Comparison between quantum entanglement and a classical version

[Adel Makram]

What I understood from Quantum Entanglement (QE), is that measuring the spin of one of two entangled particles in one location gives the spin of the other particle in other location no matter how far is the later. What I can also understand is that the same concept is applicable in the classical physics. For example, if we have a way to create two spinning tops at one point in the time and space, they will be spinning at different directions to maintain the law of conservation of angular momentum. Now measuring the direction of one of them in one location ( by just watching it), gives information about the direction of the other spinning top in the other location. So what is the difference between both QE and the classical version?

 

[Mentz114]

In QM the two tops can be in a mixed state before the measurement is made. Thus until one is actually measured neither is known with certainty. If the states are also entangled then measuring either one projects the other into the opposite state (or the same state depending the setup).

I found this Wiki article very illuminating

https://en.wikipedia.org/wiki/Quantum_entanglement

 

[Adel Makram]

In QM the two tops can be in a mixed state before the measurement is made. Thus until one is actually measured neither is known with certainty. If the states are also entangled then measuring either one projects the other into the opposite state (or the same state depending the setup).

I found this Wiki article very illuminating

https://en.wikipedia.org/wiki/Quantum_entanglement

I went quickly through the article and I found that the difference between both versions ( quantum and classical) is firstly, the 2 states are in a mixed state in the QE but they are not in classical one. Secondly, the random outcome of one particle spin measured along a particular axis ( because of quantum mechanics) will maintain the anti correlation by keeping the outcome of the other particle spin in the opposite spin measured along the same axis.
While the first difference between the two versions can be canceled if we assume a zero knowledge of information about the state of spinning tops before the measurement, the second difference is really weird.

 

[bhobba]

It's different as pointed out in a famous paper by Bell:
https://cds.cern.ch/record/142461/files/198009299.pdf

In fact entanglement is the key difference between QM and standard probability theory:
http://arxiv.org/pdf/quant-ph/0101012.pdf
http://arxiv.org/abs/0911.0695

Thanks
Bill

 

[Mentz114]

I went quickly through the article and I found that the difference between both versions ( quantum and classical) is firstly, the 2 states are in a mixed state in the QE but they are not in classical one. Secondly, the random outcome of one particle spin measured along a particular axis ( because of quantum mechanics) will maintain the anti correlation by keeping the outcome of the other particle spin in the opposite spin measured along the same axis.
While the first difference between the two versions can be canceled if we assume a zero knowledge of information about the state of spinning tops before the measurement, the second difference is really weird.

Yes, it is very weird. As bhobba has pointed it is a profound difference.

But QT is non-local so it does not require any additional mind stretching.

 

[DirkMan]

But QT is non-local

To quote Unruh , "exactly in what sense is it non-local?"

 

[Heinera]

To quote Unruh , "exactly in what sense is it non-local?"

In the operational sense that "non-local" means it predicts a violation of Bell's inequality. I guess most use the phrase "non-local" in that sense, without implying anything more specific than that (which would anyway take us to interpretations).

 

[Strilanc]

> So what is the difference between both QE and the classical version?

In quantum entanglement, you can interfere and generally "mix" the various cases to get interesting results. This prevents you from just assuming it was one of the cases beforehand, like you can with classical correlations.

- You can't do superdense coding with classical correlations, but you can with an entangled qubits.
- You can't win the Mermin-Peres game 100% of the time using classically correlated bits, but with entangled qubits you can.
- And of course the CHSH game used in Bell-inequality experiments are also an example of being able to win a game more often by using entangled qubits than you could with correlated bits.

It's difficult to give an intuitive idea of why it works. It really comes down to the way the math behaves, like the ways unitary matrices differ from stochastic matrices. I could never understand what the heck people were talking about until I started from the bottom and built upward; the high-level analogies are just too fragile.

 

[stevendaryl]

Bell's theorem is all about the testable difference between quantum entanglement and any analogous classical correlation.

Classically, if two measured values A and B are obtained far enough apart so that there is no causal influence of one measurement on the other, then any correlations between those values must be due to some unknown facts ("hidden variables") common to both measurements. A classical version of EPR might look like this:

1. You produce a pair of particles. One particle goes to Alice, another goes to Bob.
   
2. Alice picks a direction \vec{a}
   
3. Bob picks a direction \vec{b}
4. Alice measures the spin \vec{s_A} of her particle relative to \vec{a}, and writes A=+1 if \vec{s_A} \cdot \vec{a} > 0. Otherwise, she writes down A=-1
5. Bob measures the spin \vec{s_B} of his particle relative to \vec{b}, and writes B=+1 if \vec{s_B} \cdot \vec{b} > 0. Otherwise, he writes down B=-1
6. Then they compare the results A and B

They find that the results are correlated, in the sense that if \vec{a} = \vec{b}, then A = -B

That's a classical correlation. It is easily explained using "hidden variables". You just assume that \vec{s_A} and \vec{s_B} are fixed at the time of the creation of the pair of particles in such a way that \vec{s_A} = - \vec{s_B}. The pair (\vec{s_A}, -\vec{s_A}) is the "hidden variable".

What Bell showed is that the correlations predicted by quantum mechanics cannot be explained by any such hidden-variable model (unless we allow faster-than-light influences, or back-in-time influences, or some other exotic possibility).

 

[Adel Makram]

Bell's theorem is all about the testable difference between quantum entanglement and any analogous classical correlation.

