Understanding bell's theorem: why hidden variables imply a linear relationship?

[billschnieder]

No.
Because you are making the assumption that somehow the LHV knows on beforehand what the polarizer settings will be, which of course is non-locality.

And where did you get such a ridiculously wrong idea from, in anything I said?

 

[billschnieder]

So again, can you phrase your argument in terms of the concepts and terminology we've been discussing all this time?

OK, let us start by you deriving the inequality you brought up, quoted in my previous post to you.

 

[DevilsAvocado]

And where did you get such a ridiculously wrong idea from, in anything I said?

The ridiculous idea:

However if we calculate the "ab" term from the first photon pair, the "ac" term from the second photon pair and "bc" from the third pair (ie, a1b1, a2c2, b3c3), is it reasonable to expect the inequality |ab + ac| - bc ≤ 1 to be obeyed?

Similarly if we calculate the "ab" correlation from the first set of photons, the "ac" correlation from the second set of photons and the "bc" correlation from the third set of photons (ie a1b1, a2c2, b3c3), is it reasonable to expect the inequality |C(a,b) + C(a,c)| - C(b,c) ≤ 1 to be obeyed?

--> I say the answer to the above two questions is NO. <--

If you could just explain in fairly simple English exactly how three (3) different photon pair/sets are going to violate Bell’s inequality without CHANGING their properties between the three (3) different photon pair/sets?

It doesn’t work, does it?

Without change you will get THE SAME result as “one pair/set”, right?

Now, what are you going to use as ‘parameter’ for your change? I’m all ears...

 

[billschnieder]

The ridiculous idea:
If you could just explain in fairly simple English exactly how three (3) different photon pair/sets are going to violate Bell’s inequality without CHANGING their properties between the three (3) different photon pair/sets?

Why do they have to change anything? They are three different pairs, each pair is free to have values +1 and -1 for each of the outcomes.

Without change you will get THE SAME result as “one pair/set”, right?

Wrong. Again you are surely confused about something, there is no changing involved. For a single a pair, you have 8 possible outcome combinations, ALL of which satisfy the inequality. For three separate pairs, you have 64 possible outcome combinations, 16 of which violate the inequality.

 

[DevilsAvocado]

The same ridiculous idea:

For three separate pairs, you have 64 possible outcome combinations, 16 of which violate the inequality.

I’m just wondering... of 64 possible outcome combinations you have the grand opportunity to get it right – and violate the inequality – in one out of four = 25% ... what are you going to do with the 75% that blow your dream into pieces?? Dismiss them as “illegal”??

 

[billschnieder]

I’m just wondering... of 64 possible outcome combinations you have the grand opportunity to get it right – and violate the inequality – in one out of four = 25% ... what are you going to do with the 75% that blow your dream into pieces?? Dismiss them as “illegal”??

A single violation is enough to disqualify the inequality as a valid rule for three separate pairs. What part of that don't you understand? If you want to keep the inequality for three separate pairs, it is you who will have to design a system which actively eliminates the 25% "illegal" cases.

 

[DevilsAvocado]

A single violation is enough to disqualify the inequality as a valid rule for three separate pairs.

Please Bill, you can’t be serious... you are going to cherry pick 16 combinations out of 64 and stop there and convince yourself that this is science/math...

Gosh... I thought you knew probability theory. I can win the lottery every day with your “magical formula”...

I guess this is getting way out in “Outer Crackpot Space”. I’m sorry Bill, my time is unfortunately not infinite and this is going absolutely nowhere (except to OCS)...

Take care and have a nice summer.

 

[lugita15]

But I've already answered this question. I've already agreed that if the relative frequency of "something" is the same in three disjoint sets, then the relative frequency of that same "something" will be the same in the union of those three disjoint sets..

Well, you did seem to agree initially. But then when I asked you whether all three relative frequencies have the same value for s that they do for p, q, and r, you said no. Well, if you don't think all three relative frequencies have the same value for s as they do for p, q, and r, then that means you think that at least one of them DOESN'T have the same value for s as it does for p, q, and r. In other words, you must think the relative frequency of something can be different for the union even though it's the same for the three sets. Where am I going wrong here?

