Hidden Variable: The Role of Genetics and Parents in Twins

[beda pietanza]

Tell me if the analogy holds:

Twins are generated by black eye fathers and green eyed mothers


Let it be the "green eye" or the "black eye" our observable variable.
Only identical twins are chosen.


The twins A and B are sealed, each one, in a closed box,
we have to predict the color of the eyes of twin B by opening the box
of twin A: we don't know anything about the genetic laws and about the
color of the eyes of the parents.


If we only open a stream of A boxes the statistic of the outcome is a
random distribution of 50% of green eyes and 50% of black eyes.


When we open the box A we know that B will have the same color as A.


Of course the correlation is preexistent at our opening the twin box
A , nevertheless the prediction is only possible after we open the twin A
box not before.


This is a clear case of hidden variable: the genetic and the parent
eyes.


please comment.


best regards

beda pietanza

 

[JesseM]

What makes quantum physics different is that the correlations cannot be explained by "local hidden variables", ie the idea that the correlations can be explained by imagining that the source of the two particles creates them in correlated states which guarantee they'll give the same results when you make the same measurements, as in your example. The statistics in quantum physics violate "Bell inequalities", and it's been proven by Bell's Theorem that this is incompatible with local hidden variables; your example would not violate any of these inequalities. You might want to check out the thread non-locality, where in post #10 I gave this example:

Here's a closer analogy. Suppose we have a machine that generates pairs of scratch lotto cards, each of which has three boxes that, when scratched, can reveal either a cherry or a lemon. We give one card to Alice and one to Bob, and each scratches only one of the three boxes. When we repeat this many times, we find that whenever they both pick the same box to scratch, they always get opposite results--if Bob scratches box A and finds a cherry, and Alice scratches box A on her card, she's guaranteed to find a lemon.

Classically, we might explain this by supposing that there is definitely either a cherry or a lemon in each box, even though we don't reveal it until we scratch it, and that the machine prints pairs of cards in such a way that the "hidden" fruit in a given box of one card is always the opposite of the hidden fruit in the same box of the other card. If we represent cherries as + and lemons as -, so that a B+ card would represent one where box B's hidden fruit is a cherry, then the classical assumption is that each card's +'s and -'s are the opposite of the other--if the first card was created with hidden fruits A+,B+,C-, then the other card must have been created with the hidden fruits A-,B-,C+.

The problem is that if this were true, it would force you to the conclusion that on those trials where Alice and Bob picked different boxes to scratch, they should find opposite fruits on at least 1/3 of the trials. For example, if we imagine Bob's card has the hidden fruits A+,B-,C+ and Alice's card has the hidden fruits A-,B+,C-, then we can look at each possible way that Alice and Bob can randomly choose different boxes to scratch, and what the results would be:

Bob picks A, Alice picks B: same result (Bob gets a cherry, Alice gets a cherry)

Bob picks A, Alice picks C: opposite results (Bob gets a cherry, Alice gets a lemon)

Bob picks B, Alice picks A: same result (Bob gets a lemon, Alice gets a lemon)

Bob picks B, Alice picks C: same result (Bob gets a lemon, Alice gets a lemon)

Bob picks C, Alice picks A: opposite results (Bob gets a cherry, Alice gets a lemon)

Bob picks C, Alice picks picks B: same result (Bob gets a cherry, Alice gets a cherry)

In this case, you can see that in 1/3 of trials where they pick different boxes, they should get opposite results. You'd get the same answer if you assumed any other preexisting state where there are two fruits of one type and one of the other, like A+,B+,C-/A-,B-,C+ or A+,B-,C-/A-,B+,C+. On the other hand, if you assume a state where each card has the same fruit behind all three boxes, like A+,B+,C+/A-,B-,C-, then of course even if Alice and Bob pick different boxes to scratch they're guaranteed to get opposite fruits with probability 1. So if you imagine that when multiple pairs of cards are generated by the machine, some fraction of pairs are created in inhomogoneous preexisting states like A+,B-,C-/A-,B+,C+ while other pairs are created in homogoneous preexisting states like A+,B+,C+/A-,B-,C-, then the probability of getting opposite fruits when you scratch different boxes should be somewhere between 1/3 and 1. 1/3 is the lower bound, though--even if 100% of all the pairs were created in inhomogoneous preexisting states, it wouldn't make sense for you to get opposite answers in less than 1/3 of trials where you scratch different boxes, provided you assume that each card has such a preexisting state with "hidden fruits" in each box.

But now suppose Alice and Bob look at all the trials where they picked different boxes, and found that they only got opposite fruits 1/4 of the time! That would be the violation of Bell's inequality, and something equivalent actually can happen when you measure the spin of entangled photons along one of three different possible axes. So in this example, it seems we can't resolve the mystery by just assuming the machine creates two cards with definite "hidden fruits" behind each box, such that the two cards always have opposite fruits in a given box.