OK Corral: Local versus non-local QM [Archive]

Feb19-07, 07:33 PM

Preamble: I have yet to find a reason to abandon my support of a LOCAL interpretation of QM; especially in relation to EPRB. It is therefore my hope that we might here bring the issue to a head. Given the frequent verbal misunderstandings across the LOCAL--NONLOCAL divide, it is also my hope that we might resolve the issue in the beautiful (and usually clearer) language and logic of mathematics.

PS: I suggest we constrain our discussion on this thread to the EPR-Bohm experiment with spin-half particles; it being the source of Bell's (1964) theorem. So:

On another thread, I (wm) wrote:

Dear vanesch, I appreciate your defence of MWI but it still does not make sense to me (though I am a hardened LOCALIST):

1. You say: such as non-conspiracy in nature which requires random events to be statistically independent if there is no causal link somewhere, in one way or another.

I thought that the experimental results reveal statistical (= LOGICAL) dependence? Which is what I would expect, given that the ''entangled particles'' (created by the conservation of angular momentum) represent some of the highest correlations that we can generate (ie, the particle-state is spherically symmetric). Given then that the measuring-devices are also HIGHLY correlated by their differential setting: the detected correlations appear to be LOCAL and non-mysterious to me. What am I missing?

2. To help my understanding of MWI: Let us say that I come to Paris (to discuss MWI with you) and we agree to toss a coin to decide who pays for dinner. As the coin arcs through the air, I guess we agree that it is in a superposition? On coming to rest, the coin reveals a definite result.

It seems to me that you and I remain physically in the world with the definitive physical result (the world where we expect to have dinner): AND that no other real world exists. Why then the need for MW in MWI?

Thanks, wm

DrChinese replied:

The measuring devices in Bell tests are sometimes set while the entangled particles are mid-flight. Therefore their settings cannot be causally correlated.

The mystery is this: was the outcome of a particular detector setting determined when the particle pair was created? If you say YES, then you run afoul of Bell: because then the particles must be carrying enough "answers" to match all possible detector settings (Bell shows this cannot be true). If you say NO, then how does the answer at one point get transmitted to the other point? (Emphasis added.)

1. Causally correlated?? DrC, note that I said that the detector settings are correlated by their diifferential setting. If detector A on the left is set at a (a unit vector) and detector B' on the right is set at b' (another unit vector), then the detector settings are correlated by a function of (a, b'): that is, by a function of the differential setting.

For example: If we took cos(a, b') as the correlation function; +1 indicates a parallel setting; -1 indicates an antiparallel correlation; etc.

Of course the correlation that counts is that established by the respective detector-settings at the instant of arrival of each particle: set in any manner of your choosing.

2. Was the outcome of a particular detector setting determined when the particle pair was created?? The outcome was determined when the particle and the detector interacted LOCALLY.

So could I ask you to provide the detailed maths by which you derive the EPRB correlation function? So that I might see where you believe that nonlocality enters (in the language of maths.)?

3. How does the answer at one point get transmitted to the other point? ??

Well, in that we have not introduced Alice and Bob to the discussion thus far: it does not get transmitted! How could it?

Trust this all helps to convince you of the need for your mathematical derivation of the EPRB correlation function,

With best regards, wm

Hurkyl

Feb19-07, 07:46 PM

Why then the need for MW in MWI?
I can answer this one.

There is a lot of information in the universe. There is information about Alice's laboratory, there is information about Bob's laboratory, there is information about the Andromeda galaxy, etc.

Alice can measure anything in her laboratory, but she can't (directly) measure anyting in Bob's laboratory. Nor can she directly measure anything in the Andromeda galaxy.

When we take the state of the universe and discard the information that isn't (directly) accessible to Alice, what remains is what we call Alice's world.

In particular, it isn't like Star Trek where we have a parallel "universe" in which our evil twins live.

If the entire universe consisted of the pair of photons in the experiment, then the state of the universe might be

(3/5) |01> + (4/5) |10>

Alice's world is the first coordinate. If we take the partial trace along the second coordinate to obtain the state of Alice's world, we get the statistical mixture

36% chance of |0>
64% chance of |1>

Similarly, Bob's world is the statistical mixture

64% chance of |0>
36% chance of |1>

Feb20-07, 12:44 AM

I can answer this one.

There is a lot of information in the universe. There is information about Alice's laboratory, there is information about Bob's laboratory, there is information about the Andromeda galaxy, etc.

Alice can measure anything in her laboratory, but she can't (directly) measure anyting in Bob's laboratory. Nor can she directly measure anything in the Andromeda galaxy.

When we take the state of the universe and discard the information that isn't (directly) accessible to Alice, what remains is what we call Alice's world.

In particular, it isn't like Star Trek where we have a parallel "universe" in which our evil twins live.

If the entire universe consisted of the pair of photons in the experiment, then the state of the universe might be

(3/5) |01> + (4/5) |10>

Alice's world is the first coordinate. If we take the partial trace along the second coordinate to obtain the state of Alice's world, we get the statistical mixture

36% chance of |0>
64% chance of |1>

Similarly, Bob's world is the statistical mixture

64% chance of |0>
36% chance of |1>

Dear hurkyl, I appreciate this answer very much!

BUT I wonder: How long before someone comes to clobber us in this common belief. Dare I summarise that belief as: Many private local worlds, one common local universe?

AND I wonder that I've not seen such a clear delineation of the MW in MWI before? That is: I wonder if dedicated MWIers will accept it?

For, if I'm not mistaken, you allow that the overlap in Alice and Bob's private worlds may increase when they get on the phone to discuss their respective results: in full accord with locality.

And in full accord with the common-sense response to DrC's question re information (answers) about Alice's result (in Alice's world) being transferred to Bob (in Bob's world); and vice-versa: There being no virtual/mysterious/magical/non-local/Star-Trek universes built from virtual Alices and Bobs ... ad infinitum.

Hoping I've understood you correctly, many thanks, wm

Demystifier

Feb20-07, 03:25 AM

Demystifier

Feb20-07, 03:42 AM

Let me also explain why I claim that correlations are nonlocal. If two particles spacially separated are correlated, then this correlation is nonlocal. Clearly, in the EPR(B) setup this is the case. It is also clear that the Schrodinger equation allows entangled wave functions without local interactions between particles. In that regard see also
http://arxiv.org/abs/quant-ph/0304031

vanesch

Feb20-07, 06:35 AM

I've been giving my PoV on this issue at least a dozen of times, so I'm not going to do so here again...

Demystifier

Feb20-07, 06:51 AM

I've been giving my PoV on this issue at least a dozen of times, so I'm not going to do so here again...
Then let me put a summary of my view of your PoV. It is MWI and local and is in agreement with what I said above. Namely, it involves a fundamental role of conscious observers, but their role is not *clearly* explained. Instead, their role remains somewhat mysterious. In other words, just as I said above, you save locality by avoiding a *clear* answer to the question - what is QM about?!

But of course, it does not mean that your PoV is wrong. It may be a fundamental property of consciousness that it is uncomprehensible and logically unexplainable. In that case, QM can indeed be local, while apparent nonlocality can be merely an artefact of misleading attempts to find a non-existing complete comprehensible model of nature.

Demystifier

Feb20-07, 07:33 AM

Let me also say the following. If QM is both local and comprehensible, then there must exist a formulation of QM in which this locality is manifest in *every* equation. In particular, it should not contain many-particle wave functions that cannot be reduced to a collection of single-particle wave functions. But no one ever constructed such a formulation of QM. Thus, I conclude that QM is either nonlocal or noncomprehensible (or possibly both). This conjecture is nothing but a somewhat stronger variant of the well-known *theorem* (Bell, Hardy, ...) that QM is either nonlocal or nonrealistic (or possibly both).

But if you accept the option that QM is noncomprehensible, then you come to the teritory of religion-like arguments. Whenever scientists give a logical argument against some religion dogma, religious people reply that the logical argument does not disprove anything, because God is not comprehensible by humans. I respect such arguments as long as they are used consistently, but then physicists advocating locality of QM should not pretend that they find QM fully comprehensible. Locality is compatible with QM in the same way as religion is compatible with science.

Demystifier

Feb20-07, 07:56 AM

DrChinese

Feb20-07, 08:51 AM

vanesch

Feb20-07, 09:18 AM

What I cannot understand is WHY some physicists so desperately seek for a local interpretation of QM? What is so sacred about locality? Before getting familiar with Einstein theory of relativity, (almost) nobody had problems with accepting the Newton nonlocal theory of gravity or the Coulomb nonlocal law of electrostatics. So, why locality could not be only an approximative principle valid only in a restricted domain? Why many physicists find so dangerous or even unacceptable to have nonlocal laws of physics?

The one and only reason is the one you give yourself: relativity. Its basic principles require locality, and it is difficult to restore all the confirmed results of relativity in a natural way (I mean: following from some principle, and not one by one put in by hand to make things come out AS IF) if you violate locality in interaction.

After all, the idea is to distill some fundamental principles on which to build physics. If we do not do this, and allow for a "deus ex machina" each time that we *seem* to have a principle, but which we'd like to violate somehow, which makes things come out nevertheless AS IF that principle were valid, then you can just as well go all the way: there are no laws of physics. There's just a big bag of events, and there happen to be correlations between them, but there's no real reason for that, there's no structure behind it, and if we find some, then that's a funny coincidence. What we think are cause-effect relationships are just coincidences, and their apparently systematic relationship is just by good (or bad) fortune. Tomorrow, things may be totally different. Or not.

So, if we want to avoid the above viewpoint (which, after all, could be "true"), we'd better try to find rigorous principles, and be wary each time we find funny relationships that don't seem to rely on such a principle. This quest is not guaranteed to work, of course. But when we HAVE such a principle, and we have no absolute reason to reject it, we shouldn't reject it carelessly! Now, the relativity principle seems to be ok: there's no known, observed violation of anything that's derived from it, and it is extremely powerful in explaining a lot of stuff. So that's why people are very reluctant to dump it for no good reason but just some bad feeling about certain interpretations of quantum theory.

Now, that said, Newton himself had a lot of troubles with his own "action-at-a-distance" ! In his principia, he states:

That one body may act upon another at a distance through a vacuum without the mediation of anything else, by and through which their action and force may be conveyed from one another, is to me so great an absurdity that, I believe, no man who has in philosophic matters a competent faculty of thinking could ever fall into it

So, locality does have some attractiveness, apart from its necessity in relativity.

Demystifier

Feb20-07, 09:40 AM

Vanesch, I agree, relativity looks as a nice and general physical principle with which we could start.
But the existence of objective reality also looks as a nice and even *more general* physical principle with which we could start as well.
On the other hand, the Bell theorem proves that at least one of these two nice principles is wrong.
Isn't it more natural to retain the more general principle and to crucify the less general one?
In addition, if you abandon the general principle of the existence of objective reality, then the existence of spacetime may not be objectively real as well, which means that even the principle of relativity (or locality) may not be objectively real, so you may lose *both* nice principles, which is certainly not what you want. (For an expanded version of this idea see also
http://arxiv.org/abs/quant-ph/0607057 )

Hurkyl

Feb20-07, 09:52 AM

That is: I wonder if dedicated MWIers will accept it?
I'm not sure about the semantics anymore: I (thought I) had gotten it straight from Wikipedia, but I don't see it there anymore. MWIers will certainly agree with what I said, but they might disagree that's the definition of a "world".

Schrodinger's Dog

Feb20-07, 09:56 AM

There are local and nonlocal *interpretations* of quantum mechanics (QM).
But what can we say about QM (non)locality independently on the interpretation?
We can say that a (many-particle) state is described by a wave function of the form psi(x_1,...,x_n), i.e., that, in general, there is no wave function for each particle, but that there is only one wave function describing all particles together. Isn't that an interpretation-independent sign of NONlocality?

Of course, if you start to deal with interpretations, you may say that quantum physics is only about correlations, and that only the correlations are the entities which are nonlocal. But if the only entities quantum mechanics is about are nonlocal, then obviously quantum mechanics IS nonlocal, isn't it?

And if QM is *not* only about the correlations, then what is it about? Whatever *clear* answer to that question you choose, it seems that you cannot avoid the conclusion that this thing must be nonlocal. The only way to save locality is to advocate an interpretation that avoids a clear answer to the question - what is QM about?!

Or to presume that your interpretation of QM is wrong in some as yet undefined way. Not that I think it is definitely, but it's a possibility; that only our practically limited understanding of what exactly is going on, is clouding our theoretical approach to what is actually going on, in other words we're making incorrect assumptions.

On the other hand of course we could just assume that QM is correct and there is something enigmatic(not necessarily a hidden variable as we might think of it, but something intrinsic we're missing) Which with our current technology we cannot reveal, that would make sense of the seemingly unusual or non common sense ideas.

Perhaps we need to decouple ourselves from current reason and accept that probability or energy and matter may well work in as yet incomprehensible ways we cannot yet grasp at our current level. Maybe we're unfortunate enough to be living in a time where our reach exceeds our grasp?

I honestly have no idea, and probably will not have any more the more I come to learn about QM in depth.

Demystifier

Feb20-07, 10:05 AM

Schrodinger's Dog

Feb20-07, 10:19 AM

Schrodinger's Dog, but there *is* something very general that is proved rigorously by the Bell theorem:
If the results of measurements are manifestation of *some* objective reality (*whatever* this reality might be), then this reality must be nonlocal. This generality of the Bell theorem expressed by my words "some" and "whatever" is what makes this theorem so strong.

Oh I agree given what we currently know the Bell theorem makes a pretty strong case for non locality, that's not something I would presume to argue about as it's a pretty air tight thought experiment, the only thing I suggested is that it might not necessarily be based on a true picture, and therefore we could be making flawed assumptions, which of course is entirely speculatory.

We assume that there is something based on our logic system at work, without knowing if there is something else going on we can't possibly account for given our current technology, for example, it's like Newton looking at light without understanding it's duality so he claims it's a particle, and then Young claims it must be a wave because of his experiment, both are wrong, because both don't have the bigger picture.

This doesn't mean there is some hidden variable or locality exists with QM, but I think it's possible that QM could be mistaken in some key area so I don't discount anything without having the 100% picture. Obviously this is just an example where nothing is 100% air tight, particularly when we don't have all the books in our collection.

I'm not disagreeing with anything, just keeping an open mind.

If Bell's theorem is not falsifiable then it is not a theorem :smile:

Tomsk

Feb20-07, 10:54 AM

complexPHILOSOPHY

Feb20-07, 11:33 AM

vanesch

Feb20-07, 11:45 AM

Schrodinger's Dog, but there *is* something very general that is proved rigorously by the Bell theorem:
If the results of measurements are manifestation of *some* objective reality (*whatever* this reality might be), then this reality must be nonlocal. This generality of the Bell theorem expressed by my words "some" and "whatever" is what makes this theorem so strong.

Yes, this is about correct (if you make the implicit hypothesis of no super-determinism, but which I think is necessary to make if we are going to draw ANY conclusion from observation): one has to choose between giving up the objective (and unique!) existence of outcomes of measurement, OR non-locality. Now, given that we can more or less cope with the first without hurting the known principles of physics (namely, MWI), but that, when accepting non-locality, we have to throw out the whole formal machinery of relativity, I would say that the requirement of objective and unique outcomes is for sure an esthetic and philosophically very satisfying one, but not absolutely strictly required by what we know, formally, about physics. However, the second (locality), is.

So it seems that Bell + experimental confirmation (which is suggestively strong, but I'm not sure it is 100% watertight) makes us choose between:
- a philosophically satisfying concept, that "what we see, IS there, and ONLY that" OR
- the formal principle of relativity.

I would say that as a human being, I'd go for the first, but as a physicist, I go for the second. Because the first principle doesn't bring me any FORMAL confort, while the second one is formally useful... at least as long as there isn't another principle from which we can get the *results* of relativity without its founding principle (which is the 4-dim spacetime manifold).

vanesch

Feb20-07, 11:53 AM

complexPHILOSOPHY

Feb20-07, 12:00 PM

I love that :!!) :approve:
It is exactly my critique on Rovelli's RQM.

The argument regarding 'the pink elephant,' deconstructs objectivity into a collective agreeance of subjective perceptions and it seems rather sound.

I hate you Berkely.

vanesch

Feb20-07, 12:40 PM

The argument regarding 'the pink elephant,' deconstructs objectivity into a collective agreeance of subjective perceptions and it seems rather sound.

The point is: in order for the :"we all agree over the pink elephant" to make any progress in the EPR-Bell dilemma, we need to say that at a certain point, Alice and her friends all agree on a pink elephant, Bob and his friends all agree on a blue elephant, and Alice and her friends all agree on the fact that Bob hasn't seen an elephant. Because if they do, they can assign a probability distribution to a hidden variable ("what Bob and his friends have seen"), and Bell bites us.

complexPHILOSOPHY

Feb20-07, 01:02 PM

DrChinese

Feb20-07, 01:25 PM

...So it seems that Bell + experimental confirmation (which is suggestively strong, but I'm not sure it is 100% watertight)...

Just a note for anyone who thinks this means that QM might not be "right": the leeway on this is VERY small - multiple standard deviations of agreement between QM and actual results.

Also, the comparison point for local realistic theories is severely problematic as well; because there are no such theories currently in existence that come close to actual results. I.e. the closest you can come to actual results - and still be within the Bell Inequality - does not match to any existing local realistic interpretation. That would be (more or less) a straight line function which agrees with QM at theta=0, 45 and 90 degrees but is slightly different at all other angles. (QM, of course, predicts the COS^2 function and that is what is observed.) So, a local realistic theory not only needs to match the Bell Inequality, it also needs to show how it gets there as a function of theta. This is not so easy, as the Bell Inequality is essentially 1-(theta/90 degrees) for correlated Alice and Bob. You then need "experiment error" and bias to account for the difference between this and actual results.

So don't think for a second that experimental problems in Bell tests will allow you to return to local realistic positions. This avenue creates as many problems as it purports to solve.

Feb20-07, 05:23 PM

Just a note for anyone who thinks this means that QM might not be "right": the leeway on this is VERY small - multiple standard deviations of agreement between QM and actual results.

Also, the comparison point for local realistic theories is severely problematic as well; because there are no such theories currently in existence that come close to actual results. I.e. the closest you can come to actual results - and still be within the Bell Inequality - does not match to any existing local realistic interpretation. That would be (more or less) a straight line function which agrees with QM at theta=0, 45 and 90 degrees but is slightly different at all other angles. (QM, of course, predicts the COS^2 function and that is what is observed.) So, a local realistic theory not only needs to match the Bell Inequality, it also needs to show how it gets there as a function of theta. This is not so easy, as the Bell Inequality is essentially 1-(theta/90 degrees) for correlated Alice and Bob. You then need "experiment error" and bias to account for the difference between this and actual results.

So don't think for a second that experimental problems in Bell tests will allow you to return to local realistic positions. This avenue creates as many problems as it purports to solve.

DrC, surely you have mis-spoken? Or are badly mistaken? Even just plain wrong?

You write: ''So, a local realistic theory not only needs to match the Bell Inequality, it also needs to show how it gets there as a function of theta.''

Here is the background to my questioning:

1. QM is correct.

2. Bell-tests are correct.

So, in response to your claim above, I say:

3. The only theories that need to match the Bell Inequality are those which meet the Bellian realism conditions.

4. Bellian realism is satisfied by such stable classical objects as dirty socks, down-hill skiers, books in libraries.

5. So could I repeat an earlier question and again request your personal derivation of the EPRB (spin-half) correlations? That is, a derivation that you understand intimately. (Noting that this is NOT a request for you or anyone else to derive the Bellian inequalities.)

Thanks, wm

Feb20-07, 05:32 PM

The point is: in order for the :"we all agree over the pink elephant" to make any progress in the EPR-Bell dilemma, we need to say that at a certain point, Alice and her friends all agree on a pink elephant, Bob and his friends all agree on a blue elephant, and Alice and her friends all agree on the fact that Bob hasn't seen an elephant. Because if they do, they can assign a probability distribution to a hidden variable ("what Bob and his friends have seen"), and Bell bites us.

Help please! The OP requested the discussion here be in the context of EPRB with spin-half particles.

Is there some reason why we need to consider elephants and wombats? The subject is already beset and riddled with linguistic problems. And, given that, I personally tend to avoid further confusion via extraneous literary elements.

PS: For the record, I understand that wombats in Australia anciently were as big as elephants. Did John Bell know that?

Sincerely, wm

Feb20-07, 05:36 PM

I love that :!!) :approve:
It is exactly my critique on Rovelli's RQM.

Can you point me to your critique (or a summary thereof), please?

wm

Feb20-07, 05:57 PM

I have a question- how can you test for realism? Is it falsifiable? You would say, realism is a false assumption if everybody does not 'see the same elephant', but if 'everybody hears everybody else stating that they see the same elephant they see', then you can't then know if the elephant has an objective reality. It certainly seems that the elephant is an objective thing, say from A's point of view, especially if B is saying it's an elephant, but a completely different consistent set of obsevations is perfectly possible- B seeing a wombat, and hearing A say that it's a wombat.

Locality is certainly falsifiable, and has been tested, so it would surely make more sense to get rid of realism than locality.

I'd like to encourage you in the view that: It surely makes more sense to get rid of pseudo-realism (= limited realism = Bellian realism) than locality.

Thus Bell once strongly endorsed a derivation of his inequalities by d'Espagnat (Sci. Am. November 1979). In it you find this move:

"These conclusions require a subtle but important extension of the meaning assigned to the notation A+. Whereas previously A+ was merely one possible outcome of a measurement made on a particle, it is converted by this argument into an attribute of the particle itself.'' (Emphasis added.)

In my view, most quantum objects are perturbed by ''measurement'' and that is why Bellian Inequalities are breached by quantum objects! Bellian realism being of very limited validity.

