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

[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?

 

[lugita15]

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?

Obviously I know that we're talking about a set containing a lot of different photon pairs. That's the reason why I said "Let us denote the ith photon pair in s by Pi". And then I made a statement about each photon pair Pi in the set. And then I used that to make a statement about sets, namely {Pi : Mi(-30,30)} ⊆ {Pi : Mi(-30,0) or Mi(0,30)}.

So where am I going wrong? What makes you think that my argument doesn't work "when you now have three different pairs (1,2,3)"? Not only do I have 3 different photon pairs, I have many more photon pairs than that.

 

[lugita15]

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?

Yes, I agree that none of these three sets are disjoint from one another.

 

[wle]

So I'm not sure what you mean by "round" in anything I described.

Same thing as you mean by "pair", except I prefer the more agnostic language: the point of Bell's theorem is to prove certain constraints on the statistics allowed by locally causal theories. It's not a result specifically about entangled photons.

You can find the argument laid out clearly in three posts #270, #345 and #346.

I thought I already got the point from your post #346. Let's consider the scenario and model you propose: we measure three photon pairs. All the possible outcomes for each pair are predetermined by lists $\lambda_{i} = (a_{i}, b_{i}, c_{i},\; i=1,2,3$, and we consider the case where Alice and Bob measure the $ab$ term on the first pair, the $ac$ term on the second, and $bc$ on the third. You want to pick terms in the list such that the Bell correlator
$$S_{123} = a_{1} b_{1} + a_{2} c_{2} - b_{3} c_{3}$$
attains its maximal possible value of 3. No problem: just take, for instance, $a_{1} = a_{2} = b_{1} = b_{3} =c_{2} = 1$ and $c_{3} = -1$. This already fixes six of the nine variables you have to play around with.

If Alice and Bob instead decide to first measure $bc$, then $ab$, and finally $ac$ in that order, then you can use the remaining three free variables to maximise the relevant Bell correlator
$$S_{231} = a_{2} b_{2} + a_{3} c_{3} - b_{1} c_{1}$$
by setting $b_{2} = 1$ and $a_{3} = c_{2} = -1$. But now all the variables are fixed, and you still have four more orders in which Alice and Bob could measure the three terms, which now have fixed values:
$$\begin{eqnarray} S_{312} &=& -3 \,, \\ S_{132} &=& 1 \,, \\ S_{213} &=& 1 \,, \\ S_{321} &=& 1 \,. \end{eqnarray}$$
So in this hypothetical scenario, Alice and Bob only observe a violation if they measure the three terms in one of two specific orders ($ab$, $ac$, $bc$, or $bc$, $ab$, $ac$).

Of course, something like this was inevitable simply because
$$\begin{eqnarray} S_{123} + S_{231} + S_{312} &=& S_{111} + S_{222} + S_{333} \leq 3 \,, \\ S_{132} + S_{213} + S_{321} &=& S_{111} + S_{222} + S_{333} \leq 3 \,. \end{eqnarray}$$
This is what I meant by "tuning". Arrange the hidden variables in such a way that Alice and Bob will see a Bell violation for a particular sequence of measurements they could perform, and you will inevitably sabotage the correlator for other sequences of measurements they could just as well perform.

 

[billschnieder]

Obviously I know that we're talking about a set containing a lot of different photon pairs. That's the reason why I said "Let us denote the ith photon pair in s by Pi". And then I made a statement about each photon pair Pi in the set. And then I used that to make a statement about sets, namely {Pi : Mi(-30,30)} ⊆ {Pi : Mi(-30,0) or Mi(0,30)}.

So where am I going wrong? What makes you think that my argument doesn't work "when you now have three different pairs (1,2,3)"? Not only do I have 3 different photon pairs, I have many more photon pairs than that.

Again you are not understanding.

When you derived the relationship, you started by relating Mi(-30,30) to Mi(-30,0) and Mi(0,30). So that relationship applies to each individual photon pair in your set s, which could be as large as you like. But you also admit that the same relationship does not apply if between different pairs of photons, in other words if i=1 for the first term, and i=2 for the second term and i=3 for the third term. You agree that the relationship "M1(-30,30) implies M2(-30,0) or M3(0,30)" is not correct, even though pairs 1,2,3 might be within the same set "s".

So then you also must agree that the relationship "{Pi : Mi(-30,30)} ⊆ {Pj : Mi(-30,0) or Mk(0,30)}." is invalid, where the ith photon is drawn from a set p, and the jth photon is drawn from a set q, and the kth photon is drawn from a set r, and p, q, and r are disjoint sets. Do you agree?