I agreed to this question already. However, to make things clear, please can you prove this inequality by deriving it, it will be evident later why this is important and since you brought it up, let it be on you to derive it from a single set "s"

All right, for any photon pair in s, if f(-30,30) is equal to (1,-1) or (-1,1), then either f(-30,0) is equal to (1,-1) or (-1,1), or f(0,30) is equal to (1,-1) or (-1,1).

Therefore, the number of photon pairs in s for which f(-30,30) is equal to (1,-1) or (-1,1) is less than or equal to number of photon pairs in s for which f(-30,0) is equal to (1,-1) or (-1,1), plus the number of photon pairs in s for which f(0,30) is equal to (1,-1) or (-1,1). Dividing both sides by the total number of photon pairs in s, it follows that the relative frequency in s of f(-30,30) being equal to (1,-1) or (-1,1) is less than or equal to the relative frequency in s of f(-30,0) being equal to (1,-1) or (-1,1), plus the relative frequency in s of f(0,30) being equal to (1,-1) or (-1,1).

 

[billschnieder]

Please Bill, you can’t be serious... you are going to cherry pick 16 combinations out of 64 and stop there and convince yourself that this is science/math...

I don't what to make any assumptions about your ability to understand simple logic so I will explain again and hopefully you will understand this time. If somebody argues that A ≤ B, and I produce one instance in which A > B, the argument is disproved. Do you understand this simple logic? So if out of 64 possible outcomes A > B for 16 of them, those are 16 different reasons, each powerful enough to disprove the argument on it's own, but together providing overwhelming evidence that without a doubt, the original argument is false. There is no cherry picking involved. It is the person making the original argument who will now have to cherry pick and say, "but A ≤ B is valid only for these 48 possibilities, in which those 16 are not allowed". Do you understand this simple logic?

If you do not understand even this, then there is no hope left. Maybe if you read DrC's signature, you might understand it.

 

[billschnieder]

Well, you did seem to agree initially. But then when I asked you whether all three relative frequencies have the same value for s that they do for p, q, and r, you said no.

Obviously, this is not an accurate representation of my disagreement. We are discussing a topic with may subtle elements and it pays to be proper in characterizing the components of the discussion. When you talk of "all three relative frequencies" it could mean f(-30,0) in p, f(-30, 0) in q and f(-30, 0) in r. But it could mean f(-30,0) in p, f(-30,30) in q and f(0, 30) in r. So you may be thinking that I disagree with one thing, when actually I disagree with something else and it has appeared to me at times that you are going back and forth between them without realizing the difference. For this reason, I would advice and appreciate that you adopt a more terse notation since your current one of talking about "f(-30,30) is equal to (1,-1) or (-1,1)" can easily obscure subtle issues. Since we are talking about mismatch at (-30, 30), I would suggest you simply say "M(-30,30)" if you don't mind, or any other more terse notation.

 

[billschnieder]

All right, for any photon pair in s, if f(-30,30) is equal to (1,-1) or (-1,1), then either f(-30,0) is equal to (1,-1) or (-1,1), or f(0,30) is equal to (1,-1) or (-1,1).

Why must that be the case? Please be explicit.

I would say for 3 predetermined outcomes from a single pair of photons at -30, 0, 30, if the outcome at -30 mismatches the outcome at 30, then the outcome at 0 must mismatch one of either -30 or 30. Do you agree that this is how you arrived at the above?

Therefore, the number of photon pairs in s for which f(-30,30) is equal to (1,-1) or (-1,1) is less than or equal to number of photon pairs in s for which f(-30,0) is equal to (1,-1) or (-1,1), plus the number of photon pairs in s for which f(0,30) is equal to (1,-1) or (-1,1)

This is a giant leap. Please I'll appreciate if you could show step by step how you went from the above to this inequality.

 

[DevilsAvocado]

Final comment:

then there is no hope left. Maybe if you read DrC's signature, you might understand it.

Don’t want to crush your dreams or something, but you are not Einstein and you will not get a Nobel Prize for this mess, but it doesn’t matter much.

Go have a beer or two, see a football match, listen to some good music, dance with a beautiful girl, go out and enjoy the sun, swim in a lake – do anything but this crazy activity – because it will take you nowhere = total waste of time.

Now, listen to a wise-guy-avocado!

