Quantum Mechanics with HV is Local!

[Joy Christian]

Huh? Of course individual outcomes are observed in experiments..

Provide a reference of an experimental paper proving that individual outcomes like A = +1 or B = -1 are observed in an EPRB type experiment. You won't be able to.

No, both QM predictions and experimental evidence says that all four combinations (+1,+1), (-1,-1). (+1,-1), and (-1,-1) will be observed for any fixed pair of settings (with exception of equal or exactly opposite settings). These are observables, and not correlations.

That is incorrect. Only coincidences between A and B are observed in the experiments, amounting to measuring their product AB = either +1 or -1.

***

Here's what Gregor Weihs' Phys. Rev paper on his experiment says:

7. COINCIDENCE EVALUATION
After a measurement run was completed, either for a certain time or for a maximal
number of data points on each side, the data were written to the computers’ hard drives
on each side. Anyone could then later examine the data and draw their own conclusions.
We decided to take the files and extract the photon time-tags for which there was a
coincident detection on the other side.

Notice that there are two computers, one for Alice and one for Bob. They each recorded individual measurements for Alice and Bob. Coincidences are worked out later by comparing the data on these two computers.

Congratulations to the authors and to you. If what you are interpreting is correct, then the authors of this experiment and you have just disproved quantum mechanics.

Quantum mechanics is a statistical theory. It does not predict individual outcomes in any experiment. If you disagree with this, then please provide a detailed prediction using quantum mechanics, of individual outcomes like A = +1 or B = -1 in an EPRB type experiment, or in any experiment for that matter. A textbook reference to an elementary derivation will do.

Now you are claiming that the authors of this experiment actually observed individual outcomes like A = +1 and B = -1. Thus you are claiming that the authors of this experiment have observed something that is not predicted by quantum mechanics. In that case, they should be awarded a Nobel Prize at once, for refuting quantum mechanics by observing something that goes beyond quantum mechanics.

***

[FrediFizzx]

Joy, yes the experimenters always just observe clicks on a detector event by event for A and B individually.

[Joy Christian]

Joy, yes the experimenters always just observe clicks on a detector event by event for A and B individually.

If so, then the experiments are disproving quantum mechanics for decades. In fact, only "coincident counts" are observed between the events A and B, and even that only statistically.

Just think about this for a second. How does quantum mechanics predict anything? What are the basic rules for predicting anything in quantum mechanics? Do the rules say that if you make a measurement of spin along a direction n then you will obtain a result +1? No. That is not what quantum mechanics says. It says that if you make a measurement of spin along direction n, then there is a certain probability that you will obtain a result +1, depending on the quantum state of the spin and the Born rule for extracting probabilities form that quantum state.

In any case, if anyone disagrees with me, then they should be able to provide a simple quantum mechanical prediction for individual outcomes like A = +1 or B = -1, from some textbook.

A definite prediction of outcomes like A = +1 or B = -1 is a business of hidden variable theories, not of quantum mechanics. And to date, no experiment has observed anything beyond the predictions of quantum mechanics.

***

[FrediFizzx]

Theories don't work like experiments. There is someone else we know that thinks theories should work like experiments.

[Joy Christian]

Theories don't work like experiments. There is someone else we know that thinks theories should work like experiments.

If so, then experiments have been disproving quantum mechanics for decades. Quantum mechanics does not predict individual outcomes (if anyone disagrees, they can provide evidence to the contrary from a textbook). But according to you, Heinera and jreed, experiments observe individual outcomes. If so, then that is a clear-cut evidence that quantum mechanics is wrong.

***

[gill1109]

Theories don't work like experiments. There is someone else we know that thinks theories should work like experiments.

If so, then experiments have been disproving quantum mechanics for decades. Quantum mechanics does not predict individual outcomes (if anyone disagrees, they can provide evidence to the contrary from a textbook). But according to you, Heinera and jreed, experiments observe individual outcomes. If so, then that is a clear-cut evidence that quantum mechanics is wrong.

***

Quantum mechanics tells us probabilities of clicks. In an EPR-B experiment with 100% detection rate, QQ tells us the four probabilities of ++, +-, -+ and -- for any given pair of settings. Experimenters run the experiment for a while and count the numbers of outcomes of each of the four types, for each pair of setting values. They divide by the total number of "time-slots" for each pair of setting values which the experimenter has used.

In the CHSH type experiments, for each time-slot, the experimenter chooses between two settings completely at random independently on both sides of the experiment. At the end of the day the experimenter has got 16 counts of ++, +-, -+, -- for each of the four pairs of settings 11, 12, 21, and 22

Take a look at my recent slides https://www.slideshare.net/gill1109/yet-another-statistical-analysis-of-the-data-of-the-loophole-free-experiments-of-2015-revised which includes a nice table of numbers from the 2015 Vienna experiment (built around the Eberhard J inequality rather than the CHSH S, but actually the two are equivalent under the no-signalling assumption).

