Comparison between quantum entanglement and a classical version

In summary: But in the case of quantum entanglement, the correlations are still there--they're just due to the fact that the particles are related in a particular way.In summary, quantum entanglement is a profound way to maintain correlations between particles even when they are far apart.
  • #1
Adel Makram
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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?
 
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  • #2
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
 
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  • #3
Mentz114 said:
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.
 
  • #5
Adel Makram said:
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.
 
  • #6
Mentz114 said:
But QT is non-local
To quote Unruh , "exactly in what sense is it non-local?"
 
  • #7
DirkMan said:
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).
 
  • #8
> 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.
 
  • #9
Bell's theorem is all about the testable difference between quantum entanglement and any analogous classical correlation.

Classically, if two measured values [itex]A[/itex] and [itex]B[/itex] 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 [itex]\vec{a}[/itex]
  3. Bob picks a direction [itex]\vec{b}[/itex]
  4. Alice measures the spin [itex]\vec{s_A}[/itex] of her particle relative to [itex]\vec{a}[/itex], and writes [itex]A=+1[/itex] if [itex]\vec{s_A} \cdot \vec{a} > 0[/itex]. Otherwise, she writes down [itex]A=-1[/itex]
  5. Bob measures the spin [itex]\vec{s_B}[/itex] of his particle relative to [itex]\vec{b}[/itex], and writes [itex]B=+1[/itex] if [itex]\vec{s_B} \cdot \vec{b} > 0[/itex]. Otherwise, he writes down [itex]B=-1[/itex]
  6. Then they compare the results [itex]A[/itex] and [itex]B[/itex]
They find that the results are correlated, in the sense that if [itex]\vec{a} = \vec{b}[/itex], then [itex]A = -B[/itex]

That's a classical correlation. It is easily explained using "hidden variables". You just assume that [itex]\vec{s_A}[/itex] and [itex]\vec{s_B}[/itex] are fixed at the time of the creation of the pair of particles in such a way that [itex]\vec{s_A} = - \vec{s_B}[/itex]. The pair [itex](\vec{s_A}, -\vec{s_A})[/itex] 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).
 
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  • #10
stevendaryl said:
Bell's theorem is all about the testable difference between quantum entanglement and any analogous classical correlation.

Classically, if two measured values [itex]A[/itex] and [itex]B[/itex] 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 [itex]\vec{a}[/itex]
  3. Bob picks a direction [itex]\vec{b}[/itex]
  4. Alice measures the spin [itex]\vec{s_A}[/itex] of her particle relative to [itex]\vec{a}[/itex], and writes [itex]A=+1[/itex] if [itex]\vec{s_A} \cdot \vec{a} > 0[/itex]. Otherwise, she writes down [itex]A=-1[/itex]
  5. Bob measures the spin [itex]\vec{s_B}[/itex] of his particle relative to [itex]\vec{b}[/itex], and writes [itex]B=+1[/itex] if [itex]\vec{s_B} \cdot \vec{b} > 0[/itex]. Otherwise, he writes down [itex]B=-1[/itex]
  6. Then they compare the results [itex]A[/itex] and [itex]B[/itex]
They find that the results are correlated, in the sense that if [itex]\vec{a} = \vec{b}[/itex], then [itex]A = -B[/itex]

That's a classical correlation. It is easily explained using "hidden variables". You just assume that [itex]\vec{s_A}[/itex] and [itex]\vec{s_B}[/itex] are fixed at the time of the creation of the pair of particles in such a way that [itex]\vec{s_A} = - \vec{s_B}[/itex]. The pair [itex](\vec{s_A}, -\vec{s_A})[/itex] 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.
 
  • #11
Adel Makram said:
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 [itex]\alpha[/itex] (relative to the x-axis) and Bob chooses orientation [itex]\beta[/itex], then they will get the same result (both spin-up or both spin-down) with probability [itex]sin^2(\beta - \alpha)[/itex], and will get opposite results with probability [itex]cos^2(\beta - \alpha)[/itex]. The details of what is going on in the measurement process seem irrelevant; the only thing that is relevant is the angles [itex]\alpha[/itex] and [itex]\beta[/itex].
 
