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

[vanesch]

A stream consists of elements, and an element and detector setting gives an outcome/result. We do not assume we have any further knowledge about elements. Just that the stream characteristics when different combinations of detector settings are used, lead to different results and correlations.

Can we model your setup as follows ?

We have a sequence of mathematical objects of your choice, drawn from a set S:

s: N -> S: i -> s(i).

For each element s of S, we have 6 operators A(1), A(2), A(3), B(1), B(2) and B(3) which can act upon s, and which produce a result + or -. There's no need in making the operators stochastical, as the elements of S can contain all random numbers one needs.

There are two mappings, a and b, from N into the set {1,2,3}:

a: N -> {1,2,3}: i -> "alice choice"
b: N -> {1,2,3}: i -> "bob choice"

The 3-some (s,a,b) gives us an "experimental result", which is given by the sequence of points:
exp: N -> {+,-} x {+,-}: i -> (A(a(i)) s(i) ; B(b(i)) s(i))

That is, we have the table of paired results by Alice and Bob, which is a list of pairs ++, +-, -+ and --, which is given by the act of A(a(i)) (with i-th choice by Alice) onto the element s(i) of our stream etc...

A priori, there only needs to be a definition of A(a(i)) and B(b(i)) on the i-th element of s ; that is, the other outcomes do not need to be defined.

It is the stream exp(i) which needs to satisfy the correlation and boundary conditions for sufficiently long series 1,2,3...,i,...

However, given that it is an assumption in the Bell derivation that the source (which is the mapping s) doesn't know in advance what are going to be the "drawn operators", the above condition needs to be satisfied:
- for one and the same s mapping
- for ALL possible mappings a and b.

This assumption is the fundamental assumption of Bell. If we might change the s-mapping for each different mapping a or b (which would mean that somehow, the "source knew of the choices of the detectors"), or if we had to pick mapping b AS A FUNCTION OF MAPPING a and s, (which would mean that b somehow knew of the a-choice) or if the operators A(1), ... were function of i or of a or of b, then there would be "communication" between the alice and the source, or bob and the source, or alice and bob, which is, BY ASSUMPTION, not supposed to happen.

It is this which allows us to:
1) have the same operators A(1), ... B(3) for all elements of s.
2) to require that the same single mapping s must give same statistics for ALL choices of a and b.

The proof is now rather simple:
consider first the series:
a: i -> 1 for all i
b: i -> 1 for all i

We now have to have that A(1) s(i) = B(1) s(i) for all i.
So this means that for all i:
A(1) s(i) is defined, B(1) s(i) is defined, and moreover they are equal.
We can hence do away with the operator B(1), because it is the same as A(1) for the given s(i).
We also have to have that for half of the i-values, A(1) s(i) = + and for half of the i-values, A(1) s(i) = -.

Similar for the choice:
a: i -> 2 for all i
b: i -> 2 for all i.

etc...

We can hence conclude:
for all i: the three operators A(1), A(2) and A(3) must be defined for s(i), moreover, they are identical to the B(1), B(2) and B(3) values for s(i), hence no need for a specific different notation.

Given the first N elements of s(i) (N big enough to be statistically significant), we can hence write down a table, which contains the number of times we have, in the set {s(1), s(2), ... s(N)}

1) the case A(1)s(i) = + AND A(2) s(i) = + AND A(3) s(i) = +

Say that it is N1

2) the case A(1)s(i) = + AND A(2) s(i) = + AND A(3) s(i) = -

say that it is N2

...

8) the case A(1)s(i) = - AND A(2) s(i) = - AND A(3) s(i) = -

say that it is N8

N1 + N2 + N3 + ... + N8 = N of course...

and we're back to our initial table of 8 possibilities, from which we derive a Bell inequality for the numbers Ni, and from which we derive an equality for the observed correlations if we make one more assumption:

That the series a and the series b are randomly picked streams which are uncorrelated with s.

This is the assumption against superdeterminism. It is nothing else but the assumption that "randomly and independently" picking a "polarizer direction" picks out a fair sample of the overall population, and hence allows us to find the correct "population correlation" and is not biased in a way.

 

[calamero]

Hi Calamero, and welcome to PhysicsForums.

Thanks Dr Chinese

QM does not "ignore" the history. There simply isn't any. Each and every observation puts a particle into a new eigenstate. Once that happens, the particle has no "memory" of earlier states. I know that does not sound reasonable. Neither is it "reasonable" that a pair of entangled particles do NOT have definite spin at the time they are created.

So thers no such thing as evolution in nature?..in order to evolve then nature must have regard to past events correct? Qm is meant to describe nature / reality?

That is why Bell's Theorem is so important. I will repeat the generally accepted conclusion of Bell:

No physical theory of local Hidden Variables can ever reproduce all of the predictions of Quantum Mechanics.

When bell said that he is talking from the viewpoint of qm where after doing calculations of probabilities etc so many outcomes are realized..The hidden variables are only attached to nature by him to explain qm results that may themselves be wrong..Thats the problem he's asking nature to match the results when qm itself admits it doesn't know everything so that argument doesn't hold?

Keep in mind that experiments support the predictions of Quantum Mechanics and do NOT support the predictions of local Hidden Variables.

see above

You must abandon your assumptions about either locality (speed of light is an upper limit) or realism (particles have definite attributes at all times).

Yes i accept no particles have definite attributes all the time becuse people say quanta is the minimum observable action that one can observe so its fuzzy because its "alive" say...BUT forget everything else for now, i know that if i find out what a is i know what b is ..THATS A FACT?...it will never change?...so if we are dealing in facts then surely that is the only one qm has given us to date?...ok I am sure this more but loolkking at it in this expirement..thats the only fact we assertained so far..the rest is theory

I hope this helps.

Yep thanks drchinese its gave me a bit of confidence:)

 

[DrChinese]

Doc, a quick question re realism: I understand that some stellar sources produce ''unpolarised light'' -- which (of course) can become readily polarised.

Is this a classical example of ''particles NOT having definite attributes at all times''?

Thanks, wm

No, because the stellar light is still a collection of photons. You don't know the polarization of any particular one, just as you don't know the polarization of individual photons from a collection of entangled particles in a Bell test.

 

[DrChinese]

1. So thers no such thing as evolution in nature?..in order to evolve then nature must have regard to past events correct? Qm is meant to describe nature / reality?

2. When bell said that he is talking from the viewpoint of qm where after doing calculations of probabilities etc so many outcomes are realized..The hidden variables are only attached to nature by him to explain qm results that may themselves be wrong..Thats the problem he's asking nature to match the results when qm itself admits it doesn't know everything so that argument doesn't hold?

3. Yes i accept no particles have definite attributes all the time becuse people say quanta is the minimum observable action that one can observe so its fuzzy because its "alive" say...BUT forget everything else for now, i know that if i find out what a is i know what b is ..THATS A FACT?...it will never change?...so if we are dealing in facts then surely that is the only one qm has given us to date?...ok I am sure this more but loolkking at it in this expirement..thats the only fact we assertained so far..the rest is theory

1. Particles evolve, yes. But an observation erases prior history and makes that particle otherwise indistinguishable from others with a similar observed attribute.

2. Bell holds. That doesn't mean QM is perfect. All it means is that LOCAL HIDDEN VARIABLE THEORIES are "impossible" to reconcile with experiment. QM, on the other hand, is fully consistent with experiment.

3. Here too, there is much more to the picture than readily meets the eye. Sure, QM predicts if we learn about A then we also learn something about B. But it also says that WHAT we learn about B is functionally dependent on what we learn about A. Now that's amazing, because A and B are far apart! When you say the rest is theory, you are missing the point: that the standard QM theory makes correct predictions while other theories (classical, hidden variable, etc.) do not.

 

[calamero]

1. Particles evolve, yes. But an observation erases prior history and makes that particle otherwise indistinguishable from others with a similar observed attribute.

2. Bell holds. That doesn't mean QM is perfect. All it means is that LOCAL HIDDEN VARIABLE THEORIES are "impossible" to reconcile with experiment. QM, on the other hand, is fully consistent with experiment.

3. Here too, there is much more to the picture than readily meets the eye. Sure, QM predicts if we learn about A then we also learn something about B. But it also says that WHAT we learn about B is functionally dependent on what we learn about A. Now that's amazing, because A and B are far apart! When you say the rest is theory, you are missing the point: that the standard QM theory makes correct predictions while other theories (classical, hidden variable, etc.) do not.

Thanks drchinese

1..OK ill find out more :)

2..Ok bells theory dismisses any ideas about hidden variables such as each particle holds within itself an action for every possible outcome that lies ahead...i can't say with certainty obviously but i do don't think there is hidden variables like that myself.BUT it doesn't prove that the process of finding out the state of something at a actually changes the outcome at b because these 2 points are already there at that time of measurement..ie discovering that at point a something is 3R doesn't change point b into 3r..point b was always 3r or 3g or 3y its just that we didnt know it at the time ,in a sense point b is all 3 states changing so fast that we only observe the mist but once we know what point a is it stops everything at a frame of reference just say and its then we can say with certainty what b is not because a has affected b but soley becuse we've learned that's how quata behaves at that particular reference frame..no magic just reality?

