Comparison between quantum entanglement and a classical version

  • #51
https://en.wikipedia.org/wiki/Bell's_theorem
From wikipedia: "With the measurements oriented at intermediate angles between these basic cases, the existence of local hidden variables could agree with a linear dependence of the correlation in the angle but, according to Bell's inequality could not agree with the dependence predicted by quantum mechanical theory, namely, that the correlation is the negative cosine of the angle."
Let`s filtering data from the experiment where Alice particle is spin-up(+) and Bob particle is down (-) relative to +z-axis where Alice aligns here detector. Up to here there is no difference between the classical theory and the quantum theory. The difference of the correlation according to the Wikipedia comes from the measurement where classically, the correlation is linear with the angle of Bob detector but quantum non-locally, the correlation is a function of the negative cosine of the angle.
 
Last edited:
Physics news on Phys.org
  • #52


In this video a nice demonstration of QE and how hidden variables theory yields different results in 5/9 of randomly chosen detectors direction ( 3 directions in this video) while QM predicts only 50% different outcomes.

But suppose there exists plans of spin-pairs along all possible angles which can be represented by f(θ) provided that when Alice measures spin-up along one direction, Bob measures spin-up along the opposite direction. f(θ) will be then a hidden variable and if it extends to all θ (from 0 to 2π), then measuring spin along randomly chosen directions in both locations will give different outcomes in only 50% provided that f(θ) is also a random function ( white noise). So a hidden variable of white noise replicates the Quantum Theory prediction.!
 
Last edited:
  • #53
Adel Makram said:
But suppose there exists plans of spin-pairs along all possible angles which can be represented by f(θ) provided that when Alice measures spin-up along one direction, Bob measures spin-up along the opposite direction. f(θ) will be then a hidden variable and if it extends to all θ (from 0 to 2π), then measuring spin along randomly chosen directions in both locations will give different outcomes in only 50% provided that f(θ) is also a random function ( white noise). So a hidden variable of white noise replicates the Quantum Theory prediction.!

That's exactly what Bell's theorem proves is impossible. Letting f(\theta) be random instead of deterministic sounds like it is more general, but it actually isn't. The exact same inequalities apply, and those inequalities are violated by QM.
 
  • #54
I will go through the proof of Bell`s theorem.

But one question, is the difference in the correlation between both theories ( QM and local hidden variables) due to how the experimenter calculate the probability of measuring the spin along a given direction? (Wikipedia says that the correlation is linear with the angle in the classical theory and a function of the negative of the cosine in QT).

In other words, is the difference in the correlation due to the ways different measurements probabilities are calculated or due to superposition states of the spins?

Will the superposition of states can be virtually eliminated by filtering the data where the Alice spin (+)?. For if the correlation is still a function of the negative cosine of the angle despite that the superposition is now eliminated, then the only way to explain it is by how measuring the spin is calculated.
 
Last edited:
  • #55
Adel Makram said:
I will go through the proof of Bell`s theorem.

But one question, is the difference in the correlation between both theories ( QM and local hidden variables) due to how the experimenter calculate the probability of measuring the spin along a given direction? (Wikipedia says that the correlation is linear with the angle in the classical theory and a function of the negative of the cosine in QT).

In other words, is the difference in the correlation due to the ways different measurements probabilities are calculated or due to superposition states of the spins?

Will the superposition of states can be virtually eliminated by filtering the data where the Alice spin (+)?. For if the correlation is still a function of the negative cosine of the angle despite that the superposition is now eliminated, then the only way to explain it is by how measuring the spin is calculated.
The point of Bell's inequality is that it talks about correlations which are directly observable. Alice and Bob each measure a spin in some direction alpha, beta, and get to see an outcome +/-1. They repeat this a number of times. The experimentally observed correlation is the average of the product of the outcomes. So it is equal to the number of times that Alice and Bob's outcomes were equal, minus the number of times it was unequal, divided by the total number of repetitions. No "theory" is being used to calculate measurement probabilities. It is all about relative frequencies in many repetitions of the same experiment.
 
  • #56
gill1109 said:
The point of Bell's inequality is that it talks about correlations which are directly observable. Alice and Bob each measure a spin in some direction alpha, beta, and get to see an outcome +/-1. ...
The point is that they don't actually measure an orientation, they observe the results of an interaction and make the mistake of assuming it is a measurement of orientation. The information available from the result is a denial, not a bivalent determination. All you can say of a photon emerging from the A channel of an analyzer, is that the initial state was not a polarization state exactly aligned with B - which is a denial, all orientations that are not exactly B are possible, e.g. circular, elliptical, and linear in any direction but B.

