Bell experiment would somehow prove non-locality and information FTL?

[JesseM]

Yes; its called determinism. That is why, in a Bell-test, when the detectors have the (say) same settings, the outcomes are identical (++, ++, --, ++, --, ...) (with no evidence of DrC's HUP). NEVERTHELESS, each perturbed particle (with its revealed observable) now differs from its pre-measurement state (with its often-hidden beable).

If the results follow in a determistic way from the choice of measurement + the preexisting state, what difference does it make if the revealed observable differs from the preexisting state? In terms of the proof it's interchangeable...if we say the preexisting state is {A+, B+, C-}, you could either say that this means the particle's state was spin-up on axis A, spin-up on axis B, and spin-down on axis C, and that the measurement just reveals these preexisting spins, or you define the state {A+, B+, C-} to mean "the particle is in a state X such that if it is perturbed by a measurement on the A-axis, the deterministic outcome will be that it is measured to be spin-up; if it is perturbed by a measurement on the B-axis, the deterministic outcome will be that it is measured to be spin-up; and if it is perturbed by a measurement on the C-axis, the determistic outcome will be that it is measured to be spin-down." Note that this second definition doesn't make any assumptions about what state X was actually like, just that the combination of the preexisting state X and a given measurement Y will always determistically lead to the same outcome.

 

[heusdens]

Yes; its called determinism. That is why, in a Bell-test, when the detectors have the (say) same settings, the outcomes are identical (++, ++, --, ++, --, ...) (with no evidence of DrC's HUP). NEVERTHELESS, each perturbed particle (with its revealed observable) now differs from its pre-measurement state (with its often-hidden beable).

AND NOTE: Prior to one or other measurement, our knowledge of the state is generally insufficient for us to avoid a probablistic prediction; here 50/50 ++ XOR --. So, from locality, determinism is the underlying mechanism that delivers such beautifully correlated results from randomly delivered twins; HUP notwithstanding.

HUP stands for?

 

[wm]

If the results follow in a determistic way from the choice of measurement + the preexisting state, what difference does it make if the revealed observable differs from the preexisting state? In terms of the proof it's interchangeable...if we say the preexisting state is {A+, B+, C-}, you could either say that this means the particle's state was spin-up on axis A, spin-up on axis B, and spin-down on axis C, and that the measurement just reveals these preexisting spins, or you define the state {A+, B+, C-} to mean "the particle is in a state X such that if it is perturbed by a measurement on the A-axis, the deterministic outcome will be that it is measured to be spin-up; if it is perturbed by a measurement on the B-axis, the deterministic outcome will be that it is measured to be spin-up; and if it is perturbed by a measurement on the C-axis, the determistic outcome will be that it is measured to be spin-down." Note that this second definition doesn't make any assumptions about what state X was actually like, just that the combination of the preexisting state X and a given measurement Y will always determistically lead to the same outcome.

The difference is that one view is true and helpful, the other misleading and confusing:

The difference is that the post-measurement state is ''manufactured'' from the pre-measurement state by the chosen observation. One pre-state, differing manufacturing processes, differing post-states. Example: if I deliver vertically-polarised photons to you, you can manufacture various alternative polarisations from the choice of detector setting (manufacturing process).

Did the delivered (pristine, pre-measurement, virginal) photons carry this countable-infinity of polarisations BEFORE you ''measured'' (= manufactured = processed) them?

 

[wm]

HUP stands for?

Heisenberg's Uncertainty Principle.

 

[heusdens]

Maybe this is a very naive attempt, but what if we just create 3 random streams

(that is: each stream is random of it self, and in relation to the other, so that neither one can predict any of the data of the same stream, or of the other stream, nor can one when combining any two streams or even all 3 streams, extract any usefull data from it).

of data, which are labelled a,b,c, corresponding to detector settings A,B,C of Alice and Bob. If Alice picks A and Bob picks A, they get the same data, likewise for B and for C. However if Alice picks a different setting as Bob, they get random outcomes. Does that match the criteria for breaking the inequality, or not?

Does anybody comment this?

Looks like if Alice and Bob choose unequal detector setting they now get random values, so with probability of equal (or unequal) values of 50%.

This is larger then we expect from Bell Inequality (1/3) ?

 

[JesseM]

Maybe this is a very naive attempt, but what if we just create 3 random streams

(that is: each stream is random of it self, and in relation to the other, so that neither one can predict any of the data of the same stream, or of the other stream, nor can one when combining any two streams or even all 3 streams, extract any usefull data from it).

of data, which are labelled a,b,c, corresponding to detector settings A,B,C of Alice and Bob. If Alice picks A and Bob picks A, they get the same data, likewise for B and for C. However if Alice picks a different setting as Bob, they get random outcomes. Does that match the criteria for breaking the inequality, or not?

No, it won't violate either inequality. Suppose at the moment Bob is picking a letter, his computer is receiving + from stream a, + from stream b, and - from stream c (meaning that if he types A he'll see + on the screen, if he types B he'll see +, and if he types C he'll see -). This means that at the same moment Alice's computer must be getting - from her stream a, - from her stream b, and + from her stream c.

You could represent this by saying that at this moment, Bob's computer is primed in state {a+, b+, c-} based on its data streams, and Alice's computer is primed in state {a-, b-, c+} based on its own streams. Likewise, at another moment the Bob's computer could be primed in state {a-, b+, c-} and Alice's would be primed in state {a+, b-, c+}. This is just like the assumption that each particle they receive has a definite spin on all three axes.

So, consider the inequality:

(number of trials where Bob's computer primed in a state that includes a+ and b-) plus (number of trials where Bob's computer primed in a state that includes b+ and c-) is greater than or equal to (number of trials where Bob's computer primed in a state that includes a+ and c-)

If you think about it, you should be able to see why this must be true. On every trial where Bob's computer is "primed in a state that includes in a state that includes a+ and c-", the computer must either be primed in the state {a+, b+, c-} or in the state {a+, b-, c-}, there aren't any other possibilities. If it's primed in the state {a+, b-, c-}, then this must also be a trial in which it's "primed in a state that includes a+ and b-", so it contributes to the number on the left side of the inequality as well as the number on the right side. But if it's primed in the state {a+, b+, c-}, then this must be a trial in which it's "primed in a state that includes b+ and c-", so it still contributes to the left side of the inequality. Either way, every trial that contributes 1 to the right side of the inequality must also contribute 1 to the left side, so the number on the left side must always be greater than or equal to the right side.