Classically, if two measured values A and B are obtained far enough apart so that there is no causal influence of one measurement on the other, then any correlations between those values must be due to some unknown facts ("hidden variables") common to both measurements. A classical version of EPR might look like this:

1. You produce a pair of particles. One particle goes to Alice, another goes to Bob.
   
2. Alice picks a direction \vec{a}
   
3. Bob picks a direction \vec{b}
4. Alice measures the spin \vec{s_A} of her particle relative to \vec{a}, and writes A=+1 if \vec{s_A} \cdot \vec{a} > 0. Otherwise, she writes down A=-1
5. Bob measures the spin \vec{s_B} of his particle relative to \vec{b}, and writes B=+1 if \vec{s_B} \cdot \vec{b} > 0. Otherwise, he writes down B=-1
6. Then they compare the results A and B

They find that the results are correlated, in the sense that if \vec{a} = \vec{b}, then A = -B

That's a classical correlation. It is easily explained using "hidden variables". You just assume that \vec{s_A} and \vec{s_B} are fixed at the time of the creation of the pair of particles in such a way that \vec{s_A} = - \vec{s_B}. The pair (\vec{s_A}, -\vec{s_A}) is the "hidden variable".

What Bell showed is that the correlations predicted by quantum mechanics cannot be explained by any such hidden-variable model (unless we allow faster-than-light influences, or back-in-time influences, or some other exotic possibility).

So, the quantum modification of the same experiment could be; instead of labeling a definite + or - for A and B, Alice and Bob define a probability of finding + or - relative to their measurement directions. In addition, corresponding readings will always yield a correlation which means each time Alice writes A=+1 Bob writes B=-1.

If this is the case, then the source of puzzle is not in the characteristic of the particles ( entanglement) but rather in the way of measuring them.

 

[stevendaryl]

So, the quantum modification of the same experiment could be; instead of labeling a definite + or - for A and B, Alice and Bob define a probability of finding + or - relative to their measurement directions. In addition, corresponding readings will always yield a correlation which means each time Alice writes A=+1 Bob writes B=-1.

If this is the case, then the source of puzzle is not in the characteristic of the particles ( entanglement) but rather in the way of measuring them.

The particular predictions of quantum mechanics for spin-1/2 twin-pair version of EPR is this: If Alice and Bob both use spin detectors oriented in x-y plane, and Alice chooses orientation \alpha (relative to the x-axis) and Bob chooses orientation \beta, then they will get the same result (both spin-up or both spin-down) with probability sin^2(\beta - \alpha), and will get opposite results with probability cos^2(\beta - \alpha). The details of what is going on in the measurement process seem irrelevant; the only thing that is relevant is the angles \alpha and \beta.

 

[Nugatory]

So, the quantum modification of the same experiment could be; instead of labeling a definite + or - for A and B, Alice and Bob define a probability of finding + or - relative to their measurement directions. In addition, corresponding readings will always yield a correlation which means each time Alice writes A=+1 Bob writes B=-1.

Try it... You will not be able to construct such a model, and have the probabilities produce correlations that match the quantum mechanical predictions, unless you allow for the probability that Alice gets a given result on her particle to vary with the direction that Bob chooses to measure on - not just the result that Bob gets, but the direction he gets it on.

It sounds like it's possible, and people have spent enormous amounts of time on ever more complicated models that try to avoid this basic weirdness of quantum entanglement... but it cannot be done.

 

[Adel Makram]

and will get opposite results with probability cos^2(\beta - \alpha).

So according to the example of the hidden variable you gave where both Alice and Bob orient their detectors along the same direction, the quantum interpretation of probability of opposite results =cos^2(\beta-\alpha) will be 1 which will be equivalent to the hidden variable theory result. I mean there will no be any sort of weirdness if both observers align their detectors in the same direction? If the answer is no, then the source of general weirdness of QE is in the way both observers align their detectors not in the idea of superposition.

 

[DrChinese]

So according to the example of the hidden variable you gave where both Alice and Bob orient their detectors along the same direction, the quantum interpretation of probability of opposite results =cos^2(\beta-\alpha) will be 1 which will be equivalent to the hidden variable theory result. I mean there will no be any sort of weirdness if both observers align their detectors in the same direction? If the answer is no, then the source of general weirdness of QE is in the way both observers align their detectors not in the idea of superposition.

When the detectors are aligned the same, there is no obvious "weirdness". That much you say is correct. But that does not mean that superposition is not "weirdness", it only means that its "weirdness" does not show up in that example. At most any other relative angle setting, there will be an inequality that violates a classical explanation.

In other words: superpositions of particles that are space-like separated will demonstrate quantum non-locality at all angles. A few of those angles won't seem weird, but others will.

 

[Mentz114]

When the detectors are aligned the same, there is no obvious "weirdness". That much you say is correct. But that does not mean that superposition is not "weirdness", it only means that its "weirdness" does not show up in that example. At most any other relative angle setting, there will be an inequality that violates a classical explanation.

In other words: superpositions of particles that are space-like separated will demonstrate quantum non-locality at all angles. A few of those angles won't seem weird, but others will.

I agree.
It seems that to calculate the quantities in the CHSH inequality one needs two settings (directions) on each detector, as in the Stern-Gerlach experiment. In that case, as you say, some settings give violations. I could never work out how, with only one setting on the detectors one can do more than measure correlations and hope to get $\pm 1$.