 

[billschnieder]

Same thing as you mean by "pair", except I prefer the more agnostic language: the point of Bell's theorem is to prove certain constraints on the statistics allowed by locally causal theories. It's not a result specifically about entangled photons.

But the point is that the constraint on the statistics was not derived from the statistics but from the specific properties of entangled photons contributing to the statistics.

 

[lugita15]

So then you also must agree that the relationship "{Pi : Mi(-30,30)} ⊆ {Pj : Mi(-30,0) or Mk(0,30)}." where the ith photon is drawn from a set p, and the jth photon is drawn from a set q, and the kth photon is drawn from a set r, and p, q, and r are disjoint sets. Do you agree?

I don't even quite know what that expression means. But if the left hand side is a set of photon pairs in p and the right hand side is a set of photon pairs in q, obviously the statement is wrong, because p and q are disjoint.

By the way, I just wanted to make sure you noticed my post #392, because it was a reply to an edit you made to one of your posts.

 

[billschnieder]

Let's consider the scenario and model you propose: we measure three photon pairs. All the possible outcomes for each pair are predetermined by lists $\lambda_{i} = (a_{i}, b_{i}, c_{i},\; i=1,2,3$, and we consider the case where Alice and Bob measure the $ab$ term on the first pair, the $ac$ term on the second, and $bc$ on the third.
...
$$\begin{eqnarray} S_{312} &=& -3 \,, \\ S_{132} &=& 1 \,, \\ S_{213} &=& 1 \,, \\ S_{321} &=& 1 \,. \end{eqnarray}$$
So in this hypothetical scenario, Alice and Bob only observe a violation if they measure the three terms in one of two specific orders ($ab$, $ac$, $bc$, or $bc$, $ab$, $ac$).

You've described one scenario in which they could obtain a violation but as you know that is not the only one. Besides, if Alice measures 123 for one pair, how do you make sure that they also must measure 213 and 321 in order to compensate for the violation? This is what I explained in post #351 about a specific assumption that must be made for the inequality to apply to disjoint sets. So I'm not sure I get your argument, it appears you are agreeing with me but it is written as a disagreement.

 

[billschnieder]

I don't even quite know what that expression means. But if the left hand side is a set of photon pairs in p and the right hand side is a set of photon pairs in q, obviously the statement is wrong, because p and q are disjoint.

OK. Then if you agree that 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, and you agree that the relationship {Pi : Mi(-30,30)} ⊆ {Pj : Mi(-30,0) or Mk(0,30)}" is wrong, where each term is from a disjoint set, then you must also agree that:

The relative frequencies (lets use R) Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" can not all be equal to the corresonding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively. Do you agree?

 

[wle]

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

Of course.

Do you agree that the correct inequality for 3 separate pairs has a maximum of 3 on the RHS?

Obviously. I just filled in the blanks in one of your own examples in my last post.

One is enough to demonstrate that the inequality is not valid for 3 separate pairs.

Nobody has any reason to care that the inequality is not valid for 3 separate pairs. The validity and usefulness of Bell's theorem is not dependent on that.

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.

But if, based on a statistical analysis assuming locality, I work out that there is a less than one-in-a-billion chance of getting more than 1.001 when I estimate the Bell correlator on 10,000 photon pairs, and I actually get 1.5 when I do the experiment, I'm not going to think locality is very plausible.

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.

I never said it was.

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

It's not an assumption. It's an inevitable feature of any locally causal model. I've just illustrated this with your own model.

 

[wle]

But the point is that the constraint on the statistics was not derived from the statistics but from the specific properties of entangled photons contributing to the statistics.

No, the constraint on the statistics is derived from a very black box definition of locality that Bell introduced. From that point of view, entangled photons are interesting only as an example of a system we can manipulate that can produce statistics that violate those constraints.

 

[lugita15]

The relative frequencies (lets use R) Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" can not all be equal to the corresonding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively. Do you agree?

No, I don't agree with that at all. I believe that Rs(-30,0)=Rp(-30,0)=Rq(-30,0)=Rr(-30,0), and similarly for (0,30) and (-30,30).

And why do I believe that? Because if the relative frequency of something is the same for three different sets, then it is the same for their union, a statement I thought you agreed with.

 

[wle]

You've described one scenario in which they could obtain a violation but as you know that is not the only one.

If you're referring to the fact that I only considered one set of lists, then it was only intended as an example, and it's clear you're going to get similar results with any other set of lists anyway.