Garden's paper presents a mathematical analysis that shows the when the information is a denial then violations of Bell's inequality are expected.
From what I can see, she actually proves that when the information obtained is a denial, then Bell's theorem does not apply.

Second, the examples presented so far begin by assuming that the information obtained from an interaction is a bivalent value, and thus ensure that Bell's theorem applies, thereby going into a circular argument. By constraining the type of 'classical ' interaction to one where the result is that the instrument samples a value, like an analog to digital converter, one constrains the problem to comply with Bell's theorem and thus make sure the a result consistent with a preconceived notion about the nature of the problem.

If you assume that evaluations are bivalent (i.e. A means NOT B) only then, do you need to create some ad-hoc non-local effect, in order to make a deterministic system reproduce the results of QM. The conclusion that non-locality is required for systems to reproduce the results of quantum mechanics is nothing more than an ad-hoc assumption to rescue the initial assumptions about the nature of the information obtained from a 'classical' interaction.
 
  • #57
Nugatory said:
I don't know that I'd say the theorem "fails", as the theorem claims to and does preclude a large class of theories: informally, we say that it precludes all "local realistic hidden variable theories". This bivalence property, along with counterfactual definiteness, is part of what the informal speakers mean when they informally say "realistic", so the fact that rejecting bivalence allows the inequalities to be violated is consistent with Bell's theorem.

Thus (unless I'm misunderstanding your argument) you have successfully demolished the straw man claim that Bell's theorem requires rejecting locality but left the claim that is actually being made, namely that Bell's theorem requires rejecting at least one of locality and the complex of properties that we informally call "realism", untouched.I've seen many. For a trivial example, consider the hypothesis that when Alice makes her measurement, a superluminal pixie is created, and this pixie travels to Bob's in-flight particle and twists its spin to point opposite to whatever Alice measured. I've deliberately chosen this example to be absurd, but it is consistent with the experimental results, and it gets that way by being explicitly non-local - experiments have shown that normal subluminal pixies won't work when the detection events are spacelike separated.
All your pixie is, is an ad-hoc mechanism, an unfalsifiable 'epicycle', not actually consistent with either classical or quantum theory. Which proves my point, the only device you can think of is some contrivance, nobody has come up with something any more explicit. To claim that the magic pixie, and other contrivances, is explicit is to do nothing more concocting a device to override the action of supposed hidden variables, and magically make the system do whatever QM does, 'magically' doesn't cut it as explicit science.

My point is that you are forced to create ad-hoc mechanisms if Bell's theorem is assumed to apply universally to all possible classes of interaction model. However when you examine Bell's theorem you find it depends on making assumptions as to the nature of the information that is obtained from observing an interaction - that the information is a bivalent determination. The paper I quoted (Garden) shows that when the information available is a denial (which is consistent with the problem) then Bell's theorem fails to apply.
Experiments confirm that QM's predictions are correct. The ad-hoc mechanisms would only rescue the type of 'classical' models where a measurement results in a bivalent determination - the conclusion you can reach from Bell's theorem is that bivalent assessments from interactions are ruled out.

Rachel Garden's paper is anything but informal, it explicitly proves that the assumption of bivalence is necessary to prove Bell's theorem. The real "straw man"
is the insistence of bivalence, without it you cannot prove Bell's theorem. Bivalence is a necessary condition and supposition about magic pixies is not.
 
  • #58
jknm said:
Rachel Garden's paper is anything but informal, it explicitly proves that the assumption of bivalence is necessary to prove Bell's theorem. The real "straw man" is the insistence of bivalence, without it you cannot prove Bell's theorem. Bivalence is a necessary condition and supposition about magic pixies is not.

I can't make a judgment about Garden's paper, since I can't read it, but your description of it sounds completely wrong.
 
  • #59
jknm said:
explicitly proves that the assumption of bivalence is necessary to prove Bell's theorem.
Of course it is, but who is arguing otherwise? Bell's theorem can be succinctly stated as "If you make certain assumptions, then a particular inequality will hold". The theorem is important and interesting because experiments have shown that this inequality is violated, so therefore we can reasonably conclude that one or more of the assumptions is false.

