Spin difference between entangled and non-entangled

[Nugatory]

What a surprise. But that doesn't sound right, how did anyone come up with the idea that classically those silver atoms would just bunch up around the middle as if there isn't any external magnetic field at all?

You are misunderstanding that picture. The "classical prediction" doesn't show the atoms bunching up in the middle as if there were no external magnetic field, it shows the atoms spreading out in the direction of the field as some of them are deflected more than others.

That's what classical E&M predicts, and it's how larger charged rotating objects behave in an inhomogeneous magnetic field.

I think proper classical modeling would show bunching up and down just as is measured. I think both up and down silver atoms would actually end up with their magnetic north pole aligned vertically downwards, and so whether they will go up or down would depend more on their initial position and direction when entering the external magnetic field than on their original magnetic dipole orientation. Wouldn't it?

No. The classical analysis says that when the particles first enter the field they are subject to very different forces according to the direction of their initial magnetic moments. That causes them to spread out initially. Even if they eventually align themselves with the field, by then they're already spread out.

 

[Alien8]

You are misunderstanding that picture. The "classical prediction" doesn't show the atoms bunching up in the middle as if there were no external magnetic field, it shows the atoms spreading out in the direction of the field as some of them are deflected more than others.

Ok, but that's not much better. Would you agree how much the two bunches separate depends on particles velocity?

That's what classical E&M predicts, and it's how larger charged rotating objects behave in an inhomogeneous magnetic field.

Why do you say "charged rotating objects" instead of "permanent magnets"?

I just did an experiment myself. I have a bunch of little spherical permanent magnets about 5mm in diameter, which I let fall between two bigger (2cm) cylindrical magnets, and each of them ended up sticked to one of the two magnets. I see only two variables can influence what magnet they will stick to, initial position where from I let go of them and their magnetic vector orientation. I think initial position matters more, but it's hard to tell because who knows how quickly little magnetic balls can rotate in their free fall to align with the external magnetic field. It looks like rather complex situation to calculate, actually.

No. The classical analysis says that when the particles first enter the field they are subject to very different forces according to the direction of their initial magnetic moments. That causes them to spread out initially. Even if they eventually align themselves with the field, by then they're already spread out.

Yes, spread out, but how much is what makes the difference. So for example, what does it take for a little ball magnet to pass between two magnets straight through without being deflected towards either of them? It seems kind of impossible to me, given slow enough speed or strong enough magnetic fields.

 

[DrChinese]

... So that model predicts the red graph.

You are right that the graph is for spin-1/2, I actually didn't even look at the scale.

The issue is that the graph is a readout of a DIFFERENCE between 2 measurement settings. So first you must say whether your model is intended to be rotationally invariant. The graph is for such models. Yours is if the original spin vector S is randomly oriented across some series of trials. So Alice and Bob obviously won't know that orientation.

Let's assume Alice and Bob are both set at 0 degrees and there is no classical interaction related to their settings. The red graph predicts anti-correlation. But that doesn't occur in those cases in which S is oriented at 90 degrees. Alice's overlap produces a 50-50 outcome for +1 and -1, and Bob's overlap produces a 50-50 outcome for -1 and +1. So in those cases, there is NO correlation at all. At other angles, there is varying anti-correlation. When you integrate across all possible S, you get correlation varying from -.5 to +.5 - which is NOT the red line. (And perhaps I am not following your model correctly at this point, not entirely sure.)

