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What about quantum entanglement is suprising, and what makes it a strictly quantum effect?.

Suppose we have an particle of spin 0 which decays into two other particles, each with either spin up or down. Supposedly the particles are "entangled". The angular momentum of the system is conserved, so measuring one particle's spin as up would mean if we measure the other particle's spin, it must be down, and vice versa. Why is this suprising, and what does this have to do with entanglement or information about the measurement performed passing from one particle to the other.

As far as I can see, it's just momentum conservation. And I don't see what's quantum about this either, as we could do the same thing with classical particles (say an object which breaks into two parts) and the momentum measurements would be anti-correlated.

Qmechanic
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math_lover
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  • Someone might have a more eloquent answer, but essentially the weird features of entangled states are that they do not factor into (Particle A is in state 1)(Particle B is in state 2) which would be a satisfactory classical description. Instead entangled states look like (One of A/B is in state 1, then other is in 2). There are factored states as well, but they're much rarer and they would behave differently in terms of interference effects. – jacob1729 Jun 01 '19 at 19:04
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    I don't find the answer given there to be satisfactory. – math_lover Jun 01 '19 at 19:22
  • Though it might seem trivial and that entanglement is a consequence of conservation of momentum, but wait, there is more to it. Have you considered the non-locality of space which is inherent to quantumness of your entanglement experiment that you have described(above)? You aren't surprised by that aren't you? – Fracton Jun 01 '19 at 19:23
  • what do you mean by non locality. could you identify the problem with the momentum conservation explanation ? – math_lover Jun 01 '19 at 19:25
  • @JoshuaBenabou If you think your question is a duplicate but don't like the other answers, specify why you don't like it! Because to me, it seems to answer exactly what you're asking. – knzhou Jun 01 '19 at 19:57
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  • because the example of a "more complicated experiment" where thr socks logic doesn't work is some thought experiment with knobs which is poorly described. for example, how does he calculates the various probabilites he mentions. secondly he says such an experiment demonstrates quantum weirdness. however, as i understand it the so called weirdness was initially a prediction of the theory, so what wete thr founders of quantum mechanics actually debating before the experiments finally settled the debate? – math_lover Jun 01 '19 at 20:04
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    If you don't find the answers to a question satisfactory, the proper action is to offer a bounty explaining why not, not to ask the same question again. – tparker Jun 01 '19 at 20:18
  • I agree, just for up and down this is not surprising. But then for the same particles it holds in any basis. So if you perfom spin measurement not in Z basis, but instead (both) in X or Y basis you get the same correlations. Or any other direction. If you think in classical terms this implies you shd be able to prepare a sample with precise Z and precise X spin. But this is fundsmentally impossible as you know. – lalala Jun 01 '19 at 20:38
  • Related: https://physics.stackexchange.com/q/446974, which includes a derivation of the CHSH inequality, a relatively simple example of a Bell inequality. The correlation described in the OP is not surprising. What is surprising (from the perspective of classical physics) is that relationships like the CHSH inequality can be violated. – Chiral Anomaly Jun 01 '19 at 23:29
  • Try looking at my answer here: https://physics.stackexchange.com/a/330571/4993 – WillO Jun 02 '19 at 00:43
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    @JoshuaBenabou: Unfortunately the surprising thing about entanglement isn't this easy to describe. It's the fact that the joint state of a set of entangled particles is not the aggregate of their individual states (their joint relationship has an effect too), just like how the potential energy of a system is not the aggregate of their individual potential energies (the joint configuration has an effect too). Bell created an experiment that demonstrated this has physical implications. Try the bottom of page 18 here – user541686 Jun 02 '19 at 06:16
  • I am not sure why my comment about this question to be a duplicate is being ignored (0 upvote compared to kzhou's suggestion which has now 4 upvotes). It contains many answers to the very same question, one answer has over 80 upvotes. What's more, kzhou's link is actually a duplicate of the question in the link I provide (posted in Feb. 2013 vs August 2013 for kzhou's link). – untreated_paramediensis_karnik Jun 02 '19 at 12:41

4 Answers4

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Maybe a good demonstration of why entanglement is so puzzling is the Mermin-Peres magic square game. There are three players, two of whom (A and B, say) have entangled states. A and B are on the same side, and you are on the other side.

A and B are allowed to communicate and arrange their strategy in advance, but they cannot communicate once the game is in progress.

There is a $3 \times 3$ grid. You can ask A for any column of the grid (but just one), and you can ask B for any row of the grid (but just one). The rules are that A and B must assign 0s and 1s for their cells in the grid, they must agree on the cell where the row and column intersect, and the number of 1s in a column is always odd, but the number of 1s in a row is always even.

For example, if you ask for column 1 and row 2, they might return:

A      B

1xx    xxx  
0xx    011  
0xx    xxx

If they have entangled states, A and B can always win.

"Simple," you say, "A and B have agreed on which cells have 0s and 1s in advance, and they just return those values."

But if that's their strategy, does the master grid have an even or an odd number of 1's?

