Spin difference between entangled and non-entangled

[Alien8]

What exactly entangled-spin pairs do in Stern–Gerlach experiment that ordinary particles pairs with opposite magnetic dipole moments do not?

http://upload.wikimedia.org/wikipedia/en/thumb/e/e2/Bell.svg/600px-Bell.svg.png

What equation represents the blue line and what equation if for the red line?

 

[atyy]

The experiment is a test of the Bell inequalities. An explanation is given in http://arxiv.org/abs/quant-ph/0205171. That article plots a version of the different curves in Fig. 4, with the equations for the entangled and unentangled cases given by Eq. 10 and 19 respectively.

The Bell inequalities assume that the experimental results are causally explained by local hidden variables and local measurement settings. The causal relationship assumed by the Bell inequalities is diagrammed in Fig. 19 of http://arxiv.org/abs/1208.4119. However, the predictions of quantum mechanics are incompatible with the causal relationship of Fig. 19. In other words, quantum mechanics predicts that the Bell inequalities will be violated by experiments. Experimental results thus far are consistent with the predictions of quantum mechanics. Some alternative causal relationships are diagrammed in Fig. 25, 26 and 27, and are known as non-locality, superdeterminism and retrocausation. Because the non-locality alternative is usually considered the most natural for a scientific theory that can be used by human beings to make predictions, it is often said that quantum mechanics is nonlocal.

 

[Alien8]

The experiment is a test of the Bell inequalities.

The first question is about particle spin and magnetic dipole moment. If particle A has a magnetic dipole moment vector mA=(1,1,1) and particle B has mB(-1,-1,-1), then if particle A is measured along x,y, or z axis it will be north up, and if particle B is measured along -x,-y, or -z axis it will also be north up, or spin up, that is they will be correlated because they were created correlated. So how are two ordinary particles with opposite magnetic dipole moments different than two quantum entangled particles with opposite spins?

An explanation is given in http://arxiv.org/abs/quant-ph/0205171. That article plots a version of the different curves in Fig. 4, with the equations for the entangled and unentangled cases given by Eq. 10 and 19 respectively.

Eq. 10:
P_{VV}(\alpha,\beta) = 1/2 * cos^2(\beta - \alpha)

What quantum mechanics principle or equation is this prediction related to (derived from)?

 

[Nugatory]

What exactly entangled-spin pairs do in Stern–Gerlach experiment that ordinary particles pairs with opposite magnetic dipole moments do not?

There is a subtle problem with the way you've asked this question. A quantum mechanical property like spin has no definite value until is measured, so when you say "a particle pair with opposite magnetic dipole moments" you're really saying "a pair of particles that we just measured on the same axis and found one to be up and the other to be down". Unless and until we've made that measurement, there's no such thing as an "ordinary particle pair with opposite magnetic dipole moments".

The difference between an entangled pair and a non-entangled pair is that we know that if we measure them both on the same axis, we will always get opposite results with an entangled pair but we won't always get opposite results with a non-entangled pair.

It's also important to remember that after we've measured them, there is no difference whatsoever between the pairs that had been entangled and unentangled pairs that just happened to give opposite results. After the measurement, they're all just what you said: "ordinary pairs with opposite magnetic dipole moments". But that's after the measurement; before we make the measurement none of them have definite spins.

 

[Nugatory]

What quantum mechanics principle or equation is this prediction related to (derived from)?

In quantum mechanics, the result of measuring any observable will be an eigenvalue of the operator corresponding to that observable. This formula is derived from the relationship between the operators corresponding to measurements of the spin along different axes.

I understand that this is a completely unhelpful answer... But until you have a basic familiarity with the math behind QM there is no way of answering your question, and the only way of acquiring that familiarity is to spend some serious quality time with a decent textbook.

 

[Alien8]

There is a subtle problem with the way you've asked this question. A quantum mechanical property like spin has no definite value until is measured, so when you say "a particle pair with opposite magnetic dipole moments" you're really saying "a pair of particles that we just measured on the same axis and found one to be up and the other to be down". Unless and until we've made that measurement, there's no such thing as an "ordinary particle pair with opposite magnetic dipole moments".

I understand QM interpretation. I don't understand what is supposed to be classical interpretation and how are they different.

There's no such thing as an "ordinary particle pair with opposite magnetic dipole moments"? Is there some reason we can not make neutrons orient their magnetic moments in a specific desired direction and then send them off to be measured?