 

[lugita15]

Obviously, this is not an accurate representation of my disagreement. We are discussing a topic with may subtle elements and it pays to be proper in characterizing the components of the discussion. When you talk of "all three relative frequencies" it could mean f(-30,0) in p, f(-30, 0) in q and f(-30, 0) in r. But it could mean f(-30,0) in p, f(-30,30) in q and f(0, 30) in r. So you may be thinking that I disagree with one thing, when actually I disagree with something else and it has appeared to me at times that you are going back and forth between them without realizing the difference. For this reason, I would advice and appreciate that you adopt a more terse notation since your current one of talking about "f(-30,30) is equal to (1,-1) or (-1,1)" can easily obscure subtle issues. Since we are talking about mismatch at (-30, 30), I would suggest you simply say "M(-30,30)" if you don't mind, or any other more terse notation.

OK, I'm fine with abbreviating "f(θ1,θ2) equals (1,-1) or (-1,1)" as M(θ1,θ2). Let me restate what I said in terms of this notation.

You agreed that the relative frequency of M(-30,0), the relative frequency of M(0,30), and the relative frequency of M(-30,30), are the same for p, q, and r. However, you disagreed that the relative frequency of M(-30,0), the relative frequency of M(0,30), and the relative frequency of M(-30,30), are the same for s as they are in p, q, and r. That means that you believe that the relative frequency of either M(-30,0), M(0,30), or M(-30,30) has a different value for s than it does for p, q, and r. Do I have that right?

 

[lugita15]

Why must that be the case? Please be explicit.

I would say for 3 predetermined outcomes from a single pair of photons at -30, 0, 30, if the outcome at -30 mismatches the outcome at 30, then the outcome at 0 must mismatch one of either -30 or 30. Do you agree that this is how you arrived at the above?

Yes, that's exactly how I arrived at it. (Except of course I'd replace "the outcome at" with "the outcome you would get at".) I was only talking about a single photon pair in that quote.

This is a giant leap. Please I'll appreciate if you could show step by step how you went from the above to this inequality.

OK, so we know that for each photon pair in s, M(-30,30) implies M(-30,0) or M(0,30). Therefore, the set of photon pairs in s for which M(-30,30) is a subset of the set of photon pairs in s for which M(-30,0) or M(0,30). And this implies that the number of photon pairs in s for which M(-30,30) is less or equal to the number of photon pairs in s for which M(-30,0) or M(0,30). And then the number of photon pairs in s for which M(-30,0) or M(0,30) is less than or equal to the number of photon pairs in s for which M(-30,0), plus the number of photon pairs in s for which M(0,30). Therefore, the number of photon pairs in s for which M(-30,30) is less than or equal to the number of photon pairs in s for which M(-30,0), plus the number of photon pairs in s for which M(0,30).

 

[billschnieder]

OK, so we know that for each photon pair in s, M(-30,30) implies M(-30,0) or M(0,30). Therefore, the set of photon pairs in s for which M(-30,30) is a subset of the set of photon pairs in s for which M(-30,0) or M(0,30).

You keep jumping steps directly to the inequality. Is there a specific reason why? In other words, why is the set of pairs in s for M(-30,30) only a subset of pairs in s for "M(-30,0) or M(0,30)". Please be explicit.

 

[billschnieder]

Final comment:

Really! I doubt it.

do anything but this crazy activity – because it will take you nowhere = total waste of time.

Translation: "My mind is made up, don't confuse me with the facts!"

 

[lugita15]

You keep jumping steps directly to the inequality. Is there a specific reason why? In other words, why is the set of pairs in s for M(-30,30) only a subset of pairs in s for "M(-30,0) or M(0,30)". Please be explicit.

Sorry, I think you're having a misunderstanding about what subset means. A subset is allowed to equal the whole set. I think the term you're thinking of is proper subset.

A set A is said to be a subset of a set B if a being an element of A implies that a is an element of B. In our case, for any given photon pair in s, if M(-30,30), then M(-30,0) or M(0,30). Therefore, by definition of subset the set of photon pairs for which M(-30,30) is a subset of the set of photon pairs for which M(-30,0) or M(0,30).

 

[billschnieder]

Sorry, I think you're having a misunderstanding about what subset means. A subset is allowed to equal the whole set. I think the term you're thinking of is proper subset.