3..Didnt mean to but think above covers that

calamero

What is weird is what happens in a delayed choice double slit experiment involiving electrons..people say that the mere fact that we "know" what slit an electron passed thu is enough to collapse the interference pattern / wave function even if we locked the results away in a safe for 6 months the pattern at the detector changes according to whether we look at the detector results 1st or the slit results..That begs the questions Is nature trying to hide something from us or is the collapse a result of the way maths interperts the data

ps..These are just thoughts of mines with no real study or argument to take them futher but at least I've put them out there for debate so thanks for that!

ok I am off to do a lot of reading etc so no more from me till I am a bit closer to where you are drchinese!:)

 

[kvantti]

All information in quantum systems is, notwithstanding Bell's theorem, localised. Measuring or otherwise interacting with a quantum system S has no effect on distant systems from which S is dynamically isolated, even if they are entangled with S.

http://arxiv.org/abs/quant-ph/9906007

What do you people have to say about Deutsch's paper?

 

[Anonym]

Calamero:“sorry i won't jump into posts again till i understand things a bit better but that how i feel and not dumb..history has a part to play here wether qm needs to incoperate somehow or nature itself carries some about with it”.

Calamero, All My Respect!

N.Bohr as quoted by Aage Peterson (1968):

“… the unambiguous interpretation of any measurement must be essentially framed in terms of the classical physical theories, and we may say that in this sense the language of Newton and Maxwell will remain the language of physicists for all time. Even when the phenomena transcend the scope of classical physical theories, the account of the experimental arrangement and recording of observations must be given in plain language, suitably supplemented by technical physical terminology. This is a clear logical demand, since the very word experiment refers to a situation where we can tell others what we have done and what we have learned.”

However, the Jewish Rabbies say that the prophecy is a business of fools.

Calamero:” Einstein beleived that everything in nature has a set value ..even if he didnt know the value of something at the time the thing has a value and that value never changes..”

If you are interested to know what were A. Einstein “believs”, here you are:

1.A. Einstein to M.Born (12.5.1952)(sorry for the translation mistakes, they are all mine):
”Did you see as Bohm (as by the way de Broglie 25 years ago) believs that the quantum theory may be interpreted deterministically otherwise? It, in my view, cheap consideration, however, your judgement is better indeed.”

2.A. Einstein to M.Born (12.10.1953)(sorry for the translation mistakes, they are all mine):
”For the planned in your honour collection of papers I wrote “physical” childish song, which slightly confused Bohm and de Broglie. It purpose is to demonstrate that your statistical interpretation of quantum mechanics is not necessary… May be, you will be pleased. We all, everybody, apparently are responsible for our sepun(aru) bubbles.”

3.A. Einstein dialog with W. Heisenberg (spring 1926) as quoted by W.Heisenberg (sorry for the translation mistakes, they are all mine):

3.1W.Heisenberg “It is reasonable to include in the theory only observable quantities…”

3.2A. Einstein:”Are you seriously assume that the physical theory includes only observable quantities?”

3.3W.Heisenberg:”But you used this idea in foundation of your relativity theory? You emphesized that it is impossible to talk about absolute time since this absolute time is impossible to observe…”

3.4A. Einstein:”Probably, I used the philosophy of that kind, but it nevertheless rubbish. Or, speaking more carefully, the remembering of what we are really observe and what we do not, has probably some heuristic value. However, from the principal point of view, the attempt to formulate the theory based only on observable quantities is completely nonsense. Because in the reality everything that happens are just an opposite. Only the theory itself can decide what is and is not observable. You see, the observation, generally speaking, is very complicated notion…”

This is the definition of the A.Einstein notion of the reality. It is ignored by the majority of the professional physicists. It is substituted by something that pervert completely the A.Einstein understanding. Your definition above seems to me identical.

Dear Calamero,

if you are interested to see the physically rigorous and unambiguous prove that A. Einstein was correct in all four his “believs” read:

1.Johann v. Neumann “Mathematische Grundlagen der Quantenmechanik”.
2.John A. Wheeler and Wojciech H. Zurek “Quantum Theory and Measurement”
3.Leonard Mandel and Emil Wolf “Optical coherence and quantum optics”.
(EM is the only relevant fundamental interaction when you consider double-slit, Bob-Alice flirt, BI,BM, etc).

As I already stated in the “Local QM? MWI, RQM, QFT, LQM, + ?” session of the PF:
There is no legitimation for M.Born statistical interpretation any more and it may be removed from the formalism of the Quantum Theory. QT is the local field theory of massive waves. No interpretation required.

Quote:
Originally Posted by Anonym

Dear Karl,

One very famous physicist (level compatible with A. Einstein) wrote:” We are facing here the perennial question whether we physicists do not go beyond our competence when searching for philosophical truth.”

The philosophy is far beyond my competence...

Dear Anonym,

This material is far off thread and belongs elsewhere, perhaps in the philosophy section. You should probably review the forum posting guidelines.

Regards,

-DrC

Dear DrChinese,

I am not a philosopher. I am only the integrator of the contributions made by the others and I am translator them into the language of mathematics which is strictly the area of my competence. Without request, I am referring to the invention of seven physically relevant mathematical languages presented in the paper entitled “ Quantum Mechanics of Non-Abelian Waves”.

Dear Karl,

I hope you meet my criticism properly, I did not intend to hurt you, it is just a matter of scientific discussion. By the way, V.A. Fock, head of Department of the Theoretical Physics where I was educated, in “Under banner of markcism”, 1, 149 (1938) expressed similar to yours point of view. Fortunately, his expected travel to Solovki turn to be gedankenexperiment. You may find the description of that event on p.92-94 in highly colour book, written by triplet of Russians V.V. Belokurov, O.D. Timofeevska and O.A. Chrustalev, “Quantum Teleportation-usual miracle”. In addition, on p.167 of that book as epigraph to the chapter entitled “Lifes and reproduction of the Schrödinger’s cats or what does it mean quantum ensemle” you may find reference to the hidden co-author of my recent paper (quant-ph/0606121,it is intentionally made readable for the mathematically unequiped scientists).

Daniel Gleekstein.

 

[vanesch]

http://arxiv.org/abs/quant-ph/9906007

What do you people have to say about Deutsch's paper?

I didn't read the paper, but given that it is Deutch, I can guess what he writes (and what I regularly claim here too :-)
In the MWI view (of which Deutch is a fervent - sometimes a bit too much! - defender), there is indeed absolutely no "non-local" effect to the EPR setup. The clue is that there is also not a single outcome at Alice and Bob during their measurements, and that the correlations are in fact only "decided about" upon their meeting, which they have to. It is only upon the "meeting" (which is an act by which Bob measures Alice's state - and vice versa), that the correlations make sense. But at that point, they became local information. It is also through this same scheme that we understand why the EPR effect cannot be used as "superluminal signalling".

 

[kvantti]

I didn't read the paper, but given that it is Deutch, I can guess what he writes (and what I regularly claim here too :-)
In the MWI view (of which Deutch is a fervent - sometimes a bit too much! - defender), there is indeed absolutely no "non-local" effect to the EPR setup. The clue is that there is also not a single outcome at Alice and Bob during their measurements, and that the correlations are in fact only "decided about" upon their meeting, which they have to. It is only upon the "meeting" (which is an act by which Bob measures Alice's state - and vice versa), that the correlations make sense. But at that point, they became local information. It is also through this same scheme that we understand why the EPR effect cannot be used as "superluminal signalling".

Actually, there is no mention of MWI in the paper.
They use the Heisenberg picture to show that, in any case, quantum theory is a local theory:

Given that quantum theory is entirely local when expressed in the Heisenberg picture, but appears nonlocal in the Schrödinger picture, and given that the two pictures are mathematically equivalent, are we therefore still free to believe that quantum theory (and the physical reality it describes) is nonlocal?
We are not - just as we should not be free to describe a theory as "complex" if it had both a simple version and a mathematically equivalent complex version. The point is that a "local" theory is defined as one for which there exists a formulation satisfying the locality conditions that we stated at the end of Section 1 (and a local reality is defined as one that is fully described by such a theory).
If we were to classify theories as nonlocal whenever it was possible to reformulate them in terms of nonlocal quantities (say, p +q and p - q, where p and q are local to A and B respectively), then no theory would qualify as local.
Moreover, although the Schrödinger picture disguises the locality of quantum physical processes, all our results could also, with sufficiently careful analysis, be obtained using the Schrödinger picture.

 

[DrChinese]

BUT it doesn't prove that the process of finding out the state of something at a actually changes the outcome at b because these 2 points are already there at that time of measurement..ie discovering that at point a something is 3R doesn't change point b into 3r..point b was always 3r or 3g or 3y its just that we didnt know it at the time ,in a sense point b is all 3 states changing so fast that we only observe the mist but once we know what point a is it stops everything at a frame of reference just say and its then we can say with certainty what b is not because a has affected b but soley becuse we've learned that's how quata behaves at that particular reference frame..no magic just reality?

Glad to see you asking the questions, Calamero.

First, no one claims to know what exactly is actually going on at the quantum level. QM only makes predictions on what can be observed.

On the other hand, Bell's Theorem is that the predictions of QM are incompatible with the "reasonable" assumptions of locality and hidden variables (realism). The description of the "mist" you provide is "reasonable", but is specifically excluded by Bell's Theorem UNLESS there is a faster-than-light (FTL) mechanism we don't know about.

The reason is that B's result is affected by the choice of measurement at A. The statistics show it (via the Bell Inequality). If the B measurement was NOT affected by the choice of measurement at A, then we would get a different set of correlations. We don't see it in the cases in which A and B are measured identically (RR or GG or YY). But we DO see it in the other cases (RG or GY or YR), which are the cases that contribute to the Bell Inequality. In other words, there is a measurable statistical bias that goes against the "reasonable" point of view, but magically happens to exactly agree with the QM point of view.