The difference between an entangled pair and a non-entangled pair is that we know that if we measure them both on the same axis, we will always get opposite results with an entangled pair but we won't always get opposite results with a non-entangled pair.

If we send today 100 neutrons to Alice each of which with definite magnetic moment orientation (Xi,Yi,Zi) that we select at random, and if tomorrow we send 100 neutrons to Bob each of which with definite magnetic moment orientation (-Xj,-Yj,-Zj), then how are we not supposed to always get opposite result for i=j pairs if we measure them along the same axis?

 

[Alien8]

In quantum mechanics, the result of measuring any observable will be an eigenvalue of the operator corresponding to that observable. This formula is derived from the relationship between the operators corresponding to measurements of the spin along different axes.

I understand that this is a completely unhelpful answer... But until you have a basic familiarity with the math behind QM there is no way of answering your question, and the only way of acquiring that familiarity is to spend some serious quality time with a decent textbook.

I'm actually looking for a historical answer. When did "entanglement" become a part of QM, what is the actual "entanglement" equation and what experiment was it originally inferred from?

 

[Nugatory]

I'm actually looking for a historical answer. When did "entanglement" become a part of QM, what is the actual "entanglement" equation and what experiment was it originally inferred from?

It's been there pretty much from the beginning, but it took a while (decades!) for the implications to be fully appreciated. If you're looking for a historical answer and don't feel like grinding through a few years of differential equations and linear algebra on the way then I recommend "The Age of Entanglement" by Louisa Gilder.

The idealized entanglement thought experiment is a single particle with zero spin that decays into two particles flying in opposite directions; conservation of angular momentum requires that if we measure the spins of the two daughter particles on the same axis the sum must also be zero. The fundamental equation of QM, the Schrodinger equation, applies to the entire system, so as far as the math of QM is concerned, we have a single system with zero net angular momentum before and after the decay. Entanglement (this detector registered spin-up; therefore the entire system is in a state such that the other detector will register spin-down if we perform that measurement) follows from there.

 

[stevendaryl]

From Wikipedia on the subject:

The counterintuitive predictions of quantum mechanics about strongly correlated systems were first discussed by Albert Einstein in 1935, in a joint paper with Boris Podolsky and Nathan Rosen.[1] In this study, they formulated the EPR paradox (Einstein, Podolsky, Rosen paradox), a thought experiment that attempted to show that quantum mechanical theory was incomplete. They wrote: "We are thus forced to conclude that the quantum-mechanical description of physical reality given by wave functions is not complete."[1]

However, they did not coin the word entanglement, nor did they generalize the special properties of the state they considered. Following the EPR paper, Erwin Schrödinger wrote a letter (in German) to Einstein in which he used the word Verschränkung (translated by himself as entanglement) "to describe the correlations between two particles that interact and then separate, as in the EPR experiment."[15] He shortly thereafter published a seminal paper defining and discussing the notion, and terming it "entanglement." In the paper he recognized the importance of the concept, and stated:[2] "I would not call [entanglement] one but rather the characteristic trait of quantum mechanics, the one that enforces its entire departure from classical lines of thought."

 

[Nugatory]

Is there some reason we can not make neutrons orient their magnetic moments in a specific desired direction and then send them off to be measured?

We can. We take a beam of neutrons and pass them through a S-G apparatus oriented in a partcular direction, and then if we only want the spin-up direction we only use the neutrons in the part of the beam that was deflected upwards. But that's a measurement; the things did not have any orientation before then (and that is actually an experimentally verifiable fact - google for "Bell's Theorem" and "Alain Aspect").

There is no way of producing a beam of neutrons that is known to be in a particular spin orientation without somehow "processing" them to get them into the desired state. The neat thing about entangled particles is that no matter what orientation we process one member of the pair into, we know that a measurement of the other member will produce the opposite result; in effect we're always "processing" the entire pair.

Let's say I want to produce two beams of neutrons, both spin-up, but going in opposite directions. So I set up two S-G devices, one on each side of my neutron source, orient them both vertically, and then block off the downwards-deflected beam from both. If my neutron source is generating entangled pairs, then I will never (except by random chance when multiple pairs arrive at the SG devices at the same time) find a neutron passing through the left-hand device at the same moment that a neutron passes through the right-hand device. Many neutrons will make it through the left-hand SG device, and many neutrons will make it through the right-hand device, but for each pair, only one member will make it through.

