High School Preserving local realism in the EPR experiment

[stevendaryl]

Sure, that's as in the real experiment. So I still don't get what you are after in making a computer simulation. If it's simulates the probabilities predicted by QT by input of the code it will produce precisely what QT predicts (provided there's no bug in the code). Isn't this just tautological?

It's not tautological, because it's false. You can't simulate QT this way. That's an immediate consequence of Bell's theorem.

 

[vanhees71]

Ok, I'm not understanding this claim. Let's leave it at that.

 

[entropy1]

Sure, that's as in the real experiment. So I still don't get what you are after in making a computer simulation. If it's simulates the probabilities predicted by QT by input of the code it will produce precisely what QT predicts (provided there's no bug in the code). Isn't this just tautological?

If you manage to correctly simulate the experiment without reference to the relative angle of the detectors, I don't know how you do it!

 

[vanhees71]

Of course, you have to precisly simulate the probabilities given for the assumed experimental setup. Of course, I have to use the relative angle of the detectors to calculate these probabilities, and thus this information goes into the code.

 

[Mentz114]

If you manage to correctly simulate the experiment without reference to the relative angle of the detectors, I don't know how you do it!

Isn't it sufficient to project the second measured photon ( say B) into the angle of A's polarizer if A gets a hit ?
If A's detector does not click ( no transmission) what can we say about B's alignment ?
Also, I'm not sure if the pair are produced with a random alignment or some prepared value.

I'm referring to what goes in the code, of course.

 

[entropy1]

Isn't it sufficient to project the second measured photon ( say B) into the angle of A's polarizer if A gets a hit ?

If A gets a hit, we know its angle. But how does B know if A got a hit?

 

[Mentz114]

If A gets a hit, we know its angle. But how does B know if A got a hit?

B does not need to know that. The paired photon may have changed state after the first measurement. That is what entanglement is supposed to mean.

In the simulation that is the only thing that differs from a random setup.

 

[vanhees71]

Well, in the usual setup the measurement of B doesn't do anything on A. To ensure this you make the measurement acts spacelike separated. That's what's so mind boggling about these stronger-than-classically-possible correlations described by an almost trivial math. As easy is the math as mind boggling are the implications, particularly if you stick to traditional collapse assumptions of the early days! The worrying immediately stops if you simply accept that nature is inherently probabilistic and that there is a very successful formalism called QT that tells you probabilities and only probabilities.

 

[stevendaryl]

Of course, you have to precisly simulate the probabilities given for the assumed experimental setup. Of course, I have to use the relative angle of the detectors to calculate these probabilities, and thus this information goes into the code.

But the relative angle is not known until the last minute. The situation is the following:

You have three devices: C, which is a source of message pairs, simulating photon pair production, and A and B, which simulate the measurement events.

• Every "round", C sends out a pair of messages, m_A to A and m_B to B.
• After the messages are sent, but before they are read, settings for A and B are chosen, independently. The settings are two angles, \theta_A and \theta_B.
• Device A determines an output, R_A(\theta_A), which is either +1 or -1, based on the message received from C and the setting \theta_A.
• Similarly, device B determines an output, R_B(\theta_B) based on its message and setting.
• Over many, many rounds, we can gather statistics for the correlation: \langle R_A(\theta_A) R_B(\theta_B) \rangle as a function of the pair of settings, \theta_A, \theta_B.

Bell's inequality implies that |\langle R_A(\theta_A) R_B(\theta_B) \rangle| \leq 2, no matter what algorithms are used by A, B, and C, as long as

1. There are no communications among A, B, C other than those specified.
2. The settings \theta_A and \theta_B for each round are unpredictable by C.

On the other hand, if instead of C sending messages, it generates a pair of entangled photons, and sends one to A and one to B, then you can violate the inequality. (Inside A and B, instead of a computer algorithm, you have polarizing filters and photon detectors, and each sends out +1 if the photon passes through the filter at the orientation specified by \theta_A or \theta_B.

 

[entropy1]

Because, in a computer, this would have to be a HV-setup, it would not reproduce the desired result.

 

[Mentz114]

Well, in the usual setup the measurement of B doesn't do anything on A. To ensure this you make the measurement acts spacelike separated. That's what's so mind boggling about these stronger-than-classically-possible correlations described by an almost trivial math. As easy is the math as mind boggling are the implications, particularly if you stick to traditional collapse assumptions of the early days! The worrying immediately stops if you simply accept that nature is inherently probabilistic and that there is a very successful formalism called QT that tells you probabilities and only probabilities.

I don't know if this is addressed to me but - I'm not worried nor ever have been by any of the great 'problems' in QT !

Like you I believe if we can calculate probabilities we can simulate any experiment.

Entanglement changes the probabilities a lot and it is easy enought to show this.

