I have thought about quantum entanglement for some time, and I still don't quite understand the reasoning behind the conclusion that entangled particles somehow can communicate their state to each other instantaneously, even though they are separated by a substantial distance (e.g., the communication would happen faster than the speed of light). From what I gather, the assumption is that each entangled particle is in a superposition of all possible states, and that upon observation of one of the particles, it immediately (and randomly?) converges to only one of the permissible states. If the second (entangled) particle is observed, at the same or at a different time, the second particle is observed to have the same state as the first particle. Is this interpretation correct?

Based on the first two answers that I have seen, I will add somewhat to my question. Yes, I realize that quantum entanglement seems to violate Einstein's laws, or he wouldn't have called it "spooky action at a distance". And no, I don't want a lesson in tensor calculus as my answer. The jist of my question is: What observational evidence and logic leads physicists to conclude that there is "spooky action at a distance", rather than conclude that the individual particles are somehow correlated with each other?

Round 3: There is approximately a 50% probability that I am about to get to the bottom of the answer that I am looking for, but (in the finest quantum tradition), I'm not sure yet. Anyway, here goes. Based on LENGTHY replies here (for those who went to the effort, thanks), and a bit of background research, I wasn't aware that the quantum state of two entangled particles could not be described independently. I'm not sure why this is, but given this "fact", it makes perfect sense that the physicists from several decades ago didn't like the conclusions that they were drawing.

Is it correct that the quantum states of two entangled quantum particles cannot be independently specified? Or is this a case where I can specify either one of the particles, but to specify them as entangled I necessarily have to specify both together?

Chris Drost - thanks for the clarification. The occasional text that refers to "communication between particles", "spooky action at a distance" (yes, this quote is decades old), etc., now makes a good deal more sense in the context of the mathematical framework that you described.

  • $\begingroup$ possible duplicate of Does entanglement not immediately contradict the theory of special relativity? $\endgroup$ Commented Jul 29, 2015 at 18:47
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    $\begingroup$ No, this interpretation is not correct. There is no such thing as "the state of a particle". The entire two-particle system is in a state which (by definition of entanglement) is not a tensor product of two individual particle states. Perhaps instead of "thining about quantum entanglement for some time" you should try "reading about quantum entanglement for some time". $\endgroup$
    – WillO
    Commented Jul 29, 2015 at 18:51
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    $\begingroup$ David, please do not vandalize posts, even if they are your own. $\endgroup$
    – HDE 226868
    Commented Jul 29, 2015 at 21:40
  • $\begingroup$ "Is it correct that the quantum states of two entangled quantum particles cannot be independently specified?" You can always partially describe either in terms of a state matrix; for entangled systems this yields a state matrix which cannot be written as $|\psi\rangle\langle\psi|$ for any $\psi$, hence it's a "mixed state," not a "pure state". Furthermore when you describe the system with the two matrices you "miss" the strange correlations. On the plus side, mixed states look more classical than pure ones, so you get to trade in wave-particle duality in exchange for the correlations. $\endgroup$
    – CR Drost
    Commented Jul 29, 2015 at 22:17

2 Answers 2


I still don't quite understand the reasoning behind the conclusion that entangled particles somehow can communicate their state to each other instantaneously, even though they are separated by a substantial distance

This isn't correct, they occupy a joint state.

From what I gather [...] upon observation of one of the particles, it immediately (and randomly?) converges to only one of the permissible states.

This also isn't correct. Measurements happen continuously. And they happen by interaction, but since it is a joint state of entangled particles, the interaction in one region changes the joint state of the entangled particles.

To answer your question, I'll first address how to measure a single particle. Then I'll show how to describe a joint state in sufficient detail that you can see how measurements work on a joint system. Then we can discuss the particular entanglement where they spins are setup to be the same.


Let's go over measurement. Let's say a beam is traveling in the positive y direction and the beam has some thickness in the x and z directions and goes through a region of magnetic field where the magnetic field is in the z direction, is inhomogeneous in the z direction, and only is strong for one brief region near $y=0.$ I just described the setup of a Stern-Gerlach device.