PS: As I recall, Bell said he could do no better than d'Espagnat!

wm

DrChinese

Feb20-07, 08:31 PM

DrC, surely you have mis-spoken? Or are badly mistaken? Even just plain wrong?

You write: ''So, a local realistic theory not only needs to match the Bell Inequality, it also needs to show how it gets there as a function of theta.''

Here is the background to my questioning:

1. QM is correct.

2. Bell-tests are correct.

So, in response to your claim above, I say:

3. The only theories that need to match the Bell Inequality are those which meet the Bellian realism conditions.

4. Bellian realism is satisfied by such stable classical objects as dirty socks, down-hill skiers, books in libraries.

5. So could I repeat an earlier question and again request your personal derivation of the EPRB (spin-half) correlations? That is, a derivation that you understand intimately. (Noting that this is NOT a request for you or anyone else to derive the Bellian inequalities.)

Thanks, wm

I still don't understand what you are asking, especially about the derivation of EPRB. What is the issue about photons vs. electrons? Photons are a lot easier to discuss and that is what all tests are about.

Bell's theorem is NOT satisified by dirty socks and library books (specifically those things do NOT violate Bell's Inequality because they are classical in nature). A Bell local + realisitic theory will never match Bell test experimental results.

You say 1. and 2. above are things you accept, so why are you still talking about local realism?

Feb20-07, 09:22 PM

I still don't understand what you are asking, especially about the derivation of EPRB. What is the issue about photons vs. electrons? Photons are a lot easier to discuss and that is what all tests are about.

Bell's theorem is NOT satisified by dirty socks and library books (specifically those things do NOT violate Bell's Inequality because they are classical in nature). A Bell local + realisitic theory will never match Bell test experimental results.

You say 1. and 2. above are things you accept, so why are you still talking about local realism?

1. In that EPRB was the source of Bell (1964), and in that we both accept that the experimental results would agree with QM, if the experiment was done, I am interested in your personal derivation of the EPRB correlation.

2. Can you provide such? Because your words are (to me) so unclear and confusing that I get lost. I am much less likely to get lost when I see you derive the EPRB correlation in mathematical terms that you understand and commit to.

3. See here (again) how my wording is twisted by you: I said that Bell realism is satisfied by dirty socks and library books. THAT IS: They satisfy Bellian Inequalities. Then YOU say: Bell's theorem is not satisfied by dirty socks and library books.

4. Bellian inequalities are satisfied by the Bellian realism of dirty socks (= the realism from which it was derived) and Bellian inequalities are breached by quantum particles because they are not like dirty socks.

5. YOU ASK: Given my acceptance of QM and Bell-test results, why am I still talking about local realism? BECAUSE it is Bellian (limited, constrained) realism that we should reject; it being not valid in general. (See my recent note on this thread about the shifty (''subtle'') move by d'Espagnat regarding A+; a move endorsed by Bell and many others ... and rejected by me. Do you personally accept it for quantum particles?)

6. So let's see your derivation of the EPRB (spin-half) correlations and take it from there. OK?

I might be wrong, and my words no better than yours: BUT Maths is the best logic (so let's see yours)! wm

JesseM

Feb20-07, 11:20 PM

DrChinese

Feb20-07, 11:49 PM

1. In that EPRB was the source of Bell (1964), and in that we both accept that the experimental results would agree with QM, if the experiment was done, I am interested in your personal derivation of the EPRB correlation.

2. Can you provide such? Because your words are (to me) so unclear and confusing that I get lost. I am much less likely to get lost when I see you derive the EPRB correlation in mathematical terms that you understand and commit to.

3. See here (again) how my wording is twisted by you: I said that Bell realism is satisfied by dirty socks and library books. THAT IS: They satisfy Bellian Inequalities. Then YOU say: Bell's theorem is not satisfied by dirty socks and library books.

4. Bellian inequalities are satisfied by the Bellian realism of dirty socks (= the realism from which it was derived) and Bellian inequalities are breached by quantum particles because they are not like dirty socks.

5. YOU ASK: Given my acceptance of QM and Bell-test results, why am I still talking about local realism? BECAUSE it is Bellian (limited, constrained) realism that we should reject; it being not valid in general. (See my recent note on this thread about the shifty (''subtle'') move by d'Espagnat regarding A+; a move endorsed by Bell and many others ... and rejected by me. Do you personally accept it for quantum particles?)

6. So let's see your derivation of the EPRB (spin-half) correlations and take it from there. OK?

I might be wrong, and my words no better than yours: BUT Maths is the best logic (so let's see yours)! wm

What EPR-B correlations are you taking about?

You need to make a specific statement and let's discuss that. I will be glad to discuss any side of Bell's Theorem you want to discuss. I have already referenced several derivations of Bell on my web pages, so I am not sure what you are asking.

To make it clear: I advocate a standard reading of EPR/Bell/Aspect. Bell realism is as limited - or not - as you care to view it. There are those who refer to it as "naive realism" but I personally reject that description (as would Einstein, who was a realist of the same vein).

The math of the Bell realism assumption is simple: assume the simultaneous existence of pre-determined values for 3 non-commuting spin operators (A, B and C). Then prepare a table which shows these 8 permutations when measured as up/down (electrons) and the relative percentages. You will find that it is not possible to create such a table AND have it agree to experiment UNLESS you put negative percentages in some spots.

If you think we should reject "Bellian realism" and accept "locality": I think that is a perfectly sensible interpretation and have no issue with it. But I doubt that most folks will conclude that local realism is still a viable option just because "Bellian realism" is too "limited". If you can come up with an acceptable alternative definition of realism, I would be interested in seeing it.

DrChinese

Feb20-07, 11:59 PM

In my view, most quantum objects are perturbed by ''measurement'' and that is why Bellian Inequalities are breached by quantum objects! Bellian realism being of very limited validity.

So what if an observation perturbs a system under study? That in no ways explains anything, and it certainly does not explain Bell test results. This is pure hand-waving, and is just as true in the classical world.

Feb21-07, 12:03 AM

wm, would you agree that it would be impossible to violate the Bell inequalities classically if one obeyed all the conditions specified in the proof of Bell's theorem (including the condition that the state of the objects/signals emitted by the source be statistically independent of the detector settings), even if one allowed the measurements to modify the state of the objects/signals recieved by the two experimenters?

Hi JesseM,

Would I agree that it is impossible to violate the Bell inequalities classically if one obeyed all the conditions [see* below] specified in the proof of Bell's theorem (including the condition that the state of the objects/signals emitted by the source be independent of the detector settings), even if one allowed the measurements to modify the state of the objects/signals received by the two experimenters?

No; I would not agree.

*But (to be sure we agreeing on the question), I would surely like you to spell out all the conditions, especially any that you see relating to the move in Bell's (1964) maths which no experiment can confirm. I refer to the unnumbered equations between his (14) and (15).

PS: (1) I have not forgotten an old question of yours and have been waiting a reply from a central authoritative source.

(2) Is it not fascinating that Bell should leave unnumbered the most crucial equations in his paper? Did he have doubts? Remember (as I understand the position): He [like me] did not like his theorem!

So let's be sure of the conditions; regards, wm

Feb21-07, 12:20 AM

What EPR-B correlations are you taking about?

You need to make a specific statement and let's discuss that. I will be glad to discuss any side of Bell's Theorem you want to discuss. I have already referenced several derivations of Bell on my web pages, so I am not sure what you are asking.

To make it clear: I advocate a standard reading of EPR/Bell/Aspect. Bell realism is as limited - or not - as you care to view it. There are those who refer to it as "naive realism" but I personally reject that description (as would Einstein, who was a realist of the same vein).

The math of the Bell realism assumption is simple: assume the simultaneous existence of pre-determined values for 3 non-commuting spin operators (A, B and C). Then prepare a table which shows these 8 permutations when measured as up/down (electrons) and the relative percentages. You will find that it is not possible to create such a table AND have it agree to experiment UNLESS you put negative percentages in some spots.

If you think we should reject "Bellian realism" and accept "locality": I think that is a perfectly sensible interpretation and have no issue with it. But I doubt that most folks will conclude that local realism is still a viable option just because "Bellian realism" is too "limited". If you can come up with an acceptable alternative definition of realism, I would be interested in seeing it. (Emphasis added.)

1. DrC, EPRB, EPR-B stands here for EPR-BOHM. (EPR-Bell is written for EPR-BELL; and I am not seeking a derivation of Bellian Inequalities.) So I am sincerely seeking your derivation of the related correlation:

(1) CORRELATION (EPR-Bohm; spin-half particles) = -a.b'

per terms in OP.

2. Thank you for the realisation that one can drop Bellian-realism and RETAIN LOCALITY. That exactly summarises my position.

3. That alternative definition of realism (called CLR = Common-sense local realism) is on my website (known to you). I'd welcome some critique of it before throwing it in here.

4. So: Could I see your maths for EPR-Bohm, please?

Thanks, wm

Feb21-07, 12:32 AM

So what if an observation perturbs a system under study? That in no ways explains anything, and it certainly does not explain Bell test results. This is pure hand-waving, and is just as true in the classical world.

Hand-waving?

I had the impression that Bell thought (counter-factually) that an unmeasured system had the property that would have been revealed IF that system had been measured.

That is why he endorsed ''the d'Espagnat move'' mentioned by me here earlier. Thus:

I'd like to encourage you in the view that: It surely makes more sense to get rid of pseudo-realism (= limited realism = Bellian realism) than locality.

Thus Bell once strongly endorsed a derivation of his inequalities by d'Espagnat (Sci. Am. November 1979). In it you find this move:

"These conclusions require a subtle but important extension of the meaning assigned to the notation A+. Whereas previously A+ was merely one possible outcome of a measurement made on a particle, it is converted by this argument into an attribute of the particle itself.'' (Emphasis added.)

In my view, most quantum objects are perturbed by ''measurement'' and that is why Bellian Inequalities are breached by quantum objects! Bellian realism being of very limited validity.

PS: As I recall, Bell said he could do no better than d'Espagnat!

wm

DrC, Are you saying that Bellian Inequalities are based on measurement perturbation?

Regards, wm

JesseM

Feb21-07, 12:39 AM

Hi JesseM,

Would I agree that it is impossible to violate the Bell inequalities classically if one obeyed all the conditions [see* below] specified in the proof of Bell's theorem (including the condition that the state of the objects/signals emitted by the source be independent of the detector settings), even if one allowed the measurements to modify the state of the objects/signals received by the two experimenters?

No; I would not agree.

*But (to be sure we agreeing on the question), I would surely like you to spell out all the conditions, especially any that you see relating to the move in Bell's (1964) maths which no experiment can confirm. I refer to the unnumbered equations between his (14) and (15). Well, on the previous thread I already attempted to spell out all the conditions I thought were relevant, in post #133 (http://www.physicsforums.com/showpost.php?p=1215310&postcount=133): do you agree or disagree that if we have two experimenters with a spacelike separation who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like), then if they always get opposite answers when they make the same measurement on any given trial, and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement with no violations of locality allowed (and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event, as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation), then the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3 I guess I should note that when I say "the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial", this refers to the assumption that the detailed state of the objects/signals sent out by the source, a state which we assume implies a predetermined answer to every measurement (otherwise I don't see any way of explaining how both experimenters always get the same answer when they make the same measurement), is not in any way correlated with or informed by the experimenters' choice of detector settings on that trial.

Also, note that I am not making any assumption that when they make a measurement of a property, they are simply revealing a property which was already present in the state before measurement. I only assume that the state before measurement + the choice of detector setting determines the outcome of the measurement completely. For example, if the experimenter measures a particle on axis A and gets the result "spin-up", this need not imply the particle was somehow in a spin-up state on axis A before it was measured; it just implies that the state of the particle before measurement was such that it was guaranteed that if the detector was on setting A on the measurement, the result would come back "spin-up". Again, without assuming this sort of determinism, there seems to be no way that you could explain how both experimenters always get the same result when they make the same measurement, and still satisfy all the conditions I describe above. Would you agree, at least, with this necessity for determinism in the outcome given both the state of the object/signal emitted by the source on a trial and the choice of detector setting, if the experimenters do indeed get the same result on every trial where they choose the same setting, and the object/signal is a purely classical one, and all my conditions above are being obeyed? PS: (1) I have not forgotten an old question of yours and have been waiting a reply from a central authoritative source. Ultimately it is not really important whether any given physicist remembered to include the condition I mentioned in their statements of Bell's theorem or not (although I've shown that several do in their papers); all that's really important is my claim that if you include that condition, along with others I mention, then it is impossible to violate any Bell inequalities classically, but possible to violate them in quantum physics (we are, I hope, debating the physical question of whether quantum results are compatible with local realism, not the historical question of whether Bell or any other particular physicist remembered to state all the relevant conditions in their proofs). If you disagree, then you should be able to come up with a classical experiment where this condition and the other ones I mentioned are all obeyed, yet some Bell inequality is violated; your previous example involving classical polarized light and the source being "yoked" to Alice's detector setting obviously does not obey all my conditions.

DrChinese

Feb21-07, 12:42 AM

DrC, Are you saying that Bellian Inequalities are based on measurement perturbation?

I am saying that perturbation from measurement has nothing to do with explaining why quantum systems violate Bell Inequalities. It is either because a) realism is a bad assumption; or b) locality is a bad assumption.

Your statement about A+ has nothing to do with this. What Bell thought about his theorem in later years does not prove anything anyway.

DrChinese

Feb21-07, 12:47 AM

1. DrC, EPRB, EPR-B stands here for EPR-BOHM. (EPR-Bell is written for EPR-BELL; and I am not seeking a derivation of Bellian Inequalities.) So I am sincerely seeking your derivation of the related correlation:

(1) CORRELATION (EPR-Bohm; spin-half particles) = -a.b'

per terms in OP.

2. Thank you for the realisation that one can drop Bellian-realism and RETAIN LOCALITY. That exactly summarises my position.

3. That alternative definition of realism (called CLR = Common-sense local realism) is on my website (known to you). I'd welcome some critique of it before throwing it in here.

4. So: Could I see your maths for EPR-Bohm, please?

Thanks, wm

1. I know what EPR-B stands for. I have no idea of the context.

2. Good.

3. I do not know what common sense realism is. Bell's Realism is pretty common sense to most people.

4. What math are you talking about? Are you talking about the predictions of QM for spin 1/2 particles?

vanesch

Feb21-07, 01:03 AM

I only assume that the state before measurement + the choice of detector setting determines the outcome of the measurement completely. For example, if the experimenter measures a particle on axis A and gets the result "spin-up", this need not imply the particle was somehow in a spin-up state on axis A before it was measured; it just implies that the state of the particle before measurement was such that it was guaranteed that if the detector was on setting A on the measurement, the result would come back "spin-up". Again, without assuming this sort of determinism, there seems to be no way that you could explain how both experimenters always get the same result when they make the same measurement, and still satisfy all the conditions I describe above.

Indeed. One could illustrate this with the following observation:
Imagine two people, Alice in New York, and Bob in Tokyo, throwing each 1000 times a dice in the following way. They can choose, for each of their 1000 trials, to use a red, a blue or a green box at there disposal ; then they throw the dice in the box of their choice, and write down the outcome and the color of the box they chose.

Note that, if Bob picks the red box for his 52th throw, then he will never know what he would have gotten if instead he'd have picked the green one. And if he next picks the green one, that will not be his 52th, but his 53th throw.

The funny thing now, is when Bob and Alice come together, that they find out that each time that, by coincidence, they picked the same color, well, they also got the same outcome ! Of course, in advance, they cannot know for which throws they will pick the same color, and they cannot determine the outcome. But they simply see that in those particular cases WHEN they pick the same color, then they ALWAYS obtain the same outcomes.

Now, this funny correlation would be totally incomprehensible if there were not some "common origin" or "some action at a distance" between the dice, right ? And if we exclude the last possibility, then we would be looking at some very funny phenomenon. It would be black magic, until we proposed some MECHANISM by which both dice would somehow, in advance, know what to set as a result as a function of the color of the box. One would go and look at the producer of the dice: maybe he put some very complicated mechanism inside each of them.

People who dismiss any "a priori" determinism of the outcomes in an EPR experiment, ought to feel totally comfortable with the above situation, under the motto: "correlations happen".

Feb21-07, 01:13 AM

1. I know what EPR-B stands for. I have no idea of the context.

2. Good.

3. I do not know what common sense realism is. Bell's Realism is pretty common sense to most people.

4. What math are you talking about? Are you talking about the predictions of QM for spin 1/2 particles?

In reply, by number:

1. EPR-Bohm has two spin-half particles in the singlet state; correlation as previously given here.

2. Good.

3. I told you where to find a definition of CLR (= common-sense local realism) but you do not look? Bell's realism is common-sense? Particles unperturbed ... that A+ ''d'Espagnat move'' again? Are you saying that this Bell-endorsed move is of no consequence?

I say: Drop such ''nonsense'' and such ''fiddles'' and retain LOCALITY?

4. I am talking about you deriving the EPR-Bohm correlation of -a.b' (per terms in OP) so that I can better understand the realism that you hold to; or the locality that you reject; and the terminology that you support mathematically.

If you do the maths, I am presuming we might agree re the terms and come to some agreement about the validity (or otherwise) of LQM (Local QM).

Was it Feynman who said: Do the maths, or risk parrotting the errors of others?

wm

Demystifier

Feb21-07, 04:05 AM

Wm was right: This really is OK corral. :biggrin:

DrChinese

Feb21-07, 10:28 AM

3. I told you where to find a definition of CLR (= common-sense local realism) but you do not look? Bell's realism is common-sense? Particles unperturbed ... that A+ ''d'Espagnat move'' again? Are you saying that this Bell-endorsed move is of no consequence?

I say: Drop such ''nonsense'' and such ''fiddles'' and retain LOCALITY?

If you want to push a new definition of realism, bring it out where we can discuss it. But I don't see the purpose of a new definition when the current one is so well accepted. After all, it is exactly what Einstein would have expected.

Bell's realism is common sense, that is why Bell's Theorem is so important. If it did not match up to something most people can understand, it would not be as important.

I don't know why you are making a point about some "move" you are saying Bell endorsed. For all I know, he endorsed Richard Nixon (this is a joke, because he was not an American). The point is that Bell's Theorem stands as written and is generally accepted as such. There has been much debate about whether "hidden variables" exist or not, and if so, whether they are intrinsic particle attributes. If you deny Bell realism (as I am prone to do), none of that matters.

1. As to the math of the realism requirement, the usual presentation is essentially as follows:

1 >= P(A, B, C) >= 0

where A, B and C are 3 simultaneously "real" hidden variables (or attributes, or measurement setting outcomes).

2. As to the correlation in an EPR-B setup with electrons, the usual formula for matches (both up, or both down) is:

p(Match) = sin^2(\Theta/2)

The only significant difference for electrons versus photons being that there is the factor of 1/2 applied to electrons to adjust for being a spin 1/2 particle, while photons have a factor of 1 being a spin 1 particle. Also, entangled photons pairs are usually created by either Type I or Type II PDC. Type II gets a sin^2 function for matches while Type I has the cos^2 function.

complexPHILOSOPHY

Feb21-07, 11:10 AM

Allow me to interject for a moment to pose a question for philosophical clarity. If the term 'common sense realism' is being used in the context that I suppose it might, are you referencing Thomas Reid?

That is the only 'common sense realism' that I am familiar with.

EDIT: Now that I read through some of the posts, I don't think you are referencing the philosopher. However, when I search for "Common Sense Local Realism" on google, I get returned back to physicsforums.

This leads me to believe that you should just tell me what it is, since google directed me here MY FRIEND!!!! <3333

PAYCEEEE HOMIES.

Feb21-07, 01:50 PM

Well, on the previous thread I already attempted to spell out all the conditions I thought were relevant, in post #133 (http://www.physicsforums.com/showpost.php?p=1215310&postcount=133): I guess I should note that when I say "the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial", this refers to the assumption that the detailed state of the objects/signals sent out by the source, a state which we assume implies a predetermined answer to every measurement (otherwise I don't see any way of explaining how both experimenters always get the same answer when they make the same measurement), is not in any way correlated with or informed by the experimenters' choice of detector settings on that trial.

Also, note that I am not making any assumption that when they make a measurement of a property, they are simply revealing a property which was already present in the state before measurement. I only assume that the state before measurement + the choice of detector setting determines the outcome of the measurement completely. For example, if the experimenter measures a particle on axis A and gets the result "spin-up", this need not imply the particle was somehow in a spin-up state on axis A before it was measured; it just implies that the state of the particle before measurement was such that it was guaranteed that if the detector was on setting A on the measurement, the result would come back "spin-up". Again, without assuming this sort of determinism, there seems to be no way that you could explain how both experimenters always get the same result when they make the same measurement, and still satisfy all the conditions I describe above. Would you agree, at least, with this necessity for determinism in the outcome given both the state of the object/signal emitted by the source on a trial and the choice of detector setting, if the experimenters do indeed get the same result on every trial where they choose the same setting, and the object/signal is a purely classical one, and all my conditions above are being obeyed?

Ultimately it is not really important whether any given physicist remembered to include the condition I mentioned in their statements of Bell's theorem or not (although I've shown that several do in their papers); all that's really important is my claim that if you include that condition, along with others I mention, then it is impossible to violate any Bell inequalities classically, but possible to violate them in quantum physics (we are, I hope, debating the physical question of whether quantum results are compatible with local realism, not the historical question of whether Bell or any other particular physicist remembered to state all the relevant conditions in their proofs).

If you disagree, then you should be able to come up with a classical experiment where this condition and the other ones I mentioned are all obeyed, yet some Bell inequality is violated; your previous example involving classical polarized light and the source being "yoked" to Alice's detector setting obviously does not obey all my conditions.