My first run of 400 produced these quartets (count, coincidences, $\chi^2$ contribution, expected value = count/2 )

A₁B₁₌( 105, 48, 0.38, 52.5)
A₁B₂₌( 103, 25, 13,2, 51.5)
A₂B₁₌( 100, 77, 14.5, 50)
A₂B₂₌( 92, 56, 2.17, 46)

The sum of the $\chi^2$ terms is 30.5 which is in the very small percentile of the $\chi^2$ distribution with 3 dof.

Clearly allowing the first projection to set both photons has skewed the probabilities drastically compared to the unentangled set up.

 

[vanhees71]

But the relative angle is not known until the last minute. The situation is the following:

View attachment 217546

You have three devices: C, which is a source of message pairs, simulating photon pair production, and A and B, which simulate the measurement events.

• Every "round", C sends out a pair of messages, m_A to A and m_B to B.
• After the messages are sent, but before they are read, settings for A and B are chosen, independently. The settings are two angles, \theta_A and \theta_B.
• Device A determines an output, R_A(\theta_A), which is either +1 or -1, based on the message received from C and the setting \theta_A.
• Similarly, device B determines an output, R_B(\theta_B) based on its message and setting.
• Over many, many rounds, we can gather statistics for the correlation: \langle R_A(\theta_A) R_B(\theta_B) \rangle as a function of the pair of settings, \theta_A, \theta_B.

Bell's inequality implies that |\langle R_A(\theta_A) R_B(\theta_B) \rangle| \leq 2, no matter what algorithms are used by A, B, and C, as long as

1. There are no communications among A, B, C other than those specified.
2. The settings \theta_A and \theta_B for each round are unpredictable by C.

On the other hand, if instead of C sending messages, it generates a pair of entangled photons, and sends one to A and one to B, then you can violate the inequality. (Inside A and B, instead of a computer algorithm, you have polarizing filters and photon detectors, and each sends out +1 if the photon passes through the filter at the orientation specified by \theta_A or \theta_B.

Sure, I'm aware of this, but no matter when you choose the orientation (even in the last femtosecond before the envisaged photon hits the detector) you know the probabilities you have to simulate, when using this setup to calculate the corresponding probabilities (or expectation values). The probabilities are known of course; it's what's predicted by QT, and that's what's simulated with a correspondingly programmed MC simulator.

 

[stevendaryl]

Sure, I'm aware of this, but no matter when you choose the orientation (even in the last femtosecond before the envisaged photon hits the detector) you know the probabilities you have to simulate,

I'm not exactly sure what your point is. Yes, we know the probabilities for joint detection. The question is whether you can simulate the detection process in the way that I have sketched out and get those probabilities. The answer is "no".

When you choose the orientation makes a difference, because if the orientation was planned in advance, then device C could use that information to reproduce the predictions of QM.

 

[stevendaryl]

I don't really know what you (@vanhees71) mean by a "Monte Carlo" simulation for this experiment, but the only way you can reproduce the predictions of quantum mechanics for this case is if the settings for the detectors are known by the simulation code. In other words, by cheating (according to the rules laid out).

 

[Mentz114]

I don't really know what you (@vanhees71) mean by a "Monte Carlo" simulation for this experiment, but the only way you can reproduce the predictions of quantum mechanics for this case is if the settings for the detectors are known by the simulation code. In other words, by cheating (according to the rules laid out).

https://en.wikipedia.org/wiki/Monte_Carlo_method

 

[stevendaryl]

https://en.wikipedia.org/wiki/Monte_Carlo_method

But in a Monte Carlo simulation, the inputs are generated, as well as the outputs, which means that the inputs (the detector settings, in this case) are known in advance.

 

[DrChinese]

Of course, I have to use the relative angle of the detectors to calculate these probabilities, and thus this information goes into the code.

As mentioned in post #25: you cannot simulate entanglement if Alice doesn't know Bob's choice of measurement setting, and Bob doesn't know Alice's choice of measurement setting. So yes, if you circumvent that, you "cheat". That is to say that using the relative angle in the code to generate the answer is a cheat. If you don't cheat, you can't even hand pick data sets that match the probability predictions of QM.

So hopefully we are all in agreement on this point, which was what was in error in the OP's original code - which made use of the relative angle.

 

[forcefield]

As mentioned in post #25: you cannot simulate entanglement if Alice doesn't know Bob's choice of measurement setting, and Bob doesn't know Alice's choice of measurement setting. So yes, if you circumvent that, you "cheat". That is to say that using the relative angle in the code to generate the answer is a cheat. If you don't cheat, you can't even hand pick data sets that match the probability predictions of QM.

So hopefully we are all in agreement on this point, which was what was in error in the OP's original code - which made use of the relative angle.

Isn't it enough if "Charlie" (or "C") knows both measurement settings as said by @stevendaryl in post #43 ?

 

[entropy1]

Isn't it enough if "Charlie" (or "C") knows both measurement settings as said by @stevendaryl in post #43 ?