 

[stevendaryl]

So according to the example of the hidden variable you gave where both Alice and Bob orient their detectors along the same direction, the quantum interpretation of probability of opposite results =cos^2(\beta-\alpha) will be 1 which will be equivalent to the hidden variable theory result. I mean there will no be any sort of weirdness if both observers align their detectors in the same direction? If the answer is no, then the source of general weirdness of QE is in the way both observers align their detectors not in the idea of superposition.

You have to actually go through the mathematics to see why that number, cos^2(\beta - \alpha), cannot be explained through normal local mechanisms. It can be explained through a nonlocal mechanism easily enough:

1. Initially, the spin directions for Alice's and Bob's particles are completely undetermined.
2. When Alice measures the spin of her particle, she randomly gets \pm 1, with 50/50 probability of each outcome.
3. If Alice measures +1 at direction \alpha, then Bob's particle "collapses" to the state with spin direction \phi = \alpha - \pi.
4. If Alice measures -1 at direction \alpha, then Bob's particle "collapses" to the state with spin direction \phi = \alpha.
5. Later, when Bob measures the spin of his particle at direction \beta, he gets +1 with probability cos^2(\frac{\beta - \phi}{2}) and -1 with probability sin^2(\frac{\beta - \phi}{2}).

This nonlocal model (the "collapse" interpretation) explains the probabilities, but at the cost of nonlocality; Bob's particle's state changes instantaneously when Alice finishes her measurement.

[edit: probabilities in 5 corrected; originally there was a missing factor of 2]
[edit 2: sign of \phi in 3 was changed]

 

[bohm2]

. So what is the difference between both QE and the classical version?

You may find these papers comparing classical versus quantum entanglement worth reading as they point out the limits of classical analogues of entanglement:

Brownian Entanglement
http://arxiv.org/pdf/quant-ph/0412132v1.pdf
https://staff.fnwi.uva.nl/t.m.../BrownianEntanglement-1-10.ppt

A classical analogy of quantum mechanical entanglement is presented, using classical light beams. The analogy can be pushed a long way, only to reach its limits when we try to represent multiparticle, or nonlocal, entanglement. This demonstrates that the latter is of exclusive quantum nature.

A Classical Analogy of Entanglement
http://www.science.uva.nl/research/aplp/eprints/Spr98.pdf

Entanglement in classical Brownian motion
https://esc.fnwi.uva.nl/thesis/centraal/files/f292681290.pdf

 

[Adel Makram]

Later, when Bob measures the spin of his particle at direction \beta, he gets +1 with probability cos^2(\beta - \phi) and -1 with probability sin^2(\beta - \phi).
This nonlocal model (the "collapse" interpretation) explains the probabilities, but at the cost of nonlocality; Bob's particle's state changes instantaneously when Alice finishes her measurement.

I am confused after reviewing this post compared with yours #11.
Suppose Bob chose his detector to be aligned at angle α which is the same as Alice (α=β). Now, the probability of Bob measuring +1 is cos²(β-Φ)=cos²(α-Φ)=cos²(α-π+α)=cos²(2α-π). So the probability of getting the same result (+1) ≠ 0 unless α=π/2. In fact if α is chosen to be 0, then that probability is 1.
In post #11, the probability of getting the same result is sin²(β-α)=0?

 

[Nugatory]

I am confused after reviewing this post compared with yours #11.

There's a missing division by two in steps 4 and 5 - it's a typo. Try this instead:
1) Initially, the spin directions for Alice's and Bob's particles are completely undetermined
2) When Alice measures the spin of her particle, she randomly gets ±1, with 50/50 probability of each outcome.
3) If Alice measures +1 at direction $\alpha$ then Bob's particle "collapses" to the state with spin direction $\phi=\pi-\alpha$
4) If Alice measures -1 at direction α, then Bob's particle "collapses" to the state with spin direction $\phi=\alpha$
5 )Later, when Bob measures the spin of his particle at direction $\beta$, he gets +1 with probability $\cos^2\frac{\beta−\phi}{2}$ and -1 with probability $\sin^2\frac{\beta−\phi}{2}$.

These are the rules for spin-entangled electrons. You'll also see people explaining Bell's theorem using experiments with polarization-entangled photons instead; the argument is the same but that division by two is not there (maximum anti-correlation effect is found at 90 degrees instead of 180 degrees). However, the all-important dependence on $\beta-\alpha$ is still there - that's what makes the result at one detector change when we change the setting of the other detector.

 

[Nugatory]

What I understood from Quantum Entanglement (QE), is that measuring the spin of one of two entangled particles in one location gives the spin of the other particle in other location no matter how far is the later. What I can also understand is that the same concept is applicable in the classical physics. For example, if we have a way to create two spinning tops at one point in the time and space, they will be spinning at different directions to maintain the law of conservation of angular momentum. Now measuring the direction of one of them in one location ( by just watching it), gives information about the direction of the other spinning top in the other location. So what is the difference between both QE and the classical version?

You might want to give this paper a try: http://www.quantum3000.narod.ru/papers/edu/cakes.pdf

Although it argues by analogy, it does a pretty good job of identifying the essential difference between quantum and classical models, and doesn't require understanding the QM treatment of the singlet state as the price of admission.