Of course, Bob only finds out the state of one of his three streams on each trial. But under the assumption that Alice's stream at a given moment is always the opposite of Bob's, if Bob types A and gets + while Alice types B and gets +, that implies that Bob's computer must have been primed in a state that includes a+ and b-. So assuming Alice and Bob both pick letters randomly with equal frequencies, we can rewrite the inequality in terms of their actual measurements as:

(probability that Bob types A and gets + while Alice types B and gets +) plus (probability that Bob types B and gets + while Alice types C and gets +) is greater than or equal to (probability that Bob types A and gets + while Alice types C and gets +).

If the computers decide their output based on your data-stream method, the inequality above should be satisfied over a large number of trials.

And to understand why the other inequality I mentioned (the one saying that if they pick different letters, the probability of their getting opposite answers should be greater than or equal to 1/3) will be satisfied too using your datastream method, here's a slight modification of my post #10:

if we imagine Bob's computer is primed in state {a+, b-, c+} and Alice's computer is primed in state {a-, b+, c-} then we can look at each possible way that Alice and Bob can randomly choose different letters to type, and what the results would be:

Bob picks A, Alice picks B: same result (Bob gets a +, Alice gets a +)

Bob picks A, Alice picks C: opposite results (Bob gets a +, Alice gets a -)

Bob picks B, Alice picks A: same result (Bob gets a -, Alice gets a -)

Bob picks B, Alice picks C: same result (Bob gets a -, Alice gets a -)

Bob picks C, Alice picks A: opposite results (Bob gets a +, Alice gets a -)

Bob picks C, Alice picks picks B: same result (Bob gets a +, Alice gets a +)

In this case, you can see that in 1/3 of trials where they pick different letters, they should get opposite results. You'd get the same answer if you assumed any other heterogeneous primed state where there are two of one outcome and one of the other, like {a+, b+ c-}/{a-, b-, c+} or {a+, b-, c-}/{a-, b+, c+}. On the other hand, if you assume a homogenous primed state where each of the three datastreams tells the computer to give the same output, like {a+, b+, c+}/{a-, b-, c-}, then of course even if Alice and Bob pick different keys to type they're guaranteed to get opposite results on their screen probability 1. So if you imagine that when multiple primed states are generated by the datastreams over the course of many trials, on some fraction of trials there are inhomogoneous primed states like {a+, b-, c-}/{a-, b+, c+} while on other trials you have homogoneous primed states like {a+, b+, c+}/{a-, b-, c-}, then the probability of getting opposite answers when they type different letters should be somewhere between 1/3 and 1. 1/3 is the lower bound, though--even if the datastreams were made in such a way that the computer was in inhomogenous primed states in 100% of the trials, it wouldn't make sense for Alice and Bob to get opposite answers in less than 1/3 of trials where they typed different letters.

 

[JesseM]

Does anybody comment this?

Looks like if Alice and Bob choose unequal detector setting they now get random values, so with probability of equal (or unequal) values of 50%.

This is larger then we expect from Bell Inequality (1/3) ?

The Bell inequality says that if they type different letters (choosing different streams under your method), the probability that they'll get opposite answers should be greater than or equal to 1/3. It's only violated if the probability ends up being smaller than 1/3.

 

[JesseM]

The difference is that one view is true and helpful, the other misleading and confusing:

The difference is that the post-measurement state is ''manufactured'' from the pre-measurement state by the chosen observation. One pre-state, differing manufacturing processes, differing post-states. Example: if I deliver vertically-polarised photons to you, you can manufacture various alternative polarisations from the choice of detector setting (manufacturing process).

Did the delivered (pristine, pre-measurement, virginal) photons carry this countable-infinity of polarisations BEFORE you ''measured'' (= manufactured = processed) them?

But the question is irrelevant to the proof the Bell inequality. As long as you assume that the result of each possible measurement is predetermined, and that this includes the fact that the two experimenters are predetermined to get opposite spins if they measure the same axis, then Bell inequalities follow from this, there's nothing in the proof that requires you to assume you are just measuring a preexisting spin on that axis without perturbing it.

 

[wm]

But the question is irrelevant to the proof the Bell inequality. As long as you assume that the result of each possible measurement is predetermined, and that this includes the fact that the two experimenters are predetermined to get opposite spins if they measure the same axis, then Bell inequalities follow from this, there's nothing in the proof that requires you to assume you are just measuring a preexisting spin on that axis without perturbing it.

Consider Bell (1964) and identify the un-numbered equations between (14) and (15) as (14a), (14b), (14c).

1. I'd welcome your detailed comment on Bell's move from (14a) to (14b), bound by your claim that there's nothing in the proof that requires you to assume you are just measuring a pre-existing spin on that axis without perturbing it.

2. That is, inter alia, please specify the FUNCTIONS A and B that satisfy Bell's move.

Thanks, wm

 

[JesseM]

Consider Bell (1964) and identify the un-numbered equations between (14) and (15) as (14a), (14b), (14c).

1. I'd welcome your detailed comment on Bell's move from (14a) to (14b), bound by your claim that there's nothing in the proof that requires you to assume you are just measuring a pre-existing spin on that axis without perturbing it.

2. That is, inter alia, please specify the FUNCTIONS A and B that satisfy Bell's move.

Thanks, wm

I don't have this paper--is it online? In any case, if you're not worried about perfect mathematical rigor it's quite easy to prove that various versions of Bell's inequality must hold if particles have predetermined responses to all measurements and locality is obeyed--can you identify a flaw in the short proofs I gave in post #126 to heusdens, for example? Based on these sorts of simple proofs, I'm confident that even if Bell assumed measurements simply revealed preexisting spins (or whatever variable is being measured) in his original proof, it would be a fairly trivial matter to modify the proof to remove this assumption.