Good luck in your reading, and please keep participating in threads like this one.

 

[heusdens]

Can we model your setup as follows ?

We have a sequence of mathematical objects of your choice, drawn from a set S:

s: N -> S: i -> s(i).

For each element s of S, we have 6 operators A(1), A(2), A(3), B(1), B(2) and B(3) which can act upon s, and which produce a result + or -. There's no need in making the operators stochastical, as the elements of S can contain all random numbers one needs.

There are two mappings, a and b, from N into the set {1,2,3}:

a: N -> {1,2,3}: i -> "alice choice"
b: N -> {1,2,3}: i -> "bob choice"

The 3-some (s,a,b) gives us an "experimental result", which is given by the sequence of points:
exp: N -> {+,-} x {+,-}: i -> (A(a(i)) s(i) ; B(b(i)) s(i))

That is, we have the table of paired results by Alice and Bob, which is a list of pairs ++, +-, -+ and --, which is given by the act of A(a(i)) (with i-th choice by Alice) onto the element s(i) of our stream etc...

A priori, there only needs to be a definition of A(a(i)) and B(b(i)) on the i-th element of s ; that is, the other outcomes do not need to be defined.

It is the stream exp(i) which needs to satisfy the correlation and boundary conditions for sufficiently long series 1,2,3...,i,...

However, given that it is an assumption in the Bell derivation that the source (which is the mapping s) doesn't know in advance what are going to be the "drawn operators", the above condition needs to be satisfied:
- for one and the same s mapping
- for ALL possible mappings a and b.

This assumption is the fundamental assumption of Bell. If we might change the s-mapping for each different mapping a or b (which would mean that somehow, the "source knew of the choices of the detectors"), or if we had to pick mapping b AS A FUNCTION OF MAPPING a and s, (which would mean that b somehow knew of the a-choice) or if the operators A(1), ... were function of i or of a or of b, then there would be "communication" between the alice and the source, or bob and the source, or alice and bob, which is, BY ASSUMPTION, not supposed to happen.

It is this which allows us to:
1) have the same operators A(1), ... B(3) for all elements of s.
2) to require that the same single mapping s must give same statistics for ALL choices of a and b.

The proof is now rather simple:
consider first the series:
a: i -> 1 for all i
b: i -> 1 for all i

We now have to have that A(1) s(i) = B(1) s(i) for all i.
So this means that for all i:
A(1) s(i) is defined, B(1) s(i) is defined, and moreover they are equal.
We can hence do away with the operator B(1), because it is the same as A(1) for the given s(i).
We also have to have that for half of the i-values, A(1) s(i) = + and for half of the i-values, A(1) s(i) = -.

Similar for the choice:
a: i -> 2 for all i
b: i -> 2 for all i.

etc...

We can hence conclude:
for all i: the three operators A(1), A(2) and A(3) must be defined for s(i), moreover, they are identical to the B(1), B(2) and B(3) values for s(i), hence no need for a specific different notation.

Given the first N elements of s(i) (N big enough to be statistically significant), we can hence write down a table, which contains the number of times we have, in the set {s(1), s(2), ... s(N)}

1) the case A(1)s(i) = + AND A(2) s(i) = + AND A(3) s(i) = +

Say that it is N1

2) the case A(1)s(i) = + AND A(2) s(i) = + AND A(3) s(i) = -

say that it is N2

...

8) the case A(1)s(i) = - AND A(2) s(i) = - AND A(3) s(i) = -

say that it is N8

N1 + N2 + N3 + ... + N8 = N of course...

and we're back to our initial table of 8 possibilities, from which we derive a Bell inequality for the numbers Ni, and from which we derive an equality for the observed correlations if we make one more assumption:

That the series a and the series b are randomly picked streams which are uncorrelated with s.

This is the assumption against superdeterminism. It is nothing else but the assumption that "randomly and independently" picking a "polarizer direction" picks out a fair sample of the overall population, and hence allows us to find the correct "population correlation" and is not biased in a way.

The point I tried to make is that the stream, which is emitted from the source, might actually contain much more elements then get detected, depending on detector settings.
Now if I understand your formalized version well, you do not admit that, since you assume every element of the stream that is emitted, is also detected ? (correct me if I'm wrong).
And that could never explain our results, I agree with that.

Further, I do not understand the conditions named N1..N8, since they use (combinations of) detector settings which are not possible/allowed.
We can either use one detector, in 3 possible settings. Each detector setting is used only once. So, if we inspect one element of stream s, s(i), it can either have a result with detector setting A(1) OR A(2) OR A(3), but not more as one at once.
A combination of outcomes : A(1) s(i) = + ; A(2) s(i) = + ; A(3) s(i) = +; is therefore not an allowed combination, since it never occurs!
It is an important point to make, I guess!

We could however write a possible combination as:
A(1) s(i) = + ; A(2) s(j) = + ; A(3) s(k) = +;

where i <> j <> k

That is, each result at ONE side of the detector, can only be a result for ONLY ONE element of the stream s, not more then one. If stream element s(i) is used for ONE setting of dector ALice, then it can never be used for any other detector setting, and the same holds true for Bob's detector.

For TWO detectors, we allow of course (and must allow) simultanious outcomes for TWO detectors on the SAME element of the stream s, but only ONE at each side. These are the correlated outcomes.

* * *​

But clearly, I do not make my point very well, since I did not give a significantly elaborated mathematical formulation of how precisely it can be done. I just made the speculation it can be done, but did not proof it. I am still struggling with it.


What I (assume) to be a key issue is that randomness and some correlated behaviour are not absolute opposites. Like for instance, the digital expresion of the number pi is on one hand a random pattern of digits, on the other hand it is a very precise description of a number with precise properties.
Likewise a stream of 'elements' might be constructed in such a way that the stream itself is random, but still contains substreams which have very precise properties. You don't see the non-random aspect when measuring only at one detector.

 

[JesseM]

In this abstract form, there is no spacelike separation any more.
The stream doesn't have foreknowledge of detector settings, rather the other way, the detector combination selects a (sub)stream.

Well, if you're trying to contradict Bell's theorem with a classical example, there has to be a spacelike separation between measurement-events, because that's one of the conditions the theorem requires in order to guarantee the Bell inequalities are not violated classically.

But your answer doesn't make much sense to me in any case--are you sure you understand the meaning of the term "spacelike separation" in physics? If two events have a spacelike separation, all that means is that it would be impossible for a signal have left from the place and time of one event and reached the place and time of the other event, assuming the signal could not travel faster than light. So, for example, if I make a measurement on earth, and 1 year later you make a measurement 1.5 light-years away from Earth (as measured in some inertial reference frame, like the Earth's rest frame), it would be impossible for news of the result of my measurement to have reached you by the time you made your measurement, so these two measurement-events have a spacelike separation. On the other hand, if you made a measurement 1 year later but only 0.8 light-years away, you could have already learned the result of my measurement by the time you made yours (so you might adjust your detector setting based on that result), so in this case the measurement-events do not have a spacelike separation, instead they have a timelike separation.

If the source sends the streams at the speed of light in opposite directions, and Alice and Bob both make their measurements at the same time but different locations, this is enough to guarantee the two measurement-events have a spacelike separation. Does this clarify things?

The element of the stream (not string) that reaches the detectors of Alice and Bob is the same.

Sorry, I thought you were talking about streams of information, which can be represented as a string of digits.

We don't know what they contain...just that they contain some element, which is input for a detector outcome for each side.

OK, but the question is, given that Alice has just received a particular set of elements from the source, and that her detector is set to 3, is that enough to completely determine whether she gets a + or a -, or is there an additional random aspect, so that even if you had two trials on which the source sent Alice an identical set of elements, and Alice had her detector set to 3 on both trials, she might get + on one trial and - on the other? If there were any randomness, then it seems you can't guarantee that each time Bob and Alice have the same detector setting they'll get the same answer, even if the source sent an identical set of elements to each one.

What's more, the source doesn't know in advance on which trials they'll both choose setting 2. So if it's true on trials where they both choose setting 2 that the source always sends out signals with properties that make it 100% certain they'll both get a +, or signals with properties that make it 100% certain they'll both get a -, then this must be true on all trials, even the ones where they don't in fact choose to measure using setting 2. So we can say that every signal sent out by the source must either be of "type 2+" (meaning that its properties are such that if the detector is set to 2, the result is guaranteed to be +) or of "type 2-" (meaning on setting 2 you're guaranteed to get -). The source always sends both a signal of type 2+, or it sends them both a signal of type 2-; this is the only way to explain how they always get the same answer when they both choose setting 2, without violating local realism.

The source does not know the settings, but it is rather the effect of choosing detector combinations that filters the stream in such a way that detector outcomes show non-random behaviour.
So your assumption is not justified.

I think you misunderstood what I was saying, I didn't say that the source *does* know the settings in advance, the only assumption I'm making above is that once the source has sent the streams to Alice and she has chosen her setting, there is no additional random aspect (see my earlier comments)--either the streams have a combination of elements that make it guaranteed that if she chooses setting 2 she'll get a +, or the streams have a combination of elements that make it guaranteed that if she chooses setting 2 she'll get a -. There are no combinations of elements that the source sends out such that if she chooses setting 2, she has a 70% chance of getting + and a 30% chance of getting a -; knowing the combination and her choice of detector setting completely determines the results (but of course, Alice herself does not know the combination, and we don't need to either, all we need to know is that any given combination of elements would have only a single possible outcome for each possible detector setting). Do you disagree with this?

