Bell Imposed -0 and +0 Bounds on the CHSH Correlator

[minkwe]

I just posted the following on RW in response to Jay,

Jay R. Yablon
AB + AB’ + A’B – A’B’ (1)
Each of the four terms is an element of reality. Nobody disagrees.

I do not disagree either. I simply say it is not sufficiently clear to say each of those terms is an element of reality. I say the discussion will go much more smoothly if you identify and everyone agrees what kind of reality they represent -- *possibilities* or *actualities* there is a risk of losing sight of very important subtleties if that distinction is not made clear at this point.

Jay R. Yablon
But only one of those four terms can ever be measured in one trial if that trial truly represents EPR. The other three terms may be reality, but we cannot measure them, so they are unknowable reality. The disputes between Joy and others all boil down to what this means for Bell.

Rather, I would say all those terms A, A’, B, B’ are *possibilities*. But since the A contradicts A’, and B contradicts B’, their linear combination in the form

AB + AB’ + A’B – A’B’ (1)

is NOT a *possibility* since it amounts to accepting mutually exclusive possibilities as simultaneously true — a contradiction. I think this is Joy’s point, in my own words.

Jay R. Yablon
Others say it means nothing, because the other three are still elements of reality by postulate, and that this postulate also renders irrelevant that the other three terms are not and cannot be measured.

As I hope you now see, these “others” can’t reasonably say that, if we were clear from the beginning about distinguishing *possibilities* from *actualities*. Just like it is obviously unreasonable to admit the simultaneous logical truth of two mutually contradictory statements.

Jay R. Yablon
Forgetting about all of this, I pose only two questions:
First: is the SUM in (1) itself an “element of reality”? But since I do not even want to try to define “element of reality” at this moment, let me borrow from Star Trek and call each term in (1) a “tribble” and ask whether the whole sum in (1) is a “tribble,” knowing that at the end I will set:
“tribble” = “element of reality” (2)

The problem does not go away because you use “tribble” it is only made worse, so long as you do not identify that there are two types of “tribbles”. You give wiggle room later, for someone to intentionally or mistakenly switch from one type to another at the expense of clarity. There is a type of “tribble” for which the sum (1) will also be a “tribble” and there is a a different type of “tribble” for which the sum (1) is not a “tribble”. This is the subtlety being missed. If I follow along, then if A, A’, B, B’ are “tribbles”, I must interpret tribble later as
“tribbles” = “possibilities”
and at no point in the future is it allowed to use
“tribbles” = “actualities”
With that in mind, I would answer that the sum (1) is not a “tribble” since it involves a logical contradiction, and the maximum value of the sum(1) is 2.

Jay R. Yablon
AB + AB’ + A’B – A’B’ = 2 = 2*(H+T) (3)
where either H=1 and T=0, or H=0 and T=1. Mutually exclusive. Because of this orthogonality, we can also discuss this with two state vectors (1 0) and (0 1). And we all know that this 2 turns into the CHSH bound |2|.
So, now my first question metamorphises to this: is 2*(H+T) a tribble, because that is the sum of three tribbles minus the fourth tribble. If you permit me to divide out the 2, then the question is more simply stated:
Is 1 = H+T a tribble, or is not a tribble?

if H, T are mutually exclusive tribbles, then H + T is not a “tribble”. But as I’ve been trying hard to explain, (H, T) in your discussion are not always mutually exclusive. If H, T are head/Tail counters as you say, then they are not mutually exclusive and are therefore NOT “tribbles”! So without the clarity, it may appear reasonable (although it is absolutely not), for somebody to go from your coin toss analogy to suggest that the (1) is also a tribble by glossing over details.

For example, If 0 means down, and 1 means up, then a single coin toss can give you (H=0, T=1) both of which are *actualities*. And because *actualities* are necessarily consistent with each other, H + T = 1 is also an *actuality*. Now if I permit you to be obscure with your “tribble” terminology, then you might think because you called H, T mutually exclusive possibilities, your H + T represents a combination of tribbles similar to (1). But that is false. For your H/T counters, the mutually exclusive possibilities for a single coin are not H or T, they are (H=0, T=1) or (H=1, T=0), and your sum (H + T) is not a sum of those, therefore H + T from H/T counters is not a linear combination of mutually exclusive possibilities in the same manner as (1). And therefore you can’t reasonably draw a conclusion about the “tribbleness” of (1) by using the tribbleness of the expression (H+T=1, where H, T are H/T counters). You have fallen prey to the trap I’ve been trying to warn you about. H/T counters are *actualities* not *possibilities*!