 

[vanesch]

Yes, but the point that you miss then that any detectable object is spatially spread, this would then mean also that it contains "independent objects" - by your same reasoning! - rather as one object! So, if on that account the world is treated, independent objects wouldn't exist!

Indeed, relativistically, extended solid objects do not exist as single entity - except for hypothetical lightlike objects such as strings. But your "solid body" does not have a relativistic meaning as such ; it must always be seen as a bound state of smaller objects.
One reason this can be seen is that otherwise, the extended object would be able to transmit signals faster than light. Relativity requires that the speed of sound in a body must be smaller than the speed of light, which implies finite elasticity and hence "independent parts" which can move wrt each other.

So when can we know wether an object - any object at all! - can be treated as one object, or as a constellation of independent objects?

The "one object" treatment is always an approximation, and it depends on the situation at hand to know whether the approximation will yield good enough results for the application at hand.

Read the longer post which refutes MWI on logic grounds.

It's erroneous, as I showed...

1 cloud + 1 cloud might equal 1 cloud, that is the clouds themselves may merge, and what we previously saw as two separate cloud, becomes one new cloud.
Still in numbers/abstract form, 1+1=2 still applies, only the underlying reality we speak about, does not hold on to this formality.
This is of course because what for logic is a requirement, that we can speak of independend and seperable "objects", is not a requirement for the world itself.

There is no requirement in logic to "talk about independent objects" or whatever. Logic is about the truth of falsehood of statements.
Logic is rather: I saw two clouds. I saw a dog. Hence, the statement: "I saw two clouds and I saw a dog" is also true. It doesn't imply any kind of conservation law about statements of physical objects.

For example you could make a logical valid statement about an object and from the dynamics of the situation you could describe it's motion, which would incorporate making statements about where in the world the object would need to be found at any given time.
So this formaly would then state that an object at some given time would either be at location x, or not be a location x, but not both or something else.
So, there you already see the limitations of such formalism.

This is again not a requirement of logic, but of the hypotheses that are build into the physical theory at hand, and if the result turns out not to be correct, the hypotheses have a problem.
You need to make many hypotheses even to be able to make the statement that to an object is associated a point in an Euclidean space. This doesn't need to be so at all, but you can make that hypothesis. You also need to make the hypothesis that this association is unique. If you then find that you should assign two points in a Euclidean plane to a single object, then this simply means that your hypothesis of this unique association was erroneous. That's btw what quantum theory does.

The question then is: is there a complete and consistent description of the world possible at all, in which what we recognize on abstract/formal and mathematical grounds as true, also is true in the real world?

Nobody knows. The only thing we know is that there are abstract/formal mathematical theories which give us rather good results concerning statements of observation. It would probably be naive to think that they are ultimately correct, but they are good enough. It now seems that we have a mathematical model (build upon standard logic) which gives us a prediction of EPR situations, which seems to be in agreement with observation. As such, all we can do, is done.

The result can be more generalized to formalized systems.
But I'm not exactly sure about what constraints the formalized or formalizable system must have.

That's simple: it must make correct predictions of observations ! That's the single one and only requirement.

No, this is completely wrong, in the sense that the limitations of formal logic were discovered long time before quantum mechanics showed us these paradoxes.

But quantum mechanics doesn't show us any "paradox" in the sense that we derive two different and contradicting results WITHIN QUANTUM THEORY. Quantum theory (which is a formal theory based upon standard logic) gives us unambiguously the correct result, which is also experimentally observed. This shows us that there is no "problem of logic" there. We only have difficulties *believing* what quantum mechanics tells us, that's all. So we try to explain it DIFFERENTLY, with EXTRA REQUIREMENTS. And THEN we run into troubles.

So given that a formal theory based upon standard logic gives us the correct predictions, it would be a strange reaction to want to change standard logic (or the theory that makes the correct predictions). It simply means that we misjudged the explanatory power of the theory at hand, by thinking that it was "just a statistical tool but which must have some or other underlying mechanism". It is the last statement which fails.

 

[wm]

I don't have this paper--is it online? In any case, if you're not worried about perfect mathematical rigor it's quite easy to prove that various versions of Bell's inequality must hold if particles have predetermined responses to all measurements and locality is obeyed--can you identify a flaw in the short proofs I gave in post #126 to heusdens, for example? Based on these sorts of simple proofs, I'm confident that even if Bell assumed measurements simply revealed preexisting spins (or whatever variable is being measured) in his original proof, it would be a fairly trivial matter to modify the proof to remove this assumption.

1. I think the paper is available from DrC's website. I'm happy to wait for you to access it

2. I suggest we stick with it (ie, Bell 1964) because it is well-known that Bell's theorem is applicable to dirty-socks, down-hill skiers, computers, students, ... .

3. So (I hope), the question we are addressing is this: Why is Bell's theorem (BT) invalid for the original case-study (EPRB; ie Bell 1964) and similar settings??

4. That is: Why are Bellian inequalities FALSE in the settings that BT was designed to illuminate? (Hint: Examine those totally simplistic settings in which it holds!)

If this is not OK, or you have a better idea, let me know. Regards, wm

 

[JesseM]

2. I suggest we stick with it (ie, Bell 1964) because it is well-known that Bell's theorem is applicable to dirty-socks, down-hill skiers, computers, students, ... .

Although I tailored the short proofs I gave above to a particular thought-experiment, it's quite trivial to change a few words so they cover any situation where two people can measure one of three properties and they find that whenever they measure the same property they get opposite results. If you don't see how, I can do this explicitly if you'd like.

3. So (I hope), the question we are addressing is this: Why is Bell's theorem (BT) invalid for the original case-study (EPRB; ie Bell 1964) and similar settings??

I am interested in the physics of the situation, not in playing a sort of "gotcha" game where if we can show that Bell's original proof did not cover all possible local hidden variable explanations then the whole proof is declared null and void, even if it would be trivial to modify the proof to cover the new explanations we just thought up as well. I'll try reading his paper to see what modifications, if any, would be needed to cover the case where measurement is not merely revealing preexisting spins, but in the meantime let me ask you this: do you agree or disagree that if we have two experimenters with a spacelike separation who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like), then if they always get opposite answers when they make the same measurement on any given trial, and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement with no violations of locality allowed (and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event, as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation), then the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

 

[DrChinese]

Sorry, what does HUP stand for?