A set A is said to be a subset of a set B if a being an element of A implies that a is an element of B. In our case, for any given photon pair in s, if M(-30,30), then M(-30,0) or M(0,30). Therefore, by definition of subset the set of photon pairs for which M(-30,30) is a subset of the set of photon pairs for which M(-30,0) or M(0,30).

No. I think you are having a misunderstanding. I did not say it can not be the whole set. You said it was a subset. I'm asking you why you use the word "subset" to begin with rather than "whole set". You are implying by the choice of the word that there could be a situation in which it might not be the whole set and there might be situations in which it might be the whole set. I'm asking you to state explicitly what those situations are. In other words, I'm asking you to justify the inequality rather than just state it.

But since you did not answer, let me answer the question and you can tell me if you agree or disagree.

({M(-30, 0)} U {M(0, 30)}) * {M(-30,0)&M(0, 30)} = {M(-30,30)}

In words, the set in s where M(-30,30) is the difference of the set in s where "M(-30,0) or M(0,30)" and the set in s where "M(-30,0) and M(0, 30)". In the case where the set with "M(-30,0) and M(0, 30)" is null, the set where M(-30, 30) will be the same as the set where "M(-30,0) or M(0,30)". In other words the inequality implies that the set in s where "M(-30,0) and M(0, 30)" is either null or non-empty. Do you agree?

This is a crucial missing piece in your description as will be evident later.

 

[lugita15]

But since you did not answer, let me answer the question and you can tell me if you agree or disagree.

({M(-30, 0)} U {M(0, 30)}) * {M(-30,0)&M(0, 30)} = {M(-30,30)}

In words, the set in s where M(-30,30) is the difference of the set in s where "M(-30,0) or M(0,30)" and the set in s where "M(-30,0) and M(0, 30)". In the case where the set with "M(-30,0) and M(0, 30)" is null, the set where M(-30, 30) will be the same as the set where "M(-30,0) or M(0,30)".

Yes, I agree with this, although I didn't need it to derive the inequality.

 

[wle]

Thought I'd dive in here:

In other words, the relationship we derived from properties within a single pair of photons does not apply between properties from 3 different pairs of photons.

That only works because (referring to the notation used in derivations of Bell inequalities), you've not only allowed the past information \lambda to change from one round of a Bell test to the next, but you've given a model where the hidden variables are specifically tuned for the measurement settings in each round.

Suppose I add a subscript i to the individual correlators, referring to the correlator's expectation value in the ith round of a Bell experiment. Then if I understand you correctly, you're saying that Bell's theorem would only establish something like

C_{i}(\bar{a}, \bar{b}) + C_{i}(\bar{a}, \bar{c}) - C_{i}(\bar{b}, \bar{c}) \leq 1 \,, \quad \forall i \,. \quad(*)
But in an actual Bell test, Alice and Bob might measure (\bar{a}, \bar{b}) in the first round, (\bar{a}, \bar{c}) in the second round, and (\bar{b}, \bar{c}) in the third. There's no particular reason the inequality should hold for correlators from different rounds, so it is entirely possible that

C_{1}(\bar{a}, \bar{b}) + C_{2}(\bar{a}, \bar{c}) - C_{3}(\bar{b}, \bar{c}) \nleq 1 \,.
If that's your point, then it's true but it doesn't invalidate Bell's theorem or render it untestable. The reason is that Alice and Bob's measurements are supposed to be chosen randomly, so you don't know in advance which measurements they're going to perform in each round. This matters because, while you can arrange for the Bell inequality to be violated for a particular sequence of measurements with a locally causal model, you can't arrange for it to be violated for all of them.

This is easy to see. Suppose we consider a three-round Bell test where each of the three terms in the inequality is measured once^*, and we define S_{ijk} = C_{i}(\bar{a}, \bar{b}) + C_{j}(\bar{a}, \bar{c}) - C_{k}(\bar{b}, \bar{c}), with the labels ijk identifying which correlator was tested in each round (there are six possible permutations: 123, 132, 231, 213, 312, 321). Then it is entirely possible that the inequality S_{123} \leq 1 is violated, but not for instance all three of S_{123} \leq 1, S_{231} \leq 1, and S_{312} \leq 1 simply because the condition (*) above implies

S_{123} + S_{231} + S_{312} \leq 3 \,.
The average over the six possible tests, which is also the expectation value of the full Bell correlator, satisfies the inequality: \langle S_{ijk} \rangle \leq 1.

