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

[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

 

[wm]

Non-locality: Getting specific

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.

DrChinese, Discussing photon polarisation, in the interests of advancing our understanding of non-locality:

Let's allow that I am sending you photons; one at a time, one every minute (say) on an agreed line of flight.

1. What is the wave-function (mathematically)?

2. What is the wave-function (physically, or in your own words)?

3. What is the observation that collapses the wave-function?

4. What then is the collapsed wave-function?

5. Please elaborate on any non-locality that enters your maths or your thinking.

PS: Should my terminology need clarification, I'd be happy to do that before you answer.

Thanks, wm

 

[DrChinese]

DrChinese, Discussing photon polarisation, in the interests of advancing our understanding of non-locality:

Let's allow that I am sending you photons; one at a time, one every minute (say) on an agreed line of flight.

1. What is the wave-function (mathematically)?

2. What is the wave-function (physically, or in your own words)?

3. What is the observation that collapses the wave-function?

4. What then is the collapsed wave-function?

5. Please elaborate on any non-locality that enters your maths or your thinking.

PS: Should my terminology need clarification, I'd be happy to do that before you answer.

Thanks, wm

The answers depend on the source of the photons. As I indicated, there are more or less 3 types: known entangled, known polarization, and unknown polarization. Their wave functions are as shown above. If they are in a mixed state, their wave function can be collapsed by a suitable observation.

If a photon was part of an entangled pair, then an observation on either will collapse the wave function for both. Keep in mind that entangled particles should not be considered as having separate existence until their wave function is collapsed.

Once you know a photon's polarization, it retains that value until a different polarization observation is performed.

I hope this helps.

 

[wm]

NON-LOCALITY: Getting specific.

The answers depend on the source of the photons. As I indicated, there are more or less 3 types: known entangled, known polarization, and unknown polarization. Their wave functions are as shown above. If they are in a mixed state, their wave function can be collapsed by a suitable observation.

If a photon was part of an entangled pair, then an observation on either will collapse the wave function for both. Keep in mind that entangled particles should not be considered as having separate existence until their wave function is collapsed.

Once you know a photon's polarization, it retains that value until a different polarization observation is performed.

I hope this helps.

DrC, It would really be more helpful if you answered the questions one-by-one? Is that not possible?

In the way that I specified the photons, is not the answer to the Q1 (in your terms) something like this:

(A1) |Y> = h|H> + v|V>; where you define the terms?

Does that not open the way for you to continue with your specific answers to the remaining questions?

PS: I am seeking to undertand your specific conceptualisation of non-locality. Presumably (as I understand your position) you will need to entertain (in your answers) the consequences that might follow if each photon sent to you were somehow paired with another sent to someone else. (You may exclude the possibility that they are observing them.)

Regards, wm

 

[wm]

Sorry; this was responding to another thread; though it derives from this thread so I'll leave it.

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

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

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

Jesse,

1. Do you see here how long you have sentences?

2. Does not my classical model (of old) refute this Bellian-Inequality easily? Are you not giving conditions which my model meets?

3. Are you not saying (as I will let you):

a. That Alice may make a countable-inifinity of detector-settings, each delivering outcome of {+1, -1}.

b. That Bob may make a countable-infinity of detector-settings, each delivering outcomes of {+1', -1'}.

4. Anyway: Down-hill skiers, dirty-socks, books and the like will satisfy your inequality. More subtle, less wholly concrete objects will sink it for some detector combinations. Yes?

5. Is my conclusion not what vanesch has shown?

wm