Binney's interpretation of Violation of Bell Inequalities

[TrickyDicky]

This indeed is innocuous assumption. The argument is philosophical but there is not much that can be done about it as you are questioning philosophical assumption.
Practicing scientific inquiry requires that there is such a thing as "the state of knowledge at a particular time" and because our knowledge is recorded in physical matter it follows that practicing scientific inquiry requires that there is such a thing as "the state of the universe at a particular time". Do you agree?

One must define mathematically both state and particular time, this is the not innocuous part.

 

[zonde]

I don't agree, as that would require knowing that time is fundamental in all future scientific theories. However, locally causal theories are a significant class of theories. Indeed, it is often said that relativity is a theory of causality. Relativistic quantum theories show that that is not neccessarily the case.

Not sure I follow. How you can talk about future scientific theories if you don't have some idea about time that is independent from these scientific theories.

 

[TrickyDicky]

I don't agree, as that would require knowing that time is fundamental in all future scientific theories. However, locally causal theories are a significant class of theories. Indeed, it is often said that relativity is a theory of causality. Relativistic quantum theories show that that is not neccessarily the case.

Isn't QFT locally causal?

 

[stevendaryl]

I'm not looking for any alternative, just reacting to what you said.
You wrote: "Bell's starting point is the assumption that there is such a thing as "the state of the universe at a particular time", whether or not we have complete knowledge of that state. That seems like an innocuous enough assumption, independent of the details of our physical theories." And that:"that assumption, together with the predictions of quantum mechanics for EPR type experiments lead to the conclusion that there are nonlocal interactions connecting distant parts of the universe"
What I'm saying is that the way that assumption is formalized is the formalism of QM

I don't agree with that. QM is not a way of describing the state of the universe. Or at least, the claim that it describes the state of the universe is controversial. In the Many-Worlds Interpretation, the QM state is interpreted as the state of the universe. But in other interpretations, it is not considered to necessarily be an object fact about the universe.

I showed in #46 that abstracting from that assumption one can explain operationally the results of those experiments just using the non-commutative algebra of QM.

I would say that QM describes the results, but I wouldn't say that it explains them. It tells how to calculate them.

 

[TrickyDicky]

I don't agree with that. QM is not a way of describing the state of the universe. Or at least, the claim that it describes the state of the universe is controversial. In the Many-Worlds Interpretation, the QM state is interpreted as the state of the universe. But in other interpretations, it is not considered to necessarily be an object fact about the universe.

No, I was referring to defining a state of a quantum system as an element of a Hilbert space, not necessarily to the state of the universe as discussed in MW, etc.

 

[stevendaryl]

No, I was referring to defining a state of a quantum system as an element of a Hilbert space, not necessarily to the state of the universe as discussed in MW, etc.

Well, that sounds like the opposite of what you were saying. I thought you were saying that Hilbert space was the QM way of formalizing the state of the universe at a given moment.

 

[atyy]

Isn't QFT locally causal?

It depends on what one means by locally causal. In terms of the classical idea of relativity as a theory of causality, QFT is not locally causal.

 

[TrickyDicky]

Well, that sounds like the opposite of what you were saying. I thought you were saying that Hilbert space was the QM way of formalizing the state of the universe at a given moment.

Of formalizing the assumption that a system(the universe, whatever) is in a certain state at a certain time, this is QM's first postulate, and I'd say it covers the assumption you mentioned above as the starting point of Bell's theorem.

 

[TrickyDicky]

It depends on what one means by locally causal. In terms of the classical idea of relativity as a theory of causality, QFT is not locally causal.

The classical idea of SR as a theory of causality was global, not local(relativity of simultaneity), that idea must have surely been preserved in QFT?

 

[atyy]

The classical idea of SR as a theory of causality was global, not local(relativity of simultaneity), that idea must have surely been preserved in QFT?

The classical idea of causality in SR is that the causes of an event are entirely in its past light cone - that idea is not preserved in QFT. What is preserved in QFT is a weaker sense of causality - that classical information cannot be sent faster than light.