So while it is possible to observe a Bell inequality violation according to local causality, such a violation would constitute a statistical outlier. It cannot be guaranteed deterministically if Alice's and Bob's measurements are not known in advance, and the probability of a significant Bell inequality violation drops exponentially fast in the number of rounds.

The bottom line is: to get anything more than an "accidental" Bell inequality violation in a locally causal model, it's not enough just to let the hidden variable \lambda vary from one round of a Bell experiment to the next. It must also be correlated with Alice's and Bob's measurement choices (in violation of what's sometimes called the "no conspiracies" or "free will" assumption).

As an aside, is there a reason that this discussion has been based around Bell's original 1964 inequality? Refreshing my memory on it, my impression is that more modern Bell inequalities (such as CHSH) admit cleaner and more "black box" derivations.

----------

^* This is just for simplicity of exposition. Of course, there would be no way for Alice and Bob to do this without communicating with each other.

 

[DrChinese]

Thought I'd dive in here:...

Welcome to PhysicsForums, wle!

 

[billschnieder]

I would say for 3 predetermined outcomes from a single pair of photons at -30, 0, 30, if the outcome at -30 mismatches the outcome at 30, then the outcome at 0 must mismatch one of either -30 or 30.

Yes, that's exactly how I arrived at it. (Except of course I'd replace "the outcome at" with "the outcome you would get at".) I was only talking about a single photon pair in that quote.

Clearly then, you obtain your inequality by extending a relationship which exists between "outcomes you would get" for a single pair of photons to a set of such photons. That is you started with

M(-30,30) implies M(-30,0) or M(0,30)

What if you start with 3 different pairs of photons (1,2,3). Do you now agree that the following statement is not necessarily true:

M1(-30,30) implies M2(-30,0) or M3(0,30)

And consequently do you agree that the inequality you obtained by extending the relationship from a single pair to a single set is not necessarily true for three separate pairs and consequently 3 separate sets?

If you agree to the above two questions, do you agree that for the relationship for a single pair to be true for 3 different pairs, you MUST make an additional assumption? If you agree, I would like you to state that particular assumption and explain why it changes the relationship from one which is not necessarily true to one which is necessarily true.

 

[billschnieder]

That only works because (referring to the notation used in derivations of Bell inequalities), you've not only allowed the past information \lambda to change from one round of a Bell test to the next, but you've given a model where the hidden variables are specifically tuned for the measurement settings in each round.

I do not agree with this characterization. There is no specific tuning involved in anything I've described. I've been talking about individual pairs, 3 different pairs, a single set, and three distinct sets. So I'm not sure what you mean by "round" in anything I described.

Suppose I add a subscript i to the individual correlators, referring to the correlator's expectation value in the ith round of a Bell experiment. Then if I understand you correctly, you're saying that Bell's theorem would only establish something like

C_{i}(\bar{a}, \bar{b}) + C_{i}(\bar{a}, \bar{c}) - C_{i}(\bar{b}, \bar{c}) \leq 1 \,, \quad \forall i \,. \quad(*)

That is not an accurate characterization of my argument. You can find the argument laid out clearly in three posts #270, #345 and #346. To obtain the inequality for a single set, I started with the relationship within a single pair and extended it to the single set just like Bell did. In other words, the inequality appears to be a relationship between statistics, but it is in fact an arithmetic identity in a single pair that has been extended to a single set.

 

[billschnieder]

This matters because, while you can arrange for the Bell inequality to be violated for a particular sequence of measurements with a locally causal model, you can't arrange for it to be violated for all of them.

So you agree that the inequality is not a valid inequality for three separate pairs? Do you agree that the correct inequality for 3 separate pairs has a maximum of 3 on the RHS? Nobody claimed that the inequality must be violated for every 3 pairs. One is enough to demonstrate that the inequality is not valid for 3 separate pairs. If you want to apply it to 3 separate pairs, you must make an additional assumption about the 3 separate pairs. As concerns 3 separate sets, there is a long way between ≤1 and ≤3, and 1.1 is a violation just as much as 3. If somebody were claiming (though nobody is) a maximal violation of 3, then you could make the argument that it should be violated everytime.

