Commonsense local realism refutes Bell's theorem

[gill1109]

As for my proposed experiment, you and Gill should really read the two relevant papers before going about spewing nonsense about my work all over the internet.

As far as I know, Michel hasn't read these papers yet. He did have serious objections to calculating all the correlations on the same set of particles.

Back on topic, Watson's paper ...

There is a lot of misunderstanding about the equation

E(a, b) = int A(a, lambda) B(b, lambda) rho(lambda) d lambda.

Part of the problem is writing the integral using the notation appropriate when lambda is an element of Euclidean space and rho(lambda) a probability density function (typical physics notation of 50 years ago). Today, a mathematician might write

E(a, b) = int A(a, lambda) B(b, lambda) dP(lambda)

where P is a probability measure on a measurable space.

But anyway, this key equation is not a *definition* but a theorem, though a rather basic and easy theorem, belonging to the theory of local hidden variables and to classical probability theory (whether Bayesian or frequentist).

E(a, b) stands for the mean value of products of outcomes of measurements of spin in the directions a and b of infinitely many similarly prepared pairs of particles. ("State" = "preparation"). (I take a frequentist interpretation of probability here. If you have a Bayesian interpretation that is fine: the mean of a variable is just the fair price of a lottery ticket whose prize is the outcome of the variable. Probability is your uncertainty as to its value, prior to the experiment in which the variable is observed once).

The integral on the right hand side is the result of (1) assuming a local hidden variables model and (2) the law of large numbers (if you are a frequentist; if you are a Bayesian you have another theorem to take care of this basic building block of the theory: the relation between expectations and probabilities).

The local hidden variables theory says that preparing a new pair of particles and measuring a and b is like: picking lambda at random according to the probability measure P; getting to see the values A(a, lambda) and B(b, lambda).

Given functions A and B and a probability measure the mathematician is free to study a new composed function such as
A(a, lambda)B(b, lambda) - A(a, lambda)B(b', lambda) + ... + ... and integrate over lambda with respect to the probsbility measure P. Because this new function is everywhere -2 or +2, its average lies in between those bounds.

There is no suggestion anywhere of measuring different things at the same time. There are some functions, there is a trivial logical bound, there is some calculus and writing an integral of a sum as a sum of integrals.

No voodoo, just plain model + calculus. If you believe the model represents reality, then model deductions about things that are observable in some way or another, should fit too. If they don't fit, the model is inappropriate.

It's all rather simple as long as one careful distinguishes physical reality from mathematical models of parts of it.

[minkwe]

Tsirelson's beautiful inequality is indeed in a sense "just geometry" but I think that if you look at the proof you will see it is a little harder than you think. After all, the Hilbert spaces can have arbitrary dimension. It is not a theorem about spin half particles.

There is no Hilbert space necessary to derive 2 sqrt 2. The proof is a lot easier than you think. If you want I can show you how it is done (in a different thread), much simpler without any Hilbert space confusion, maybe you will have another eureka moment from it.

The shortest proof I know is this one
https://en.wikipedia.org/wiki/Tsirelson's_bound#Tsirelson_bound_for_the_CHSH_inequality
I would be surprised if you can beat that ...

Richard, I see that you are frantically trying to drown the fact that I just proved you were confused and mixing strongly objective and weakly objective interpretations of expectation values. All the rambling about long litigated and settled issues won't change the fact that you mix incompatible interpretations errorneously. I will simply ignore those ramblings as I have better things to do right now.

Before I take my break however, as concerns Tsirelson's bound:

[1] S = E(a − b) + E(a − b') + E(a' − b) − E(a' − b')
Can be written as
[2] S = f (x) + f (y) + f (z) − f (w), w, x, y, z ∈ [0, 2π], and −1 ≤ f(.) ≤ 1

If where w, x, y and z were independent variables, the upper bound of S would be 4 as we've established elsewhere. But inspection of [1] shows that they are not because we can express w in terms of the other three variables w = y + z -x to get

[3] S = f (x) + f (y) + f (z) − f (y + z − x)

At the extrema of S, the partial derivatives of S are all zero:
f'(x) + f'(y + z − x) = f'(y) − f'(y + z − x) = f'(z) − f'(y + z − x) = 0, therefore y = z and f'(y) = f'(z) = -f'(x). Substituting in [3] we get

