23

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?

Andreas K.
  • 2,156

8 Answers8

12

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.

Greg Graviton
  • 5,157
  • 1
    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 07:45
  • @Marek: Fair enough. I've reworded it a bit, did it become better? – Greg Graviton Jul 15 '11 at 11:47
  • 3
    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
  • Are you saying that if Alice measures her $S_x$ and Bob his $S_y$ and if Bob tells his result to Alice she still can't be sure for her $S_y$? – Andreas K. Jul 15 '11 at 13:25
  • I have another question that bothers me. Alice and Bob are far away from each other so how can we be sure that Alice's direction of measurement, lets say x, and Bob's x-direction are really the same? Their orientation on the space can be such that Alice's x-direction is Bob's y-direction. Or the absolute direction doesn't matter? – Andreas K. Jul 15 '11 at 13:32
  • @ANKU: Exactly, Alice will still be unsure of her $S_y$. This seems to contradict the (anti)correlation, but it is required by the uncertainty principle. – Greg Graviton Jul 15 '11 at 13:47
  • @ANKU: Your second comment is a non-issue. Alice and Bob have plenty of time and resources to prepare their apparatuses. In particular, they can make sure that they face in the same directions. For instance, they could start very close to each other and then walk apart without changing orientation. Their spatial distance is of negligible importance. – Greg Graviton Jul 15 '11 at 13:51
  • 1
    @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. – Andreas K. Jul 15 '11 at 16:04
  • @ANKU: It's a good analogy for the correlation, yes. The core of the paradox is basically just that $S_x$ and $S_z$ are complementary quantities. – Greg Graviton Jul 15 '11 at 16:49
  • I am a bit lost on the supposed paradox, even as Marty Green tries to spell it out below. If I understand, Alice (A) and Bob (B) both get different balls, neither knows what color (Spin) each of their balls are. Each ball can be black (Spin X-direction: $S_x = +1 $) or white (Spin Y-direction: $S_y= -1$). I think I am a but confused on bullet 3, If A measures $S_x$ to be +1, so why would B be measuring $S_x$? He has is own ball? – Relative0 Jan 17 '16 at 01:16
  • @Relative0 The balls have two properties (spin $S_x$ and spin $S_y$), not just one (color). Be careful about $x$ and $y$. The paradox is that each property is perfectly correlated between the balls, but measuring property $y$ of the 2nd ball after measuring property $x$ of the 1st ball does not give you information about property $y$ of the first ball! – Greg Graviton Jan 17 '16 at 09:50
  • Imagine her surprise, then, when she tries to confirm her conclusion by measuring $S_y$ herself and obtaining +1 in 50% of the cases - can you elaborate this procedure a bit better? Wouldn't Bob akso measure -1 (perfect anti-correlation) for $S_y$ in these same 50% of cases? Or you are saying that Alice can measure $S_y$ on that same first article over and over again? – Lou Mar 31 '16 at 08:46
  • @LousyCoder You have to be careful about when to measure what. In this case, the protocols are: A) First, Alice measures $S_y$. Then, Bob measures $S_y$. B) First, Alice measures $S_x$. Then, Bob measures $S_y$. Then, Alice measures $S_y$. The paradox is that in the situation A), both measurements of $S_y$ will be perfectly anticorrelated, but in situation B), the measurements of $S_y$ will not be correlated anymore. – Greg Graviton Mar 31 '16 at 12:10
  • Thanks, so it's the same particle then, I wasn't sure. The problem with this example is that Alice somehow did "interact" with her particle, so the "spooky" action is not as clear as I would hope it would be. In other words, this concrete example wouldn't rule out local hidden variables. – Lou Mar 31 '16 at 13:54
5

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

But let us say Bertlmann 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.

Norbert Schuch
  • 20,338
Ron Maimon
  • 1
3

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.

AndyS
  • 1,053
2

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.)

Marty Green
  • 4,079
  • 1
  • 21
  • 27
  • To me, this doesn't make it clear what you think is wrong with Greg's answer. Could you add some explanation about what parts are nonsensical? – B T Mar 07 '16 at 01:09
  • Because classically, it is perfectly normal for 50% of the light which gets through a vertical polarizer to then also get through a 45 degree offset polarizer. I think Greg is just wrong when he claims a particle that gets through an x polarizer will always be blocked by a y polarizer. – Marty Green Mar 07 '16 at 02:12
  • That makes sense, I agree – B T Mar 07 '16 at 02:13
1

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.

Siyuan Ren
  • 4,932
  • 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? – Alan Rominger Jul 15 '11 at 04:34
1

I have the gut feeling that Greg's answer doesn't describe a situation that isn't able to be explained classically, since putting two detectors in series by definition is not measuring the two axes of spin simultaneously (and so all bets are off essentially, and this is just as true classically).

My understanding of why entanglement can't be explained classically is a Stern-Gerlach experiment where the direction of measurement of each detector is randomly chosen out of 3 possible directions. Each direction is 120 degrees offset from the other two. When this experiment is run on entangled electrons, the results are that the spin is the different 50% of the time. If there were no "action at a distance" (the concept of entanglement), we would expect the experiment to show the spins as different 5/9 of the time, because combinatorics.

Here's a video explaining this further: https://www.youtube.com/watch?v=ZuvK-od647c

B T
  • 792
0

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.

Chetan Waghela
  • 575
-1

Technically your experiment and analogy both work. But the difference is that your analog requires exponential information to store the states of entangled superpositions of different outcomes.

With two balls, this is not a big deal. But with hundreds or thousands of (qu)bit states to combine (ie. balls in your example), this requires exponential states.

All quantum computations can be simulated on a classical computer, so that means you could simply "set up" your entanglement. And this would work. But you pay for it in not having a continuous state space of superpositions, real entanglement, and interference.

Michael Tamillow
  • 99