 

[Alien8]

From Wikipedia on the subject:

If the concept of "entanglement" didn't exist before Einstein and co. invented it for their EPR paper, then what were they referring to, what is it they based their premise on? In other words, how could there be QM prediction if there was no prior QM theory which to base that prediction on?Now, looking at Wikipedia I think this might be the answer:

According to quantum mechanics, under some conditions, a pair of quantum systems may be described by a single wave function, which encodes the probabilities of the outcomes of experiments that may be performed on the two systems, whether jointly or individually. At the time the EPR article discussed below was written, it was known from experiments that the outcome of an experiment sometimes cannot be uniquely predicted. An example of such indeterminacy can be seen when a beam of light is incident on a half-silvered mirror. One half of the beam will reflect, the other will pass. If the intensity of the beam is reduced until only one photon is in transit at any time, whether that photon will reflect or transmit cannot be predicted quantum mechanically.
http://en.wikipedia.org/wiki/EPR_paradox

A beam of light incident on a half-silvered mirror has 50% probability to pass through, so we may describe those photons with the same wave function. That seem to be the original empirical root I'm asking about, which gave birth to the whole "entanglement" concept. It seems like "entanglement" is the same thing as "equal probability". But this doesn't quite explain if these photons are supposed to be in entangled state before or after they interact with the mirror.

 

[DrChinese]

What exactly entangled-spin pairs do in Stern–Gerlach experiment that ordinary particles pairs with opposite magnetic dipole moments do not?

http://upload.wikimedia.org/wikipedia/en/thumb/e/e2/Bell.svg/600px-Bell.svg.png

What equation represents the blue line and what equation if for the red line?

You have already received some answers which may or may not directly address you questions. First, there has never really been a classical theory which matches the red line. The red line represents the CLOSEST any classical theory could every come to the QM prediction for entangled particle pairs, shown in blue. The red represents a boundary condition, in other words.

Second, you asked about the difference between entangled particles and particle pairs which are have opposite spins. The difference can easily be seen in their spin statistics. The first has "entangled state" statistics in which measurements will be completely anti-correlated at ANY angle chosen. The second has "product state" statistics in which measurements will be variably anti-correlated at ANY angle chosen. The predictions are different, and experiment matches predictions on both.

Historically, EPR (1935) did not know much about entanglement other than a few key basics. I don't believe experimental versions appeared until much later. Even theoretical treatments didn't go far until the 1950's (Bohm comes to mind). There was so much going on in QM during this period that it took a while to put all the pieces together on entanglement. Bell's Theorem probably another big turning point, although that too took years to fully sink in.

But the essential point is the there is a conservation factor and there is an indistinguishability factor. Classical particles are distinguishable in all respects even when they obey conservation as a pair. This "detail" makes all the difference.

I would skip the reference to half-silvered mirrors when considering entanglement, as that is more of an example of single particle superposition. Entangled particles are in a superposition, true enough.

 

[atyy]

I'm not sure whether this is the historical route, but a simple way to get entanglement from elementary quantum mechanics is to consider a system of two partcles, eg. the electrons in a helium atom.

For two distinguishable particles, entanglement means that the wave function cannot be written as a product state.

For one particle, an arbitrary wave function $\psi(x)$ can be written as a superposition of basis states:

$\psi(x) = \sum\limits_{n} \phi_{n}(x)$.

For two particles, an arbitrary wave function $\psi(x,y)$ can be written as a superposition of basis states:

$\psi(x,y) = \sum\limits_{n,m} \phi_{n}(x)\phi_{m}(y)$.

So entanglement arises from the fact that for the two particles basis states are built of products of the one particle basis, and that the general state is formed by a superposition of basis states.

 

[atyy]

In Pauli's Nobel lecture http://www.nobelprize.org/nobel_prizes/physics/laureates/1945/pauli-lecture.html he mentions that already in 1926, Heisenberg wrote papers about wave function symmetrization for identical particles. That requires the idea that the two-particle Hilbert space is the tensor product of the one particle Hilbert spaces. So the placement of entanglement within the quantum formalism goes back to within a year or two of quantum mechanics proper. Of course, many consequences were only worked out later. I should also say that in the previous post, I only talked about entanglement for distinguishable particles. For identical particles, one needs a different definition. There seem nowdays to be many refinements in the classification of entanglement, depending on what operations one is interested in.