[4] Se = f(x) + 2f(y) − f(2y − x)

For even f(.), f'(x) = f'(-x) = f'(y). At the extrema therefore, y = -x and we can reduce [4] further to

[5] Se = 3f(y) - f(3y)

Therefore the maximum of S = cos(a, b) + cos(a, b') + cos(a', b) - cos(a', b') is also the maximum of the function

3cos(y) - cos(3y)

Which is 2 sqrt 2. http://www.wolframalpha.com/input/?i=Ma ... os%283x%29

Like I told you, there is nothing quantum about Tsirelson's bound, nor is any Hilbert space involved.

[gill1109]

... as concerns Tsirelson's bound:

[1] S = E(a − b) + E(a − b') + E(a' − b) − E(a' − b')
Can be written as
[2] S = f (x) + f (y) + f (z) − f (w), w, x, y, z ∈ [0, 2π], and −1 ≤ f(.) ≤ 1

If where w, x, y and z were independent variables, the upper bound of S would be 4 as we've established elsewhere. But inspection of [1] shows that they are not because we can express w in terms of the other three variables w = y + z -x to get

[3] S = f (x) + f (y) + f (z) − f (y + z − x)

At the extrema of S, the partial derivatives of S are all zero:
f'(x) + f'(y + z − x) = f'(y) − f'(y + z − x) = f'(z) − f'(y + z − x) = 0, therefore y = z and f'(y) = f'(z) = -f'(x). Substituting in [3] we get

[4] Se = f(x) + 2f(y) − f(2y − x)

For even f(.), f'(x) = f'(-x) = f'(y). At the extrema therefore, y = -x and we can reduce [4] further to

[5] Se = 3f(y) - f(3y)

Therefore the maximum of S = cos(a, b) + cos(a, b') + cos(a', b) - cos(a', b') is also the maximum of the function

3cos(y) - cos(3y)

Which is 2 sqrt 2. http://www.wolframalpha.com/input/?i=Ma ... os%283x%29

Like I told you, there is nothing quantum about Tsirelson's bound, nor is any Hilbert space involved.

Tsirelson's bound is a bound on four "correlations" defined as follows:

E(a, b) := trace(rho X(a) otimes Y(b)) where rho is a joint state of a bipartite quantum system (a trace 1 non-negative self-adjoint operator) and X(a) and Y(b) are two self-adjoint operators on the two subsystems each with spectrum contained in {-1, +1}.

Perhaps you should read Tsirelson's paper to find out what it is actually all about.

[minkwe]

... as concerns Tsirelson's bound:

[1] S = E(a − b) + E(a − b') + E(a' − b) − E(a' − b')
Can be written as
[2] S = f (x) + f (y) + f (z) − f (w), w, x, y, z ∈ [0, 2π], and −1 ≤ f(.) ≤ 1

If where w, x, y and z were independent variables, the upper bound of S would be 4 as we've established elsewhere. But inspection of [1] shows that they are not because we can express w in terms of the other three variables w = y + z -x to get

[3] S = f (x) + f (y) + f (z) − f (y + z − x)

At the extrema of S, the partial derivatives of S are all zero:
f'(x) + f'(y + z − x) = f'(y) − f'(y + z − x) = f'(z) − f'(y + z − x) = 0, therefore y = z and f'(y) = f'(z) = -f'(x). Substituting in [3] we get

[4] Se = f(x) + 2f(y) − f(2y − x)

For even f(.), f'(x) = f'(-x) = f'(y). At the extrema therefore, y = -x and we can reduce [4] further to

[5] Se = 3f(y) - f(3y)

Therefore the maximum of S = cos(a, b) + cos(a, b') + cos(a', b) - cos(a', b') is also the maximum of the function

3cos(y) - cos(3y)

Which is 2 sqrt 2. http://www.wolframalpha.com/input/?i=Ma ... os%283x%29

Like I told you, there is nothing quantum about Tsirelson's bound, nor is any Hilbert space involved.

Tsirelson's bound is a bound on four "correlations" defined as follows:

E(a, b) := trace(rho X(a) otimes Y(b)) where rho is a joint state of a bipartite quantum system (a trace 1 non-negative self-adjoint operator) and X(a) and Y(b) are two self-adjoint operators on the two subsystems each with spectrum contained in {-1, +1}.