 

[calamero]

Calamero:“sorry i won't jump into posts again till i understand things a bit better but that how i feel and not dumb..history has a part to play here wether qm needs to incoperate somehow or nature itself carries some about with it”.

Calamero, All My Respect!

Thanks Anonym it appreciated!

If you are interested to know what were A. Einstein “believs”, here you are:

3.A. Einstein dialog with W. Heisenberg (spring 1926) as quoted by W.Heisenberg (sorry for the translation mistakes, they are all mine):

3.1W.Heisenberg “It is reasonable to include in the theory only observable quantities…”

3.2A. Einstein:”Are you seriously assume that the physical theory includes only observable quantities?”

3.3W.Heisenberg:”But you used this idea in foundation of your relativity theory? You emphesized that it is impossible to talk about absolute time since this absolute time is impossible to observe…”

3.4A. Einstein:”Probably, I used the philosophy of that kind, but it nevertheless rubbish. Or, speaking more carefully, the remembering of what we are really observe and what we do not, has probably some heuristic value. However, from the principal point of view, the attempt to formulate the theory based only on observable quantities is completely nonsense. Because in the reality everything that happens are just an opposite. Only the theory itself can decide what is and is not observable. You see, the observation, generally speaking, is very complicated notion…”

This is the definition of the A.Einstein notion of the reality. It is ignored by the majority of the professional physicists. It is substituted by something that pervert completely the A.Einstein understanding. Your definition above seems to me identical.

Dear Calamero,

if you are interested to see the physically rigorous and unambiguous prove that A. Einstein was correct in all four his “believs” read:

1.Johann v. Neumann “Mathematische Grundlagen der Quantenmechanik”.
2.John A. Wheeler and Wojciech H. Zurek “Quantum Theory and Measurement”
3.Leonard Mandel and Emil Wolf “Optical coherence and quantum optics”.
(EM is the only relevant fundamental interaction when you consider double-slit, Bob-Alice flirt, BI,BM, etc).

Ok will do , “Quantum Theory and Measurement” looks intersting :)

As I already stated in the “Local QM? MWI, RQM, QFT, LQM, + ?” session of the PF:
There is no legitimation for M.Born statistical interpretation any more and it may be removed from the formalism of the Quantum Theory. QT is the local field theory of massive waves. No interpretation required.

Quote:
Originally Posted by Anonym

Daniel Gleekstein.

Cheers again Anonym and drchinese for the advice and helps!

calamero

 

[NateTG]

Can we model your setup as follows ?

...
That the series a and the series b are randomly picked streams which are uncorrelated with s.

This is the assumption against superdeterminism. It is nothing else but the assumption that "randomly and independently" picking a "polarizer direction" picks out a fair sample of the overall population, and hence allows us to find the correct "population correlation" and is not biased in a way.

If we start with local realism as an assumption, we can restrict the local hidden state to something like:
H=\{+.-\} \times \{+,-\} \times \{+,-\}

From the fair sampling assumption you get as far as assigning probability to assigning probabilities to subsets that restrict two of the three dimensions. For example (for appropriate angle selections):
P(\{&lt;+,+,-&gt; &lt;+,+,+&gt;\})=.25

My question is, how do you justify the assumption that singleton subsets have well-defined probabilities?

 

[vanesch]

The point I tried to make is that the stream, which is emitted from the source, might actually contain much more elements then get detected, depending on detector settings.

That's correct. I guess you try to point to the "detector efficiency loophole" in the Bell tests.

Now if I understand your formalized version well, you do not admit that, since you assume every element of the stream that is emitted, is also detected ? (correct me if I'm wrong).
And that could never explain our results, I agree with that.

Indeed, that's the point.

Further, I do not understand the conditions named N1..N8, since they use (combinations of) detector settings which are not possible/allowed.
We can either use one detector, in 3 possible settings. Each detector setting is used only once. So, if we inspect one element of stream s, s(i), it can either have a result with detector setting A(1) OR A(2) OR A(3), but not more as one at once.

That's true. However, it was an ASSUMPTION that the stream s was independent of the detector settings. As such, I *assumed* that we had to have, for the same mapping s, and for ALL POSSIBLE DIFFERENT streams a and b, the right statistics. I used special choices for the streams a and b (namely, constant detector settings!) to show that for each of these cases, we needed to have a result from each of the operators A(1), A(2), A(3), ...

This was part of the assumption that the stream s must be un-knowledgeable of the choice of a and b that is going to happen, and cannot be "by coincidence" correlated with it (hypothesis of no superdeterminism). So for one and the same mapping s, we need to "be prepared" for all thinkable a and b series. This is a fundamental assumption in Bell. As such, the answers must be "pre-disposed" for all thinkable a and b.

A combination of outcomes : A(1) s(i) = + ; A(2) s(i) = + ; A(3) s(i) = +; is therefore not an allowed combination, since it never occurs!
It is an important point to make, I guess!

As said, it is an *assumption* that the generated events (the mapping s) is independent of any choice of a and b, and hence, the SAME mapping s must work for all thinkable a and b. What I simply did, was to choose first a = 1 all the time, then a = 2 all the time, then a = 3 all the time (not successively, but as alternative mappings). Given that the same s must work for all of them, the operators A(1), A(2) and A(3) must have a result for each s(i), EVEN THOUGH WE CANNOT MEASURE THIS. Again, the reason is again that the very same mapping s must work for all thinkable a and b (although in an actual experiment, we will only be able to make one definite choice for the a stream and the b stream).

That is, each result at ONE side of the detector, can only be a result for ONLY ONE element of the stream s, not more then one. If stream element s(i) is used for ONE setting of dector ALice, then it can never be used for any other detector setting, and the same holds true for Bob's detector.

True. This is what makes that quantum theory can violate Bell. But it was an ASSUMPTION that the stream had to be "ready" no matter what was going to be the stream of choices a and b (and hence had to be ready for all of them).

But clearly, I do not make my point very well, since I did not give a significantly elaborated mathematical formulation of how precisely it can be done. I just made the speculation it can be done, but did not proof it. I am still struggling with it.

If you DROP this hypothesis, then you CAN obtain a correct statistical description violating Bell's inequalities: the very proof is that the statistical description that comes out of quantum mechanics does exactly that.

However, in a LOCAL REALIST view without superdeterminism, the ONLY WAY to obtain perfect anti-correlations is that the *potential* outcomes are pre-determined. This is exactly the assumption that FOR THE SAME MAPPING s, we must be able to consider all POSSIBLE streams a and b, and not just the one we will experimentally realize.

What I (assume) to be a key issue is that randomness and some correlated behaviour are not absolute opposites. Like for instance, the digital expresion of the number pi is on one hand a random pattern of digits, on the other hand it is a very precise description of a number with precise properties.

But there is no difficulty. Imagine a deity, in a classical block universe, which has pre-determined everything which will happen. So the 4-dim block universe has been filled by our deity with events. Some of them are the measurements done by Alice and Bob. Our deity can have disposed the events in such a way, that the quantum correlations happen to be correct. But note that in such a world, there are no laws of nature (except the fortuous correlations between events which are there because of our deity's desires)). All we see as "regularities" in the universe is just due to a specific desire of disposing events in a certain way. There is no causality, there is no locality. We may be deluded in thinking that this was to be so, but in fact there is not one single piece of "mechanism" in nature, just "dispositions of events in 4-dim spacetime". This may coincide with quantum predictions (say, until the 4th of march, 2015, when everything changes because from that timeslice onward, our deity has arranged events differently).
This is a realistic model (in that there is an ontology: a bag of events). The "laws of nature" we observe are just correlations within dispositions of events, without any further meaning.

What we have here is a form of superdeterminism. You can also call it "conspiracy". Indeed, by disposing events in a certain way, or in another way, the deity can make you believe anything she wants. You will think that there are causal relationships expressed by statistical relationships, or even by deterministic relationships, but there is nothing of the kind. Just events, scattered through a 4-dim manifold. Given that there are no mechanisms in nature, the discussion of whether these mechanisms are local or not are moot.

When you accept this, then anything goes, and no theorem will be derived which tells you that this or that kind of correlation is not possible. So any theorem of the kind will already assume that this kind of superdeterminism is not true.

As such, if you say that "the source is independent of the CHOICE of the setting of a", then that means that there must exist a POTENTIAL ANSWER to the question of what would have happened IF we made the choice to measure A(1) or A(2) or A(3). That doesn't mean that we acknowledge that it is physically possible to measure them simultaneously, but there must exist a statistical description of WHAT WOULD HAVE RESULTED IF, as long as we assume that there exists a kind of causal mechanism that determines (even stochastically) the outcome, and that it is not a deity that has scattered the results "like that" (which is an assumption we made).

 

[vanesch]

Actually, there is no mention of MWI in the paper.
They use the Heisenberg picture to show that, in any case, quantum theory is a local theory:

I read the paper now. Right, there is no EXPLICIT mentioning of any MWI. However, ALL operations are unitary operations, and the "correlation measurement" is obtained by a unitary operation (lower part of p 12), bringing together the "quantum version" of Alice and Bob into a quantum gate.
This is nothing else but the MWI version of a Bell-EPR setup, without explicitly saying so: the QUANTUM result of the measurement is transported (that is, the TWO terms in the superposition are kept in the "measurement quantum bit").

