Comparison between quantum entanglement and a classical version

morrobay

Gold Member
674
91
Initially, the spin directions for Alice's and Bob's particles are completely undetermined.
  1. When Alice measures the spin of her particle, she randomly gets [itex]\pm 1[/itex], with 50/50 probability of each outcome.
  2. 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].
  3. If Alice measures -1 at direction [itex]\alpha[/itex], then Bob's particle "collapses" to the state with spin direction [itex]\phi = \alpha[/itex].
  4. 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]


After Bob makes his one spin measurement later at direction β what additional measurement is made that detects that his particle had collapsed to state with spin direction Φ (2. or 3. above) that is then applied in the two formulas ? If there is no additional measurement the how is Φ determined ?
 
Last edited:

Nugatory

Mentor
12,189
4,655
After Bob makes his one spin measurement later at direction β what additional measurement is made that detects that his particle had collapsed to state with spin direction Φ (2. or 3. above) that is then applied in the two formulas ? If there is no additional measurement the how is Φ determined ?
##\phi## is calculated from Alice's result, not measured. After we calculate it, we use it to further calculate the probabilities that when Bob makes his measurement he will get spin-up or spin-down. Those probabilities are what we do measure, and the measured values agree with the calculated and predicted values.

After Bob's measurement his particle is no longer in the state ##\phi## - it is either spin-up or spin-down along ##\beta##, the axis he set his detector to. We know that his particle was in the state ##\phi## before the measurement from the probabilities of getting spin-up or spin-down in the measurement.
 

morrobay

Gold Member
674
91
##\phi## is calculated from Alice's result, not measured. After we calculate it, we use it to further calculate the probabilities that when Bob makes his measurement he will get spin-up or spin-down. Those probabilities are what we do measure, and the measured values agree with the calculated and predicted values.

After Bob's measurement his particle is no longer in the state ##\phi## - it is either spin-up or spin-down along ##\beta##, the axis he set his detector to. We know that his particle was in the state ##\phi## before the measurement from the probabilities of getting spin-up or spin-down in the measurement.
Are the two Φ states that Bobs particle collapses to ( from 2 and 3 in post 76 ) QM axioms? I am sure not questioning the simple : Φ = α -π
And from the second paragraph : How is the state Φ defined ? Is it defined in terms of relation to detector setting β ?
From post #75 if Alice gets +1 at detector setting α then Bobs particle collapses to 1/2 (1 - β ⋅ α ) Then does this equal Φ = α - π ?
And is Bobs particle affected twice :
(1). After Alice gets ± 1 Bobs particle collapses to state Φ
(2) After Bob makes measurement particle is no longer in state Φ

* I am not questioning the spin up spin down measurement results or the sin and cos probability formulas that are in agreement.
 
Last edited:

Nugatory

Mentor
12,189
4,655
Are the two Φ states that Bobs particle collapses to ( from 2 and 3 in post 76 ) QM axioms?
It's not an axiom of quantum mechanics, but it comes from the calculation of how the initial entangled state changes when it interacts with Alice's detector. And of course that calculation is based on the laws of QM.
And from the second paragraph : How is the state Φ defined ? Is it defined in terms of relation to detector setting β ?
Pick some direction (usually straight up and down) and call it zero degrees. Then ##\alpha##, ##\beta##, and##\phi## are just angles relative to that direction. It doesn't matter what you call zero degrees as long as you're consistent about it.

From post #75 if Alice gets
+1 at detector setting α then Bobs particle collapses to 1/2 (1 - β ⋅ α ) Then does this equal Φ = α - π ?
You are misunderstanding #75. After Alice's measurement, Bob's particle collapses to either ##\phi=\alpha## (Alice measured -1) or ##\phi=\alpha-\pi## (Alice measured +1).
And is Bobs particle affected twice :(1). After Alice gets ± 1 Bobs particle collapses to state Φ
(2) After Bob makes measurement particle is no longer in state Φ
Yes. The state ##\phi## is a superposition of "+1 along angle ##\beta##" and "-1 along angle ##\beta##" (except in the special cases ##\alpha=\beta## and ##\alpha=\beta+\pi## where Alice's and Bob's detectors are either aligned or exactly opposite) and Bob's measurement causes that superposition to collapse into one of those two states.
 
