Heinera wrote:And let me just add a question to minkwe; what is the point that this shuffling is supposed to explain?
minkwe[ wrote:Heinera wrote:And let me just add a question to minkwe; what is the point that this shuffling is supposed to explain?
Heine, like I've explained to you, address all 8 points, and you'll get the point. Anything else will be ignored.
The upper bound of 2, depends on the fact that the terms are dependent. It is a contradiction to claim that the same upper bound applies to independent sets.
Jochen wrote:OK, so let's get this over with. Point 1: I agree that for the hidden variable value assignments, every row is in fact bounded by 2. Point 2: I agree that if you separate out the data into your 2xN list, any reshuffling of rows does not change the expectation value (though of course, you can't then recombine the shuffled lists without undoing the shuffling, at least in such a way that the new 4xN list is just a row-shuffled form of the original). Can we now move on?
Jochen wrote:minkwe wrote:3. Do you agree that for 4 independent particle pairs , the expression
is false, and the correct upper bound should be , and by the same logic of the argument in point (1) above, the correct upper bound for averages is
For independent particle pairs, it is indeed the case that they are not bound by the value 2, but from there, it does not follow that hence, .
Jochen wrote:the inequality can be extended to averages because no average of a set of values can't be greater than the maximum value of that set. So you have a set of values, all below two, and immediately you know that the average value will also be below two.
Jochen wrote:Consider two independent dice, together with the observables T=value on top of the die after a throw, and B=value on the bottom after a throw. For each throw of a single die, then B + T = 7. This also holds in the average: . But for a single throw of both, the value on top of one plus the value at the bottom of the other is not 7, but bounded above by 12. So if you keep throwing both die, and record the value on top of one, and the value at the bottom of the other, does it then follow that ? (Hint: the answer is no.)
Jochen wrote:minkwe wrote:4. You have claimed that the CHSH expression,
where represents a single set of particle pairs, is statistically equivalent to
where represent disjoint independent sets of particle pairs. Because according to you,
and and
Yes I've claimed that, but you can scratch the 'according to you', since it's just according to basic statistics: given the same probability distribution , then indeed the correlation between two different experimental runs will be the same, because it's just derived from that distribution.
Jochen wrote:5. Do you agree that the expressionmeans that for every single individual pair of particles in the series which produced the outcome pair , there is an equivalent pair of particles in the series, such that as the number of particles approaches infinity, it should be possible to find a function , which rearranges the sequence of outcomes to match the sequence of , so that we have the same numbers of +1's and -1's and the same pattern of occurrences of those numbers? Yes or no?
This is just what the expectation value means: for large number of measurements n, so in order to yield the same EV, there need to be approximately the same numbers of +1 outcomes and of -1 outcomes. However, equality of these numbers only holds for , and for any two finite experiment runs i and j, both and , as well as and , will generally differ. Hence, in (almost) all real world experiments, such a function won't exist.
Jochen wrote:6. Do you agree that from point (4) above (which you believe), it follows that there must also exist other functions, , and . Such that after applying the functions, and , where the prime , represents the fact that the sequence of outcomes has been rearranged using the appropriate function? That is, the equivalent sets of outcomes are for all practical purposes "identical". Yes or no?
Again, only for the case, which is never attained in experiments.
Jochen wrote:7. Do you agree that if points (5) and (6) are true, then you can apply the same argument from point (1),to measurements performed on 4 disjoint sets of particle pairs , precisely because after rearranging, you will get
Which can be factorized just like in point (1) to
since and
There's two problems here. The first one is the erroneous nature of the argument in point 1, which I've already noted.
The second is that even if the reordering functions did exist (which they won't in general), then you can't perform such reorderings and expect everything to stay the same: the correlations are generally not preserved under such a reordering.
