Spin difference between entangled and non-entangled

[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".

 

[Alien8]

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.

Are you saying Malus' law starts to deviate from cos^2 with lower light intensities?

 

[Nugatory]

Are you saying Malus' law starts to deviate from cos^2 with lower light intensities?

No. I am saying, as others have said about 48,786 times already in this thread, that photons are not classical electromagnetic waves so there is no particular reason to expect them to act according to an empirical observation about classical electromagnetic waves.

Light obeys Malus's law, photons don't, and there is no contradiction here.

 

[Alien8]

Light obeys Malus's law, photons don't, and there is no contradiction here.

Photons are quanta of light like H2O molecules are quanta of water. Light ought to do what photons do just like water does what H2O molecules do. It's the same phenomena, one macroscopic and the other microscopic description of the same thing.

Malus' law works out light intensity percentage. Intensity is energy per unit time per unit area, so given the same energy (frequency) of each photon QM translates this directly to amount of photons. I don't know what is intensity supposed to be in terms of classical physics, but "number of EM waves" sounds kind of awkward. It seems to me it's fair to say Malus' law, and "light intensity" in general, is much closer to QM than is to classical physics.

 

[Nugatory]

Photons are quanta of light like H2O molecules are quanta of water. Light ought to do what photons do just like water does what H2O molecules do. It's the same phenomena, one macroscopic and the other microscopic description of the same thing.

You'll see that analogy occasionally in the pop-sci press, but as with all such analogies it is terribly misleading if you take it too literally. You will have to put it out of your mind and replace it with something better before you will be able to understand QM as deeply as you clearly want to.

Give Feynmann's "QED: The strange theory of light and matter" a try; it also uses analogies, but at least they're good ones.

 

[Nugatory]

I don't know what is intensity supposed to be in terms of classical physics...

It's the amplitude of the wave, which is to say the magnitude of the electrical and magnetic fields at their peaks. If the electrical field of a given wave at the point $x$ and time $t$ is $sin(kt-vx)$, then $2sin(kx-vt)$ is a wave with the same frequency and wavelength but greater intensity.

It would be easy to think that "there are more photons in the second wave", but that would be a mistake. The second wave is more likely to deliver more energy when it interacts with matter (no surprise, as the fields involved are stronger). This energy will appear in fixed size amounts at single points within the area exposed to the radiation, and when this happens we say "a photon hit there". We would be better off saying and thinking "a photon appeared there".

 

[miosim]

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.

It seems that the large number of photons should be identical to result of the large statistical sample of individual photons.
The article in wiki (see below) talks about Malus' Law and Bell theorem, but I am not sure if this article conforms or rejects this link.

http://en.wikipedia.org/wiki/Local_hidden_variable_theory

"Optical models deviating from Malus' Law
If we make realistic (wave-based) assumptions regarding the behavior of light on encountering polarisers and photodetectors, we find that we are not compelled to accept that the probability of detection will reflect Malus' Law exactly."

However this isn't my main concern. My biggest problem with the Bell theorem is that I don't understand the origin of the red line: how the unlimited scenarios of possible complex hidden variables is transformed into two simple strait lines. Unfortunately I wasn't able to find the comprehensive for a layman description of this transformation.
http://upload.wikimedia.org/wikipedia/en/thumb/e/e2/Bell.svg/600px-Bell.svg.png

 

[Nugatory]

My biggest problem with the Bell theorem is that I don't understand the origin of the red line: how the unlimited scenarios of possible complex hidden variables is transformed into two simple straight lines. Unfortunately I wasn't able to find the comprehensive for a layman description of this transformation.

If you haven't already found DrChinese's web page, give it a try: http://www.drchinese.com/Bells_Theorem.htm

 

[DrChinese]

Miosim,

http://www.drchinese.com/David/Bell_Theorem_Negative_Probabilities.htm

A. Figure 3 at the above link is very similar to the graph you presented, but instead for photons. The analogy is as follows: what does QM predict for matches for an angle difference between Alice and Bob's setting (blue line); what realistic prediction comes closest to QM (red line). The Red Line must not lead to logical contradictions such as negative probabilities, probabilities in excess of 100%, violations of CHSH etc.

