billschnieder said:
(continued from the last post)
So far we have dealt with pairs, just like Bell up to his equation (14). Let us then, following in Bell's footsteps introduce the third variable (see page 406 of his original paper).
Bell said:
It follows that c is another unit vector
E(a,b) - E(a,c) = -\int d\lambda \rho (\lambda )[A(a,\lambda )A(b,\lambda )-A(a,\lambda )A(c,\lambda )]
using (1), whence
\left | E(a,b)-E(a,c) \right |\leq \int d\lambda \rho [1 - A(b,\lambda)A(c,\lambda )]
The second term on the right is E(b,c), whence
1 + E(b,c) >= |E(a,b) - E(a,c)| ... (15)
Note a few things here: Bell factorizes at will within the integral. ρ(λ) is a factor of every term under the integral. That is why I explained in my previous detailed post that ρ(λ) must be the same for all three terms.
And I explained in #1213 that it doesn't make any sense to use these equations as the
reason why ρ(λ) should be the same in all three terms, since the equations he writes down for E(a,b) and E(b,c) and E(a,c) are not meant to be
definitions of the expectation values, but rather conclusions about how the expectation values can be written down in a universe that obeys local realist laws along with the no-conspiracy assumption. See everything in post #1213 starting with the paragraph that begins "I don't understand how you can say..."
Anyway,
if we accept Bell's
physical argument that in a local realist universe we should be able to write the expectation values as follows:
E(a,b) = -\int d\lambda\rho (\lambda )A(a,\lambda )A(b,\lambda )
E(a,c) = -\int d\lambda\rho (\lambda )A(a,\lambda )A(c,\lambda )
E(b,c) = -\int d\lambda\rho (\lambda )A(b,\lambda )A(c,\lambda )
...
then we can see why the factorization he does in the equations you wrote above should be justified. But he does need to make that physical argument to justify it.
Also, there is some ambiguity in what you mean when you say "ρ(λ) must be the same for all three terms", I discussed this at the start of post #1214. I was interpreting it just as a statement that the "true" or "objective" probability distributions on different values of λ (which would give the frequencies of different values of λ that would be expected in the limit as the number of trials went to infinity) should not depend on the detector settings. If you mean something different, like that the actual finite run of trials on each detector setting should involve the same frequencies of different values of λ, then I disagree that Bell's equation implies anything of the sort since it only deals with "true" probabilities and not empirical results, but again see post #1214 for the detailed discussion on this point.
billschnieder said:
Secondly, Bell derives the expectation value term E(b,c) by factoring out the corresponding A(b,.) and A(c,.) terms from E(a,b) and E(a,c). Therefore, E(b,c) does not contain different A(b,.) and A(c,.) terms but the exact same ones present in E(a,b) and E(a,c).
I don't know why you have replaced terms like A(b,λ) with notation like A(b,.)--easier to type, or some deeper significance? Anyway, Bell is assuming that for any given value of λi, A(a,λi) is the same regardless of whether the other detector was on setting b or setting c, and so forth for A(b,λi) and A(c,λi). In other words, the result at a given detector depends only on that detector's setting and the value of all hidden variables on that trial, it doesn't depend on the other detector's setting (and we wouldn't expect it to in a local realist universe!) Is this all you're saying, or do you think the factorization has some further implications?
billschnieder said:
In other words, in order to obtain all three expectation values E(a,b), E(a,c) and E(b,c), we ONLY need three lists of outcomes corresponding to A(a,.), A(b,.), A(c,.) or in simpler notation, we only need a single list of triples [(a',b',c')] to calculate all terms for
1 + <b'c'> >= |<a'b'> - <a'c'>|
No, again it seems like you are confusing theoretical terms with empirical results. E(a,b) doesn't depend on what results we got on any finite series of trials, it's the "true" expectation value that can be defined as
E(a,b) = (+1*+1)*P(detector with setting a gets result +1, detector with setting b gets result +1) + (+1*-1)*P(detector with setting a gets result +1, detector with setting b gets result -1) + (-1*+1)*P(detector with setting a gets result -1, detector with setting b gets result +1) + (-1*-1)*P(detector with setting a gets result -1, detector with setting b gets result -1)
Where each of the P's represents the "true" or "objective" probability for that pair of results, as distinguished from the fraction of some finite number of trials where that pair of results was seen (as always, in frequentist terms the objective probabilities would be the fraction of trials with that pair of results in the limit as the number of trials goes to infinity).
billschnieder said:
So then, we are destined to obtain this inequality for any list of triples of two valued variables (or outcomes of two-valued functions) were the allowed values are (+1 or -1), no matter the physical, metaphysical or mystical situation generating the triples.
But that's not the situation with Bell's theorem. Rather, with Bell's theorem we have three runs with different combinations of detector settings (a,b), (b,c) and (a,c), and considering
the average from each run. Bell is showing that if we know the true expectation values for each individual run, in a local realist universe they should obey:
1 + E(b,c) >= |E(a,b) - E(a,c)|
Since each expectation value is for a
different run, even if you assume that every iteration of every run is determined by a set of triples, you can't derive the above equation from arithmetic alone since each expectation value would deal with a
different collection of triples. So, you do need to consider the "physical, metaphysical or mystical situation generating the triples". And once you are convinced that the above equation should hold for the true expectation values, then by the law of large numbers you can conclude that if you do 1000 trials on each run, in a local realist universe you are astronomically unlikely to see a violation of the following inequality on your data:
1 + (average for product of results on the run with settings b and c) >=
|(average for product of results on the run with settings a and b) -
(average for product of results on the run with settings a and c)|
billschnieder said:
Suppose now that we generate from our list of triples, three lists of pairs corresponding to [(a',b')], [(a',c')] and [(b',c')], we can simply calculate our averages and be done with it. It doesn't matter if the order of pairs in the lists are randomized so long as the pairs are kept together. In this case, we can still sort them as described in my previous detailed description, to regenerate our list of triples from the three lists of pairs.