 

[Alien8]

Let's say I want to produce two beams of neutrons, both spin-up, but going in opposite directions. So I set up two S-G devices, one on each side of my neutron source, orient them both vertically, and then block off the downwards-deflected beam from both. If my neutron source is generating entangled pairs, then I will never (except by random chance when multiple pairs arrive at the SG devices at the same time) find a neutron passing through the left-hand device at the same moment that a neutron passes through the right-hand device. Many neutrons will make it through the left-hand SG device, and many neutrons will make it through the right-hand device, but for each pair, only one member will make it through.

That's how it works, but I'm asking how classical physics fails to explain it. What is spin for QM in classical mechanics is fully 3-dimensional magnetic vector field which can have arbitrary orientation and simultaneous definite magnitude component along x,y, and z axis.

According to that this is how I suppose classical theory should work, please point out the step where it goes astray:

1. We can send neutrons A to Alice with precisely defined, randomly chosen, magnetic moment orientation in 3D space, which can be represented with 3D vectors (x,y,z)

2. The next day for each neutron A we can send neutron B to Bob with the opposite magnetic moment orientation, which can be represented with 3D vector (-x,-y,-z)

3. Therefore if we measure both corresponding neutrons along the same axis we will always measure the opposite spin simply because they were sent with that particular orientation to begin with

 

[atyy]

A very good introduction to the difference between classical and quantum mechanics regarding entanglement is given by http://arxiv.org/abs/1303.3081. The key is the derivation of the Bell inequalities. Under a number of assumptions, one can show that no classical strategy of pre-established agreement (such as a process that always prepares both spins in the same direction) can violate the Bell inequalities. Yet quantum mechanics predicts the violation of the Bell inequalities.

Two features of quantum mechanics that are not present in classical mechanics that enable it to violate the Bell inequalities are non-commuting observables, and entanglement.

 

[stevendaryl]

I don't know whether anyone conclusively answered the question as to what the red and blue curves represent in the figure.

The blue curve is the quantum prediction for correlation between measurements of the two particles in a spin-1/2 twin pair EPR experiment. If Alice measures the spin of one particle along axis a and Bob measures the spin of the other particle along axis b, then the correlation E(a,b) is the average value, over many trials, of AB where A is \pm 1, depending on whether Alice measures spin-up or spin-down, and B is \pm 1, depending on whether Bob measures spin-up or spin-down.

The quantum prediction is: E(a,b) = -cos(\theta) where \theta is the angle between a and b. So it's -1 at \theta = 0 and 0 at \theta = 90^o

So what's the red line? It's a little bit misleading to call that the classical prediction. It's the prediction of a particular classical model. There are lots of different possible classical models, and they make different predictions. But the specific model that that's a graph for is one describe by Bell in his discussion of EPR. It has the nice feature that E(a,b) = -1 when \theta = 0 and E(a,b) = 0 at \theta = 90^o and E(a,b) = +1 at \theta = 180^o, just like the quantum predictions. This classical model is the following:

Assume that when a twin pair is produced, there is, attached to the particles, an associated spin vector \vec{S} pointing in a random direction.

Alice then chooses an axis \vec{a} to measure the spin relative to. She gets A= +1 if the angle between \vec{S} and \vec{a} is less than 90°. She gets A = -1 if the angle is more than 90°.

Bob chooses an axis \vec{b}, and gets B = \mp 1 depending on whether the angle between \vec{b} and \vec{S} is less than or more than 90°. (The opposite of the rule for Alice.)
​

With this classical rule, if \vec{a} = \vec{b}, (relative angle \theta = 0), they will always get opposite results, for a correlation of -1. If \vec{a} = -\vec{b} (relative angle \theta = 180^o), they will always get the same result, for a correlation of +1. If \vec{a} and \vec{b} are at right angles (relative angle \theta = 90^o), they will get the same result 50% of the time, and opposite results 50% of the time, for a correlation of 0.

So this classical model makes the same predictions as QM for the cases of relative angles of 0^o, 90^o, 180^o, but makes different predictions for other angles.

 

[stevendaryl]

That's how it works, but I'm asking how classical physics fails to explain it. What is spin for QM in classical mechanics is fully 3-dimensional magnetic vector field which can have arbitrary orientation and simultaneous definite magnitude component along x,y, and z axis.

