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Consider that we have two balls, one white and one black, and two distant observers A and B with closed eyes. We give the first ball to the observer A and the second ball to the observer B. The observers don't know the exact color (state) of their balls, they know only the probability of having one or another color, until they look at them (measure). If the observer A looks at his ball he will see its color, which is white, so he immediately knows the color of the second ball. Lets call this “classical entanglement”.

My question is: What is the difference between this “classical entanglement” and the quantum entanglement, for example, of two entangled electrons with opposite spins states? Can this analogy be used to explain the quantum entanglement?

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Your "classical entanglement" is not a physical phenomenon. Is is a product of the rules of that game. (two balls, one black, one white) Quantum properties like eg spins can change on each "ball" –  Georg Jul 15 '11 at 8:27
    
@Georg: Paint exists. Classical properties like ball color can be changed too. Is it better to replace the balls with two coins, one heads and one tails? –  wnoise Jul 15 '11 at 10:05
    
Of course, flipping two coins is a real classic analogon. –  Georg Jul 15 '11 at 10:21
    
This does not answer your question fully, but you can sure use quantum tic tac toe as a great analogy for quantum entanglement (also superpositioning and collapse). –  user4503 Jul 15 '11 at 15:08
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@Georg: Of course paint only exists in the macroscopic world. But this is supposed to be a classical analogy. Why is it a flaw of a classical analogy that it uses classical concepts? The setup in both cases is the same: you have a property that two objects have that is setup so that one is the opposite of the other, but you don't know the individual state of one. Measuring one should let you know the state of the other. And indeed, it does let you know. The difference is the wider basis freedom you have in quantum mechanics, the existence of off-diagonal elements in the density matrix. –  wnoise Jul 16 '11 at 19:06

7 Answers 7

up vote 5 down vote accepted

Quantum entanglement is different from the "classical entanglement" in the following way:

  • In your example, each ball has only one property of interest, namely "color $\in$ {white, black}".
  • In the traditional examples of quantum entanglement, each ball (particle) has two properties of interest, namely "spin in x-direction $S_x$" and "spin in y-direction $S_y$". Moreover, the properties are complementary, i.e. you can't actually measure them simultaneously to arbitrary precision.
  • Still, in the entangled state, each of these properties alone is perfectly (anti-)correlated: If A makes a measurement of $S_x$ and obtains the value, say +1, then B will always obtain -1 if he also measures $S_x$. Similar for -1 and +1 and for the other spin direction $S_y$.

The paradox, now, is the following:

Suppose that Alice measures $S_x$ and obtains, say +1. But observer Bob measures $S_y$, obtaining, say +1. Exhulted, Alice proclaims that she has managed to measure two complementary properties simultaneously! After all, her measurement gave her the spin in x-direction, +1, while Bob's measurement allows her to conclude that the particle also has spin -1 in y-direction.

Imagine her surprise, then, when she tries to confirm her conclusion by measuring $S_y$ herself and obtaining +1 in 50% of the cases.

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The second bullet point makes it sound like the ball has precisely two properties which is a gross oversimplification of the actual situation (which is that there is a continuum of related properties). –  Marek Jul 15 '11 at 7:45
    
@Marek: Fair enough. I've reworded it a bit, did it become better? –  Greg Graviton Jul 15 '11 at 11:47
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Let me play devil's advocate for a moment: why is what you described not simply a result of "measurement == preparation" as opposed to anything to do with entanglement, per say? –  genneth Jul 15 '11 at 12:32
    
Yeah, sounds fine to me now. +1 –  Marek Jul 15 '11 at 12:34
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@Greg Graviton, thanks. But I still think that the "classical entanglement" is a good analogy, but only for one spin-projection direction, because as you said the classical ball has only one property (color). But I will agree with you that for more properties (spin-projection directions) there is no classical analogy, even if you paint each ball with more colors etc., it will not work. –  Andyk Jul 15 '11 at 16:04

In classical physics when we have a composite system that consists of two (or more) subsystems, then the state of that system is given by a combination of the states of the individual subsystems. The black-and-white-ball system you mentioned is an example of that; the state of the composite system is that one ball is black and the other white, defined solely by the fact that the state of one subsystem is that the ball is white, and the state of the other that the ball is black.

In quantum mechanics there are cases where things behave like in classical physics, and cases where they don't. When we have a composite system in quantum mechanics, there are two options for its state (wavefunction):

  1. it can be given just by knowing the states of the individual subsystems, or
  2. it may be its own thing, i.e. a state that is defined for the composite system and which cannot be decomposed into states of the individual subsystems.

The first option has a classical counterpart, like your example, and has nothing to do with quantum entanglement. A quantum state with the property of option 1 is called a product state in the quantum-physics literature.

The second option is what is called an entangled state, and has no counterpart in classical physics. It displays the idea of entanglement clearly: it is the phenomenon where the state of a system is not given by some combination of the states of the individual subsystems that make it up. In fact, all you can define for entangled systems is the state of the composite system, you cannot assign states to the individual subsystems. Clearly, that can never happen in classical physics.

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I am dismayed that Greg's answer continues to get upvotes because I think it is badly off target. I would like to note first that quantum entanglement experiments are not done with electrons but rather with photons. So I have taken the liberty of translating Bob's article word-for-word into the language of photons. It is often said that the case of entangled photons corresponds perfectly to the case of electrons except that all the angles are divided by 2: instead of perfect anti-correlation at 180 degrees, it becomes perfect anti-correlation at 90 degrees, etc. I hope I have transformed Greg's argument correctly; and I believe that it becomes transparently nonsensical when put in these terms.