Heinera wrote:(And by the way, four sets generated by a LHV model are obviously not statistically independent).
minkwe wrote:Now instead of a 4xN spreadsheet, you have a 8xN spreadsheet, with columns Since the pairs are independent of each other, the maximum value of the expression is 4 and by the same logic which you stated yourself above, the average of the values of set is bounded above by the maximum value each member in the set can attain. This is completely uncontroversial and your claim that it does not follow that is clearly misguided.
minkwe wrote:Jochen wrote:Consider two independent dice, together with the observables T=value on top of the die after a throw, and B=value on the bottom after a throw. For each throw of a single die, then B + T = 7. This also holds in the average: . But for a single throw of both, the value on top of one plus the value at the bottom of the other is not 7, but bounded above by 12. So if you keep throwing both die, and record the value on top of one, and the value at the bottom of the other, does it then follow that ? (Hint: the answer is no.)
I will show you exactly how the example fails
* In your example, for the toss of a single die, <B> + <T> = 7 is an absolute equality (absolute upper and lower bound) which can never be violated, not even by a single iota. If you toss the single die, 1 billion times, you will never violate the number 7 for any individual throw (B+T), nor will you ever violate the number 7 for the averages <B> + <T>. If anyone claims to have violated it by even 1x10^-50, then there is a mathematical error involved. This is the meaning of the sign in mathematics.
minkwe wrote:* Now you say that if two separate independent dice are tossed many times, reading B from one and T from the other, you think that the equality <B> + <T> = 7 should still apply. But this is quite naive.
minkwe wrote:It contains a hidden assumption that the two separate tosses of two independent dice have exactly the same probability distributions for the outcomes B, T.
minkwe wrote:If instead of doing those measurements on the same die, they decide to toss 2 separate independent dice, it will be foolish to claim that the relationship continues to hold simply because the two dice are of the same type! Obviously, even if the dice are identical, the fact that you have 2 separate independent dice affords more degrees of freedom for the experimenters and therefore the expectation values calculated from a single die will generally not agree with those from 2 independent dice.
minkwe wrote:Your example misses the mark. My point (3) remains valid , even better I would say there is a non-zero probability that for 4 independent particle pairs.
Alice | Bob
A | C | B | D
[+1]| +1 | -1 | [-1]
[-1]| -1 |[-1] | +1
+1 |[-1] | +1 | [+1]
.
.
.
-1 |[-1] |[-1] | -1
[+1]| +1 |[+1] | +1
[+1]| -1 | -1 | [+1]
minkwe wrote:As shown above in my response to your response to point (3), you are mistaken. It is not always the case that and and . Simply because the nature of the experiment may not allow you to sample the same distribution.
minkwe wrote:Given that you have since changed your response to point (1) from "disagree" to "agree", you will need to amend your responses to point (7) which is obviously based on a misunderstanding, before I respond any further.
minkwe wrote:3. Do you agree that for 4 independent particle pairs , the expression
is false, and the correct upper bound should be , and by the same logic of the argument in point (1) above, the correct upper bound for averages is
Jochen wrote:Minkwe's wrote:7. Do you agree that if points (5) and (6) are true, then you can apply the same argument from point (1),to measurements performed on 4 disjoint sets of particle pairs , precisely because after rearranging, you will get
Which can be factorized just like in point (1) to
since and
There's two problems here. The first one is the erroneous nature of the argument in point 1, which I've already noted. The second is that even if the reordering functions did exist (which they won't in general), then you can't perform such reorderings and expect everything to stay the same: the correlations are generally not preserved under such a reordering.minkwe wrote:8. Do you see now that in order forto be true, it must be the case that and . But both and have already been rearranged independently of each other, and since any rearrangement will shuffle both outcomes in the set of pairs any new rearrangement to make agree with will undo the previous rearrangements; the same for and . Therefore for measurements on 4 independent disjoint sets to obey the the inequality, you need several different independent sorting functions to be dependent. In other words, the assumption that 4 independent sets of particle pairs should obey the inequality is a contradiction and your claim in point (4) above fails. The upper bound of 2, depends on the fact that the terms are dependent. It is a contradiction to claim that the same upper bound applies to independent sets
The only rearrangements you can validly perform on the measurement tables are row permutations;...
Jochen wrote:Anything I'm going to say is contained in my previous response, where I've shown that only row permutations can't change the value of the CHSH quantity.
Jochen wrote:minkwe wrote:* Now you say that if two separate independent dice are tossed many times, reading B from one and T from the other, you think that the equality <B> + <T> = 7 should still apply. But this is quite naive.