(there are six possible permutations: 123, 132, 231, 213, 312, 321). Then it is entirely possible that the inequality S_{123} \leq 1 is violated, but not for instance all three of S_{123} \leq 1, S_{231} \leq 1, and S_{312} \leq 1 simply because the condition (*) above implies

S_{123} + S_{231} + S_{312} \leq 3 \,.
The average over the six possible tests, which is also the expectation value of the full Bell correlator, satisfies the inequality: \langle S_{ijk} \rangle \leq 1.

But you do realize that
\langle S_{123} + S_{231} + S_{312}\rangle \neq \langle S_{123}\rangle + \langle S_{231}\rangle + \langle S_{312}\rangle read that as not necessarily equal.

So the above argument is invalid unless you are making an additional assumption that whenever there is a violation in one of correlators, there will be a compensating correlation in the other correlators, and how is that possible without action at a distance. In otherwords, for your argument to be valid, you must assume action at a distance/conspiracy.

 

[BenjaminTR]

So you agree that the inequality is not a valid inequality for three separate pairs?

[...]

Bell's inequality is about three separate expectation values, not three separate pairs. The results you actually measure for three pairs need not match the expectation values.

 

[billschnieder]

Bell's inequality is about three separate expectation values, not three separate pairs. The results you actually measure for three pairs need not match the expectation values.

Have you been paying attention to anything that has been discussed so far? From your statement above, it appears you haven't been paying close attention to the details of the discussion. See posts #270, #345 and #346 for a summary of the argument.

 

[billschnieder]

As an aside, is there a reason that this discussion has been based around Bell's original 1964 inequality? Refreshing my memory on it, my impression is that more modern Bell inequalities (such as CHSH) admit cleaner and more "black box" derivations.

The arguments I've presented apply equally well to 4-term CHSH inequalities. I've been using the 3 term version partly because the original argument put forth by lugita15 involved 3 terms and partly because it is less cumbersome to discuss the 3-term version.

 

[wle]

Welcome to PhysicsForums, wle!

Thanks!

 

[lugita15]

What if you start with 3 different pairs of photons (1,2,3). Do you now agree that the following statement is not necessarily true:

M1(-30,30) implies M2(-30,0) or M3(0,30)

Yes, obviously it would be wrong to assume that. But I'm not assuming that at all. Let me go through the logic again.

Let us denote the i^th photon pair in s by Pi. Then for each i, if Mi(-30,30), then Mi(-30,0) or Mi(0,30). Therefore, {Pi : Mi(-30,30)} ⊆ {Pi : Mi(-30,0) or Mi(0,30)}, and thus the number of elements in {Pi : Mi(-30,30)} is less than or equal to the number of elements in {Pi : Mi(-30,0) or Mi(0,30)}.

 

[billschnieder]

Yes, obviously it would be wrong to assume that.

I'm happy you see that.

But I'm not assuming that at all. Let me go through the logic again.
Let us denote the i^th photon pair in s by Pi. Then for each i, if Mi(-30,30), then Mi(-30,0) or Mi(0,30). Therefore, {Pi : Mi(-30,30)} ⊆ {Pi : Mi(-30,0) or Mi(0,30)}, and thus the number of elements in {Pi : Mi(-30,30)} is less than or equal to the number of elements in {Pi : Mi(-30,0) or Mi(0,30)}.

You are not understanding. What you have above is fine and good for a single pair Pi and therefore for a single set. There is no question about that. The only question is when you now have three different pairs (1,2,3). Are you saying in your argument, you will never apply this relationship above to three separate pairs? Are you saying you will never apply inequalities derived from the above relationship to three disjoint sets?

Do you agree that the set {Pi: Mi(-30,30)} in s, is not disjoint from the set {Pi: Mi(0,30)} in s nor are those two previous sets disjoint from the set {Pi: Mi(0,30)} in s? In other words, starting with a single set "s", it is impossible for {Pi: Mi(-30,30)}, {Pi: Mi(0,30)} and {Pi: Mi(0,30)} to be three disjoint sets?