So assume you have a setup like that. The result is that an incoming beam might be one beam coming in purely in the y direction with a superposition of two spin states (a state with a positive z component of spin and a different state with a negative z component of spin, i.e. the z component of the spin is positive or negative respectively in those two state). As the beam goes through, the Schrödinger equation (which is how things evolve in time in quantum mechanics) predicts that the beam splits into two beams, one angled up in the z direction and one that angles down in the z direction. But the spin becomes purely up in one of the beams and becomes purely down in the other beam.

A general superposition of spin states for a spin half particle can be visualized by a unit 3d real vector, so the beam can come in with any spin vector, but one outgoing beam has a spin vector that points straight up and the other outgoing beam has a spin vector that points straight down.

So you can imagine visualizing this process by pulling out a piece of paper and having a positive $z$ axis going to the top of the paper and the positive y axis going to the right. Then draw two parallel horizontal lines on the left side that's the $z$ width on the incoming beam. Then near $y=0$ angle the top line upwards and the bottom one downwards and add a sideways V so that the whole thing looks like a Y that fell down to the right. So to the right there are two beams each has a spin vector that points in the $\pm \hat z$ direction.

That's one particle, and measuring in just the z direction. Imagine picking a vertical cross section like $y=-2$ you would see some thickness in the z direction and a spin vector pointing any direction whatsoever.

Then you could pick a vertical cross section like $y=+2$ and you would see some thickness in the z direction and a spin vector pointing like $+\hat z$ then farther down you would see some thickness in the z direction and a spin vector pointing like $-\hat z.$

For slices in between you'd see some overlap as the beam spreads before breaking into two and as the spin vectors starts to polarize along the streamlines that end up in the two branches.

So practice seeing this without using the y axis just imagine a movie where over time a line segment in the z direction gets longer and longer and longer and then separates into two pieces that are then moving away from each other and imagine the spin polarizes itself so that when the pieces are separate they each have a spin in the $\pm \hat z$ direction.

Joint states

Why did we go through so much work? Because the wavefunction of quantum mechanics is not a wave in space when you have more than one particle. $\Psi=\Psi(t,x,y,z)$ only works for one particle. For two particles you have $\Psi=\Psi(t,x_1,y_1,z_1,x_2,y_2,z_2).$

This is probably news to you, the wavefunction of quantum mechanics doesn't assign a complex number to locations in space it assigns complex numbers to configurations of all the particles. So for instance $(t,x_1,y_1,z_1,x_2,y_2,z_2)$ corresponds to the configuration of particles at time $t$ where particle one is at $(x_1,y_1,z_1)$ and particle two is at $(x_2,y_2,z_2).$ So if $\Psi(t,0,0,0,10m,0,0)$ is nonzero, that (very roughly) corresponds to a probability of particle one at the origin and particle two being 10m over in the $x$ direction.

So the wavefunction can describe whole configurations of all the particles.

So now imagine you want to measure two particles. Each can be a beam heading in the positive y direction each with some initial thickness in the x direction and the z direction, just have one beam near $x=0m$ and the other beam be near near $x=10m$ so they are parallel to each other. We are going to ignore the $x$ and $y$ directions because we need to visualize two $z$ directions.

So imagine at $t=0$ that the two beams are jointly a square in the $z_1,z_2$ plane. Let's draw that on our paper with with $z_1$ going right and $z_2$ going up. So left-right tells us $z_1,$ which is the $z$ coordinate of particle one and up-down tells us $z_2,$ which is the z$ $coordinate of particle two. Knowing a point on the piece of paper tells us the $z$ coordinate of both particles, we specify the location of both particles by specifying a point on this sheet of paper.

If we measure just particle one then the square gets longer in the left right direction, becomes a rectangle then develops a vertical line somewhere down the middle and then the two new rectangles start to move left and right away from each other.

If we measure just particle two then the square gets taller in the up down direction, becomes a rectangle then develops a horizontal line somewhere across the middle and then the two new rectangles start to move up and down away from each other.

