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Why are Bell's inequalities violated?

by JK423
Tags: bell, inequalities, violated
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bohm2
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Jan16-13, 08:58 AM
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Quote Quote by nanosiborg View Post
I think that something other than nonlocality will eventually answer the thread question.
There seems to be only 3 options based on assumptions made by Bell:

1. Non-locality
2. Anti-realism
3. Superdeterminism (no freedom of choice)
DrChinese
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Jan16-13, 09:10 AM
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Quote Quote by harrylin View Post
Some rhetorical questions deserve an answer. Extreme claims require extreme evidence. Thus I require the same level of scientific proof as for perpetuum mobilae.
Well, first of all, the observed results match theory. That isn't true of perpetual motion machines.

Second, I guess you answered the question about why you don't demand the same proof for relativity. The answer is that what is "extreme" is subjective (to you). You consider relativity "reasonable" in light of experimental proof but falsification of local realism "unreasonable" in light of experimental proof. Ergo you essentially conclude that which you sought to prove.

The reason I called it a rhetorical question is because of this point. If you are a local realist in 2013, you aren't going to let evidence affect your viewpoint. So no point in trying to answer the question.
harrylin
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Jan16-13, 09:56 AM
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Quote Quote by DrChinese View Post
Well, first of all, the observed results match theory. That isn't true of perpetual motion machines.

Second, I guess you answered the question about why you don't demand the same proof for relativity. The answer is that what is "extreme" is subjective (to you). You consider relativity "reasonable" in light of experimental proof but falsification of local realism "unreasonable" in light of experimental proof. Ergo you essentially conclude that which you sought to prove.

The reason I called it a rhetorical question is because of this point. If you are a local realist in 2013, you aren't going to let evidence affect your viewpoint. So no point in trying to answer the question.
For sure perpetual motion machines match the theory of such experimenters, or so they claim (I did look into a few of them, so I know what I'm talking about). And not surprisingly, those experimenters reply in quite the same manner, claiming that no evidence is ever good enough for the scientific community - however, that's not true; only the requirements are extremely strict. But thanks for clarifying why your question was rhetorical.
DrChinese
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Jan16-13, 10:30 AM
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Quote Quote by harrylin View Post
For sure perpetual motion machines match the theory of such experimenters, or so they claim (I did look into a few of them, so I know what I'm talking about).
Ah, that is still not the same at all. Bell involves no modification to preexisting, accepted theory that forms our common ground. (I assume you don't disagree with QM pre-Bell.) So you can't really object that it hypothesizes new effects. There aren't any. The only new thing is realizing that QM did predict entanglement, and thus you can have a field day with experiments in that vein that are derived from orthodox QM. Of course, orthodox QM itself is falsifiable on a variety of fronts.

On the other hand, PMM advocates ARE proposing modifications to pre-existing, accepted theory (so there is not common ground). Those should be falsifiable if they are to be useful (otherwise they would be "ad hoc").

In my book, you pick and choose what evidence you accept, in order to be consistent with your pre-ordained conclusion.
JK423
#23
Jan16-13, 01:06 PM
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Another question came up..
The CHSH quantity,
[itex]S_j = A_j\left( {{a_1}} \right)B_j\left( {{b_1}} \right) + A_j\left( {{a_1}} \right)B_j\left( {{b_2}} \right) + A_j\left( {{a_2}} \right)B_j\left( {{b_1}} \right) - A_j\left( {{a_2}} \right)B_j\left( {{b_2}} \right)[/itex],
where [itex]A_j\left( {{a_i}} \right) = \pm 1[/itex] and [itex]B_j\left( {{b_i}} \right) = \pm 1[/itex], and j denoting a particular photon pair,
is always [itex]{S_j} = \pm 2[/itex], for any measurement result A and B.
When we take the mean value over all photon pairs, [itex]\,\left\langle S \right\rangle = \frac{1}{N}\sum\limits_{i = 1}^N {{S_j}} [/itex] we find it to be bounded, i.e.
[itex] - 2 \le \,\left\langle S \right\rangle \le 2[/itex].
This quantity is bounded whatever the values of A and B for the photon pairs.

