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.