There is no guarantee that the Red Line is the ONLY possible line to compare to the Blue line. But at least it is "sorta close" to the QM prediction. And you can easily see that it does NOT match the QM expectation except at zero and a few other special cases.

-------

B. In my example at the above link, I choose A=0, B=67.5, C=45 degrees. Keeping A and C constant, and varying B only, gives the following function:

f(B)=(cos^2(B-0)+sin^(45-0)-cos^2(B-45))/2 which corresponds to the sum of the likelihood of 2 particular realistic cases occurring. These 2 cases have NO relevance in QM (they don't exist), and are simply an arbitrary formula otherwise - like the CHSH inequality. But f(B) is negative for any value between 45 degrees and 90 degrees exclusive, which means the QM prediction (if matched by a local realistic theory somehow) would predict a negative likelihood for those cases (which is obviously absurd).

C. On the other hand, if you plugged the red line prediction in instead, you would NOT get negative values. But it would not agree with experiment, because it is different than the QM expectation (because the Blue line and Red line are different).

-------

To summarize: if the Blue line is the LR prediction, you get a variance with experiment. If the Blue line were to be the LR prediction, you get absurd results (negative probabilities). But again, agreeing with what you are saying, the Blue line is only ONE of the "unlimited scenarios of possible complex hidden variables". But every one of those will reduce to either B. or C. eventually, so you always return to the same point.

 

[miosim]

DrChinese,

I read your site and the links you provided, but I am still didn't find the answer I am looking for: correspondence between realistic prediction and the red line. I am looking for an explanation that I can accept but after critical evaluation. However if this explanation requires mathematical background I would accept my limitations and gave up.

Thankscorresponds

 

[stevendaryl]

DrChinese,

I read your site and the links you provided, but I am still didn't find the answer I am looking for: correspondence between realistic prediction and the red line. I am looking for an explanation that I can accept but after critical evaluation. However if this explanation requires mathematical background I would accept my limitations and gave up.

Thanks
corresponds

The red line is not the only possibility for a local realistic theory. It's just that it is the prediction for a very specific locally realistic model that agrees with QM at the points $\theta = 0^o, 90^o, 180^o, 270^o, 360^o$.

As DrChinese pointed out this particular model is not very plausible, for other reasons.

The model is this:

• When an electron/positron pair is produced, there's an associated spin-vector $\vec{S_e}$ associated with the electron, and the opposite spin-vector $\vec{S_p}$ associated with the positron, where $\vec{S_p} = -\vec{S_e}$
• $\vec{S_e}$ for each electron is chosen completely randomly; it is just as likely to point in any direction.
  
• When the spin of a particle with spin-vector $\vec{S}$ is measured relative to an axis $\vec{X}$, the result is +1 if the angle between $\vec{S}$ and $\vec{X}$ is less than $90^o$, and -1 if the angle is more than $90^o$.

That's all there is to the model. It's a deterministic model, in the sense that the outcome for any measurement is a deterministic function of the spin-vector of the particle being measured. If one experimenter, Bob, measures the spin of his particle in direction $\vec{b}$, and a second experimenter, Alice, measures the spin of her particle in direction $\vec{a}$, then:

• If $\vec{a} = \vec{b}$ (so the angle between them is $0^o$), then they always get opposite results, because $\vec{S_e}$ is always opposite $\vec{S_p}$. So the correlation is -1.
  
• If $\vec{b} = -\vec{b}$ (so the angle between them is $180^o$), then they always get the same results, because $\vec{S_e}$ is always opposite $\vec{S_p}$, and so are their measurement orientations. So the correlation is +1.
• Halfway between, when the angle is $90^o$ or $270^o$, there is no correlation at all, so the correlation is 0.

This model fails at other angles, but it works for the 5 easy angles: 0, 90, 180, 270, 360.