Thought Experiment: Exploring Violations of Bell Inequalities

[Johan0001]

TL;DR Summary

Does non locality follow from bell tests

I have been thinking about the Violation of bell inequalities , trying to justify how non locality can be determined from violation of bell tests.
I have been through Dr. Chinese page which has partially convinced me that there can be no hidden variables , but I need to understand what is wrong with the following thought experiment , to get better clarity.

With analogy to the Alain Aspect experiment of entangled photon creation, I would like to Propose a Factory which produces pairs of Dice in mass production , very small ones if you would prefer to include QM effects.

As these dice are pumped out of the factory their values are read when they land on the ground.

So let's just look at one permutation , the probability of each die landing on 6. Which is 1/36 or 2.77%.

Now if after many repetitive Measurements I find that this probability has been violated , say we consistently measure 10%.

What conclusion can I make with this.

1. My first and most obvious is that the paired dice are not exactly equivalent, there are slight differences in their creation.
The factory is producing inconsistent pair , one of them is possibly not uniformly dense on one of the sides inside the die.
Although in total each pair does have a consistent total mass when produced , they are not identical.
Thus leading to inconsistent(correlated) readings.
What I'm saying here ,is that it is not a hidden variable , but the value of the variable itself that is slightly different than the other
counterpart die.
Could this analogy not be an alternative to to Entangled photons, if not Why?.

2. There is inconsistency in the way that the dice are being thrown or measured.
I guess in the photon pairs this would equate to fair sampling loophole

3. The dice are some how entangled , and this is a quantum phenomenon.
Which seems to be the case with photon pair production.

How does my analogy differ from the Quantum wavefunction approach of instantaneous collapse, and what am i missing?

Johan

 

[PeroK]

You are proposing that the particles produced (by a certain experiment) vary in their fundamental properties. There is no evidence for that. You can carry out simple experiments with photons or other particles to convince yourself that you have a uniform production process.

This is all established in experiment long before you get to entanglement: you do not get a mixture of all different types of electrons or other particles. Elementary particles are, literally, identical. They do come out in all shapes and sizes.

 

[Johan0001]

Elementary particles are, literally, identical. They do come out in all shapes and sizes.

But perhaps their total energy/frequency of photon are slightly different.
Photons as smeared out in free space not point particles, is this correct?
I know I'm going off topic , but if we adopt string theory for example , are we saying that entangled photons have exactly the same vibrations on a much smaller scale, which make them indistinguishable.

You can carry out simple experiments with photons or other particles to convince yourself that you have a uniform production process.

What is the evidence that these elementary particles are uniform.
If you could send me some references i would appreciate it.

Regards

Johan

 

[DrClaude]

1. My first and most obvious is that the paired dice are not exactly equivalent, there are slight differences in their creation.
The factory is producing inconsistent pair , one of them is possibly not uniformly dense on one of the sides inside the die.
Although in total each pair does have a consistent total mass when produced , they are not identical.
Thus leading to inconsistent(correlated) readings.
What I'm saying here ,is that it is not a hidden variable , but the value of the variable itself that is slightly different than the other
counterpart die.

I don't see how this is different from a hidden variable. The difference in the dice can be encoded in the hidden variable.

 

[DrClaude]

But perhaps their total energy/frequency of photon are slightly different.
Photons as smeared out in free space not point particles, is this correct?
I know I'm going off topic , but if we adopt string theory for example , are we saying that entangled photons have exactly the same vibrations on a much smaller scale, which make them indistinguishable.
What is the evidence that these elementary particles are uniform.
If you could send me some references i would appreciate it.

Regards

Johan

Entanglement does not require identical particles.

 

[PeroK]

What is the evidence that these elementary particles are uniform.
If you could send me some references i would appreciate it.

Any textbook on QM discusses "identical" or "indistinguishable" particles. This is required to align QM with observation! See, for example:

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

Or, consult any undergraduate textbook, such as Griffiths or Sakurai.

Moreover, the whole basis of modern particle physics rests on indistinguishability and symmetry. The modern theory of particle physics would simply never have developed without this theoretical and experimentally verified indistinguishability of elementary particles.

 

[Dale]

What is the evidence that these elementary particles are uniform.
If you could send me some references i would appreciate it.