The only way to get the red line is if ALL possible outcomes (for each angle setting) are pre-determined and fixed prior to measurement. There can be no probability relating to an interaction with Alice or Bob. So it might look something like the following:

S oriented at 17 degrees (changes from pair to pair):

A@17 degrees, B@17 degrees: + -
A@18 degrees, B@18 degrees: + -
A@19 degrees, B@19 degrees: - +
A@20 degrees, B@20 degrees: + -
A@21 degrees, B@21 degrees: + -
...
A@105 degrees, B@105 degrees: - +
A@106 degrees, B@106 degrees: + -
A@107 degrees, B@107 degrees: - +
A@108 degrees, B@108 degrees: + -
A@109 degrees, B@109 degrees: - +
...
A@194 degrees, B@194 degrees: - +
A@195 degrees, B@195 degrees: + -
A@196 degrees, B@196 degrees: - +
A@197 degrees, B@197 degrees: - +

This allows Alice and Bob to always get the same results at the same settings. Of course, what is above is a full blown local hidden variables model and if that is the effect of the Bloch sphere model, then I would agree with you.

 

[DrChinese]

And in case I didn't make clear in my post #33, the purpose of my variations in the hidden variables outcomes was to reproduce (as closely as possible) the cos(theta) function that also shows up for known orientation of S. Obviously, the Bloch sphere model doesn't look anything like that.

 

[Nugatory]

Some responses below, but I also have to point out that you're asking fewer questions and arguing more. PhysicsForums is here to help people understand established science, not to argue its correctness. So far your questions indicate that you understand very little of either the classical or the quantum mechanical physics involved in an S-G experiment; we can help with that, but not if you're going to argue.

Would you agree how much the two bunches separate depends on particles velocity?

yes, as well as whole bunch of other things: the gradient of the inhomogeneous magnetic field, the time that the particles spend in it, their mass, the strength of their magnetic moment, probably some other stuff that I've overlooked.

Why do you say "charged rotating objects" instead of "permanent magnets"?

A habit of being precise... We know that there isn't really a little permanent magnet embedded inside the particle so I don't talk as if there is.

I just did an experiment myself. I have a bunch of little spherical permanent magnets about 5mm in diameter, which I let fall between two bigger (2cm) cylindrical magnets, and each of them ended up sticked to one of the two magnets.

Your initial velocity is too small and the gradient of your magnetic field is too weak (both by many orders of magnitude) to produce a measurable classical S-G effect.

 

[billschnieder]

I see only two variables can influence what magnet they will stick to, initial position where from I let go of them and their magnetic vector orientation. I think initial position matters more, but it's hard to tell because who knows how quickly little magnetic balls can rotate in their free fall to align with the external magnetic field. It looks like rather complex situation to calculate, actually.

Yes, spread out, but how much is what makes the difference. So for example, what does it take for a little ball magnet to pass between two magnets straight through without being deflected towards either of them? It seems kind of impossible to me, given slow enough speed or strong enough magnetic fields.

It seems to me if the reorientation happens very fast at the moment they enter the field, then there will be very little spread of the two bunches, since you pretty much end up with only two possible orientations for the majority of the flight through the field, even classically.

 

[Alien8]

Some responses below, but I also have to point out that you're asking fewer questions and arguing more. PhysicsForums is here to help people understand established science, not to argue its correctness. So far your questions indicate that you understand very little of either the classical or the quantum mechanical physics involved in an S-G experiment; we can help with that, but not if you're going to argue.

I knew S-G magnets were part of 1/2 spin entanglement experiments, I didn't know about quantization thing. I'm just talking, expressing my point of view according to what I currently know, so it can be corrected or expanded upon by kind people who know better. Perhaps you see it as argument because I'm trying to be concise. It's just questions really, I have lots of questions.

yes, as well as whole bunch of other things: the gradient of the inhomogeneous magnetic field, the time that the particles spend in it, their mass, the strength of their magnetic moment, probably some other stuff that I've overlooked.

I could only guess. It would be very interesting to see actual calculation that leads to the conclusion those silver atoms would bunch up around the middle instead of to separate away from it. If you know of some link where I can read about it please let me know.

 

[Alien8]

It seems to me if the reorientation happens very fast at the moment they enter the field, then there will be very little spread of the two bunches, since you pretty much end up with only two possible orientations for the majority of the flight through the field, even classically.