 

[morrobay]

There's a missing division by two in steps 4 and 5 - it's a typo. Try this instead:
1) Initially, the spin directions for Alice's and Bob's particles are completely undetermined
2) When Alice measures the spin of her particle, she randomly gets ±1, with 50/50 probability of each outcome.
3) If Alice measures +1 at direction $\alpha$ then Bob's particle "collapses" to the state with spin direction $\phi=\pi-\alpha$
4) If Alice measures -1 at direction α, then Bob's particle "collapses" to the state with spin direction $\phi=\alpha$
5 )Later, when Bob measures the spin of his particle at direction $\beta$, he gets +1 with probability $\cos^2\frac{\beta−\phi}{2}$ and -1 with probability $\sin^2\frac{\beta−\phi}{2}$.

These are the rules for spin-entangled electrons. You'll also see people explaining Bell's theorem using experiments with polarization-entangled photons instead; the argument is the same but that division by two is not there (maximum anti-correlation effect is found at 90 degrees instead of 180 degrees). However, the all-important dependence on $\beta-\alpha$ is still there - that's what makes the result at one detector change when we change the setting of the other detector.

So later Bob makes two ± 1 spin measurements ( sin and cos functions from step 5 ) for both β's
The first two measurements at detector angle setting β from step 3.
Plus second two measurements at detector angle setting β from step 4.?
Also can you clarify: 5) Later when Bob measures the spin of his particle at direction β
Ie. If t₀ is "collapse " of Bobs particle. then what is time t₁ , measurement ?

 

[Nugatory]

So later Bob makes two ± 1 spin measurements ( sin and cos functions from step 5 ) for both β's
The first two measurements at detector angle setting β from step 3.
Plus second two measurements at detector angle setting β from step 4.?

For every pair, there's only one measurement by Alice and only one measurement by Bob. Step 3 is what happens if Alice's one and only measurement of her particle has the result +1 and step 4 is what happens if it has the result -1. At step 5 Bob performs one measurement on his particle that also that comes up -1 or +1; the $\cos^2$ and $\sin^2$ expressions are the probabilities of getting one result or the other (remember that $\sin^2\theta=1-\cos^2\theta$, as you'd expect of the probabilities of two mutually exclusive outcomes).

Also can you clarify: 5) Later when Bob measures the spin of his particle at direction β
Ie. If t₀ is collapse of Bobs particle. then what is time t₁ , measurement?

"Later" here means that Alice's measurement came first, so $t_1\gt{t}_0$.

 

[Adel Makram]

There's a missing division by two in steps 4 and 5 - it's a typo. Try this instead:
4) If Alice measures -1 at direction α, then Bob's particle "collapses" to the state with spin direction $\phi=\alpha$
5 )Later, when Bob measures the spin of his particle at direction $\beta$, he gets +1 with probability $\cos^2\frac{\beta−\phi}{2}$ and -1 with probability $\sin^2\frac{\beta−\phi}{2}$.

Thank you. It works now. $\cos^2\frac{\beta−\phi}{2}$ is mathematically equivalent to $\sin^2{\beta−\alpha}$

 

[Adel Makram]

I would like also to know what is the meaning of " if Alice measures +1 at a direction α?"
Provided that α is the direction of the detector relative to +x-axis, does this mean the spin is now in the direction α? I mean the set up of the experiment takes into account the general polarization of spins along y-direction ( either spin up along +y or spin down along -y) or it considers the spin reaching the detector in random direction in xy plane?

 

[Heinera]

I would like also to know what is the meaning of " if Alice measures +1 at a direction α?"
Provided that α is the direction of the detector relative to +x-axis, does this mean the spin is now in the direction α? I mean the set up of the experiment takes into account the general polarization of spins along y-direction ( either spin up along +y or spin down along -y) or it considers the spin reaching the detector in random direction in xy plane?

The direction here is meant to be the orientation of the detector. She either measures +1 or -1 with that orientation, and that is all we need to take into account.

 

[Nugatory]

I would like also to know what is the meaning of " if Alice measures +1 at a direction α?"
Provided that α is the direction of the detector relative to +x-axis, does this mean the spin is now in the direction α?

Yes, and this is one of the major ways in which quantum spin is unlike its classical analogue. No matter what direction you choose to measure the spin of a spin-1/2 particle along there are only two possible outcomes, +1 and -1 (to within a constant multiple - they're actually $\pm\hbar/2$). After the measurement, any subsequent measurement of that particle along that direction will produce the same result and a measurement on a different direction will yield the same or different result with probabilities given by the $\sin^2\frac{\alpha-\beta}{2}$ rule.

That is, measuring the particle along a given direction causes it to jump into a state in which the spin is aligned with (if we measured +1) or against (if we measured -1) that direction.

I mean the set up of the experiment takes into account the general polarization of spins along y-direction ( either spin up along +y or spin down along -y) or it considers the spin reaching the detector in random direction in xy plane?

There is no "general polarization" here. No matter what direction and what direction Alice chooses to measure, she has an equal chance of getting +1 and of getting -1. It's easiest to do this experiment when the direction of measurement is perpendicular to the direction of particle travel, but it's not necessary - Alice could have chosen to measure the spin parallel to the direction of travel of her particle if she had wanted.

 

[morrobay]

For every pair, there's only one measurement by Alice and only one measurement by Bob. Step 3 is what happens if Alice's one and only measurement of her particle has the result +1 and step 4 is what happens if it has the result -1. At step 5 Bob performs one measurement on his particle that also that comes up -1 or +1; the $\cos^2$ and $\sin^2$ expressions are the probabilities of getting one result or the other (remember that $\sin^2\theta=1-\cos^2\theta$, as you'd expect of the probabilities of two mutually exclusive outcomes)..