Say that the choice of the angles [itex]a_i[/itex],[itex]b_i[/itex] is not random but they are correlated to each other. I don't see how this inequality could be violated by a local and realistic model.. Can you help? I am trying to understand how this *loophole* works..
DrChinese
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Jan16-13, 01:23 PM
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Quote Quote by JK423 View Post
Another question came up..
The CHSH quantity,
[itex]S_j = A_j\left( {{a_1}} \right)B_j\left( {{b_1}} \right) + A_j\left( {{a_1}} \right)B_j\left( {{b_2}} \right) + A_j\left( {{a_2}} \right)B_j\left( {{b_1}} \right) - A_j\left( {{a_2}} \right)B_j\left( {{b_2}} \right)[/itex],
where [itex]A_j\left( {{a_i}} \right) = \pm 1[/itex] and [itex]B_j\left( {{b_i}} \right) = \pm 1[/itex], and j denoting a particular photon pair,
is always [itex]{S_j} = \pm 2[/itex], for any measurement result A and B.
When we take the mean value over all photon pairs, [itex]\,\left\langle S \right\rangle = \frac{1}{N}\sum\limits_{i = 1}^N {{S_j}} [/itex] we find it to be bounded, i.e.
[itex] - 2 \le \,\left\langle S \right\rangle \le 2[/itex].
This quantity is bounded whatever the values of A and B for the photon pairs.

Say that the choice of the angles [itex]a_i[/itex],[itex]b_i[/itex] is not random but they are correlated to each other. I don't see how this inequality could be violated by a local and realistic model.. Can you help? I am trying to understand how this *loophole* works..
The word "loophole" is not usually used in this context, as it has a rather different meaning altogether.

A local realistic model will always have S<=2, do you see that? (Since S is between -2 and +2.)

However, experiments typically give a value of S>2, often in the 2.2 to 2.4 range depending on the particulars of the setup and efficiency. The usual value given for the QM predicted theoretical value is about 2.7 (again this varies somewhat depending on assumptions). So quantum theory and experiment are in reasonable agreement, but are at odds with predictions based on local realistic assumptions.

Does that address your question?
JK423
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Jan16-13, 01:38 PM
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Thank you for your answer, although i don't think i understood your point.

You are saying that a local realistic model will always give |S|<=2.
So does this hold for any choice of the angles?
There is no demand for random non-correlated choice of angles?
Then what is all the fuss about free will of the observers etc and superdeterminism?

I thought that by using correlations between the choice of the angles, you could make a local realistic model violate the inequality.
Am i wrong?
nanosiborg
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Jan16-13, 02:15 PM
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Quote Quote by bohm2 View Post
There seems to be only 3 options based on assumptions made by Bell:

1. Non-locality
2. Anti-realism
3. Superdeterminism (no freedom of choice)
Assuming either superdeterminism or nonlocality is not informative.

The answer is in the realm of anti-realism, which has to do with modeling restrictions.
Nugatory
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Jan16-13, 02:25 PM
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Quote Quote by JK423 View Post
You are saying that a local realistic model will always give |S|<=2.
So does this hold for any choice of the angles?
There is no demand for random non-correlated choice of angles?
Then what is all the fuss about free will of the observers etc and superdeterminism?

I thought that by using correlations between the choice of the angles, you could make a local realistic model violate the inequality.
Am i wrong?
At the risk of putting words in DrChinese's mouth (but if I get it wrong we'll find out), I expect that he meant "non-conspiratorial local realistic model". If the choice of angles is deterministic and the source knows what the determining rule is, or if the source is allowed to influence the choice of direction, then the source can produce pairs tailored to produce any results it pleases.

We tend to leave out the qualifier because the conspiratorial models are either uninteresting or reduce to some form of superdeterminism, or both.
DrChinese
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Jan16-13, 02:34 PM
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Quote Quote by JK423 View Post
Thank you for your answer, although i don't think i understood your point.

a) You are saying that a local realistic model will always give |S|<=2.
So does this hold for any choice of the angles?
There is no demand for random non-correlated choice of angles?

b) Then what is all the fuss about free will of the observers etc and superdeterminism?

c) I thought that by using correlations between the choice of the angles, you could make a local realistic model violate the inequality.
Am i wrong?
There are a batch of different issues in your comments. I will do my best to address.

a) Yes, LR models always yield |S|<=2 and that is for any choice of angles.

c) The LR model predicts results that are consistent with the inequality. However, those results do not happen in experimental situations. So the real world violates the inequality, not the LR model.

b) OK, the idea of superdeterminism is something I routinely criticize as non-scientific. But some folks I respect think it is worth mentioning, so I will attempt to describe the argument in as objective terms as able. I am answering this after c) so you can read c) again as needed.

The idea is that the angle settings that we think we are freely selecting correspond to ones in which the inequality will be violated, but that we are actually choosing ones in which this result was predetermined to violate the inequality even though the inequality is NOT really violated. So the results are predetermined, and further the results were predetermined in such a way to be misleading.

Imagine we are playing 3 card monte with queen and 2 jacks. You are trying to pick the queen each hand. I bend the queen card (but not the other 2) and shuffle them around. You pick the bent card and I turn it over to reveal a jack. You thought you freely chose the card but I fooled you ('cause I am sneaky). Every time we do it, you pick a jack. Eventually you conclude there is no queen. But actually there was one, you just didn't pick it.