You say that as if perfect magnetic alignment would make the force towards up equal the force towards down. Even theoretically if the little ball magnet was going right in between the two big magnets we still have Earnshaw's theorem which I think says there would be no equilibrium configuration for any inverse-square law forces. And then, as soon as it goes astray a little, it gets pulled more where it leaned to, and so more and more. Isn't that how it works? By the way, do you think these little ball-magnets of mine actually move in a spiral fashion until they align their magnetic vectors with the external field?

 

[stevendaryl]

You are right that the graph is for spin-1/2, I actually didn't even look at the scale.

The issue is that the graph is a readout of a DIFFERENCE between 2 measurement settings. So first you must say whether your model is intended to be rotationally invariant. The graph is for such models. Yours is if the original spin vector S is randomly oriented across some series of trials. So Alice and Bob obviously won't know that orientation.

Let's assume Alice and Bob are both set at 0 degrees and there is no classical interaction related to their settings. The red graph predicts anti-correlation. But that doesn't occur in those cases in which S is oriented at 90 degrees.

If \vec{S} is chosen randomly to be in any direction, then the probability that the angle between \vec{S} and \vec{a} is EXACTLY 90° is zero. Sets of measure zero are irrelevant in computing correlations.

My rule is that if \vec{S} makes an angle of less than 90° relative to Alice's orientation \vec{a}, then Alice gets +1. Otherwise, she gets -1. For Bob, it's the opposite: if \vec{S} makes an angle of less than 90° relative to Bob's orientation \vec{b}, then Bob gets -1. Otherwise, he gets +1.

So if Alice and Bob's orientations are the same, then either Alice gets +1 and Bob gets -1, or Alice gets -1 and Bob gets +1. So the product A(\vec{a}, \vec{S}) B(\vec{b}, \vec{S}) = -1, no matter what \vec{S} is. (The exception being the set of measure zero where \vec{S} makes an angle of exactly 90° relative to \vec{a}.

If Alice and Bob's orientations are in opposite directions, then Alice and Bob will always get the same result, regardless of the value of \vec{S} (again, except on a set of measure 0), so the product A(\vec{a}, \vec{S}) B(\vec{b}, \vec{S}) = +1, no matter what \vec{S} is.

If Alice and Bob's orientations are at 90°, then there are 4 possibilities, all of which are equally likely:
(1) A(\vec{a}, \vec{S}) = +1, B(\vec{a}, \vec{S}) = +1
(2) A(\vec{a}, \vec{S}) = +1, B(\vec{a}, \vec{S}) = -1
(3) A(\vec{a}, \vec{S}) = -1, B(\vec{a}, \vec{S}) = +1
(4) A(\vec{a}, \vec{S}) = -1, B(\vec{a}, \vec{S}) = -1

The correlation in that case is 0.

 

[stevendaryl]

A few more details about the linear model.

The proposed rule for Alice's outcome A(\vec{a}, \vec{S}), where \vec{a} is Alice's orientation, and \vec{S} is the hidden variable (a spin vector), is this:

A(\vec{a}, \vec{S}) = sign(\vec{a}\cdot\vec{S})

where sign(x) is +1 or -1 depending on whether x is positive or negative.

Bob's outcome B(\vec{b},\vec{S}), where \vec{b} is Bob's chosen orientation, is the opposite:

B(\vec{a}, \vec{S}) = sign(\vec{b}\cdot\vec{S})

Now, we can characterize a vector \vec{S} by two numbers: \theta = the angle between \vec{S} and the projection of \vec{S} onto the planet containing \vec{a} and \vec{b}, and \phi, the angle between the projection of \vec{S} onto that plane and the vector \phi. To compute A(\vec{a},\vec{S}) and B(\vec{b},\vec{S}), only \phi is relevant. (There are a few cases for which \theta and/or \phi is undefined, but let's ignore those, since they are a set of measure zero.)

