4 ms·

I feel I'll never get to understand entanglement (I hope it's like monads :P). If it's not "spooky" but just half an Oreo... then what is so special about it an

by kaoD 12y ago

I feel I'll never get to understand entanglement (I hope it's like monads :P). If it's not "spooky" but just half an Oreo... then what is so special about it and why would physicist study it then?

stouset 12y ago
The difference is that when you separate a quantum Oreo, until you actually observe which side has the creme on it, the universe hasn't decided yet. Once you observe one half, the universe instantaneously picks one of the two halves at random to contain creme, and one to be empty. We know it must be this way due to statistics. However, that's the point where the Oreo analogy breaks down.
- FranOntanaya 12y ago
  How do we know the universe hasn't decided?
  - stouset 12y ago
    You and I decide to test whether or not the universe has decided. We have Nabisco mail us quantum oreos, three at a time and we each choose randomly to measure one of them (this is analogous to measuring the spin of an electron in a randomly-chosen axis). 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 If we run this experiment many, many times, we should converge upon (2/8 * 9/9) + (6/8 * 5/9) = 2/3 of the time our measurements agree. This would be consistent with a model of "preprogrammed" quantum Oreos. This is not what we see in practice. When we actually run this experiment, our measurements agree exactly 50% of the time which can only be explained if the quantum Oreos are not preprogrammed.
Xcelerate 12y ago
I have never found a good, simple analogy. The best I've got is still a mathematical explanation, but I think it can make it "click" for some people: 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.)
- yetihehe 12y ago
  It makes sense, but I've never understood how do we know that photons don't already have this polarization at start, how do we know it's value is defined only on detection?
svedlin 12y ago
David Mermin wrote a classic paper that illustrates the strange statistical properties of entanglement in a way that doesn't require any physics knowledge. http://web.pdx.edu/~pmoeck/lectures/Mermin%20shorter.pdf 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/Flash/Mermin/Mermin.html http://faraday.physics.utoronto.ca/PVB/Harrison/BellsTheorem...
- po 12y ago
  That Mermin PDF is pretty interesting thanks!
yen223 12y ago
Are you saying quantum entanglement is like a burrito?
tzs 12y ago
Let's play an imaginary game. It will be you and a friend as a team against the house, represented by me and my two assistants. Here's how the game works. 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 Note that if the random bits are both 1, then you two do your basis and his are 67.5 degrees apart. On the other three cases, they will only be 22.5 degrees apart. 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.

by kaoD 12y ago

I feel I'll never get to understand entanglement (I hope it's like monads :P). If it's not "spooky" but just half an Oreo... then what is so special about it and why would physicist study it then?

stouset 12y ago

The difference is that when you separate a quantum Oreo, until you actually observe which side has the creme on it, the universe hasn't decided yet. Once you observe one half, the universe instantaneously picks one of the two halves at random to contain creme, and one to be empty. We know it must be this way due to statistics. However, that's the point where the Oreo analogy breaks down.

FranOntanaya 12y ago
How do we know the universe hasn't decided?
- stouset 12y ago
  You and I decide to test whether or not the universe has decided. We have Nabisco mail us quantum oreos, three at a time and we each choose randomly to measure one of them (this is analogous to measuring the spin of an electron in a randomly-chosen axis). 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 If we run this experiment many, many times, we should converge upon (2/8 * 9/9) + (6/8 * 5/9) = 2/3 of the time our measurements agree. This would be consistent with a model of "preprogrammed" quantum Oreos. This is not what we see in practice. When we actually run this experiment, our measurements agree exactly 50% of the time which can only be explained if the quantum Oreos are not preprogrammed.

Xcelerate 12y ago

I have never found a good, simple analogy. The best I've got is still a mathematical explanation, but I think it can make it "click" for some people: 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.)

yetihehe 12y ago
It makes sense, but I've never understood how do we know that photons don't already have this polarization at start, how do we know it's value is defined only on detection?

svedlin 12y ago

David Mermin wrote a classic paper that illustrates the strange statistical properties of entanglement in a way that doesn't require any physics knowledge. http://web.pdx.edu/~pmoeck/lectures/Mermin%20shorter.pdf 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/Flash/Mermin/Mermin.html http://faraday.physics.utoronto.ca/PVB/Harrison/BellsTheorem...

po 12y ago
That Mermin PDF is pretty interesting thanks!

yen223 12y ago

Are you saying quantum entanglement is like a burrito?

tzs 12y ago

Let's play an imaginary game. It will be you and a friend as a team against the house, represented by me and my two assistants. Here's how the game works. 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 Note that if the random bits are both 1, then you two do your basis and his are 67.5 degrees apart. On the other three cases, they will only be 22.5 degrees apart. 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.