 

[stevendaryl]

Of formalizing the assumption that a system(the universe, whatever) is in a certain state at a certain time, this is QM's first postulate, and I'd say it covers the assumption you mentioned above as the starting point of Bell's theorem.

Well, I disagree. I don't think that the QM state can be interpreted as the "state of the universe at a certain time", precisely because of scenarios such as EPR. Initially, you have the state in a superposition |\psi\rangle = \frac{1}{\sqrt{2}}(|U D\rangle - |D U \rangle). Alice measures the spin of the first particle and finds it spin-up. So what is the state of the system now? If you say that it collapses to |U D\rangle, well that's a nonlocal change. If you say that it doesn't collapse, then you have Many Worlds.

 

[Derek Potter]

Well, I disagree. I don't think that the QM state can be interpreted as the "state of the universe at a certain time", precisely because of scenarios such as EPR. Initially, you have the state in a superposition |\psi\rangle = \frac{1}{\sqrt{2}}(|U D\rangle - |D U \rangle). Alice measures the spin of the first particle and finds it spin-up. So what is the state of the system now? If you say that it collapses to |U D\rangle, well that's a nonlocal change. If you say that it doesn't collapse, then you have Many Worlds.

Many Worlds as popularized depicts it as parallel universes which certainly makes THE state of THE universe pretty meaningless. But in the improved version of Relative State Formulation there remains just one universe in a pure state. The pure state can, of course be decomposed into a superposition of orthogonal observer-worlds. EPR is explained locally in such a model.

 

[Derek Potter]

The classical idea of causality in SR is that the causes of an event are entirely in its past light cone - that idea is not preserved in QFT. What is preserved in QFT is a weaker sense of causality - that classical information cannot be sent faster than light.

Can QFT can be reformulated in MW terms so that it remains local and SR-causal?

 

[TrickyDicky]

Well, I disagree. I don't think that the QM state can be interpreted as the "state of the universe at a certain time",

I said state of the system, it's anyone's choice what that system is.

precisely because of scenarios such as EPR. Initially, you have the state in a superposition |\psi\rangle = \frac{1}{\sqrt{2}}(|U D\rangle - |D U \rangle). Alice measures the spin of the first particle and finds it spin-up. So what is the state of the system now? If you say that it collapses to |U D\rangle, well that's a nonlocal change.

You are calling this collapse nonlocal, I prefer to call it simply non-unitary, because as I explained in #46 its consequences on entangled two-states systems can be described without any reference to locality, reality or causality. And yes this is in contradiction with the rest of the formalism as everyone knows.
Also note that the original EPR involved entanglement of continuous observables that have been shown(see Popper's experiment) not to exhibit correlations contrary to what happens in the two-state systems like spin or polarization that Bell's theorem exemplify.

 

[Pat71]

Also note that the original EPR involved entanglement of continuous observables that have been shown(see Popper's experiment) not to exhibit correlations contrary to what happens in the two-state systems like spin or polarization that Bell's theorem exemplify.

Doesn't Strekalov's "Ghost Interference" experiment of 1995 counter this claim? PRL74p3600.pdf

"...Another experiment which (unknowingly) implements Popper's test in a conclusive way, has actually been carried out. Its results are in contradiction with Popper's prediction..." (Tabish Qureshi, "Analysis of Popper's experiment and its realization," 2012): ptp.oxfordjournals.org/content/127/4/645.short

Or, similarly: www.arxiv.org/abs/1206.1432

 

[atyy]

Can QFT can be reformulated in MW terms so that it remains local and SR-causal?

I don't know. However, the Many-Worlds interpretation certainly escapes being nonlocal via the Bell theorem, because the Bell theorem assumes that only one outcome is realized. That is among the well-known loopholes including superdeterminism and retrocausation, which I usually don't mention just to be concise.