Where does this line form? It depends on the original spin. If it was all spin up coming in, then it forms on the far edge, i.e. the whole beam just deflects without forming two beams. If the incoming spin is an equally sized superposition of spin up and spin down then the line forms right in the exact middle and equally sized beams go left and right. For spin that is more spin up than spin down the line forms so the two beams have relative sizes related to how much more spin up the incoming beams had compared to spin down. (For experts, the L2 norm of each beam adds up to the L2 norm of the incoming beam with the sizes relative to the L2 size of the projection of the original spin onto the spin eigenstates.)

So now we understand what measurement look like, and we can visualize in real time what happens in a real measurement. This is great (and most people don't bother to learn this skill).


So let's look at an entangled state where the spins are entangled to have the same spin as each other. To do this, it happens to require that the beam splits into two equal beams, that's because for each particle it is a special kind of equal mixture. This is a technicality, it just means that those vertical and horizontal lines that appear will be in the middle for the first measurement.

So what happens if you measure particle one first? Recall that the square gets longer in the left right direction, becomes a rectangle then develops a vertical line somewhere down the middle and then the two new rectangles start to move left and right away from each other. But in this case the line forms right in the middle. And remember that the two separated beams have the spin get polarized? Well, since the spins are entangled both spins become up in one rectangle and both spins become down in the other rectangle. But the rectangles didn't get taller, to the person standing by the other beam it still looks like a regular width beam. And the other person hasn't measured spin so they still don't know the spin and they don't know anything has happened. Now if you measure the second beam, each of those rectangles just starts deflecting up or down because each of them is purely spin up or purely spin down, and that is what a spin measurement does on something that is all spin up (it deflects it up) or all spin down (it deflects it down). So you end up with two squares in the $z_1,z_2$ plane. One square in the upper left, and one square in the lower right corner. The first measurement spread it put in one direction, split it in the middle and then separated the squares. The second measurement second deflected each of the two squares.

What if you measured the spin of particle two first? Then the initial square grows taller in the up down direction becomes a rectangle then develops a horizontal line right across the exact middle and then the two new rectangles start to move up and down away from each other. But then remember that the two separated beams have the spin get polarized? Well, since the spins are entangled both spins become up in one square and both spins become down in the other square. But the rectangle didn't get any wider, so to the person standing by the other beam it still looks like a regular width beam. Now if you measure the first beam, each of those squares just starts deflecting left or right because each of them is purely spin up or purely spin down, and that is what a spin measurement does on something that is all spin up (it deflects it right) or all spin down (it deflects it left). So you end up with two squares in the $z_1,z_2$ plane.

In both cases the squares are in the upper-right corner of the $z_1,z_2$ plane and the lower-left corner of the $z_1,z_2$ plane. This is because it was entangled to have the same spin. If they were entangled to have opposite spins they would have ended up in the upper-left and the lower-right.

What if you measure both beams at the same time? It evolves from one square into two squares in those opposite corners as the spin polarizes to have the spins be correlated.

No one gets information. Each person just sees an initial beam with some thickness and then later sees some beams that have been deflected. You can't see the spin. So you don't know who is doing what first, and you don't know that the results were correlated until you tell each other about your results and you notice that yours when left (spin down) when theirs went down (spin down).

And you always do the same thing. You have a beam with thickness, you separate it into pieces based on how much spin up versus spin down it has, and then thin it and stretch it and break it into two pieces each of which has a totally polarized spin $\pm \hat z.$ Whether the other person did something first doesn't change what happens to you, and isn't detectable anyway.

The two particles just had a joint state and you have to deal with that because that is the way the world is. It is unfortunate that people try to oversimplify. On the other hand, reading through this answer and doing all the things I said is a lot of work, and most people don't care that much. If all you want to do is compute the relative frequencies that you get certain results and certain correlations, less work is involved.

For further details of how the different measurements can't see what is going on elsewhere, consider the case where you measured particle one first. Then a vertical line appeared and two squares separated in the left right direction. What does that look like to the person by the second beam? Nothing. All the changes happened to the $z$ coordinate of particle one, to someone that can only see particle two any motion left-right is undetectable. That whole getting longer and separating and moving left and right was all about the configuration where the only thing changing was the coordinate of the particle you can't see. But what about the spin? That changed, right? But you can't see spin. In fact these separating beams is how you detect spin.

