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.

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    $\begingroup$ Possible duplicate of Can entanglement be explained as a consequence of conservation laws? $\endgroup$ – knzhou Jun 1 '19 at 19:10
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    $\begingroup$ I don't find the answer given there to be satisfactory. $\endgroup$ – Joshua Benabou Jun 1 '19 at 19:22
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    $\begingroup$ Possible duplicate of Quantum Entanglement - What's the big deal? $\endgroup$ – AccidentalBismuthTransform Jun 1 '19 at 20:04
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    $\begingroup$ 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. $\endgroup$ – tparker Jun 1 '19 at 20:18
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    $\begingroup$ @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 $\endgroup$ – user541686 Jun 2 '19 at 6:16

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?

  • $\begingroup$ "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. $\endgroup$ – Tanner Swett Jun 2 '19 at 4:47
  • $\begingroup$ @Tanner Swett: that's correct. Let me clarify that. $\endgroup$ – Peter Shor Jun 2 '19 at 10:03
  • $\begingroup$ 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. $\endgroup$ – Joshua Benabou Jun 2 '19 at 11:08
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    $\begingroup$ @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. $\endgroup$ – Peter Shor Jun 2 '19 at 11:34

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.

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    $\begingroup$ 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? $\endgroup$ – Steve Jun 1 '19 at 23:20
  • $\begingroup$ 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? $\endgroup$ – candied_orange Jun 2 '19 at 3:30
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    $\begingroup$ 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. $\endgroup$ – Jonathon Reinhart Jun 2 '19 at 4:31
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    $\begingroup$ 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? $\endgroup$ – user541686 Jun 2 '19 at 9:35
  • $\begingroup$ @JonathonReinhart not quite right; see added material in my answer. $\endgroup$ – Andrew Steane Jun 2 '19 at 15:14

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.

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    $\begingroup$ 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? $\endgroup$ – Joshua Benabou Jun 1 '19 at 20:16
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    $\begingroup$ third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements? $\endgroup$ – Joshua Benabou Jun 1 '19 at 20:19
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    $\begingroup$ 3) 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 en.wikipedia.org/wiki/Hidden-variable_theory $\endgroup$ – Hilmar Jun 1 '19 at 22:11
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    $\begingroup$ @Hilmar Bell's theorem does not show that entangled particles can communicate instantaneously, if by 'communicate' you mean 'communicate information'. No-communication theorems make it clear that the failure of local realism does not imply instantaneous communication of information. I think your answer should be edited to choose a different word than 'communicate'. $\endgroup$ – gabe Jun 1 '19 at 22:54
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    $\begingroup$ @Hilmar, are you aware of the "hydrodynamic quantum analogs"? A number of quantum mechanical phenomena have now been explained with conventional physics (and perhaps more importantly, with conventional physical philosophy, and not Bohr's Copenhagen nonsense). $\endgroup$ – Steve Jun 1 '19 at 23:28

A very good question. As I see it, the amazement regarding entanglement is really a consequence of the Copenhagen interpretation of quantum mechanics (which is rather absurd in my opinion, but let's not get into that now) and how we usually talk about it. This is how pretty much every physicist talks about and teaches quantum mechanics, so let's not think too hard about it an look at what entanglement means in this interpretation.

I should probably clarify that I am talking about non-relativistic quantum mechanics here. Many things become a lot less clear once you transition to QFT and, as far as I know, there is no clear ontology in QFT at all. I should also add that I will only ever talk about measuring the locations of particles because 1. I want to avoid the word "state" in order to not confuse some property of the particle we are trying to measure with the state of the wave function (before or after the measurement) and 2. because location (i.e. a point in the phase space) really is the only thing we can measure, other properties like "the spin of a particle" are inferred from this measurement using the theory.

So, in the Copenhagen interpretation particles have no location unless we measure it: The measurement itself makes the wave function collapse and the location of the particle is created in that process. When two particles $A$ and $B$ are entangled, measuring the location of particle $A$ will also collapse the wave function of particle $B$, no matter how far away from $A$ (and thus the measurement) it is. This is surprising because it is a non-local effect and it occurs without any direct interaction with particle $B$. The collapse of $B$'s wave function is thought of as a real event because it could have collapsed into a different location if measured before the measurement of $A$.

Now, while this is what I think makes most people marvel at entanglement, there are some obvious problems here. Classical concepts are carelessly applied to the problem. In reality, there is no "wave function of $B$" in the theory, the is just The Wave Function. A collapse is always a highly non-local event and we can never measure just a single particle, because particles don't exist; we can only preform measurements on the whole system. All the magic of entanglement arises from the magic the measurement process is imbued with by the Copenhagen interpretation.

In summary, there really is nothing particularly surprising about entanglement at all. It is just an expression of the fact that the state of the wave function along one axis is not symmetrical along some other axis (i.e. the possible combinations of locations of some coordinates have been linked by some process to exclude certain combinations).

I hope this has not become all too convoluted. It is not so easy to explain the problems of this particular thought inside the Copenhagen interpretation without touching on the more fundamental issues with this interpretation. Please let me know what needs clarification.

  • $\begingroup$ Using entanglement, one can find the factors of an integer of $n$ digits by an algorithm (Shor's algorithm) whose operation time scales as $n^3$ whereas existing classical computer algorithms require $\exp(n^{1/3})$ (approximately). This suggests entanglement is computationally powerful. It is certainly remarkable and it can also be used for various other tasks, such as transmitting a general state by teleportation. Nothing surprising? $\endgroup$ – Andrew Steane Jun 2 '19 at 18:18
  • $\begingroup$ Well, no. From the perspective of classical physics, sure. But the fact that a different physical theory enables us to build different machines that are capable of tasks that could not be solved in an older theory should not be surprising in and of itself. Inside the theory of quantum mechanics, an entangled state is not a particularly special or surprising state. It is only when we (vainly) try to understand it inside a different theory that it becomes surprising. $\endgroup$ – schtandard Jun 3 '19 at 9:10
  • $\begingroup$ Perhaps, but the question here was precisely about what is different about QM and why isn't it just ang. mom. conservation etc. Also, the 'different theory' here is not just classical mechanics, but every model falling in the category called local realism. I am content to carry on feeling surprised unless and until we get a better grasp of this, which will probably require a better understanding of spacetime (i.e. quantum gravity or something like it). $\endgroup$ – Andrew Steane Jun 3 '19 at 9:26

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