 

[Rade]

If we start with local realism as an assumption, we can restrict the local hidden state to something like:
H=\{+.-\} \times \{+,-\} \times \{+,-\}

Any reason why not this ?:
H=\{+.-\} \times (\{+,-\} \times \{-,+\})

 

[NateTG]

Any reason why not this ?:
H=\{+.-\} \times (\{+,-\} \times \{-,+\})

Apparently I made a typo, it should have been:
H=\{+,-\} \times \{+,-\} \times \{+,-\}
(Note that there is a comma rather than a period in the first set.)
\times denotes the cartesian product.
I don't understand your question, since, aside from the comma/period typo, the two are trivially isomorphic sets.

 

[vanesch]

I don't understand your question, since, aside from the comma/period typo, the two are trivially isomorphic sets.

Yes, that was also what I was wondering about...

 

[DrChinese]

1. If we start with local realism as an assumption, we can restrict the local hidden state to something like:

H=A \{+,-\} \times B \{+,-\} \times C \{+,-\}

2. From the fair sampling assumption you get as far as assigning probability to assigning probabilities to subsets that restrict two of the three dimensions. [/tex]

A couple of comments:

1. For anyone who does not already see how the realism assumption is present in Bell's Theorem, NateTG has nicely laid this out for us to see. I have added the A/B/C to map it to Bell's usage.

With A/B/C you have 3 hypothetical hidden variables, which would lead to 2^3 permutations, or 8. You could hypothesize A/B/C/D, A/B/C/D/E, or generally as follows:

H=A\{+,-\} \times B\{+,-\} \times C\{+,-\} \times D\{+,-\} \times E\{+,-\} ...

However, you only need the A/B/C that NateTG provided to do the trick with Bell's Theorem.


2. A minor quibble : observing 2 of 3 settings is not the "fair sampling" assumption. In Bell's Theorem, this is expressed by Bell's words as follows:

"The vital assumption [2] is that the result B for particle 2 does not depend on the setting a, of magnet for particle 1, nor A on b."

This is actually the locality assumption, and Bell referenced a quote from Einstein himself (1949) to make this clear:

"But on one supposition we should, in my opinion, absoultely hold fast: the real factual situation of the system S2 is independent of what is done with the system S1, which is spatially separated from the former."


3. By the way, the fair sampling assumption is the idea that detected sample (photon pairs) are representative of the entire universe of pairs in an experiment. Some local realists have objected that the subset actually detected is intrinsicly biased to yielding a result consistent with the predictions of QM. Without discussing the merits of this idea: as the % of detected pairs approaches 100% (so it is no longer a subset), the fair sampling assumption becomes superfluous.

-DrC

 

[NateTG]

That's not the cleanest notation. The way you wrote it, A could be some kind of uniary set operation. It would be better to do something like:
H \equiv \{+,-\}^{\{A,B,C\}}
(which I was tempted to put into the post above which discusses equivalent sets.)

 

[NateTG]

2. A minor quibble : observing 2 of 3 settings is not the "fair sampling" assumption.

Apparently the earlier post was unclear:
Let's assume (for the sake of discussion) that we have a local hidden variable theory - so at the point where our EPR particles split, they have hidden states \vec{h} and -\vec{h}. Where \vec{h} \in H and
H=\{&lt;-,-,-&gt;,&lt;-,-,+&gt;,&lt;-,+,-&gt;,&lt;-,+,+&gt;,&lt;+,-,-&gt;,&lt;+,-,+&gt;,&lt;+,+,-&gt;, &lt;+,+,+&gt;\}

Now, we know that h \in H with probability 1, and that h \in \null with probability 0. Stipulating the fair sampling assumption, we can experimentally determine that the probability that h \in \{&lt;+,-,-&gt;,&lt;+,-,+&gt;,&lt;+,+,-&gt;,&lt;+,++&gt;\} some real number p_{&lt;+,*,*&gt;} \in [0,1] by measuring particle \alpha directly, and the probabilities for the other 5 symetric subsets can also be tested.
Similarly, it's possible, by measuring both particles to determine the probability p_{&lt;+,+,*&gt;} that h \in \{&lt;+,+,-&gt;,&lt;+,+,+&gt;\} or any of the other 11 subsets symetric to this one.

However, it's impossible to measure p_{&lt;+,+,+&gt;}, that is, the probability that h \in \{&lt;+,+,+&gt;\}, experimentally, but Bell's theorem assumes that it is a well defined real value i.e. that p_{&lt;+,+,+&gt;} \in [0,1].

In a classical regime, where it's possible to make non-perturbing measurements, it's trivially possible to measure the concurrence of two things, so coincidence probabilities are well-defined, which validates the assignment of probabilities to these singletons. However, in a quantum setting, it's necessary to assume, without experimental justification, that these singleton probabilities exist and are well-defined for Bell's Theorem to be valid.

 

[JesseM]

However, it's impossible to measure p_{&lt;+,+,+&gt;}, that is, the probability that h \in \{&lt;+,+,+&gt;\}, experimentally, but Bell's theorem assumes that it is a well defined real value i.e. that p_{&lt;+,+,+&gt;} \in [0,1].

But in a classical world, can't you assume that everything really is in a single definite state at all times, even if you can't measure that state? It's not necessary for the proof that we actually know the probability of these states, just there is some definite probability, which could be known by an imaginary omniscient observer who knows the complete state of the universe at every moment.

This criticism would make more sense if you were directing it at some specific hidden-variables theory which claimed to be a valid theory of physics, but since Bell's theorem is just trying to show that local hidden variable theories can't work, I don't think it's a weakness of the argument that even if they did "work" in the sense of not being ruled out a priori by violations of Bell inequalities, they still wouldn't "work" as good theories of physics since they would involve quantities which could never be measured by experiment.

 

[NateTG]

But in a classical world, can't you assume that everything really is in a single definite state at all times, even if you can't measure that state? It's not necessary for the proof that we actually know the probability of these states, just there is some definite probability, which could be known by an imaginary omniscient observer who knows the complete state of the universe at every moment.

Although, for the usual notion of probability, the coincidence of two states with well-defined probabilities always has a well-defined probability; the assumption of a definite state does not infer a well-defined probability for that state.

For example, under the usual notions, the probability that a random real number chosen from the interval [0,1] is less than \frac{1}{2}, provided that it is a rational number is undefined. (This is effectively a division by zero.) On the other hand, the probability that a random real number from [0,1] is rational and less than \frac{1}{2} is zero.

 

[heusdens]

Well, if you're trying to contradict Bell's theorem with a classical example, there has to be a spacelike separation between measurement-events, because that's one of the conditions the theorem requires in order to guarantee the Bell inequalities are not violated classically.

Well, my "experiment" does not take place in a spatial organized way, it is just a very abstract experiment (Gedanken experiment). For that reason probably it could show to conflict the Bell inequality, but for that same reason does not qualify as a "real" experiment, that could somehow break the inequality with a classical set-up.

But your answer doesn't make much sense to me in any case--are you sure you understand the meaning of the term "spacelike separation" in physics? If two events have a spacelike separation, all that means is that it would be impossible for a signal have left from the place and time of one event and reached the place and time of the other event, assuming the signal could not travel faster than light. So, for example, if I make a measurement on earth, and 1 year later you make a measurement 1.5 light-years away from Earth (as measured in some inertial reference frame, like the Earth's rest frame), it would be impossible for news of the result of my measurement to have reached you by the time you made your measurement, so these two measurement-events have a spacelike separation. On the other hand, if you made a measurement 1 year later but only 0.8 light-years away, you could have already learned the result of my measurement by the time you made yours (so you might adjust your detector setting based on that result), so in this case the measurement-events do not have a spacelike separation, instead they have a timelike separation.
If the source sends the streams at the speed of light in opposite directions, and Alice and Bob both make their measurements at the same time but different locations, this is enough to guarantee the two measurement-events have a spacelike separation. Does this clarify things?

Yes, I'm aware that in any real setup (wether classical or quantum) this is of importance; I'm aware of the difference between space-like and time-like seperations in SR, which has to do with the cone of worldlines; if A and B are on points well within (and including) each other cones, they are time-like seperated, and if not, they are space-like seperated.

Sorry, I thought you were talking about streams of information, which can be represented as a string of digits.

The actual stream of "information", what is encoded in the elements of the stream, might be anything that can be expressed mathematically (if it can be shown that anything that is expressable mathematically can be represented in digits in a unique way, then that may well be the case).

OK, but the question is, given that Alice has just received a particular set of elements from the source, and that her detector is set to 3, is that enough to completely determine whether she gets a + or a -, or is there an additional random aspect, so that even if you had two trials on which the source sent Alice an identical set of elements, and Alice had her detector set to 3 on both trials, she might get + on one trial and - on the other? If there were any randomness, then it seems you can't guarantee that each time Bob and Alice have the same detector setting they'll get the same answer, even if the source sent an identical set of elements to each one.

There could not likely be two trials in which the source sends out the same stream, it would be very improbable (would depend on the sample size, of course). And further, the detector settings determine what part of the stream actually get detected.