Of course there are can be states and settings where some outcomes have probability 0 or 1. In that case QM does predict individual outcomes. Check out the famous Lucien Hardy proof of Bell's theorem without inequalities.

[FrediFizzx]

Sure, we all know that the experiments validate QM. However, Hardy's proof of Bell's theorem has been proven wrong by the topic of this thread and accompanying paper.

[Joy Christian]

Check out the famous Lucien Hardy proof of Bell's theorem without inequalities.

Check out how all sixteen predictions of the Hardy state are reproduced local-realistically by my 3-sphere model, thereby refuting the Hardy proof of Bell's theorem without inequalities.

Lucien was flabbergasted that I was able to reproduce, exactly, all sixteen predictions of the quantum state that is named after him. Note that I wrote that paper that back in 2009.

In any case, we were discussing the standard EPRB type experiments. In that case, quantum mechanics does not predict individual outcomes but only probabilities for those outcomes.

***

[Heinera]

In any case, we were discussing the standard EPRB type experiments. In that case, quantum mechanics does not predict individual outcomes but only probabilities for those outcomes.
***

No one has disputed that. But when QM, for a given pair of settings, predicts non-zero probabilities for all four outcome combinations, then eventually all four will be observed for that pair of settings.

[Joy Christian]

In any case, we were discussing the standard EPRB type experiments. In that case, quantum mechanics does not predict individual outcomes but only probabilities for those outcomes.
***

No one has disputed that. But when QM, for a given pair of settings, predicts non-zero probabilities for all four outcome combinations, then eventually all four will be observed for that pair of settings.

That does not really help your argument. The QM HV model we are discussing here reproduces the correlation E(a, b) = -a.b exactly. Mathematically that is equivalent to the sum of all four probabilities for the outcomes ++, --, +- and -+. If those four different outcomes for a given pair of settings were not predicted by the model, then it would not be possible to derive the correlation E(a, b) = -a.b exactly. For details, see eq. (44) of this paper: https://arxiv.org/abs/1405.2355. Note that first and second lines of this equation are mathematically equivalent.

***

[Heinera]

The QM HV model we are discussing here reproduces the correlation E(a, b) = -a.b exactly. Mathematically that is equivalent to the sum of all four probabilities for the outcomes ++, --, +- and -+. If those four different outcomes for a given pair of settings were not predicted by the model, then it would not be possible to derive the correlation E(a, b) = -a.b exactly.
***

So since the functions A and B in the model obviously do not predict four different outcomes (because the hidden virable is only binary), something is clearly wrong.

[Joy Christian]

The QM HV model we are discussing here reproduces the correlation E(a, b) = -a.b exactly. Mathematically that is equivalent to the sum of all four probabilities for the outcomes ++, --, +- and -+. If those four different outcomes for a given pair of settings were not predicted by the model, then it would not be possible to derive the correlation E(a, b) = -a.b exactly.

So since the functions A and B in the model obviously do not predict four different outcomes (because the hidden variable is only binary), something is clearly wrong.

What is wrong is that you have been ignoring my previous replies. This is a QM version of the 3-sphere model. The functions A and B do predict all four different outcomes given a, b, and lambda, and dictated by the S^3 geometry. If they didn't, then E(a, b) = -a.b could not have been reproduced. As long as you keep ignoring S^3 geometry, you will continue to be puzzled.

***

[Heinera]

What is wrong is that you have been ignoring my previous replies. This is a QM version of the 3-sphere model. The functions A and B do predict all four different outcomes given a, b, and lambda, and dictated by the S^3 geometry. If they didn't, then E(a, b) = -a.b could not have been reproduced. As long as you keep ignoring S^3 geometry, you will continue to be puzzled.
***

I am looking only at the mathematical expressions (8) and (9) that defines A and B. If there is some magic performed by the S^3 geometry it is certainly not reflected in those mathematical expressions.

[Joy Christian]

What is wrong is that you have been ignoring my previous replies. This is a QM version of the 3-sphere model. The functions A and B do predict all four different outcomes given a, b, and lambda, and dictated by the S^3 geometry. If they didn't, then E(a, b) = -a.b could not have been reproduced. As long as you keep ignoring S^3 geometry, you will continue to be puzzled.
***

I am looking only at the mathematical expressions (8) and (9) that defines A and B. If there is some magic performed by the S^3 geometry it is certainly not reflected in those mathematical expressions.