All the second person can see is that they have a beam and that later they can get two beams. They have no idea whether the beam is separated in the direction of some other particle that is far away from them. One person can only see one particle, the one with an $x$ coordinate near them (I know we weren't drawing the $x$ coordinates but that how you know which particle you are measuring, one person was near $x=0m$ the other person was near $x=10m$ and each had just one particle near them).

So one person can only see the up down direction in our $z_1,z_2$ plane and the other person can only see the left right direction in our $z_1,z_2$ plane. So they can't send information to each other.

Which is great, since for other reasons we happen to know that simultaneous is not an objective thing.

edit to respond to edited (entirely new) question

Your new question about spooky-action versus correlation is getting it completely backwards. It is the assumption that there is some background thing that is correlated that leads to the silly idea that the background things have to be spooky. In particular someone might imagine (incorrectly) that there is a secret function from unit vectors $\hat n$ to the set $\{+1,-1\}$ and that this secret function determines which detector goes off if you place a detector in each outgoing beam path. But because of a feature I didn't bring up it turns out that secret function would have to change what it does here based on what happens over there. The idea was brought up to ridicule quantum mechanics by people that thought quantum mechanics was wrong. Later we did the experiments and quantum mechanics was confirmed. So it's actually a bit silly to study spooky action unless you are into science history of the learn-about-predictions-that-were-wrong. And it is the opposite of learning about how quantum mechanics works. So assuming that there were things with correlations leads to those things being spookily correlated (or quantum mechanics being wrong, which we now know is not the case). It is not popular now (and hasn't been for decades).

The reason it is considered a spooky action rather than a mere correlation is because of an entirely different thing. Something I din't need to bring up in the previous answer about how entanglement works to produce the same results. But remember that the spooky action view has not been popular in decades (except in textbooks, which feel they have to mention it for historical reasons, so in textbooks it may never die). Namely, spooky action needs a real effect, which is related to the fact that you can choose to measure in many different directions not just the $z$ direction. These different measurements are complementary.

Complementary Measurements

Experimentally, just rotate your device a quarter turn in the $x-z$ plane to measure in the $x$ component of spin or rotate your device half as much as that to measure the $\frac{\sqrt{2}}{2}\hat x + \frac{\sqrt{2}}{2}\hat z$ component of spin. Or bend it with a magnetic and then put it through the machine so that it measures the $y$ component.

If you did this to both of the particles, nothing is different, the math is the same. But you could measure the $z$ component of one and the $x$ component of the other. If so, you expect a certain correlations (not the 100% correlation we had when both measured the $z$ component). The correlations are explained fine by what I described above, in this case each rectangle becomes two squares and you end up with one in each corner. In general, the horizontal or vertical line is placed based on how much of the two possibilities you had. And the correlations are determined by how often you get each of the four boxes, which is proportional to the sizes (really, the L2 weight) of those boxes. So nothing is weird. Nothing is spooky.

They are complementary because doing one comes at the expense of doing another. If you did two measurements (beam splitting) in the $\pm\hat z$ direction the second one is deflected only one direction. That repeatability is the only (scant) justification for calling it a measurement rather than a polarization. But if you do a beam splitting in the $\pm\hat z$ direction and then do a beam splitting in the $\pm\hat x$ then each beam is split into two.

Clearly these interactions (beam splitting) are not both 100% determined by some unknown X and also not changing that thing X. They are either changing something, or are determined by something else. In reality from my first answer you know that they are determined by a combination of where in the incoming beam you were (upper part goes up, lower part goes down) and the spin vector (to determine where the dividing line is), and then both of those things change as the spin vector becomes polarized and the new outgoing beam(s) gets deflected.

That's why it isn't popular now. But we still carry around the baggage of former years. In fact that is why we call it the "z component" of the spin. Really we are polarizing a beam by making the outgoing beams be polarized to have a spin vector of $\pm \hat n$ for some real unit vector $n.$ We are not measuring a preexisting component of the spin vector, we are splitting a beam.