I think you misunderstood what I was saying, I didn't say that the source *does* know the settings in advance, the only assumption I'm making above is that once the source has sent the streams to Alice and she has chosen her setting, there is no additional random aspect (see my earlier comments)--either the streams have a combination of elements that make it guaranteed that if she chooses setting 2 she'll get a +, or the streams have a combination of elements that make it guaranteed that if she chooses setting 2 she'll get a -. There are no combinations of elements that the source sends out such that if she chooses setting 2, she has a 70% chance of getting + and a 30% chance of getting a -; knowing the combination and her choice of detector setting completely determines the results (but of course, Alice herself does not know the combination, and we don't need to either, all we need to know is that any given combination of elements would have only a single possible outcome for each possible detector setting). Do you disagree with this?

Yes, but the stream contains more information then the correlated results.
Only, depending on detector settings, not all that information shows up in the results when measured. This is the crucial point.

But to this point, I think it is not worth discussing any more since:
a. This abstract model misses an aspect to make it a sensible "classical" setup which is in accordance with the prescriptions (space-like separation of the detectors).
b. I have not come up with an actual "mechanism" for explaining the behaviour of the "experiment", but just assumed such a "mechanism" could in theory exist (even if we were just providing the adequate mathematical expressions)

So, until I have come up with something to fix these omissions, we are not discussing something concrete, and we end up mere speculating.
If Bell's prediction is right of course we could not do it. I have not shown this wrong, and perhaps I can't.

 

[DrChinese]

Yes, but the stream contains more information then the correlated results.

Actually, a stream of entangled photons does not contain any information at all. Each member of each pair is in a purely random state of polarization, and so by definition it cannot contain any information.

(The pairs themselves of course have an interesting property because they are entangled, which does allow you to learn about the detector settings when the results are correlated later on.)

 

[JesseM]

Although, for the usual notion of probability, the coincidence of two states with well-defined probabilities always has a well-defined probability; the assumption of a definite state does not infer a well-defined probability for that state.

For example, under the usual notions, the probability that a random real number chosen from the interval [0,1] is less than \frac{1}{2}, provided that it is a rational number is undefined.

Isn't this a consequence of having an infinite number of possible outcomes, with "rationality/nonrationality" dependent on knowing an infinite number of digits for every outcome? In this situation it's not clear that the idea of picking a "random real" between 1 and 0, with each outcome being either definitely rational or definitely non-rational, makes sense in the first place...it would depend on your views on some tricky philosophy-of-math issues, I think. On the other hand, in the proof of Bell's theorem it can be assumed that every state emitted by the source must either fall into the category of "would yield a + if measured on detector setting 2" or "would yield a - if measured on detector setting 2", since we know that whenever both experimenters do use detector setting 2 they always get definite opposite answers, and the source has no way of knowing which trials the experimenters will use which detector settings in advance.

So if we assume that on each trial the hidden variables were definitely either of type +2 or -2, then there must be some definite ratio (number of trials where state was of type +2)/(total number of trials), and presumably the ratio would have to approach a single value in the limit as the number of trials goes to infinity. Even if there was a physical process which as an outcome generated an object with an infinite number of properties which could be encoded as a string of 1's and 0's (and with each individual property being equally like to be 1 or 0), and if the experimenter would get a + on setting 2 whenever this string was rational and a - on setting 2 whenever it was irrational, the ratio of one type of outcome to another would have to be well-defined for any given series of trials...again, the problem is that we can't be sure if it even makes sense to imagine an observable measurement whose outcome is dependent on whether an infinite random string is rational or irrational, it's a bit like imagining an observable measurement whose outcome is dependent on whether the continuum hypothesis is true or false (a question which may not even have an objective answer depending on your views of the philosophy of mathematics).

edit: Also, since the set of rational numbers between 0 and 1 is countable, the notion of randomly picking a rational number between 0 and 1 should be equivalent to the notion of picking a random integer from 0 to infinity, with each one equally likely. This clearly leads to nonsensical conclusions; for example, if you pick two such numbers randomly and put each one in an envelope, then open one and look at it, you should conclude that there is a 100% probability the second envelope will contain a larger number, regardless of which envelope you opened first, since there are an infinite number of integers larger than the first one you looked at and only a finite number smaller. Indeed, in probability it is forbidden to have a uniform distribution on all the numbers from 0 to infinity.

 

[MeJennifer]

Each member of each pair is in a purely random state of polarization, and so by definition it cannot contain any information.

Actually we cannot claim that at all, that is completely wrong.

We simply do not know how those states of polarization are determined, it could be random but that is certainly not the only option.

We cannot determine how they get their values, but what we can do is repeat the same experiment and make predictions based on the statistics. But in no way does that prove that each experiment is random.

 

[DrChinese]

Actually we cannot claim that at all, that is completely wrong.

We simply do not know how those states of polarization are determined, it could be random but that is certainly not the only option.

We cannot determine how they get their values, but what we can do is repeat the same experiment and make predictions based on the statistics. But in no way does that prove that each experiment is random.

Huh? (Maybe we are talking about 2 different things...)

Of course the polarization of Bell test photon pairs is demonstrably random. Let's consider Type II PDC, for example. One member of the pair is vertically polarized, the other member is hortizonally polarized. You simply never know which is which. Because of that, it is not possible to encode any information in the stream.

Now you might believe there is a "prior" cause that we simply do not know about, and that might seem a reasonable hypothesis at first glance. Except for one minor detail. The correlated polarization statistics for photon pairs that are not entangled are different that ones that are entangled. IF there were a prior "cause" that we could potentially tap into - and therefore use as a mechanism for encoding a bit of information - then the pair would no longer be represented by a single wave function. It would now be represented by 2 separated wave functions. Doesn't matter if we actually look at the prior cause or not. IF we could have, then that would be enough to eliminate the entangled state.

If a pair of photons is polarization entangled, then the polarization is fully random (to whatever degree of randomness you choose to measure) and NO information can be encoded in the stream.

QED.

 

[MeJennifer]

Huh?

Of course the polarization of Bell test photon pairs is demonstrably random. Let's consider Type II PDC, for example. One member of the pair is vertically polarized, the other member is hortizonally polarized. You simply never know which is which. Because of that, it is not possible to encode any information in the stream.

Now you might believe there is a "prior" cause that we simply do not know about, and that might seem a reasonable hypothesis at first glance. Except for one minor detail. The correlated polarization statistics for photon pairs that are not entangled are different that ones that are entangled. IF there were a prior "cause" that we could potentially tap into - and therefore use as a mechanism for encoding a bit of information - then the pair would no longer be represented by a single wave function. It would now be represented by 2 separated wave functions. Doesn't matter if we actually look at the prior cause or not. IF we could have, then that would be enough to eliminate the entangled state.

If a pair of photons is polarization entangled, then the polarization is fully random (to whatever degree of randomness you choose to measure) and NO information can be encoded in the stream.

You seem to mixup the question if nature is fundamentally random with the ability for us to determine that.

QED.

The only QED we have here is that you cannot admit you made a mistake.

 

[NateTG]

...So if we assume that on each trial the hidden variables were definitely either of type +2 or -2, then there must be some definite ratio (number of trials where state was of type +2)/(total number of trials), and presumably the ratio would have to approach a single value in the limit as the number of trials goes to infinity.

That what you'd think, but why? And, how does that compare with the other possibilities?

 

[wm]

Actually, a stream of entangled photons does not contain any information at all. Each member of each pair is in a purely random state of polarization, and so by definition it cannot contain any information.

(The pairs themselves of course have an interesting property because they are entangled, which does allow you to learn about the detector settings when the results are correlated later on.)

DrC, I'm wondering if the wording here is confusing?

1. Is it not the case that each PAIR is in an identical (random) state? Such that a test on one twin reveals information about a property of the other (correlated) twin; that is, information about how the untested twin will respond to a specific test?

So the stream does encode information; though that information is not encoded by us?

2. Also: It seems to me that you learn about the detector settings later on; by directly comparing detector settings. Not by comparing test results?

So other information that you learn later on (when the test-results are correlated) is about each twinned-pair's response to each specific pair of detector settings?

wm

 

[DrChinese]

You seem to mixup the question if nature is fundamentally random with the ability for us to determine that.


The only QED we have here is that you cannot admit you made a mistake.

LOL.

I didn't say nature was fundamentally random, because no one knows this for a fact. I said that all of the following are true:

1. It is not possible to encode any information in a polarization entangled pair of photons. This statement was a response to a comment by heusdens. If you know a way to do this, please let me know.

2. It is impossible to distinguish the polarization values of a set of such photons from that of a truly random source. This makes it clear that your statement is merely conjecture, as the evidence points the other way. IE if it walks like a duck, and quacks like a duck, it is truly random.

3. It is possible to distinguish a stream of polarization entangled pairs of photons from a stream which is not polarization entangled. This is a simple consequence of experiment, and goes to show that in fact there is no information (i.e. cause) that we are simply "missing" (because if there were, the entanglement would immediately disappear).

You are welcome to dismiss the science of this, if you prefer to operate at a semantical level.

 

[DrChinese]

DrC, I'm wondering if the wording here is confusing?

1. Is it not the case that each PAIR is in an identical (random) state? Such that a test on one twin reveals information about a property of the other (correlated) twin; that is, information about how the untested twin will respond to a specific test?

So the stream does encode information; though that information is not encoded by us?

2. Also: It seems to me that you learn about the detector settings later on; by directly comparing detector settings. Not by comparing test results?

So other information that you learn later on (when the test-results are correlated) is about each twinned-pair's response to each specific pair of detector settings?

wm

You cannot transmit information in such a stream (at least using an entangled attribute).