Perhaps you should read Tsirelson's paper to find out what it is actually all about.

I just showed you that you were wrong about Tsirelson's bound. There is nothing quantum or Hilbert space required to derive it. The proof is clearly laid out above for all to see. You have nothing to say because you are surprised that I provided a much simpler and clearer derivation than you've ever seen.

[Joy Christian]

I just showed you that you were wrong about Tsirelson's bound. There is nothing quantum or Hilbert space required to derive it. The proof is clearly laid out above for all to see. You have nothing to say because you are surprised that I provided a much simpler and clearer derivation than you've ever seen.

Michel, your proof is not a proof of Tsirelson's bound.

Tsirelson's bound is about any conceivable physical correlations between any conceivable set of physical systems. In your proof you choose a specific function,

f(x) = cos(x),

to get your final result. That is no good. One has to show that no choice of the function f(x) can break the bound. That is a nontrivial result. It requires a much more sophisticated approach (although exactly in agreement with your main conceptual point that Tsirelson's bound is about geometry and has little or nothing to do with quantum mechanics or Hilbert space per se). A completely general geometric proof of the bound is derived in this paper (see also the Appendix C of this paper).

[minkwe]

Michel, your proof is not a proof of Tsirelson's bound.

Tsirelson's bound is about any conceivable physical correlations between any conceivable set of physical systems. In your proof you choose a specific function,

f(x) = cos(x),

to get your final result. That is no good.

Joy, I disagree, with this. In your proof, you also use a specific function (eqns 70 ...). Tsirelson himself used a specific description. But all of that is peripheral, as they simply ensure that you end up with a cosine function in the end. The key features that determine the upper bound are:
1) The fact that we have only three independent variables not 4. If we had 4 independent variables, the bound cannot be less than 4.
2) The properties of the specific function. In this case, cos(x) gives an upper bound of 2 sqrt 2. If you use cos^2(x) instead, you will have an upper bound of 1+sqrt(2).
3) The result 3f(x) - f(3x) gives you a recipe for calculating the upper bound for any function f(x). The only requirement being that the function be periodic, even, and has a range of [-1, 1] (as demanded by the EPR scenario)

One has to show that no choice of the function f(x) can break the bound. That is a nontrivial result. It requires a much more sophisticated approach (although exactly in agreement with your main conceptual point that Tsirelson's bound is about geometry and has little or nothing to do with quantum mechanics or Hilbert space per se). A completely general geometric proof of the bound is derived in this paper (see also the Appendix C of this paper).

It is trivial to find a function meeting those criteria for which the maximum of f(x) + f(y) + f(z) − f(w) > 2 sqrt(2) (with x, y, z, w all independent). Candidates are (cos(x), cos^2(x), sign(cos(x)), etc, pretty much all functions meeting those criteria will have an upper bound of 4 for 4 independent variables w, x, y, z). For 3 independent variables as is the case in the CHSH correlator, I didn't see that your proof (or Tsirelson's) was a proof showing that such a function is impossible in general. In fact, it is possible to find such a function for which the maximum of 3f(x) - f(3x) is greater than 2 sqrt(2).

For example, a quick investigation reveals that the function f(x) = sign(cos(x)), which appears to satisfy all those requirements, gives a maximum of 4 for 3f(x) - f(3x). Therefore it appears 2 sqrt (2) is not a general upper bound.

[Joy Christian]

Joy, I disagree, with this. In your proof, you also use a specific function (eqns 70 ...).

In my proof the situation is more involved. I start with the summary of well known results relating the only four possible normed division algebras and the only four parallelizable spheres, namely S^0, S^1, S^3, and S^7. This is discussed in the preamble of section IV, entitled "UPPER BOUND IS SET BY THE MAXIMUM POSSIBLE TORSION IN THE SET OF ALL POSSIBLE OUTCOMES" (i.e., possible within S^7). The "specific function (eqns 70 ...)" is then simply a measure of torsion within S^7, and the maximum of the torsion sets the upper bound of 2 sqrt(2). So the specific function is not specific at all, but is necessitated by the geometrical structure of S^7.