Let \alpha be the angle between \vec{a} and \vec{b}. There are two cases to consider:

Case 1: 0 < \alpha< \pi/2

Case 2: pi > \alpha> \pi/2

As shown in the figure, in Case 1, there are 4 regions of interest:

1. \alpha - \pi/2 < \phi < \pi/2. In this region, A(\vec{a},\vec{S}) = +1 and B(\vec{b},\vec{S}) = -1
2. \pi/2 < \phi < \alpha + \pi/2. In this region, A(\vec{a},\vec{S}) = -1 and B(\vec{b},\vec{S}) = -1
3. \alpha + \pi/2 < \phi < 3\pi/2. (Note: 3\pi/2 is the same angle as -\pi/2). In this region, A(\vec{a},\vec{S}) = -1 and B(\vec{b},\vec{S}) = +1
4. - \pi/2 < \phi < \alpha -\pi/2. In this region, A(\vec{a},\vec{S}) = +1 and B(\vec{b},\vec{S}) = +1

In regions 1 and 3, A(\vec{a},\vec{S})B(\vec{b},\vec{S}) = -1
In regions 2 and 4, A(\vec{a},\vec{S})B(\vec{b},\vec{S}) = +1

If \phi is chosen randomly, then the fraction of time that it will be in regions 1 or 3 is given by: P_{1,3} = \dfrac{2(\pi - \alpha)}{2\pi} = 1 - \dfrac{\alpha}{\pi}

The fraction of time that \phi will be in regions 2 or 4 is given by: P_{2,4} = \dfrac{2\alpha}{2\pi} = \dfrac{\alpha}{\pi}

So the correlation E(\vec{a}, \vec{b}) is -1 \cdot (1 - \dfrac{\alpha}{\pi}) + 1 \cdot \dfrac{\alpha}{\pi} = -1 + \dfrac{2\alpha}{\pi}

The case with \alpha > \pi/2 can be figured out analogously, but I'm too tired to do it.

 

[DrChinese]

A few more details about the linear model...

...but I'm too tired to do it.

OK, I see your angle on it. No disagreement.

It doesn't pass the sniff test on the usual S-G stats for a known S, I was thinking we wanted something reasonable on that too. But what you present does match the red line.

 

[Alien8]

Tracked down how's classical prediction supposed to work out:
http://www.toutestquantique.fr/#magnetisme

So it is assumed magnetic dipole orientation would somehow stay fixed along the whole journey through the external magnetic field. That's not what I see when I experiment with my magnets, the first thing they seem to want to do is to rotate in alignment with an external field. Based on what physics would anyone expect the little magnet on B and C image would not flip its south (white) pole upwards towards the external north (blue) pole? Also, if the magnet on image A started at a bit lower position, would it not get attracted downwards throughout its whole trajectory and end up below the center green line?

 

[DrChinese]

...That's not what I see when I experiment with my magnets, the first thing they seem to want to do is to rotate in alignment with an external field. Based on what physics would anyone expect the little magnet on B and C image would not flip its south (white) pole upwards towards the external north (blue) pole? Also, if the magnet on image A started at a bit lower position, would it not get attracted downwards throughout its whole trajectory and end up below the center green line?

Your magnet is a large system. It is completely classical.

Quantum systems won't behave like that at all. And when you ask about classical predictions for quantum systems, you really are asking about something historical.

When you pass a quantum particle through an S-G device, you get a spin measurement. When you pass it through a second device oriented at a different angle, you get another spin measurement. It does NOT act like a little magnet at all.

And as said before, the stats are completely different. A "little magnet" (per your example, which is not directly comparable) would orient itself closest to the first measurement device all of the time. A particle showing its spin would orient itself closest to the first measurement device cos(theta) of the time.

 

[Nugatory]

So it is assumed magnetic dipole orientation would somehow stay fixed along the whole journey through the external magnetic field. That's not what I see when I experiment with my magnets, the first thing they seem to want to do is to rotate in alignment with an external field.

This is because, as I said earlier, both the velocity of your magnets and the gradient of your magnetic field are many orders of magnitude too small to produce the effect that you're looking for.