1. If the polarization values are NOT assigned by us, how would we encode anything? heusdens was commenting - really a side comment as I read it - that there was information in the stream but all of it was not usable. I was merely making a point that such a stream consists of random values. To encode information, you would choose to polarize a photon as vertical or horizontal and then establish bit values (0 or 1) to correspond with what the receiver detects. If you try to do this, you will NOT have polariztion entanglement. This is both per theory and per experiment.

2. Sure, you learn about the detector settings when you compare the results later. In fact, I specifically said that... "(The pairs themselves of course have an interesting property because they are entangled, which does allow you to learn about the detector settings when the results are correlated later on.)" So no disagreement there.

Hopefully, we are on the same page.

 

[MeJennifer]

It is impossible to distinguish the polarization values of a set of such photons from that of a truly random source. This makes it clear that your statement is merely conjecture, as the evidence points the other way. IE if it walks like a duck, and quacks like a duck, it is truly random.

My statement merely conjecture?
It seems I have to remind you who actually made what statement.

Your statement:

"Each member of each pair is in a purely random state of polarization, and so by definition it cannot contain any information. "

I answered:

We simply do not know how those states of polarization are determined, it could be random but that is certainly not the only option.

We cannot determine how they get their values, but what we can do is repeat the same experiment and make predictions based on the statistics. But in no way does that prove that each experiment is random.

And now my statement is supposedly "merely conjecture"
Anyway it demonstrates my prior point, you can't admit you made a mistake and furthermore when caught with one you attempt to wiggle yourself out of it.

Anyway there seems to be no point in cluttering this topic any further. Feel free to PM me.

 

[wm]

You cannot transmit information in such a stream (at least using an entangled attribute).

1. If the polarization values are NOT assigned by us, how would we encode anything? heusdens was commenting - really a side comment as I read it - that there was information in the stream but all of it was not usable. I was merely making a point that such a stream consists of random values. To encode information, you would choose to polarize a photon as vertical or horizontal and then establish bit values (0 or 1) to correspond with what the receiver detects. If you try to do this, you will NOT have polariztion entanglement. This is both per theory and per experiment.

2. Sure, you learn about the detector settings when you compare the results later. In fact, I specifically said that... "(The pairs themselves of course have an interesting property because they are entangled, which does allow you to learn about the detector settings when the results are correlated later on.)" So no disagreement there.

Hopefully, we are on the same page.

DrC, it seems that we are not yet on the same page re this small matter.

1. I did NOT say that we can transmit information is such a stream.

Rather: Accepting the existence of the twin-streams; one stream correlated with the other ... then, from our interrogation of one stream, we learn something [ = information] about the other stream.

This correlative information was encoded in the streams by virtue of the creation of each pair (each set of twins) in the (highly-correlated) singlet-state.

2. Now BIG disagreement/misunderstanding here: It is by NO property of the PAIRS that we learn of the detector settings. We learn that by direct observation of each setting; or by direct communication; IRRESPECTIVE of the PAIRS.

I'm a bit surprised at our entanglement here, wm

 

[wm]

No, because the stellar light is still a collection of photons. You don't know the polarization of any particular one, just as you don't know the polarization of individual photons from a collection of entangled particles in a Bell test.

Clarification of terminology, please:

Question: Is there such a thing as an unpolarised photon?

OR: When we see the phrase (in physics papers) ''unpolarised photon''; this is always to be interpreted as ''photon with polarisation unknown''?

Thanks, wm

 

[DrChinese]

Anyway there seems to be no point in cluttering this topic any further.

The door is that way, for those that want to make an unwarranted assertion that they prefer not to defend...

 

[JesseM]

That what you'd think, but why? And, how does that compare with the other possibilities?

Well, you'd agree that for any finite number of trials, there must be a definite ratio, right? It seems to me that the only way that the ratio would not converge to a single value as the number of trials approached infinity is if there were some time-dependence in the hidden variables, like if the first 10 were 2+, the next 100 were 2-, the next 1000 were 2+, the next 10000 were 2-, and so forth. But if there were any time-dependent pattern, then when using the same measurement setting on trial after trial you would notice some sort of pattern in the +'s vs. the -'s, it wouldn't look like a completely random string.

Anyway, it seems to me that the proof of Bell's theorem does not actually depend on saying anything about the probabilities of different hidden states, just that there is a definite hidden state on each trial, and that there is no statistical correlation between the choice of detector settings and the hidden state. Suppose each hidden state must definitely either be of type 1+ (meaning if you use detector setting 1 you're guaranteed to get a + with that hidden state) or 1-, and likewise each hidden state must definitely be of type 2+ or 2-, and must definitely either be of type 3+ or 3-. In this case we can make the following statement about the number of trials on which the hidden state is of one type or another, a statement which has nothing to do with probabilities:

Number(state of type 1+ and 2-) + Number (state of type 2+ and 3-) greater than or equal to Number(state of type 1+ and 3-)

This has to be true, because on every trial where the state was of type 1+ and 3-, it must have been either of type 2+ or 2-, and if it was of 2- this would also add one to Number(state of type 1+ and 2-), while if it was of type 2+ it would add one to Number(state of type 2+ and 3-). So, every trial which adds one to the number on the right side of the inequality must add one to the number on the left side to, showing the left side must be greater than or equal to the right side.

From this non-probabilistic statement, it shouldn't be too hard to show that if the choice of detector settings is random on each trial and is uncorrelated with the hidden state, and Alice and Bob always get the same result when they choose the same detector setting, this leads naturally to the conclusion that

Probability(Alice chooses setting 1, gets + and Bob chooses setting 2, gets -) + Probability(Alice chooses setting 2, gets + and Bob chooses setting 3, gets -) greater than or equal to Probability(Alice chooses setting 1, gets + and Bob chooses setting 3, gets -)

 

[DrChinese]

wm,

There are the following definitions:

Polarized photon: polarization known (or could have been known).

Unpolarized photon: polarization not known (and could not have been known). In the terminology of QM, and following the Heisenberg Unvertainty Principle, the polarization is not definite until observed. You do NOT need to accept this explanation to use the term "unpolarized" as the term merely means that no polarization observation has been performed yet.

Polarization entangled photon: an unpolarized photon that shares a wave function with one or more particles.

 

[NateTG]

Well, you'd agree that for any finite number of trials, there must be a definite ratio, right? It seems to me that the only way that the ratio would not converge to a single value as the number of trials approached infinity is if there were some time-dependence in the hidden variables, like if the first 10 were 2+, the next 100 were 2-, the next 1000 were 2+, the next 10000 were 2-, and so forth. But if there were any time-dependent pattern, then when using the same measurement setting on trial after trial you would notice some sort of pattern in the +'s vs. the -'s, it wouldn't look like a completely random string.

Assuming local realism, any finite number of trials should, indeed, produce a ratio. However, for a non-stochastic process, the central limit theorem does not apply, and, as a consequence, there is no confidence interval.

Now, you'd think leads to a problem because you could have one person observing at one frequency (say, every trial) and another person observing at a different frequency (like pairs of trials), leading to a contradiction, but the quantity we're talking about is unobservable.

Anyway, it seems to me that the proof of Bell's theorem does not actually depend on saying anything about the probabilities of different hidden states, just that there is a definite hidden state on each trial, and that there is no statistical correlation between the choice of detector settings and the hidden state.

The process for deriving Bell's Inequality involves adding or subtracting the probabilities for hidden states which is only sensible if those probabilities are well defined. I would be very interested and surprised to see any derivation of Bell's Inequality that only involves experimentally measurable probabilities.

Number(state of type 1+ and 2-) + Number (state of type 2+ and 3-) greater than or equal to Number(state of type 1+ and 3-)

Sure.

From this non-probabilistic statement, it shouldn't be too hard to show that if the choice of detector settings is random on each trial and is uncorrelated with the hidden state, and Alice and Bob always get the same result when they choose the same detector setting, this leads naturally to the conclusion that

Probability(Alice chooses setting 1, gets + and Bob chooses setting 2, gets -) + Probability(Alice chooses setting 2, gets + and Bob chooses setting 3, gets -) greater than or equal to Probability(Alice chooses setting 1, gets + and Bob chooses setting 3, gets -)

With the usual assumptions about experimental bias and detector choices, doesn't that work out to:
\frac{1}{16}+\frac{1}{16} \geq \frac{1}{16}

Which hardly seems contradictory to me.

 

[JesseM]

Assuming local realism, any finite number of trials should, indeed, produce a ratio. However, for a non-stochastic process, the central limit theorem does not apply, and, as a consequence, there is no confidence interval.

Now, you'd think leads to a problem because you could have one person observing at one frequency (say, every trial) and another person observing at a different frequency (like pairs of trials), leading to a contradiction, but the quantity we're talking about is unobservable.

Can you give an example of a non-stochastic process or algorithm that would generate a series of + and - results, where the pattern of results shows no time-dependence, and yet the ratio of +'s to -'s would not approach some fixed ratio as the number of trials goes to infinity? I'm having trouble imagining how this would work.

The process for deriving Bell's Inequality involves adding or subtracting the probabilities for hidden states which is only sensible if those probabilities are well defined.

I don't think it necessarily does--as I suggested, you can just talk about the number of different hidden states in a large set of trials, and then the only probabilistic assumption you need is the assumption that the choice of detector settings on each trial is random and there is no correlation between the detector setting and the hidden state emitted by the source.