Those assumptions include locality (in the sense that the results of a measurement cannot be influenced by events outside of the past light cone of the measurement event), counterfactual definiteness, and the property we're calling "bivalence" in this thread. So there you have it: At least one of the assumptions of locality, counterfactual definiteness, and "bivalence" does not match the way the universe works. There is room for interesting discussions about which of these assumptions are to be rejected, but you have to accept the correctness of the theorem before you have a logical basis for rejecting any of them.

(Conversely, if you want to show that Bell's theorem has "failed" you would prove, presumably by example, the existence of something that the theorem says is impossible: a theory that is local and "bivalent" and counterfactually definite. Garden's paper , as you're describing it, doesn't do that; instead it shows that rejecting bivalence is one way of reconciling experimental results with the correctness of Bell's theorem).
 
  • Like
Likes Mentz114
  • #60
jknm said:
The point is that they don't actually measure an orientation, they observe the results of an interaction and make the mistake of assuming it is a measurement of orientation. The information available from the result is a denial, not a bivalent determination. All you can say of a photon emerging from the A channel of an analyzer, is that the initial state was not a polarization state exactly aligned with B - which is a denial, all orientations that are not exactly B are possible, e.g. circular, elliptical, and linear in any direction but B.
Alice and Bob each toss a fair coin and set a setting on a measurement device to one of two possible values. Something happens inside a black box, and out comes a binary outcome which we conventionally encode as +/-1. Later we correlate the four streams of binary values: setting Alice, setting Bob, outcome Alice, outcome Bob.

There is no assumption about any orientations at all. There is no assumption about polarization.

In the recent experiment http://arxiv.org/abs/1508.05949 the black box contains a Nitrogen-Vacancy imperfection in a diamond ...

By the way anyone who would like to see the paper by Rachel Wallace Garden, please find my email address on internet and send me a regular email.

Please notice: the bivalence is *enforced* by the experimental design (if we are talking about a loophole-free CHSH-Bell type experiment). It is a macroscopic feature of the laboratory arrangement. It is not an optional assumption about the underlying physics.
 
  • #61
gill1109 said:
No "theory" is being used to calculate measurement probabilities. It is all about relative frequencies in many repetitions of the same experiment.

But in order to theoretically explain the correlation between different outcomes, a theorem must exist about how the measurement probability is calculated.

post #16
gill1109 said:
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}).
QT allows Bob to measure his particle in (+) direction with with probability cos^2(\frac{\beta - \phi}{2}). But I don't know how Bob would measure it according to the classical hidden variables theory which is for me a decisive point in understanding Bell`s theorem.
 
  • #62
Adel Makram said:
QT allows Bob to measure his particle in (+) direction with with probabilit). But I don't know how Bob would measure it according to the classical hidden variables theory which is for me a decisive point in understanding Bell`s theorem.

Alice sends her particle through her Stern-Gerlach device, Bob sends his particle through his Stern-Gerlach device, and they each write down the direction in which the particle was deflected. That's the measurement part, and it's not done according to any theory - we're just gathering data, to see if it matches the predictions made by the various theories. After they've done a large number of pairs, they get together and compare notes, see whether the correlations in their two lists of measurements violate Bell's inequality.
(This is also the first time that the ##\sin^2\frac{\alpha-\beta}{2}## rule will appear - they didn't need it to make their measurements).

If the inequality is violated, then the experiment we've just done tells us that we can reject any local realistic hidden variable theorem, because Bell's theorem shows that no local realistic hidden variable theory can produce results that violate the inequality.
 
  • Like
Likes Adel Makram
  • #63
I mean how to derive the dependence of correlations ( quantum and classical) on the angle between the detectors.
 
  • #64
Adel Makram said:
I mean how to derive the dependence of correlations ( quantum and classical) on the angle between the detectors.
According to classical theory, many different correlation functions are possible. According to quantum theory, many *more* correlation functions are possible. Much more. Quantum theory is richer than classical theory.

I wrote a short paper exploring what are the possible correlation functions according to classical theory: http://arxiv.org/abs/1312.6403
 
  • Like
Likes Adel Makram
  • #65
  • #67
There is an unstated principle in Bell's derivation, which is known as Reichenbach's Common Cause Principle. Basically, this says that if two random variables are correlated, then there must be a common cause for each. Mathematically, if you have two contingent events A and B, and

P(A \& B) \neq P(A) \cdot P(B)

then there must be some cause C in the common causal past of A and B such that

P(A \& B | C) = P(A | C) \cdot P(B | C)

In other words, if you knew all the facts in the past relevant to A and B, then the probabilities would factor. This gives rise to Bell's assumption about hidden variables:

P(A \& B) = \int P(\lambda)\ d\lambda\ P(A, \lambda) \cdot P(B, \lambda)

My understanding is that this principle isn't logically necessary, although it is tacitly assumed in almost all reasoning about cause and effect.
 