The probability that Alice measures her particle spin-up P1=##\frac{(1+\cos(\theta1))}{2}## and the probability that Bob measures his particle spin-up P2= ##\frac{(1-\cos(\theta2))}{2}## where ##\theta1## and ##\theta2## are angles of Alice and Bob detectors, respectively.


Classically, if the spin direction is a random variable and the outcomes of Alice and Bob measurement are independent (because they depend only on their detectors angles which are freely chosen), then the joint probability of 2 outcomes is a product of 2 probabilities of Alice and Bob measurements. Pclassical=P1P2= ##\frac{(1+\cos(\theta1))(1-\cos(\theta2))}{4}##.


On the other hand, Quantum Theory sets up a definite probability of having both spin up P12= ##\frac{(1-cos(\theta1-\theta2))}{4}##. The last term represents a coefficient of state lup>lup> while the probability of similar outcomes Psimilar= ##\frac{(1-\cos(\theta1-\theta2))}{2}##as it equals the sum of coefficients of the states lup>lup> and ldown>ldown>.


Comparing the two probabilities yields the probability Psimilaris larger than Pclassical where the ratio Psimilar/Pclassical>1is plotted in a 3D graph using Wolframealpha ( see that attached picture).


The conclusion is the probability of similar outcomes, or the probability of having both spins-up, according to the QT is larger than the joint probability of classical theory as the quantum probability depends on ##\theta1-\theta2## which corresponds to an interaction between Alice and Bob according o the collapse interpretation.


My opinion is; that interaction is not the only way to explain what happened. It is only manifested when we assume that Alice measurement changes Bob spin state which makes Bob now measures his spin-up with a probability ##\frac{(1-\cos(\theta2))}{2}##. It is only manifested when we assume that Alice has come to a definite state of her spin and she gives the ball in the Bob playground to measure his spin. But if we let both observers have two probabilities of measuring their particles, spins-up, then the outcome of their measurements depends only on the angle between their two detectors ##\theta1-\theta2)##. As if there is some sensor in the space that knows the angles between 2 space-like located detectors. Therefore, no need to say that Alice particle jumps into a state with the same direction of her detector or Bob spin state changed according to what Alice measures, instead, we can say that the probabilities of both outcomes depends on the angle between their detectors. The second idea indicates a sort of space sensor (variable, metric or whatever) which does not appear in the Minkowiski 4-space equation but at the same time won`t affect the quantum prediction. So it is not a classical hidden variable.
 

Attachments

Last edited:

Nugatory

Mentor
12,189
4,655
As if there is some sensor in the space that knows the angles between 2 space-like located detectors.
Not only as if there is a such a sensor, but also that sensor can influence the measurement results at both detectors. Any theory that requires such a sensor is pretty much by definition non-local. That's the one of the points of the exercise: No local realistic hidden variable theory will work.
 
Not only as if there is a such a sensor, but also that sensor can influence the measurement results at both detectors. Any theory that requires such a sensor is pretty much by definition non-local. That's the one of the points of the exercise: No local realistic hidden variable theory will work.
Then do we need to rewrite the special relativity to include an additional space metric that has a non-local effect but doesn't refute the invariance of the speed of the light?
 

Nugatory

Mentor
12,189
4,655
Then do we need to rewrite the special relativity to include an additional space metric that has a non-local effect but doesn't refute the invariance of the speed of the light?
No. The conclusion that you can draw from these Aspect/Bell experiments is that collapse interpretations don't play well with special relativity.

The essential premises behind special relativity are that the laws of physics are the same in all inertial frames and that the speed of light is invariant. From these we can conclude that if it is logically necessary that A happened before B (for example, because B was caused by A) then A must be somewhere in B's past lightcone and the two events are timelike-separated. If we make the further reasonable assumption that departure on a journey is logically required to happen before arrival at the end of the journey, we see that the departure event must be somewhere in the past lightcone of the arrival event, and therefore that the travel speed must be less than the speed of light. Conversely, if the speed exceeded the speed of light, the departure event would be outside the past lightcone of the arrival event and some observers would find that the journey illogically ended before it started. That's the argument against faster-than-light travel.