Dear JesseM,

1. It seems to me that some of your parenthetic comments (''otherwise I don't see any way of explaining how both experimenters always get the same answer when they make the same measurement'') would be helped by your doing the maths in detail so that you understand every step.

2. This is not to say that my maths will always be correct; nor that your words are unintelligible. But more and more I find that those who offer ''almost non-intelligible maths'' (or none at all) are those whose words I struggle most to comprehend.

3. As for determinism: I am most certainly that way inclined! Take any anti-parallel detector settings in EPRB and the detectors punch out identical (++) XOR (--) results till kingdom come.

4. You have completely missed the question that I await an answer on. Let me answer it now, without the external input that I was hoping for: It is my view that Bell (dissatisfied with his own theorem) was open to any hidden-variable theory; local or non-local. However, in my view, non-local hidden-variables are so trivial as to be unworthy of the great man. For (it seems to me) one postulates that a measurement reveals a non-local hidden-variable in one wing of the experiment AND THEN that revelation is non-locally transmitted to the other wing. UGH!

5. Please recall that my earlier CLASSICAL experiment classically refuted Bellian Inequalities, not Bell's theorem. That is, that CLASSICAL experiment complied with the (plus/minus one) conditions used to derive the CHSH etc Inequalities.

6. I'd hoped that DrC would have detailed his maths for the EPRB-correlations. Then my proposed more general CLASSICAL refutation of Bell's Theorem (meeting the more general conditions that you point to, and responding thereto) could have been posted as a counter-point. The hope was that we could see and discuss where DrC needed non-locality and where I thought that I did not!

PS: Is there a simple spot on PF where I can pick up on LaTeX? (My search revealed too much.) Though I'll probably post in a simpler but wholly adequate fashion.

7. So (to be clear): That earlier classical experiment of mine was directed at Bellian Inequalities only. The next is also wholly classical, but seeks to meet the more general Bellian conditions and establish the EPRB correlation -a.b' (per OP) in response to our discussion.

Thanks, and cheers, wm

Feb21-07, 02:08 PM

If you want to push a new definition of realism, bring it out where we can discuss it. But I don't see the purpose of a new definition when the current one is so well accepted. After all, it is exactly what Einstein would have expected.

Bell's realism is common sense, that is why Bell's Theorem is so important. If it did not match up to something most people can understand, it would not be as important.

I don't know why you are making a point about some "move" you are saying Bell endorsed. For all I know, he endorsed Richard Nixon (this is a joke, because he was not an American). The point is that Bell's Theorem stands as written and is generally accepted as such. There has been much debate about whether "hidden variables" exist or not, and if so, whether they are intrinsic particle attributes. If you deny Bell realism (as I am prone to do), none of that matters.

1. As to the math of the realism requirement, the usual presentation is essentially as follows:

1 >= P(A, B, C) >= 0

where A, B and C are 3 simultaneously "real" hidden variables (or attributes, or measurement setting outcomes).

2. As to the correlation in an EPR-B setup with electrons, the usual formula for matches (both up, or both down) is:

p(Match) = sin^2(\Theta/2)

The only significant difference for electrons versus photons being that there is the factor of 1/2 applied to electrons to adjust for being a spin 1/2 particle, while photons have a factor of 1 being a spin 1 particle. Also, entangled photons pairs are usually created by either Type I or Type II PDC. Type II gets a sin^2 function for matches while Type I has the cos^2 function.

1. I'll advance my definition of CLR (common-sense local realism) in the hope that you (and Einstein) will find it agreeable.

2. I'm heartened in this regard, noting with some cameraderie that you too are ''prone to deny Bell realism''. Let me add (from my perspective) you are moving in the direction of a growing band.

3. Thus: In my opinion: If you study the Bell-endorsed d'Espagnat move (re the subtle change re A+ -- see earlier posts), then you'll see more clearly what Bellian realism entails. I think it involves a clear denial of measurement perturbation; and I'm not sure that I know where Bell moved away from that position.

4. I'm not too happy with your maths here; though I too am not a mathematician. I'll post my maths ideas (based on my view of common-sense) to see if I can move you further in that direction.

Best regards, wm

DrChinese

Feb21-07, 02:10 PM

Dear JesseM,

1. ... would be helped by your doing the maths in detail so that you understand every step.

6. I'd hoped that DrC would have detailed his maths for the EPRB-correlations....

We already understand the math of Bell. If you have something to say, say it. It is getting old wondering when you are going to drill into a specific point. I have presented standard treatments of Bell, and so far there is no point of disagreement other than you don't seem to like Bell's realism assumption. If you want to replace his with your own, you will need to make a convincing argument of the benefit of doing so because otherwise there will be no interest in pursuing the matter.

DrChinese

Feb21-07, 02:23 PM

1. I'll advance my definition of CLR (common-sense local realism) in the hope that you (and Einstein) will find it agreeable.

2. I'm heartened in this regard, noting with some cameraderie that you too are ''prone to deny Bell realism''. Let me add (from my perspective) you are moving in the direction of a growing band.

3. Thus: In my opinion: If you study the Bell-endorsed d'Espagnat move (re the subtle change re A+ -- see earlier posts), then you'll see more clearly what Bellian realism entails. I think it involves a clear denial of measurement perturbation; and I'm not sure that I know where Bell moved away from that position.

4. I'm not too happy with your maths here; though I too am not a mathematician. I'll post my maths ideas (based on my view of common-sense) to see if I can move you further in that direction.

Best regards, wm

1. Einstein said of realism: "I think that a particle must have a separate reality independent of the measurements. That is: an electron has spin, location and so forth even when it is not being measured." That is Bell realism to a tee, and I don't see how you are going to do Bell & Einstein one better. I can tell you for sure that I am not interested in a semantics debate on realism. There needs to be a connection to the physics.

2. I haven't moved anywhere on this subject. :smile:

3. This has nothing to do with anything. I assume there is measurement perturbation, the question is what is the significance of it? Is it physical? Does it have non-local impact? etc.

4. Sorry professor! You are always welcome to add your improvements.

-DrC

DrChinese

Feb21-07, 02:28 PM

3. Re: ''no point of disagreement other than you don't seem to like Bell's realism assumption''. I just wanted to check that you ALSO are moderately sympathetic to this point of disagreement? Yes?

No, not in the least. Bell's realism is well-defined, and it is reasonable to reject it as an assumption (while keeping locality as an assumption). That does NOT mean it is a poor definition - it is a good one and that is a big part of why Bell's Theorem is so strong!

A lot of people reject realism as an assumption, and I am hardly the first. Don't confuse the "assumption" with the "definition of the assumption" - they are entirely different.

vanesch

Feb21-07, 02:37 PM

That is the only 'common sense realism' that I am familiar with.

I don't think it is a technical term. Look at my post #42. The "common sense realism" is the idea that these correlations between dice throws of Alice and Bob "just cannot be" if there is no common causal origin.
It comes down to the idea that, when correlations occur where they are not a priori expected for some logical reason, there must be some causal link (by common origin, or by influence), and that this causal link must be found in some ontologically existing mechanism.
The reason to adhere to it, is that it is our main (and only) technique of inference and empirical enquiry.

Indeed, if you throw a switch, and a light goes on, and you throw it again, and the light goes out, and you do this 20 times, you expect there to be some kind of ontological mechanism to exist which explains this. It doesn't necessarily mean that the switch causes the light to go on. A technician might observe you through a camera, and steer the light as you throw a non-connected switch. Or worse, he might switch on and off a light, and "send you some brain waves" which make you throw the switch. Or even better, when you walked into the room, a scanner might have found out the exact state of your brain, and calculated at what moments you will throw the switch. One extravagant explanation even worse than the other. But you prefer that, over: well, correlations happen. There's no relationship. This desire of explaining correlations is, I think, what is meant by "common sense realism": there must be something real, which is the mechanism which explains the correlations.

Feb21-07, 03:00 PM

I don't think it is a technical term. Look at my post #42. The "common sense realism" is the idea that these correlations between dice throws of Alice and Bob "just cannot be" if there is no common causal origin.
It comes down to the idea that, when correlations occur where they are not a priori expected for some logical reason, there must be some causal link (by common origin, or by influence), and that this causal link must be found in some ontologically existing mechanism.
The reason to adhere to it, is that it is our main (and only) technique of inference and empirical enquiry.

Indeed, if you throw a switch, and a light goes on, and you throw it again, and the light goes out, and you do this 20 times, you expect there to be some kind of ontological mechanism to exist which explains this. It doesn't necessarily mean that the switch causes the light to go on. A technician might observe you through a camera, and steer the light as you throw a non-connected switch. Or worse, he might switch on and off a light, and "send you some brain waves" which make you throw the switch. Or even better, when you walked into the room, a scanner might have found out the exact state of your brain, and calculated at what moments you will throw the switch. One extravagant explanation even worse than the other. But you prefer that, over: well, correlations happen. There's no relationship. This desire of explaining correlations is, I think, what is meant by "common sense realism": there must be something real, which is the mechanism which explains the correlations. (Emphasis added.)

YES! And going one step deeper: We expect there to be something real which accounts for the correlata.

Such is enlightened human nature, perhaps?

Recalling that the preparation of the highly-correlated singlet state (with its spherical symmetry), produces one of the most highly correlated states (= highly correlated correlata) that we know of.

wm

Feb21-07, 03:12 PM

No, not in the least. Bell's realism is well-defined, and it is reasonable to reject it as an assumption (while keeping locality as an assumption). That does NOT mean it is a poor definition - it is a good one and that is a big part of why Bell's Theorem is so strong!

A lot of people reject realism as an assumption, and I am hardly the first. Don't confuse the "assumption" with the "definition of the assumption" - they are entirely different.

Sorry Doc, but I'm lost and confused again. Beyond belief!

<<<Here is my assumption: LEFT BLANK.

Please, dear Professor, Do not confuse my assumption with the definition of my assumption.

Ah (light dawns): perhaps you DrC are relying on non-local transmission of my assumption.>>>

My problem! But to say ''a lot of people reject realism'' without in any way qualifying the realism of which you speak ... well that continues to be beyond me.

For now, I think it best that I find my old maths ... and maybe become (with hard study) a mathematician.

Believing, as I do, that: Maths is the best logic; and I've much to learn = comprehend.

Respectfully, wm

DrChinese

Feb21-07, 03:23 PM

DrC,

1. Interesting: I see no mention of perturbation in your citation?

But it is specifically addressed in my definition of ''common-sense local realism'': <link removed> Would you (Einstein, Bell) disapprove? (PS: AND NB: I'm sure it can be improved!)

2. To quote my erstwhile friend: ''If you deny Bell realism (as I DrChinese am prone to do), none of that matters.''
Sorry; English is not my first language. Your proneness (= lying horizontal?) I misunderstood.

3. Non-local impact??? What would that be? Peres and many others poo-poo any physicality with non-locality (I seem to recall). I am with them re physicality, since (as I before wrote to you?): How can an abstract non-physical wave not in space-time influence a concrete object in space-time?

1. The issue of perturbation by the measurement apparatus can be looked at any way you want to. It really does not affect the mathematical description of realism, which is:

1 >= P(A, B, C) >= 0

The reason is that A, B and C individually are elements of reality, because they can be predicted in advance in a Bell test. The issue is not whether the measurement somehow distorts the results, it is whether these elements of reality (EPR) exist simultaneously independently of the ACT of observation.

2. There are 2 primary assumptions in Bell's Theorem, and Bell test results lead us to deny at least one of them. You are fully justified in rejecting either one, it is simply a matter of personal preference. My preference happens to be to accept locality and deny realism, but the reverse is an acceptable position. The unjustified conclusion is to reject neither.

3. Non-locality is an issue of vital relevance. It is one of the 2 primary assumptions in Bell's Theorem, and cannot be dismissed lightly.

DrChinese

Feb21-07, 03:26 PM

My problem! But to say ''a lot of people reject realism'' without in any way qualifying the realism of which you speak ... well that continues to be beyond me.

Again, the definition is as follows:

1 >= P(A, B, C) >= 0

This is what Bell realism is, and it remains an accepted standard definition.

Feb21-07, 03:31 PM

Sorry Doc, but I'm lost and confused again. Beyond belief!

<<<Here is my assumption: LEFT BLANK.

Please, dear Professor, Do not confuse my assumption with the definition of my assumption.

Ah (light dawns): perhaps you DrC are relying on non-local transmission of my assumption.>>>

My problem! But to say ''a lot of people reject realism'' without in any way qualifying the realism of which you speak ... well that continues to be beyond me.

For now, I think it best that I find my old maths ... and maybe become (with hard study) a mathematician.

Believing, as I do, that: Maths is the best logic; and I've much to learn = comprehend.

Respectfully, [B]wm

This is a preliminary draft from wm, for critical comment, please. It responds to various requests for a classical derivation of the EPR-Bohm correlations which would nullify Bell's theorem. It's off the top of my head; and a more complex denouement might be required (and can be provided) to satisfy mathematical rigour:

(Figure 1) D(a) -<- w(s) [Source] w'(s') ->- D'(b')

Two objects fly-apart [w with property s (a unit-vector); w' with property s' (a unit-vector)] to respectively interact with detectors D (oriented a, an arbitrary unit vector) and D' (oriented b', an arbitrary unit vector). The detectors D (D') respectively project s (s') onto the axis of detector-orientation a (b').

Let w and w' be created in a state such that

(1) s + s' = 0; say, zero total angular momentum. That is:

(2) s' = -s.

Then the left-hand result is a.s and the right-hand result is s'.b'; each a dot-product.

To derive the related correlation, we require (using a recognised notation http://en.wikipedia.org/wiki/Column_vector ), with < ... > denoting an average:

(3) <(a.s) (s'.b')>

(4) = - <(a.s) (s.b')>

(5) = - <[(ax ay az) (sx, sy, sz)] [(sx sy sz) (bx', by', bz')]>

(6) = - (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz')

(7) = - (ax ay az) <s.s> (bx', by', bz')

(8) = - (ax ay az) <1> (bx', by', bz')

(9) = - a.b'

(10) = - cos (a, b').

Let s and s' be classical angular-momenta. Then (to the extent that we meet all the Bell-theorem criteria) the result is a wholly classical refutation of Bell's theorem. [It is Bell's constrained realism that we reject; thereby maintaining the common-sense locality clearly evident above.)

E and OE! QED?

Critical comments most welcome, (though I'll be away for a day or so),wm

JesseM

Feb21-07, 03:57 PM

1. It seems to me that some of your parenthetic comments (''otherwise I don't see any way of explaining how both experimenters always get the same answer when they make the same measurement'') would be helped by your doing the maths in detail so that you understand every step. I don't agree this would help, there is very little in the way of "math" here, and trying to write this in a more formal way would provide no conceptual illumination. But maybe it would help if I spelled out the reasoning and assumptions a little more clearly.

1. Do you disagree that in a classical world, if the results of two measurements exhibit a 100% correlation, this must be either because one measurement determined the other, or because the results of both measurements were determined by some other event or events?

2. Do you disagree if you have a 100% probability that two measurements using the same settings always give the same results as one another on repeated trials, and the two measurements have a spacelike separation and we assume no possibility of FTL signals, then the only way to explain the perfect 100% correlation in a classical world is to assume that on each trial the results were predetermined by some event or events (presumably the emission of both signals/objects from a common origin at the source) which lie in the past light cone of both measurement-events? (if you disagree, can you suggest an alternative explanation?)

3. Do you disagree that if there was any random element to the results of either experimenter's measurement on a given trial where they both use the same settings (and I'm only talking about randomness in the outcome an experimenter will get if we know both his detector setting and the precise state of the signal/object they're measuring--the original event or events which determined the state of both object/signals at the source may still have a random element), then the probability they both get the same answer could not be 100%?

If you agree that the answers must have been predetermined on the trials where they both picked the same detector setting, then if we also add the assumption that this predetermining event or events could not in any way be affected by information about what detector settings each experimenter will choose (note that this condition can be assured in a classical universe obeying locality if each experimenter chooses their setting randomly shortly before the measurement, so that a signal moving at the speed of light would not have time to travel from the event of an experimenter picking a setting to the event of the other experimenter making their measurement), then this means the answers must have been predetermined on *every* trial, even the ones where they pick different settings. 2. This is not to say that my maths will always be correct; nor that your words are unintelligible. But more and more I find that those who offer ''almost non-intelligible maths'' (or none at all) are those whose words I struggle most to comprehend. If you have difficulty comprehending anything, could you try to explain what point is confusing you? 3. As for determinism: I am most certainly that way inclined! Take any anti-parallel detector settings in EPRB and the detectors punch out identical (++) XOR (--) results till kingdom come. But I wasn't just asking if you were "inclined" this way, I was asking if you agreed it would be impossible to explain the results any other way in a classical universe obeying locality where there is no way for one measurement to causally affect the other. If you disagree or are not sure, please go through my various questions about the need for determinism above to see if you disagree with any individually. 4. You have completely missed the question that I await an answer on. Let me answer it now, without the external input that I was hoping for: It is my view that Bell (dissatisfied with his own theorem) was open to any hidden-variable theory; local or non-local. However, in my view, non-local hidden-variables are so trivial as to be unworthy of the great man. For (it seems to me) one postulates that a measurement reveals a non-local hidden-variable in one wing of the experiment AND THEN that revelation is non-locally transmitted to the other wing. UGH! You say "you have completely missed the question that I await an answer on", but you didn't actually say what this question was. And again, I am not really interested in historical questions about Bell's opinions; I am interested in trying to show that it is possible to prove that quantum results absolutely rule out local hidden variables, which you disagreed with earlier when I asked you about it. 5. Please recall that my earlier CLASSICAL experiment classically refuted Bellian Inequalities, not Bell's theorem. OK, but why do you think this is relevant? I'm not aware of any physicist in history who denied that it's trivial to violate the inequalities classically if you are allowed to violate the conditions of Bell's theorem; on the other thread I showed you a very simple way of doing this in a question-and-answer game where I get to hear both questions before answering "yes" or "no". PS: Is there a simple spot on PF where I can pick up on LaTeX? (My search revealed too much.) Though I'll probably post in a simpler but wholly adequate fashion. Yes, see the sticky thread Introducing LaTeX Math Typesetting (http://www.physicsforums.com/showthread.php?t=8997) at the top of the "Math & Science Tutorials" forum. 7. So (to be clear): That earlier classical experiment of mine was directed at Bellian Inequalities only. The next is also wholly classical, but seeks to meet the more general Bellian conditions and establish the EPRB correlation -a.b' (per OP) in response to our discussion. When you say "the more general Bellian conditions", do you mean the conditions of Bell's theorem, including the one I mentioned that there can be no correlation between the state of the signals/objects emitted by the source and the experimenters' choice of detector settings on each trial? If so, please present it--I'm quite confident you are either missing one of the conditions of Bell's theorem, or that your example does not actually violate any of the inequalities.

vanesch

Feb21-07, 03:59 PM

Then the left-hand result is a.s and the right-hand result is s'.b'; each a dot-product.

The problem is: the outcome is not a continuous quantity! It is a discrete quantity, with PROBABILITY equal to the numbers you give, with a shift. You only give the expectation values of the outcomes, but the trick is that each individual outcome is a +1 or a -1, and not a continuous value in between both (although their expectation is of course).

So the correlation is not found by taking the expectation of the product of their expectation values, but rather by taking the product of the outcomes (the +1 or -1 for each), and weighting that with the relative probabilities for this to happen ASSUMING that, whatever probability distribution is given on the A-side (as a function of the local setting and the local unit vector) for the +1 and the -1, it is INDEPENDENT of the probability distribution on the B-side (as a function of the local setting and the local unit vector there).

DrChinese

Feb21-07, 04:33 PM

Let s and s' be classical angular-momenta. Then (to the extent that we meet all the Bell-theorem criteria) the result is a wholly classical refutation of Bell's theorem.

wm,

1. This is basicly akin to saying "let's assume Bell's Theorem is wrong", which is hand-waving. You have to provide us something that yields results consistent with QM AND is local AND meets the realism requirement. You can't just say you have accomplished this because s and s' are classical.

2. Specifically, what are the expected probabilities for the 8 permutations:

A+B+C+
A+B+C-
...
A-B-C-

You will find that you cannot fill in such a table with non-negative numbers and still match QM. In other words, you have ignored the realism requirement entirely.

-DrC

JesseM

Feb21-07, 04:39 PM

Feb21-07, 04:47 PM

Feb21-07, 04:55 PM

Yeah, what Vanesch said. If the value a.s represents the probability of the left detector getting result +1, and (1 - a.s) is the probability of the left detector a getting -1, and s'.b' is the probability of the right detector getting +1, and (1 - s'.b') is the probability of the right detector getting -1, then presumably the expectation value for the correlation would be:

(a.s)*(s'.b') + (1 - a.s)*(1 - s'.b') - (a.s)*(1 - s'.b') - (1 - a.s)*(s'.b')

or

4*(a.s)*(s'.b') - 2*(a.s + s'.b') + 1

Dear JesseM, this is a bit rushed, BUT:

If I anywhere have probabilities going negative, JUST SHOOT ME!

a.s is a dot product that make take on values from -1 to +1. It cannot be a probability in my classical maths.

I haven't look at the rest of your post. I will (later) if you want me to?