I do not exactly conform myself to any of such explenations, because as for one thing, they basically shift the problem to some other department of physics, without resolving it (we would in some of these explenations for example have to reconsider relativity since it undermines it basic premisses, or otherwise undermine other basic premisis about our understanding of the world, or introduce arbitrary new phenomena, like many worlds, etc.).

So, actually I am trying to figure things out in a more substantial way.

The refereces to dialectics was meant to give a clue to this, because dialectics tries to escape from this one-sidedness of these formal mathematical explenations, and instead give a full picture of what can be regarded as truth.

[ Perhaps not everyone is happy with that, cause dialectics is not specifically related to quantum physics, and such discussions are meant to occur in the forums meant for philosophic topics, yet most of such threads are rather worthless, since most topics are rather un concrete. ]

Sorry, I ususally give my abbreviations...

HUP = Heisenberg Uncertainty Principle.

As to more "substantial" treatments... we are all interested in this as well. I caution you that it is an error to think that this has not been explored in depth by many folks. If you check the Preprint Archives you will find hundreds of papers like Travis Norsen's (well not exactly like...) - in the past year alone - that look at Bell from every conceivable angle. Most of these papers claim to add new insight, but precious few are likely to be remembered years from now.

Also, please keep in mind that there are forum guidelines regarding personal theories. You keep referring to "dialectics" which I have never seen mentioned in regards to Bell's Theorem. Unless this has some direct bearing on this thread topic, I would recommend staying away from this.

 

[DrChinese]

To bring clarity to the discussion, let's allow that there is: naive realism, strong realism, EPR realism, Bell realism, Einstein realism, ... (In my view: naive realism = strong realism = EPR realism = Bell realism = silliness.)

Now I am not aware that Einstein ever endorsed the other versions, so could you let me have the full quotes and sources that you rely on?

Note: We are not looking for Einstein's support of ''pre-measurement values'' BUT for the idea that measurement does NOT perturb the measured system (for that is the implicit silliness with EPR, Bell, etc). Einstein (1940, 1954) understood that the wave-function plus Born-formula related to the statistical prediction of ''measurement ourcomes'' and NOT pre-measurement values.

Einstein of course supported what you call "pre-measurement values". This is because he said:

"I think that a particle must have a separate reality independent of the measurements. That is: an electron has spin, location and so forth even when it is not being measured. I like to think that the moon is there even if I am not looking at it."

I cannot find a date and exact source.

 

[DrChinese]

DrC, could you please expand on this interesting position? As I read it, you have a local comprehension of Bell-test results? (I agree that there is such, but did not realize that you had such.)

However, your next sentence is not so clear: By definition, an observable is observation dependent, so you seem to be saying that there are no underlying quantum beables?? There is no ''thing-in-itself''??

Personally: I reject BM on the grounds of its non-locality; and incline to endorse MWI with its locality, while rejecting its need for ''many worlds''.

I tend to support a rejection of realism rather than a rejection of locality (in order to reconcile with Bell's Theorem). I do not know if there are beables, but there definitely are observables. I do not, for instance, if there is a one-to-one mapping of observables to beables. My guess would be that there is not, since there can be nearly an infinite number of observables for a single particle,

I definitely do not agree that it is a closed question (i.e. by definition) an observable must be observer dependent. That is one of the questions we seek the answer to. I happen to think it is, but I do not expect others to necessarily agree with this position. I believe that the Heisenberg Uncertainty Principle essentially calls for this position.

 

[DrChinese]

Yes; its called determinism. That is why, in a Bell-test, when the detectors have the (say) same settings, the outcomes are identical (++, ++, --, ++, --, ...) (with no evidence of DrC's HUP). NEVERTHELESS, each perturbed particle (with its revealed observable) now differs from its pre-measurement state (with its often-hidden beable).

Ah, but the Heisenberg Uncertainty Principle (HUP) is quite present in such cases! Note that we cannot learn MORE information than the HUP allows about one particle by studying its entangled twin!

 

[DrChinese]

1. I think the paper is available from DrC's website. I'm happy to wait for you to access it

EPR, Bell & Aspect: The Original References (in PDF Format)

 

[DrChinese]

Does anybody comment this?

Looks like if Alice and Bob choose unequal detector setting they now get random values, so with probability of equal (or unequal) values of 50%.

This is larger then we expect from Bell Inequality (1/3) ?

As JesseM has pointed out: you actually get values as low as 25% in actual Bell test situations - not the 50% you imagine. The reason is that there is (anti)correlation between the results of unequal detector settings. So I hope we all see this point.

I have a web page that shows the cases (AB, BC, AC as red/yellow/blue) and may help anyone to visualize the situation:

Bell's Theorem with Easy Math

 

[heusdens]

As JesseM has pointed out: you actually get values as low as 25% in actual Bell test situations - not the 50% you imagine. The reason is that there is (anti)correlation between the results of unequal detector settings. So I hope we all see this point.

I have a web page that shows the cases (AB, BC, AC as red/yellow/blue) and may help anyone to visualize the situation:

Bell's Theorem with Easy Math

Yeah, the example I gave was obviously wrong, cause I didn't get the equality right and made a too simple approach.

I want to perform some experiments in thought and get everyone's reaction to it. (and please correct me where I get things wrong).

We have the (formally described) setup of two detectors , which each have a setting of A,B,C and each setting (of which only one at a time is used) produces a value as '+' or '-'. Right?
We then also have this source that produces output to both detectors.
We have no prior idea about how the source and detector setting correspond to output values.

First let us imagine, there wasn't a source at all. We then have two 'devices' with settings A,B and C, that produce either a '+' or a '-' when set. Like in the previous case, only one of A, B or C for both devices can be set.
The devices are now some kind of blackbox. We don't know anything about the internals of it, if either the result is produced from something within, or if there is some signal going in. Anyway we get a result.