However, Alice's and Bob's measurements are not related in this way; it is not logically necessary that one of them happen first for all observers. We've been talking as if Alice measured first, and when she gets spin-up this "causes" Bob's wave function to collapse to the spin-down state, but that's just a (non-relativistic) way of imagining what's going on. If the two measurement events are spacelike-separated, then some observers moving relative to us will look at the exact same set of facts and say that Bob measured first and got spin-down, "causing" Alice's particle to collapse into the spin-up state before she measured it. Because both descriptions are equally valid, it is clear that using the word "cause" in either of them has to be a mistake... But it's not a mistake in QM, it's a mistake that we're making as we try to imagine what's going on in terms of a wavefunction collapse that transmits a causal influence.
 

morrobay

Gold Member
674
91
It's not an axiom of quantum mechanics, but it comes from the calculation of how the initial entangled state changes when it interacts with Alice's detector. And of course that calculation is based on the laws of QM.

Pick some direction (usually straight up and down) and call it zero degrees. Then ##\alpha##, ##\beta##, and##\phi## are just angles relative to that direction. It doesn't matter what you call zero degrees as long as you're consistent about it.

You are misunderstanding #75. After Alice's measurement, Bob's particle collapses to either ##\phi=\alpha## (Alice measured -1) or ##\phi=\alpha-\pi## (Alice measured +1).

Yes. The state ##\phi## is a superposition of "+1 along angle ##\beta##" and "-1 along angle ##\beta##" (except in the special cases ##\alpha=\beta## and ##\alpha=\beta+\pi## where Alice's and Bob's detectors are either aligned or exactly opposite) and Bob's measurement causes that superposition to collapse into one of those two states.
No. The way that I was defining things, [itex]\vec{\beta}[/itex] is the orientation of Bob's detector, and [itex]\vec{S_B}[/itex] is the spin state of Bob's particle. If [itex]\vec{\beta} = \vec{S_B}[/itex], then Bob measures spin-up with 100% probability.

In the "collapse" interpretation, if Alice measures spin-up along direction [itex]\vec{\alpha}[/itex], then [itex]\vec{S_B}[/itex] "collapses" so that [itex]\vec{S_B} = -\vec{\alpha}[/itex]. So in that case, [itex]\frac{1}{2}(1+\vec{\beta} \cdot \vec{S_B}) = \frac{1}{2}(1-\vec{\beta} \cdot \vec{\alpha})[/itex].
Thanks . That clears up phi. Regarding post #75 when Alice measures spin up along direction α then Bobs particle collapses to SB = - α
The other definition in that case is Bobs particle collapsing to Φ = α - π . So I should have correctly asked how are these two definitions equated in a numerical example ?
α = 30ο
β = 80ο
Φ = - 150ο ?
Also if probability of Bob measuring spin down = sin2 (β - Φ)/2)
And this formula is equated to 1/2 ( 1 - β ⋅ α ) above then can someone show this numerically also ?
 

Nugatory

Mentor
12,189
4,655
Regarding post #75 when Alice measures spin up along direction α then Bobs particle collapses to SB = - α
Bob's particle does not collapse to ##-\alpha##. It collapses into either spin-up along the direction ##\alpha## (Alice found spin-down along direction ##\alpha##) or spin-up along the direction ##\alpha-\pi## (Alice found spin-up along direction ##\alpha##). Note that spin-up along the direction ##\alpha## is the same thing as spin-down along the direction ##\alpha-\pi## and vice versa, so all we're saying here is that Bob's particle collapses to the opposite of what Alice measured.
The other definition in that case is Bobs particle collapsing to Φ = α - π . So I should have correctly asked how are these two definitions equated in a numerical example ?
α = 30ο
β = 80ο
Φ = - 150ο ?
Yes, that is correct. Of course -150 is the same angle as +210, so we could have just as easily said that ##\phi=\alpha+\pi##. The important thing is that ##\phi## points in exactly the opposite direction from the spin that Alice measured. If Alice set her detector to +30 and measured spin-up, then her particle has collapsed into the state in which its spin is pointing in the 30o direction and Bob's has collapsed into the state in which its spin is pointing in the -150 (or +210) direction. If Alice set her detector to +30 and measured spin-down, then her particle has collapsed into the state in which its spin is pointing in the -150 (or +210) direction and Bob's has collapsed into the state in which its spin is pointing in the +30 direction.
 
