Jarek wrote:I have taken this Bell inequality from http://www.johnboccio.com/research/quan ... /paper.pdf
which says its source is: J. Preskill, lecture notes at http://www.theory.caltech.edu/people/preskill/ph229
Anyway, it is mathematically impossible to violate also this one if assuming a probability distribution among such 8 possibilities ( {0,1}^3 for binary variables ABC) - as in the diagram, sum over all P(A=B) + P(A=B) + P(A=C) cases covers all 8 possibilities, hence sums to at least 1.
However, as in QM or MERW ( https://en.wikipedia.org/wiki/Maximal_E ... andom_Walk ), for violation I have assumed Born rules: probability is normalized square of amplitude.
This square is the problem - it leads to violation of Bell, but MERW shows natural intuition where it comes from: assuming Boltzmann distribution among paths, probability toward one direction (past or future) is the amplitude, to randomly get a given value we need to "draw it" from both directions - getting the squares, violating Bell.
Jarek wrote:Could you elaborate?
We know well (also from experiments) that Bell inequalities are violated in QM - because of the Born rule (squares).
The violations I have written also use exactly the same squares for this purpose.
If you don't trust these extremely sophisticated calculations (denominator in example is 1^2 + 2^2 + 2^2 + 1^2 = 10), just run calculations which I have attached ... or point something suspicious there?
Jarek wrote:Could you elaborate?
We know well (also from experiments) that Bell inequalities are violated in QM - because of the Born rule (squares).
The violations I have written also use exactly the same squares for this purpose.
If you don't trust these extremely sophisticated calculations (denominator in example is 1^2 + 2^2 + 2^2 + 1^2 = 10), just run calculations which I have attached ... or point something suspicious there?
Jarek wrote:Hi Fred and Joy,
Whatever we will call it: violation or exceeding, I have also shown it for the original Bell inequalities a few posts up.
Sure inequalities as a mathematical theorem cannot be violated, unless we change the assumptions - and this is exactly what is done here or in QM: inequalities are derived for standard probability theory, while example of their violation is for a nonstandard one: with these very controversial squares - exactly as in QM Born rules.
My example is not very long or complex (bottom of the diagram at the top) and I would gladly add explanations if it's not clear - beside the general impossibility remarks, could you maybe refer to it, what do you think is the problem there?
Jarek wrote:"Now suppose that there are actually local hidden variables that provide a complete description of this system, and the quantum correlations are to arise from a probability distribution governing the hidden variables. Then, in this context, the Bell inequality is the statement
P(A=B) + P(B=C) + P(A=C) >= 1
b]This is satisfied by any probability distribution for the three coins because no matter what the values of the coins, there will always be two that are the same[/b]. But in quantum mechanics,
P(A=B) + P(B=C) + P(A=C) = 3/4"
FrediFizzx wrote:Thanks for the good explanation. Do you think this mathematical madness will ever end? Doesn't seem like it if people keep propagating in paper after paper.
minkwe wrote:Here is the key you are missing: The outcomes of the A coin used in the P(A=B) calculation is the very same outcome of the A coin used in the P(A=C) calculation. Similarly, the B outcomes used in the P(A=B) are the very same ones used in the P(B=C) calculation, and the same applies for the C coin outcomes in P(A=C). This is the ONLY scenario in which the above inequality is valid.
Jarek wrote:minkwe wrote:Here is the key you are missing: The outcomes of the A coin used in the P(A=B) calculation is the very same outcome of the A coin used in the P(A=C) calculation. Similarly, the B outcomes used in the P(A=B) are the very same ones used in the P(B=C) calculation, and the same applies for the C coin outcomes in P(A=C). This is the ONLY scenario in which the above inequality is valid.
Assume any (hidden) probability distribution on 8 scenarios of values for 3 binary variables.
Now measure two of them (example of taking https://en.wikipedia.org/wiki/Marginal_distribution ), using natural:
(*) probability of alternative of (non-overlapping) events is sum of individual probabilities.
Here Pr(AB) = sum_C Pr(ABC).
This way, as in the diagram in my first post, you get:
P(A=B) + P(B=C) + P(A=C) = 1 + 2P(A=B=C=0) + 2P(A=B=C=1)
so it is at least 1
However, QM can get lower values, if we want to reproduce such violation/exceeding, we need a different rule, e.g. Born's:
(**) probability of alternative of (non-overlapping) events is proportional to sum of square of amplitudes (being square roots of probabilities of individual events)
|psi| = sqrt(probabitiy)
Pr(union) ~ (sum of sqrt(probability) )^2
I don't know why do you think it is related to some "cyclicity"?
It is just about marginalization: assuming a probability distribution on a larger set, then marginalizing it to a smaller set, e.g. using (*): Pr(AB) = sum_C Pr(ABC).
I have recently worked on the details of getting (**) instead of (*) in MERW for the violation and it has turns out a bit more complicated: https://www.dropbox.com/s/ax35hvxrorx72ff/bell_MERW.pdf
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