Best, wm

JesseM

Feb21-07, 05:02 PM

Dear JesseM, this is a bit rushed, BUT:

If I anywhere have probabilities going negative, JUST SHOOT ME! Sorry, I didn't realize that with the way you allow s and a to vary, a.s could be negative; but see below. a.s is a dot product that make take on values from -1 to +1. It cannot be a probability in my classical maths. But all the Bell inequalities I know of are based on the assumption that each measurement can only have two distinct outcomes, like "spin-up" and "spin-down". So, I was assuming that when the detector projects s onto a using the dot product, the resulting value is then used as a probability to display one of the two possible results, which I assign values +1 and -1 (following the convention used in the CHSH inequality (http://en.wikipedia.org/wiki/CHSH_Bell_test) which we've discussed before). I hadn't noticed that a.s could be negative, but you could always remedy this by making the probability equal to (1/2)(a.s) + 1/2, which will be a value between 0 and 1. Obviously this would mean you'd have to modify my equations above...if (1/2)*(a.s) + 1/2 is the probability of the left detector getting result +1, and 1 - [(1/2)*(a.s) + 1/2] = 1/2 - (1/2)*(a.s) is the probability of the left detector getting result -1, and (1/2)*(s'.b') + 1/2 is the probability of the right detector getting result +1, and 1 - [(1/2)*(s'.b') + 1/2] = 1/2 - (1/2)*(s'.b') is is the probability of the right detector getting result -1, then the expectation value for the product of their two results is:

[(1/2)*(a.s) + 1/2]*[(1/2)*(s'.b') + 1/2]
+ [1/2 - (1/2)*(a.s)]*[1/2 - (1/2)*(s'.b')]
- [(1/2)*(a.s) + 1/2]*[1/2 - (1/2)*(s'.b')]
- [1/2 - (1/2)*(a.s)]*[(1/2)*(s'.b') + 1/2]

Surprisingly, this all seems to simplify to an expectation value of (a.s)*(s'.b'), which is what you were calculating in the first place! Were you making this assumption about the probabilities all along, or is it just lucky? Either way, I think the math in your proposed proof that this is equal to -cos(a, b') is incorrect, see my next post for more on that point. Also note that if you make this assumption about probabilities, the results of the two detectors will not be perfectly correlated when they pick the same setting, so if you're out to challenge Bell's theorem, you can only look at inequalities like the CHSH inequality which do not make any assumption about perfect correlations with identical settings.

Alternately, you might avoid probabilities by saying that on any trial where the value of a.s was greater than or equal to -1 and smaller than 0, the experimenter will see the result spin-down (assigned a value of -1), and on any trial where the value of a.s was greater than or equal to 0 and smaller than or equal to 1, the experimenter will see the result spin-up (assigned a value of +1). Then you could say the same applies to s'.b', and calculate the expectation value of the product of their two results; but again, it would be something different than -cos (a, b'). Quickly diagramming the problem leads me to think that if s is equally likely to have any angle, then if (a, b') represents the angle between a and b' in degrees, the probability that they both get the same spin would be (a, b')/180, and the probability they get opposite spins would be [180 - (a, b')]/180, so the expectation value for the product of their results would be (a, b')/180 - [180 - (a, b')]/180, or [(a, b')/90] - 1.

Either way, I think you need to fix it so each experimenter can only get two discrete results on a given trial. If you know of any Bell inequalities that do not assume each measurement can have only one of two possible results, then please give the name of the inequality you're thinking of, or a link giving the mathematical formulation of the inequality.

JesseM

Feb21-07, 06:45 PM

To derive the related correlation, we require (using a recognised notation http://en.wikipedia.org/wiki/Column_vector ), with < ... > denoting an average:

(3) <(a.s) (s'.b')>

(4) = - <(a.s) (s.b')>

(5) = - <[(ax ay az) (sx, sy, sz)] [(sx sy sz) (bx', by', bz')]>

(6) = - (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz')

(7) = - (ax ay az) <s.s> (bx', by', bz')

(8) = - (ax ay az) <1> (bx', by', bz')

(9) = - a.b'

(10) = - cos (a, b').
To add to the issue I brought up in my previous post about the need for only two possible outcomes, there also seems to be an error in your math here. Let's say a = 0 degrees, b' = 60 degrees, and s = 90 degrees. Since they all are of unit length, the dot product of any of these two vectors is just the cosine of the angle between them. So from (4) we have - (a.s) (s.b') = - cos(90) * cos(30) = 0. But from (9) we have have - a.b' = - cos(60) = -0.5, so (4) does not seem to be equal to (9). Steps (5) and (6) in your proof don't make sense to me--in (5), is that supposed to be two column vectors multiplied by each other? The dot product is supposed to be a row vector times a column vector, not a column vector times a column vector. If you avoid vector notation and just write out both dot products from (4) in terms of components, it seems to me (5) would be something like this:

- (a_x * s_x + a_y * s_y + a_z * s_z)*(b'_x*s_x + b'_y*s_y + b'_z*s_z)

But this is not equal to -(a_x * b'_x + a_y * b'_y + a_z * b'_z), even if you stipulate that (s_x * s_x + s_y * s_y + s_z * s_z) = 1. It seems like you got the rules for the dot product confused with the rules for multiplication, you can't say that (a.s)*(s.b') is equivalent to (s.s)*(a.b').

Feb22-07, 06:50 PM

The problem is: the outcome is not a continuous quantity! It is a discrete quantity, with PROBABILITY equal to the numbers you give, with a shift. You only give the expectation values of the outcomes, but the trick is that each individual outcome is a +1 or a -1, and not a continuous value in between both (although their expectation is of course).

So the correlation is not found by taking the expectation of the product of their expectation values, but rather by taking the product of the outcomes (the +1 or -1 for each), and weighting that with the relative probabilities for this to happen ASSUMING that, whatever probability distribution is given on the A-side (as a function of the local setting and the local unit vector) for the +1 and the -1, it is INDEPENDENT of the probability distribution on the B-side (as a function of the local setting and the local unit vector there).

Dear vanesch (and with respect: DrC and JesseM and some others as well).

This is a bit rushed as I am in a meeting BUT:

1. I'd like to point out that I began with the exact equation that Bell used [1964; equation (3)]. I get the identical result also: - a.b'.

2. NB: At the moment I have limited my derivation to that which I offered: A wholly LOCAL and CLASSICAL derivation of the EPRB correlation. That is, I have derived the limit to which your derivation must tend in accord with Bohr's Correspondence Principle.

3. I hope we might agree on the following important point: Since the space-like experimental results were derived by me in terms of high-school maths, AND without any reference whatsoever to non-locality, there must be an equivalent QM derivation equally devoid of non-locality.

4. So: May I ask you again to provide your fundamental derivation of the EPRB correlation (ie, from first principles; and preferably in the terms of the OP), beginning with Bell's equation (just as I did)?

5. I request this of you because you are a PF MENTOR and because SCIENCE ADVISER DrC has not been able to derive it and SCIENCE ADVISER JesseM is a bit confused on my mathematics (but I will sort that out soon: noting for now that there are no errors in my maths, so far as I can see from my high-school text on vector-analysis).

6. Your derivation will not be wasted as I am keen to learn. HOWEVER: If you will not be providing this important derivation; could you please point to where I might find a detailed version; preferably one that complies with your own local interpretation of QM?

7. Not to muddy the waters any further now: I respectfully suggest that there are other matters in your post which may be presented differently and more clearly. Not to be addressed now because they may be clarified when I see your EPRB derivation.

Thank you, and sincerely, wm

DrChinese

Feb22-07, 09:54 PM

5. I request this of you because you are a PF MENTOR and because SCIENCE ADVISER DrC has not been able to derive it and SCIENCE ADVISER JesseM is a bit confused on my mathematics (but I will sort that out soon: noting for now that there are no errors in my maths, so far as I can see from my high-school text on vector-analysis).

wm,

You can pick a single item out of Bell's paper, and quote it out of context and it still won't mean anything. You might consider toning down your claims a bit until you see them all the way through.

I will repeat what I have stated previously: there is a mathematical requirement that you are skipping entirely, and that is the requirement of realism. If you ignore that, you are missing the entire point of Bell. That requirement is that there is a real probability of a specified outcome of observations at settings A, B and C which has a value between 0 and 1. You do not need the formula you are tinkering with to derive Bell's Theorem, as Mermin has shown.

JesseM

Feb22-07, 10:26 PM

1. I'd like to point out that I began with the exact equation that Bell used [1964; equation (3)]. I get the identical result also: - a.b'. In physics it is important to understand what physical quantity the terms in an equation stand for. In Bell's paper a and b represent possible angles of the stern-gerlach device used to measure the spin of the two particles, and these measurements will always yield one of two results, which in the "Formulation" paragraph on p. 1 of the paper you refer to (http://www.drchinese.com/David/Bell.pdf) he labels as +1 and -1. The "expectation value" refers to the average expected value of the product of the two measurements, which would be:

(+1)*probability(angle a yields +1, angle b yields +1) +
(+1)*probability(angle a yields -1, angle b yields -1) +
(-1)*probability(angle a yields +1, angle b yields -1) +
(-1)*probability(angle a yields -1, angle b yields +1)

This experiment is not one where the result of each measurement is an arbitrary real number between -1 and +1, and where the expectation value is the average value of the product of these two real numbers, as you seem to assume in your example. Again, Bell is assuming that each measurement always yields one of two results which are assigned values +1 and -1, so when you multiply the two values you always get the result -1 or +1 on any given trial; the expectation value refers to the average this product over many trials.

As I pointed out in a previous post, if you assume that each experimenter has a device which projects the vector s onto their own angle (either a or b'), like a.s, and then this continuous value is used to determine the probability (1/2)*(a.s) + 1/2 that the experimenter will get a +1 result on that trial or a -1 result, then it actually does work out that the expectation value for the product of their results will end up being a.s*s'.b' as you had in your attempted proof. But again, in this case you don't have a guarantee that when they pick the same angle they always get opposite results on a given trial, so Bell's theorem would only rule out inequalities which don't include this assumption, like the CHSH inequality. 2. NB: At the moment I have limited my derivation to that which I offered: A wholly LOCAL and CLASSICAL derivation of the EPRB correlation. That is, I have derived the limit to which your derivation must tend in accord with Bohr's Correspondence Principle. I don't see how the correspondence principle would imply that the expectation value for an experiment in which each experimenter can get any result between +1 and -1 on a given trial would be identical to the expectation value for an experiment in which each experimenter can only get one of two results, either +1 or -1. Is this what you're claiming here? 3. I hope we might agree on the following important point: Since the space-like experimental results were derived by me in terms of high-school maths, AND without any reference whatsoever to non-locality, there must be an equivalent QM derivation equally devoid of non-locality. If you were indeed able to reproduce the result that the expectation value is -cos(a, b) in a purely classical experiment, where on each trial each experimenter gets either +1 or -1 and the expectation value is for the average of the products of their two answers, and your classical experiment obeyed the conditions of Bell's theorem like the source not having foreknowledge of the detector settings, then yes, this would show that QM was compatible with local hidden variables. The problem is you didn't do this--you seem to assume that each experiment can yield a continuous spectrum of values rather than just +1 or -1, and even if you make the assumption I mentioned above where the probability of getting +1 is (1/2)*(a.s) + 1/2, so that the expectation value is indeed just a.s*s'.b', there seems to be an error in your "high school math", since this is not equal to -cos(a, b). 4. So: May I ask you again to provide your fundamental derivation of the EPRB correlation (ie, from first principles; and preferably in the terms of the OP), beginning with Bell's equation (just as I did)? You're looking for a derivation of why quantum mechanics predicts that the expectation value is -a.b? Why would this be useful, since here we are just trying to figure out whether this expectation value can be reproduced in a classical experiment? 5. I request this of you because you are a PF MENTOR and because SCIENCE ADVISER DrC has not been able to derive it and SCIENCE ADVISER JesseM is a bit confused on my mathematics (but I will sort that out soon: noting for now that there are no errors in my maths, so far as I can see from my high-school text on vector-analysis). Yes, please state whatever theorems from your vector textbook you are making use of in your proof. But in the meantime, could you please check the math on my example of a = 0 degrees, b' = 60 degrees, and s = 90 degrees? Do you disagree that in this case, - a.s*s.b' = - cos(90)*cos(30) = - (0)*(0.866) = 0, while - cos(a, b') = - cos(60) = -0.5? If you agree with my math on this example, then it seems clear there must be an error in your proof somewhere, unless I misunderstood what you claimed to have proved. 6. Your derivation will not be wasted as I am keen to learn. HOWEVER: If you will not be providing this important derivation; could you please point to where I might find a detailed version; preferably one that complies with your own local interpretation of QM? Have you ever studied the basics of QM? Derivations of probabilities and expectation values have nothing to do with one's interpretation, they basically just involve finding state vector representing the quantum state of the system, expanding it into a weighted sum of eigenvectors of the operator representing the variable you want to measure (energy, for example), and then the square of the complex amplitude for a given eigenvector represents the probability that you'll get a given value when you measure that variable (the value corresponding to a particular eigenvector is just the eigenvalue of that vector). And of course, once you know the probability for each possible value, the expectation value is just the sum of each value weighted by its probability. If you're not familiar with the general way probabilities and expectation values are derived in QM, then a specific derivation of the expectation value for the spins of two entangled electrons won't make much sense to you. And like I said, the derivation itself would have nothing to say about locality or nonlocality, it's just when you apply Bell's theorem to the predictions of QM that you see they are not compatible with local hidden variables.

edit: by the way, if you are familiar with calculations in QM, you can look at this page (http://www.mathpages.com/home/kmath521/kmath521.htm) for a nearly complete derivation. What they derive there is that if q represents the angle between the two detectors, then the probability that the two detectors get the same result (both spin-up or both spin-down) is sin^2 (q/2), and the probability they get opposite results (one spin-up and one spin-down) is cos^2 (q/2). If we represent spin-up with the value +1 and spin-down with the value -1, then the product of their two results when they both got the same result is going to be +1, and the product of their results when they got different results is going to be -1. So, the expectation value for the product of their results is:

(+1)*sin^2 (q/2) + (-1)*cos^2 (q/2) = sin^2 (q/2) - cos^2 (q/2)

Now, if you look at the page on trigonometric identities here (http://math2.org/math/trig/identities.htm), you find the following identity:

cos(2x) = cos^2 (x) - sin^2 (x)

So, setting 2x = q, this becomes:

cos(q) = cos^2 (q/2) - sin^2 (q/2)

Multiply both sides by -1 and you get:

sin^2 (q/2) - cos^2 (q/2) = - cos (q)

This fills in the final steps to show that the expectation value for the product of their results will be the negative cosine of the angle between their detectors.

JesseM

Feb22-07, 11:44 PM

will repeat what I have stated previously: there is a mathematical requirement that you are skipping entirely, and that is the requirement of realism. When you say wm is not satisfying the requirement of realism, are you referring to the idea that wm's example involves negative probabilities? If you look at wm's post #65, I don't think that's the problem--while it's true that the dot product of the vector s sent from the source and the vector a representing the experimenters measurement setting can be negative, I don't think wm intended for this dot product to be a probability in the first place. Rather, it seems to me that wm has just failed to realize that Bell was assuming that each experiment could only yield two discrete results, spin-up or spin-down; wm is instead imagining an experiment where each experiment can yield a continuous infinity of results between -1 and 1, and the "expectation value" he's calculating is for the product of the real number that each experimenter gets as a result.

vanesch

Feb22-07, 11:58 PM

I moved the exchange between ttn and I to a new thread (what we see is bogus in MWI), because it started to hijack this one...

DrChinese

Feb23-07, 12:35 AM

When you say wm is not satisfying the requirement of realism, are you referring to the idea that wm's example involves negative probabilities? ...

JesseM,

No, I am not referring to that.

Looking at Bell as a reference, so we're talking about the same thing:

1. wm's "- a . b" deal is the QM expectation value, referenced as (3). It really doesn't matter how you get to this, the key is that you know that it reduces to cos theta for spin 1/2 particles. Obviously, for photons it is a slightly different formula. If this was all there was to it, then Bell would have stopped after (7) or so - and wouldn't have too much.

2. Bell then goes to great pains to show that the mapping of (2) to (3), with A and B, WILL work. I.e. that a hidden A and hidden B is possible, and will get you to the QM predictions if you need it to. So we still don't have much.

3. Then around (14), Bell introduces the realism requirement: the mapping with hidden variables does not extend to an A, B AND C! He does not label it as the "realism requirement", that is something I label it as because it is present in EVERY proof of Bell's Theorem one way or another. It is the assumption - requirement - that there simultaneously be an A, B and C to discuss. If there isn't, then there is no (15) which is the Inequality.

Why do we need this assumption? Because without it we can't see the real problem that occurs when wm says:

"(4) = - <(a.s) (s.b')> becomes (9) = - a.b' "

Clearly this is the original classical hidden variable idea in disguise, which Bell says is "no difficulty". But this doesn't work if there is a c too, and Bell somehow figured that out. (Amazing accomplishment to me...)

4. In my proofs of Bell's Theorem, I always make this assumption *explicit*. I will repeat that without simultaneous A, B and C: there is no Bell's Theorem. You can reformulate the theorem in many ways, such as my negative probabilities version (http://drchinese.com/David/Bell_Theorem_Negative_Probabilities.htm) and my version that follows Mermin (http://drchinese.com/David/Bell_Theorem_Easy_Math.htm). These versions substitute easier math, or at least an easier notation for most people to follow - and is built around one specific counter-example. Bell presents a more general proof and then picks a specific example (ac=90 degrees, ab=bc=45 degrees is how I read it) to show the issues.

-DrC

JesseM

Feb23-07, 12:55 AM

It is the assumption - requirement - that there simultaneously be an A, B and C to discuss. But why do you think wm's example violates this assumption of realism? In his example, if one experimenter can pick for his measurement setting a one of three possible angles a = A, a = B, or a = C, then if you know the angle of the vector S sent out by the source, you can determine in advance the value of A.S if the experimenter picks angle A, and the value of B.S if the experimenter picks B, and the value of C.S if the experimenter picks C. Of course these values may be any real number between -1 and 1 depending on the angles involved, whereas in the experiments Bell is dealing with, "local realism" means that if you know the hidden state of the particle, then you can determine in advance whether each of the three detector angles will yield either the result -1 or +1, with no other possibilities.

DrChinese

Feb23-07, 01:08 AM

But why do you think wm's example violates this assumption of realism? In his example, if one experimenter can pick for his measurement setting a one of three possible angles a = A, a = B, or a = C, then if you know the angle of the vector S sent out by the source, you can determine in advance the value of A.S if the experimenter picks angle A, and the value of B.S if the experimenter picks B, and the value of C.S if the experimenter picks C. Of course these values may be any real number between -1 and 1 depending on the angles involved, whereas in the experiments Bell is dealing with, "local realism" means that if you know the hidden state of the particle, then you can determine in advance whether each of the three detector angles will yield either the result -1 or +1, with no other possibilities.

It is not that he violates it, it is that he has not included it. You can see that his "formula" is simply the early part of Bell's paper. So nothing has happened! It is as if he showed 1+1=2 and says that proves classical reality. He is trying to assert that classical local hidden variables is equivalent to the predictions of QM, which we already know is completely wrong.

In other words: his formula may have some issues with it, but no one is really doubting that Bell's (2) and (3) can be made to work together as long as you limit it to considering a and b. Bell himself says exactly that! And then he introduces c, and that leads immediately to the Inequality.

So when you include the reality assumption, you get the Inequality. The Inequality is violated in nature; therefore one of the assumptions is wrong. The assumptions are locality and realism; and one of these needs to be thrown out.

P.S. Besides, you can't do what you are saying about A.S, B.S and C.S. - this is precisely what Bell shows. The reason is that these 3 cannot be made to be internally consistent. I.e. the relationship A.S to B.S, A.S to C.S, and B.S to C.S won't work.

JesseM

Feb23-07, 01:20 AM

In other words: his formula may have some issues with it, but no one is really doubting that Bell's (2) and (3) can be made to work together as long as you limit it to considering a and b. But in wm's notation a and b don't represent 2 particular angles, they're just variables representing arbitrary angles chosen by the left and right detectors. You could easily say that the left detector has a choice between 3 possible angles a=A, a=B, and a=C, and likewise that the right detector has a choice between the same 3 possible angles b=A, b=B, and b=C. When you talk about Bell saying that A and B are not enough, but that you also need to consider C, don't A, B, and C represent three particular detector settings that each experimenter can choose between?

Feb23-07, 04:20 AM

To add to the issue I brought up in my previous post about the need for only two possible outcomes, there also seems to be an error in your math here. Let's say a = 0 degrees, b' = 60 degrees, and s = 90 degrees. Since they all are of unit length, the dot product of any of these two vectors is just the cosine of the angle between them. So from (4) we have - (a.s) (s.b') = - cos(90) * cos(30) = 0. But from (9) we have have - a.b' = - cos(60) = -0.5, so (4) does not seem to be equal to (9). Steps (5) and (6) in your proof don't make sense to me--in (5), is that supposed to be two column vectors multiplied by each other? The dot product is supposed to be a row vector times a column vector, not a column vector times a column vector. If you avoid vector notation and just write out both dot products from (4) in terms of components, it seems to me (5) would be something like this:

- (a_x * s_x + a_y * s_y + a_z * s_z)*(b'_x*s_x + b'_y*s_y + b'_z*s_z)

But this is not equal to -(a_x * b'_x + a_y * b'_y + a_z * b'_z), even if you stipulate that (s_x * s_x + s_y * s_y + s_z * s_z) = 1. It seems like you got the rules for the dot product confused with the rules for multiplication, you can't say that (a.s)*(s.b') is equivalent to (s.s)*(a.b').

JesseM, I appreciate both your sticking-with-me and your going-after-me BUT it seems your maths is a bit rusty! As far as I can see, there is no error in my maths; and there is more to come.

Thanks also for the cross-references that you give me.

Now, to the maths. Does this help you:

1. Note first that a, b', s and s' are unit-vectors, NOT angles.

2. And Yes; the dot product between the unit-vectors is the cosine of the angular difference.

3. (5) looks OK to me. I don't see two column vectors multiplied.

4. Note that in your equational comparison, you appear to have overlooked the fact that one expression is an ensemble average, initially over two fixed unit-vectors and two random (but opposite) unit-vectors (of infinite variety).