Besides, if Alice measures 123 for one pair, how do you make sure that they also must measure 213 and 321 in order to compensate for the violation?

They don't. If they happen to measure 123 then, in this particular scenario they get a very violent Bell violation despite the fact the model is local. That's their end result. If they were planning on writing a paper, then that's what they publish: they got 3. Same if they measured 231. But not if they measured 312, 132, 213, or 321. So if they pick the measurements completely randomly out of this set then according to this local model there's a 1 in 3 chance that they observe a Bell violation, and a 2 in 3 chance that they don't. This is in an experiment where the Bell correlator is estimated from measurement results on three photon pairs. The point is, it's not going to be a 1 in 3 chance for an experiment performed on, say, 10,000 photon pairs.

So I'm not sure I get your argument, it appears you are agreeing with me but it is written as a disagreement.

I'm saying that yes, a Bell inequality can be violated even according to a local theory, but it's not a problem if the chance of that happening becomes vanishingly small when the experiment is performed on a very large number of photon pairs.

If you're doing a Bell test on 10,000 entangled photon pairs and quantum physics says 1.5, does it really matter whether locality says < 1 or < 1.0001 except for some extremely low probability?

 

[billschnieder]

No, the constraint on the statistics is derived from a very black box definition of locality that Bell introduced. From that point of view, entangled photons are interesting only as an example of a system we can manipulate that can produce statistics that violate those constraints.

That is where you are mistaken, the constraints on the statistics exist precisely because those same constraints exist in the individual pairs. Let us verify this using Bell's original derivation http://www.drchinese.com/David/Bell_Compact.pdf and you can follow along:

Bell's expression (2) is an integral of the paired product of individual outcomes, then on Page 406, leading up to equation 15, he writes P(a,b) - P(a,c) on the LHS but note what is happening on the RHS, he is factoring out individual outcomes in precisely the way I did earlier for individual pairs. First he factors out A(a,λ) within the integral, and then he factors out A(b,λ) and as a result of this factorization, he creates the A(b,λ)A(c,λ) product within the integral which is then integrated to obtain the P(b,c) term. The factoring is a pivotal step! Obviously, the relationship between the statistics P(a,b), P(a,c) and P(b,c) is derived by considering the relationship between the individual outcomes being integrated.

 

[billschnieder]

This factorization step already implies that there are only three "lists" of outcomes [A(a,λi)], [A(b, λi)], and [A(c,λi)], which have been recombined and integrated over to obtain the terms P(a,b), P(a,c) and P(b,c). This is of course what Bell intended in the derivation because he simply assumes that another vector c exists along side a and b.

However, if you now want to use 3 correlations P(a1,b1), P(a2,c2) and P(b3,c3), obtained by recombining 6 lists [A(a1,λi)], [A(b1, λi)], [A(a2,λj)], [A(c2, λj)], [A(b3,λk)], and [A(c3,λk)], while maintaining the validity of the inequality derived for 3 lists, you MUST make one of the following two assumptions:

1. assume that [A(a1,λi)]=[A(a2, λj)] and [A(b1,λi)]=[A(b3, λk)] and [A(c2,λi)]=[A(c3,λk)]
2. assume that, whenever b3c3 = -1, there must be a corresponding "tuning" such that |a1b1 + a2c2| < 2 and when ever |a1b1 + a2c2| = 2, there must be a corresponding "tuning" such that b3c3 = 1.

These are the only two situations under which the inequality derived for a single set as Bell did, would apply to three different sets. It is not a choice, you do not have to like it, but by taking correlations obtained from three disjoint sets and comparing with the inequality, you ARE IN FACT making one or both of those assumptions whether you see it or not.

If you still disagree, I will ask you to derive the inequality starting with 3 different disjoint sets of photons or 6 lists of outcomes[A(a1,λi)]=[A(a2, λj)] and [A(b1,λi)]=[A(b3, λk)] and [A(c2,λi)]=[A(c3,λk)].

If you can derive the inequality from this starting point, without making one of those two assumptions then you will have proven that I am wrong on this point. If you prefer you can start with the statistics P(a1,b1), P(b3,c3) and P(a2,c2) and derive the inequalities. But we both know that you can't, because it is not possible. You will arrive at an inequality with 3 on the RHS, and you know this.

 

[billschnieder]

No, I don't agree with that at all. I believe that Rs(-30,0)=Rp(-30,0)=Rq(-30,0)=Rr(-30,0), and similarly for (0,30) and (-30,30).

