Joy Christian wrote:***
Below is the central derivation of the EPR-Bohm correlation from my latest paper, which predicts exactly what quantum mechanics predicts for such correlations.
Anyone still concerned about my use of "realistic", please tell us where in my derivation there is "non-reality", non-locality", or lack of determinism creeping in:
guest1202 wrote:...To save readers the trouble of going back to that post, here in brief is the definition of a "1202realistic" model. We consider a situation in which observer Alice has a choice of two measurements, and , each yielding +1 or -1. Another oberver Bob similarly has a choice of measurements and . Quantum mechanics gives us the probability that if Alice uses measurement and Bob uses , Alice obtains result , and Bob obtains. Here each of and equals either +1 or -1. Quantum mechanics also gives three other similarly labeled probabilities and .
Let us write for the outcome which Alice obtains when she selects measurement and obtains result +1, with similar notation for the other possible measurement selections. Classically, besides the above probabilities given by quantum mechanics, we could also speak, for example, of the probability
that , and .
Then the four probabilities given by quantum mechanics would be marginals of .
For example, we would have
The definition of "1202realistic" is that there exists such a whose marginals are the four probability distributions given by quantum mechanics. This is the definition of the "realistic" part of "local-realistic" which is used in almost 100% of the papers on the subject known to me, though it's often presented differently (see below). Its intuitive meaning is that Alice and Bob are measuring things which are "real" in the sense that they have an existence independent of any measurement. An example is whether or not one of the "quantum cakes" of Kwiat/Hardy (discussed in another post) has risen earlier than usual halfway through the baking. Classically, we feel comfortable talking about that possibility whether or not we have opened the oven to check.
I am quite sure that Dr. Christian's model cannot be 1202realistic. The reason is that there is a mathematical theorem with a very simple proof that for to exist with given marginals , etc., the marginals have to satisfy the CHSH inequalities, and the marginals of quantum mechanics don't necessarily satisfy them. The proof of this theorem has been checked by thousands of mathematicians and physicists. I have never seen it questioned, including in the papers of Dr. Christian...
Do you have an online reference for this theorem [that given probability distributions
are marginals of some if and only if they satisfy the CHSH inequalities]? However, it may not be necessary as you seem to be confusing the marginals as coming from QM then you say right here that the QM marginals "...don't necessarily satisfy them". Quite frankly, it is all a matter of whether the terms in the Bell-CHSH inequality are dependent or independent. If you make it so that the QM marginals are dependent, then they will satisfy Bell-CHSH. See the argument in this paper.
guest1202 wrote:
...To save readers the trouble of going back to that post, here in brief is the definition of a "1202realistic" model. We consider a situation in which observer Alice has a choice of two measurements, and , each yielding +1 or -1. Another oberver Bob similarly has a choice of measurements and . Quantum mechanics gives us the probability that if Alice uses measurement and Bob uses , Alice obtains result , and Bob obtains. Here each of and equals either +1 or -1. Quantum mechanics also gives three other similarly labeled probabilities and .
FreddiFizzx asks:
How does QM give those probabilities? Those are given by EPR via the Bell-CHSH argument.
guest1202 wrote:I'm not trying to be pedantic about the language. I mention it because I have often been puzzled by some regular posters' use of "independence". Often, I simply can't figure out what they are trying to say. If they want their posts to be read and understood, they ought to realize that outside of technical probability theory, the term "independent" is susceptible to many meanings, and which one they are using should be made explicit.
guest1202 wrote:This post will discuss questionable mathematical exposition
in Dr. Christian's latest paper "Macroscopic Observability of Spinorial Sign Changes: A Simplied Proof"
His equation (27) contains an expression of the general form
(27)
where and are functions whose exact form will be irrelevant to the
discussion. The next equation (28) appears to transform (27) into
(28)
At a minimum, this is not standard mathematical exposition.
If interpreted literally, it is wrong because s cannot approach a and b
simultaneously unless a = b (and in the context a=b does not necessarily hold).
I haven't been able to think of a reasonable interpretation of (28).