I didn't see that your proof (or Tsirelson's) was a proof showing that such a function is impossible in general. In fact, it is possible to find such a function for which the maximum of 3f(x) - f(3x) is greater than 2 sqrt(2).

Indeed. We now know that 2 sqrt(2) bound can be broken, and the strongest possible correlations within the bound of 4 can be easily achieved. But it turns out that no physical experiment to date has ever exhibited correlations stronger than 2 sqrt(2). This fact cries out for explanation. The "proofs" of the bound are thus explanations rather than proofs, of why is it that the bound 2 sqrt(2) is never broken in the experiments, when it is logically quite possible to break it (see, for example, and for fun, the fourth plot in this simulation: http://rpubs.com/jjc/16567).

You are suggesting that you have found another explanation of why the bound of 2 sqrt(2) is not broken in Nature. Your explanation depends on the requirements that the function f(x) is periodic and even in [-1, +1]. But most correlations observed in Nature are not periodic or even. So there are open questions in your explanation.

[minkwe]

You are suggesting that you have found another explanation of why the bound of 2 sqrt(2) is not broken in Nature. Your explanation depends on the requirements that the function f(x) is periodic and even in [-1, +1]. But most correlations observed in Nature are not periodic or even. So there are open questions in your explanation.

The requirement to be periodic comes from rotational invariance of the EPRB experiment. The requirement to be even comes from the fact that there is no preferred reference frame. If Alice and Bob swap angles, they should get the same results, E(a-b) should give the same result as E(b-a). The requirement that the function has a range [-1,+1] comes from the the fact that the outcomes of the experiment are only permitted to have values in the range [-1, +1] (by definition). So that the correlation function can not exceed those bounds. So in the specific scenario of the EPRB experiment the functions f(x) must be periodic and even. Any model of the EPRB experiment must obey those requirements. Of course the analysis does not apply to any functions which do not satisfy those requirements. But then such functions cannot be models of the EPRB experiment.

[Joy Christian]

Of course the analysis does not apply to any functions which does not satisfy those requirements. But then such functions cannot be models of the EPRB experiment.

This brings us back to my main point. Your proof is fine as far as the EPRB experiments are concerned. But the significance of Tsirelson's explanation (and also of my explanation---and there are other explanations based on the uncertainty principle and so on) is that it extends to all possible physical (read: "quantum mechanical") correlations, not just the EPRB correlations. As you note, EPRB correlations are about the rotationally invariant state. But what about the rotationally non-invariant states, like the Hardy state or the GHZ states? None seem to break the bound of 2 sqrt(2), and that is the mystery we are trying to explain.

[minkwe]

But the significance of Tsirelson's explanation (and also of my explanation---and there are other explanations based on the uncertainty principle and so on) is that it extends to all possible physical (read: "quantum mechanical") correlations, not just the EPRB correlations. As you note, EPRB correlations are about the rotationally invariant state. But what about the rotationally non-invariant states, like the Hardy state or the GHZ states? None seem to break the bound of 2 sqrt(2), and that is the mystery we are trying to explain.

If I may ask, what is E(a,b) for the GHZ and Hardy states?

[Joy Christian]

But the significance of Tsirelson's explanation (and also of my explanation---and there are other explanations based on the uncertainty principle and so on) is that it extends to all possible physical (read: "quantum mechanical") correlations, not just the EPRB correlations. As you note, EPRB correlations are about the rotationally invariant state. But what about the rotationally non-invariant states, like the Hardy state or the GHZ states? None seem to break the bound of 2 sqrt(2), and that is the mystery we are trying to explain.

If I may ask, what is E(a,b) for the GHZ and Hardy states?

Good question.

Hardy state is very complicated (it is asymmetrical). 16 different functions E(a, b) are needed to describe it fully. See, for example, page 14 of this paper: http://arxiv.org/pdf/0904.4259v4.pdf.

The GHZ states, on the other hand, are 3- and 4-particle states. So in general they are described by functions of the form E(a, b, c) and E(a, b, c, d), which can be reduced to E(a, b) by fixing (or ignoring) c and d, for example [cf. equation (99) of the above paper].

It is well known, however, that no matter what the quantum state of the system is, and no matter how we manipulate it, the bound 2 sqrt(2) is never broken.