Last edited:

Nugatory

Mentor
12,189
4,655
Also if probability of Bob measuring spin down = sin2 (β - Φ)/2)
And this formula is equated to 1/2 ( 1 - β ⋅ α ) above then can someone show this numerically also ?
That's just basic trig identities. ##\sin^2x=\frac{1}{2}(1-\cos{2}x)## so if you set ##x=(\beta-\phi)/2## and remember that ##\vec{\beta}\cdot\vec{\alpha}=\cos(\beta-\alpha)## because they're both unit vectors you'll get there.
 
But it's not a mistake in QM, it's a mistake that we're making as we try to imagine what's going on in terms of a wavefunction collapse that transmits a causal influence.
And that what made EPR-experiment a paradox.
But if we adopt the simple interpretation that only the difference between detectors angles matters, no paradox appears. In this interpretation, faster than light speed is not required because a causal influence on Bob spin-state from Alice`s wave-function collapse is not required too. What is required is a universal "space awareness" which is a knowledge about the difference between detectors angles in two space-like events. As if there is an additional term in 4D Minkowski space equation that when switching to "quantum mode" cancels the time metric and reduced it to Galilean 3D space equation. That makes me think that EPR-experiment is a real test for special relativity not for quantum theory.
 
Last edited:
159
55
That makes me think that EPR-experiment is a real test for special relativity not for quantum theory.
It's not even a real test for SR, because there is no way Alice and Bob can use the quantum correlations for any kind of instant communication. If Bob looks at his results in isolation, he can't infer anything about Alice's setting. It is only when they compare their results and compute the correlation that they can decide that something non-classical has taken place. So, since the quantum results can't be used for signalling from Alice to Bob or vice versa, they are perfectly consistent with special relativity.
 
It's not even a real test for SR, because there is no way Alice and Bob can use the quantum correlations for any kind of instant communication. If Bob looks at his results in isolation, he can't infer anything about Alice's setting. It is only when they compare their results and compute the correlation that they can decide that something non-classical has taken place. So, since the quantum results can't be used for signalling from Alice to Bob or vice versa, they are perfectly consistent with special relativity.
May be I meant, there must be a more general law of physics than SR and Minkowski space equation which does not only deal with light transmission but with quantum information as well.
 

Nugatory

Mentor
12,189
4,655
And that what made EPR-experiment a paradox.
But if we adopt the simple interpretation that only the difference between detectors angles matters, no paradox appears
There is no EPR "paradox" - no apparent contradictions appear anywhere in the EPR argument. The EPR argument is that QM is incomplete, that there is an as-yet-undiscovered local, realistic, and deterministic hidden variable theory that explains why quantum mechanics works (similarly to way that the macroscopic randomness we see when we spin a roulette wheel is explained by classical mechanics and statistical methods applied to incompletely specified initial conditions). Given what was known at the time, that was a reasonable enough expectation - it took the best part of three decades for Bell to discover the impossibility of explaining QM with such a theory.

What is required is a universal "space awareness" which is a knowledge about the difference between detectors angles in two space-like events. As if there is an additional term in 4D Minkowski space equation that when switching to "quantum mode" cancels the time metric and reduced it to Galilean 3D space equation. That makes me think that EPR-experiment is a real test for special relativity not for quantum theory.
Now you're just babbling. What is this "4d Minkowski space equation"? What is a "time metric"? What is "quantum mode"?
 

Mentz114

Gold Member
5,372
270
And that what made EPR-experiment a paradox.
But if we adopt the simple interpretation that only the difference between detectors angles matters, no paradox appears. In this interpretation, faster than light speed is not required because a causal influence on Bob spin-state from Alice`s wave-function collapse is not required too. What is required is a universal "space awareness" which is a knowledge about the difference between detectors angles in two space-like events. As if there is an additional term in 4D Minkowski space equation that when switching to "quantum mode" cancels the time metric and reduced it to Galilean 3D space equation. That makes me think that EPR-experiment is a real test for special relativity not for quantum theory.
I don't think that is correct. Introducing an extra dimension ( if that is what you propose ) has the same effect as a NLHV theory but no physical justification.
 