First we inspect individual data from the detectors.

For both detectors and for every setting we use we get + or - in equal amounts, that is the chance of having either + or - is 50% (or .5).

{is this assumption correct?}

Now we inspect results from both detectors, and see how they compare.

First remarkable thing:

I.

If the detectors have an equal setting, then the results are either ++ or --. The positive correlation (= same result from detectors) is 100% (or 1).

{Question:
a. can we still assume that each individual detector produces a really random result?
b. does the correlation only hold for exactly simultanious results?
c. the cances of either ++ or -- are 50% each ??

It would be weird if b and/or c does not hold and a still holds...
}

Second remarkable thing:

II.

If the detectors have an unequal setting, then we find that results of +- and -+ , that is negative correlation (= unequal result from detectors) happening with a change of 25% (or .25).

{Same questions as above, but for c now: chances of +- or -+ are 50% each??}

Now how can we explain this??

We first try to find independent explenations for the separate remarks.

First let us look at I. (detector settings equal)

We can make all kind of suggestions about how this could be the case.

For example, we could assume that both detectors have an exactly the same algorithm with which to produce the data. Each data separate is a random result, but the results of both sides are always the same. The algorithm works because what the detectors still have in common is time and possible also other easily overlooked common sources (external light source, common to both observers, and other such common sources).

{the assumption is here that if we take the detector results 'out of sync', for instance data of detector 1 at time t and data of detector 2 at time t + delta t (delta t>0) this results aren not produced; -- is that a realistic assumption??}

A less trivial approach is to suspect that detector 1 has received a signal from detector 2, and knows that it setting is the same, somehow, and can produce the positive correlation result. The signal need not be instantanious to explain it (if the signal contains the timestamp). The weird thing also for this explenation is that it breaks the symmetrie, since we could also suppose detector 2 somehow gets a signal from detector 1. If we assume symmetry, both signals would occur for this explenation. But then how could we get the correlation as we see, based on both the signals? In the a-symmetric case, we would have no trouble to find a possibility for correlation, since then only one detector would have to adjust to produce the corresponding output. It is more difficult this can happen for the symmetric case (both adjustments would cancel out), but if we assume that the setup is symmetric, we have to assume just that. This however can then be showed to be equal to the case in which both detectors receive a simultanious signal that the detector settings match, so both detectors can make equal and simultanious adjustments. This is like postulating that exactly in the middle (in order for simultanious arrival) between those detectors is a receiver/transmitter, that receives the detector signals, and transmit back wether they are equal or not.Now we look at II. (detector settings unequal)

We get a .25 chance that we have unequal results (+- or -+).
This is same as a .75 chance for having equal results (++ or --).

Both detectors individually (if I assume correctly) still produce random results, but they results from both detectors are now equal in 3 out of 4 on average, which is the same as that they are unequal in 1 out of 4 on average.

In principle we can now suppose that same kind of things that were supposed to explain the outcomes in the previous case, also happen here, with the exception that the output that is generated is not always ++ or --, but only in 3 out of 4 cases.
This is then just assuming a different algorithm to produce that result.Now we try to combine explenations I and II.

For each explenation I and II seperately we could assume that something purely internal generated the outcomes. But if I and II occur, we have no way of explaining this.
So, this already urges us to assume that the detector states (settings) are getting transmitted to a common source, exactly in the middle (that is in the orthogonal plane which intersects the line between both detectors in the middle point of that line).

If we can also verify that in the experiment (by placing detectors very far away) this hypothetical signal speed is like instantanious, by changing the dector setting simultaniously, and get instant correlations.
To cope with that, the hypothetical assumption is that this is like a signal that travels back in time from the detector to a common source, and travels forward in time to the detector.

Conclusion:
Although we did not setup this imaginary experiment with this common source, it already follows from the results of this experiment, that such a common source must be assumed, which communicates back and forth between the detectors.

 

[DrChinese]

I.

If the detectors have an equal setting, then the results are either ++ or --. The positive correlation (= same result from detectors) is 100% (or 1).

{Question:
a. can we still assume that each individual detector produces a really random result?
b. does the correlation only hold for exactly simultanious results?
c. the cances of either ++ or -- are 50% each ??

It would be weird if b and/or c does not hold and a still holds...
}

Second remarkable thing:

II.

If the detectors have an unequal setting, then we find that results of +- and -+ , that is negative correlation (= unequal result from detectors) happening with a change of 25% (or .25).

{Same questions as above, but for c now: chances of +- or -+ are 50% each??}

a. Each detector sees a random pattern, always.
b. There is a coincidence window used for detections, and the setup is calibrated so the middles of the windows are equivalent. But it does not actually matter at all if the detections are simultaneous.
c. The likelihood of + or - at any detector is always 50%. Just to be specific: when we have Type I PDC entangled photons, then we have perfect correlations (at identical settings). So you get ++ or -- almost all of the time.

In the II.c. case, there are 4 permutations: ++/-- and +-/-+ ) i.e. matches and non-matches. Matches can drop as low as 25% for certain settings (usually specified as 0, 120, 240 degrees).

 

[DrChinese]

Now how can we explain this??

We first try to find independent explenations for the separate remarks.

First let us look at I. (detector settings equal)

We can make all kind of suggestions about how this could be the case.

1. For example, we could assume that both detectors have an exactly the same algorithm with which to produce the data. Each data separate is a random result, but the results of both sides are always the same. The algorithm works because what the detectors still have in common is time and possible also other easily overlooked common sources (external light source, common to both observers, and other such common sources).