According to that this is how I suppose classical theory should work, please point out the step where it goes astray:

1. We can send neutrons A to Alice with precisely defined, randomly chosen, magnetic moment orientation in 3D space, which can be represented with 3D vectors (x,y,z)

2. The next day for each neutron A we can send neutron B to Bob with the opposite magnetic moment orientation, which can be represented with 3D vector (-x,-y,-z)

3. Therefore if we measure both corresponding neutrons along the same axis we will always measure the opposite spin simply because they were sent with that particular orientation to begin with

The way that spin measurements work is that you are only able to measure one component of the spin. Spin is a vector \vec{S} (or can be thought of as a vector) with 3 components, but you can only measure the spin relative to a chosen axis \vec{a}. And that answer always gives the answer +1/2 or the answer -1/2. That fact by itself shows that quantum spin is very different from a classical vector. If a classical vector \vec{S} is pointing in the x-direction, and I measure its component in the y-direction, I'll get zero. Quantum-mechanically, you never get zero for a spin-1/2 particle, you always get \pm 1/2 (in units of \hbar).

 

[DrChinese]

That's how it works, but I'm asking how classical physics fails to explain it.

...

3. Therefore if we measure both corresponding neutrons along the same axis we will always measure the opposite spin simply because they were sent with that particular orientation to begin with

And as I mentioned in my post, the statistical predictions are quite different between the classical situation you describe and the entangled stats. The classical is a product state. Let's take your example:

I have 2 electrons oriented at x=+ and x=- respectively. If I measure them later at a 90 degree angle to x, I get NO correlation.

On the other hand, entangled electrons would show perfect anti-correlation at that measurement angle.

 

[Alien8]

I'm not sure whether this is the historical route, but a simple way to get entanglement from elementary quantum mechanics is to consider a system of two partcles, eg. the electrons in a helium atom.

For two distinguishable particles, entanglement means that the wave function cannot be written as a product state.

For one particle, an arbitrary wave function $\psi(x)$ can be written as a superposition of basis states:

$\psi(x) = \sum\limits_{n} \phi_{n}(x)$.

For two particles, an arbitrary wave function $\psi(x,y)$ can be written as a superposition of basis states:

$\psi(x,y) = \sum\limits_{n,m} \phi_{n}(x)\phi_{m}(y)$.

So entanglement arises from the fact that for the two particles basis states are built of products of the one particle basis, and that the general state is formed by a superposition of basis states.

That's it. So let's put it in perspective. For example take ordinary unpolarized sunlight incident on some polarizer. Half of the photons will pass through, so each of them may be represented with the same wave (probability) function. Does that mean they were all entangled before, or only those that went through are entangled now, or neither?

The thing is that's all still only about one light beam interacting with only one polarizer, where does it say this can be applied to two separate polarizers?

 

[DrChinese]

So what's the red line? It's a little bit misleading to call that the classical prediction. It's the prediction of a particular classical model...

Actually, I have never seen a model put forth that reproduces the red line. I think I could program one. But it would have the obvious issue that it could not reproduce Malus for single photons.

 

[DrChinese]

Does that mean they were all entangled before, or only those that went through are entangled now, or neither?

Neither.

"For two distinguishable particles, entanglement means that the wave function cannot be written as a product state. "

This may be a correct statement, but it can be a bit confusing too. There is a known system value, but you cannot distinguish the particles on the basis they are entangled on. The problem is in the meaning of the words - the equations usually dispense with that.

 

[atyy]

That's it. So let's put it in perspective. For example take ordinary unpolarized sunlight incident on some polarizer. Half of the photons will pass through, so each of them may be represented with the same wave (probability) function. Does that mean they were all entangled before, or only those that went through are entangled now, or neither?

The thing is that's all still only about one light beam interacting with only one polarizer, where does it say this can be applied to two separate polarizers?