[Ben6993]

gill1109 wrote on Mon Jun 02, 2014 3:38 am

Ben6993 wrote:

It is impossible to make a 4XN table of observables to beat the CHSH limit. So it is impossible for nature to beat the CHSH limit in laboratory space, else the outcomes of the experiment could be put into such a table.

This is totally irrelevant. ...

Hi Richard

I don't doubt for a minute that your work on inequalites is fine. And interesting too, though I have simply taken your equations to be correct insofar as they apply to flatland.

The usual graphical display showing the -cosine curve is a plot of a function of the counts of observable outcomes (y axix) versus theta (x axis). So the y axis is definitely in both flatland and roundland as observed results for A and B are true in both places. (If I can put aside quantum erasers which might undo A and B? I don't know much about them.) But the x axis is only in flatland. The x axis is usually shown as from 0 to 2π, but the complete cycle for rotations in roundland is from 0 to 4π. I don't understand Alice and Bob's angles very much in that everyone seems happy for them to be constrained in flatland's 0 to 2π.

When angle b is established by Bob as 45 deg in flatland, it could be either 45 or 45 + 360 deg in roundland. If 45 deg gave B=1 then 45+360 deg would give B = -1. So it matters what the angle in roundland is, i.e. which cover of space is being used. What I do not understand is if Bob's apparatus is set up initially as 45+360 deg, will it switch randomly to and fro from 45 to 45+360 during the experiment, maybe depending on which particle does the detecting. (As I have said before, the roundland world is difficult for me to understand.) Therefore, the X axis at 45 degrees shows a flatworld conflation of the two separate angles of roundland: 45 and 45+360 degrees.

[gill1109]

The usual graphical display showing the -cosine curve is a plot of a function of the counts of observable outcomes (y axix) versus theta (x axis). So the y axis is definitely in both flatland and roundland as observed results for A and B are true in both places. (If I can put aside quantum erasers which might undo A and B? I don't know much about them.) But the x axis is only in flatland. The x axis is usually shown as from 0 to 2π, but the complete cycle for rotations in roundland is from 0 to 4π. I don't understand Alice and Bob's angles very much in that everyone seems happy for them to be constrained in flatland's 0 to 2π.

Experimenters set angles on detectors which run from 0 to 2 pi. Alice and Bob each choose an angle, two particles are emitted, there is a +/-1 observed at either station. Afterwards data are binned (by angles) and counted and we get to observe a *surface* E(a, b) equal to the number of equal minus the number of unequal outcome divided by the total number of outcomes, at settings a and b.

Everything that comes out of the experiment is in flatland and the experimenters make flatland pictures of them. They observe E(a, b) is approximately equal to cos(a - b). The task of physics is to explain why that can be so.

[Ben6993]

Yes, apologies: the x-axis is a-b and that can take one value in flatland corresponding to four different angles in roundland: a1-b1, a1-b2, a2-b1, a2-b2; where 1 denotes an angle in space cover 1 (0 to 2π) and 2 denotes an angle in cover 2 ( 2π to 4π).

[minkwe]

Already in the first lecture of QM 101 it is demonstrated that QM reduces to classical mechanics for macroscopic bodies.

I'm sure you were also told where the boundary is between microscopic and macroscopic is. They didn't, tell you? You should have asked exactly what the difference was between microscopic and macroscopic. Then you wouldn't be making statements like the above.

http://phys.org/news78650511.html

[gill1109]

Yes, apologies: the x-axis is a-b and that can take one value in flatland corresponding to four different angles in roundland: a1-b1, a1-b2, a2-b1, a2-b2; where 1 denotes an angle in space cover 1 (0 to 2π) and 2 denotes an angle in cover 2 ( 2π to 4π).

You want to paste Bob's choice of measurement settings from 0 to 2 pi at the end of Alice's to get one axis going from 0 to 4 pi ????

Why ever would you want to do that? Alice and Bob each choose one setting, each between 0 and 2 pi, independently of one another. We need the square [0, 2 pi]^2. Not the interval [0, 4 pi].

[Ben6993]

gill1109 wrote on Fri Jun 06, 2014 3:30 am

Ben6993 wrote:

Yes, apologies: the x-axis is a-b and that can take one value in flatland corresponding to four different angles in roundland: a1-b1, a1-b2, a2-b1, a2-b2; where 1 denotes an angle in space cover 1 (0 to 2π) and 2 denotes an angle in cover 2 ( 2π to 4π).