Oh... but the "magic" of the geometry of S^3 is manifestly built-in in the expressions (8) and (9). Let me spell it out for you: Note that (8) and (9) involve products like (sigma.a)(sigma.s), where sigma is the Pauli matrix and a and s are ordinary vectors. Now (sigma.a) and (sigma.s) are isomorphic to pure quaternions, and their product (sigma.a)(sigma.s) is a unit quaternion. But a set of unit quaternions is just the parallelizable 3-sphere I am talking about. So the 3-sphere geometry is built-in in the definitions of the measurement functions A(a, h) and B(b, h).

***

[Heinera]

Oh... but the "magic" of the geometry of S^3 is manifestly built-in in the expressions (8) and (9). Let me spell it out for you: Note that (8) and (9) involve products like (sigma.a)(sigma.s), where sigma is the Pauli matrix and a and s are ordinary vectors. Now (sigma.a) and (sigma.s) are isomorphic to pure quaternions, and their product (sigma.a)(sigma.s) is a unit quaternion. But a set of unit quaternions is just the parallelizable 3-sphere I am talking about. So the 3-sphere geometry is built-in in the definitions of the measurement functions A(a, h) and B(b, h).
***

And the functions A(a, h) and B(b, h) are deterministic and can only produce two combinations.

[Joy Christian]

Oh... but the "magic" of the geometry of S^3 is manifestly built-in in the expressions (8) and (9). Let me spell it out for you: Note that (8) and (9) involve products like (sigma.a)(sigma.s), where sigma is the Pauli matrix and a and s are ordinary vectors. Now (sigma.a) and (sigma.s) are isomorphic to pure quaternions, and their product (sigma.a)(sigma.s) is a unit quaternion. But a set of unit quaternions is just the parallelizable 3-sphere I am talking about. So the 3-sphere geometry is built-in in the definitions of the measurement functions A(a, h) and B(b, h).
***

And the functions A(a, h) and B(b, h) are deterministic and can only produce two combinations.

What is wrong is that you have been ignoring my previous replies. This is a QM version of the 3-sphere model. The functions A and B do predict all four different outcomes given a, b, and lambda, and dictated by the S^3 geometry. If they didn't, then E(a, b) = -a.b could not have been reproduced. As long as you keep ignoring S^3 geometry, you will continue to be puzzled.

***

[gill1109]

Reproduction of the quantum correlations is no big deal. You can do it with conventional Hilbert space apparatus. You can do it with quaternions or with bigger Clifford algebras. The big issue is - can you do it in a local way? Subject to the usual restrictions on the co-domain of the functions A(a, lambda) and B(b, lambda). If lambda only takes on two different values +/-1 (fair coin tosses) you trivially cannot do it. As Heinera already remarked several times...

Today I am in the UK not terribly far from Oxford. Like to meet up, Joy? I've sent you an email with more detailed info.

BTW I chatted quite a lot with Lucien Hardy last week. He's a great guy.

[Heinera]

The functions A and B do predict all four different outcomes given a, b, and lambda, and dictated by the S^3 geometry. If they didn't, then E(a, b) = -a.b could not have been reproduced.
***

This only shows that there is a glaring inconsistency in the model. The expressions for A and B can only produce two outcome combinations (which anyone can check), yet you somehow claim E(a, b) = -a.b.

[Joy Christian]

The functions A and B do predict all four different outcomes given a, b, and lambda, and dictated by the S^3 geometry. If they didn't, then E(a, b) = -a.b could not have been reproduced.
***

This only shows that there is a glaring inconsistency in the model. The expressions for A and B can only produce two outcome combinations (which anyone can check), yet you somehow claim E(a, b) = -a.b.

The inconsistency is in your understanding of what the model is all about. I am not going to keep repeating my replies over and over again. For the last time, all four combinations of outcomes, ++, --, +- and -+ are possible. They come about, locally, because of the geometry of the 3-sphere. If they did not, then the correlation could not have been E(a, b) = -a.b.

***

[gill1109]

The functions A and B do predict all four different outcomes given a, b, and lambda, and dictated by the S^3 geometry. If they didn't, then E(a, b) = -a.b could not have been reproduced.
***

This only shows that there is a glaring inconsistency in the model. The expressions for A and B can only produce two outcome combinations (which anyone can check), yet you somehow claim E(a, b) = -a.b.

The inconsistency is in your understanding of what the model is all about. I am not going to keep repeating my replies over and over again. For the last time, all four combinations of outcomes, ++, --, +- and -+ are possible. They come about, locally, because of the geometry of the 3-sphere. If they did not, then the correlation could not have been E(a, b) = -a.b.

***

You can *say* what you like. Your *formulas* tell another story.