 

[TrickyDicky]

Doesn't Strekalov's "Ghost Interference" experiment of 1995 counter this claim? PRL74p3600.pdf

"...Another experiment which (unknowingly) implements Popper's test in a conclusive way, has actually been carried out. Its results are in contradiction with Popper's prediction..." (Tabish Qureshi, "Analysis of Popper's experiment and its realization," 2012): ptp.oxfordjournals.org/content/127/4/645.short

Or, similarly: www.arxiv.org/abs/1206.1432

No, it counters Popper's original claim that QM's indeterminacy(HUP) was falsified by a certain outcome of his thought experiment. Both Shih's 1999 and more recent 2015(see http://phys.org/news/2015-01-popper-againbut.html) papers agreed on this too.

 

[TrickyDicky]

It's not clear to me either, and I'm interested in all possible alternatives to explaining EPR. If anything, you're quite laconic, can you point to a paper which shares your views? Or if you really have a personal new idea, but maybe don't have enough skills to fully expand on it, at least lay down a formal pointer to your interpretation in which noncommutative algebra in some way eliminates the ftl action problem?

Edited scenario as the initial description allowed confusion

That correlations of two-state systems is unrelated to locality considerations can be easily shown with this example: Take a book and call the cover the +> state and the back cover the -> state when the book is oriented parallel to the x axis, and similarly when it is oriented parallel to the y and z axes, now you can truly claim that after an arbitrary 3-dimensional rotation of the book(involving rotations in different axes) the +> and -> states when measured in the diffrent orientations for x, and z axes are correlated in a way that violates Bell's inequalities. No so called "local realistic hidden variables theory" can explain the correlation of the cover and the back of the book. Spooky?

 

[Derek Potter]

the up> and down> states are correlated in a way that violates Bell's inequalities. No so called "local realistic hidden variables theory" can explain the correlation of the cover and the back of the book. Spooky?

Neither spooky nor true. 100% anti-correlation does not violate the Bell inequality.

 

[stevendaryl]

You are calling this collapse nonlocal, I prefer to call it simply non-unitary, because as I explained in #46 its consequences on entangled two-states systems can be described without any reference to locality, reality or causality.

If the state collapses for Bob as well as for Alice when Alice performs her measurements, then the collapse is nonlocal, affecting distant parts of space simultaneously. If the state only collapses for Alice, then it can't be a state of the universe, it must be relative to the observer.

 

[stevendaryl]

That correlations of two-state systems is unrelated to locality considerations can be easily shown with this example: Take a book and call the cover the up> state and the back cover the down> state, now you can truly claim that after an arbitrary 3-dimensional rotation of the book(involving rotations in different axes) the up> and down> states are correlated in a way that violates Bell's inequalities. No so called "local realistic hidden variables theory" can explain the correlation of the cover and the back of the book. Spooky?

What you're describing IS a local realistic hidden variables theory, and it doesn't violate Bell's inequalities.

 

[Derek Potter]

Hardly hidden !

 

[stevendaryl]

What you're describing IS a local realistic hidden variables theory, and it doesn't violate Bell's inequalities.

To expand on this, Bell's inequalities involve two adjustable parameters, \alpha and \beta, which can take on values a_1, a_2, b_1, b_2 and two possible measurements A and B, each of which produces a result \pm 1. You perform measurements at a variety of settings in order to come up with 4 numbers:

1. E(a_1, b_1)
2. E(a_1, b_2)
3. E(a_2, b_1)
4. E(a_2, b_2)

where E(a,b) is the average value of the product AB of measurement results, restricted to those measurements for which \alpha = a and \beta = b. Locality comes into play when you have a situation where the settings \alpha and \beta that can be made independently at spacelike separations, and where the measurements A and B also take place at spacelike separations. In that situation, Bell showed that a local realistic theory predicts a certain inequality holding among the 4 numbers.

In your (Tricky Dicky's) toy model, it's clear that you are interpreting the measurement as seeing the front versus back of the book. But to apply Bell's inequality to the model, you have to two independent parameters \alpha and \beta that can be chosen at a spacelike separation. Presumably these are supposed to be the rotations of the book along two different axes? But that can't be done independently, at a spacelike separation.

So I don't see how your toy model relates to Bell's inequality at all.

 

[TrickyDicky]

Neither spooky nor true. 100% anti-correlation does not violate the Bell inequality.