At least now it is clear on which point we disagree. Or is it?

And why do I believe that? Because if the relative frequency of something is the same for three different sets, then it is the same for their union a statement I thought you agreed with.

I agree with this statement. But this is not the same statement as what we disagree with, nor does this statement prove or disprove what we disagree with. Can't you see that?

I said:

The relative frequencies Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" CAN NOT ALL BE EQUAL to the corresponding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively.

You are saying that the relative frequencies Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" ARE ALL EQUAL to the corresponding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively.


I think an example of such a set is in order.

 

[lugita15]

I agree with this statement. But this is not the same statement as what we disagree with, nor does this statement prove or disprove what we disagree with. Can't you see that?

No, I'm afraid I can't see that. The logic is really simple. We know that Rp(-30,0)=Rq(-30,0)=Rr(-30,0), so the relative frequency of M(-30,0) is the same for p, q, and r. Therefore it's the same for their union. So Rs(-30,0) is also equal to the same thing. (I can prove this if you like.) Where am I going wrong?

You are saying that the relative frequencies Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" ARE ALL EQUAL to the corresponding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively..

Yes, and the reason I'm saying that is because you agreed that Rp(-30,0), Rq(-30,0), and Rr(-30,0) are the same, and similarly for (0,30) and (-30,30).

 

[BenjaminTR]

Ok, how about this. Bill, if you think an LHV theory can match the predictions of QM, let's see a set of particles that does it.
QM predicts that 75% of pairs will differ when set at (-30,30), 25% will differ when measured at (-30,0) and 25% will differ when measured at (0,30). We want three disjoint sets that all obey these predictions separately, and thus whose union obeys them. Thus, consider 12 pairs, divided into groups a, b, and c, each with 4 pairs. For each pair, you just need to say whether it would give 1 or -1 at -30 degrees, 1 or -1 at 0 degrees and 1 or -1 at 30 degrees.
In other words, we just need 12 ordered triples from {-1,1}x{-1,1}x{-1,1} such that 9 of them differ between first and third place, 3 differ between first and second, and 3 differ between second and third. Then pick groups a, b and c so that 3 of their pairs differ between first and third place, 1 differ between first and second, and 1 differ between second and third.
Should be easy.

 

[billschnieder]

No, I'm afraid I can't see that.

Rs(-30,30) = Rs(-30,0) + Rs(0,30) - 2* Rs((-30,0)&(0, 30)), the equality
Rs((-30,0)&(0, 30)) >= 0 ... (*)
therefore Rs(-30,30) <= Rs(-30,0) + Rs(0,30), your inequality

therefore Rs((-30,0)&(0, 30)) = 0.5 * [Rs(-30,0) + Rs(0,30) - Rs(-30,30)] >= 0


If Rp=Rs(-30,30)=0.75, Rq=Rs(-30,0)=0.5, and Rr=Rs(0, -30)=0.5 ... (*)
then Rs((-30,0)&(0, 30)) = 0.5 * [0.5 + 0.5 - 0.75] = -0.125 < 0

You have two contradictory assumptions (*). If Rs((-30,0)&(0, 30)) >= 0 as you assumed when you derived the inequality, then it must be the case that the three correlations Rp(-30,30), Rq(-30,0), and Rr(0,30) CAN NOT ALL BE EQUAL to the three correlations Rs(-30,30), Rs(-30,0) and Rs(0,30). This is obvious because if the sets p, q, r are disjoint then R((-30,0)&(0, 30)) is meaningless, since M(-30,0) is measured in one set and M(0, 30) is measured in a disjoint set from the first. So your argument boils down to the tautological conclusion that the meaningless relative frequency is negative. In other words, you assumed that p, q, r were not disjoint (presence of Rs((-30,0)&(0, 30)) in the derivation), and then later assumed that they were disjoint --> violation.

 

[billschnieder]

Ok, how about this. Should be easy.

It should be easy for you to derive Bell's inequalities starting from 3 disjoint sets. How about that?

QM predicts that 75% of pairs will differ when set at (-30,30), 25% will differ when measured at (-30,0) and 25% will differ when measured at (0,30).

Wrong. QM does not predict that for the same set. QM predicts that for 3 disjoint sets, and those percentages do not violate the inequality derived from 3 disjoint sets. If you disagree, derive the inequality directly from 3 disjoint sets and show that such an inequality is still violated by QM.