Ok thanks, I tried to edit out "measurements" with sin and cos functions but it was too late. I intended to convey that those functions are applied to the two Φ,s ( the angles Bobs particle collapses to after Alice gets ± 1 ) to get expected outcomes. ++,--,-+,+-
Can someone give numerical values to Φ ,α, β in steps 3,4.5 post # 19 :
Φ = π - α
β - α

* When all questions here have been answered can we go over this subject with photons ?

 

[Adel Makram]

That is, measuring the particle along a given direction causes it to jump into a state in which the spin is aligned with (if we measured +1) or against (if we measured -1) that direction.

If I assume that the spin before the measurement is oriented along any possible direction in 3D space (like a 3D vector with the tip draws a 3D sphere and the base of the spin is at the origin of the sphere). Now if the measurement causes it to jump into the direction of the detector, what direction should the other spin is then aligned? Geometrically the tip of the second particle should lie along a 2D circle which satisfies π-α. This means the direction ( spin) of the second particle is still not yet defined after measuring the spin of the first particle ( although it is only confined to a circle not a sphere).

 

[Nugatory]

Now if the measurement causes it to jump into the direction of the detector, what direction should the other spin is then aligned?

If Alice's detector points in a given direction, then after the measurement either Alice's particle has its spin aligned in that direction and Bob's particle will collapse into a state in which its spin is aligned in the opposite direction (Alice measured +1, Bob will measure -1), or Alice's particle has its spin aligned opposite to that direction and Bob's particle will collapse into a state in which its spin is aligned with that direction (Alice measured -1, Bob will measure +1).

Write the orientation of Alice's detector as a vector $\vec{v}$. After the measurement, one of the particles will have its spin aligned with $\vec{v}$ and the oher will have its spin aligned with $-\vec{v}$.

 

[Nugatory]

Can someone give numerical values to Φ ,α, β in steps 3,4.5 post # 19 :
Φ = π - α
β - α

You only need values for $\alpha$ and $\beta$ because if you have these you can calculate $\phi$.

Alice can choose any angle she wants for $\alpha$ and Bob can choose any angle he wants for $\beta$ (we could even replace both Alice and Bob with computers running random number generators). The point of the exercise is that no matter what values are chosen for $\alpha$ and $\beta$, the $\cos^2$ and $\sin^2$ formulas will give you the probability of them both getting or not getting the same result on their measurements.

 

[Adel Makram]

In order for Bob`s particle to be collapsed into a state in the opposite direction of Alice`s particle as in the post:

If Alice's detector points in a given direction, then after the measurement either Alice's particle has its spin aligned in that direction and Bob's particle will collapse into a state in which its spin is aligned in the opposite direction.

A change should be made in the form of $\phi=\pi+\alpha$ in the post:

3) If Alice measures +1 at direction $\alpha$ then Bob's particle "collapses" to the state with spin direction $\phi=\pi-\alpha$.

Because for the same frame of reference of both observers, the opposite to the direction of any angle $\alpha$ relative to the +x direction is $\pi+\alpha$

 

[stevendaryl]

In order for Bob`s particle to be collapsed into a state in the opposite direction of Alice`s particle as in the post:

A change should be made in the form of $\phi=\pi+\alpha$ in the post:

Because for the same frame of reference of both observers, the opposite of the direction of any angle $\alpha$ in the direction of +x is $\pi+\alpha$

Yes, you're right. Perhaps a better description of the "collapse interpretation" of EPR is in terms of orientation vectors:

• Alice chooses a measurement direction \vec{\alpha}. (Normalized so that |\vec{\alpha}| = 1)
  
• Bob chooses a measurement direction \vec{\beta}. (Normalized so that |\vec{\beta}| = 1)
  
• Let \vec{S_A} be the spin of Alice's particle. (Normalized so that |\vec{S_A}| = 1)
  
• Let \vec{S_B} be the spin of Bob's particle. (Normalized so that |\vec{S_B}| = 1)
  
• Alice measures "spin-up" or "spin-down" with 50/50 chance of each.
  
• If Alice measures "spin-up", then afterward, \vec{S_A} = \vec{\alpha} and \vec{S_B} = - \vec{\alpha}.
• If Alice measures "spin-down", then afterward, \vec{S_A} = -\vec{\alpha} and \vec{S_B} = + \vec{\alpha}.
• Later, Bob measures "spin-up" with probability \frac{1}{2}(1 + \vec{\beta} \cdot \vec{S_B}), and spin-down with probability \frac{1}{2}(1 - \vec{\beta} \cdot \vec{S_B}).

The last line is using the fact that if the angle between \vec{\beta} and \vec{S_B} is \varphi, then cos^2(\frac{\varphi}{2}) = \frac{1}{2}(1 + cos(\varphi)) = \frac{1}{2}(1 + \vec{\beta} \cdot \vec{S_B}) and sin^2(\frac{\varphi}{2}) = \frac{1}{2}(1 - cos(\varphi)) = \frac{1}{2}(1 - \vec{\beta} \cdot \vec{S_B})

 

[stevendaryl]

• If Alice measures "spin-up", then afterward, \vec{S_A} = \vec{\alpha} and \vec{S_B} = - \vec{\alpha}.
• If Alice measures "spin-down", then afterward, \vec{S_A} = -\vec{\alpha} and \vec{S_B} = + \vec{\alpha}.