One of the early evidences is in thermodynamics, in particular the entropy would be completely different for a system of distinguishable particles and a system of indistinguishable particles.

Recall that entropy is based on counting the number of micro states that are consistent with a given macro state. In that counting process if particles are distinguishable then two states which swap two particles are two different states and each must be counted. If the particles are indistinguishable then swapping two particles does not change the state, the two states are the same state and that state must be counted only once.

It turns out that the entropy of the distinguishable statistics would not follow the basic laws of thermodynamics. In fact, entropy would not be an extensive property. So any experiment confirming the usual laws of thermodynamics and especially entropy is an experiment showing that particles are indistinguishable.

 

[atyy]

1. My first and most obvious is that the paired dice are not exactly equivalent, there are slight differences in their creation.
The factory is producing inconsistent pair , one of them is possibly not uniformly dense on one of the sides inside the die.
Although in total each pair does have a consistent total mass when produced , they are not identical.
Thus leading to inconsistent(correlated) readings.
What I'm saying here ,is that it is not a hidden variable , but the value of the variable itself that is slightly different than the other
counterpart die.
Could this analogy not be an alternative to to Entangled photons, if not Why?.

2. There is inconsistency in the way that the dice are being thrown or measured.
I guess in the photon pairs this would equate to fair sampling loophole

As @DrClaude says above, both of these are effectively hidden variables - they are additional factors that are hidden.

 

[Nugatory]

Could this analogy not be an alternative to to Entangled photons,

No, because this analogy doesn't capture the fundamental non-classicality of entanglement. Even if we assume for the sake of argument that the two particles might be distinguishable (the issue discussed at length above) there's a deeper problem, but we may have to refer back to Bell's actual paper to see it.

We create pairs of dice that are asymmetrical in some correlated way. These aren't fair dice, so there's no particular reason to expect them to obey fair dice statistics: any given double doesn't have to appear on one out of every thirty-six throws and any given non-double doesn't have to appear on one out of fifteen throws.

However, that correlation is not non-locality in the sense of Bell's theorem (just as the fair-dice correlations also aren't). We can calculate the correlation for the two dice as $P()=\int d\lambda_a d\lambda_b A(\lambda_a)B(\lambda_b)$ where $\lambda_a$ and $\lambda_b$ are the probability distributions for everything known and unknown at station A and B respectively; that is, the behavior of the die at station A doesn't depend on what happens at station B and vice versa. The two dice may be different somehow, and that may affect their behavior when they land, but these differences are captured in the $\lambda_{a,b}$ functions.

However, the quantum mechanical formula for the probability of opposite results depends on the angle between the two detectors, so if we use that formula we cannot split the "everything known and unknown at station A and B respectively" into separate functions $\lambda_a$ and $\lambda_b$; the calculation always relies on something from both sides. That's a well-known and uncontroversial result from quantum mechanics, and it is the key thing that the dice analogy fails to capture.

(I've taken some liberties with Bell's notation here because of the difference between the two-dice problem and the singlet spin measurement problem: we don't have detector settings so the entire state is captured by the $\lambda$ variable. If someone wants to improve on my notation, please do - my feelings won't be hurt)

 

[Johan0001]

Thank you guys ,lot of information above to ponder on.
Will do some homework ..and revert back.

 

[Johan0001]

I don't see how this is different from a hidden variable. The difference in the dice can be encoded in the hidden variable.

Hi DrClaude

What was trying to compare with this analogy , is that the density of the two die is not uniform.
The density variable is not hidden , but they may be different in the individual die.
Similairly the energy of each pair of photons , not being a hidden variable, but slightly different.
Which may cause the bell violation at 45 degrees.

 

[PeroK]

Hi DrClaude

What was trying to compare with this analogy , is that the density of the two die is not uniform.
The density variable is not hidden , but they may be different in the individual die.
Similairly the energy of each pair of photons , not being a hidden variable, but slightly different.
Which may cause the bell violation at 45 degrees.

The Bell violation is a violation of classical probability laws. It doesn't matter how the dice are configured, if the configuration is determined at the point of creation, then classical probability applies.

QM on the other hand uses complex probability amplitudes. This results in correlations not possible with classical probabilities.