Do you see that in this case, if there is something influencing you that you are not aware of, you might come to a false conclusion? This is the *analogy* to the superdeterminism argument. So all you have to do is acknowledge that IF your choice was somehow influenced with bias and you were not aware of it, THEN you could come to the wrong conclusion. This is superdeterminism.

----------------------

Keep in mind that I vehemently deny that random angle choices has anything to do with a Bell test OTHER THAN to enforce strict Einsteinian separability. That was demonstrated in 1998 by Weihs et al. In a normal Bell test, you do not need to do ANYTHING other than show that the cos^2(theta) relationship predicted by QM is consistent with your results. The multiple angles thing is not needed at all and tends to confuse everyone. The reason is that Bell demonstrated that LR theories will not be able to yield datasets consistent with QM's cos^2(theta). So if QM is right (confirmed experimentally) then LR cannot be.

You don't need to know anything about Bell's proof or inequality if you simply try to construct a local realistic dataset of your own (I supplied some example angles at the beginning of this thread). Just try to create a dataset and you will see it cannot ever match the cos^2(theta) at those angles. If QM is experimentally right about the cos^2 relationship, then LR is logically excluded.
DrChinese
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Jan16-13, 02:42 PM
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Quote Quote by Nugatory View Post
At the risk of putting words in DrChinese's mouth (but if I get it wrong we'll find out), I expect that he meant "non-conspiratorial local realistic model". If the choice of angles is deterministic and the source knows what the determining rule is, or if the source is allowed to influence the choice of direction, then the source can produce pairs tailored to produce any results it pleases.

We tend to leave out the qualifier because the conspiratorial models are either uninteresting or reduce to some form of superdeterminism, or both.
That's true, but I equate:

Conspiracy <=> Superdeterminism <=> Hand of god

in the sense that all of these are "outs". Of course these apply equally for all theories: evolution, cosmology, relativity, etc. I have no idea why any of these should be included as an qualification for a scientific discussion. You don't say "the universe is 13.7 billion years old UNLESS it is really 4000 years old and there is superdeterminism at work." I think everyone understands that if we are ALL being hoodwinked, then all bets are off on anything we might think we know.
JK423
#30
Jan16-13, 03:02 PM
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Quote Quote by Nugatory View Post
At the risk of putting words in DrChinese's mouth (but if I get it wrong we'll find out), I expect that he meant "non-conspiratorial local realistic model". If the choice of angles is deterministic and the source knows what the determining rule is, or if the source is allowed to influence the choice of direction, then the source can produce pairs tailored to produce any results it pleases.

We tend to leave out the qualifier because the conspiratorial models are either uninteresting or reduce to some form of superdeterminism, or both.
Hmm i think that this helped! Thank you.
So, if i understand correctly:

The reason why a local realistic (non-conspirational) model will always satisfy Bell's inequality is due to the factorization of [itex]S_j[/itex] (second line):
[itex]\begin{array}{l}
{S_j} = {A_j}\left( {{a_1}} \right){B_j}\left( {{b_1}} \right) + {A_j}\left( {{a_1}} \right){B_j}\left( {{b_2}} \right) + {A_j}\left( {{a_2}} \right){B_j}\left( {{b_1}} \right) - {A_j}\left( {{a_2}} \right){B_j}\left( {{b_2}} \right) \\
= {A_j}\left( {{a_1}} \right)\left( {{B_j}\left( {{b_1}} \right) + {B_j}\left( {{b_2}} \right)} \right) + {A_j}\left( {{a_2}} \right)\left( {{B_j}\left( {{b_1}} \right) - {B_j}\left( {{b_2}} \right)} \right) \\
= \pm 2 \\
\end{array}[/itex]. (1)
In order to violate Bell's inequality, this factorization should not be possible.
For example, non-locality would force the following change:
[itex]{A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _1}} \right) \to A_j\left( {{a_1},{\beta _1}} \right) \cdot {B_j}\left( {{a_1},{\beta _1}} \right)[/itex],
[itex]{A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _2}} \right) \to A_j\left( {{a_1},{\beta _2}} \right) \cdot {B_j}\left( {{a_1},{\beta _2}} \right)[/itex], (2)
etc.
This change allows the violation of the inequality because in general it's
[itex]A_j\left( {{a_1},{\beta _1}} \right) \ne A_j\left( {{a_1},{\beta _2}} \right)[/itex], (3)
preventing the factorization as in (1), so [itex]S_j=2[/itex] won't be always true and a violation of Bell's inequality is possible.