5. I'm pretty sure that I did not mix the rules?

6. I'm pretty sure that the maths is spot-on. But please have a look at the above comments and let me know. I've more to come but I would like to take it correct-step by correct-step.

Thanks again, in haste for now, wm

Feb23-07, 04:36 AM

But why do you think wm's example violates this assumption of realism? In his example, if one experimenter can pick for his measurement setting a one of three possible angles a = A, a = B, or a = C, then if you know the angle of the vector S sent out by the source, you can determine in advance the value of A.S if the experimenter picks angle A, and the value of B.S if the experimenter picks B, and the value of C.S if the experimenter picks C. Of course these values may be any real number between -1 and 1 depending on the angles involved, whereas in the experiments Bell is dealing with, "local realism" means that if you know the hidden state of the particle, then you can determine in advance whether each of the three detector angles will yield either the result -1 or +1, with no other possibilities.

JesseM, DrC says/claims that I violate realism (UNQUALIFIED)::: DESPITE THEIR BEING MULTITUDINOUS VARIETIES and me repeatedly requesting that he be specific).

As far as I am aware, I DO NOT violate the realism specifically defined by me. (I'll post it later). Rather, I use it (the most general common-sense local realism) IN THAT I allow specifically the measurement outcome to be a consequential perturbation of the particle-detector interaction.

The s and s' are real (and random); the a and b' are arbitrary (as they should be). The consequential projection is real. QED it seems to me?

For now, I'll leave it to you two to come to some agreement.

Regards, wm

Feb23-07, 04:43 AM

It is not that he violates it, it is that he has not included it. You can see that his "formula" is simply the early part of Bell's paper. So nothing has happened! It is as if he showed 1+1=2 and says that proves classical reality. He is trying to assert that classical local hidden variables is equivalent to the predictions of QM, which we already know is completely wrong.

In other words: his formula may have some issues with it, but no one is really doubting that Bell's (2) and (3) can be made to work together as long as you limit it to considering a and b. Bell himself says exactly that! And then he introduces c, and that leads immediately to the Inequality.

So when you include the reality assumption, you get the Inequality. The Inequality is violated in nature; therefore one of the assumptions is wrong. The assumptions are locality and realism; and one of these needs to be thrown out.

P.S. Besides, you can't do what you are saying about A.S, B.S and C.S. - this is precisely what Bell shows. The reason is that these 3 cannot be made to be internally consistent. I.e. the relationship A.S to B.S, A.S to C.S, and B.S to C.S won't work.

Limited comment: On the point of which Bellian assumption to reject, I'd like to suggest that if you studied the implication of Bell's maneuver in his unnumbered equations THEN you might just find it easy to reject Bellian realism and retain locality.

Quick suggestion only, wm PS: It works for me!

JesseM

Feb23-07, 08:24 AM

1. Note first that a, b', s and s' are unit-vectors, NOT angles.

2. And Yes; the dot product between the unit-vectors is the cosine of the angular difference. Yes, that's why I talked about angles rather than vectors, no other feature of the vector is relevant here. 3. (5) looks OK to me. I don't see two column vectors multiplied. Then your notation is unclear to me. What exactly do (ax ay az), (sx, sy, sz), (sx sy sz), and (bx', by', bz') represent, if not 4 column vectors? Of course, it's all right if they are, as long as you understand that the dot product is not equal to the product (in terms of matrix multiplication) of two column vectors, it's equal to the product of a row vector and a column vector...could you rewrite your proof so that you always include both the symbol you've been using for a dot product (.) and the symbol for multiplication (*), to distinguish between them? For instance, I assume that in (5) when you write

(5) = - <[(ax ay az) (sx, sy, sz)] [(sx sy sz) (bx', by', bz')]>

presumably this would be

(5) = - <[(ax ay az).(sx, sy, sz)]*[(sx sy sz).(bx', by', bz')]>

correct? But then when you write for (6)

(6) = - (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz')

Would this be

(6) = - (ax ay az).<(sx, sy, sz).(sx sy sz)>.(bx', by', bz')

or

(6) = - (ax ay az)*<(sx, sy, sz).(sx sy sz)>*(bx', by', bz')

Or what? Neither makes sense to me. Again, it really seems to me you are mixing up the rules for the dot product and ordinary multiplication here. 4. Note that in your equational comparison, you appear to have overlooked the fact that one expression is an ensemble average, initially over two fixed unit-vectors and two random (but opposite) unit-vectors (of infinite variety). OK, but I don't see that you really took that into account in your proof either. If the two angles of the detectors are, in radians, a and b, then the expectation value when you allow s to take any angle \theta between 0 and 2pi should be:

- \frac{1}{2\pi} \int_{0}^{2\pi} cos(\theta - a)*cos(\theta - b) \, d\theta

Are you claiming that this would work out to - cos (a - b)? If you enter Cos[x - a] Cos[x - b] into the integrator (http://integrals.wolfram.com/index.jsp), you get something fairly complicated:

(x*Cos[a - b])/2 + (-(Cos[2*x]*Sin[a + b])/2 + (Cos[a + b]*Sin[2*x])/2)/2

Calculating (expressionabove(x=2pi) - expressionabove(x=0)) gives:

(2pi*Cos[a - b])/2 + (-(Sin[a + b])/2)/2
- (-(Sin[a + b])/2)/2

Huh, so the integral actually does work out to pi*Cos[a - b], and if you multiply by the factor of -1/2pi outside the integral it comes out to -1/2 * Cos[a - b], only off from what you had by a factor of 1/2. Someone should check my math, and I still don't think your own proof makes sense, but this would at least suggest your end result is almost right. Even so, I still don't see this as a counterexample to what Bell proved about the impossibility of local hidden variables explaining quantum results, since Bell was assuming in his 1964 paper that each experimenter only sees +1 or -1 on each trial, and that when they choose the same angle they get opposite results (do you disagree that he was making these assumptions?) On the other hand, you seem to assume each experimenter can get a continuous result between -1 and +1, and even if you adopt my suggestion of having the number a.b be the basis for a probability (1/2)(a.b) - (1/2) of getting +1, you still would not ensure that the experimenters always get opposite results when they pick the same angle (anyway, if I'm right about the extra factor of 1/2 in front of the -cos(a-b), then you're not duplicating the quantum expectation value exactly, so you may not even violate the inequality in the 1964 paper in the first place, I'd have to check that). Most of the Bell inequalities I know of depend on the assumption that experimenters get opposite (or identical) results on the same measurement setting, this is the key reason for the conclusion of determinism which I talked about in an earlier post...the only exception I know of is the CHSH inequality (http://en.wikipedia.org/wiki/CHSH_Bell_test), but I don't think you'll find 4 angles a, b, a' and b' such that (-1/2)cos(a, b) − (-1/2)cos(a, b′) + (-1/2)cos(a′, b) + (-1/2)cos(a′, b′) is not between 2 and -2, which is what you'd need to violate that inequality.

DrChinese

Feb23-07, 09:30 AM

JesseM, DrC says/claims that I violate realism (UNQUALIFIED)::: DESPITE THEIR BEING MULTITUDINOUS VARIETIES and me repeatedly requesting that he be specific).

As far as I am aware, I DO NOT violate the realism specifically defined by me. (I'll post it later).

I will repeat for the Nth time: it is not that you violate the realism requirement, the problem is that you IGNORE it. It is quite specific as I have said previously: you must consider A, B and C and that leads to 8 permutations (2^3). These 8 permutations cannot be modelled with probabilities in the range 0 to 1 inclusive (0% to 100%) at certain angles (such as Bell's example A=0, B=45, C=90 degrees for spin 1/2 particles, or Mermin's example A=0, B=120, C=240 for spin 1 particles). No matter how you try, you will never be able to fill out the following table with values (for the ?) that match experiment (the ratio of any 2 columns must match the QM prediction) AND are non-negative:

Case A B C %
----- -- -- -- -----
[1] + + + ?
[2] + + - ?
[3] + - + ?
[4] + - + ?
[5] - + + ?
[6] - + - ?
[7] - - + ?
[8] - - - ?

It works like this: you can present a hundred derivations and examples that support classical local realism, and you will be in exactly the same spot as Einstein and Bohr were in circa 1935 - a pissing match. But it only takes a single counter-example to refute any theory, and that is what Bell presented in 1965. So you can ignore the realism requirement - which I have challenged you above on - and you will learn nothing about why Bell's Theorem is so important.

On the other hand, QM does not include the realism requirement. Therefore A, B and C do not need to exist simultaneously. Therefore, there are only 2^2 permutations. I can satsify the table for this quite simply (as Bell explains in the early sections of his paper):

Case A B %
----- -- -- -----
[1] + + QM expection value
[2] + - QM expection value
[3] - + QM expection value
[4] - - QM expection value

The QM expectation value will be one thing for spin 1/2 particles, another thing for spin 1 particles, the cases will add to 100%, and all values will be non-negative.

So in conclusion: looking at examples which support your local realism hypothesis is a waste of time, since you must address Bell's counter-example and it is impossible to refute that. As a result, we conclude that either the Einstein (=Bell) locality or the Einstein (=Bell) realism assumption must be rejected; and which you choose to reject is a matter of personal interpretation.

DrChinese

Feb23-07, 09:42 AM

Even so, I still don't see this as a counterexample to what Bell proved about the impossibility of local hidden variables explaining quantum results, ...

JesseM,

You are missing the big picture on this. It is not that wm is presenting a counterexample to Bell. wm is presenting an example of local realism, which Bell had already presented a counterexample to. You should be able to see that wm is simply replicating what Bell himself has already shown to be true: that looking at versions of local realism with A and B will work. But if you extend that same logic into a situation with A, B and C, it does NOT work. Don't let the derivation wm presents fool you, because it does not disprove Bell in any way.

-DrC

JesseM

Feb23-07, 09:47 AM

I will repeat for the Nth time: it is not that you violate the realism requirement, the problem is that you IGNORE it. It is quite specific as I have said previously: you must consider A, B and C and that leads to 8 permutations (2^3). wm does not specifically name 3 angles A, B, and C, but I think this was "left as an exercise for the reader" as it were. If he was correct that he had a classical example mirroring the conditions of the quantum experiment, and the expectation value for the product of the two results for any two angles X and Y was -cos(X-Y), then this is identical to the quantum expectation value, so it should be trivial to pick three specific angles A, B, C which would violate an inequality stated in terms of expectation values like the CHSH inequality (although in the specific case of the CHSH inequality you only need 2 possible angles for each detector), since we know all these inequalities can be violated in QM. Of course, as I've said before, his classical example does not mirror the conditions of the quantum experiment since he has more than two possible results. Also, my calculations above suggest the expectation value in his classical experiment would actually be -(1/2)cos(X-Y).

In any case, I agree that wm's argument would certainly be a lot clearer if he picked some specific choices of angle for each detector, and then explained which specific Bellian inequality he thinks will be violated in his experiment with those choices of angles. You are missing the big picture on this. It is not that wm is presenting a counterexample to Bell. wm is presenting an example of local realism, which Bell had already presented a counterexample to. Bell didn't present a counterexample to local realism, he presented a general proof that local realism could never work (although I suppose you could say he did this by picking an example of a quantum experiment which could never be replicated in a universe obeying local realism). So if wm was able to come up with a classical example which replicated all Bell's conditions and also violated an inequality, this would be a counterexample to Bell's proof, just like if you tried to give a proof that there was no prime number larger than 13, I could present 17 as a counterexample. Don't let the derivation wm presents fool you, because it does not disprove Bell in any way. Of course I agree with this, but for different reasons (again, because he does not replicate the conditions of the experiment were each experimenter can only get one of two possible results, either spin-up or spin-down, and also because his proof of the expectation value seems to be incorrect).

DrChinese

Feb23-07, 10:45 AM

1. wm does not specifically name 3 angles A, B, and C, but I think this was "left as an exercise for the reader" as it were.

2. Bell didn't present a counterexample to local realism, he presented a general proof that local realism could never work (although I suppose you could say he did this by picking an example of a quantum experiment which could never be replicated in a universe obeying local realism).

3. So if wm was able to come up with a classical example which replicated all Bell's conditions and also violated an inequality, this would be a counterexample to Bell's proof, just like if you tried to give a proof that there as no prime number larger than 13, I could present 17 as a counterexample.

1. That "exercise for the reader" IS Bell's Theorem. wm is asserting that A and B work, therefore it works in all situations. That is roughly like saying all prime numbers are even (because you only looked at cases that agree with your hypothesis). Make no mistake: wm is simply advocating traditional local realism. I went to his web page to make sure, and yup, there it is as big as day. He calls it "common sense realism" but it is local hidden variables with no new anything. He is simply acting as if Bell's Theorem is not valid.

2. I would definitely agree with your representation on this.

3. He doesn't consider the A/B/C condition. It is not possible to provide a counter-example to Bell, because Bell is itself a counter-example. The only way to disprove Bell would be to show that the counter-example is flawed.

For example, consider the "theory" that there no primes larger than 13. Bell comes along and says, whoa! what about 17? Now wm comes along and say Bell is wrong, look at 2, 3, 5, 7, 11, 13 as my proof. No, he must show that 17 is NOT a prime to make his case.

Now, because he has seen Bell's Theorem, Mermin (and many others, I just use him as an example) now knows the trick: there are certain specific situations (such as 17, 19, etc.) that are counter-examples. So Mermin can construct a very simple counter-examples to explain the situation, and that is his classic "Is the moon there when nobody looks? Reality and the quantum theory / Physics Today (April 1985) " (http://space.virgilio.it/baldazzi69@tin.it/papers/mermin_moon.pdf)

A review of that shows it as a fine counter-example to the original theory (local realism). And guess what? wm now must prove this wrong too, because it too is a counter-example to be contended with.

And what else? Now that they are armed with the "trick", these rather bright guys Greenberger, Horne and Zeilinger come up with yet another counter-example to local realism. And guess what? wm must prove this wrong too.

So my point is simple: there is no such thing as a counter-example to a counter-example, the counter-example must actually be proven wrong. And in this case, we now have multiple counter-examples to consider. So the burden of (dis)proof has grown exponentially larger.

JesseM

Feb23-07, 11:30 AM

1. That "exercise for the reader" IS Bell's Theorem. wm is asserting that A and B work, therefore it works in all situations. I think you're confusing the issue by using A and B to represent both specific angles and general variables representing arbitrary angles chosen by each detector. It would be simpler if you said that a was an arbitrary angle chosen by the left detector, b an arbitrary angle chosen by the right one, then you could have A, B, and C be specific choices of angles for either detector.

What wm attempted to do was give a general proof that for arbitrary angles a and b, in his classical experiment the expectation value for the product of the two results would be -cos(a - b). This would cover all specific angles you chould choose--for example, if a=B and b=C, then the expectation value for a large set of trials with these angles would be -cos(B - C); if a=C and b=A, then the expectation value for a large set of trials with these angles would be -cos(C - A); and so forth. I disagree that "Bell's theorem" primarily revolves around picking specific angles, if that's what you mean by "That 'exercise for the reader' IS Bell's Theorem". The proof involves finding an inequality that should hold for arbitrary angles under local realism; then it's just a fairly simple final step to note that the inequality can be violated using some specific angles in some specific quantum experiment, but this last step is hardly the "meat" of the theorem.

For example, look at the CHSH inequality. This inequality says that if the left detector has a choice of two arbitrary angles a and a', the right detector has a choice of two arbitrary angles b and b', then the following inequality should be satisfied under local realism:

-2 <= E(a, b) - E(a, b') + E(a', b) + E(a', b') <= 2

Now, suppose wm were correct that he had a classical experiment satisfying the conditions of Bell's theorem such that the expectation value E(a, b) would equal -cos(a - b). In this case it we could pick some specific angles a = 0 degrees, b = 0 degrees, a' = 30 degrees and b' = 90 degrees; in this case we have E(a, b) = - cos(0) = -1, E(a, b') = -cos(90) = 0, E(a', b) = -cos(30) = -0.866, and E(a', b') = -cos(60) = -0.5. So E(a, b) - E(a, b') + E(a', b) + E(a', b') would be equal to -1 - 0 - 0.866 - 0.5 = -2.366, which violates the inequality. The hard part was the proof that the expectation value was -cos(a - b), just as in QM; once we have this expectation value, it's a pretty trivial exercise for the reader to find some specific angles which allow the inequality to be violated, just as in QM. Again, the problem here is that wm did not actually replicate the conditions assumed in Bell's theorem, where each measurement can only yield two possible answers rather than a continuous spectrum of answers, and also his derivation of the expectation value seems to be flawed, my math suggested the expectation value would actually be E(a, b) = (-1/2)*cos(a - b). 3. He doesn't consider the A/B/C condition. It is not possible to provide a counter-example to Bell, because Bell is itself a counter-example. The only way to disprove Bell would be to show that the counter-example is flawed. I don't understand what you mean by "counter-example" here. Bell provides a general proof that a certain inequality can never be violated under local realism, a statement of the form "for all experiments obeying local realism and satisfying certain conditions, this inequality will be satisfied". Logically, any statement of the form "for all X, Y is true" can be disproved with a single counterexample of the form "there exists on X such that Y is false". And that's what wm tried to do--find a single example of a local realist experiment which would satisfy Bell's conditions and yet violate an inequality. But he did it incorrectly, because he didn't satisfy the conditions, and his math for the expectation value was wrong anyway, with the correct expectation value I don't think you could violate any inequality using his experiment. For example, consider the "theory" that there no primes larger than 13. Bell comes along and says, whoa! what about 17? Now wm comes along and say Bell is wrong, look at 2, 3, 5, 7, 11, 13 as my proof. No, he must show that 17 is NOT a prime to make his case. But I disagree, wm came along and tried to show a classical example that would satisfy Bell's conditions and yet give an expectation value which, with the correct choice of angles, could violate an inequality (like my choice of angles for the CHSH inequality above). If he had actually satisfied Bell's conditions and if his calculation of the expectation value were correct, this would disprove Bell's theorem; but of course he didn't do this, and since I can follow Bell's theorem and see that it is logically airtight, I am totally confident he'll never be able to do this, just like I'm confident no one will find a counterexample to the statement "there are no even prime numbers larger than 2". A review of that shows it as a fine counter-example to the original theory (local realism). And guess what? wm now must prove this wrong too, because it too is a counter-example to be contended with. Well, in what sense is this a counter-example to local realism, as opposed to a general proof that local realism cannot replicate quantum predictions? Again, when I use the word counter-example, I'm thinking of disproving a statement of the form "for all X, Y is true" by coming up with an example of the form "there exists a particular X such that Y is false". I guess you could say that if one agrees with Bell's theorem, then local realism makes the prediction that "for all experiments satisfying X conditions, inequality Y will be satisfied". And in this case, QM can give an example of the form "here's an experiment satisfying X conditions which violates inequality Y", thus proving QM is incompatible with local realism. But the problem here is that wm believes there's a flaw in Bell's theorem, so he does not agree that local realism makes the prediction "for experiments satisfying X conditions, inequality Y will be satisfied" in the first place; he's trying to disprove Bell's theorem by showing that local realism can also give an example of the form "here's an experiment satisfying X conditions which violates inequality Y". As a general approach to disproving Bell's theorem this makes sense, it's just that he thinks he's found such an example but he actually hasn't, because his example does not actually satisfy the X conditions of Bell's theorem (specifically the one about each experiment yielding one of two possible answers), and also his math for the expectation value is wrong, with the correct expectation value I'm not sure he could violate any Bellian inequality even if you ignore the first issue.

enotstrebor

Feb23-07, 11:31 AM

Case A B C %
----- -- -- -- -----
[1] + + + ?
[2] + + - ?
[3] + - + ?
[4] + - + ?
[5] - + + ?
[6] - + - ?
[7] - - + ?
[8] - - - ?

.

Just a quick note that not all of the above cases are always valid even in a classical correlation scenario. (e.g [2] and [7] can be illegal simultaneous events)

Second, one can give classical examples which also violates Bell's inequality, when one does not have the correct probability model. For example, a probability model based on behavior (making assumptions about the physics or cause of the behavior) can results in the violation of Bell's inequality.

So in conclusion: looking at examples which support your local realism hypothesis is a waste of time, since you must address Bell's counter-example and it is impossible to refute that.

What follows is speculative but raises the question of assumptions.

One of the implicit assumptions in Bell type inequalities is that "we" truly understand the physics rather than only understanding the mathematics.

For example, the average photon behavior is determined by a single "phase" variable. There is no physics understanding, on an individual photon basis, of why some photons pass through a polarizer/analyzer and others don't.

The photon's polarization properties indicates a bi-vectored object. What has not been considered is that the use of a single "phase" (an average interaction variable) in the present behavioral model does not fully describe the physics that is occurring at the analyzer/polarizer in deciding which photon's pass. It only gives the "on average phase based value".

But if the photon is bi-vectored one would actually expect that the analyzer/polarizer interaction on an individual photon basis (or individual pair basis) should be determined by two "phases" or a phase (e.g. the average of the two vectors) and a second parameter (the spread of the two vectors about the phase average).

If this is the case, then Bell's approach of adding hidden variables externally is asking the wrong question (?resulting in the wrong answer?).

If it is the "spread" rather than the phase (average of the vectors) that is fundamental to the passage through the polarizer/analyzer then the observed probability of observing a correlated pair (correlation via the hidden phase/spread aspect) can be different than the on average single "phase" would predict. Effectively the pair passes or does't pass changing the pair probability verus the average left or average right polarizar probability.

Fundamentally, Bell's inequality and associate interpretation of EPR assumes no hidden physics. A. O. Barut in a published paper essentially pointed in this direction with respect to spin 1/2 particles.