The measurement settings are not fixed in principle; they can be changed at the last moment.

 

[forcefield]

The measurement settings are not fixed in principle; they can be changed at the last moment.

I know that - are you saying that I can't simulate entanglement without changing measurement settings at the last moment ?

 

[entropy1]

I know that - are you saying that I can't simulate entanglement without changing measurement settings at the last moment ?

The principle is that the settings are unknown (till the last moment). If they are fixed and conveyed (to C) we have a special case IMHO.

 

[Mentz114]

But in a Monte Carlo simulation, the inputs are generated, as well as the outputs, which means that the inputs (the detector settings, in this case) are known in advance.

This is hard to explain because I cannot see the problem. We know that A and B will have definate settings before they project their photon. It does not matter when they get them as long as the selection is random (independent).

The simulation works stepwise. Produce a random orientation $\theta_0$ ( 1 random used ). Now assume that A's photon reaches the polarizer set to $\theta_A$. The probability of passing we know is $\cos(\theta_A-\theta_0)^2$. Now draw another RN to see if it passes.
If it passes we can say the alignment of photon B is $\theta_A$ and now we can calculate if it will pass B's polarizer which depends on B's setting.

It works. Statistically the results are obviously violating the expectations.

 

[DrChinese]

I know that - are you saying that I can't simulate entanglement without changing measurement settings at the last moment ?

If your algorithm for determining the result of A's measurement requires knowledge of B's setting, or if you need to know A's setting to determine B's results: then you are not using a separable algorithm. Thus the "cheat" and it is not simulating local realism. If you allow the "cheat", you CAN simulate QM/entanglement. But that is the only way.

 

[DrChinese]

This is hard to explain because I cannot see the problem. We know that A and B will have definate settings before they project their photon. It does not matter when they get them as long as the selection is random (independent).

The simulation works stepwise. Produce a random orientation $\theta_0$ ( 1 random used ). Now assume that A's photon reaches the polarizer set to $\theta_A$. The probability of passing we know is $\cos(\theta_A-\theta_0)^2$. Now draw another RN to see if it passes.
If passes we can say the alignment of photon B is $\theta_A$ and now we can calculate if it will pass B's polarizer.
It works.

This would simulate QM entanglement, because of your usage of $\theta_A$ to calculate B's outcome. Without that information, you can't get agreement with the predictions of QM. So if the A and B algorithms are separate, per Bell you can't get that agreement.

 

[Mentz114]

This would simulate QM entanglement, because of your usage of $\theta_A$ to calculate B's outcome. Without that information, you can't get agreement with the predictions of QM. So if the A and B algorithms are separate, per Bell you can't get that agreement.

Yep. It is possible to simulate entanglement. Only the (simulated) projection of both photons is required to do this. A's setting is not 'known' by B - it is carried by the photon and affects B's result.

 

[entropy1]

Yep. It is possible to simulate entanglement. Only the (simulated) projection of both photons is required to do this. A's setting is not 'known' by B - it is carried by the photon and affects B's result.

If the wavefunction (which is what you mean, I think) is transporting information from A to B, wouldn't we have manifest non-locality?

 

[Mentz114]

If the wavefunction (what is what you mean, I think) is transporting information from A to B, wouldn't we have manifest non-locality?

When either of the photons is projected into a definate polarization state the other must also be in that state. Experiments seem to show that the separation is irrelevant. So information has gone from the first projected photon to the other ( it is said ) but I think they are just always are in the same state - a shared field.

 

[entropy1]

When either of the photons is projected into a definate polarization state the other must also be in that state. Experiments seem to show that the separation is irrelevant. So infoemation has gone from the first projected photon to the other.

I am not so sure myself; there is no temporal ordening of the detections; A is not before B, nor B before A in a spacelike separated setting. So there is no 'transporting' in any definite direction.

 

[DrChinese]

I am not so sure myself; there is no temporal ordening of the detections; A is not before B, nor B before A in a spacelike separated setting. So there is no 'transporting' in any definite direction.

In the simulation, you can do it as you like (A before B or whatever). It is interesting that in real life, as you say, there is no apparent definite direction. And the predictions are the same regardless of which measurement occurs first.

 

[stevendaryl]

This is hard to explain because I cannot see the problem. We know that A and B will have definate settings before they project their photon. It does not matter when they get them as long as the selection is random (independent).

The simulation works stepwise. Produce a random orientation $\theta_0$ ( 1 random used ). Now assume that A's photon reaches the polarizer set to $\theta_A$. The probability of passing we know is $\cos(\theta_A-\theta_0)^2$. Now draw another RN to see if it passes.
If it passes we can say the alignment of photon B is $\theta_A$ and now we can calculate if it will pass B's polarizer which depends on B's setting.

It works. Statistically the results are obviously violating the expectations.

I don't know exactly what it is that you are describing here. This is the answer to what question?