No, as long as we're talking about expectation values, the equality applies exactly; however, in every real experiment with a finite number of runs, it will of course apply only approximately. The confusion between the two is the same problem you also have in connecting Bell's theoretical work with actual experiment.
minkwe wrote:There was a notation typo in the last example of my previous post, I used (B, T) instead of (A, B) in the inequalities. It should have read:
If I toss a die such that it settles with the 5 exposed sides facing up(U), north(N), south(S), east(E) and west(W). I ask two experimenters (Alice and Bob) to pick a direction from the exposed sides to observe and the outcomes which we will denote as (A) and (B) respectively, one after the other. Let us suppose that the act of reading the die destroys the chosen face and makes it unavailable for the second experimenter to observe the same face. Once a direction is picked, say N, the remaining experimenter can only pick one of the remaining 4 directions U, S, E, W. Now based on the description of the experiment, we derive a certain inequality . For the purpose of this illustration, it does not matter what is. Now it is obvious that the two expectation values <A> and <B>, are not independent in a number of ways. First, the fact that the paired outcomes (A,B) are both faces of the same die carries with it a specific kind of dependence due to symmetry. But in addition, there is a dependence because whatever Alice picked is unavailable to Bob and vice versa. Therefore the relationship will include those dependencies. If instead of doing those measurements on the same die, they decide to toss 2 separate independent dice, it will be foolish to claim that the relationship continues to hold simply because the two dice are of the same type! Obviously, even if the dice are identical, the fact that you have 2 separate independent dice affords more degrees of freedom for the experimenters and therefore the expectation values calculated from a single die will generally not agree with those from 2 independent dice.
Heinera wrote:I agree with this. But the premise
minkwe wrote:Could you please tell us what Quantum Mechanics predicts for the 4 expectation values ?
minkwe wrote:So again, unless you want to concede my argument, address points (7) and (8) specifically, then I will respond to the remaining issues you have raised and show you how my original argument already anticipated and disposed of all your claims.
minkwe wrote:Heinera wrote:I agree with this. But the premise
You don't know what models we are discussing, not having read or understood points (1) to (8). The example you are now responding to, was addressing a very specific error Jochen made. You've missed the point completely because instead of addressing the points one by one, you have jumped in the middle and misunderstood the purpose of the example.
So please, once again, I ask you to address them. There is no point discussing with you otherwise. If you want to understand the argument, you have to engage it point by point.
minkwe wrote:A question you claimed was meaningless and had no answer. A question you haven't answered ever since. Jochen proceeded to claim that the QM predictions were , , and , respectively, the same as the QM predictions for because he believes and and .
Jochen wrote:As such, the question is meaningless. You could ask 'What does Quantum Mechanics predict for the 4 expectation values if the measurements are carried out on a singlet state, and the measurements are those in the standard CHSH setting?', in which case the prediction would be , , and , respectively; but in different settings, wildly different outcomes are possible. Does that help?
Jochen wrote:Point 7: if all you do is row rearrangings, I'm OK with it, provided I understand everything correctly. Point 8: if all you do is row rearrangings, I'm OK with that, provided I understand everything correctly. Can we now please quit this silliness and return to the actual argument?
Heinera wrote:There is no point in me addressing all eight points, since I basically agree with all of them. So what?
minkwe wrote:This is not silliness, unless you mean your actions. You claim point (7) fails because of the errorneous nature of point (1). Then you later recant and agree with point (1), and yet have the audacity to accuse me of silliness because I give you the opportunity to fix your response to point (7). It is you who is being silly, given what you've argued here so far, you will not agree to point (8) if you've read it carefully and understood it. Because it hammers the final nail into your argument. So again I give you another opportunity to read it carefully and make sure you understand it before you admit that is correct, which effectively ends the discussion.
minkwe wrote:Heinera wrote:There is no point in me addressing all eight points, since I basically agree with all of them. So what?
Good! You agree with all the 8 points! That wasn't so hard that it needed almost a week to extract.
Point (8) proves that point (4) is false. In other words, the QM predictions for are not the same as the QM predictions for. In other words, you guys (including Bell) have been using the wrong QM predictions to claim violation.
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