{the assumption is here that if we take the detector results 'out of sync', for instance data of detector 1 at time t and data of detector 2 at time t + delta t (delta t>0) this results aren not produced; -- is that a realistic assumption??}

A less trivial approach is to suspect that detector 1 has received a signal from detector 2, and knows that it setting is the same, somehow, and can produce the positive correlation result. The signal need not be instantanious to explain it (if the signal contains the timestamp). The weird thing also for this explenation is that it breaks the symmetrie, since we could also suppose detector 2 somehow gets a signal from detector 1. If we assume symmetry, both signals would occur for this explenation. But then how could we get the correlation as we see, based on both the signals? In the a-symmetric case, we would have no trouble to find a possibility for correlation, since then only one detector would have to adjust to produce the corresponding output. It is more difficult this can happen for the symmetric case (both adjustments would cancel out), but if we assume that the setup is symmetric, we have to assume just that. This however can then be showed to be equal to the case in which both detectors receive a simultanious signal that the detector settings match, so both detectors can make equal and simultanious adjustments. This is like postulating that exactly in the middle (in order for simultanious arrival) between those detectors is a receiver/transmitter, that receives the detector signals, and transmit back wether they are equal or not.


Now we look at II. (detector settings unequal)

2. We get a .25 chance that we have unequal results (+- or -+).
This is same as a .75 chance for having equal results (++ or --).

Both detectors individually (if I assume correctly) still produce random results, but they results from both detectors are now equal in 3 out of 4 on average, which is the same as that they are unequal in 1 out of 4 on average.

In principle we can now suppose that same kind of things that were supposed to explain the outcomes in the previous case, also happen here, with the exception that the output that is generated is not always ++ or --, but only in 3 out of 4 cases.
This is then just assuming a different algorithm to produce that result.


3. Now we try to combine explenations I and II.

For each explenation I and II seperately we could assume that something purely internal generated the outcomes. But if I and II occur, we have no way of explaining this.

4. So, this already urges us to assume that the detector states (settings) are getting transmitted to a common source, exactly in the middle (that is in the orthogonal plane which intersects the line between both detectors in the middle point of that line).

If we can also verify that in the experiment (by placing detectors very far away) this hypothetical signal speed is like instantanious, by changing the dector setting simultaniously, and get instant correlations.
To cope with that, the hypothetical assumption is that this is like a signal that travels back in time from the detector to a common source, and travels forward in time to the detector.

Conclusion:
5. Although we did not setup this imaginary experiment with this common source, it already follows from the results of this experiment, that such a common source must be assumed, which communicates back and forth between the detectors.

1. Actual experiments are designed to rule out other sources.

2. It actually should be .25 for matches and .75 for mismatches if we have 1.0 for the matches in your I. case. This comes from the cos^2 rule with 120 degrees as the difference in settings.

3. This is correct, we need to consider both of these together.

4. Good, you are seeking possible explanations for the results. And now we find ourselves considering new physical phenomena not otherwise known... such as backwards in time signaling and hypothetical effects derived from previously unknown sources. But these have severe theoretical problems too, since they only appear for entangled particles.

5. It is not a requirement that there is a common source, but that is certainly one possibility.

Your general line of approach is definitely improving.

 

[wm]

Although I tailored the short proofs I gave above to a particular thought-experiment, it's quite trivial to change a few words so they cover any situation where two people can measure one of three properties and they find that whenever they measure the same property they get opposite results. If you don't see how, I can do this explicitly if you'd like. I am interested in the physics of the situation, not in playing a sort of "gotcha" game where if we can show that Bell's original proof did not cover all possible local hidden variable explanations then the whole proof is declared null and void, even if it would be trivial to modify the proof to cover the new explanations we just thought up as well. I'll try reading his paper to see what modifications, if any, would be needed to cover the case where measurement is not merely revealing preexisting spins, but in the meantime let me ask you this: do you agree or disagree that if we have two experimenters with a spacelike separation who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like), then if they always get opposite answers when they make the same measurement on any given trial, and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement with no violations of locality allowed (and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event, as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation), then the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

Dear Jesse,

1. YES; I agree that an experiment with your boundary conditions can deliver the results you claim. But you seem not to agree that a wholly classical experiment with the same boundary conditions can deliver a different result?

2. Could I therefore take up your offer and ask you to present the case where Alice and Bob get identical results for identical settings? (I think it will help us all, especially heusdens, being in my experience easier to discuss and follow than the ''opposite'' case.)

3. Then, given your interest in the physics of the situation, could I ask you to use more maths in your presentation? (That's a long sentence of yours, there, above)

4. So would you be happy to deliver the following inequality:

(1) P(BC = +1|bc) - P(AC = -1|ac) - P(AB = +1|ab) less than or equal to 0?

Here P(BC = +1|bc) denotes the probability of Alice and Bob getting the same result (+1, +1 or -1, -1) under the respective test conditions b and c.

5. (1) is just my way of attempting to standardise the way BT is presented. And I'm happy to accept most boundary conditions; say, consistent with common-sense.

Regards, wm

 

[wm]

Einstein of course supported what you call "pre-measurement values". This is because he said:

"I think that a particle must have a separate reality independent of the measurements. That is: an electron has spin, location and so forth even when it is not being measured. I like to think that the moon is there even if I am not looking at it."

I cannot find a date and exact source.

OK. The point I make is that Einstein realism nowhere prohibits a ''measurement'' perturbing the pristine system.

So (in my view) when he says something like: Physical reality exists independent of substantion and perception he is not prohibiting measurement perturbation, a phenomenon well-known throughout classical and quantum physics. wm

 

[wm]

I tend to support a rejection of realism rather than a rejection of locality (in order to reconcile with Bell's Theorem). I do not know if there are beables, but there definitely are observables. I do not, for instance, if there is a one-to-one mapping of observables to beables. My guess would be that there is not, since there can be nearly an infinite number of observables for a single particle,

I definitely do not agree that it is a closed question (i.e. by definition) an observable must be observer dependent. That is one of the questions we seek the answer to. I happen to think it is, but I do not expect others to necessarily agree with this position. I believe that the Heisenberg Uncertainty Principle essentially calls for this position.

OK; we differ: I reject Bell realism and (so) locality remains unfettered.

To me there must be beables (from BEING), for how else do we deliver observables? By ''deliver'' I allow that sometimes we may deliver the beable ''unperturbed'' (say, charge) and most generally we deliver the beable perturbed (say, polarisation).

By ''observer dependent'' I meant the process whereby a beable becomes an observable. No process, no observable: a closed question, I'd like to think.