The elephant in the room is this. Why do you always use the QM prediction for three disjoint sets when comparing with the inequality derived for a single set. Why don't you use the QM prediction for a single set while comparing with an inequality from a single set, or why don't you compare the QM prediction for 3 disjoint sets with the equivalent inequality derived for 3 disjoint sets? Why? Why? Why?

You may find the answer here:
Foundations of Physics Letters, Vol 15, No 5 (2002)
http://arxiv.org/pdf/quant-ph/0211031

 

[morrobay]

It should be easy for you to derive Bell's inequalities starting from 3 disjoint sets. How about that?



Wrong. QM does not predict that for the same set. QM predicts that for 3 disjoint sets, and those percentages do not violate the inequality derived from 3 disjoint sets. If you disagree, derive the inequality directly from 3 disjoint sets and show that such an inequality is still violated by QM.

The elephant in the room is this. Why do you always use the QM prediction for three disjoint sets when comparing with the inequality derived for a single set. Why don't you use the QM prediction for a single set while comparing with an inequality from a single set, or why don't you compare the QM prediction for 3 disjoint sets with the equivalent inequality derived for 3 disjoint sets? Why? Why? Why?

You may find the answer here:
Foundations of Physics Letters, Vol 15, No 5 (2002)
http://arxiv.org/pdf/quant-ph/0211031

The other elephant in the room is that n[y-z+] + n[x-y-] ≥ n[x-z+] is derived from values with parallel detector settings where P(α|λ| is considered deterministic. But then the inequality is compared against outcomes for non parallel detector settings: 45° , 90°, 135° where P(α|λ| can be non deterministic with respect to detector setting and state of particle at time of measurement.

 

[wle]

These are the only two situations under which the inequality derived for a single set as Bell did, would apply to three different sets. It is not a choice, you do not have to like it, but by taking correlations obtained from three disjoint sets and comparing with the inequality, you ARE IN FACT making one or both of those assumptions whether you see it or not.

You're substituting the point I made with a strawman. I am not making any such assumptions. That should have been clear to you since I explicitly agreed that the Bell correlator in a Bell test could violate the local bound you'd find in a textbook, even according to a local model.

But so what? Suppose it can be shown that the chance of a significant Bell violation becomes vanishingly unlikely in a Bell test involving measurements on thousands or millions of entangled pairs. Suppose I'm doing a CHSH test and expecting something close to the quantum bound of $2 \sqrt{2}$. Does it really matter that locality says "the probability of getting more than $2 + \varepsilon$ is less than a billion for some very small $\varepsilon$", instead of just "you should get less than 2"? That is the question you should be asking yourself.

 

[billschnieder]

I am not making any such assumptions.

You are. You have no other legitimate reason to compare correlations from 3 sets with an inequality derived from a single set. If you reject the only possible assumptions that allow you to make that comparison, then you can't continue to make the comparison. Period!

That should have been clear to you since I explicitly agreed that the Bell correlator in a Bell test could violate the local bound you'd find in a textbook, even according to a local model. But so what?

So you are not allowed to use the inequality, unless you assume that the model is tuned to avoid situations where violations occur. Derive the inequality from 3 sets and show that it is still violated.

 

[BenjaminTR]

[...]
Wrong. QM does not predict that for the same set. QM predicts that for 3 disjoint sets, and those percentages do not violate the inequality derived from 3 disjoint sets. If you disagree, derive the inequality directly from 3 disjoint sets and show that such an inequality is still violated by QM.

[...]

It's true the predictions are for three disjoint sets. However, if the expected fraction of disagreeing pairs measured at (-30,0) is 1/4, the LHV theory must either say that 1/4 of all photon pairs would disagree if measured in that direction, or that the photons we measure are systematically biased, i.e. no fair sampling. Without one of these two assumptions, the expected value is not 1/4. Ruling out conspiracies, we have to pick the first option, that 1/4 of the pairs would disagree at (-30,0), whether we measure that or not.

 

[wle]

You are. You have no other legitimate reason to compare correlations from 3 sets with an inequality derived from a single set. If you reject the only possible assumptions that allow you to make that comparison, then you can't continue to make the comparison. Period!

If you think I am making such assumptions, then instead of making vague accusations I invite you to point out where, exactly, I use such an assumption in for instance my [POST=4432843]post #393[/POST], and how the actual argument I actually made is affected by it.

So you are not allowed to use the inequality, unless you assume that the model is tuned to avoid situations where violations occur. Derive the inequality from 3 sets and show that it is still violated.

You are demanding something completely unnecessary. It is not necessary for the inequality to hold for all 3 sets. If you insist that that alone makes experimental Bell tests meaningless then you are knocking down a strawman.

