3 ms·

I have a good analogy for entanglement, which will hopefully help you see the difference between classical (anti)correlations and quantum (anti)correlations. S

by ivan_ah 12y ago

I have a good analogy for entanglement, which will hopefully help you see the difference between classical (anti)correlations and quantum (anti)correlations. 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> and the |Psi-> state which is a 50-50 superposition of |01> and |10>: 1/√2*|01> - 1/√2*|10> a.k.a |Psi-> In both cases if you ask the 0 or 1 question, the answers will be anti-correlated, but you can ask the |Psi-> state any other anti-podal question (e.g. are you |0>+|1> or |0>-|1>?) and the answers will always be anti-correlated. 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#Properties http://en.wikipedia.org/wiki/Joint_quantum_entropy#Propertie...

darkmighty 12y ago
I take his oreo in space question and raise you the rotating oreo in space question. 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).
- SEMW 12y ago
  Sure. And for what you described, that works fine. 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.
  - darkmighty 12y ago
    Nice answer, thank you. Rose my interest to study some quantum mechanics to figure out how exactly this can happen.

by ivan_ah 12y ago

darkmighty 12y ago

I take his oreo in space question and raise you the rotating oreo in space question. 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).

SEMW 12y ago
Sure. And for what you described, that works fine. 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.
- darkmighty 12y ago
  Nice answer, thank you. Rose my interest to study some quantum mechanics to figure out how exactly this can happen.