It is only when they compare their results and compute the correlation that they can decide that something non-classical has taken place.
But the quantum correlation is valid whether they compare or not, isn't it? Like two men (European and Asian) know they have two different time zones whether they compare their watches or not.
 
159
55
But the quantum correlation is valid whether they compare or not, isn't it? Like two men (European and Asian) know they have two different time zones whether they compare their watches or not.
The quantum correlation is valid in the sense that this is the outcome QM predicts for the computation of the correlation. If you are contemplating what the prediction is for an uncomputed correlation, then we are moving towards methaphysical territory. My point was that if Alice and Bob don't compare their results and never compute the correlation, they could just as well explain their isolated results with a classical theory, so no instant signalling is possible.
 
My point was that if Alice and Bob don't compare their results and never compute the correlation, they could just as well explain their isolated results with a classical theory, so no instant signalling is possible.
This is not my point. My point is quantum correlation will not involve faster than light signaling therefore it will not invalidate SR. But quantum entanglement ( spooky action) requires a more generalized theory than SR because it doesn't necessary involve signal transmission. For example, GR is more general than SR because if the mass energy tensor =0, GR reduced to SR but this does not mean that SR is invalid. It means it is not general enough to explain the physics if there is a gravitational mass.
 

Simon Phoenix

Science Advisor
Gold Member
291
224
So in your opinion, how to explain the outcomes of Alice Bob spins-experiment and quantum correlation based on causality or based on anything?
I don't understand your question.

One way to phrase the locality condition for this experiment is that results 'here' do not depend on settings 'there'. This is true in QM too (if it weren't true we'd be able to use this to construct a FTL communication scheme).

Another way to see this is the following. We'll consider 2 experiments
(1) Alice is given a spin-1/2 particle uniformly randomly selected to be spin up or spin down
(2) Alice is given one spin-1/2 particle from a maximally entangled pair of spin-1/2 particles
She receives a particle but is not told whether it is from experiment (1) or experiment (2)

There is no measurement Alice can do on her particle alone that will enable her to determine whether it is experiment (1) or experiment (2).
 
159
55
This is not my point. My point is quantum correlation will not involve faster than light signaling therefore it will not invalidate SR. But quantum entanglement ( spooky action) requires a more generalized theory than SR because it doesn't necessary involve signal transmission. For example, GR is more general than SR because if the mass energy tensor =0, GR reduced to SR but this does not mean that SR is invalid. It means it is not general enough to explain the physics if there is a gravitational mass.
But why should quantum entanglement require a more generalized theory than SR, as long as it already is consistent with SR? No FTL communication is possible with quantum entanglement, in agreement with SR.
 
But why should quantum entanglement require a more generalized theory than SR, as long as it already is consistent with SR? No FTL communication is possible with quantum entanglement, in agreement with SR.
Ok. So how SR explains the non locality then?
 

Simon Phoenix

Science Advisor
Gold Member
291
224
Ok. So how SR explains the non locality then?
What non-locality?

Could you explain precisely what it is about QM that you see as non-local?

There is nothing Alice can do locally that in any way affects, superluminally, the results of a distant measurement - this is true in QM too. So what is it you're seeing as being 'non-local'?
 
159
55
Ok. So how SR explains the non locality then?
If by "non locality" you mean the violation of Bell's inequality, then SR does not need to explain it, since it is consistent with SR. The closest you come to an explanation is in QM itself, from the fact that the wave function of the two particles depends on both, even when they are spatially separated. But QM gives no deeper explanation than that, something which is left to the interpretations. You could have an interpretation that says that QM is a quasi-classical model that requires instant signalling in a hidden layer, but of a kind that could never be used for proper communication (so that SR would still hold). But it would only be pure speculation, since no empirical test could uncover this hidden layer.
 

Related Threads for: Comparison between quantum entanglement and a classical version

Replies
1
Views
2K
Replies
1
Views
1K
Replies
15
Views
3K
Replies
7
Views
1K
Replies
13
Views
3K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top