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billschnieder said:Hello Travis,
Hi Bill, thanks for the thoughtful questions about the actual article! =)
By "statistical regularities" do you mean simply a probability distribution Pα1,α2(A1,A2) exists? Or are you talking about more than that.
Nothing more. But of course the real assumption is that this probability distribution can be written as in equations (3) and (4). In particular, that is where the "no conspiracies" and "locality" assumptions enter -- or really, here, are formulated.
What if instead you assumed that λ did not originate from the source but was instantaneoulsy (non-locally) imparted from a remote planet to produce result A2 together with α2, and result A1 together with α1. How can you explain away the suggestion that the rest of your argument, will now prove the impossibility of non-locality?
I don't understand. The λ here should be thought of as "whatever fully describes the state of the particle pair, or whatever you want to call the 'data' that influences the outcomes -- in particular, the part of that 'data' which is independent of the measurement interventions". It doesn't really matter where it comes from, though obviously if you have some theory where it swoops in at the last second from Venus, that would be a nonlocal theory.
But mostly I don't understand your last sentence above. What is suggesting that the rest of the argument will prove the impossibility of non-locality? I thought the argument proved the inevitability of non-locality!
To make the following clear, I'm going to fully specify the implied notation in the above as follows:
|C(\mathbf a,\mathbf b|\lambda)-C(\mathbf a,\mathbf c|\lambda)|+|C(\mathbf a',\mathbf b|\lambda)+C(\mathbf a',\mathbf c|\lambda)|\le2,
You've misunderstood something. The C's here involve averaging/integrating over λ. They are in no sense conditional/dependent on λ. See the equation just above where CHHS gets mentioned, which defines the C's.
Which starts revealing the problem. Unless all terms in the above inequality are defined over the exact same probability measure. The above inequality does not make sense. In other words, the only way you were able to derive such an inequality was to assume that all the terms are defined over the exact same probability measure P(λ). Do you agree?
No. You are confusing the probability P_{\alpha_1,\alpha_2}(\cdot|\lambda) with P(\lambda). You first average the product A_1 A_2 with respect to P_{\alpha_1,\alpha_2}(\cdot|\lambda) to get E_{\alpha_1,\alpha_2}(A_1 A_2 | \lambda). Then you average this over the possible λs using P(λ).
Maybe you missed the "no conspiracies" assumption, i.e., that P(λ) can't depend on \alpha_1 or \alpha_2.
(a) Now since you did not show it explicity in the article, I presume when you say Bell's theorem contradicts quantum theory, you mean, you have calculated the LHS of the above inequality from quantum theory and it was greater than 2. If you will be kind as to show the calculation and in the process explain how you made sure in your calculation that all the terms you used were defined over the exact same probability measure P(λ).
I don't understand. The QM calculation is well-known and not controversial. You really want me to take the time to explain that? Look in any book. But I have the sense you know how the calculation goes and you're trying to get at something. So just tell me where you're going. Your last statement makes no sense to me. In QM, λ is just the usual wave function or quantum state for the pair; typically we assume that this can be completely controlled, so P(λ) is a delta function. But in QM, you can't do the factorization that's done in equation (4). It's not a local theory. (Not that you need Bell's theorem to see/prove this.)
(b) You also discussed how several experiments have demonstrated violation of Bell's inequality, I presume by also calculating the LHS and comparing with the RHS of the above. Are you aware of any experiments in which experimenters made sure the terms from their experiments were defined over the exact same probability measure?
No, the experiments don't measure the LHS of what you had written above. What they can measure is the C's as we define them -- i.e., involving the averaging over λ.
(5) Since you obviously agree that non-contextual hidden variables are naive and unreasonable, let us look at the inequality from the perspective of how experiments are usually performed. For this purpose, I will rewrite the four terms obtained from a typical experiment as follows:
C(\mathbf a_1,\mathbf b_1)
C(\mathbf a_2,\mathbf c_2)
C(\mathbf a_3',\mathbf b_3)
C(\mathbf a_4',\mathbf c_4)
Where each term originates from a separate run of the experiment denoted by the subscripts. Let us assume for a moment that the same distribution of λ is in play for all the above terms. However, if we were to ascribe 4 different experimental contexts to the different runs, we will have the terms.
C(\mathbf a,\mathbf b|\lambda,1)
C(\mathbf a,\mathbf c|\lambda,2)
C(\mathbf a',\mathbf b|\lambda,3)
C(\mathbf a',\mathbf c|\lambda,4)
Where we have moved the indices into the conditions. We still find that each term is defined over a different probability measure P(λ,i), i=1,2,3,4 , where i encapsulates all the different conditions which make one run of the experiment different from another.
Therefore could you please explain why this is not a real issue when we compare experimental results with the inequality.
Yes, for sure, if P(λ) is different for the 4 different (types of) runs, then you can violate the inequality (without any nonlocality!). The thing we call the "no conspiracies" assumption precludes this, however. It is precisely the assumption that the distribution of λ's is independent of the alpha's.
So I guess your issue is just what I speculated above: you do not accept the reasonableness of "no conspiracies", or didn't realize this assumption was being made. (I doubt it's the latter since we drum this home big time in that section especially, and elsewhere.)