gill1109 wrote:By the way, I do believe myself that quantum correlations are probably consequences of the spinorial properties of spacetime.
Jarek wrote:Richard,
this discussion is not only about QM, but also classical field theories like electromagnetism or general relativity: if physics solves them using Euler-Lagrange, then current state of the field is local realistic "hidden variable" - satisfies Bell's assumptions leading to inequalities violated by physics - contradiction.
While QM might have dozens of philosophical interpretations (due to collapses caused by not including environment), Lagrangian formalism is pure math - doesn't leave so much freedom, wanting to solve a model we can use Euler-Lagrange or the least action principle ... what other options are there? Which one is used by physics?
gill1109 wrote:I don't see the contradiction. Bell's theorem is also pure math (or can be seen as pure math). ...
FrediFizzx wrote:Bell's "theorem" is NOT pure math.
It is physically impossible for anything to violate the inequalities.
That should be your first clue that something is wrong with the "theorem".
Heinera wrote:minkwe wrote:I've been saying something similar here for years. Counterfactual outcomes are not statistically independent of actual outcomes.
There is nothing in the proof of Bell's theorem that requires that counterfactual outcomes must be statistically independent of actual outcomes.
Jarek wrote:FrediFizzx wrote:Bell's "theorem" is NOT pure math.
So maybe let's focus on simpler Mermin's:
Pr(A=B) + Pr(A=C) + Pr(B=C) >= 1
which is nearly "tossing 3 coins, at least 2 are equal".
Its derivation doesn't need any ambiguous "locality", "realism", just that "there exists Pr(ABC) probability distribution" assumption:
Pr(A=B) = P(000) + P(001) + P(110) + P(111)
Pr(A=B) + Pr(A=C) + Pr(B=C) = 2P(000) + 2P(111) +sum_ABC P(ABC) = 2P(000) + 2P(111) + 1 >= 1
Isn't it just a simple pure math?
It is physically impossible for anything to violate the inequalities.
Here in page 4 is its violation with QM formalism: https://arxiv.org/pdf/1212.5214
That should be your first clue that something is wrong with the "theorem".
Indeed, somehow the assumptions used to derive this inequality are nonphysical.
Here the only used assumptions is just: there exists P(ABC) probability distribution.
But how such obvious assumption could not be true???
So Ising model is an example of situation where this assumption is not true - instead of single probability distribution P(ABC), from symmetry we have two amplitudes psi(ABC) as in QM - to be added over unmeasured variables, then multiplied to get the probabilities, as in Born rule: https://physics.stackexchange.com/quest ... mble-ising
In a recent thread, I tried to show the error in Bell/CHSH type inequalities and their alleged violation by QM and experiments. Some have been unable to understand the argument. I will attempt again to illustrate the issue. We will use an analogy of coin-tossing which maintain all the key features of the Bell-CHSH discussion. To my knowledge, nobody has presented the argument in this manner before. I will present the argument in two parts.
Feel free to ask clarifying questions as we go along. If there is interest, I may even discuss exactly how all the loopholes fit into this analogy, and why they are irrelevant, including super-determinism, and show exactly the flaw in Gill's most recent paper.
Part 1: The Bell argument
Assumptions:
1. Coins are local realistic
2. Coins have only two possible outcomes H = +1, T=-1
Derivation of the equality (cf inequality):
- Let A be the outcome we get when we toss a single coin.
- We assume that the other outcome we could have gotten also exists, even though we did not get it, call it B.
- Tossing a single coin produces either H or T and not both. Possible outcomes for AB are (HT, TH)
- Therefore A + B = 0, and E(A) + E(B) = 0 for all local realistic coins.
QM predictions:
- QM predicts E(A) = E(B) = 0.25.
Bell's theorem:
- Since from QM E(A) + E(B) = 0.5 =/= 0, it means QM is not local realistic.
Aspect-type experiments:
- We need to test experimentally whether QM or local realism is correct. Unfortunately we can only read one outcome at a time from a single coin. However, if we toss two coins, we can still obtain accurate estimates of E(A) and E(B) in the form of <A> and <B>, where A is the outcome we get from the first coin, and <B> is the outcome we get from the second coin. We should get similar results because the two coins are drawn from the same population.
- After the experiment we observe <A> = <B> = 0.25, <A> + <B> = 0.5, exactly what QM predicted.
- Therefore QM is correct and local realism is wrong.
Part 2: The unraveling
--------------------------
The argument I have made many times is that the reason for the violation is due to an erroneous substitution of actual results from two separate systems, into an expression derived using actual & counter-factual results from a single system. It sounds so simple but why has this been difficult to understand so far? This discussion between Heinera and Richard in a related thread illustrates why:Heinera wrote:
So if the original correlations (all computed on the whole set) didn't violate the CHSH inequality (CHSH<2), and the correlations computed on four disjoint random subset would not change much, we can now conclude that the four latter correlations would still not significantly violate the CHSH inequality, since term by term, they are approximately equal to the original correlations?gill1109 wrote:
Yes Heinera, you are home. They might violate it a little, but in all probability they won't violate it by much
Let us translate these innocent looking argument to our coin toss system:
If we measure <A> on a fair sample, of the population, we should get almost the same result as the population.