Jay R. Yablon
The CHSH sum itself contains no accompanying statement as to whether we are asking what this sum may be in advance of a measurement, or what it actually was after a measurement. That is part of what we all need to fill in.

Here I disagree, it does by implication. *possibilities* vs *actualities* correspond to *prediction* vs *measurement*. The output of predictions are *possibilities* and the outputs of measurements are *actualities*. Again all of this confusion will be cleared-up at once if you would seriously consider my clarifying suggestions to use *possibilities* and *actualities* instead of *elements of reality* or *tribble*.

Jay R. Yablon
So I am asking everybody to tell me if 1 = H + T ought to be or ought not to be regarded as a tribble. But if it is not a tribble, then the sum in (1) is not a tribble, even though it combines four individual tribbles.

I would answer that your language has unfortunately not been precise enough to accomplish your noble goal, which I share.

[FrediFizzx]

That Is very good. Hopefully it will get Jay straight and maybe some lurkers.

[Dirkman]

To me, 1=H+T would be an element of reality if H and T exist at the same time. And for a coin or a slip its obviously true. H is on one side and T is on the other side, both at the same time. And with this assumption you get a CHSH bound of 2. So when you switch to QM, the bound is violated meaning H doesnt coexist with T, so there's no such thing as...superposition ? Hmmm...I must be wrong somehow.

[Joy Christian]

***
My reply to Jay's question on Retraction Watch: Is "1" in “1 = H+T” an element of reality in the EPR sense (where H = 1 is heads and T = 1 is tails in a single coin toss)?

There are only two possible answers: Either 1 in “1 = H+T” is an element of reality, or it is not.

But the correct answer depends on what is meant by “+” in “1 = H+T.” If by “+” you mean the “exclusive or”, as is usually the case for a coin toss, then the answer is: Yes, 1 is an element of reality. But if what is meant by “+” in “1 = H+T” is something other than the “exclusive or” (such as the “and” in the CHSH sum), then the answer is: No, 1 is NOT an element of reality.

***

[minkwe]

***
My reply to Jay's question on Retraction Watch: Is "1" in “1 = H+T” an element of reality in the EPR sense (where H = 1 is heads and T = 1 is tails in a single coin toss)?

There are only two possible answers: Either 1 in “1 = H+T” is an element of reality, or it is not.

But the correct answer depends on what is meant by “+” in “1 = H+T.” If by “+” you mean the “exclusive or”, as is usually the case for a coin toss, then the answer is: Yes, 1 is an element of reality. But if what is meant by “+” in “1 = H+T” is something other than the “exclusive or” (such as the “and” in the CHSH sum), then the answer is: No, 1 is NOT an element of reality.

***

Exactly Joy,

If H = 1 means Head is UP and H = 0 means Head is down, and the same for T, then from a single coin toss, H + T = 1 is an element of reality, since H and T are not mutually exclusive possibilities, they are just Head/Tail counters for *actual* results obtained. But if H = 1, T=1 are the mutually exclusive possibilities, then H + T, is not an element of reality for a single coin toss because only one of those outcomes (H=1) or (T=1) can be obtained, not both.

The CHSH sum is more similar to the latter than the former. I think there are serious flaws in the coin-toss analogy.

[minkwe]

I boil down the argument to the claim that:

A sum of mutually exclusive *possibilities* is a meaningless quantity that can't be reasonably compared with a sum of mutually compatible *actualities*.

Here is a proof.

For the single coin toss with mutually exclusive possibilities (H=1, T=1), the sum of *possibilities* is H + T = 2
However, we find that after tossing a single coin we may obtain actual results results (H=1, T=0) or actual results (H=0, T=1). Therefore the sum of *actualities* H + T = 1
Therefore H + T = 2 is not a *possibility*

I think this is what Joy means when he says the set of *possibilities* is not closed-under addition. But the set of *actualities* is.

QED

[Joy Christian]

The CHSH sum is more similar to the latter than the former. I think there are serious flaws in the coin-toss analogy.

I agree. But Jay is stubbornly following the coin-toss analogy. Eventually he will see its limitations himself.

PS: I just noticed your proof above. That is all there is to it.