Spooky action

Some people thought the only take away was "order matters for complementary experiments."

So the idea persisted that maybe there were secret functions that tell you the outcome of the "first" measurement. And that lead some people to think that maybe with entanglement you could sneak two things in as "first".

So if you imagined (incorrectly) that these beam splitting interactions are measuring the component of some unknown secret vector then you can assume (incorrectly) that there are three prexisting numbers (the outcomes of spin measurements in the three independent directions) that are revealed by the "first" beam splitting experiment. And then you get predictions that disagree with what we see. So it is wrong.

That isn't spooky. The spookiness is to assume that there a magician playing three card monty with the three numbers, so they are "real" but the magician is changing them here based on what goes on there. The term was coined by someone that thought the predictions of quantum mechanics on spatially separated particles was wrong. We've done those experiments now, and the predictions of quantum mechanics are confirmed. So what was wrong wasn't quantum mechanics what was wrong was assuming that these beam splitting measurements reveal (and are determined by) prexisting components of some secret set of three numbers.

And I have to be very carefully here, for spin 1/2 there is a spin (bi)vector and the magnitude of it's components do determine how big the two separated beams are, but (1) the spin (bi)vector is known, it is equivalent to the state $\alpha|+\rangle+\beta|-\rangle$ and (2) the spin (bi)vector just determines the sizes of the two split beams, both beams are created. The kind of spin components that the spooky action imagined was an unknown thing that tells you which of the two outgoing beams will set off a particle detector if you placed a particle detector along both outgoing beams.

So studying spooky action is an exercise in learning bad (experimentally disproven) physics made by people that made predictions that were wrong, people bent on trying to designed to make quantum mechanics look bad. They honestly thought quantum mechanics was bad. And they wanted to clear mental space to make room to make new theories. But quantum mechanics wasn't wrong.

Instead you, yes you, could learn how to understand the correct predictions that quantum mechanics makes. For that, you can read my first answer about measurements, how they work, joint states, how they work, and entanglement and how it is about measurements on joint states.

If you do that, you see that the part of the beam that ends up left or right depends on whether it came from the left or right part of the incoming of the beam that was on the left or the right, something that is (1) unknowable, and (2) already the whole story so we don't need magic results that are spookily changed to explain exactly the dynamics involved.

The dynamics is the Schrödinger equation. You can track the evolution of the wavefunction according to the Schrödinger equation for the actual experimental setup (like I did) without appeal to anything else (spooky or otherwise). Yes, it is nonlocal. The wavefcuntion is defined on configuration space, it isn't a wave in space. You don't need anything other than having a function that has configuration space as the domain, and has a spin valued range that evolves according to the Schrödinger equation. If you want to then ask what determined a detector going off in a certain rectangle, you can trace that evolution all the way through the system to the initial wavefunction, and the magnitude of the wave determines the relative frequency of the detector going off.

The Schrödinger equation isn't spooky, so nothing spooky is going on. You can pretend that something spooky is going on if you pretend that there are secret data that says which detector will go off in which rectangle, and then realize that to get the right answers, those secret data (that you can't ever know) must change in a spooky manner. But why would you do that? Is it important that unknown, unknowable things that allegedly determine things that experimentally seem random are also changing behind our back when they were unknown and unknowable anyway.

Is it correct that the quantum states of two entangled quantum particles cannot be independently specified? Or is this a case where I can specify either one of the particles, but to specify them as entangled I necessarily have to specify both together?

I want to say yes to both except I don't understand why there is an or at the beginning of the second one. The literal definition of entangled is that you cannot write the joint state as a product of a separate state for each particle, by definition.


One way to facilitate this discussion is to think of what's classically forbidden but quantumly permissible. My favorite so far is a game that I call Betrayal. Let me explain that in this answer.

Betrayal: Game Rules

The players are a cooperative three-person team, they will either all win or they will all lose.

