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I am not a physicist so, first of all I like to apologise if I my question is stupid. I just want to know where I am going wrong here.

But if I have understood anything about the entanglement principle; it is that two entangled particles have opposite characteristics that is kept regardless of distance: e.g. if one of them is detected to have a up spin then other is bound to have a down spin. My question is does that hold only when they are detected at the same time, the synchronicity of which must be quite difficult to achieve? Given such small tolerances how can we be sure if both measurements are synchronous?

None the less if that is verified to be correct then I don't see it being that spooky. All we are saying is that when a wave particle is split into two wave particles they remain in sync although mirrored through time (be it a short time). i.e. they are not really communicating with one another through some as yet unknown dimension beyond time and space or something like it.

But if the measurement (detection) doesn't have to be at the same time then it is really spooky because it implies that detecting one part fixes the other part is a definite state which again if I have understood it correctly is contrary to the superposition principle which claims that the state of a particle can not be determined until it is detected. If we can know the state of a qubit before it can is detected then all sorts of things should become possible or at least theoretically, such as instantaneous communication because then we can select the qubits based on their known states.

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  • $\begingroup$ What exactly is your question? -- In any case, the fact that the outcomes are opposite is not the point about entanglement. The same holds for classical randomness. If you want to know what is really special about entanglement, try to read Section 4.2.1 in theory.caltech.edu/~preskill/ph229/notes/chap4_01.pdf (you don't have to be a physicist to understand it!). $\endgroup$ – Norbert Schuch Jul 29 '15 at 14:20
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    $\begingroup$ As a sidenote, please note that the Superposition principle does NOT say that "the state of a particle can not be determined until it is detected". $\endgroup$ – 299792458 Jul 29 '15 at 14:33
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It does seem like you have some misconceptions. You don't have to measure them simultaneously, in fact the whole idea of "being simultaneous" turns out to be subjective and observer dependent in relativity.

But the real issue is that there are many measurements you can do. For instance you could measure the z component of spin, or you could measure the y component of spin, or you could measure the x component of spin. But you can't measure all three at once, even for a single particle.

Let's go over measurement. Let's say a beam is travelling 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. 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 one of the beam.

A general superposition of spin states for a spin half particle can be visualized by a unit 3d vector, so it 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 fells down to the right. So to he right there are two beams each has a spin that points in the $\pm z$ 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 stream lines that end up in the two branches.

So practice seeing this with the y axis just imagine a movie where over time a line segment in the z direction gets 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.

Why did we go to so much work? Because the wavefunction is another 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 at say one beam is near $x=0m$ and the other is 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 it is initially 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.

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 coning in 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 visual 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.

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. 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.

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 rectangle and both spins become down in the other rectangle. But the rectangles didn't get wider, 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 rectangles 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.

Now, you talked about signalling.

Let's say 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 of particle one, to someone that can only see particle one 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 were 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.

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that two entangled particles have opposite characteristics that is kept regardless of distance: e.g. if one of them is detected to have a up spin then other is bound to have a down spin.

Firstly: as your description shows, the relevant correlation is pair by pair. It must be (ideally) unambiguous whether a given detection indication of the one analyzer and a given detection indication of the other analyzer were due to the same one (specific) pair of "entangled particles", or instead due to one "partner" of one (specific) pair and the other "partner" of another (disjoint, independent) pair.

In contrast it is (as good as) not relevant at all whether the two analyzers under consideration were at rest to each other (such that simultaneity or dis-simultaneity of their individual indications might be determined at all); or even that each indication of the one analyzer detecting one particle had been simultaneous to indication of the other analyzer detecting the corresponding "partner" particle of the same one (specific) pair of "entangled particles".

Secondly, your description is still a bit misleading; instead:
if, in the course of a suitably large number of trials (i.e. having analyzed a suitably large number of pairs of "entangled particles") it had been found that in each trial

  • if one particle had been detected by the one analyzer as having had "up spin" then the corresponding "partner" particle of the same pair had been detected by the other analyzer as having had "down spin", or (likewise)

  • if one particle had been detected by the one analyzer as having had "down spin" then the corresponding "partner" particle of the same pair had been detected by the other analyzer as having had "up spin",

then it follows (as the actual result of the measurement) that the "up/down" labelling of the one analyzer and the "up/down" labelling of the other analyzer had been "aligned", in the set of trials under consideration (at least to within an uncertainty of $\Delta \theta \approx \text{ArcSin}[~\frac{1}{N}~]$ for a set of $N$ trials under consideration).

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