And YES, HUP seems applicable (even proof) of my position where quanta are involved. For how can a quantum be given or received without change = perturbation; and how can there be ''observation'' without a quantum change? wm

 

[wm]

Ah, but the Heisenberg Uncertainty Principle (HUP) is quite present in such cases! Note that we cannot learn MORE information than the HUP allows about one particle by studying its entangled twin!

Doc, this seems a strange use of the HUP? It had not occurred to me that you were using it that way.

And surely you are not correct? By testing one particle I can observe its pristine reaction to an a setting. By testing the other particle I can observe its pristine reaction to a b setting. I have THUS learned something MORE about each twin!

Given one particle only, HUP says this is impossible: and I agree; its that quanta again; a particle is pristine ONCE ONLY. BUT: Given two, its surely common-sense that we learn MORE about each?

Am I missing something here? wm

 

[JesseM]

Dear Jesse,

1. YES; I agree that an experiment with your boundary conditions can deliver the results you claim. But you seem not to agree that a wholly classical experiment with the same boundary conditions can deliver a different result?

What do you mean? I didn't suggest a specific experiment, just a general idea of two experimenters doing two measurements where they each have a choice of three measurement settings A,B,C, and there are only two possible results + or - on each measurement. I was asking if you agreed the two inequalities I gave must be satisfied for any classical experiment in which they always get opposite results when they make the same measurement, not if they can be satisfied. If you think they could be violated for a classical experiment, can you present an example, or just point out where you see an error in the proofs I gave?

2. Could I therefore take up your offer and ask you to present the case where Alice and Bob get identical results for identical settings? (I think it will help us all, especially heusdens, being in my experience easier to discuss and follow than the ''opposite'' case.)

Well, if Alice and Bob get identical results for identical settings, this is different from what happens when you measure spins of entangled particles on the same axis, which always have opposite spins on that axis. But sure, I can come up with some inequalities for this case. If they always get the same result for the same setting, then under the conditions I described above, the following inequalities must be obeyed:

* Probability(Bob measures A and gets +, Alice measures B and gets -) plus Probability(Bob measures B and gets +, Alice measures C and gets -) greater than or equal to Probability(Bob measures A and gets +, Alice measures C and gets -)

* When they pick different settings, the probability they get identical results must be greater than or equal to 1/3.

Do you think there would be any classical situation where locality is obeyed but either of these equalities could be violated? Do you see a flaw in the proofs I gave that these inequalities cannot be violated?

3. Then, given your interest in the physics of the situation, could I ask you to use more maths in your presentation? (That's a long sentence of yours, there, above)

Can you tell me which part of the sentence you'd like to see elaborated? Most of it was just discussing the precise conditions of what I mean by a "local hidden variables theory", but if you just think in terms of a classical situation that obeys locality, you'll almost certainly be obeying those conditions. Another shorthand way of stating the condition is that we're assuming the reason both experimenters get identical results for the same setting is because the two objects/signals they receive were prepared in some state where the outcome of each possible measurement was predetermined, and the states are such that they are predetermined to get identical results with the same setting. Again, if you present an example, if it violates one of the conditions I discussed, I'll explain why, but as long as you think in classical terms it's unlikely there'll be a problem.

4. So would you be happy to deliver the following inequality:

(1) P(BC = +1|bc) - P(AC = -1|ac) - P(AB = +1|ab) less than or equal to 0?

Here P(BC = +1|bc) denotes the probability of Alice and Bob getting the same result (+1, +1 or -1, -1) under the respective test conditions b and c.

So I assume P(AC = -1) mean the probability that they each get different results? This is somewhat confusing notation since you're using + and - both for individual results and for whether they both get the same or different results--would you mind if we use S for same and D for different instead? Also, when you say "under respective test conditions b and c", do you just mean that Bob uses the measurement setting B and Alice uses the measurement setting C? Assuming this is what you meant, then yes, I agree your inequality must be satisfied for any local classical experiment. The reason is that in order for (BC = S | bc) (using my S and D notation) to be satisfied, the thing they are measuring must be in one of the following four types of predetermined states:

1. A+ B+ C+
2. A- B+ C+
3. A+ B- C-
4. A- B- C-

(here a predetermined state of type A+ B- C- just means any state in which it is predetermined that if the experimenter chooses setting A she'll get +, if she chooses setting B or C she'll get -).

Now, notice that if it's in predetermined state 2 or 3, it will also satisfy (AC = D| ac), while if it's in predetermined state 1 or 4, it will also satisfy (AB = S| ab). So, any possible predetermined state that satisfies the first must also satisfy one of the other two (and either or both of the other two could be satisfied without satisfying the first, as with a predetermined state of type A+ B+ C-), so this makes it clear that P(BC = S | bc) must be less than or equal to P(AC = D| ac) + P(AB = S| ab) under any local hidden variables theory, regardless of whether measurement disturbs the state or not.

 

[wm]

What do you mean? I didn't suggest a specific experiment, just a general idea of two experimenters doing two measurements where they each have a choice of three measurement settings A,B,C, and there are only two possible results + or - on each measurement. I was asking if you agreed the two inequalities I gave must be satisfied for any classical experiment in which they always get opposite results when they make the same measurement, not if they can be satisfied. If you think they could be violated for a classical experiment, can you present an example, or just point out where you see an error in the proofs I gave? Well, if Alice and Bob get identical results for identical settings, this is different from what happens when you measure spins of entangled particles on the same axis, which always have opposite spins on that axis. But sure, I can come up with some inequalities for this case. If they always get the same result for the same setting, then under the conditions I described above, the following inequalities must be obeyed:

* Probability(Bob measures A and gets +, Alice measures B and gets -) plus Probability(Bob measures B and gets +, Alice measures C and gets -) greater than or equal to Probability(Bob measures A and gets +, Alice measures C and gets -)

* When they pick different settings, the probability they get identical results must be greater than or equal to 1/3.

