Suppose you want to construct a pair of yes-no-question-answering boxes. The goal is that, however one box answers, the other box will answer the opposite way. The answers produced must be anti-correlated, but I'll just write correlated because it's 4 chars less!
Classical correlations are like a pair of boxes that consistently succeed for some specific question. (e.g. repeatedly asking the same question of any one box, will produce 50% yes and 50% no answers, with the other box answering exactly oppositely)
Quantum correlations are like a pair of boxes that consistently answer anticorrelatedly for any question. For example, if you ask the boxes the question left-or-right? one will answer left, the other right. If you ask up-or-down? one will say up the other down, front-or-back? --- anticorrelated. And so on for any question. Whether that's spooky, I don't know...
To throws some math into this, let's denote |x>, the state of a quantum system, and denote |xy> the joint state of a pair of quantum systems, one in state x, the other in y.
Consider the difference between a statistical mixture:
50% of time produce |01> and 50% of time produce [10>
1/√2*|01> - 1/√2*|10> a.k.a |Psi->
But who has seen this quantum stuff? Isn't |Psi-> just some math scribble on a piece of paper? In fact, |Psi-> is quite real---it is the spin state of the electrons in the unexcited Helium atom. We know there are 2 electrons, and we know they have opposite spins, but we don't know anything else about them. Hence if a He atom were to explode and the electons fly to distant sides of the universe, we still won't know what direction the individual spins are in, the only thing we know is their total spin is 0 -- hence they have opposite spin directions. This is how we get the for any question" aspect of q-entanglement.
It really makes sense if you think about these things as information, and q-measurements as answering well posed questions. From the point of view of information, a system is entangled whenever you know more about the system as a whole than about its parts: http://en.wikipedia.org/wiki/Joint_quantum_entropy#Propertie...
You take a rapidly rotating oreo, break it in half and send the halves alway at ~c. Each experimenter asks: "Is the oreo, projected on an axis, is oriented core up or face up?"
For any axis chosen, answers will be anti-correlated, and of course the probability that they observe each is 50% (if I understood correctly, your |0>+|1>/|0>-|1> is equivalent to a 45 degree oreo decison axis).
But now, what if instead of measuring both oreos on the same axis, we measure then on slightly different axes?
In particular, let C(a) be the correlation coefficient between the orientation of the oreo measured along the z axis and its orientation measured at an angle a to the z axis.
You can derive a limit on the correlation that can be seen between any two angles (a and b, assume for simplicity coplanar with the z axis) - assuming the state of each oreo is a function of local variables like its speed and axis of rotation:
|C(a) - C(b)| - C(b - a) <= 1
For example, if you pick a and b both = 0 (measure everything along the positibe z axis, like you do in your comment), then the left hand side is just |(-1) - (-1)| - (-1) = 1, i.e. the QM prediction for that is perfectly replicable by tasty, tasty oreos.
But here's the thing: quantum mechanics breaks that limit for some angles. E.g. try it with a = pi/3, b = 2pi/3. Left hand side comes out as 1.5.
Informally, the absolute best correlation between a and b you can get with oreos is perfect linear correlation - what the inequality above is = 1. But with QM, you get a sine wave!
So rotating oreos seem like a good analogy to entanglement at first, but it's misleading. Turns out, neither rotating oreos nor any other classical system of local hidden variables can actually replicate the results of entanglement in QM. That's (a simplified form of) what Bell proved.
David Kaiser isn't even really a "tech reporter", he's one of the physicists behind the experiment described in the article, which addresses exactly what you're complaining about.
1. You and your friend can consult with each other before the game starts and agree on a strategy. You can build or obtain any equipment you think will help you play the game.
2. You and your friend, and any equipment you bring, meet me at the place where the game starts. I have two spaceships waiting. You and one of my assistants will get into spaceship A and go off into space. Your friend and my other assistant will go into spaceship B and go off into space. You and your friend can bring along whatever equipment you want.
3. My assistants will take you to separate locations in space. The locations will be at rest with respect to each other, and several light minutes apart. My assistants will establish synchronized clocks between the two locations, and then play starts. I'll only describe what goes on at your location. The same thing is going on at your friend's location.
4. There will be 1000 rounds, and each round lasts one minute. In each round my assistant will use a random number generator to generate a single bit, 0 or 1, and will tell you the bit. You can assume that this random number is truly random. There is no known way to predict it, and there is no correlation between it and the random number generator at the other location. The rounds start on a strict schedule so that you and your friend start round N on your respective ships at the same time.
5. After you are told the random number, you must generate a single bit, 0 or 1, by any means you wish. You can come up with it in your head. You can use any equipment you brought. You can read it from a list of predetermined bits you and your friend generated before the game started. You can try to communicate with your friend (or with anyone else) if you brought communications equipment, and use the results of that communication--but keep in mind you are several light minutes from your friend.
6. After the 1000th round, my assistant will fly you back to Earth where I will be waiting. My assistants will each give me the list of random numbers and responses. I will then compute your team score by adding up your scores for each round. Scoring for each round is as follows:
• If you both received a 1 in the round from the random number generators, then your team scores 1 point if you and your friend gave DIFFERENT bit values for that round. I.e., if the random bits were both 1, then you score if you gave 0 and your friend gave 1, or if you gave 1 and your friend gave 0.
• Otherwise, your team scores 1 point if you and your friend gave the SAME bit value for that round.
Your win prizes if your score is above 800.
Question: what is your plan?