Also, the variables are assumed to be "hidden" because they there is no evidence for them. If the variables are not hidden, then you can measure them directly. In the case of your dice, you simply look at the dice and see that they are not all the same. EPR were forced to postulate "hidden" variables that no one had ever seen or measured in order to provide an alternative to QM. If, like you, then had postulated non-hidded properties, then these properties could simply be measured directly.

PS I suspect you have not really understood QM or the Bell inequality. I would focus on understanding how probabilities in QM (based on complex probability amplitudes) differ from classical probabilities. For example, quantum interference relies on probability amplitudes cancelling; whereas, classical probabilities can only add.

 

[Johan0001]

Also, the variables are assumed to be "hidden" because they there is no evidence for them. If the variables are not hidden, then you can measure them directly. In the case of your dice, you simply look at the dice and see that they are not all the same. EPR were forced to postulate "hidden" variables that no one had ever seen or measured in order to provide an alternative to QM. If, like you, then had postulated non-hidded properties, then these properties could simply be measured directly.

Yes I follow your reasoning above.
And you are correct if you can't measure the difference you can assume it being hidden.

In my analogy above , if I start to record differences, (in the expected Dice probabilities) I would tend to go for my first conclusion , that the Dice are not entirely indistinguishable , and that I just can measure this difference yet.
However after many iterations of recorded values , I can "deduce" that there is a difference in the variable.

To jump from this conclusion, to entanglement and correlation / non locality and realism still eludes me.

Johan

 

[PeroK]

To jump from this conclusion, to entanglement and correlation / non locality and realism still eludes me.

Johan

It's all about the mathematical calculations. Bell's theorem only resolves the issue because of the precise and specific nature of QM probability calculations.

If, when you dig into the details, QM didn't predict a stronger correlation than was classically possible, then Bell's theorem would be worth nothing. Bell's theorem is:

1) Do classical probability and get a limit (*) on correlations between classically determined variables.

2) Do QM probability calculations and predict a stronger correlation.

3) Do a test and if you get stronger correlations than 1) allows, then we cannot have classically determined values.

(*) Note that the limit applies to any distribution of classically determined variables. If you understand this, then it doesn't matter what form these variables take.

In the end, it's all about hard calculations, not woolly ifs and buts.

 

[Johan0001]

1) Do classical probability and get a limit (*) on correlations between classically determined variables.

2) Do QM probability calculations and predict a stronger correlation.

3) Do a test and if you get stronger correlations than 1) allows, then we cannot have classically determined values.

(*) Note that the limit applies to any distribution of classically determined variables. If you understand this, then it doesn't matter what form these variables take.

Yes agreed again , QM probabilities predict a stronger Correlation. Which we cannot deny.
But in my view ALL probabilities (Classical or QM) have at their root Realism/ hard coded variables , by which they are approximated over repetition , and deduced.
To explain the violation is i guess at the core of QM today.

QM probabilities does not do magic , there are underlying reasons for these results, my dice analogy could be just one of many reasons I suppose, or am i still missing it.

Johan

 

[PeroK]

But in my view ALL probabilities (Classical or QM) have at their root Realism/ hard coded variables , by which they are approximated over repetition , and deduced.

That's a personal theory that is not supported by any theoretical or experimental evidence. PF is not the place to discuss mistaken personal theories.

 

[Nugatory]

QM probabilities does not do magic , there are underlying reasons for these results, my dice analogy could be just one of many reasons I suppose, or am i still missing it.

There could be an underlying reason for the results. Bell’s theorem does not preclude that possibility; it precludes the possibility that the underlying reason can be a theory that is both local and realistic.
Here “local” means that the probability at either station can be determined just from what is going on at that station, including the non-uniform weight distribution and other properties of the dice at that station. “Realistic” means that in principle we could determine the properties of any single die if we measure and examine it properly; for example we can find the non-uniform weight distribution by measuring the exact center of gravity. (Note that these are informal hand-waving definitions, so it’s not worth quibbling over them. The math in Bell’s paper is a more precise statement of what is precluded).

Your dice analogy is both local (what happens at one station doesn’t affect what happens at the other station) and realistic (once we’ve created the two dice in a pair, they are what they are) so is not a good analogy for the sort of theory that might explain the quantum mechanical results.

 

[f95toli]

But perhaps their total energy/frequency of photon are slightly different.
Photons as smeared out in free space not point particles, is this correct?
I know I'm going off topic , but if we adopt string theory for example , are we saying that entangled photons have exactly the same vibrations on a much smaller scale, which make them indistinguishable.