See edited scenario, there is no 100% anti-correlation. The example just shows that rotations in 3D are non-commutative and the Bell inequalities follow a commutative algebra so they must be logically violated by non-commutative rotations.

 

[TrickyDicky]

In your (Tricky Dicky's) toy model, it's clear that you are interpreting the measurement as seeing the front versus back of the book. But to apply Bell's inequality to the model, you have to two independent parameters \alpha and \beta that can be chosen at a spacelike separation. Presumably these are supposed to be the rotations of the book along two different axes? But that can't be done independently, at a spacelike separation.

So I don't see how your toy model relates to Bell's inequality at all.

The math of the theorem just refers to outcomes and whether their correlation can be explained with the inequalities or not with Bell inequalities.
Would you see how the toy model relates to Bell if I allowed for the book to have arbitrary spacelike thickness(ignoring problems with born rigidity for the sake of the thought experiment as Bell's theorem doesn't have it as assumption)?

 

[stevendaryl]

There is no 100% anti-correlation, of course one has to allow to you can correlate the cover in two different rotated axes
See edited scenario, there is no 100% anti-correlation. The example just shows that rotations in 3D are non-commutative and the Bell inequalities follow a commutative algebra so they must be logically violated by non-commutative rotations.

As I said, Bell's inequality was developed for a scenario in which there are two adjustable parameters \alpha and \beta that can be made independently at a spacelike separation, and two measurements A and B that can also be made at a spacelike separation. I don't see how your toy model has anything to do with Bell's inequality. What are the adjustable parameters? If they are the choices of rotation angles along two different axes, then they certainly cannot be made independently at a spacelike separation.

I also don't know what you mean by saying that "Bell's inequalities follow a commutative algebra". I don't see that there is algebra involved, at all. Where in the derivation of his inequality is there an assumption made about commutativity?

 

[TrickyDicky]

I edited #84 and answered in #85 while you posted.

 

[stevendaryl]

The math of the theorem just refers to outcomes and whether their correlation can be explained with the inequalities or not with Bell inequalities.
Would you see how the toy model relates to Bell if I allowed for the book to have arbitrary spacelike thickness(ignoring problems with born rigidity for the sake of the thought experiment as Bell's theorem doesn't have it as assumption)?

I could see how it relates to Bell if you could point out two adjustable parameters that can be made independently at a spacelike separation.

His inequalities are about a situation where you have two observers, far from each other, traditionally called "Alice" and "Bob" (by me, anyway).

1. Alice chooses a detector setting, either a_1 or a_2
2. Alice performs a measurement, getting result A = +1 or A = -1
3. Meanwhile, far away, Bob chooses a detector setting, either b_1 or b_2
4. He gets a measurement result B = \pm 1

You perform the experiment many times, and get statistics for E(a,b),which is the average of the product AB of the two results, restricted to those trials where Alice chooses a and Bob chooses b

So, if you want to relate your toy model to Bell, you need to say what are the settings that Alice and Bob choose. In order to say that your toy model violates Bell's inequality, you need to compute the values E(a,b) for your model.

 

[TrickyDicky]

As I said, Bell's inequality was developed for a scenario in which there are two adjustable parameters \alpha and \beta that can be made independently at a spacelike separation, and two measurements A and B that can also be made at a spacelike separation. I don't see how your toy model has anything to do with Bell's inequality. What are the adjustable parameters? If they are the choices of rotation angles along two different axes, then they certainly cannot be made independently at a spacelike separation.

I also don't know what you mean by saying that "Bell's inequalities follow a commutative algebra". I don't see that there is algebra involved, at all. Where in the derivation of his inequality is there an assumption made about commutativity?

Axes x, y and z are spacelike so measurement referred to them are independent of spacelike separation.
The Bell's inequalities imply that + and - outcomes are related by commutative relations.

 

[TrickyDicky]

These are the inequalities:Number of <+ <+ results with apparatus A on x and apparatus B on y ≤ number of <+ <+ results with apparatus A on x and apparatus B on z plus number of <+< - result of apparatus A on y and apparatus B on z.

I call apparatus A and B Alice and Bob respectively simultaneously ascertaining if they are seeing the back or the cover of the book.