You can win 75% of the rounds with a very simple strategy: always pick 0. You'll win the rounds where the random bits were 00, 01, or 10, and lose when they were 11. That turns out to be the best you can with non-quantum methods.
If you use entanglement you can do better. You prepare 1000 pairs of qubits, numbered 1 to 1000. The two qubits in each pair are entangled in what is called a Bell state. It's an equal superposition of the two states "both qubits are 0" and "both qubits are 1". What this means is that if you take these two qubits and you measure one and your friend measures the other, AND YOU MEASURE IN THE SAME BASIS (I'll explain that in a moment), you will either both get 0 or both get 1.
What I mean by measure in the same basis is this. Suppose your qubits are implemented by, say, polarized photons. A photon can be polarized at any angle from 0 to 360 (I'm going to use degrees instead of radians because I think that will be more comfortable for most). To make qubits out of this, you might decide that 0 is represented by polarization at 0 degrees, and 1 by polarization at 90 degrees. To see if a qubit is a 0 or a 1, you could try to pass it through a polarizing filter set to 90 degrees. If it comes through, it was a 1, and if it is blocked it was a 0. If we are using 0/90 degrees for our qubits, we say we are using the 0 degree basis.
What happens if you pass that qubit through a polarizing filter at 45 degrees? Half the time it will pass, and half the time it will be blocked. So a qubit that has a definite value in the 0 basis can be any value with equal probability in the 45 degree basis. How about if we measure using a 22.5 degree basis? It can still be any value, but now there is about an 85% chance it will have the same value as in the 0 degree basis. In other words, if someone generates a qubit with a particular binary value in the 0 degree basis, and you measure it in the 22.5 degree basis, 85% of the time you'll get the same value that the person who made it set it to in their basis.
In general, if we have two qubits in the same state, and you measure one of them in one basis, and I measure the other in a basis that is T degrees different from yours, we will get the same measurement with probability cos(T)^2. E.g., if we use the same basis (T=0), then we get the same measurement all the time. If my basis is 90 degrees from yours, we'll get opposite measurements. If my basis was 45 degrees from yours, half the time my measurement would agree with yours and half the time it would not.
Here is how you use your prepared qubits to improve your game outcome. As I said, you take pairs of qubits and put that system of two qubits into an equal mix of the states "both qubits are 0" an "both qubits are 1". When you head off to space, you take one of the qubits from each pair, and your friend takes the other.
For round N, after you get the random bit from my assistant, you take your qubit for round N, and you measure it to see if it is a 0 or a 1. If my assistant gave you a 0 bit, you measure in the 0 degree basis. If my assistant gave you a 1 bit, you measure in the 45 degree basis. You report the result of your measurement as your bit.
Your friend measures in the 22.5 degree basis if he gets a 0 random bit, and he measures in the -22.5 degree basis if he gets a 1 random bit. He reports the result of his measurement as his bit. Here's a diagram showing the angles you each use depending on the random bit you receive:
1---- friend -----0
| |
-22.5 0 22.5 45
| |
0----- you --------1
So, in all cases except both random numbers being 1, which means that you want to pick the same bit as your friend, your measurements will agree with probability cos(22.5)^2 = 0.854.
In the case where both random numbers are 1, which means you and your friend want to NOT agree, you will agree with probability cos(67.5)^2 = 0.146, which means you will disagree with probability 0.854.
Net result: your probability of winning a given round is 0.854, which is quite a bit better than the 0.75 that is the best you can get with a non-quantum approach.
Try to duplicate this with a "half an Oreo" model of entanglement, and it won't work.
We know it must be this way due to statistics. However, that's the point where the Oreo analogy breaks down.
Now, if we're measuring them three at at time, there are only two plausible possibilities: one of us has all three cremes (this happens 2 out of every 8 trials) or one of us has two cremes (this happens 6 out of every 8 trials). If one of us has all the cremes, it doesn't matter which one each of us independently chooses to measure; we will both have the same outcome. If one of us has only two of the cremes (e.g., [M]e has two and [Y]ou have one), we agree 5/9ths of the time
M M Y
M 1 1 0
M 1 1 0
Y 0 0 1
You have two photon detectors that you can rotate; one is in New York and one is in Paris. Two photons are entangled and one is sent to New York for measurement, and one is sent to Paris for measurement. What you find is that while each photon's polarization measurement is random, if you take the "random" values you get for both and compute their correlation over many trials, this value is a function of the difference in angles between the photon detector in New York and the photon detector in Paris.
But here's the kicker: each photon detector can be rotated while the photons are in flight, before they get there. So how is it that the correlation is a function of both photon detector angles then? Especially considering it is light that you're measuring, and nothing can travel faster than light. That's why it's called spooky action at a distance.
(Edit: Let me know if this makes any sense. I'm in a continual search for the simplest explanation of this phenomenon.)
http://web.pdx.edu/~pmoeck/lectures/Mermin%20shorter.pdf
"A simple device is described, based on a version of Bell's inequality, whose operation directly demonstrates some of the most peculiar behavior to be found in the atomic world. [...] The extraordinary implications of its behavior should be evident to anyone."
A quick search also turned up this animation that demonstrates Mermin's device:
http://faraday.physics.utoronto.ca/PVB/Harrison/BellsTheorem...
They always talk as if it's a spooky interaction but the metaphor falls down. Here's an example: I take too halves of an Oreo cookie and send them in opposite directions. Then I use a detector to measure how much filling is on one half and instantly know how much was on the other half. Was that spooky?
I understand that at the physics level it's not like this, but the metaphors are never satisfyingly 'spooky' to me as I can easily think up a classical explanation.