Also, if the magnet on image A started at a bit lower position, would it not get attracted downwards throughout its whole trajectory and end up below the center green line?

You've missed something important here - this an inhomogenous magnetic field and the gradient, which is what matters, points in same direction on both sides of the center green line. So the direction of deflection for a given magnetic moment is the same.

 

[Alien8]

You've missed something important here - this an inhomogenous magnetic field and the gradient, which is what matters, points in same direction on both sides of the center green line. So the direction of deflection for a given magnetic moment is the same.

I don't think I've missed anything. This is what I'm talking about:

http://link.springer.com/article/10.1007/s10701-009-9338-1
...This study reveals a mechanism which modifies continuously the orientation of the magnetic dipole of the atom in a very short time interval, at the entrance of the magnetic field region.

That was not easy to find, so I guess it's either not well known or not popular for some reason. I thought it was obvious.

 

[DrChinese]

Alien8,

The subject of this thread is about entangled particle spin. What further questions do you have about that?

The reference you provided in post #45 is not really a suitable reference for discussion of the S-G mechanism, as it concludes contrary to generally accepted scientific opinion. It is certainly not suitable for discussion in this thread. As you are relatively new here, you may not be fully familiar with posting guidelines:

https://www.physicsforums.com/showthread.php?t=414380

Thanks.

 

[Nugatory]

That was not easy to find, so I guess it's either not well known or not popular for some reason.

It's not popular because it's not generally accepted. Here's one response.

Neither article, however, is relevant to the mistake you're making: you're assuming a magnetic field that points towards the bottom magnet near the bottom magnet, points towards the top magnet near the top magnet, and switches directions somewhere in the middle. That's not an S-G experiment: You need a magnetic field that points in the same direction throughout the region that the beam is moving, but changes strength - and by a lot.

 

[Alien8]

Alien8,

The subject of this thread is about entangled particle spin. What further questions do you have about that?

The OP question is about entangled vs non-entangled, as in quantum vs classical, in order to understand where and how classical physics fails to explain EPR observations in 1/2 spin experiments. To my great surprise it turned out it fails before it even begins, with a single S-G analyzer, so we were unable to compare any further, but it was a necessary detour because it is a big news to me.

Back to inequalities then. I can find decent descriptions of how experiments with photons are performed, what is measured, what is calculated, and so on, but for 1/2 spin experiments I'm not sure anymore if I really know how it's supposed to go. I thought measurements were taken only along two axis orthogonal to initial trajectory and not with any arbitrary angles.

Can you explain what this means:

http://en.wikipedia.org/wiki/Bell's_theorem

Original Bell's inequality... This inequality is however restricted in its application to the rather special case in which the outcomes on both sides of the experiment are always exactly anticorrelated whenever the analysers are parallel.

 

[Nugatory]

I can find decent descriptions of how experiments with photons are performed, what is measured, what is calculated, and so on, but for 1/2 spin experiments I'm not sure anymore if I really know how it's supposed to go. I thought measurements were taken only along two axis orthogonal to initial trajectory and not with any arbitrary angles.

In principle, it's pretty much the same thing except that you're measuring deflected up versus deflected down when a particle encounters a Stern-Gerlach device set at some angle, instead of a absorbed versus not absorbed when a photon encounters a polarizer set at some angle. The only major difference is that with photon polarization perfect anti-correlation happens when the angle between the detectors on the two side is 90 degrees and with entangled spin 1/2 particles it happens at 180 degrees; this just means that where you see a $\theta$ in the polarization formulas, you'll often see a $\theta/2$ in the corresponding formula for the spin 1/2 case.

In practice, it is easier and less expensive to produce entangled photon pairs than entangled particle pairs so you see experiments done with photon pairs more often.

Can you explain what this means:

That's saying that the original form of Bell's inequality is used to analyze experiments in which you have a choice of the same three angles on both sides. In a particle-spin experiment they might be 0, 60, and 120 degrees; for a photon polarization experiment we'd use 0, 30, and 60 degrees.