They will be put through some number $N \gg 1$ of similar challenges; let's say this will be a thousand or so. They are permitted to have rest breaks and to take a few days to complete these challenges, and during the breaks they are permitted to meet again to coordinate their actions. The team must pass all challenges to win the prize, which is some lavish trillions-of-dollars prize that anyone would be happy to divvy-up three ways. So it's a year or two of work but you get handsomely rewarded. (We might allow a smaller $N$ or a couple mistakes to correct for human error and unpredictable gamma rays, but it'll probably still be a pretty strenuous ordeal.)

The challenges have the same general structure but come in 2 general types, with 4 sorts of challenges overall: one "control" challenge and three "traitor" challenges, which only differ in which member of the team is unwittingly being coerced into betraying their friends.

Challenge structure

During a challenge, each person enters one of the rooms. The salient detail about the rooms is that each room contains a view-screen, a button labeled 0, and a button labeled 1. Players may take scientific equipment into the room with them as well if they choose, but their only interaction with the central computer must be the room and their only allowed interaction with the room itself is to read what's on the screen and hit one of the two buttons in the allotted time -- anything else is cheating. The exact details of the rooms are left a little open for further specification; probably some day they will be spaceships travelling 20 light seconds away from a central point; and perhaps they will use one of the partially-homomorphic encryption strategies, whatever.

Once the rooms are prepared, a central computer chooses one of the 4 challenges uniformly at random (control, person 1 is a traitor, person 2 is a traitor, person 3 is a traitor). The computer transmits a certain goal to each of these three screens. The experimenters then have only ten seconds to hit exactly one of the two buttons, which will transmit an unforgeable signal to the central computer. The central computer will carefully validate the timing of the responses, and sum together the appropriate digits. Each challenge is won or lost depending only on whether the sum of the 3 buttons pressed.

Control challenges

The computer transmits to all players, "make the sum even." The team passes the challenge only if the sum is, in fact, even.

Traitor challenges

There are three of these, one for each player. In each one, that player is unwittingly coerced to be a traitor against his/her friends, and those friends, knowing that one of their friends is a traitor (but not who it is), must gracefully recover. Here is how we accomplish this.

The computer transmits to the traitor, "make the sum even." The computer transmits to the other two players, "make the sum odd." The team passes only if the sum is, in fact, odd.

No classical solution

It's not too hard to see that the "easy" way to make the sum even during control rounds is for all of the players to answer 0, and the "odd" analogue is for everyone to answer 1, and that if only 2 people answer 1 while the other answers 0 then the one answering 0 has indeed "betrayed" the others. Here's a simple proof that no classical strategy has a sure solution to all 4 challenges: so in the limit as $N \gg 1$ we know that the success probability goes to 0.

In classical probability, we can fully specify a strategy for any challenge with a joint probability distribution on six random variables, $A_e, A_o, B_e, B_o, C_e, C_o$, corresponding to what our players Alice, Bob, and Carol do when they are asked to make the sum even or odd respectively. In fact for classical probability we can ask for all 6 of these numbers simultaneously up-front, then reveal what questions we were going to ask, and the success rate won't change. Therefore, for a sure solution, you must be able to choose these 6 numbers to simultaneously satisfy 4 conditions:$$\begin{align}A_e + B_e + C_e \equiv 0 & ~~\text{(modulo 2)},\\ A_o + B_o + C_e \equiv 1 & ~~\text{(modulo 2)}, \\ A_o + B_e + C_o \equiv 1 & ~~\text{(modulo 2)}, \\ A_e + B_o + C_o \equiv 1 & ~~\text{(modulo 2)}. \\ \end{align}$$Adding all 4 expressions together trivially gets us $$2~\left(A_e + B_e + C_e + A_o + B_o + C_o\right) \equiv 3 ~~\text{(modulo 2)},$$which is impossible as it describes an even number having the same remainder when divided by 2 as an odd number does. Consequently, at any given time, one of these must not be true and we must select it one quarter of the time, so the success probability has an upper bound of 3/4 and after 1000 trials we find a probability of about 10-125 of success. (We could make the game a little more attractive by requiring only 100 trials but charging $1 to play the game; then with a trillion-dollar prize your expected earnings are only about 30 cents per play, so we make a net profit from probabilistic adversaries while quantum systems can pass with a much higher error tolerance.)