  • #12
Adel Makram said:
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.
 
  • #13
stevendaryl said:
and will get opposite results with probability [itex]cos^2(\beta - \alpha)[/itex].
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 =[itex]cos^2(\beta-\alpha)[/itex] 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.
 
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  • #14
Adel Makram said:
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 =[itex]cos^2(\beta-\alpha)[/itex] 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.
 
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  • #15
DrChinese said:
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##.
 
  • #16
Adel Makram said:
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 =[itex]cos^2(\beta-\alpha)[/itex] 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, [itex]cos^2(\beta - \alpha)[/itex], 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 [itex]\pm 1[/itex], with 50/50 probability of each outcome.
  3. If Alice measures +1 at direction [itex]\alpha[/itex], then Bob's particle "collapses" to the state with spin direction [itex]\phi = \alpha - \pi[/itex].
  4. If Alice measures -1 at direction [itex]\alpha[/itex], then Bob's particle "collapses" to the state with spin direction [itex]\phi = \alpha[/itex].
  5. Later, when Bob measures the spin of his particle at direction [itex]\beta[/itex], he gets +1 with probability [itex]cos^2(\frac{\beta - \phi}{2})[/itex] and -1 with probability [itex]sin^2(\frac{\beta - \phi}{2})[/itex].
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 [itex]\phi[/itex] in 3 was changed]
 
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  • #17
Adel Makram said:
. 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 [Broken]
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
 
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  • #18
stevendaryl said:
Later, when Bob measures the spin of his particle at direction [itex]\beta[/itex], he gets +1 with probability [itex]cos^2(\beta - \phi)[/itex] and -1 with probability [itex]sin^2(\beta - \phi)[/itex].
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?
 
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  • #19
Adel Makram said:
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.
 
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  • #20
Adel Makram said:
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.
 
  • #21
Nugatory said:
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 ?
 
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  • #22
morrobay said:
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##.
 
  • #23
Nugatory said:
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}##
 
  • #24
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?
 
  • #25
Adel Makram said:
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.
 
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  • #26
Adel Makram said:
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.
 
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  • #27
Nugatory said:
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 ?
 
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  • #28
Nugatory said:
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).
 
  • #29
Adel Makram said:
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}##.
 
  • #30
morrobay said:
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.
 
  • #31
In order for Bob`s particle to be collapsed into a state in the opposite direction of Alice`s particle as in the post:
Nugatory said:
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:
Nugatory said:
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##
 
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  • #32
Adel Makram said:
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 [itex]\vec{\alpha}[/itex]. (Normalized so that [itex]|\vec{\alpha}| = 1[/itex])
  • Bob chooses a measurement direction [itex]\vec{\beta}[/itex]. (Normalized so that [itex]|\vec{\beta}| = 1[/itex])
  • Let [itex]\vec{S_A}[/itex] be the spin of Alice's particle. (Normalized so that [itex]|\vec{S_A}| = 1[/itex])
  • Let [itex]\vec{S_B}[/itex] be the spin of Bob's particle. (Normalized so that [itex]|\vec{S_B}| = 1[/itex])
  • Alice measures "spin-up" or "spin-down" with 50/50 chance of each.
  • If Alice measures "spin-up", then afterward, [itex]\vec{S_A} = \vec{\alpha}[/itex] and [itex]\vec{S_B} = - \vec{\alpha}[/itex].
  • If Alice measures "spin-down", then afterward, [itex]\vec{S_A} = -\vec{\alpha}[/itex] and [itex]\vec{S_B} = + \vec{\alpha}[/itex].
  • Later, Bob measures "spin-up" with probability [itex]\frac{1}{2}(1 + \vec{\beta} \cdot \vec{S_B})[/itex], and spin-down with probability [itex]\frac{1}{2}(1 - \vec{\beta} \cdot \vec{S_B})[/itex].
The last line is using the fact that if the angle between [itex]\vec{\beta}[/itex] and [itex]\vec{S_B}[/itex] is [itex]\varphi[/itex], then [itex]cos^2(\frac{\varphi}{2}) = \frac{1}{2}(1 + cos(\varphi)) = \frac{1}{2}(1 + \vec{\beta} \cdot \vec{S_B})[/itex] and [itex]sin^2(\frac{\varphi}{2}) = \frac{1}{2}(1 - cos(\varphi)) = \frac{1}{2}(1 - \vec{\beta} \cdot \vec{S_B})[/itex]
 