I am not exactly sure of the answer here. So let me give my best guess. In ordinary sunlight, the photons are not entangled. So extremely naively I can treat the "wave function" of N photons in sunlight as a product state ψ(x₁, x₂, ... x_N) = ø(x₁)ø(x₂)...ø(x_N). Now, of course this is wrong, because if I am a little less naive I will say photons are identical bosons so the "wave function" must be symmetrized, and cannot be written as a product state. And indeed the definition that an entangled state is one that cannot be written as a product state is one that only applies to distinguishable particles. There is the further problem, that my naive treatment of photons used a photon wave function with position coordinates. Because the photon is inherently relativistic, the treatment requires quantum field theory. So to treat entanglement in photons perfectly correctly requires understanding two issues (1) entanglement for identical particles, and (2) quantum field theory.

So, I wish to stress that my definition of entanglement as a state that cannot be written as a product state only applies to (A) entanglement for non-identical particles, and (B) non-relativistic quantum mechanics. One problem with this definition is that although the state is not a product state for a particular choice of basis, how do we know that there is no other basis in which the state is a product state? We need a tool that will quantify entanglement in a basis independent way. One such tool is the entanglement entropy, which is the von Neumann entropy of the reduced density operator corresponding to one subsystem in an entangled pair.

Regarding the entanglement of identical particles, the discussions in these papers seem very reasonable to me. But I believe the most useful concepts are still being researched.
http://arxiv.org/abs/1009.4147
http://arxiv.org/abs/1302.3509
http://arxiv.org/abs/quant-ph/0206135

 

[Alien8]

The way that spin measurements work is that you are only able to measure one component of the spin. Spin is a vector \vec{S} (or can be thought of as a vector) with 3 components, but you can only measure the spin relative to a chosen axis \vec{a}. And that answer always gives the answer +1/2 or the answer -1/2. That fact by itself shows that quantum spin is very different from a classical vector.

Spin looks more like 1-dimensional projection of 3D vector. Do you mean to say spin vector is not directly proportional to magnetic dipole vector?

If a classical vector \vec{S} is pointing in the x-direction, and I measure its component in the y-direction, I'll get zero. Quantum-mechanically, you never get zero for a spin-1/2 particle, you always get \pm 1/2 (in units of \hbar).

That's like expecting to throw a permanent bar magnet between two other magnets without it ending up attracted to one of the two sides. Wouldn't the first thing magnetic dipole moment want to do is to flip its north-south vector in alignment with the north-south vector of the magnetic field it is moving through?

 

[DrChinese]

Wouldn't the first thing magnetic dipole moment want to do is to flip its north-south vector in alignment with the north-south vector of the magnetic field it is moving through?

OK, so what is the probability that a North becomes North again when the new field is oriented ø degrees from the first? Cos(ø), right?

So 2 classical spins would yield a prediction which is a product of 2 such functions. That's gives product state statistics. That is a different prediction that what you get with 2 entangled particles.

 

[DrChinese]

In other words...

As long as you look at very particular examples, you may not see any particular difference between entangled particles vs. particles which are not entangled. But when you venture past those special cases, the differences are very clear.

 

[Nugatory]

According to that this is how I suppose classical theory should work, please point out the step where it goes astray:

1. We can send neutrons A to Alice with precisely defined, randomly chosen, magnetic moment orientation in 3D space, which can be represented with 3D vectors (x,y,z)

2. The next day for each neutron A we can send neutron B to Bob with the opposite magnetic moment orientation, which can be represented with 3D vector (-x,-y,-z)

If I prepare a bunch of neutrons in the spin-up state along a particular axis, and prepare another bunch of neutrons in the spin-down state along that axis (let's say zero degrees from vertical, just to be definite) , and then pair them up as you're describing here, then...

3. Therefore if we measure both corresponding neutrons along the same axis we will always measure the opposite spin simply because they were sent with that particular orientation to begin with

Yes, if we measure them on that particular axis we will always get opposite results, just as we would with entangled pairs that were not prepared in this fashion. But suppose that instead they set their detectors to some other angle? Say for example, 45 degrees from vertical?

With your stream of known-up and known-down particles, they will sometimes get the same result for both particles; we know this from experiments sending particles of known (by previous measurement) spin one one axis into S-G devices aligned on a different axis. With entangled pairs, they will always get opposite results: one particle spin-up on the 45-degree axis and the other particle spin-down on the that axis, and never both up or both down. Furthermore, that will be true for any angle that they choose.

Thus, there is only one axis on which your stream of known-up/known-down particles will behave like a stream of entangled pairs.

 

[stevendaryl]

Actually, I have never seen a model put forth that reproduces the red line. I think I could program one. But it would have the obvious issue that it could not reproduce Malus for single photons.