If we measure <B> on a fair sample, of the population, we should get almost the same result as the population.
Just because we measured <A> on one fair sample, and <B> on a different fair sample, does not change the fact that E(A) + E(B) = 0 for the population. You might violate it a little due to experimental error, but it shouldn't matter whether we use a single population or two disjoint fair samples of the population. Then the expectation values from the two coins should still not significantly violate the expression we obtained from the single coin
So what is wrong with this argument:
-------------------------------------------
First, let us present a counter-example which shows that the QM prediction is fully consistent with local-realism:
- We have a local realistic coin with a (H,T) probability distribution of [0.625,0.375]. E(A) = E(B) = 0.625(1)+0.375(-1) = 0.25, E(A) + E(B) = 0.5, fully agreeing with the experiments and QM.
- What about our local-realistic relationship E(A) + E(B) = 0 which we derived. Is it still valid for our local-realistic coins? Yes of course, provided we throw only single coins. Let us throw the coins on a glass table and read A from above, and B from below, and create a spreadsheet. Each row of the spreadsheet will necessarily sum to 0, and therefore <A> + <B> = 0. Agreeing with E(A) + E(B) = 0.
- How can the same local-realistic coins produce <A> + <B> = 0 in one experiment and <A> + <B> = 0.5, in another experiment? Because values calculated on two coins are not the same as values calculated on a single coin.
- Further illustration of the problem:
-------------------------------------------
Most people do not understand that counter-factual results do not have the same correlations as actual results. Take my coins for example, if the probability of H is 0.625, then the counter-factual probability for H can not also be 0.625. To illustrate even further, the actual result is what we get, the counter-factual result is what we did not get but could have gotten. If the likelihood of getting H is 0.625, then of course the likelihood of not getting H is 1-0.625 = 0.375. So the counter-factual probability is (1 - actual probability).
Failure to understand this elementary point is at the root of Bells theorem.
In the CHSH expression, S = E(a, b) − E(a, b′) + E(a′, b) + E(a′ b′), only [two] of the expectation values is actual, and the others are counterfactual. Bell assumed erroneously that each of those correlations should get the same functional form as the actual ones, not realizing that because they are counterfactual, we must flip the probability distributions.
If the actual E(a,b) = -cos(a-b)**2, then the counterfactual E(a,b) must be flipped to 1 - cos(a-b)**2.
minkwe wrote:
In the CHSH expression, S = E(a, b) − E(a, b′) + E(a′, b) + E(a′ b′), only [two] of the expectation values is actual, and the others are counterfactual. Bell assumed erroneously that each of those correlations should get the same functional form as the actual ones, not realizing that because they are counterfactual, we must flip the probability distributions.
If the actual E(a,b) = -cos(a-b)**2, then the counterfactual E(a,b) must be flipped to 1 - cos(a-b)**2.
gill1109 wrote:Exactly! A. Fine long ago showed the mathematical equivalence of the satisfaction of all Bell inequalities with the existence of a joint probability distribution reproducing all observable marginal distributions.
minkwe wrote:Heinera wrote:minkwe wrote:I've been saying something similar here for years. Counterfactual outcomes are not statistically independent of actual outcomes.
There is nothing in the proof of Bell's theorem that requires that counterfactual outcomes must be statistically independent of actual outcomes.
There absolutely is.
Heinera wrote:Well, why don't you put your effort where your mouth is and come up with a local HV model?
minkwe wrote:Heinera wrote:Well, why don't you put your effort where your mouth is and come up with a local HV model?
Huh? You serious? Do you have anything useful to contribute to the discussion rather than perpetual trolling?
Jarek wrote:Even simpler - you just need to dissatisfy this single assumption: "there exists joint probability distribution on Omega".
It is really easy with Ising model thanks to using nearly the same math (Feynman -> Boltzmann path ensemble) - instead of probability on Omega, as in QM states are defined by amplitudes on Omega, which have to be added for unmeasured variables, then multiplied to get probabilities.
minkwe wrote:* The CHSH expression E(AB) − E(AB') + E(A'B) + E(A'B') <= 2, contains a mixture of two actual expectation values and two counter-factual expectation values.
gill1109 wrote:One of my favourite short proofs of Bell's theorem goes as follows. Sorry, it is a bit long.
Joy Christian wrote:minkwe wrote:* The CHSH expression E(AB) − E(AB') + E(A'B) + E(A'B') <= 2, contains a mixture of two actual expectation values and two counter-factual expectation values.
I don't undersatnd this. There can be only one actual expectation value among the four, and three conuterfactual expectaion values.
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