Here is how it goes:

"Quantum entanglement is different from the "classical entanglement" in the following way: In your example, each ball has only one property of interest, namely "color ∈ {white, black}". In the traditional examples of quantum entanglement, each ball (particle) has two properties of interest, namely "ability to penetrate a vertical polarizer" and "ability to penetrate a 45 degree offset polarizer". Moreover, the properties are complementary, i.e. you can't actually measure them simultaneously to arbitrary precision.

"Still, in the entangled state, each of these properties alone is perfectly (anti-)correlated: If Alice sees a particle penetrate her vertical polarizer, then Bob will always see a particle blocked. And similarly if they both offset their polarizers 45 degrees.

"The paradox, now, is the following: Suppose that Alice measures sees a particle go through her vertical polarizer. But observer Bob offsets his polarizer 45 degrees and also sees a particle go through. Exhulted, Alice proclaims that she has managed to measure two complementary properties simultaneously! After all, her measurement showed her the particle had the ability to go through a vertical polarizer, while according to Bob's measurement her particle should with certainty be blocked by a 45 degree offset polarizer.

"Imagine her surprise, then, when she tries to confirm her conclusion by putting an offset polarizer in series with her vertical polarizer and finding that the particle which got through the first polarizer still gets through the second polarizer fifty percent of the time, even when Bob simultaneously detects a particle penetrating his offset polarizer."

I hope I have not done Greg's answer any injustice, but I believe this is what it leads to. (I should also mention that I totally disagree with the premise of starting off with colored balls as the classical "straw man". But that's another question.)

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None of these answers actually explain anything. "Imagine Alice's surprise when 50% photons still get through?" Why do they? How does that make an analogy?

Maybe a better explanation would be something along very simple maths lines. Imagine photons having one or more numerical "states". Entangling two photons creates a "state" analogous to multiplying their state values together. Then, upon measurement, an attribute in one photon is found to be, or forced to be, a particular value, lo, the remote photon takes on the appropriate values to ensure that the entangled product remains constant.

Also, talk of an "observation" is always confusing, since it implies non-interaction. I was just observing. But in fact, in order to observe, you need to interact at the quamtum level with the entangled system, or in some way impose a constraint on the "quantum wave" of possibilities/probablilities. A photon strikes a molecule in your eye, or whatever. In fact, you, or part of you, joins the entagled system and you become part of it. Entanglement looks one way "from the outside" and another way once it is frozen in time by you becoming part of it, seeing it from the inside as it were. The so-called collapse of the wave function. It's more a shift in perspective I think.

In the case of "Imagine her surprise, then, when she tries to confirm her conclusion by measuring Sy herself and obtaining +1 in 50% of the cases.". Surely this is because the first observation was actually an interaction. The photon has now become entangled with the observation equipment, and is thus not now constrained purely by Bob's measurement.

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It is somewhat analogous, but the analogy fails because of Bell's theorem. Bell uses a similar analogy with "Berkleman's socks", when you see one sock is white, and assuming they match, you learn the other is white instantly.

But let us say Berkleman has three feet, and tries to wear three matching socks, but he doesn't always succeed. When we see the sock on the left foot, we know that the sock on the middle foot has a 99% probability to be matching. Suppose also that the middle foot is 99% of the time the same as the right foot.

Using these two assumptions, you can conclude that the right foot can only be different from the left foot in at most 2% of the cases. This is intuitive--- the number of cases where there is a mismatch between the left and middle plus the number of cases where there is a mismatch between middle and right is always more than the number of cases where there is a mismatch between left and right. Convince yourself of this. This is called "Bell's inequality".

For entangled paticles, the measurements of "up" and "down" have the property that the quantity associated with the middle-sock can be 99% correlated with the left and right foot, but the left and right foot values are only 96% correlated. This is a violation of Bell's inequality. This means that it is not like socks. If it were like socks, the socks would have to change color in response to what you see, nonlocally, faster than the speed of light.

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No, quantum entanglement is not in any way analogous to your “classical entanglement”, or in more technical term, local hidden-variable theory.

To explain this, certain level of quantum mechanics and mathematics must be involved, but in a nutshell, such theory has been definitively invalidated by experiments.

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To squeeze a more meaningful conversation out of this, the example only had the AB observers "observing" when they looked at the object - meaning that when not looking they still had photons/particles incident on them. If we have macro objects that were almost or completely casually isolated for meaningful periods of time, would quantum weirdness become a possibility? –  AlanSE Jul 15 '11 at 4:34

Your analogy is very correct I was thinking on same kines that if we have two graph papers and a carbon paper, if I draw a line on one paper it will create a similarly correlated line on another, suppose if someone doesnt see what is made on paper and then they check what comes out they can seee they are perfectly correlated. Also, you can see that now if I rub a line on one of the papers and create a new one, it will not create a change on another. This is also provided by non-signalling theorem, it states that local unitary transformation cannot change a state of other entangled particle, hence superluminal communication is not possible. I feel the whole confusion stars because, in quantum mechanics we say nature is fundamentally probabilistic, if it is so, then it raises a question how come two particles know how to exactly correlate to each other. This is an open question, intrinsic randomness in quantum world, if this is proved to be wrong then your analogy will exactly apply to it. This was the exact statement if epr for hidden variables, apart from that violation of bells inequalities gave an upper edge to copenhagen interpretationists. Still, I feel that there are loopholes. Also, Teleportation is possible at todays world, Connect 2 3d printers and there it is a cloner as well as a teleporter, but again this question of intrinsic randomness comes up, so until that is not solved, and I feel it needs a mere restatement with corrections, the debates will go on.

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Also, check quantum eraser experiment grad.physics.sunysb.edu/~amarch –  Chetan Waghela Oct 18 '13 at 12:55

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