Well, then I will make it more explicit. From the CH paper we have the following expression right before their eq. (4),
-1 ≤ p1(λ, a)p2(λ, b) - p1(λ, a)p2(λ, b') + p1(λ, a')p2(λ, b) + p1(λ, a' )p2(λ, b') - p1(λ, a') - p2(λ, b) ≤ 0
Now we will label each of the probability terms with A, A', B, and B' so we have,
-1 ≤ AB - AB' + A'B + A'B' - A' - B ≤ 0
So it is easy to see that the terms AB and AB' depend on the same A. And so forth with the other terms. IOW, the terms in the inequality are dependent on elements of each other and that is what makes the upper bound of 0 mathematically impossible to violate by anything. Now from the "Violating" Clauser-Horne... paper that I linked to above, you will see that the QM probabilities can be formulated in such a way as to have that same dependency between the terms thus satisfying Bell-CH and not violating it. I am sure an equivalent could be done for Joy's local-realistic model. So as I said before, the proof of the theorem doesn't matter. But thanks for the references anyways.
FrediFizzx
Independent Physics Researcher
This post will discuss questionable mathematical exposition
in Dr. Christian's latest paper "Macroscopic Observability of Spinorial Sign Changes: A Simplied Proof"
His equation (27) contains an expression of the general form
(27)
where and are functions whose exact form will be irrelevant to the
discussion. The next equation (28) appears to transform (27) into
(28)
At a minimum, this is not standard mathematical exposition.
guest1202 wrote:We use the notation A = B to mean that whenever Alice measures in direction a and Bob measures in direction b, they always obtain the same result (both +1 or both -1 ). If they always obtain opposite results (+1 for Alice and -1 for Bob or -1 for Alice and +1 for Bob), we write " A = (not B)". Similarly for A' = B, for which Alice measures in direction a` and Bob in direction b, etc. We will use a Popescu/Rohrlich box for which:
A = B, B = A' , A' = B', and B' = (not A ) .
Readers who are worried that this might be contradictory, should consult the previous post for reassurance. In a classical world, it would be contradictory, but quantum mechanics permits similar (though not identical) situations. ...
guest1202 wrote:FrediFizzx wrote:Well, then I will make it more explicit. From the CH paper we have the following expression right before their eq. (4),
-1 ≤ p1(λ, a)p2(λ, b) - p1(λ, a)p2(λ, b') + p1(λ, a')p2(λ, b) + p1(λ, a' )p2(λ, b') - p1(λ, a') - p2(λ, b) ≤ 0
Now we will label each of the probability terms with A, A', B, and B' so we have,
-1 ≤ AB - AB' + A'B + A'B' - A' - B ≤ 0
So it is easy to see that the terms AB and AB' depend on the same A. And so forth with the other terms. IOW, the terms in the inequality are dependent on elements of each other and that is what makes the upper bound of 0 mathematically impossible to violate by anything. Now from the "Violating" Clauser-Horne... paper that I linked to above, you will see that the QM probabilities can be formulated in such a way as to have that same dependency between the terms thus satisfying Bell-CH and not violating it. I am sure an equivalent could be done for Joy's local-realistic model. So as I said before, the proof of the theorem doesn't matter. But thanks for the references anyways.
FrediFizzx
Independent Physics Researcher
I have seen such statements before in this forum, and they have always puzzled me. I can't imagine what can lead to the conclusion that "the upper bound of 0 [is] mathematically impossible to violate by anything."
Before presenting a counterexample that does violate CH, to avoid misunderstanding, I ought to straighten out the notation. In CH's equation (4) which you quote, there is no . To get the CH inequalities, one integrates over . After the integration, your term becomes , which represents the probability that Alice measures in direction a and obtains result +1, while Bob measures in direction b and also obtains +1.
The "B" at the end of the middle term represents the probability that Bob measures +1 irrespective of what Alice measures. The term "AB" is not a product of "A" and "B", though you might not have meant to imply that.
The CH inequality that you attribute to Clauser/Horne's equation (4) is actually:
-1 ≤ p12(a,b) - p12(a, b' ) + p12(a' , b) + p12(a' , b' ) - p1( a' ) - p2( b) ≤ 0 .
