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.

 

[Point Conception]

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.

 

[Point Conception]

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.