This discussion is completely irrelevant for the topic at hand. You can entangle two systems and perform Bell tests using systems that are in no way identical (solid state qubits come to mind), whether the "particles" you are working with are identical does not matter.

 

[DrChinese]

Yes agreed again , QM probabilities predict a stronger Correlation. Which we cannot deny.
But in my view ALL probabilities (Classical or QM) have at their root Realism/ hard coded variables , by which they are approximated over repetition , and deduced.
To explain the violation is i guess at the core of QM today.

QM probabilities does not do magic , there are underlying reasons for these results, my dice analogy could be just one of many reasons I suppose, or am i still missing it.

Johan

Sure, but the dice analogy doesn't actually extend to the quantum statistics. You must remember that when entangled particles are measured in the same manner (for both), the results are 100% predictable. That wouldn't happen with your dice because of the underlying variability you propose.

And in fact the only variable in the outcome statistics is the relationship of the measurement bases performed by Alice and Bob. All other variables completely cancel out!

 

[Johan0001]

This discussion is completely irrelevant for the topic at hand. You can entangle two systems and perform Bell tests using systems that are in no way identical (solid state qubits come to mind), whether the "particles" you are working with are identical does not matter.

Agreed , but the point I was trying to make previously is that the 2 "Quantum objects" which could be photons/electrons , may have slight differences in their known properties. So when they interact with their "detectors" they do not produce the same expected classical values but have minimal variances over a large ensemble of measurements which could account for the bell violation.

 

[Johan0001]

Sure, but the dice analogy doesn't actually extend to the quantum statistics. You must remember that when entangled particles are measured in the same manner (for both), the results are 100% predictable. That wouldn't happen with your dice because of the underlying variability you propose.

Why are ALL underlying HIDDEN variables/interactions excluded from a possible explanation, with the Bell paper.

This was my first assumption, of the three , I put forward, in my original post, I was posing the question as to why entanglement (spooky action) was chosen over a classical possibility , where minor variances can change the expected measurement, in a predictable way, with variances in the properties.
Surely this could be a plausible reason.

Johan

 

[DrClaude]

Why are ALL underlying HIDDEN variables/interactions excluded from a possible explanation, with the Bell paper.

The Bell inequalities set limits on correlations when using local, hidden variable models. It doesn't matter what the hidden variable is or represents, so long as all the information is local.

Quantum mechanics violates the Bell inequalities, so the results of QM cannot be reproduced by any "hidden variable" model. Experiments have shown that the predictions of QM hold.

 

[DrChinese]

This was my first assumption, of the three , I put forward, in my original post, I was posing the question as to why entanglement (spooky action) was chosen over a classical possibility , where minor variances can change the expected measurement, in a predictable way, with variances in the properties.
Surely this could be a plausible reason.

No, that's not possible. The reason - I mentioned this - is that there are perfect correlations at the same angle settings for Alice and Bob. Those "minor variances" can't be present, as they are ruled out by experiment. Remember, the setting of Alice and the setting of Bob are independent, so neither can be sure what the other one is set at.

 

[StevieTNZ]

Quantum mechanics violates the Bell inequalities, so the results of QM cannot be reproduced by any "hidden variable" model. Experiments have shown that the predictions of QM hold.

I would re-write that, for clarity, that "the results of QM cannot be reproduced by any "hidden variable" model" with "the results of QM cannot be reproduced by any "local hidden variable" model".

 

[RUTA]

Why are ALL underlying HIDDEN variables/interactions excluded from a possible explanation, with the Bell paper.

This was my first assumption, of the three , I put forward, in my original post, I was posing the question as to why entanglement (spooky action) was chosen over a classical possibility , where minor variances can change the expected measurement, in a predictable way, with variances in the properties.
Surely this could be a plausible reason.

Johan

Read this paper by David Mermin. It clarifies why "instruction sets" don't work and aligns perfectly with Bell spin states even though you don't have to know anything about QM per se to understand the paper.

 

[DrClaude]

I would re-write that, for clarity, that "the results of QM cannot be reproduced by any "hidden variable" model" with "the results of QM cannot be reproduced by any "local hidden variable" model".

Yes. I had "local" in the first paragraph and should have had it in the second also.