  2. This polarization gives the probability that it will pass through a polarizing filter, as described by Malus' law.
  3. After passing through the filter, the photon has a polarization determined by the orientation of the filter.
  4. If two photons are entangled, then a change in the polarization of one causes the polarization of the other to change instantaneously so that after one passes the filter, they continue to have equal polarizations.
  5. Presumably, after passing a filter, the entanglement of the two photons is broken.
1-3 are all local. 4 is the "collapse". You didn't explicitly mention 5, but I think it has to be true to be in agreement with experiment.
 
  • #8
Mentz114 said:
We have here a non-local model of an idealized EPR/Aspect experiment and the joint distribution of outcomes (from photon detectors) for the setup . This distribution appears to predict the outcomes of some actual experiments including violations of the CHSH and Clauser-Holt/Eberhard inequalities. That is all. Nothing is proved or disproved. If the probabilities are a fair description then so are deductions from them.

Maybe I am just getting a part of your model wrong. So let me try to understand. For your chosen settings, what is the average transmission of the first and of the second photon through the polarizer when investigated on its own without any reference to the second photon?
 
  • #9
stevendaryl said:
Okay. Let me summarize what I think your model is:
  1. Each photon has a polarization at each moment
  2. This polarization gives the probability that it will pass through a polarizing filter, as described by Malus' law.
  3. After passing through the filter, the photon has a polarization determined by the orientation of the filter.
  4. If two photons are entangled, then a change in the polarization of one causes the polarization of the other to change instantaneously so that after one passes the filter, they continue to have equal polarizations.
  5. Presumably, after passing a filter, the entanglement of the two photons is broken.
1-3 are all local. 4 is the "collapse". You didn't explicitly mention 5, but I think it has to be true to be in agreement with experiment.
Yes, that is right except I would modify 5 to read

5. If a photon fails to pass a filter or hits the detector entanglement is broken.

I wouldn't use the word 'collapse' because a photon does not have a wave function in the usual sense. But that makes no difference.
Point 3 is a projective measurement and I think 4 is the entanglement effects and is most non-local in requiring instantaneous projection at a distance.
 
  • #10
Cthugha said:
Maybe I am just getting a part of your model wrong. So let me try to understand. For your chosen settings, what is the average transmission of the first and of the second photon through the polarizer when investigated on its own without any reference to the second photon?
Do you mean what are the probabilities in the absence of entanglement ?

The first photon to reach a polarizer does not refer to the other. It is the projection of both photons that carries the first setting so the next projection is affected. ( A naive interpretation but it works here).

If there's no entanglement replace the projected angle of the first photon with the initial polarization of the photon.
 
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  • #11
Mentz114 said:
Do you mean what are the probabilities in the absence of entanglement ?

The first photon to reach a polarizer does not refer to the other. It is the projection of both photons that carries the first setting so the next projection is affected. ( A naive interpretation but it works here).

In principle I would be interested in the transmission of both photons. However, I suppose the first photon is sufficient. So if you consider the transmission by adding a +1 each time a photon gets transmitted and -1 each time it is not transmitted, what will be the average value of this quantity?
 
  • #12
Cthugha said:
In principle I would be interested in the transmission of both photons. However, I suppose the first photon is sufficient. So if you consider the transmission by adding a +1 each time a photon gets transmitted and -1 each time it is not transmitted, what will be the average value of this quantity?
Do you mean 'adding +1 every time the first photon gets transmitted ...' .

I'll have think about that.
 
  • #13
Mentz114 said:
Do you mean 'adding +1 every time the first photon gets transmitted ...' .

I'll have think about that.

Well, in this case yes. You are right.
In principle you can do the same for the second photon by calculating whether it will be transmitted or not depending on what happened to the first photon and then just adding up the +1 and -1 for the second photon.

One of the crucial points in CHSH experiments is that both of these averages sum up to 0.
 
  • #14
Cthugha said:
Well, in this case yes. You are right.
In principle you can do the same for the second photon by calculating whether it will be transmitted or not depending on what happened to the first photon and then just adding up the +1 and -1 for the second photon.

One of the crucial points in CHSH experiments is that both of these averages sum up to 0.
OK, I'll see if I can work that out. It is equivalent to saying that the total probability of transmission should 1/2, the same as the total probability of destruction.

[Edit]
@Cthugha
I added a couple of counters to the sim code and ran a few 20000 photon reps.

THIS IS WRONG
Typically number of first photons transmitted = 10000 and of those, 5000 ( second) photons transmitted.
So I guess it will satisfy the average = 0 requirement. Later I will use the probs to confirm it formally, if possible.

THIS IS NOT
I spoke too soon. It depends on the settings and deviates from 1/2 for non-symmetric settings.
 
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  • #15
Something of a breakthrough. The experiment that uses beam-splitting polarizers is much more balanced and working out the probabilities for the new setup gives
[tex]
\begin{align*}
2P_{11}(\alpha,\beta) &=\left( {\cos\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}+{\cos\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\right) \,{\cos\left( \frac{\theta_B-\theta_A}{2}\right) }^{2}\tag{A1}\\
2P_{10}(\alpha,\beta) &=\left( {\cos\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}+{\sin\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\right) \,{\sin\left( \frac{\theta_B-\theta_A}{2}\right) }^{2}\tag{A2}\\
2P_{01}(\alpha,\beta) &= \left( {\sin\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}+{\cos\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\right) \,{\sin\left( \frac{\theta_B-\theta_A}{2}\right) }^{2}\tag{A3}\\
2P_{00}(\alpha,\beta) &=\left( {\sin\left( \frac{\theta_B-\theta_0}{2}\right) }^{2}+{\sin\left( \frac{\theta_A-\theta_0}{2}\right) }^{2}\right) \,{\cos\left( \frac{\theta_B-\theta_A}{2}\right) }^{2}\tag{A4}
\end{align*}
[/tex]

The probability of a coincidence is ##P_{11}(\alpha\beta)+P_{00}(\alpha\beta)## which gives ( from equations A1 and A4) ##P_{co} ={\cos\left( \frac{\alpha-\beta}{2}\right) }^{2}##. The correlation is ##\rho(\alpha\beta) = 2P_{co}-1= \cos(\alpha -\beta)## which is in agreement with the quantum theoretical correlation.

There is a biggish violation at ##a=0,\ a'=-\pi/4,\ b=-\pi/8, b'=\pi/8## with the expected value ##2.38895##.

I think it is safe to say that these probabilities mimic exactly the quantum mechanical predictions. Case closed.

@PeterDonis
 
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Related to Can Exact Joint Probability Distributions Predict Violations in CHSH Tests?

What is a CHSH test?

A CHSH test stands for the Clauser-Horne-Shimony-Holt test, which is a type of experiment used to test for violations of Bell's inequalities in quantum mechanics.

What is a violation of a CHSH test?

A violation of a CHSH test occurs when the results of the experiment do not conform to the predictions of classical physics, indicating the presence of quantum entanglement.

What are the implications of a violation of a CHSH test?

A violation of a CHSH test provides evidence for the non-locality of quantum mechanics, which challenges our understanding of causality and locality in the universe.

How is a CHSH test conducted?

A CHSH test involves measuring the correlations between two entangled particles using four different possible combinations of measurements, known as settings, which are then compared to the predictions of classical physics.

What are the possible explanations for a violation of a CHSH test?

The possible explanations for a violation of a CHSH test include the presence of hidden variables, non-locality in quantum mechanics, or a combination of both factors.

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