3 ms·

Sometime ago I came up with a very simple reformulation of this. Imagine that your friend Alice has a set of three fair coins, numbered from 1 to 3. If you ask

Sometime ago I came up with a very simple reformulation of this. Imagine that your friend Alice has a set of three fair coins, numbered from 1 to 3. If you ask her to flip one of the coins, the other two will disappear. Also imagine that your other friend Bob, who lives far away from Alice, has an identical set of coins that are somehow "mysteriously connected" to Alice's. The whole setup can be repeated as many times as you want. You do a few experiments and find out this: 1) If you ask both Alice and Bob to flip the same numbered coin, their answers always agree. 2) If you ask Alice to flip coin 1 and Bob to flip coin 2, their answers differ 5% of the time. 3) If you ask Alice to flip coin 2 and Bob to flip coin 3, their answers differ 5% of the time. 4) If you ask Alice to flip coin 1 and Bob to flip coin 3, their answers differ 20% of the time. Why is that weird? Well, if coin 1 differs from coin 3, then at least one of them would've differed from coin 2, so the 20% case must be completely covered by two 5% cases, which seems impossible. But you can't use it for faster-than-light communication between Alice and Bob, because all coins are individually fair no matter what. This setup actually happens in our world, it can be easily reproduced in a lab to 1 digit of precision. This kind of quantum "spooky action at a distance" sits strictly between what's possible in a classical universe and what would be possible with FTL communication. More precisely, it allows two distant players to jointly win some games that wouldn't be winnable in a classical universe (even with arbitrary communication beforehand), but not all games that would've been winnable with FTL.

mrfusion 10y ago
So experiment one could be explained by a hidden variable right? I'm guessing somehow the other experiments rule out a hidden variable but I'm not seeing why.
- strmpnk 10y ago
  I don't think that explanation is clear. It takes a bit more work to get something which proves the experimental difference. One of the better explanations I've seen is regarding measuring spin with three state detectors: https://www.youtube.com/watch?v=ZuvK-od647c https://www.youtube.com/watch?v=ZuvK-od647c This would give specific probabilities to certain outcomes (not just 50/50) IF there were a local hidden variable. (Note: the word local is important there. It means information encoded and carried by or with the particles themselves.)
- cousin_it 10y ago
  These aren't descriptions of individual experiments, but rather conclusions that you got from many experiments. Let's say you do a million experiments, choosing randomly which coins to ask Alice and Bob about. In every experiment where you asked them about the same numbered coin, their answers were always the same. Among experiments where you asked Alice about coin 1 and Bob about coin 2, their answers were different 5% of the time. Etc. Here's a quick proof why the results are impossible with hidden variables. Assume that before each experiment, Alice and Bob secretly agree on the answers they will give to you for each coin. At this stage they can conspire with each other as much as they want, flip actual coins, etc. Let's designate their pre-agreed answers as A1,A2,A3,B1,B2,B3 (each is either "heads" or "tails"). Then we know these facts: 1) P(A1 = B1) = 1, likewise for A2 and A3. So we can just use A1,A2,A3 and forget about B1,B2,B3. 2) P(A1 ≠ A2) = 0.05 3) P(A2 ≠ A3) = 0.05 4) P(A1 ≠ A3) = 0.2 But whenever A1 ≠ A3 is true, one of A1 ≠ A2 or A2 ≠ A3 must be true. So P(A1 ≠ A3) must be less than or equal to P(A1 ≠ A2) + P(A2 ≠ A3). However 0.2 is greater than 0.05 + 0.05, so we've reached a contradiction. That means Alice and Bob couldn't have agreed on all three coins in advance. In fact, if you do the whole thing in a lab (with photons), it will be physically impossible to measure all three coins, only two of them. That's exactly how quantum mechanics says "you're forbidden from knowing some variables at the same time".

mrfusion 10y ago

So experiment one could be explained by a hidden variable right? I'm guessing somehow the other experiments rule out a hidden variable but I'm not seeing why.

strmpnk 10y ago
I don't think that explanation is clear. It takes a bit more work to get something which proves the experimental difference. One of the better explanations I've seen is regarding measuring spin with three state detectors: https://www.youtube.com/watch?v=ZuvK-od647c https://www.youtube.com/watch?v=ZuvK-od647c This would give specific probabilities to certain outcomes (not just 50/50) IF there were a local hidden variable. (Note: the word local is important there. It means information encoded and carried by or with the particles themselves.)
cousin_it 10y ago
These aren't descriptions of individual experiments, but rather conclusions that you got from many experiments. Let's say you do a million experiments, choosing randomly which coins to ask Alice and Bob about. In every experiment where you asked them about the same numbered coin, their answers were always the same. Among experiments where you asked Alice about coin 1 and Bob about coin 2, their answers were different 5% of the time. Etc. Here's a quick proof why the results are impossible with hidden variables. Assume that before each experiment, Alice and Bob secretly agree on the answers they will give to you for each coin. At this stage they can conspire with each other as much as they want, flip actual coins, etc. Let's designate their pre-agreed answers as A1,A2,A3,B1,B2,B3 (each is either "heads" or "tails"). Then we know these facts: 1) P(A1 = B1) = 1, likewise for A2 and A3. So we can just use A1,A2,A3 and forget about B1,B2,B3. 2) P(A1 ≠ A2) = 0.05 3) P(A2 ≠ A3) = 0.05 4) P(A1 ≠ A3) = 0.2 But whenever A1 ≠ A3 is true, one of A1 ≠ A2 or A2 ≠ A3 must be true. So P(A1 ≠ A3) must be less than or equal to P(A1 ≠ A2) + P(A2 ≠ A3). However 0.2 is greater than 0.05 + 0.05, so we've reached a contradiction. That means Alice and Bob couldn't have agreed on all three coins in advance. In fact, if you do the whole thing in a lab (with photons), it will be physically impossible to measure all three coins, only two of them. That's exactly how quantum mechanics says "you're forbidden from knowing some variables at the same time".