This is the difference between EPR and any classically similar notion of correlation. Alice performing a measurement on her particle can seemingly "force" Bob's particle to have spin \vec{\alpha} or -\vec{\alpha}. There is no way to do that, classically, unless somehow there is a faster-than-light influence from Alice to Bob.

 

[Mentz114]

This is the difference between EPR and any classically similar notion of correlation. Alice performing a measurement on her particle can seemingly "force" Bob's particle to have spin \vec{\alpha} or -\vec{\alpha}. There is no way to do that, classically, unless somehow there is a faster-than-light influence from Alice to Bob.

I've been reading an account of the Stern-Gerlach effect which attributes the final probabilites to interference. The author* uses an argument similar to the 'which path' logic in the two-slit setup. The final decomposition is $\tfrac{1}{4} + \tfrac{1}{4} \pm \tfrac{1}{2}$ where the first two terms are classical values and the last term is interference. Obviously this gives probability $\tfrac{1}{2}$ for either outcome without interference and with interference it is $0$ or $1$

This also illustrates the difference between the quantum and classical cases.

* from a free e-book by J. D. Cresser
http://www.e-booksdirectory.com/details.php?ebook=6153

 

[Adel Makram]

Because for the same frame of reference of both observers, the opposite to the direction of any angle $\alpha$ relative to the +x direction is $\pi+\alpha$

I think the previous steps of post #16 is still good if only the following correction is made on step 3
"If Alice measures +1 at direction \alpha, then Bob's particle "collapses" to the state with spin direction $\phi=\pi+\alpha$"

N.B: I made the following mistake at post #23: I used $\phi=\pi+\alpha$ (which is right and I don't know how a writing mistake gave the correct answer later on) instead of $\phi=\pi-\alpha$ and I reached to the correct statement: $cos^2(\frac{\beta-\phi}{2})$ = $sin^2(\beta-\alpha)$.

 

[Nugatory]

I made the following mistake: I used $\phi=\pi+\alpha$ (which is right and I don't know how a writing mistake gave the correct answer later on) instead of $\phi=\pi-\alpha$ and I reached to the correct $sin^2(\beta-\alpha)$.

$\alpha-\pi$, $\pi+\alpha$, and "exactly opposite to $\alpha$" are three different ways of describing the same direction, so of course you get the same result.

If you're facing northeast and you turn $\pi$ radians to the left, or $\pi$ radians to the right, or turn around and look in the opposite direction... you'll be facing southwest.

 

[Adel Makram]

$\alpha-\pi$, $\pi+\alpha$, and "exactly opposite to $\alpha$" are three different ways of describing the same direction, so of course you get the same result.

If you're facing northeast and you turn $\pi$ radians to the left, or $\pi$ radians to the right, or turn around and look in the opposite direction... you'll be facing southwest.

No, my mistake was, instead of writing $cos^2(\frac{\beta - \phi}{2})$=$cos^2(\frac{\beta - (\pi-\alpha)}{2})$=$cos^2(\frac{(\beta +\alpha)}{2}-\frac{\pi}{2})$, I wrote it as $cos^2(\frac{(\beta -\alpha)}{2}-\frac{\pi}{2})$ which led to the correct result.

 

[jknm]

Bell's theorem is all about the testable difference between quantum entanglement and any analogous classical correlation.

What Bell showed is that the correlations predicted by quantum mechanics cannot be explained by any such hidden-variable model (unless we allow faster-than-light influences, or back-in-time influences, or some other exotic possibility).

Bell did not actually show how a faster than light influence could reproduce the (very simple) correlation relationships of quantum systems. Nobody has done so, the 'influence' is left a mystery. The faster than light hypothesis is just an ad-hoc supposition that would allow a non-local hidden variable theory (such as de Broglie-Bohm) to reproduce QM. Basically, all the ad-hoc 'influence' is doing is saying that there is a magic effect that causes quantum systems to behave in accordance with observations, a kind of quantum epicycle.

Second, the term 'classical' is very misleading, the Bell inequality applies to any system where bivalent logic values are made from measurement results, the fact that Bell's conclusions about classical systems depend on assuming that classical systems must always provide this type of value is overlooked in the rush to prop-up hi hypothesis about ftl influences. The failure of the type of a hidden variable model where the wave-function is presumed to correspond to a real phenomenon that is supplemented by hidden parameters could also be a result of assuming the wave-function to represent something real (as in de Broglie-Bohm), rather than, as in (QBism) “gambling attitudes for placing bets on measurement outcomes, attitudes that are updated as new data come to light”.

Take a look at Garden. http://www.researchgate.net/publication/226368372_Logic_states_and_quantum_probabilities

Garden develops a formal logic framework consistent with the ‘logical denial’ nature of the information that quantum interactions provide, and shows that the when this type of logic is used, then violations of the Bell inequalities are expected, and the requirement of a break with of locality appears unnecessary. Garden shows that the Bell inequality relies on bivalent truth values for its proof, and that the results of quantum interactions do not allow us to make such an assessment.

By assuming that the so-called 'classical' model returns bivalent assessment, (e.g. A result of 'Up' means 'Not Down') you can prove Bell's theorem. When a result op 'Up' means that the initial state was "not exactly aligned with down." (a denial) then Bell's theorem cannot be applied.