***

[Joy Christian]

***
Since the Bell-believers are fleeing from Retraction Watch like rats from a sinking ship, hopefully the following are our final set of comments there:

Since 1 = H + T, with your identifications, is completely equivalent to the CHSH sum (1), the coin-toss quantity 1 = H + T is not an element of reality, as I have already noted.

That is just making or continuing an argument which says (5) is true because (5) is true.

OK, Jay. Here is my same old answer, but now respecting your new logic:

Since you have defined H = ( B + B’ )/2B and T = ( B – B’ )/2B ,

1 = H + T is not an element of reality, because H and T are possible outcomes of an impossible coin.

There does not exist a coin in Nature that can have H and T as two counterfactually possible outcomes.

Joy,

With the above answer (which is actually not quite your same old answer as I will explain), I believe you can declare checkmate, and that Bell’s Theorem is toppled.

***

[FrediFizzx]

Actually a stalemate; nobody truly convinced the other side they are wrong. But we knew that was going to happen from the very beginning. But perhaps some lurkers got convinced one way or the other and it seems Jay is almost convinced that Bell was wrong. But the real bottom line about Bell is,

It is mathematically impossible for anything to violate a Bell inequality!

That the Bell fanatics don't understand nor accept that is quite mind boggling. When ask to try to prove that QM or the experiments actually violate those inequalities, they hardly ever try.

[minkwe]

I won't call it a stalemate when one side runs out of moves/arguments. They keep saying nah, nah nah, without providing any arguments to support their denials. Everything they suggested so far was shutdown, they keep trying to hang on to Gill's paper as the last hope. But Gill's paper is fatally flawed. Even Gill's own earlier paper with Larsson, disproves his latest statistics paper (see equation 11 for example).

The problem here is that the ensemble on which the correlations are evaluated changes
with the settings, while the original Bell inequality requires that they stay the same. In effect,
the Bell inequality only holds on the common part of the four different ensembles ΛAC′ , ΛAD′ ,
ΛBC′ , and ΛBD′ , i.e., for correlations of the form
E(AC′ |ΛAC′ ∩ ΛAD′ ∩ ΛBC′ ∩ ΛBD′ ). (8)
Unfortunately our experimental data comes in the form
E(AC′|ΛAC′ ),
so we need an estimate of the relation of the common part to its constituents:
...
etc

They conclude correctly that Bell's inequality only applies to the common part of the 4 ensembles. They didn't realize that there is never any common ensemble. 4 disjoint ensembles have a null intersection by definition. Since particles aren't measured more than once, there can be no common ensemble! Delta is therefore zero, and Gill's inequality reduces to an upper bound of 4 as we've been saying all along. Gills earlier paper is a disproof of his later paper!

[FrediFizzx]

I won't call it a stalemate when one side runs out of moves/arguments. They keep saying nah, nah nah, without providing any arguments to support their denials. Everything they suggested so far was shutdown, they keep trying to hang on to Gill's paper as the last hope. But Gill's paper is fatally flawed. Even Gill's own earlier paper with Larsson, disproves his latest statistics paper (see equation 11 for example).

The problem here is that the ensemble on which the correlations are evaluated changes
with the settings, while the original Bell inequality requires that they stay the same. In effect,
the Bell inequality only holds on the common part of the four different ensembles ΛAC′ , ΛAD′ ,
ΛBC′ , and ΛBD′ , i.e., for correlations of the form
E(AC′ |ΛAC′ ∩ ΛAD′ ∩ ΛBC′ ∩ ΛBD′ ). (8)
Unfortunately our experimental data comes in the form
E(AC′|ΛAC′ ),
so we need an estimate of the relation of the common part to its constituents:
...
etc

They conclude correctly that Bell's inequality only applies to the common part of the 4 ensembles. They didn't realize that there is never any common ensemble. 4 disjoint ensembles have a null intersection by definition. Since particles aren't measured more than once, there can be no common ensemble! Delta is therefore zero, and Gill's inequality reduces to an upper bound of 4 as we've been saying all along. Gills earlier paper is a disproof of his later paper!

It is mind boggling crazy that this has gone on for over 50 years. Of course we claim checkmate but to a third uninterested party it is a stalemate since one side or the other didn't admit publicly that they were wrong. But we know that hard facts don't matter to Bell fanatics so they will never admit that they are wrong. Hopefully some lurkers have seen the truth.
.