Peter Shor
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  • "There are three players, two of whom (A and B, say) have entangled states." - The third player is "you", right? And A and B are on the same side, while "you" are on the opposing side. – Tanner Swett Jun 02 '19 at 04:47
  • @Tanner Swett: that's correct. Let me clarify that. – Peter Shor Jun 02 '19 at 10:03
  • Thank you for your answer Dr. Shor. It is indeed weird. After reading about how using entanglement particles can be used to win the game every time, it seems like an interesting but construction. I guess the paradoxes raised by the EPR people were simpler to arrive at. – math_lover Jun 02 '19 at 11:08
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    @Joshua: Yes, Bell's inequality was simpler to figure out, but because it involves probabilities it's more difficult to see intuitively why violations of Bell's inequality can't be explained classically. – Peter Shor Jun 02 '19 at 11:34
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You are correct that the observation you mention is not surprising, but you have not mentioned the observation that lies at the heart of entanglement. Entanglement is interesting and surprising because it is owing to entanglement that further experiments can be done, in addition to the one you mention, and it is these further experiments that exhibit the surprising features. The further experiments can be, for example, measurements of pairs of spin-half particles, but with measurements along various different directions (e.g. 0, 120, 240 degrees to the $z$ axis if they are moving along the $x$ axis), or measurements on pairs of spin-1 particles, or measurements on triples of spin-half particles. Various scenarios break the Bell inequalities, and this means the measurement outcomes are inconsistent with a description in which each particle carries its properties with it in a local way.

In this answer I am not going to repeat the Bell arguments; you can look them up if you like (e.g. try CHSH inequality). I will simply present a nice argument involving symmetry which you may find interesting.

Suppose I have a single spin in the state $| \uparrow \rangle$. Then if I rotate it through 180 degrees then it will go to the state $| \downarrow \rangle$. One can do this in the lab and measure the outcome and thus confirm that the state does change in exactly this way. So far so good.

Now prepare two spins $A$ and $B$ in the state $$ |E\rangle \equiv \frac{1}{\sqrt{2}}( | \uparrow_A\uparrow_B \rangle + | \downarrow_A \downarrow_B \rangle ) $$ This is an entangled state. Rotate the first spin: you then get $$ |R\rangle = \frac{1}{\sqrt{2}}( | \downarrow_A \uparrow_B \rangle + | \uparrow_A \downarrow_B \rangle ). $$ This is a different state, indeed it is orthogonal to the first, so the rotation certainly changed the system and one can perform measurements to confirm that it did change. Still no surprise. But see what comes next.

Now suppose you want to return the system to its initial state. You have a choice. You could rotate spin $A$ back again, undoing the change. OR you could instead approach spin $B$ and rotate that one. Then you would get $$ \frac{1}{\sqrt{2}}( | \downarrow_A \downarrow_B \rangle + | \uparrow_A \uparrow_B \rangle ) = | E \rangle . $$ Thus this rotation returns the system to its original state.

Now think very carefully about what just happened. Those two spins could be in different places, say one in Athens and one in Bermuda. But to take the system from state $R$ to state $E$ you can rotate either the Athens spin or the Bermuda spin. These two operations, taking place on either side of the Atlantic Ocean, carry the joint system between the same two places ($R$ and $E$) in its state space. Try to imagine a classical scenario where this would happen---you will not be able to. Notice especially the sequence where first an operation is applied in Athens---an operation which certainly changes the system state---and then an operation in applied in Bermuda, and the overall outcome is no net change to the joint system.

I hope you are beginning to see how amazing entanglement is.

Its amazingness will soon be put to practical use in quantum computing. It also has deep philosophical implications, because it shows that the natural world is not completely decomposable into separate bits and pieces.

Added material to respond to comments.

Several people asked for further elucidation on why this is surprising, i.e. different from classical physics, and why it would not form a means of communication.

To emphasize what is going on, compare it to flipping something ordinary such as a chair. If I flip over a chair in the kitchen, say in order to clean the floor or something, then it would be very odd to then argue that by flipping some other chair, say one in an upstairs bedroom, I could return the pair of chairs to their starting state! (The word 'flip' here is being used in an operational sense: it means "apply a rotation through 180 degrees"; and note that the joint system does change state when this rotation is applied to either subsystem on its own---it is not like rotating perfectly symmetric spheres or something like that).

No communication is possible merely on the basis of this property, because in order to determine which of the possible states one has ($E$ or $R$ in my notation) it is necessary to bring together information from the two sites ($A$ and $B$) and this pooling of the gathered information can only happen at a light-speed-limited rate.

It is not true to say that the effect of an operation at $A$ is immediately observable at $B$ (or vice versa). Rather, the effects of operations at the two locations can eventually be determined by someone in the future light cone of both.