It is, for example tacitly assumed that spin 1/2 particles have only a single spin up or single spin down state. In fact given Stern-Gerlach experiments it makes more physics sense that the particle is a spin/magnetic quadrapole with two spin planes at 90 degrees (physically orthagonal rather than QM's mathematical 180 orthagonal spin planes). For such a quadrapole particle Stern-Gerlach experimental results actually make physics sense.

Finally, it has been said that a mathematical model is correct if it produces the correct experimental result. Bell's inequalities do not. So it is equally valid to assume we have the wrong mathematical model for the experimental situation (even if we do not know how this could be) as it is to assume non-locality (even if we don't know how this could be).

Maybe our understanding of the physics of particles is not as complete as our understanding of the application of the mathematical model we call QM.

But a lack of understanding of the physics is exactly what Feynman meant when he said "No one really understands QM".

JesseM

Feb23-07, 11:50 AM

Just a quick note that not all of the above cases are always valid even in a classical correlation scenario. (e.g [2] and [7] can be illegal simultaneous events) Bell's theorem allows for arbitrary probabilities to be assigned to each possible "hidden" state, including a probability of zero. Second, one can give classical examples which also violates Bell's inequality, when one does not have the correct probability model. For example, a probability model based on behavior (making assumptions about the physics or cause of the behavior) can results in the violation of Bell's inequality. You can't give a classical example which satisfies all the conditions laid out in Bell's theorem (for example, the hidden states sent out by the source must be uncorrelated to the choice of detector settings, which in a classical universe can be ensured by having the experimenters choose their detector settings too late for one's measurement-event to lie in the future light cone of the other's choice-event) and still violates an inequality. If you think you have one, please present it. One of the implicit assumptions in Bell type inequalities is that "we" truly understand the physics rather than only understanding the mathematics. Not really, Bell's theorem is just trying to rule out a certain set of assumptions about "the physics" of the situation, assumptions which go by the name of local realism; but it isn't giving any opinion on what should be put in their place once you've ruled them out. Fundamentally, Bell's inequality and associate interpretation of EPR assumes no hidden physics. No, it simply rules out assumptions about "hidden physics" which fall into the category of local realism. You are free to believe in nonlocal hidden physics as in Bohm's interpretation of QM, for example. It is, for example tacitly assumed that spin 1/2 particles have only a single spin up or single spin down state. In fact given Stern-Gerlach experiments it makes more physics sense that the particle is a spin/magnetic quadrapole with two spin planes at 90 degrees (physically orthagonal rather than QM's mathematical 180 orthagonal spin planes). For such a quadrapole particle Stern-Gerlach experimental results actually make physics sense. If you're implying each particle could be a classical quadrupole, then no, this could not possibly explain quantum experiments which violate Bell inequalities. Finally, it has been said that a mathematical model is correct if it produces the correct experimental result. Bell's inequalities do not. You're missing the point, Bell's theorem is a sort of proof-by-contradiction; Bell showed logically that if the laws of physics respect local realism, then certain inequalities must be satisfied in certain types of experiments; since we can see the inequalities are actually violated in these types of experiments, this proves that the laws of physics do not respect local realism. It isn't supposed to tell you anything else about how the laws of physics do work.

JesseM

Feb23-07, 01:43 PM

But a lack of understanding of the physics is exactly what Feynman meant when he said "No one really understands QM". Incidentally, Feynman actually derived his own version of a proof that local hidden variables could not reproduce the results of measurement of entangled particles, in his classic lecture Simulating Physics With Computers (http://www.cs.princeton.edu/courses/archive/fall04/cos576/papers/feynman82/feynman82.pdf) where he also first brought up the idea of a "quantum computer"--see sections 5-6 on pages 6-8 of the PDF (which is on pages 476-480 of the book that the pdf is scanned from). Interestingly, Feynman does not mention Bell in this lecture, and a comment here (http://dabacon.org/pontiff/?p=920) by physicist Michael Nielsen says "If I recall correctly, in a 1986 or 1987 festschrift paper for David Bohm (proceedings edited by Basil Hiley), Feynman comes pretty close to saying that he discovered Bell’s theorem before Bell." Another physicist disagrees with his recollection of the paper, so someone would have to check it to see what Feynman actually says.

Feb23-07, 02:06 PM

1. That "exercise for the reader" IS Bell's Theorem. wm is asserting that A and B work, therefore it works in all situations. That is roughly like saying all prime numbers are even (because you only looked at cases that agree with your hypothesis). Make no mistake: wm is simply advocating traditional local realism. I went to his web page to make sure, and yup, there it is as big as day. He calls it "common sense realism" but it is local hidden variables with no new anything. He is simply acting as if Bell's Theorem is not valid.

2. I would definitely agree with your representation on this.

3. He doesn't consider the A/B/C condition. It is not possible to provide a counter-example to Bell, because Bell is itself a counter-example. The only way to disprove Bell would be to show that the counter-example is flawed.

For example, consider the "theory" that there no primes larger than 13. Bell comes along and says, whoa! what about 17? Now wm comes along and say Bell is wrong, look at 2, 3, 5, 7, 11, 13 as my proof. No, he must show that 17 is NOT a prime to make his case.

Now, because he has seen Bell's Theorem, Mermin (and many others, I just use him as an example) now knows the trick: there are certain specific situations (such as 17, 19, etc.) that are counter-examples. So Mermin can construct a very simple counter-examples to explain the situation, and that is his classic "Is the moon there when nobody looks? Reality and the quantum theory / Physics Today (April 1985) " (http://space.virgilio.it/baldazzi69@tin.it/papers/mermin_moon.pdf)

A review of that shows it as a fine counter-example to the original theory (local realism). And guess what? wm now must prove this wrong too, because it too is a counter-example to be contended with.

And what else? Now that they are armed with the "trick", these rather bright guys Greenberger, Horne and Zeilinger come up with yet another counter-example to local realism. And guess what? wm must prove this wrong too.

So my point is simple: there is no such thing as a counter-example to a counter-example, the counter-example must actually be proven wrong. And in this case, we now have multiple counter-examples to consider. So the burden of (dis)proof has grown exponentially larger.

I am happy to address any question in this thread (in that the thread was initiated by me to question some prominent views which I cannot comprehend -- having struggled hard to do so).

I accept the (exponential) burden of truth; and will get to my further questions and answers soon (-- it's just that I am a bit tied-up at the moment --) because I want to learn.

Especially do I want to learn why some see a small piece of the world differently ...

... when that small piece of interest to me can be built from high-school maths and logic (which is about the limit of my current questions and ability).

So I'd just like to get it on the record:

1. that many prior and erroneous counter-examples in my small area of interest were long-held and wrong (as shown pre-eminently by John Bell).

2. that John Bell himself was not happy with his theorem and had not given up on finding a simple constructive counter-example (as I read him).

3. I am not a John Bell, but his simple approach has motivated me to have-a-go; notwithstanding that many others have had-a-go and failed.

4. So as soon as there is some general agreement that my high-school maths so far is correct, I would like to continue in that vein to mathematically answer some of the other questions here. (That is, I will move to dichotomic outcomes A = (+, -), B' = (+', -'); since doubts and concerns about this issue are being expressed here.) PS: That will introduce standard probability theory (in line with Ed Jaynes' views) which is also among some questions here.

5. To differentiate my local realism from other versions falling under the same phrase, I call it CLR: common-sense local realism. I think that CLR is the way many scientists see the world (while many -- but probably in the minority --- think that the world cannot be seen that way).

6. For those like me, that are not verbally-minded, the simple acid test that I expect to meet is that my views will be consistent with high-school math and logic.

7. Please note that other interpretations of QM support locality; and I support locality.

Sincerely, wm

Doc Al

Feb23-07, 02:17 PM

6. For those like me, that are not verbally-minded, the simple acid test that I expect to meet is that my views will be consistent with high-school math and logic.

So when do you plan to address your math error that JesseM has taken pains to point out?

Feb23-07, 02:23 PM

wm does not specifically name 3 angles A, B, and C, but I think this was "left as an exercise for the reader" as it were. If he was correct that he had a classical example mirroring the conditions of the quantum experiment, and the expectation value for the product of the two results for any two angles X and Y was -cos(X-Y), then this is identical to the quantum expectation value, so it should be trivial to pick three specific angles A, B, C which would violate an inequality stated in terms of expectation values like the CHSH inequality (although in the specific case of the CHSH inequality you only need 2 possible angles for each detector), since we know all these inequalities can be violated in QM. Of course, as I've said before, his classical example does not mirror the conditions of the quantum experiment since he has more than two possible results. Also, my calculations above suggest the expectation value in his classical experiment would actually be -(1/2)cos(X-Y).

In any case, I agree that wm's argument would certainly be a lot clearer if he picked some specific choices of angle for each detector, and then explained which specific Bellian inequality he thinks will be violated in his experiment with those choices of angles. Bell didn't present a counterexample to local realism, he presented a general proof that local realism could never work (although I suppose you could say he did this by picking an example of a quantum experiment which could never be replicated in a universe obeying local realism). So if wm was able to come up with a classical example which replicated all Bell's conditions and also violated an inequality, this would be a counterexample to Bell's proof, just like if you tried to give a proof that there was no prime number larger than 13, I could present 17 as a counterexample. Of course I agree with this, but for different reasons (again, because he does not replicate the conditions of the experiment were each experimenter can only get one of two possible results, either spin-up or spin-down, and also because his proof of the expectation value seems to be incorrect).

Jesse, As I wrote: I will add the up/down pieces as soon as we are agreed that the maths to-date is OK.

I sent a note re some of the maths; but I'm not sure if (having read them) you still find an error in the math?

Your question about column vectors is answered in the wiki reference that was in my original post.

(The column vectors include the commas! as I recall ... but its the commas that differentiate one-way or the other in accord with HS maths.)

As I read my equations: My maths is nor defective on that count: so are there any other maths issues ... before we move on to ups and downs?

Have I missed something which needs correction? Eh? wm

Feb23-07, 02:27 PM

So when do you plan to address your math error that JesseM has taken pains to point out?

Doc Al; I am apparently blind to my math error (which happens) but I sincerely am not sure what error is being identified by Jesse on this occasion (or any one else so far for that matter).

Can you help me, please?

Thanks, wm

Doc Al

Feb23-07, 02:31 PM

Start with this one:
It seems like you got the rules for the dot product confused with the rules for multiplication, you can't say that (a.s)*(s.b') is equivalent to (s.s)*(a.b').

JesseM

Feb23-07, 02:50 PM

To add to that, I think it would help a lot if you would address my previous request to make explicit where you are using the dot product and where you are using multiplication: Then your notation is unclear to me. What exactly do (ax ay az), (sx, sy, sz), (sx sy sz), and (bx', by', bz') represent, if not 4 column vectors? Of course, it's all right if they are, as long as you understand that the dot product is not equal to the product (in terms of matrix multiplication) of two column vectors, it's equal to the product of a row vector and a column vector...could you rewrite your proof so that you always include both the symbol you've been using for a dot product (.) and the symbol for multiplication (*), to distinguish between them? For instance, I assume that in (5) when you write

(5) = - <[(ax ay az) (sx, sy, sz)] [(sx sy sz) (bx', by', bz')]>

presumably this would be

(5) = - <[(ax ay az).(sx, sy, sz)]*[(sx sy sz).(bx', by', bz')]>

correct? But then when you write for (6)

(6) = - (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz')

Would this be

(6) = - (ax ay az).<(sx, sy, sz).(sx sy sz)>.(bx', by', bz')

or

(6) = - (ax ay az)*<(sx, sy, sz).(sx sy sz)>*(bx', by', bz')

Or what? Neither makes sense to me. Again, it really seems to me you are mixing up the rules for the dot product and ordinary multiplication here. And if you want to use matrix multiplication as opposed to ordinary arithmetical multiplication, you could use the symbol x to denote that. If you do, perhaps you could also label each vector as either a row vector or a column vector like (ax, ay, az)_c or (ax, ay, az)_r. (The column vectors include the commas! as I recall ... but its the commas that differentiate one-way or the other in accord with HS maths.) Wait, so are you saying that commas vs. no commas denotes column vectors vs. row vectors? In this case (sx, sy, sz)x(sx sy sz) will not be a single number, but rather a 3x3 matrix.

Feb23-07, 02:57 PM

Start with this one:

Originally Posted by JesseM
It seems like you got the rules for the dot product confused with the rules for multiplication, you can't say that (a.s)*(s.b') is equivalent to (s.s)*(a.b'). (Emphasis added.)

DocAl, Please, and with great respect: I can't find where that is said by me ... I don't find it in the original maths; and I haven't so far found it in a later post??

Do you have a source?

PS: Early on, JesseM had some confused views on my experiment and its maths so I wonder if the error might be his?

Also: If your question implies other errors, I'd be happy to correct or comment, as appropriate. Like you, I'd like to move ahead.

wm

vanesch

Feb23-07, 03:03 PM

DocAl, Please, and with great respect: I can't find where that is said by me ... I don't find it in the original maths; and I haven't so far found it in a later post??

Do you have a source?

In your post:

http://www.physicsforums.com/showpost.php?p=1252012&postcount=55

how do you go from (4) to (7) ?

The intermediate notation is confusing, and in as much as it is interpreted by several of us here, wrong, because it seems indeed that you use the "rule" that:

(a.s)(b.s) = (s.s)(a.b)

which is not correct.

Maybe you don't use that rule, but then the notation in (5) and (6) is completely unintelligible, and the transition from (4) to (7) ununderstandable.

EDIT:
to show that this rule is not correct, it is sufficient to have a counter example, and JesseM gave you one, to which you replied that he failed to understand that s was a random vector. But that doesn't show in the notation, because in (7), we still have the expectation symbols there, which make one think that the notation inside applies for each possible unit vector s individually.

Doc Al

Feb23-07, 03:05 PM

DocAl, Please, and with great respect: I can't find where that is said by me ... I don't find it in the original maths; and I haven't so far found it in a later post??

This is getting tedious. JesseM has picked apart your math (in your post #55) line by line. Either address his concerns or it's time to shut this thread down.

JesseM

Feb23-07, 03:10 PM

I think I understand wm's mistake now. He is not using either the dot product or arithmetical multiplication, but rather matrix multiplication (which I'll represent using the symbol x); and he didn't mention it until now, but he is using the convention that (sx, sy, sz) with commas represents a column vector and (sx sy sz) without commas represents a row vector. So his mistake is in going from 6 to 7, where he treats a column vector x row vector, namely (sx, sy, sz) x (sx sy sz), as equal to the dot product of (sx, sy, sz) with itself; in fact, in matrix multiplication a 3-component column vector x a 3-component row vector gives a 3 x 3 matrix, not a scalar like the dot product. Unlike in arithmetical multiplication, in matrix multiplication order is critical.

Feb23-07, 03:16 PM

To add to that, I think it would help a lot if you would address my previous request to make explicit where you are using the dot product and where you are using multiplication: And if you want to use matrix multiplication as opposed to ordinary arithmetical multiplication, you could use the symbol x to denote that. If you do, perhaps you could also label each vector as either a row vector or a column vector like (ax, ay, az)_c or (ax, ay, az)_r. Wait, so are you saying that commas vs. no commas denotes column vectors vs. row vectors? In this case (sx, sy, sz)x(sx sy sz) will not be a single number, but rather a 3x3 matrix.

Jesse, you may be at the nub of the problem.

But from my readings, I thought that I was using a well-accepted compact notation. One that would not lead us astray!? One that I would like to stay with (for compactness).

And just to be sure, that is why I referenced the wiki page in support.

Now I have to ask: Why would you want to go down the MATRIX path when the s.s' dot product does the job?

Or have I taken an invalid route? I did it off the top of my head, and have done it often, BUT I would want to be correct mathematically.

PLEASE: This would be a point where I would appreciate some hand-holding and the detailed explanations that you are good at.

I guess the question is: What is <s.s'>? Is it other than unity?

Please be expansive here, for maybe this is where I need to apologise and bail-out?

I'm thinking not; but lots of helpful notes will help me to decide. Thanks, wm

JesseM

Feb23-07, 03:30 PM

Jesse, you may be at the nub of the problem.

But from my readings, I thought that I was using a well-accepted compact notation. One that would not lead us astray!? One that I would like to stay with (for compactness). I hadn't seen it before, so it would have helped if you mentioned it instead of just linking to the wikipedia page which had a bunch of other information on it as well. But now that I understand it's fine. Now I have to ask: Why would you want to go down the MATRIX path when the s.s' dot product does the job?

Or have I taken an invalid route? I did it off the top of my head, and have done it often, BUT I would want to be correct mathematically. Yes, you made an invalid step. The only sense in which it makes sense the row vector (ax ay az) multiplied by the column vector (sx, sy, sz) is the same as the dot product of a.s is when you are using matrix multiplication--if you don't want to be using matrix multiplication, then you can't replace a.s with (ax ay az)(sx, sy, sz) and have it make sense. But if you're using matrix multiplication, a column vector (sx, sy, sz) times a row vector (sx sy sz) is not the dot product s.s, instead it is a 3 x 3 matrix. This is because when you multiply two matrices (and a vector is a type of matrix) A and B to get a product C, the rule is that the dot product of the first row of A and the first column of B gives the number in the first row, first column of C; the dot product of the first row of A and the second column of B gives the number in the first row, second column of C; and so forth. There's a little tutorial which might help understand how it works in the second section of this page (http://www.purplemath.com/modules/mtrxmult.htm). So if you apply these rules when A is a row vector with 3 components and B is a column vector with 3 components, then A only has one row and B only has one column, meaning that C is a 1 x 1 matrix (a scalar). But if A is a column vector with 3 components and B is a row vector with 3 components, A has 3 rows and B has 3 columns, so there are 9 possible combinations when you take the dot product of a row of A and a column of B; in this case C is a 3x3 matrix with 9 components. I guess the question is: What is <s.s'>? Is it other than unity? The error is in your previous step, where you substituted <s.s> for <(sx, sy, sz)(sx sy sz)>. They are not equivalent.

DrChinese

Feb23-07, 03:31 PM

1. I disagree that "Bell's theorem" primarily revolves around picking specific angles, if that's what you mean by "That 'exercise for the reader' IS Bell's Theorem". The proof involves finding an inequality that should hold for arbitrary angles under local realism;

Well, I both agree strongly and disagree strongly. You are quite right that Bell says: IF local realism is to be accepted, THEN the Inequality must hold for ALL arbitrary settings. Logic (contranegative) dictates that this proposition is equivalent to: IF the Inequality does NOT hold for ALL arbitrary settings, THEN local realism is NOT to be accepted. Bell did not spell out this part of the argument, it must be inferred.

But Bell NEVER asserts that the Inequality FAILS for ALL possible angle settings... and it doesn't! In his (22) he says that for ac=cos(90 degrees), ab=bc=cos(45 degrees), the Inequality is violated and therefore "the quantum mechanical expectation value cannot be represented, either accurately or arbitrarily closely, in the form (2)." He has provided one specific counter-example, and that is all he needs to do to prove the contranegative above. Of course, with the formula in hand you can create other settings that will also violate the inequality.

This would certainly explain why you and I see things differently. You believe wm is providing the counter-example, while I see Bell as providing the counter-example. If you are correct, then why does Bell essentially start with wm's results (3) and (7) and then go on to say that "In this simple case there is no difficulty in the view that the result of every measurement is determined by the value of an extra variable..." and the like.

On the other hand, shortly after (22) he make the generalization (V.) that for systems where there are MORE than 2 assumed hidden variables ("dimensionality greater than two", i.e. more than just A and B), there will always be a case in which QM is incompatible with "seperable predetermination" for the "two dimensional subspaces" in which observations can actually be performed.

Please note that I keep quoting and referencing Bell's original paper (http://www.drchinese.com/David/Bell_Compact.pdf) extensively. Not once have either you disputed anything with a similar reference. I think if you look back at the paper and see the context, you may look at my argument in a little different light.

I realize that the language of Bell does not always map directly onto what we often take for granted as being Bell's Theorem. I do consider Bell precise, and the paper is (in my opinion) a masterpiece - especially considering how much ground had to be broken to get to the final result. But the paper fully stands 40+ years later.

I must say that I have gained much out of this exchange, because I always convert the argument from spin 1/2 to spin 1 particles when I am thinking about the matter. So I have had to re-read it a bit more closely to keep in the discussion. :)

Feb23-07, 03:39 PM

In your post:

http://www.physicsforums.com/showpost.php?p=1252012&postcount=55

how do you go from (4) to (7) ?

The intermediate notation is confusing, and in as much as it is interpreted by several of us here, wrong, because it seems indeed that you use the "rule" that:

(a.s)(b.s) = (s.s)(a.b)

which is not correct.

Maybe you don't use that rule, but then the notation in (5) and (6) is completely unintelligible, and the transition from (4) to (7) ununderstandable.

EDIT:
to show that this rule is not correct, it is sufficient to have a counter example, and JesseM gave you one, to which you replied that he failed to understand that s was a random vector. But that doesn't show in the notation, because in (7), we still have the expectation symbols there, which make one think that the notation inside applies for each possible unit vector s individually.

I am for sure wondering what I have missed? So, from my maths post, with explanations for the moves from (4) to (7). And more to come if I'm still not clear:

(3) <(a.s) (s'.b')>

Since s' = -s we have:

(4) = - <(a.s) (s.b')>

Since we can expand a dot product using row and column vectors http://en.wikipedia.org/wiki/Column_vector we have (in an accepted compact notation):

(5) = - <[(ax ay az) (sx, sy, sz)] [(sx sy sz) (bx', by', bz')]>

Since the vectors a and b' are constant during a given experimental run, they may be removed from the ensemble average; which we do. So:

(6) = - (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz')

The ensemble average is now over a dot product between s and s. So we write:

(7) = - (ax ay az) <s.s> (bx', by', bz')

With s.s = 1 FOR ANY AND ALL s, we move to the conclusion.

PS: My maths was offered in good faith and was not intended to be a con-job: The last person I'd want to con on this subject is myself.