Do you think there would be any classical situation where locality is obeyed but either of these equalities could be violated? Do you see a flaw in the proofs I gave that these inequalities cannot be violated? Can you tell me which part of the sentence you'd like to see elaborated? Most of it was just discussing the precise conditions of what I mean by a "local hidden variables theory", but if you just think in terms of a classical situation that obeys locality, you'll almost certainly be obeying those conditions. Another shorthand way of stating the condition is that we're assuming the reason both experimenters get identical results for the same setting is because the two objects/signals they receive were prepared in some state where the outcome of each possible measurement was predetermined, and the states are such that they are predetermined to get identical results with the same setting. Again, if you present an example, if it violates one of the conditions I discussed, I'll explain why, but as long as you think in classical terms it's unlikely there'll be a problem. So I assume P(AC = -1) mean the probability that they each get different results? This is somewhat confusing notation since you're using + and - both for individual results and for whether they both get the same or different results--would you mind if we use S for same and D for different instead? Also, when you say "under respective test conditions b and c", do you just mean that Bob uses the measurement setting B and Alice uses the measurement setting C? Assuming this is what you meant, then yes, I agree your inequality must be satisfied for any local classical experiment. The reason is that in order for (BC = S | bc) (using my S and D notation) to be satisfied, the thing they are measuring must be in one of the following four types of predetermined states:

1. A+ B+ C+
2. A- B+ C+
3. A+ B- C-
4. A- B- C-

(here a predetermined state of type A+ B- C- just means any state in which it is predetermined that if the experimenter chooses setting A she'll get +, if she chooses setting B or C she'll get -).

Now, notice that if it's in predetermined state 2 or 3, it will also satisfy (AC = D| ac), while if it's in predetermined state 1 or 4, it will also satisfy (AB = S| ab). So, any possible predetermined state that satisfies the first must also satisfy one of the other two (and either or both of the other two could be satisfied without satisfying the first, as with a predetermined state of type A+ B+ C-), so this makes it clear that P(BC = S | bc) must be less than or equal to P(AC = D| ac) + P(AB = S| ab) under any local hidden variables theory, regardless of whether measurement disturbs the state or not.

1. You have now reframed your question to: ''Do you agree the two inequalities I gave must be satisfied for any classical experiment in which they always get opposite results when they make the same measurement.'' My answer is: NO.

2. The simpler ''same settings, same results'' experiment has to do with photons.

3. You gave the general case which, if valid, must include my specific case. I suggest, since you want to understand the physics, let's be specific and use more maths.

4. SO, to ensure that I am clear on your position: You insist that all classical experiments with the new boundary conditions must satisfy the following Bellian inequality:

(1) P(BC = +1|bc) - P(AC = -1|ac) - P(AB = +1|ab) less than or equal to 0.

Here P(BC = +1|bc) denotes the probability of Alice and Bob getting the same (S) individual results (+1, +1) xor (-1, -1) under the respective test conditions b (Alice) and c (Bob); that is, the first given result is Alice's (here B); the second Bob's (here C); etc.

5. A counter-example would then be a refutation of your position AND Bell's inequality.

PS: To be very clear: I accept the boundary conditions but not the limiting assumptions that you associate with them.

Are we agreed? wm

 

[heusdens]

a. Each detector sees a random pattern, always.
b. There is a coincidence window used for detections, and the setup is calibrated so the middles of the windows are equivalent. But it does not actually matter at all if the detections are simultaneous.
c. The likelihood of + or - at any detector is always 50%. Just to be specific: when we have Type I PDC entangled photons, then we have perfect correlations (at identical settings). So you get ++ or -- almost all of the time.

This last means, with 50/50 chance I suppose??

And if it does not matter if there is coincidence, there is something I don't understand then. My reasoning would be, it matters a lot that they are exactly in sync. How could an out of sync measurements be statistically random (for every separate measurement) AND correlated with the other measurement?

That is not clear to me.

In the II.c. case, there are 4 permutations: ++/-- and +-/-+ ) i.e. matches and non-matches. Matches can drop as low as 25% for certain settings (usually specified as 0, 120, 240 degrees).

And what about the other probabilities?

++ / -- has 50/50 (relative) probablity?

+- / -+ also 50/50 (relative) probability?

(in my imaginary experiment, I just assume to be the case, as well as that the negative correlated fraction has probability of 25%)

 

[JesseM]

1. You have now reframed your question to: ''Do you agree the two inequalities I gave must be satisfied for any classical experiment in which they always get opposite results when they make the same measurement.'' My answer is: NO.

2. The simpler ''same settings, same results'' experiment has to do with photons.

3. You gave the general case which, if valid, must include my specific case. I suggest, since you want to understand the physics, let's be specific and use more maths.

4. SO, to ensure that I am clear on your position: You insist that all classical experiments with the new boundary conditions must satisfy the following Bellian inequality:

(1) P(BC = +1|bc) - P(AC = -1|ac) - P(AB = +1|ab) less than or equal to 0.

Here P(BC = +1|bc) denotes the probability of Alice and Bob getting the same (S) individual results (+1, +1) xor (-1, -1) under the respective test conditions b (Alice) and c (Bob); that is, the first given result is Alice's (here B); the second Bob's (here C); etc.

Right, assuming we've switched from the assumption that Bob and Alice always get opposite results when they perform the same measurement to your new assumption that they always get the same result when they perform the same measurement. Again though, it's really confusing to have +1 represent both a possible result of one person's measurement and an outcome where they both got the same results, so I suggest using my notation S and D instead.

5. A counter-example would then be a refutation of your position AND Bell's inequality.

PS: To be very clear: I accept the boundary conditions but not the limiting assumptions that you associate with them.

Are we agreed? wm

By "boundary conditions" you mean the things I said about the two experimenters having three properties to measure, always getting the same results when they pick the same property, and with the assumption that only classical phenomena obeying locality are involved? (This is a nonstandard use of the phrase 'boundary conditions', which usually refers to conditions on the spatial or temporal boundary of a physical system, like the system's initial conditions.) And by "limiting assumptions" you mean my claim that the various inequalities must necessarily hold true given these conditions? (Of course this was not an 'assumption', it was something I tried to give a proof for.) If so, then yes, I agree. If you think you have a counterexample, or see a flaw in the short proof I gave, please present it.