  • Like
Likes Adel Makram
  • #68
Adel Makram said:
I mean how to derive the dependence of correlations ( quantum and classical) on the angle between the detectors.

You'll find the calculation of the correlation predicted by quantum mechanics in many texts (In fact, it looks like Mentz114 and Heinera posted examples while I was composing this!).

You won't find a calculation of the correlation predicted by "the classical theory" anywhere, because there is no single "the classical theory" in this conversation. What you will find, in Bell's paper, is the proof that any theory that does not allow Bob's result to depend on Alice's choice of direction must predict a correlation that obeys Bell's inequality. The term "classical theory" means any such theory; we don't need any specific example to be able to follow Bell's argument that all such theories must obey the inequality. (This is analogous to the way that I can prove that all right triangles obey the Pythagorean theorem without having to talk about any specific right triangle - it may help me understand if I have some specific examples of right triangles that I can look at to see the Pythagorean theorem in action, but it's not necessary for the proof).
 
  • #69
stevendaryl said:
There is an unstated principle in Bell's derivation, which is known as Reichenbach's Common Cause Principle.
Although it is not explicitly stated in his original derivation, it is also not exactly cunningly and subtly concealed :smile:. He does cover this issue in his later writings.

The history is somewhat relevant here. When Bell first published, the question on the table was whether QM was incomplete in the sense that the EPR authors meant. Thus, Bell's starting point was informally "Let's assume properties that would have satisfied the EPR authors...", and it's clear that the assumptions in his original paper met that requirement. Only after Bell's somewhat shocking result (and the experimental confirmation of inequality violations) was digested did people start putting serious energy into identifying precisely what those assumptions were. At some point along the way, the conversation shifted from the initial conclusion ("Sorry EPR - we know what you want and why you want it, but it doesn't exist") to a more rigorous specification of exactly which classes of theories are precluded by Bell's theorem and the experimental discovery of inequality violations.
 
  • #70
The difference between classical and quantum entanglement is simply that in the quantum case the state of either top is unknown until it is measured whereas in the classical case it is known all along.
 
  • #71
I am really going to try and get this thread, but I still don't get how if you know there are a blue ball and a red ball in a box and you blindly remove the red one then with certainty you know the other one is blue once you know you have red in your hand.

How is this different to spin up spin down coupled particles a galaxy distance apart?
 
  • #72
houlahound said:
I am really going to try and get this thread, but I still don't get how if you know there are a blue ball and a red ball in a box and you blindly remove the red one then with certainty you know the other one is blue once you know you have red in your hand.

How is this different to spin up spin down coupled particles a galaxy distance apart?

The difference is that in the quantum case (if we think of the balls as being quantum particles - please note that this is an analogy):

1. None of the two balls seems to have a definitive color before it is measured, instead they both have a "mix" of colors, and
2. The balls can have any color "in between" blue and red, and
3. We can not predict which ball will turn up as mostly "blue" and which ball will turn up as mostly "red", so to speak.
 
  • #73
houlahound said:
I am really going to try and get this thread, but I still don't get how if you know there are a blue ball and a red ball in a box and you blindly remove the red one then with certainty you know the other one is blue once you know you have red in your hand.

How is this different to spin up spin down coupled particles a galaxy distance apart?

There's a fairly simple example game that shows the difference. Imagine instead that I give you and a friend (Bob) each a box with two drawers (drawers 1 and 2) that you can open. You and Bob can only open one of the drawers (when you open a drawer, anything in the other drawer is burned) and when you do you find either a red ball or a blue ball. You believe (or check, by playing this game many times) the following things:
  • You (and Bob) will get either a red ball or a blue ball with 50% probability.
  • If you open drawer 1 on your box and Bob opens drawer 1 on his box, you both find the same colour ball (i.e., you and Bob both get red or or both get blue).
  • If you open drawer 1 and Bob opens drawer 2, you both find the same colour ball.
  • If you open drawer 2 and Bob opens drawer 1, you both find the same colour ball.
  • If you open drawer 2 and Bob opens drawer 2, you find opposite coloured balls (i.e, you get red and Bob gets blue, or vice versa).
Problem: how do I put balls in the drawers, without knowing in advance which ones you and Bob will decide to open, in order to guarantee the results I've just described?