SO: Have I made a mistake that cannot be corrected?

If you need to use words, please be expansionary in your comments. Compact maths should be OK however.

Thanks, wm

JesseM

Feb23-07, 03:43 PM

EDIT:
to show that this rule is not correct, it is sufficient to have a counter example, and JesseM gave you one, to which you replied that he failed to understand that s was a random vector. But that doesn't show in the notation, because in (7), we still have the expectation symbols there, which make one think that the notation inside applies for each possible unit vector s individually. Another point on this is that although wm did use <> to indicate he was talking about expectation values, his proof didn't seem to depend in any way on how the vector s could vary; if the proof is valid for the case where s is equally likely to take any angle from 0 to 2pi, then it should also be valid for a case where s can only have a discrete number of possibilities, like if s had a 50% chance of being 90 degrees and a 50% chance of being 80 degrees. In this case, if a is 0 degrees and b' is 60 degrees, then the expectation value for - <a.s*s.b'> is just - {0.5*[cos(90)*cos(30)] + 0.5*[cos(80)*cos(20)]}, which is equal to -0.0816. But meanwhile, -cos(a - b') is equal to -cos(60), or -0.5. So unless wm claims to be making specific use of the fact that the angle of s is equally likely to take any angle, this example shows it is not true in general that - <a.s*s.b'> is equal to -cos(a - b').

DrChinese

Feb23-07, 03:44 PM

Just a quick note that not all of the above cases are always valid even in a classical correlation scenario. (e.g [2] and [7] can be illegal simultaneous events)

As JesseM points out, in a classical scenario ALL of the cases must be non-negative. This is the very definition of realism.

[2] and [7] are the quantum cases that are suppressed, and end up with negative probabilities for certain angle settings.

JesseM

Feb23-07, 03:59 PM

I am for sure wondering what I have missed? So, from my maths post, with explanations for the moves from (4) to (7). And more to come if I'm still not clear:

(3) <(a.s) (s'.b')>

Since s' = -s we have:

(4) = - <(a.s) (s.b')>

Since we can expand a dot product using row and column vectors http://en.wikipedia.org/wiki/Column_vector You can only expand a dot product in row and column vectors if you are using matrix multiplication; i.e. (ax, ay, az).(sx, sy, sz) is equal to (ax ay az)x(sx, sy, sz), where I am using x to represent matrix multiplication. So with the matrix multiplication made explicit, your step 5 is something like this:

(5) = - <(ax ay az)x(sx, sy, sz)x(sx sy sz)x(bx', by', bz')>

This is valid in itself. But your mistake is here: (6) = - (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz')

The ensemble average is now over a dot product between s and s. Wrong! Again, (sx, sy, sz)x(sx sy sz) is not a dot product with a single value, it is a 3 x 3 matrix whose 9 entries look like this:

sx*sx sx*sy sx*sz
sy*sx sy*sy sy*sz
sz*sx sz*sy sz*sz

SO: Have I made a mistake that cannot be corrected? Yes. I showed a counterexample to your proof in post #99 (http://www.physicsforums.com/showpost.php?p=1254272&postcount=99), and what's more, in post #76 (http://www.physicsforums.com/showpost.php?p=1253801&postcount=76) I've given a proof that the correct value for -<(a.s)*(s.b)> would be -(1/2)cos(a - b), using the rule that if the outcome of a given experiment will be some function f(x) of a parameter x whose value is between A and B and whose probability distribution is given by p(x), then the expectation value is \int_{A}^{B} p(x)*f(x) \, dx. In this case, the outcome is a function of the angle \theta, namely cos(\theta - a)*cos(\theta - b), and \theta must be equally likely to take any value from 0 to 2pi, so we must use the probability distribution p(\theta) = 1/2\pi to ensure that if we integrate the probability from 0 to 2pi, the answer is 1. Thus, the expectation value for -<(a.s)*(s.b)> must be - \frac{1}{2\pi} \int_{0}^{2\pi} cos(\theta - a)*cos(\theta - b) \, d\theta, which works out to -(1/2)cos(a - b).

DrChinese

Feb23-07, 04:11 PM

1. Bell provides a general proof that a certain inequality can never be violated under local realism, a statement of the form "for all experiments obeying local realism and satisfying certain conditions, this inequality will be satisfied".

2. Logically, any statement of the form "for all X, Y is true" can be disproved with a single counterexample of the form "there exists one X such that Y is false".

3. And that's what wm tried to do--find a single example of a local realist experiment which would satisfy Bell's conditions and yet violate an inequality.

1. I agree totally, although the language should be more like: "for all theories obeying local realism and satisfying certain conditions, this inequality will be experimentally satisfied".

2. I agree totally.

3. This is the problem: Bell's Inequality was set up for the specific purpose of showing it is violated! Bell himself provides the counter-example. But look at what you have said here: it is not the counter-example to what you say in 1.

For what wm is saying to work, he needs the proposition to be: "If the Inequality is violated and the experiment obeys local realism, then Bell's argument is wrong. That is a different argument entirely, and certainly will be disputed by anyone if quantum mechanics is involved in any way, shape or form. There are people out there trying to show this, but so far no one has succeeded. I consider it akin to tilting at windmills, but to each his own. There are still people looking for perpetual motion too...

JesseM

Feb23-07, 04:28 PM

But Bell NEVER asserts that the Inequality FAILS for ALL possible angle settings... I never claimed it did! What comment of mine are you thinking of? In fact, I said quite the opposite, that only one example of a violation of an inequality in QM is enough to disprove local realism if you accept Bell's proof as valid, and likewise one example of a violation or an inequality in a classical scenario would be enough to show the proof is flawed (although of course I don't think such a classical example will ever be found, since I think the proof is fine). Please read this comment of mine again: I guess you could say that if one agrees with Bell's theorem, then local realism makes the prediction that "for all experiments satisfying X conditions, inequality Y will be satisfied". And in this case, QM can give an example of the form "here's an experiment satisfying X conditions which violates inequality Y", thus proving QM is incompatible with local realism. But the problem here is that wm believes there's a flaw in Bell's theorem, so he does not agree that local realism makes the prediction "for experiments satisfying X conditions, inequality Y will be satisfied" in the first place; he's trying to disprove Bell's theorem by showing that local realism can also give an example of the form "here's an experiment satisfying X conditions which violates inequality Y". This would certainly explain why you and I see things differently. You believe wm is providing the counter-example, while I see Bell as providing the counter-example. I still don't understand what you mean when you use the word "counter-example", you aren't specifying a counterexample to what...perhaps you could restate your comments as I have done above, where someone offers a general statement of the form "for all X, Y is true" and someone offers a counterexample of the form "there exists an X such that Y is false". Again, in these terms, what I am saying is:

1. Bell's theorem claims to prove "for all experiments which are of the form X (referring to all the conditions on the experiment like the source not having foreknowledge of detector settings) AND which respect local realism, it is true that inequality Y will be satisfied." But then he shows that in QM, "there exists a quantum experiment of the form X such that inequality Y is violated." The point here is that if you accept his proof of the general statement, then a single counterexample from QM is enough to show that QM is incompatible with local realism.

2. wm does not accept Bell's proof in the first place, and he wants to show that it is flawed by demonstrating that in an ordinary classical universe, "there exists an experiment which is of form X AND which does respect local realism, yet which violates inequality Y." If he could indeed produce a single example like this, then he'd have proved there must be a flaw in Bell's proof, since the theorem asserts that all experiments of the form X which respect local realism must obey inequality Y.

To be clear, I of course think that wm has failed to find an experiment of form X (ie the conditions specified in proofs of Bell's theorem) which respects local realism yet violates some Bellian inequality. I of course think Bell's theorem is solid, and that no one will find a purely classical counterexample. But that's what wm is trying to do, and all I'm saying is that it makes sense as a strategy.

JesseM

Feb23-07, 04:33 PM

3. This is the problem: Bell's Inequality was set up for the specific purpose of showing it is violated! I think you're missing a key part of my 3, though: 3. And that's what wm tried to do--find a single example of a local realist experiment which would satisfy Bell's conditions and yet violate an inequality. Bell did not provide an example of a local realist experiment that violated the inequality; rather, he provided a quantum example, and thus (if you accept his proof as valid) he showed that QM is not compatible with local realism. For what wm is saying to work, he needs the proposition to be: "If the Inequality is violated and the experiment obeys local realism, then Bell's argument is wrong. Exactly! That's what I think wm is trying to do. After all, his experiment is a purely classical one that does obey local realism (it just involves sending two vectors in opposite directions, and each experimenter projects the one they get onto their own vector to get a real number between -1 and +1), yet he claims that the expectation value for the product of the two experimenter's answers will be given by -cos(a - b), and I showed that this would violate the CHSH inequality for some particular choices of detector angles.

Feb23-07, 07:01 PM

You can only expand a dot product in row and column vectors if you are using matrix multiplication; i.e. (ax, ay, az).(sx, sy, sz) is equal to (ax ay az)x(sx, sy, sz), where I am using x to represent matrix multiplication. So with the matrix multiplication made explicit, your step 5 is something like this:

(5) = - <(ax ay az)x(sx, sy, sz)x(sx sy sz)x(bx', by', bz')>

This is valid in itself. But your mistake is here: Wrong! Again, (sx, sy, sz)x(sx sy sz) is not a dot product with a single value, it is a 3 x 3 matrix whose 9 entries look like this:

sx*sx sx*sy sx*sz
sy*sx sy*sy sy*sz
sz*sx sz*sy sz*sz

Yes. I showed a counterexample to your proof in post #99 (http://www.physicsforums.com/showpost.php?p=1254272&postcount=99), and what's more, in post #76 (http://www.physicsforums.com/showpost.php?p=1253801&postcount=76) I've given a proof that the correct value for -<(a.s)*(s.b)> would be -(1/2)cos(a - b), using the rule that if the outcome of a given experiment will be some function f(x) of a parameter x whose value is between A and B and whose probability distribution is given by p(x), then the expectation value is \int_{A}^{B} p(x)*f(x) \, dx. In this case, the outcome is a function of the angle \theta, namely cos(\theta - a)*cos(\theta - b), and \theta must be equally likely to take any value from 0 to 2pi, so we must use the probability distribution p(\theta) = 1/2\pi to ensure that if we integrate the probability from 0 to 2pi, the answer is 1. Thus, the expectation value for -<(a.s)*(s.b)> must be - \frac{1}{2\pi} \int_{0}^{2\pi} cos(\theta - a)*cos(\theta - b) \, d\theta, which works out to -(1/2)cos(a - b).

Jesse, thanks for this; I like it very much. Also: Excuse my jumping in and out at the moment but I'm just grabbing bits of time hopefully to move us ahead. Which should not happen till we have agreement re my equations.

I think they are going to be just fine BUT could you tell me how you want the last correct line (whatever you deem that to be) to be written.

I ask because if I go on what's above, you look like you would like an x in three places? But would that be satisfactory?

The reason that I want this correction because all I did in my own work was to see that the matrix that resides in the middle (after all correct mathematical proceses) is just the equivalent of a unit-matrix, obtained by taking the ensemble-average inside the matrix and evaluating each element's ensemble-average.

What I'm guessing here (in a hurry) is that you would be happier to see the unit-matrix equivalent in the middle, and not my short-cut?

Sorry to be slow on the up-take; but hope you won't mind me still seeing that the fully rigorous maths will not change the answer.

As soon as I get where you want some x (and some y if necessary) I'll send in the expanded version that should then suit you; or maybe have a different error.

Thanks, as always, wm

Feb23-07, 07:36 PM

Jesse, thanks for this; I like it very much. Also: Excuse my jumping in and out at the moment but I'm just grabbing bits of time hopefully to move us ahead. Which should not happen till we have agreement re my equations.

<SNIP>

As soon as I get where you want some x (and some y if necessary) I'll send in the expanded version that should then suit you; or maybe have a different error.

Thanks, as always, wm

Jesse, what abt I do something like this? (ROUGHLY)

- <a.s * s.b'>

= - <(ax ay az) (sx, sy, sz) * (sx sy sz) (bx, by, bz)>

= - (ax ay az) <(sx, sy, sz) * (sx sy sz)> (bx, by, bz)

...

...

= - (ax ay az) [100, 010, 001] (bx, by, bz)

= - a.b'

Would that matrix representation (and the one *) be OK?

wm

JesseM

Feb23-07, 08:37 PM

Jesse, thanks for this; I like it very much. Also: Excuse my jumping in and out at the moment but I'm just grabbing bits of time hopefully to move us ahead. Which should not happen till we have agreement re my equations.

I think they are going to be just fine BUT could you tell me how you want the last correct line (whatever you deem that to be) to be written.

I ask because if I go on what's above, you look like you would like an x in three places? But would that be satisfactory? Actually, one minor physical issue occurred to me--you have the vectors as 3-vectors, but if you want to mimic the type of spin measurements made in QM, they should really be 2-vectors. This is because, when you measure spin using Stern-Gerlach magnets (http://en.wikipedia.org/wiki/Stern–Gerlach_experiment), the long axis of the magnet has to be alligned parallel to the particle's path, so you just have the freedom to rotate the magnets around this axis at any angle (this is why in discussions of Bell's theorem people often talk about each experimenter choosing 'an angle'--if they had 3 degrees of freedom, they would each have to select 2 distinct angles instead).

So, I'd amend your (5) to look like this:

(5) = - <[(ax ay)x(sx, sy)]x[(sx sy)x(bx', by')]>

Of course since the two quantities in brackets give scalars, strictly speaking the middle x could be replaced by a *, but leaving it as matrix multiplication makes it easier to go to step (6):

(6) = - (ax ay)x<(sx, sy)x(sx sy)>x(bx', by')

That's the last step in your proof I'd agree with. The reason that I want this correction because all I did in my own work was to see that the matrix that resides in the middle (after all correct mathematical proceses) is just the equivalent of a unit-matrix, obtained by taking the ensemble-average inside the matrix and evaluating each element's ensemble-average. The 2-matrix in the center actually does not work out to be a unit matrix. One thing to note is that if s is an individual unit vector, while it's true that (sx sy)x(sx, sy), i.e. the dot product of s with itself, is always 1, it's not true that the 2x2 matrix (sx, sy)x(sx sy) is always a unit matrix; for example, if sx = 0.5 and sy = 0.866, the matrix works out to be:

(0.5)*(0.5) (0.5)*(0.866)
(0.866)*(0.5) (0.866)*(0.866)

or

0.25 0.433
0.433 0.75

However, you'd probably point out that we are interested in the average expectation value of this matrix when s is allowed to take any angle from 0 to 2pi. We know that if the angle of s is \theta, then s_x = cos(\theta ) and s_y = sin(\theta ). So, the matrix would be:

\left( \begin{array}{cc} cos^2(\theta ) & cos(\theta )*sin(\theta ) \\
sin(\theta )*cos(\theta ) & sin^2(\theta ) \end{array} \right)

For each of these four components, to find the expectation value we must integrate them from 0 to 2pi, then multiply the result by (1/2pi)...see the end of my previous post for an explanation of why the expectation value of a function based on an arbitrary angle would be calculated in this way.

Using the integrator (http://integrals.wolfram.com/index.jsp), we have:

\int cos^2(\theta ) \, d\theta = (1/2)*(\theta + cos(\theta )*sin(\theta ))
\int sin(\theta )*cos(\theta ) \, d\theta = (-1/2)*cos^2(\theta )
\int sin^2(\theta ) \, d\theta = (1/2)*(\theta - cos(\theta )*sin(\theta ))

So, taking each function f(\theta ) and plugging in the limits of integration f(2pi) - f(0), the expectation value for the matrix is:

\frac{1}{2\pi} \left( \begin{array}{cc} \pi & 0 \\
0 & \pi \end{array} \right)

or:

\left( \begin{array}{cc} \frac{1}{2} & 0 \\
0 & \frac{1}{2} \end{array} \right)

So, it looks like this will end up just being another way of proving that - <a.s*s.b'> is equal to -(1/2)*cos(a - b), which I had proved earlier by just doing one big integral.

Feb23-07, 09:22 PM

Actually, one minor physical issue occurred to me--you have the vectors as 3-vectors, but if you want to mimic the type of spin measurements made in QM, they should really be 2-vectors. This is because, when you measure spin usingStern-Gerlach magnets (Stern–Gerlach_experiment), the long axis of the magnet has to be alligned parallel to the particle's path, so you just have the freedom to rotate the magnets around this axis at any angle (this is why in discussions of Bell's theorem people often talk about each experimenter choosing 'an angle'--if they had 3 degrees of freedom, they would each have to select 2 distinct angles instead).

So, I'd amend your (5) to look like this:

(5) = - <[(ax ay)x(sx, sy)]x[(sx sy)x(bx', by')]>

Of course since the two quantities in brackets give scalars, strictly speaking the middle x could be replaced by a *, but leaving it as matrix multiplication makes it easier to go to step (6):

(6) = - (ax ay)x<(sx, sy)x(sx sy)>x(bx', by')

That's the last step in your proof I'd agree with. The 2-matrix in the center actually does not work out to be a unit matrix. One thing to note is that if s is an individual unit vector, while it's true that (sx sy)x(sx, sy), i.e. the dot product of s with itself, is always 1, it's not true that the 2x2 matrix (sx, sy)x(sx sy) is always a unit matrix; for example, if sx = 0.5 and sy = 0.866, the matrix works out to be:

(0.5)*(0.5) (0.5)*(0.866)
(0.866)*(0.5) (0.866)*(0.866)

or

0.25 0.433
0.433 0.75

However, you'd probably point out that we are interested in the average expectation value of this matrix when s is allowed to take any angle from 0 to 2pi. We know that if the angle of s is \theta, then s_x = cos(\theta ) and s_y = sin(\theta ). So, the matrix would be:

cos^2(\theta ) \,\,\,\, cos(\theta )*sin(\theta )
sin(\theta )*cos(\theta ) \,\,\,\, sin^2(\theta )

\left( \begin{array}{cc} 1- cos^2(\theta ) & cos(\theta )*sin(\theta ) \\
-sin(\theta )*cos(\theta ) & sin^2(\theta ) \end{array} \right)

For each of these four components, to find the expectation value we must integrate them from 0 to 2pi, then multiply the result by (1/2pi)...see the end of my previous post for an explanation of why the expectation value of a function based on an arbitrary angle would be calculated in this way.

Using the integrator (http://integrals.wolfram.com/index.jsp), we have:

\int cos^2(\theta ) \, d\theta = (1/2)*(\theta + cos(\theta )*sin(\theta ))
\int sin(\theta )*cos(\theta ) \, d\theta = (-1/2)*cos^2(\theta )
\int sin^2(\theta ) \, d\theta = (1/2)*(\theta - cos(\theta )*sin(\theta ))

So, taking each function f(\theta ) and plugging in the limits of integration f(2pi) - f(0), and then multiplying the result by the (1/2pi) which was outside the integral, we find that the average expected value for the four components of the matrix works out to:

1/2 0
0 1/2

So, it looks like this will end up just being another way of proving that - <a.s*s.b'> is equal to -(1/2)*cos(a - b), which I had proved earlier by just doing one big integral.

Jesse, I think we're getting there.

BUT: Why do you say theta takes on values O = 2pi? It seems to me that if you want to use angles (I prefer unit vectors) then you need to integrate over 4pi (which would give you the missing factor of 2 that you're looking for)?

The 2pi divisor in you calculation would remain unchanged because the angular differences can only range thereover.

This would then give the same result as that matrix that you gave earlier: for averaging inside the matrix by observation only, you produce exactly the unit matrix.

Do you see that (by observation without any maths; maybe except in your head) your matrix reduces to the unit-matrix (1, U, I, E)? I prefer 1 (but it is part of my problem here): Should I write it as U in QM? What is best?

Thus it seems I was mathematically wrong in representing the unit-matrix as a plain 1 (= s.s) and so just writing <s.s> = <1>, etc. Actually I was more reading the equations physically and thinking my short-cut was Ok, given that the spaces were defined by the opening gambit.

I know you seem to say that mine/yours is not the unit matrix, but would you reconsider it and let me know. I cannot see why it is not; for you half make the case yourself above. PERHAPS you maybe overlooking that the ensemble-average is over an infinity of exemplars: so, on averaging,

(1) sisj = sij = dij (Kronecker)?

(In my opinion, all elements are averaging zero, except on the main diagonal which all average 1).

PS: Regarding the S-G orientation, and given I'm considering idealised experiments, and given the spherical symmetry of the classical state that I'm using: I find it easier to remain general with unit-vectors; for when we get it right there will be no error, even if the experiments today cannot match it.

Of course, practically/loosely, running along the flight-axis a little longer gives time for the spins to stabilise and form the two-peaked S-G output. But once we get our present maths correct, dichotomic outputs like S-G fall naturally from my equations via a simple realistic re-definition of what constitutes an experimental outcome. But this is getting ahead of where we are.

PS: Once I get your reply I think it's time to re-present my experiment and maths. To tidy lots of ends up? Do you agree?

Thanks, wm

JesseM

Feb23-07, 10:19 PM

BUT: Why do you say theta takes on values O = 2pi? It seems to me that if you want to use angles (I prefer unit vectors) How would you propose to integrate over all possible unit vectors without having an angle in the integral? I assume you want the probability distribution to have the property that the probability of getting a unit vector whose angle is between 0 and 20 is the same as the probability of getting a unit vector whose angle is between 20 and 40, and so forth, so the probability of an angle between A and A+C is always the same as the probability of getting one between B and B+C.