The wikipedia article goes on to describe the CHSH inequality, of which the Bell three-angle inequality is a special case. The CHSH inequality is used to analyze experiments in which you have a choice of two angles on one side and two angles on the other. For example, the Weihs experiment was done with polarized photons and angles of 0 and 45 degrees on one side, 22.5 and 67.5 on the other side.

 

[DrChinese]

1. The OP question is about entangled vs non-entangled, as in quantum vs classical, ...

2. Can you explain what this means:

1. That is not an association usually made. Classical mechanics had no entanglement, true, but EPR thought quantum entanglement might lead to an extension of QM - as opposed to a step backward to classical ideas.

2. The "special case" is the one where there are perfect correlations. Some hidden variable theories CAN explain that particular case, so there is no Bell inequality for that.

 

[Alien8]

Neither article, however, is relevant to the mistake you're making: you're assuming a magnetic field that points towards the bottom magnet near the bottom magnet, points towards the top magnet near the top magnet, and switches directions somewhere in the middle. That's not an S-G experiment: You need a magnetic field that points in the same direction throughout the region that the beam is moving, but changes strength - and by a lot.

Magnetic field lines between two opposite magnetic poles go in the same direction, like this:

Magnetic forces acting on a third magnet in between is what has two directions. If you hold a compass needle anywhere in between the two magnets it will always flip its poles in the same direction of the field lines, but if you let it go the force deciding which magnet it will stick to depends on its relative distance. If the magnets are equal the equilibrium line goes right in between. Second magnet in S-G experiment has a function to explain classical physics failure, but QM explanation should really work with one magnet only.

 

[DrChinese]

Really, your magnet analogy is not suitable for discussion of spin 1/2 systems. Here is an actual Bell test of entangled Beryllium ions. Don't be confused by the use of photons for detection, it us actually quite similar to the ideas you are touching on:

http://www.nature.com/nature/journal/v409/n6822/full/409791a0.html

By the way, the lead author (Wineland) won the Nobel for this and other groundbreaking work.

 

[Nugatory]

Magnetic field lines between two opposite magnetic poles go in the same direction, like this:
.

I'm sorry, I didn't explain that properly. The key is the inhomogeneous magnetic field with a high gradient; the force it creates acts to separate the particles in the beam before they've had time to rotate and align. And as we've already said, with your dropped magnets the gradient is far too small and the speed is far too low to produce the SG effect that you're looking for.

 

[billschnieder]

I'm sorry, I didn't explain that properly. The key is the inhomogeneous magnetic field with a high gradient; the force it creates acts to separate the particles in the beam before they've had time to rotate and align. And as we've already said, with your dropped magnets the gradient is far too small and the speed is far too low to produce the SG effect that you're looking for.

Are you saying with a large enough gradient and a fast enough speed, he will observe the SG effect for his little 5mm magnets?

 

[Alien8]

Are you saying with a large enough gradient and a fast enough speed, he will observe the SG effect for his little 5mm magnets?

For either theory increasing speed should have the same effect as using weaker magnets or moving them further apart, it should narrow down the separation.

High gradient I think refers to the gradient difference between the two magnetic fields. In terms of forces on a magnetic dipole, one external magnet gives us one-sided slope , but with two magnets we get two opposite slopes and a hill top line in between. You now try to roll a bunch of bowling balls along this hilltop and see if the result will manifest "spatial quantization". The steeper the slopes, or slower the speed, the more they will go astray.

But wait, what QM needs two slopes/magnets for? QM says all the silver atoms would be either spin up or spin down and will stay spin up or down throughout the whole interaction with the magnetic fields. So one magnet should be sufficient, say the top one - if it's spin up atom it gets attracted to the top and if spin down it gets repulsed to the bottom. Right?