I think the graphs are for the spin-1/2 case, rather than the spin-1 (photon) case. The correlation for photons goes to zero at 45°, rather than 90°.

As I said, the straight line graph is the correlation you would get if the measurement process works like this:

Assume each particle has an associated spin vector \vec{S}. Equivalently, (since the magnitude is unimportant), we can view the particle as having an associated point on the unit sphere. In twin-pair experiments, the two particles have opposite spin vectors.

When Alice chooses an orientation for her measuring device, \vec{a}, she is equivalently choosing a point on the unit sphere. Associated with here point is a hemisphere of points on the unit sphere--the set of points making an angle of less than 90° with respect to \vec{a}. If \vec{S} lies in this hemisphere, she gets +1. Otherwise, she gets -1.

Similarly, when Bob chooses an orientation \vec{b}, he is equivalently choosing a hemisphere of directions. If his particle falls in this hemisphere, he gets +1, otherwise, he gets -1.

Since Alice and Bob are getting particles with opposite spin-vectors, if \vec{S} falls in the intersection of their two hemisphere's, then Alice will get +1 and Bob will get -1. If -\vec{S} is in the intersection, then Alice will get -1 and Bob will get +1. So the fraction of time that they get opposite results is proportional to the intersection of their two hemispheres. A geometric result that I had to look up is that the area of the intersection of the hemisphere centered on \vec{a} with the hemisphere centered on \vec{b} is 2 (\pi - \theta), where \theta is the angle between \vec{a} and \vec{b} (in radians). So the fraction of times that a randomly chosen point \vec{S} on the sphere will lie in the intersection is \dfrac{2 (\pi - \theta)}{4 \pi} = 1/2 - \dfrac{\theta}{2\pi}. The probability that -\vec{S} lies in the intersection is the same. So the probability that Alice and Bob get opposite results is 1 - \dfrac{\theta}{\pi}. The probability that they get the same result is \dfrac{\theta}{\pi}. So the correlation is

E(a,b) = (Probability of same result) - (Probability of opposite results) = \dfrac{2 \theta}{\pi} - 1​

So this model predicts a correlation that is -1 when \theta=0, rises linearly to zero at \theta = \pi/2, and continues to +1 at \theta = \pi

So that model predicts the red graph.

 

[stevendaryl]

That's like expecting to throw a permanent bar magnet between two other magnets without it ending up attracted to one of the two sides. Wouldn't the first thing magnetic dipole moment want to do is to flip its north-south vector in alignment with the north-south vector of the magnetic field it is moving through?

That's an example of a classical local hidden-variable model. To flesh it out a little, assume that the electron has a "true" spin axis \vec{S}, and that the measurement device has an orientation \vec{a}. If the angle between \vec{S} and \vec{a} is small (less than 90°), then electron's axis flips to point in the direction \vec{a}. If the angle between \vec{S} and \vec{a} is large (greater than 90°), then the electron's axis flips to point in the direction that is opposite from \vec{a}.

But now what does your model predict if the electron goes through a second measurement? So you first measure it along an axis \vec{a_1} and then measure it along an axis \vec{a_2}? If the "flipping" is deterministic, then measuring +1/2 for the first measurement should imply +1/2 for the second measurement, as well, if the angle between \vec{a_1} and \vec{a_2} is less than 90°. But that is not the way actual electrons work: After getting spin-up along direction \vec{a_1}, the result of measuring the spin along direction \vec{a_2} is non-deterministic: You get spin-up with probability cos^2(\theta/2) and spin-down with probability sin^2(\theta/2), where \theta is the angle between \vec{a_1} and \vec{a_2}.

You can keep coming up with models all day, and they won't match all the predictions of quantum mechanics. We know that because Bell proved that no classical hidden-variables model matches all the predictions of quantum mechanics.

 

[Alien8]

Yes, if we measure them on that particular axis we will always get opposite results, just as we would with entangled pairs that were not prepared in this fashion. But suppose that instead they set their detectors to some other angle? Say for example, 45 degrees from vertical?

I didn't take it seriously enough, but it was suggested previously that officially classical physics already fails with the simplest Stern–Gerlach experiment and only one analyzer. If there is no classical answer for that, then there is no point going any further.

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?

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? This is completely different than what I originally thought, but I don't see anything else would classically even make any sense at this point.

 

[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.