Here p12(a, b) represents the probability that Alice measures in direction a and obtains +1 and Bob measures in direction b and obtains +1, and similarly for the other double-subscripted terms. The p1(a' ) represents the probability that Alice measures in direction a' and obtains result +1, irrespective of Bob's measurement. The p2(b) represents the probability that Bob measures in direction b and obtains results +1 regardless of what Alice does. The singly-subscripted terms refer to measurements by one of the observers (subscript 1 for Alice, 2 for Bob) without reference to the other observer. The terms with the double subscript "12" represent joint measurement probabilities.
The counterexample that I am going to present is a "Popescu/Rohlich (PR) box", which I discussed in a previous post. The PR boxes generally provide the maximum violation of Bell/type inequalities, and hence the maximum non-classicality. This is easy to believe from the fact that the very definition of PR boxes seems classically impossible.
We use the notation A = B to mean that whenever Alice measures in direction a and Bob measures in direction b, they always obtain the same result (both +1 or both -1 ). If they always obtain opposite results (+1 for Alice and -1 for Bob or -1 for Alice and +1 for Bob), we write " A = (not B)". Similarly for A' = B, for which Alice measures in direction a` and Bob in direction b, etc. We will use a Popescu/Rohrlich box for which:
A = B, B = A' , A' = B', and B' = (not A ) .
Readers who are worried that this might be contradictory, should consult the previous post for reassurance. In a classical world, it would be contradictory, but quantum mechanics permits similar (though not identical) situations.
Then
p12(a, b) = p1(a) because A = B. That is, Alice gets +1 with some probability p1(a), and Bob always gets the same result as Alice. When you work out the other terms, the middle of the inequality becomes
p12(a,b) - p12(a, b' ) + p12(a' , b) + p12(a' , b' ) - p1( a' ) - p2( b) = p1(a) - 0 + p1(a' ) + p1(a' ) - p1(a' ) - p1 (a) = p1(a`) ,
which is always strictly positive unless p1(a' ) = 0, and p1(a' ) can be chosen arbitrarily.
FrediFizzx wrote:guest1202 wrote:We use the notation A = B to mean that whenever Alice measures in direction a and Bob measures in direction b, they always obtain the same result (both +1 or both -1 ). If they always obtain opposite results (+1 for Alice and -1 for Bob or -1 for Alice and +1 for Bob), we write " A = (not B)". Similarly for A' = B, for which Alice measures in direction a` and Bob in direction b, etc. We will use a Popescu/Rohrlich box for which:
A = B, B = A' , A' = B', and B' = (not A ) .
Readers who are worried that this might be contradictory, should consult the previous post for reassurance. In a classical world, it would be contradictory, but quantum mechanics permits similar (though not identical) situations. ...
***
You are going to need to demonstrate how QM allows B = A' and A' = B' at the same time. IOW, prove that B = A' = B'.
***
I was merely demonstrating with the CH equation before eq. (4) how the dependency creeps into the inequality. It is there and you can't get rid of it. Now... you are a mathematician so you claim and what exactly is the point of an inequality that can be violated. That simply means that the inequality was invalid to start with. It is mathematical insanity to believe that an inequality of Bell's type can be violated. You are simply tricking yourself.
***
"… what exactly is the point of an inequality that can be violated. That simply means that the inequality was invalid to start with. It is mathematical insanity to believe that an inequality of Bell's type can be violated."
Gordon Watson wrote:Alas, Bell had no discussion with me [though I'm sure he would have, had I not been prematurely told that he was dead]. But his inequality is false in the context (EPRB) in which he derived it. So what 's the point …?
Any valid violation of a Bellian inequality indicates a fallacy in that inequality's derivation.
Or am I missing something?
It's a parity argument. Do you agree that in the linear maps:
ΦA→B[(0(mod2)]
ΦA→B[1(mod2)]
The first half is to the even part of A and odd part of B; the second to the odd part of B and the even part of A?
thray
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