The only theory that actually requires two magnets in order to produce separation would be the one where all the silver atoms align with the external magnetic field like a compass needle. It then becomes attraction/attraction binary system, it's different than attraction/repulsion with only one magnet and fixed spin, gives the same result but for different reasons.

The strangest thing however is that we are talking about the simplest straight forward case scenario of a magnetic dipole moving through a magnetic field. We have computers, and for some reason we are still unable to simulate this?

 

[miosim]

The question Alient8 is asking bother me too, but I am a layman and cannot fully appreciate this exchange. Instead I would like to ask few simple questions to better understand entanglement phenomenon.

Say we have a source of randomly polarized entangled photons flying in the opposite direction toward identically oriented polarizer placed at each end. Is the pair of photons that passes their corresponding polarizers are still entangled or they lost entanglement after interaction with polarizers?

 

[Nugatory]

Say we have a source of randomly polarized entangled photons flying in the opposite direction toward identically oriented polarizer placed at each end. Is the pair of photons that passes their corresponding polarizers are still entangled or they lost entanglement after interaction with polarizers?

The entanglement between the particles disappears when they have passed their polarizers.

That's one of the things that makes the problem so difficult and interesting - you only get one measurement on each member of the entangled pair, and those two measurements are all that you're ever allowed to know about the pair.

 

[miosim]

That's one of the things that makes the problem so difficult and interesting - you only get one measurement on each member of the entangled pair, and those two measurements are all that you're ever allowed to know about the pair.

Thank you Nugatory,

Few more loosely connected questions:

1. Does the result of Aspect experiment contradict with the Malus' law?

2. When we are talking about faster than light connection between entangled states, what kind of speed we are talking about; just a fraction of speed of light faster, many times faster, or instantaneous?

3. Apparently, the entanglement is one of the main attribute of the quantum mechanics. Was this phenomenon studied directly by MAINSTREAM science or only in connection with the Bell's theorem that, as I understand, was for many years on the fringes of science.

 

[Nugatory]

1. Does the result of Aspect experiment contradict with the Malus' law?

No. This and similar experiments are done with individual photons, while Malus's law describes the intensity of the classical electromagnetic waves, when there are a very large number of photons.

2. When we are talking about faster than light connection between entangled states, what kind of speed we are talking about; just a fraction of speed of light faster, many times faster, or instantaneous?

Experiments cannot ever demonstrate "instantaneous", but the can establish a lower bound on the speed of any such connection. If I recall correctly, that lower bound is at least 10,000 times $c$. (If someone else quotes a different number, chances are they're right and I'm wrong because I'm doing this from memory)

3. Apparently, the entanglement is one of the main attribute of the quantum mechanics. Was this phenomenon studied directly by MAINSTREAM science or only in connection with the Bell's theorem that, as I understand, was for many years on the fringes of science.

Your understanding is mistaken.
Entanglement was recognized as a basic consequence of quantum mechanics very early on, and experiments showing that entanglement happened as predicted by QM were being done in the 1920s. EPR was published in 1935 or thereabouts. The subject languished between then and the 1960s because no one could imagine an experiment that would test the EPR hypothesis; the two-angle experiments were the best we had, and they confirmed entanglement but did not exclude local hidden variables. Thus, it wasn't fringe; it would be more accurate to say it was so mainstream that it was boring. Bell's great inspiration was to see that three-angle measurements would permit an experimental test of these ideas - indeed that is the most important point in his original paper - and that gave the experimentalists something to work with.

 

[Alien8]

1. Does the result of Aspect experiment contradict with the Malus' law?

That's actually exactly what my question would be if instead of 1/2 spin experiments and S-G magnets I was talking about photons and polarizers. Good question.

2. When we are talking about faster than light connection between entangled states, what kind of speed we are talking about; just a fraction of speed of light faster, many times faster, or instantaneous?

Experiments are performed with Alice far enough away from Bob, so if entangled pairs are affecting each other over such distance it would have to be faster than light, hence "non-local".