 

[DrChinese]

Bell did not actually show how a faster than light influence could reproduce the (very simple) correlation relationships of quantum systems. Nobody has done so, the 'influence' is left a mystery. The faster than light hypothesis is just an ad-hoc supposition that would allow a non-local hidden variable theory (such as de Broglie-Bohm) to reproduce QM. Basically, all the ad-hoc 'influence' is doing is saying that there is a magic effect that causes quantum systems to behave in accordance with observations, a kind of quantum epicycle.

...

Garden develops a formal logic framework consistent with the ‘logical denial’ nature of the information that quantum interactions provide, and shows that the when this type of logic is used, then violations of the Bell inequalities are expected, and the requirement of a break with of locality appears unnecessary. Garden shows that the Bell inequality relies on bivalent truth values for its proof, and that the results of quantum interactions do not allow us to make such an assessment.

By assuming that the so-called 'classical' model returns bivalent assessment, (e.g. A result of 'Up' means 'Not Down') you can prove Bell's theorem. When a result op 'Up' means that the initial state was "not exactly aligned with down." (a denial) then Bell's theorem cannot be applied.

First, locality is an explicit assumption of Bell's Theorem (as is realism). So violation of a Bell inequality implies locality and/or realism is wrong. So no, there is no such ad hoc assumption of non-locality due to Bell. There are those who make such an assumption, but that is different.

Second, the bivalent nature of spin measurements has been discussed in detail. It would make more sense to discuss were it not for the existence of perfect correlations. Their existence more or less undermines that line of thinking completely.

 

[Heinera]

Bell did not actually show how a faster than light influence could reproduce the (very simple) correlation relationships of quantum systems. Nobody has done so, the 'influence' is left a mystery.

Well, not exactly; constructing a hidden variables model with ftl influences that produces the quantum correlations is utterly trivial.

See e.g. http://rpubs.com/heinera/16727

(Not that I would advise anyone to thnk about QM in terms of hidden variables and ftl communication.)

 

[Headcrash]

I see many discussions about ftl and MWI in respect of Bell, not so much on that other way out, so called superdeterminism (SD).
I guess no one likes the idea of loss of freedom.
However, I wonder if SD Is itself a misunderstanding. Let's remember the lack of an absolute frame of reference.
I suggest that, from the photon's POV, there is no time; departure and arrival are perhaps the only real definition of "simultaneous".
Then entangled particles might be an unusual(?) circumstance where three events are simultaneous: the supposed source of the entangled pair, and the two destinations. From our experince of 'time', perhaps that looks like predeterminism, but perhaps our experience is simply too limited.
I guess that puts me in the Realist camp. Kinda.

 

[Nugatory]

I suggest that, from the photon's POV, there is no time; departure and arrival are perhaps the only real definition of "simultaneous".

Even setting aside the well-known problems that arise from any attempt to analyze a physical situation "from the photon's POV", we can do Bell-type experiments with spin-entangled particles that have non-zero rest mass so do not travel at the speed of light.

 

[DrChinese]

I see many discussions about ftl and MWI in respect of Bell, not so much on that other way out, so called superdeterminism (SD).
I guess no one likes the idea of loss of freedom.

Welcome to PhysicsForums, Headcrash!

Superdeterminism (SD) and determinism are completely different. Determinism in physics is a feature of some interpretations, including Bohmian Mechanics (BM). BM is an actual theory, which reproduces the predictions of quantum mechanics. It is one of several generally accepted interpretations of Quantum Mechanics. There are plenty of advocates of BM on this forum, and I seriously doubt those who don't follow BM are worried about loss of freedom.

There is currently no theory called SD, although perhaps someone will advance one some day. It is actually the "hypothesis" that the laws of physics are local realistic; but that experiments always yield values that indicate that the laws of physics are not local realistic. It is just a fabricated concept by persons who don't like non-realism or non-locality.

There is also a hypothesis that the universe is 5 minutes old (see for example the quote by the famous philosopher Bertrand Russell). Neither this nor the SD hypothesis can ever been proven false as stated. Because it is not falsifiable, most scientists consider this metaphysics and not physics proper.

 

[Headcrash]

... we can do Bell-type experiments with spin-entangled particles that have non-zero rest mass so do not travel at the speed of light.

Oops, evidence trumps metaphysics! Thanks for that.
(And, now I recall reports of distinctly non-elementary particles (buckyballs) showing quantum-like behaviour in twin slit experiments.
Back to the drawing board for me. )

Even if it were sufficient, I'm not familiar with the "well known problems" of considering things from photons POV. Maybe I'll start here: https://www.physicsforums.com/threads/the-photons-perspective-taboo.315122/

 

[Adel Makram]

I don't understand the graph of the classical correction where the correlation at angle (45°)=-1/2 and similarly at (135°, +1/2), (225°,+1/2) and ( 315°, -1/2).
Following post#9 and suppose α=0 and β=45, then the S_B.β=cos(π+45)=-cos 45° which is less than 0. So, Bob should report -B indicative of the direction of the spin of his particle is opposite that of Alice. So the correlation should be -1 not -1/2.
The graph then should be a square wave rather than a triangle wave.

 

[Adel Makram]

You only need values for $\alpha$ and $\beta$ because if you have these you can calculate $\phi$.

