4 ms·
> The interesting effects, which EPR considered paradoxical, arise when we make measurements of both members of the pair. When we measure both members for color
by cousin_it 10y ago
> The interesting effects, which EPR considered paradoxical, arise when we make measurements of both members of the pair. When we measure both members for color, or both members for shape, we find that the results always agree. Thus if we find that one is red, and later measure the color of the other, we will discover that it too is red, and so forth. On the other hand, if we measure the shape of one, and then the color of the other, there is no correlation. Thus if the first is square, the second is equally likely to be red or to be blue.
That's not paradoxical, that's just two objects with the same random shape and the same random color. Quantum entanglement is stranger than that, it can't be explained by hidden correlated information about both objects.
Here's an example of quantum entanglement. Imagine that Alice and Bob live far from each other and can't communicate. There are three possible yes/no questions you can ask them, but each person can only answer one question. These facts are known:
1) If you ask Alice and Bob the same question, they always give opposite answers.
2) If you ask Alice question 1 and ask Bob question 2, the probability that both will say yes is 5%.
3) If you ask Alice question 2 and ask Bob question 3, the probability that both will say yes is 5%.
4) If you ask Alice question 1 and ask Bob question 3, the probability that both will say yes is 20%.
Under classical assumptions, there's no way for Alice and Bob to prepare a strategy beforehand that would lead to such probabilities. The reason is that case 4 is completely covered by cases 2 and 3, but has higher probability than the two of them combined. Here's a proof:
(4)
=> A1Y and B3Y
=> A1Y and B3Y and (A2N or A2Y) // tautology
=> A1Y and B3Y and (B2Y or A2Y) // by (1)
=> (A1Y and B2Y and B3Y) or (A1Y and A2Y and B3Y) // logical operations
=> (A1Y and B2Y) or (A2Y and B3Y) // logical operations
=> (2) or (3)
Now, as strange as it sounds, in a quantum world Alice and Bob can defeat that proof and prepare a strategy beforehand. When you ask Alice a question, she will consult her prepared photon in a box, and measure it in a way that depends on which question you asked. The measurement will mess up the photon's state, so you don't get to ask another question. Elsewhere, Bob will do the same with his photon. The two photons were prepared together, but are completely separated at the time of the game.
Before you ask, it's provable that such tricks cannot be used to send information faster than light. Quantum "spooky action" is somehow more powerful than hidden correlations, but less powerful than outright communication. It's pretty subtle and most popular accounts don't get it right.
- lmm 10y ago> it can't be explained by hidden correlated information about both objects. To be clear: it can't be explained by locally deterministic mechanisms, assuming a nondeterministic universe.
- BrandonSmithJ 10y agoI don't quite follow your proof on the last two steps; would you mind explaining? You say it's by logical operations, but I'm missing something because I see: (A && B && C) || (A && D && C) == (A && B) || (D && C)
- cousin_it 10y agoIt's =>, not ==. We need to show that (4) is contained in the union of (2) and (3).
- mh-cx 10y agoI think you misunderstood his question. I have the same problem. How do you get from this line: (A1Y and B2Y and B3Y) or (A1Y and A2Y and B3Y) to this (A1Y and B2Y) or (A2Y and B3Y) BrandonSmithJ just did some replacement to make it easier to read. Maybe he shouldn't have used the same letters to avoid confusion. Let's write it differently: Why are these terms equivalent? (u && v && w) || (u && x && w) (u && v) || (x && w)
- cousin_it 10y agoThey are not equivalent. The former term logically implies the latter term. That means the set of situations described by the former term is a subset of the situations described by the latter term. That means the probability of the former set should be less or equal than the probability of the latter set.
- mh-cx 10y agoAh, makes sense now. Thanks! Now I also get your previous comment. I've read your => as arrows instead of "equal or greater".