Quantum mechanics predicts correlations that are of the same character (though not quite as extreme) as this.
 
Last edited:
  • #74
stevendaryl said:
Later, Bob measures "spin-up" with probability \frac{1}{2}(1 + \vec{\beta} \cdot \vec{S_B}).
Do you mean spin-down? as the probability of the spin-up is \frac{1}{2}(1 - \vec{\beta} \cdot \vec{S_B}) = ## sin^2(\theta/2)## where ##\theta## is the angle between the two detectors. Also spin up means the direction of spin is along the direction of Bob`s detector.
 
  • #75
Adel Makram said:
Do you mean spin-down? as the probability of the spin-up is \frac{1}{2}(1 - \vec{\beta} \cdot \vec{S_B}) = ## sin^2(\theta/2)## where ##\theta## is the angle between the two detectors. Also spin up means the direction of spin is along the direction of Bob`s detector.

No. The way that I was defining things, \vec{\beta} is the orientation of Bob's detector, and \vec{S_B} is the spin state of Bob's particle. If \vec{\beta} = \vec{S_B}, then Bob measures spin-up with 100% probability.

In the "collapse" interpretation, if Alice measures spin-up along direction \vec{\alpha}, then \vec{S_B} "collapses" so that \vec{S_B} = -\vec{\alpha}. So in that case, \frac{1}{2}(1+\vec{\beta} \cdot \vec{S_B}) = \frac{1}{2}(1-\vec{\beta} \cdot \vec{\alpha}).
 
  • #76
stevendaryl said:
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 \pm 1, with 50/50 probability of each outcome.
  2. If Alice measures +1 at direction \alpha, then Bob's particle "collapses" to the state with spin direction \phi = \alpha - \pi.
  3. If Alice measures -1 at direction \alpha, then Bob's particle "collapses" to the state with spin direction \phi = \alpha.
  4. 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]
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:
  • #77
morrobay said:
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.
 
  • #78
Nugatory said:
##\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:
  • #79
morrobay said:
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.
 
  • #80
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

  • Ratio.png
    Ratio.png
    93.4 KB · Views: 446
Last edited:
  • #81
Adel Makram said:
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.
 
  • #82
Nugatory said:
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?
 
  • #83
Adel Makram said:
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.
 
  • Like
Likes Adel Makram
  • #84
Nugatory said:
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.

stevendaryl said:
No. The way that I was defining things, \vec{\beta} is the orientation of Bob's detector, and \vec{S_B} is the spin state of Bob's particle. If \vec{\beta} = \vec{S_B}, then Bob measures spin-up with 100% probability.

In the "collapse" interpretation, if Alice measures spin-up along direction \vec{\alpha}, then \vec{S_B} "collapses" so that \vec{S_B} = -\vec{\alpha}. So in that case, \frac{1}{2}(1+\vec{\beta} \cdot \vec{S_B}) = \frac{1}{2}(1-\vec{\beta} \cdot \vec{\alpha}).

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 ?
 
  • #85
morrobay said:
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:
  • Like
Likes morrobay
  • #86
morrobay said:
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.
 
  • Like
Likes morrobay
  • #87
Nugatory said:
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:
  • #88
Adel Makram said:
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.
 
  • Like
Likes Adel Makram and Mentz114
  • #89
Heinera said:
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.
 
  • #90
Adel Makram said:
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"?
 
  • #91
Adel Makram said:
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.
 
  • #92
Nugatory said:
There is no EPR "paradox" - no apparent contradictions appear anywhere in the EPR argument.
https://en.wikipedia.org/wiki/EPR_paradox

So in your opinion, how to explain the outcomes of Alice Bob spins-experiment and quantum correlation based on causality or based on anything?
 
  • #93
Heinera said:
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.
 
  • #94
Adel Makram said:
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.
 
  • #95
Heinera said:
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.
 
  • #96
Adel Makram said:
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).
 
  • #97
Adel Makram said:
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.
 
  • #98
Heinera said:
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?
 
  • #99
Adel Makram said:
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'?
 
  • #100
Adel Makram said:
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.
 
  • Like
Likes Mentz114

Similar threads

Replies
5
Views
1K
Replies
4
Views
3K
Replies
4
Views
1K
Replies
10
Views
2K
Replies
3
Views
1K
Replies
9
Views
1K
Replies
19
Views
2K
Replies
5
Views
2K
Back
Top