Andrew Steane
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    Why is this outcome surprising? If the outcome is determined by the state of the system - by the orientation of A and B relative to one another - then why is it surprising that rotations performed independently and yet resulting in a particular mutual state, produce a particular outcome? – Steve Jun 01 '19 at 23:20
  • I think I see @Steve s question. If I have a quarter heads up in Athens and another heads down in Bermuda the system is in a mismatched state. If I flip Athens over the system is now in a matched state. I can return to the mismatched state by flipping Athens or Bermuda. So, Andrew, how is this mundane experiment any different than what you just said about quantum entanglement? – candied_orange Jun 02 '19 at 03:30
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    Knowing practically nothing about the topic, isn't the difference that, with quantum entanglement, the change one one side of the ocean is observable on the other side? Indeed this is not the case with coins. – Jonathon Reinhart Jun 02 '19 at 04:31
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    I'm reading your example and failing to see how it doesn't imply faster-than-light communication? Alice intends to communicate to bob a bit at time T using this up/down + up/down mixture. She either flips her bit or doesn't (1 bit of information). Bob performs a measurement on his end at time T and detects whether it behaves identically to the initial system or not. (You claimed it's possible to distinguish the second state from the first, so this would be possible.) He therefore receives a bit of information faster than light. Am I misunderstanding what you wrote, or is there an error? – user541686 Jun 02 '19 at 09:35
  • @JonathonReinhart not quite right; see added material in my answer. – Andrew Steane Jun 02 '19 at 15:14
  • @AndrewSteane, the problem with "chair" analogies is that they don't really pick at the fundamental issue to be illustrated. As I understand it, there is an emitter of entangled particles, situated between two receivers - one receiver in Athens, one receiver in Bermuda. The receivers can be in one of two orientstions - let's say Left and Right. The relationship between the receivers is also a binary state - Matched or Unmatched. The two entangled particles carries some orientation, one Left and one Right (but we do not initially know which particle has which state)... (1/2) – Steve Jun 02 '19 at 16:03
  • ... what we do know however is that the entangled particles are in an Unmatched state relative to each other. Now, when a particle strikes a receiver, we know whether the relationship between orientation of particle and receiver was Matched or Unmatched. If the two receivers were in a Matched state, then it follows that one of the receivers must be in a Matched state relative to the particle it receives, and the other must be in an Unmatched state relative to the particle it receives. We know this logically because we know the particles themselves were Unmatched. Am I right so far? (2/2) – Steve Jun 02 '19 at 16:08
  • @Steve The receivers will behave as you say when they are either aligned with one another or at 90 degrees to one another, but you need other orientations (eg 120 degrees) and thus partial correlation to break the Bell inequality with a pair of spin half particles. You are exploring Bell-type arguments, which is relevant and useful; in my example I am pointing out the unusual type of overall symmetry which is on display. – Andrew Steane Jun 02 '19 at 16:32
  • @AndrewSteane, ah, then all I can say is that I haven't appreciated the significance of what you describe. Given the prevalence of theories like QM, I'm always surprised there aren't specific canonical and widely-accepted examples that illustrate the principles. – Steve Jun 02 '19 at 16:51
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In quantum mechanics, a particle doesn't really have a property like "spin up". It only has a probability of having a certain spin and the actual spin is only determined when you make a "measurement" or observation.

Let's say we park a space ship halfway between earth and moon and we shoot two entangled particles: one towards earth, one towards the moon. When they arrive we measure their spin on earth and on the moon at the same time. As expected, the spins are opposite.

The quantum mechanical interpretation says "spin only gets determined when measured". That would basically require the two particles on earth and on the moon to communicate with each other and decide on the spot, which spin to assume, so they make sure the spins come out opposite. This would require faster than light communication or, as Einstein put it "Spooky action at a distance" or violation of "locality"

This was called out by Einstein Podolsky and Rosen in the so-called ERP experiment or ERP paradox https://plato.stanford.edu/entries/qt-epr/ and a clear challenge the Copenhagen interpretation (Bohr, Heisenberg, etc.). At the time, no one could come up with setup or experiment which would result in an observable difference.

However, in the 1960s John Stuart Bell came up with Bell's theorem that predicted a measurable difference. https://en.wikipedia.org/wiki/Bell%27s_theorem. In the 1990s, Alain Aspect et al. https://en.wikipedia.org/wiki/Alain_Aspect managed to execute a meaningful experiment and it clearly showed the violation of Bell's theorem. It's been confirmed many times after this as well.

In other words, Einstein was wrong and Bohr was right: apparently entangled particles can communicate instantaneously and we have clear experimental evidence that this entanglement cannot be explained by a-priori knowledge or some hidden state variable.

Cosmas Zachos
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Hilmar
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    two things: why did the copenhagen guys believe that "an observable only gets determined when measured". second, if there is instantaneous communication, how is this not violating relativity? – math_lover Jun 01 '19 at 20:16
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    third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements? – math_lover Jun 01 '19 at 20:19
  • Since it's one of the few ways to explain quantum behavior. It's weird, but less weird that the alternatives. Read https://en.wikipedia.org/wiki/Copenhagen_interpretation .
  • – Hilmar Jun 01 '19 at 22:07
  • Yes, it violates relativity. You asked "what's so surprising about it". Answer: "it does indeed violate relativity and no one has a clue how and why, but all experimental evidence shows that it does". If that isn't surprising, I don't know what is.
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  • Hidden variable is a theory to explain quantum entanglement while maintaining locality at the same time. It's basically your initial assumption "all can be explain with conservation and conventional physics.". Doesn't work though. See https://en.wikipedia.org/wiki/Hidden-variable_theory
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