Simple quantum solution

Three players each take a qubit, a quantum bit, into the rooms with them. Quantum mechanics has not just the states $|0\rangle$ and $|1\rangle$ but hybrid states like $|+\rangle = \sqrt{1/2}\left(~|0\rangle + |1\rangle ~\right)$ and $|-\rangle = \sqrt{1/2}\left(~|0\rangle + |1\rangle ~\right)$. The rule here is that the number in front of each of the outcomes is an amplitude which you square (and take the absolute value of, if it's a complex number) to get that outcome's probability.

When I write qubit states next to each other I mean that each person measures the appropriate outcome. For example, the state $|+01\rangle = \sqrt{1/2} \left(~|001\rangle + |101\rangle ~\right)$ has a 50/50 chance of either Alice seeing 0, Bob seeing 0, and Carol seeing 1, or else Alice sees 1, Bob sees 0, and Carol sees 1. Two super-special states of this form are:$$\begin{align} |+++\rangle ~=~ &\sqrt{\frac 18} \left(~ |000\rangle + |001\rangle + |010\rangle + |011\rangle + |100\rangle + |101\rangle + |110\rangle + |111\rangle ~\right), \\ |---\rangle ~=~ &\sqrt{\frac 18} \left(~ |000\rangle - |001\rangle - |010\rangle + |011\rangle - |100\rangle + |101\rangle + |110\rangle - |111\rangle ~\right). \\ \end{align}$$In the latter, for each $|1\rangle$ we pick up a $-$ sign, so an even number of $|1\rangle$ terms gives a $+$ sign and an odd number gives a $-$ sign. We start their qubits in the specially entangled "Greenberger–Horne–Zeilinger state" of$$\begin{align}|\Psi\rangle ~=~ &\sqrt{\frac 12}\left(~|+++\rangle + |---\rangle~\right) \\ ~=~& \sqrt{\frac 14} \left(~|000\rangle + |011\rangle + |101\rangle + |110\rangle ~\right).\end{align}$$

Control challenges

The three quantum-enabled teammates just measure these three bits. In all four cases above, the sum of the bits is an even number.

Traitor challenges

The two players asked to make the sum odd apply the Hadamard-basis controlled-phase-rotation by 90 degrees, which is a complicated unitary transform allowed by quantum mechanics that maps $|+\rangle \mapsto |+\rangle $ and $|-\rangle \mapsto i~|-\rangle $, where $i$ is the imaginary base, the square root of $-1$. This is perfectly well allowed in quantum mechanics but it transforms our state to $$\begin{align}|\Psi'\rangle ~=~ &\sqrt{\frac 12}\left(~|+++\rangle + (i)^2 |---\rangle~\right) \\ ~=~ &\sqrt{\frac 12}\left(~|+++\rangle - |---\rangle~\right) \\ ~=~& \sqrt{\frac 14} \left(~|001\rangle + |010\rangle + |100\rangle + |111\rangle ~\right).\end{align}$$Now the sum of the bits is odd when we measure them. It doesn't matter that the other player, the traitor, does nothing special and thinks they're in the ordinary state.

Features highlighted by this game

Often two entangled qubits have something similar, a "Bell's inequality violation," which also has to be measured with a lot of repeated experiments and can't be reproduced with any classical theorem. Usually in quantum systems there's got to be some freedom to "classically produce" the given output by purely random luck. For some reason, the bounds tend to be a lot "looser" or harder to understand for 2 qubits than for 3 qubits.

Two entangled systems are bizarrely correlated in a way that no classical theory can explain. However, a correlation cannot transfer information: we cannot even in principle identify that the correlation exists until all of the information is brought to one central place for comparison. Until that happens you are still stuck.

There are lots of ways to apparently cheat. For example, the players could try to detect power fluctuations due to a random number generator in the signal emitted by the signal-producer, or they could potentially blast a high-power laser at the central computer, flooding its transducer circuitry and causing it to read a false positive classically. It's just as hard to secure quantum challenges as classical ones.


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