  • #33
stevendaryl said:
  • If Alice measures "spin-up", then afterward, [itex]\vec{S_A} = \vec{\alpha}[/itex] and [itex]\vec{S_B} = - \vec{\alpha}[/itex].
  • If Alice measures "spin-down", then afterward, [itex]\vec{S_A} = -\vec{\alpha}[/itex] and [itex]\vec{S_B} = + \vec{\alpha}[/itex].
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 [itex]\vec{\alpha}[/itex] or [itex]-\vec{\alpha}[/itex]. There is no way to do that, classically, unless somehow there is a faster-than-light influence from Alice to Bob.
 
  • #34
stevendaryl said:
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 [itex]\vec{\alpha}[/itex] or [itex]-\vec{\alpha}[/itex]. 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
 
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  • #35
Adel Makram said:
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 [itex]\alpha[/itex], 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)##.
 
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<h2>1. What is the difference between quantum entanglement and a classical version?</h2><p>Quantum entanglement refers to the phenomenon where two or more particles become connected in such a way that the state of one particle is dependent on the state of the other, regardless of the distance between them. In contrast, a classical version of entanglement would involve macroscopic objects and their states would be determined by classical physical laws.</p><h2>2. How does quantum entanglement work?</h2><p>In quantum entanglement, two particles are created in a way that their properties are correlated. This means that measuring the properties of one particle will instantly determine the properties of the other, regardless of the distance between them. This phenomenon is still not fully understood and is a subject of ongoing research.</p><h2>3. Can quantum entanglement be used for communication?</h2><p>While quantum entanglement allows for instantaneous correlation between particles, it cannot be used for communication in the traditional sense. This is because the state of the particles cannot be controlled or manipulated to send a specific message. However, it can be used for secure communication through quantum cryptography.</p><h2>4. How is quantum entanglement different from classical correlations?</h2><p>In classical correlations, the relationship between two particles is determined by classical physical laws, and the state of one particle can be predicted by measuring the state of the other. In quantum entanglement, the relationship between particles is based on quantum mechanics, and the state of one particle cannot be predicted by measuring the state of the other.</p><h2>5. What are the potential applications of quantum entanglement?</h2><p>Quantum entanglement has potential applications in quantum computing, quantum cryptography, and quantum teleportation. It also plays a crucial role in understanding the foundations of quantum mechanics and could potentially lead to new technologies in the future.</p>

1. What is the difference between quantum entanglement and a classical version?

Quantum entanglement refers to the phenomenon where two or more particles become connected in such a way that the state of one particle is dependent on the state of the other, regardless of the distance between them. In contrast, a classical version of entanglement would involve macroscopic objects and their states would be determined by classical physical laws.

2. How does quantum entanglement work?

In quantum entanglement, two particles are created in a way that their properties are correlated. This means that measuring the properties of one particle will instantly determine the properties of the other, regardless of the distance between them. This phenomenon is still not fully understood and is a subject of ongoing research.

3. Can quantum entanglement be used for communication?

While quantum entanglement allows for instantaneous correlation between particles, it cannot be used for communication in the traditional sense. This is because the state of the particles cannot be controlled or manipulated to send a specific message. However, it can be used for secure communication through quantum cryptography.

4. How is quantum entanglement different from classical correlations?

In classical correlations, the relationship between two particles is determined by classical physical laws, and the state of one particle can be predicted by measuring the state of the other. In quantum entanglement, the relationship between particles is based on quantum mechanics, and the state of one particle cannot be predicted by measuring the state of the other.

5. What are the potential applications of quantum entanglement?

Quantum entanglement has potential applications in quantum computing, quantum cryptography, and quantum teleportation. It also plays a crucial role in understanding the foundations of quantum mechanics and could potentially lead to new technologies in the future.

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