Incidentally, you still haven't given a direct response to a question I've posed you at least three times now:

Suppose it can be shown that the chance of a significant Bell violation becomes vanishingly unlikely in a Bell test involving measurements on thousands or millions of entangled pairs. Suppose I'm doing a CHSH test and expecting something close to the quantum bound of $2 \sqrt{2}$. Does it really matter that locality says "the probability of getting more than $2 + \varepsilon$ is less than a billion for some very small $\varepsilon$", instead of just "you should get less than 2"? That is the question you should be asking yourself.

 

[BenjaminTR]

[...]If Rp=Rs(-30,30)=0.75, Rq=Rs(-30,0)=0.5, and Rr=Rs(0, -30)=0.5 ... (*)then Rs((-30,0)&(0, 30)) = 0.5 * [0.5 + 0.5 - 0.75] = -0.125 < 0You have two contradictory assumptions (*).

[...]

This is the point of the proof. If you assume LHV and the QM correlations, you get absurd results like negative probabilities. They cannot both be right, as you have just shown.

 

[billschnieder]

This is the point of the proof. If you assume LHV and the QM correlations, you get absurd results like negative probabilities. They cannot both be right, as you have just shown.

In other words, you have assumed what you claim to be proving, that's the whole point.

 

[billschnieder]

If you think I am making such assumptions, then instead of making vague accusations I invite you to point out where, exactly, I use such an assumption in for instance my [POST=4432843]post #393[/POST], and how the actual argument I actually made is affected by it.

Ok then:

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: $\langle S_{ijk} \rangle \leq 1$.

You admit that I'm right that Bell's inequality can be violated for a 3 separate pairs even for a locally causal theory. But then you argue that when you combine many such triples, the violations will be canceled out and you will not have a violation.

In other words you are saying everytime $S_{123}$ violates the inequality, somehow, Alice and Bob must also measure the corresponding $S_{123}$, $S_{213}$ AND $S_{312}$ so that the averages $\langle S_{ijk} \rangle \leq 1$ still obey the inequality. How is that not tuning? You brought up this argument yourself, how is it a strawman?

 

[billschnieder]

You are demanding something completely unnecessary. It is not necessary for the inequality to hold for all 3 sets.

It is necessary because you claim that the data from three sets should obey the inequality and then when it doesn't you make spooky conclusions. But if the data from three sets should not be required to obey the inequality, your spooky conclusions are unfounded. So if you insist that the data from three sets must not violate the inequality, you must also be able to derive the inequality from three sets. That you do not see this obvious logic is surprising indeed.

Incidentally, you still haven't given a direct response to a question I've posed you at least three times now:

Suppose it can be shown that the chance of a significant Bell violation [using correlations obtained from three disjoint sets of photons]* becomes vanishingly unlikely in a Bell test involving measurements on thousands or millions of entangled pairs.

*[emphasis added]

I've already answered your question, perhaps not the way you liked. I've asked you to show it without assuming tuning, rather than just supposing that it can be shown. Yet you say it is unnecessary. Derive the inequality from 3 disjoint sets and show that the chance of it being violated becomes vanishingly unlikely when millions/billions of entangled pairs are measured. You can't do it, unless you make one of the two assumptions I mentioned. It is clear what you have to do to clear this up.

 

[billschnieder]

Is anyone going to answer this question:

Why oh why do you always use the QM prediction for three disjoint sets when comparing with the inequality derived for a single set. Why don't you use the QM prediction for a single set while comparing with an inequality from a single set, or why don't you compare the QM prediction for 3 disjoint sets with the equivalent inequality derived for 3 disjoint sets? Why? Why? Why?

 

[DrChinese]

Is anyone going to answer this question:

Why oh why do you always use the QM prediction for three disjoint sets when comparing with the inequality derived for a single set. Why don't you use the QM prediction for a single set while comparing with an inequality from a single set, or why don't you compare the QM prediction for 3 disjoint sets with the equivalent inequality derived for 3 disjoint sets? Why? Why? Why?

Everybody can answer why. You just dismiss the answer.

Local realism means that when examining any single stream of entangled particle pairs, you can pick any pair of angles (of 3 angles identified by Bell's Inequality) to determine attributes of the stream. According to EPR: it would be unreasonable to require that all 3 were simultaneously measurable, as you seem to want to.

I really believe this thread is going around in circles. I fail to see where you have added anything other than non-standard viewpoints anytime recently.