Now in the case of superdeterminism and conspirational models (but still local and realistic), Eq. (3) seems to hold as well. If the source knows beforehand what is to be measured, it can prepare the photon A in such a way so that e.g.
[itex]A_j\left( {{a_1},{\beta _1}} \right) = + 1[/itex] and
[itex]A_j\left( {{a_1},{\beta _2}} \right) = - 1[/itex],
being different like in (3), having the potential to lead to a violation of Bell's inequality.

Hope that i got this right!


Update: DrChinese i just saw your new post, thank you a lot for your detailed description, it clarifies lots of misconceptions that i had.
kith
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Jan16-13, 04:26 PM
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Quote Quote by DrChinese View Post
That's true, but I equate:

Conspiracy <=> Superdeterminism <=> Hand of god

in the sense that all of these are "outs". Of course these apply equally for all theories: evolution, cosmology, relativity, etc.
I'm not sure if this is really the case. In classical theories, the state of a system remains unchange by measurements, because the interaction of the observer with the system can be made arbitrarily small. So the measurement outcome doesn't depend on the past of the observer because it doesn't depend on the observer at all. In QM, the interaction between the observer leads to a physical change of the state of the system. So at least the assumption that the measurement outcome doesn't depend on the past of the observer is not as easily justifiable as in a classical theory.
Nugatory
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Jan16-13, 04:50 PM
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Quote Quote by kith View Post
So at least the assumption that the measurement outcome doesn't depend on the past of the observer is not as easily justifiable as in a classical theory.
That argument strikes me as a bit of a red herring. Yes, a superdeterministic quantum theory must include the observer whereas a deterministic theory (whether classical or merely EPR-friendly) need not. But that doesn't make superdeterminism any more palatable1; and experiment has already rejected the deterministic observer-independent theories.

1: De gustibus interpretationes non disputandum est.
kith
#33
Jan16-13, 05:17 PM
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Quote Quote by Nugatory View Post
That argument strikes me as a bit of a red herring. Yes, a superdeterministic quantum theory must include the observer whereas a deterministic theory (whether classical or merely EPR-friendly) need not.
We know as a fact that measurements on a quantum mechanical system may change the state of this system, so every fundamental theory needs to include the observer somehow. This is a key difference between classical theories and QM. It isn't an argument for superdeterminism by itself. It simply shows that the question of superdeterminism has different implications in QM than in classical theories. In classical mechanics, the physics of the system is independent of the question wether the observer is free to chose what observables he wants to measure. In QM it's not.

I am not an advocate of superdeterminism. I just replied to a statement by DrChinese with which I don't agree.
DrChinese
#34
Jan16-13, 05:47 PM
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Quote Quote by kith View Post
I'm not sure if this is really the case. In classical theories, the state of a system remains unchange by measurements, because the interaction of the observer with the system can be made arbitrarily small. So the measurement outcome doesn't depend on the past of the observer because it doesn't depend on the observer at all. In QM, the interaction between the observer leads to a physical change of the state of the system. So at least the assumption that the measurement outcome doesn't depend on the past of the observer is not as easily justifiable as in a classical theory.
That does not make sense, kith. It doesn't matter whether a theory is classical or not! That is a completely arbitrary designation.

The fact is, there is no theory - now or ever - which explains how the observer's past has anything whatsoever to do with ANY experiment. That includes QM. It is just a blind ad hoc hypothesis thrown out by a few people. So you cannot explain WHY it should apply to entanglement more (or less) than the age of the universe or measurements of c or anything else.
kith
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Jan16-13, 06:58 PM
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Quote Quote by DrChinese View Post
The fact is, there is no theory - now or ever - which explains how the observer's past has anything whatsoever to do with ANY experiment.
Usually, we have a system S and an observer O measuring some observable of the system. As soon as we consider the combined system S+O as a physical system (which may be observed by another observer O'), we acknowledge that the current state of S and O influences the evolution of the combined system. This evolution may of course include interactions between S and O. What's wrong with this kind of thinking?
nanosiborg
#36
Jan17-13, 12:42 AM
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Quote Quote by DrChinese View Post
That does not make sense, kith. It doesn't matter whether a theory is classical or not! That is a completely arbitrary designation.

The fact is, there is no theory - now or ever - which explains how the observer's past has anything whatsoever to do with ANY experiment. That includes QM. It is just a blind ad hoc hypothesis thrown out by a few people. So you cannot explain WHY it should apply to entanglement more (or less) than the age of the universe or measurements of c or anything else.
I agree with this. The assumption of superdeterminism is useless for understanding why Bell's lhv formulation is inappropriate (ie., why it produces incorrect predictions) for modeling Bell tests.


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