Probability(Alice chooses setting 1, gets + and Bob chooses setting 2, gets -) + Probability(Alice chooses setting 2, gets + and Bob chooses setting 3, gets -) greater than or equal to Probability(Alice chooses setting 1, gets + and Bob chooses setting 3, gets -)

With the usual assumptions about experimental bias and detector choices, doesn't that work out to:
\frac{1}{16}+\frac{1}{16} \geq \frac{1}{16}

Which hardly seems contradictory to me.

Can you explain how you got the 1/16 figure? What are the "usual assumptions" about detector settings? It may be that for the detector settings used in some other common version of the experiment, this inequality would not be violated, but you can violate it with the right combination of detector settings--see the section on "Applying Bell's inequality to electron spin" in this article on Bell's theorem.

 

[NateTG]

Can you give an example of a non-stochastic process or algorithm that would generate a series of + and - results, where the pattern of results shows no time-dependence, and yet the ratio of +'s to -'s would not approach some fixed ratio as the number of trials goes to infinity? I'm having trouble imagining how this would work.

It's somewhat ugly.
Consider the interval [0,1].
For example, let's say we have an evenly distributed random sequence x_n on [0,1]. Now, for A, some non-measurable subset of [0,1], define y_n as follows:
y_n=+ \rm{if} x_n \in A
y_n=- \rm{otherwise}
Now, while a particular x_n and A might lead to a y_n that has a limit frequency, that's not provably the case.

Can you explain how you got the 1/16 figure? What are the "usual assumptions" about detector settings?

I had a brain fart - I was thinking the settings were orthogonal.

 

[wm]

wm,

There are the following definitions:

Polarized photon: polarization known (or could have been known).

Unpolarized photon: polarization not known (and could not have been known). In the terminology of QM, and following the Heisenberg Unvertainty Principle, the polarization is not definite until observed. You do NOT need to accept this explanation to use the term "unpolarized" as the term merely means that no polarization observation has been performed yet.

Polarization entangled photon: an unpolarized photon that shares a wave function with one or more particles.

DrC, thanks for this. BUT I still find it confusing.

For example, how does the following fit with your definitions, please?

Gottfried and Yan (Quantum Mechanics: Fundamentals 2nd Edition, 2003; page 447): 'Hence the polarization of an arbitrary one-photon state of a given momentum is specified by a real three ''vector'' E ... and it has no polarization whatsoever when E = 0.' (Emphasis added.)

Regards, wm

 

[DrChinese]

DrC, thanks for this. BUT I still find it confusing.

For example, how does the following fit with your definitions, please?

Gottfried and Yan (Quantum Mechanics: Fundamentals 2nd Edition, 2003; page 447): 'Hence the polarization of an arbitrary one-photon state of a given momentum is specified by a real three ''vector'' E ... and it has no polarization whatsoever when E = 0.' (Emphasis added.)

Regards, wm

You'll need to provide some context. How does that quote relate to your original question? You say it is a definition, what is it a definition of? What is the thing you are really asking?

Polarized light has a simple property: it will pass through a polarizer set at an appropriate angle with nearly 100% intensity. Unpolarized light has a 50% chance of passing through that same polarizer.

 

[wm]

You'll need to provide some context. How does that quote relate to your original question? You say it is a definition, what is it a definition of? What is the thing you are really asking?

Polarized light has a simple property: it will pass through a polarizer set at an appropriate angle with nearly 100% intensity. Unpolarized light has a 50% chance of passing through that same polarizer.

Doc, I was seeking to understand the nature of single photon polarisation. You supplied some definitions and I then asked how this piece from a text-book fitted with your definitions.

So my question was:

How does the following fit with your definitions, please?

Gottfried and Yan (Quantum Mechanics: Fundamentals 2nd Edition, 2003; page 447): 'Hence the polarization of an arbitrary one-photon state of a given momentum is specified by a real three ''vector'' E ... and it has no polarization whatsoever when E = 0.' (Emphasis added.)

PS: My own understanding is that paired photons in the singlet state are not polarised and that is why the singlet-state has spherical symmetry. (Which is a far greater symmetry than symmetry about the line-of-flight axis.)

Would ''unpolarised'' photons per your definition yield this spherical symmetry?

Thanks, wm

 

[DrChinese]

1. Doc, I was seeking to understand the nature of single photon polarisation. You supplied some definitions and I then asked how this piece from a text-book fitted with your definitions.

2. PS: My own understanding is that paired photons in the singlet state are not polarised and that is why the singlet-state has spherical symmetry. (Which is a far greater symmetry than symmetry about the line-of-flight axis.)

Would ''unpolarised'' photons per your definition yield this spherical symmetry?

Thanks, wm

wm,

1. Your question probably makes perfect sense to you. But I have no idea what the quote is supposed to represent as an idea. You were asking about polarized vs. unpolarized, and you introduced this quote into the picture. I have no idea how it relates. Is it supposed to be some kind of definition? You know what you have in your mind, but I don't and so you should explain this more clearly.

2. Entangled photon pairs - the kind used for Bell tests - are unpolarized. I am not sure what you mean when you say "spherical symmetry". I guess they are, seems reasonable, and the individual members certainly exhibit a strong symmetry in every respect that I am aware of - when considered as a pair. I have never heard this term used with entanglement though, so I am trying to understand how it became a part of the discussion.

 

[wm]

wm,

1. Your question probably makes perfect sense to you. But I have no idea what the quote is supposed to represent as an idea. You were asking about polarized vs. unpolarized, and you introduced this quote into the picture. I have no idea how it relates. Is it supposed to be some kind of definition? You know what you have in your mind, but I don't and so you should explain this more clearly.
<SNIP>

Doc, I was seeking to understand the nature of photon polarisation; in essence, to understand your view of Bell's theorem. For we read (in the QM literature) of polarised and unpolarised photons.

YOU then supplied the following definitions (emphasis added):

wm,

There are the following definitions:

Polarized photon: polarization known (or could have been known).

Unpolarized photon: polarization not known (and could not have been known). In the terminology of QM, and following the Heisenberg Unvertainty Principle, the polarization is not definite until observed. You do NOT need to accept this explanation to use the term "unpolarized" as the term merely means that no polarization observation has been performed yet.

Polarization entangled photon: an unpolarized photon that shares a wave function with one or more particles.

Then I replied:

DrC, thanks for this. BUT I still find it confusing.

For example, how does the following fit with your definitions, please?

Gottfried and Yan (Quantum Mechanics: Fundamentals 2nd Edition, 2003; page 447): 'Hence the polarization of an arbitrary one-photon state of a given momentum is specified by a real three ''vector'' E ... and it has no polarization whatsoever when E = 0.' (Emphasis added.)

THAT IS: Your definition indicates that an unpolarised photon is such that ''unpolarised'' may be taken to mean that no polarization observation has been performed yet.

The textbook appears to be discussing a photon state with no polarisation whatsoever.

SO I ask:

1. How does the textbook sit with your definition?

That is: How does the textbook's ''no polarisation whatsoever'' sit with YOUR definition that an unpolarised photon is one on which ''no polarisation observation has yet been made''?

2. How does your definition sit with common-sense?

ESPECIALLY given that I can send you photons in a definite state of polarisation; photons on which ''no polarisation observation has yet been made'' by anyone. That is: Definitely polarised photons are, by your definition, unpolarised because neither YOU (nor anyone else) have yet observed them??

THE LAST SENTENCE (appearing to be directly derived from you definition) APPEARS TO ME to be confusing; even crazy!?

Hope this helps, wm

 

[DrChinese]

Doc, I was seeking to understand the nature of photon polarisation; in essence, to understand your view of Bell's theorem. For we read (in the QM literature) of polarised and unpolarised photons.

...

ESPECIALLY given that I can send you photons in a definite state of polarisation; photons on which ''no polarisation observation has yet been made'' by anyone. That is: Definitely polarised photons are, by your definition, unpolarised because neither YOU (nor anyone else) have yet observed them??

THE LAST SENTENCE (appearing to be directly derived from you definition) APPEARS TO ME to be confusing; even crazy!?

Hope this helps, wm

I don't really follow what you are saying, because you seem to be mixing and matching words. The result doesn't match too well with the common way of expressing polarization.

I DO NOT claim that: an unpolarized photon has a polarization, but that we don't know what it is.

1. An unpolarized photon is in a mixed state (where H is horizontal and V is vertical):

H> + V>

2. A polarized photon is in a pure state:

H> (in whatever basis you choose to observe it)

3. A pair of entangled photons are also in a mixed state:

H>V> + V>H> (assuming Type II PDC)

As far as I know, it takes an observation to cause the mixed state to collapse to a pure state.

 

[calamero]

hello drchinese ,anonyn,everyone
think I've had an Einstein moment were everything becomes clear .but I am shaking and scared to say what i think BUT this idea I've just had when looking at why the wave collapses at dual slit I've hit upon a different way at looking at it and guess what ..here just now a minute ago i applied the same idea to that expirement wer how does the quanta know to show both red or both green etc when the switches at the collector points are set the same but random at other times its so simple an idea i don't know if anyones thought what I am thinking before but i havnt seen it so far and it all makes sense now i can explain why that happens or I've an idea why that happens and it makes sense to me ..im new here but I've wondered about that dual slit for ages now and past year i lie awake at night thinking about it so i have gave it a lot of thought ..ill sleep on this and think about it more in other situations before asking what you all think i just wanted to come here and say something..."heres hoping NOONE has had the same idea " because if they had and you lot arnt talking about it then it must have been crap :p

calamero