See my questions and arguments about your "resorting" procedure in post #1215. First I clarified what I thought you meant by this form of "resorting" at the start of the post with a simple example, perhaps you can tell me if I've got it right or not. If I have got it right, then please address my subsequent comments and questions:
If so, I don't see how this ensures that "ρ(λi) is the same for all three terms of the inequality", or what you even mean by that. For example, isn't it possible that if the number of possible values of λ is 1000, then even though iteration #1 of the first run has been grouped in the same row as iteration #3 of the second run and iteration #2 of the third run (according to their original labels), that doesn't mean the value of λ was the same for each of these three iterations? For example, might it not have been the case that iteration #1 of the first run had λ203, iteration #3 of the second run had λ769, and iteration #2 of the third run had λ488?
As a separate issue it is of course true that if your full set of data can be resorted in this way, that's enough to guarantee mathematically that the data will obey Bell's inequality. But this is a very special case, I think it would be fairly unlikely that the full set of iterations from each run could be resorted such that every row would have the same value of a,b,c throughout, even if the data was obtained in a local realist universe that obeyed Bell's theoretical assumptions, and even if the overall averages from each run actually did obey the Bell inequality.
billschnieder said:
Now the way Bell-test experiments are usually done, is analogous to collecting three lists of pairs randomly with the assumption that these three lists are representative of the three lists of pairs which we would have obtain from a list of triple, had we been able to measure at three angles simultaneously.
Yes, that's true. Since there are only eight possible distinct triples, and the value of λ on each trial completely determines the type of triple on that trial, and we assume the true probability distribution P(λ) is the same regardless of the detector settings, then with some reasonably large number of trials (say 1000) on each run we do expect that:
Fraction of trials on first run where the hidden triple was a=+1, b=-1 and c=+1
is very close to
Fraction of trials on second run where the hidden triple was a=+1, b=-1 and c=+1
and to
Fraction of trials on third run where the hidden triple was a=+1, b=-1 and c=+1
And likewise for the fractions of the other seven types of triples that occurred on each run. Do you agree this is a reasonable expectation thanks to the law of large numbers?
billschnieder said:
And if each list was sufficiently long, the averages will be close to those of the ideal situation assumed by Bell. Again, remember that within each list of pairs actually measured, the individual pairs such as (a',b')_i measured together are assumed to have originated from a specific theoretical triple, (a',c')_j from another triple, and (b',c')_k from another triple. Therefore, our dataset from a real experiment is analogous to our three theoretical lists above, where we randomized the order but kept the pairs together while randomizing. Which means, it should be possible to regenerate our single list of triples simply by resorting the three lists of pairs while keeping the individual pairs together, as I explained previously.
Even if the data was drawn from triples, and the probability of different trials didn't depend on the detector settings on each run, there's no guarantee you'd be able to
exactly resort the data in the manner of my example in post #1215, where we were able to resort the data so that
every row (consisting of three pairs from three runs) had the same value of a,b,c throughout. You might be able to sort it so that
most rows of three pairs had the same value of a,b,c throughout, but probably not all. This would at least give a way of roughly estimating the frequencies of different types of triples, though.
billschnieder said:
If we can not do this, it means either that:
a) our data is most likely of the second kind in which randomization did not keep the pairs together or
Well, we know this does not apply in Bell tests, where every data pair is always from a single trial with a single pair of measurements on a single pair of entangled particles.
billschnieder said:
b) each list of pairs resulted from different lists of triples and/or
If the frequencies of each of the 8 types of triples differed significantly in three runs with a significant (say, 1000 or more) number of trials in each, this would imply either an astronomically unlikely statistical miracle
or it would imply that the no-conspiracy assumption is false and that the true probabilities of different triples actually does change depending on the detector settings.
billschnieder said:
c) our lists of pairs are not representative of the list of triples from which they arose
Not sure I follow what you mean here. Are you suggesting that even if we had a triple like a=+1, b=-1, c=+1 we might still get result -1 with detector setting a? If so what would be the point of assuming the data arose from triples in the first place? Remember that Bell's assumption of predetermined results on each axis came from the fact that whenever both particles were measured on the
same axis they always gave opposite results--in a local realist universe where the decisions about the two detector settings can have a spacelike separation, it seems impossible to explain this result otherwise (though some of Bell's later proofs dropped the assumption of always getting opposite or identical results when both experimenters used the same setting).
billschnieder said:
In any of these cases, Bell's inequality does not and can not apply to the data. In other words, it is simply a mathematical error to use the inequality in such situations.
No, the fact that Bell's inequality is observed not to work is empirical
evidence that one of the assumptions used in the derivation must be false, like the assumption that local realism is true (with the conclusion of predetermined triples
following from this assumption along with the observation that using the same angle always yields opposite results), or the no-conspiracy assumption. Unless you want to argue (and you probably do) that even if we assume the validity of those theoretical assumptions, this does not necessarily imply Bell's inequality should hold for the type of experiment he describes.
billschnieder said:
Also note that these represent the only scenarios in which "average value of a*b for all triples" is different from "average value of a*b for measured pairs only". And in this case, the fair sampling assumption can not hold.
What do you mean by "fair sampling assumption"?
This page says "It states that the sample of detected pairs is representative of the pairs emitted", but that could be true and Bell's inequality could still fail for some other reason like a violation of the no-conspiracy assumption.