You want to paste Bob's choice of measurement settings from 0 to 2 pi at the end of Alice's to get one axis going from 0 to 4 pi ????

Why ever would you want to do that? Alice and Bob each choose one setting, each between 0 and 2 pi, independently of one another. We need the square [0, 2 pi]^2. Not the interval [0, 4 pi].

Hi. No I did not mean it that way. (The idea made me smile though!)

No doubt that I see this in a far too imperfect way ... a full cycle of rotation for an electron is from 0 to 4pi. Full rotations of the detectors are also from 0 to 4pi. No, I am not getting confused between 4pi and 2pi and I know that 2pi is the normal complete circle in a 360 degree rotation whereas 4pi is a 720 degree rotation. So, the maximum value of a-b is 4pi, not 2pi, where say a = 0 and b = 4pi are the extreme values.


Christian's paper (Macroscopic Observability of Spinorial Sign Changes under 2 Rotations) states that "Aharonov and Susskind were able to demonstrate that in certain circumstances sign changes of spinors under 2π rotations can lead to observable effects" [Y. Aharonov and L. Susskind, Phys. Rev. 158, 1237(1967)]

I cannot remember if I saw this same point in Susskind's online entanglement lectures or in his QM lectures, and I feel sure that I have read further details online. But Susskind said that he did the experiment himself decades ago where he started with a coherent source and its interference pattern. Then he split the beam and rotated the magnets through 4pi for one half of the beam. The interference pattern changed during the rotation and did not return to the original position until after a 4pi rotation.

So, I am not sure why Alice and Bob's detectors are set to vary from 0 to 2pi rather than from 0 to 4pi.

If Bob's setting was constant then I suppose it would not matter, but when Bob's angle is being continually changed it might matter. I know nothing of the experimental details but if I were using a mechanical system and was asked to change a setting from b = 178 degrees to angle b = 2 degrees, I would be tempted just to increase the detector angle by 4 degrees rather than decrease the angle by 176 degrees. And that method of changing angle could cause a problem?

[gill1109]

So, I am not sure why Alice and Bob's detectors are set to vary from 0 to 2pi rather than from 0 to 4pi.

Because they are Stern-Gerlach magnets. Big heavy things. You can rotate them as many times as you like but most physicsts will believe that they work just the same after a rotation of 360 degrees.

Or we do the experiment with photons and polarization. You have to divide all angles by half, since your polaroid glasses work the same upside down as right way up. But different on their sides from normal (horizontal) position. If you wear sunglasses on the beach you might notice this.

Do you have polaroid glasses? Do they work differently depending on how many times you turn them round before putting them on your nose?

One pair of particles are shot off, Alice and Bob each choose one angle. They rotate the control knobs on their machines to that angle as the electrons / photons / whatever are flying through the air, the measurement is "set" and the apparatus is not moving when the particles arrive.

Then they look at what comes out (+/- 1).

Then the whole thing is repeated, lots of times, just how you like.

The detectors are not rotating while the particles are being measured.

[Mikko]

1) How many distinct lambdas are there in Bell's integral? (call that number N)

The integral is over all values of λ, as many as specified in the hidden variable model. Typically it is ℵ₁ but could be more or less.

2) Now if each particle pair measured has a distinct lambda, do you still think Bell's derivation follows?

Yes. If the probability of each value of λ is infinitesimal then each measurement has a distinct λ.

Mikko, if each measurement has a distinct lambda Bell's inequalities relevant for any experiment can not be derived. This is clearly shown in Watson's paper and the thread I suggested above. Bell both assumes that the probabilities are infinitesimal and that the same lamdas are used in the P(a,b) measurement as in the P(a,c) measurement. Review his algebra leading up to eq 15.

The suggestion by some that it doesn't matter so long as the average value of lambda is the same, is just silly and not worthy of a response.

OOPS, seems I misunderstood "each measurement". Of course all measurements of the same pair of particles have the same value of λ.

[Mikko]

my view is that the electron is only in spacetime at the two interactions (creation of the pair and measurement)

Note however that the presence of the electron at the measurement (as well as its absence at other measurements) does depend on the matter content of other spacetime events.