Alice can choose any angle she wants for $\alpha$ and Bob can choose any angle he wants for $\beta$ (we could even replace both Alice and Bob with computers running random number generators). The point of the exercise is that no matter what values are chosen for $\alpha$ and $\beta$, the $\cos^2$ and $\sin^2$ formulas will give you the probability of them both getting or not getting the same result on their measurements.

In case of both have the same angles, β-α=0, the probability of getting the same result =0 because $cos^2(\frac{(\beta-\phi)}{2})$=$sin^2(\frac{(\beta-\alpha)}{2})$

 

[jknm]

First, locality is an explicit assumption of Bell's Theorem (as is realism). So violation of a Bell inequality implies locality and/or realism is wrong. So no, there is no such ad hoc assumption of non-locality due to Bell. There are those who make such an assumption, but that is different.

Second, the bivalent nature of spin measurements has been discussed in detail. It would make more sense to discuss were it not for the existence of perfect correlations. Their existence more or less undermines that line of thinking completely.

I don't think you can dismiss the non-bivalence of spin measurements by simply alluding to 'discussion'. If they are not bivalent, then Bell's theorem fails and cannot be applied.

The fact that Bell's theorem requires locality, does not automatically mean that the only answer is non-locality, when Bell's theorem fails for other reasons.

Garden has mathematically demonstrated that the proof of Bell's theorem depends on treating spin measurements as bivalent. The classical models considered by Bell employ bivalent measurement models, so the fact that Bell's theorem applies to them is no surprise.

There is no explicit demonstration, anywhere AFAIK, as to how non-locality can be 'put in 'to a model to reproduce the results of the experiments.
In that sense, the only logical conclusion is that non-locality is nothing more than a spurious ad-hoc assumption concocted to 'explain' why quantum systems break Bell inequalities.

All it takes is for the logic assessments that we attribute to spin measurements to be a logical denial, not a bivalent assessment, and then whether or not Bell's theorem requires locality becomes irrelevant, because the theorem does not apply to the system.

 

[Nugatory]

I don't think you can dismiss the non-bivalence of spin measurements by simply alluding to 'discussion'. If they are not bivalent, then Bell's theorem fails and cannot be applied

I don't know that I'd say the theorem "fails", as the theorem claims to and does preclude a large class of theories: informally, we say that it precludes all "local realistic hidden variable theories". This bivalence property, along with counterfactual definiteness, is part of what the informal speakers mean when they informally say "realistic", so the fact that rejecting bivalence allows the inequalities to be violated is consistent with Bell's theorem.

Thus (unless I'm misunderstanding your argument) you have successfully demolished the straw man claim that Bell's theorem requires rejecting locality but left the claim that is actually being made, namely that Bell's theorem requires rejecting at least one of locality and the complex of properties that we informally call "realism", untouched.

There is no explicit demonstration, anywhere AFAIK, as to how non-locality can be 'put in 'to a model to reproduce the results of the experiments

I've seen many. For a trivial example, consider the hypothesis that when Alice makes her measurement, a superluminal pixie is created, and this pixie travels to Bob's in-flight particle and twists its spin to point opposite to whatever Alice measured. I've deliberately chosen this example to be absurd, but it is consistent with the experimental results, and it gets that way by being explicitly non-local - experiments have shown that normal subluminal pixies won't work when the detection events are spacelike separated.

 

[gill1109]

I don't think you can dismiss the non-bivalence of spin measurements by simply alluding to 'discussion'. If they are not bivalent, then Bell's theorem fails and cannot be applied.

The fact that Bell's theorem requires locality, does not automatically mean that the only answer is non-locality, when Bell's theorem fails for other reasons.

Garden has mathematically demonstrated that the proof of Bell's theorem depends on treating spin measurements as bivalent. The classical models considered by Bell employ bivalent measurement models, so the fact that Bell's theorem applies to them is no surprise.

There is no explicit demonstration, anywhere AFAIK, as to how non-locality can be 'put in 'to a model to reproduce the results of the experiments.
In that sense, the only logical conclusion is that non-locality is nothing more than a spurious ad-hoc assumption concocted to 'explain' why quantum systems break Bell inequalities.

All it takes is for the logic assessments that we attribute to spin measurements to be a logical denial, not a bivalent assessment, and then whether or not Bell's theorem requires locality becomes irrelevant, because the theorem does not apply to the system.

I would say that Bell's theorem is about (the possible mathematical-physical explanation of) *experiments* in which the experimenter observes a binary outcome. Quantum theory is usually taken to include a "measurement" part saying that we observe eigenvalues of observables, with certain probabilities. If you discard that part of QM then you can forget about Bell's theorem. But then you are disconnecting QM from the real world (it no longer makes experimental predictions).

 

[stevendaryl]

I would say that Bell's theorem is about (the possible mathematical-physical explanation of) *experiments* in which the experimenter observes a binary outcome. Quantum theory is usually taken to include a "measurement" part saying that we observe eigenvalues of observables, with certain probabilities. If you discard that part of QM then you can forget about Bell's theorem. But then you are disconnecting QM from the real world (it no longer makes experimental predictions).

It's hard to say what Rachel Wallace Garden's objection to Bell is really about without reading her paper, and I don't see a free, online copy of her paper. But by "rejecting bivalence" she might mean rejecting classical two-valued logic. In that case, I would lump her ideas into the general category of quantum logic, which in my opinion isn't a resolution to quantum weirdness, but is just a way of describing that weirdness.