Also, note that for every possible choice of angle, the components sx and sy of the unit vector with that angle are uniquely determined (ie sx = cos(angle) and sy = sin(angle)); and every possible unit vector corresponds to an angle in this way, the mapping between distinct angles and distinct unit vectors is one-to-one. then you need to integrate over 4pi (which would give you the missing factor of 2 that you're looking for)? Why do you say that? You're aware that 2pi in radians is equivalent to 360 degrees, right? And just like 380 degrees is exactly the same angle as 20 degrees, so 3pi degrees is exactly the same angle as pi degrees (and the vectors for each would have all the same components, so they'd actually be the same vector). If you integrate over 2pi, you're integrating over every possible distinct angle that s can take, and thus every possible distinct unit vector s. Do you see that (by observation without any maths; maybe except in your head) your matrix reduces to the unit-matrix (1, U, I, E)? No, it ends up being (1/2) times the unit matrix.

If you don't trust in integrals, we could try doing a numerical approximation, i.e. pick a significant number of evenly-spaced angles between 0 and 360 for s, find a.s*s.b for each angle, add them all together, and divide by the total number of angles to find the average. If we use a fairly large number of angles--say, 36 (angles at 10-degree intervals) or 72 (angles at 5-degree intervals) then this should be a pretty good approximation for the case where the angle can vary continuously, and then we can check whether the result is close to my predicted value of (1/2)cos(a - b) or close to your original predicted value of cos(a - b). Would you like me to take a shot at this and see the result?PERHAPS you maybe overlooking that the ensemble-average is over an infinity of exemplars Of course I'm not overlooking it, that's why I did an integral rather than a sum. And like I said, there's a one-to-one relationship between the set of all possible unit vectors and the set of all possible angles, that's why I integrated over every possible distinct angle. so, on averaging,

(1) sisj = sij = dij (Kronecker)? I assume sij represents the component in the ith row and jth column of the matrix, while si represents the ith element of the column vector, and sj represents the jth element of the row vector? If so, why do you think that "averaging" would give the Kronecker delta? This is simply wrong, I've already shown that when you average over all possible unit vectors s, with components s1 = cos(angle) and s2 = sin(angle), the result ends up being (1/2)*dij.PS: Regarding the S-G orientation, and given I'm considering idealised experiments, and given the spherical symmetry of the classical state that I'm using: I find it easier to remain general with unit-vectors; for when we get it right there will be no error, even if the experiments today cannot match it. I'm using unit vectors too, just unit 2-vectors rather than unit 3-vectors. If you use 3-vectors the correspondence between the classical case of projecting angles and the quantum case of measuring spins becomes murkier; what's more, all the integrals I've presented would have to be different, they'd have to be double integrals where you integrate over two different angles \theta and \phi, not single integrals where you integrate over one angle.

Feb24-07, 12:13 AM

How would you propose to integrate over all possible unit vectors without having an angle in the integral? I assume you want the probability distribution to have the property that the probability of getting a unit vector whose angle is between 0 and 20 is the same as the probability of getting a unit vector whose angle is between 20 and 40, and so forth, so the probability of an angle between A and A+C is always the same as the probability of getting one between B and B+C.

Also, note that for every possible choice of angle, the components sx and sy of the unit vector with that angle are uniquely determined (ie sx = cos(angle) and sy = sin(angle)); and every possible unit vector corresponds to an angle in this way, the mapping between distinct angles and distinct unit vectors is one-to-one. Why do you say that? You're aware that 2pi in radians is equivalent to 360 degrees, right? And just like 380 degrees is exactly the same angle as 20 degrees, so 3pi degrees is exactly the same angle as pi degrees (and the vectors for each would have all the same components, so they'd actually be the same vector). If you integrate over 2pi, you're integrating over every possible distinct angle that s can take, and thus every possible distinct unit vector s. No, it ends up being (1/2) times the unit matrix.

If you don't trust in integrals, we could try doing a numerical approximation, i.e. pick a significant number of evenly-spaced angles between 0 and 360 for s, find a.s*s.b for each angle, add them all together, and divide by the total number of angles to find the average. If we use a fairly large number of angles--say, 36 (angles at 10-degree intervals) or 72 (angles at 5-degree intervals) then this should be a pretty good approximation for the case where the angle can vary continuously, and then we can check whether the result is close to my predicted value of (1/2)cos(a - b) or close to your original predicted value of cos(a - b). Would you like me to take a shot at this and see the result? Of course I'm not overlooking it, that's why I did an integral rather than a sum. And like I said, there's a one-to-one relationship between the set of all possible unit vectors and the set of all possible angles, that's why I integrated over every possible distinct angle. I assume sij represents the component in the ith row and jth column of the matrix, while si represents the ith element of the column vector, and sj represents the jth element of the row vector? If so, why do you think that "averaging" would give the Kronecker delta? This is simply wrong, I've already shown that when you average over all possible unit vectors s, with components s1 = cos(angle) and s2 = sin(angle), the result ends up being (1/2)*dij. I'm using unit vectors too, just unit 2-vectors rather than unit 3-vectors. If you use 3-vectors the correspondence between the classical case of projecting angles and the quantum case of measuring spins becomes murkier; what's more, all the integrals I've presented would have to be different, they'd have to be double integrals where you integrate over two different angles \theta and \phi, not single integrals where you integrate over one angle.

Jesse,

1. I integrate over 4pi steradians (SOLID ANGLES, NOT PLANAR), which gives my integral the missing 2; also in the 3x3 matrix ... -> I.

2. ... which thus agrees with my evaluation of the 3x3 matrix that you generated.

3. That is: <sx*sx> = <sy*sy> = <sz*sz> = 1. All other elements average 0; the 4pi steradians being in play throughout.

4. The 3x3 unit-matrix I results.

5. Thus my -a.b' result stands OK (it seems to me).

Do we differ? And this might save me taking up your offer above, which looks a bit tedious to me.

PS: -1 </= si </= 1. -1 </= sj </= 1.

<si*si> = <sj*sj> = <{1, ..., 0} + {0, ..., 1}> = 1/2 + 1/2 = 1;

<si*sj> = <{-1, ..., 0, ..., 1}> = 0;

where { ... } indicates the infinite set of values each expression may take.

Gruss (in appreciation), wm

vanesch

Feb24-07, 12:57 AM

3. That is: <sx*sx> = <sy*sy> = <sz*sz> = 1. All other elements average 0; the 4pi steradians being in play throughout.

No, this is the mistake.

If s is a random unit vector on a sphere, then < sx*sx > works out to be 1/3.

If it is on a circle, then it works out to be 1/2.

You can easily see this:

the COMPONENTS of a unit vector can never be larger than 1, right ? They vary between -1 and 1. So their square varies between 0 and 1, right ?

Now, for something that varies between 0 and 1, to have an AVERAGE value of 1, that means that it must ALWAYS be 1.

So sx^2 must always be 1, so sx must be -1 or 1.
same for sy.

So this means that you only consider unit vectors of the form (1,1), (1,-1), (-1,1) and (-1,-1). But these are not unit vectors !!

But that means the average of the square of their components cannot be equal to 1.

Now, how to find quickly the right value ?

We know that s.s = 1.

So sx.sx + sy.sy = 1.

Take the expectation value of this equation, then we find:

< sx.sx > + < sy.sy > = 1

If we assume, by symmetry, that < sx.sx > = < sy.sy >, then they are equal to 1/2.

EDIT: btw, that expression that:

< sx.sx > + < sy.sy > = 1

by itself already proves that < sx.sx > can only be 1 if < sy.sy > = 0.

JesseM

Feb24-07, 01:08 AM

Jesse,

1. I integrate over 4pi steradians (SOLID ANGLES, NOT PLANAR), which gives my integral the missing 2; also in the 3x3 matrix ... -> I. So you're insisting on using 3-vectors rather than 2-vectors? Do you understand that the integral in 3 dimensions would have to be completely different than any of the integrals I've presented, that it would have to be a double integral over two separate angles \theta and \phi, with \theta varying from 0 to 2pi and \phi varying from 0 to pi, and using the area element dS = sin(\phi ) d\phi d\theta? Have you actually evaluated this integral for each component of the 3x3 matrix produced from (sx, sy, sz)x(sx sy sz)? If not, I have no basis for trusting your intuition that it would work out to the unit matrix, since this intuition was wrong in the case of 2-vectors. And actually evaluating this integral would be more work for me...can you first tell me whether you agree that, in the case of 2-vectors, the quantity <(sx, sy)x(sx sy)> will not work out to the unit matrix, but instead to 1/2 times the unit matrix? If you don't believe my math in the simpler 2D case, it seems to me you're unlikely to believe it in the 3D case if it doesn't work out to the unit matrix, since the integral involved is a lot more complicated in 3D. And, like I said, if my integrals in the 2D case aren't convincing to you, it might help if we actually computed a numerical approximation using a large number of evenly-spaced angles from 0 to 2pi.

edit: didn't see vanesch's argument when writing this, but it looks good to me, at least for the diagonal components of the matrix...you'd need a different argument to show the off-diagonal components are 0. PS: -1 </= si </= 1. -1 </= sj </= 1. You didn't really tell me what si and sj represent...again, are they components of column vector and row vector, with i ranging from 1-3 (in the case of a 3-vector, it'd be 1-2 for a 2-vector) and j ranging from 1-3 as well? If so, I agree that si and sj would both be between -1 and 1. However, if the vector s is equally likely to be any angle then its components are not equally likely to take any value between -1 and 1. <si*si> = <sj*sj> = <{1, ..., 0} + {0, ..., 1}> = 1/2 + 1/2 = 1; I don't understand the reason why you're taking a sum of two values between 0 and 1 there, when si*si is just the square of a single number between -1 and 1, which will be a single number between 0 and 1. In any case, if each angle is equally likely, you can't assume that si is equally likely to take any value between -1 and 1, nor can you assume that si*si is equally likely to take any value between 0 and 1, if you were making either of those assumptions.

vanesch

Feb24-07, 01:17 AM

Have you actually evaluated this integral for each component of the 3x3 matrix produced from (sx, sy, sz)x(sx sy sz)? If not, I have no basis for trusting your intuition that it would work out to the unit matrix, since this intuition was wrong in the case of 2-vectors.

For uniform distributions, unit vectors in N dimensions work out to have an expectation value of < si.si > = 1/N, simply because:

< s1.s1 > + < s2.s2 > + ... + < sn.sn > = < s.s > = 1

By symmetry (uniform distribution), all the terms are equal and hence equal to 1/N.

So in 3 dimensions, that would be 1/3.

vanesch

Feb24-07, 01:34 AM

edit: didn't see vanesch's argument when writing this, but it looks good to me, at least for the diagonal components of the matrix...you'd need a different argument to show the off-diagonal components are 0.

That's not hard either, in the case of a uniform distribution.

Consider, say, < sx.sy > = C. C is a number that is a property of the uniform distribution of vectors s. It shouldn't depend on my specific choice of coordinate axes (the description of a uniform distribution should be identical in all orthogonal coordinate axes). C is "the expectation value of the product of the first and the second coordinate".

Let me flip the y-axis, and keep the x and z. This changes everywhere sy into -sy, while keeping sx and sz.

If, in this new system, we calculate "the expectation value of the product of the first and the second coordinate", then we find
C' = < sx.(-sy) > - < sx.sy >= - C.

But we agreed that C was a number that shouldn't depend on a precise choice of axes. So we must have C = - C. Hence, C = 0.

So < sx.sy > = 0

By symmetry (flip the axes!), in general < si.sj > = 0 if i not equal to j.

DrChinese

Feb24-07, 08:22 AM

1. Bell's theorem claims to prove "for all experiments which are of the form X (referring to all the conditions on the experiment like the source not having foreknowledge of detector settings) AND which respect local realism, it is true that inequality Y will be satisfied." But then he shows that in QM, "there exists a quantum experiment of the form X such that inequality Y is violated." The point here is that if you accept his proof of the general statement, then a single counterexample from QM is enough to show that QM is incompatible with local realism.

2. wm does not accept Bell's proof in the first place, and he wants to show that it is flawed by demonstrating that in an ordinary classical universe, "there exists an experiment which is of form X AND which does respect local realism, yet which violates inequality Y." If he could indeed produce a single example like this, then he'd have proved there must be a flaw in Bell's proof, since the theorem asserts that all experiments of the form X which respect local realism must obey inequality Y.

To be clear, I of course think that wm has failed to find an experiment of form X (ie the conditions specified in proofs of Bell's theorem) which respects local realism yet violates some Bellian inequality. I of course think Bell's theorem is solid, and that no one will find a purely classical counterexample. But that's what wm is trying to do, and all I'm saying is that it makes sense as a strategy.

I think we have pretty well zeroed in on the issues, thanks. Again, referencing Bell itself :tongue: :

1. The Inequality occurs when you map local hidden variable functions into QM expectation values. Thus Bell proves: IF QM=correct predictions (3,7) AND local realism=assumed (2,14), THEN Inequality must be true (15). From the contranegative, if the Inequality is demonstrated to be false by counter-example, then either "QM=correct predictions" is false OR "local realism=assumed" is false. Bell provides such counter-example in (22). Hopefully there is no dispute about this so far.

2. Yes, I realize wm does not accept Bell's proof as valid. Specifically, he denies "IF QM=correct predictions (3,7) AND local realism=assumed (2,14), THEN Inequality must be true (15)". OK, perhaps Bell's proof itself is wrong or somehow flawed. You can't demonstrate this by any type of counter-example, you must show that there is a mistake in the proof itself. If so, which step is it?

Showing a classical setup that violates the Inequality - as a way to invalidate the proof - does nothing, because you need a QM expectation value to compare it to. How can you have a QM expectation value if it is a classical setup? These are mutually exclusive by definition! Note that since Bell's proof is equivalent to the following:

IF the Inequality is violated, THEN QM=limited validity OR local realism=bad assumption.

and therfore also:

IF the Inequality is violated AND local realism=demonstrated, THEN QM=limited validity.

This is what you would have, i.e. QM is of limited validity and doesn't apply. And it wouldn't, because it is a classical experiment. And yet the proof is still standing, intact!!

JesseM

Feb24-07, 10:45 AM

1. The Inequality occurs when you map local hidden variable functions into QM expectation values. Thus Bell proves: IF QM=correct predictions (3,7) AND local realism=assumed (2,14), THEN Inequality must be true (15). Hmm, but when you say "QM=correct predictions", you're talking about some subset of its predictions rather than all possible predictions made by QM, right? After all, one of QM's predictions is that the inequality will be violated in certain experiments! Are you just talking about the prediction that whenever both experimenters measure their particles at the same angle, they always get opposite results? From the contranegative, if the Inequality is demonstrated to be false by counter-example, then either "QM=correct predictions" is false OR "local realism=assumed" is false. Bell provides such counter-example in (22). Hopefully there is no dispute about this so far. If my guess about what you meant in the "correct predictions" step is right, then no dispute...but if it isn't, could you clarify? 2. Yes, I realize wm does not accept Bell's proof as valid. Specifically, he denies "IF QM=correct predictions (3,7) AND local realism=assumed (2,14), THEN Inequality must be true (15)". OK, perhaps Bell's proof itself is wrong or somehow flawed. You can't demonstrate this by any type of counter-example, you must show that there is a mistake in the proof itself. But that's my point, you can show a proof is flawed simply by presenting a counterexample in some circumstances. In general, if a proof makes a statement like "for all cases where X is true, Y is true", then producing a single case of the form "X is true, but Y is false" shows the proof must have a flaw somewhere, without identifying which step in the proof must be flawed.

In this case, if Bell proves something like "IF the entangled-particles experiment always produces opposite spins when the experimenters choose the same angle AND local realism=assumed, THEN Inequality must be true". If wm could come up with a purely classical way of duplicating all the results of the entangled-particles experiment, including both the fact that the experimenters always get opposite results when they pick the same angle, and also the fact that the inequality is FALSE when they pick certain different angles, and it was clear by construction that wm's experiment respected local realism, then this would be sufficient to show that Bell's general statement was false, so that there must be some flaw in a proof. Showing a classical setup that violates the Inequality - as a way to invalidate the proof - does nothing, because you need a QM expectation value to compare it to. How can you have a QM expectation value if it is a classical setup? The expectation value is simply on the spin each experimenter will find when their detector is at a given angle, which on a given trial is either spin-up (+1) or spin-down (-1). You can also look at the expectation value for the product of their two results, either the same (+1) or different (-1). Either way, you can certainly come up with a classical experiment where, on each trial, each experimenter will get either the result +1 or -1, decided based on their choice of angle combined with some classical signal or object sent from a central source, and possibly with a random element as well. If you could further set things up so that the experimenters always get opposite results when they choose the same angle, and all the conditions necessary for Bell's theorem are obeyed (like the condition that the source has no foreknowledge of what angle each experimenter will choose on a given trial), do you disagree that Bell's proof should apply in exactly the same way to this experiment, and lead you to conclude that the same inequality should be obeyed as long as the experiment does not violate local realism?

Feb24-07, 01:55 PM

That's not hard either, in the case of a uniform distribution.

Consider, say, < sx.sy > = C. C is a number that is a property of the uniform distribution of vectors s. It shouldn't depend on my specific choice of coordinate axes (the description of a uniform distribution should be identical in all orthogonal coordinate axes). C is "the expectation value of the product of the first and the second coordinate".

Let me flip the y-axis, and keep the x and z. This changes everywhere sy into -sy, while keeping sx and sz.

If, in this new system, we calculate "the expectation value of the product of the first and the second coordinate", then we find
C' = < sx.(-sy) > - < sx.sy >= - C.

But we agreed that C was a number that shouldn't depend on a precise choice of axes. So we must have C = - C. Hence, C = 0.

So < sx.sy > = 0

By symmetry (flip the axes!), in general < si.sj > = 0 if i not equal to j.

1. I hope I am wording this right, but I might be expressing a big mistake. I apologise for my lack of LaTeX (which I should fix). Perhaps if my case is hopeless that won't be necessary.

2. I agree that <sisj> = 0, where si and sj are (as I understand them) the s projections on the i and j axes. But I think that <sisi> might =1; not to dispute your result for ''routine'' unit-vectors but to see if there is not a difference that we need to take into account.

3. s (and s' = -s) is a unit-vector representing angular momentum. Such vectors transform satisfactorily infinitesimally; but not in general (I believe). Does this not mean that the 1/3 factor that is correct for routine unit-vectors may be different for vectors representing angular-momentum?

4. I have in mind Bell's (1964, equation (3)); effectively:

(1) <s.a*s'.b'> = -a.b'.

5. Would you comment? And could you provide or point-me-to a step-by-step working of this relation?

Thanks, wm

JesseM

Feb24-07, 02:12 PM

4. I have in mind Bell's (1964, equation (3)); effectively:

(1) <s.a*s'.b'> = -a.b'.

5. Would you comment? And could you provide or point-me-to a step-by-step working of this relation? That equation is a specifically quantum-mechanical one. You do not obtain it by assuming s is an angular momentum which was pointing in some definite direction before measurement, and then getting the expectation value by averaging over all possible definite angles for the angular momentum. If you would like a quantum-mechanical derivation of the fact that the expectation value for the product of the two experimenter's spins will be the negative cosine of the angle between their detectors, i.e. -a.b = -cos(a - b), then you can look at the derivation I linked to at the end of post #66 (http://www.physicsforums.com/showpost.php?p=1253427&postcount=66): by the way, if you are familiar with calculations in QM, you can look at this page (http://www.mathpages.com/home/kmath521/kmath521.htm) for a nearly complete derivation. What they derive there is that if q represents the angle between the two detectors, then the probability that the two detectors get the same result (both spin-up or both spin-down) is sin^2 (q/2), and the probability they get opposite results (one spin-up and one spin-down) is cos^2 (q/2). If we represent spin-up with the value +1 and spin-down with the value -1, then the product of their two results when they both got the same result is going to be +1, and the product of their results when they got different results is going to be -1. So, the expectation value for the product of their results is:

(+1)*sin^2 (q/2) + (-1)*cos^2 (q/2) = sin^2 (q/2) - cos^2 (q/2)

Now, if you look at the page on trigonometric identities here (http://math2.org/math/trig/identities.htm), you find the following identity:

cos(2x) = cos^2 (x) - sin^2 (x)

So, setting 2x = q, this becomes:

cos(q) = cos^2 (q/2) - sin^2 (q/2)

Multiply both sides by -1 and you get:

sin^2 (q/2) - cos^2 (q/2) = - cos (q)

This fills in the final steps to show that the expectation value for the product of their results will be the negative cosine of the angle between their detectors. As I said earlier in that post, though, this derivation won't be of much use to you unless you already have a basic familiarity with the way probabilities and expectation values are derived in QM.

Feb24-07, 02:25 PM

That equation is a specifically quantum-mechanical one. You do not obtain it by assuming s is an angular momentum which was pointing in some definite direction before measurement, and then getting the expectation value by averaging over all possible definite angles for the angular momentum. If you would like a quantum-mechanical derivation of the fact that the expectation value for the product of the two experimenter's spins will be the negative cosine of the angle between their detectors, i.e. -a.b = -cos(a - b), then you can look at the derivation I linked to at the end of post #66 (http://www.physicsforums.com/showpost.php?p=1253427&postcount=66): As I said earlier in that post, though, this derivation won't be of much use to you unless you already have a basic familiarity with the way probabilities and expectation values are derived in QM.

Jesse, I looked at them, and they are helpful. Thanks.

But what I am still hoping for is a derivation that starts where Bell starts <s.a*s'.b'> and ends where Bell ends -a.b'.

Note that I am not in any way disputing the result, since it can be derived in many ways. It's just that I think Bell's way is likely to be one of the cleanest (and clearest) for me.

wm

vanesch

Feb24-07, 02:33 PM

4. I have in mind Bell's (1964, equation (3)); effectively:

(1) <s.a*s'.b'> = -a.b'.

5. Would you comment? And could you provide or point-me-to a step-by-step working of this relation?

AAAAHH ! You are quoting from the article "On the Einstein-Podolsky-Roosen paradox" Physics 1 (1964) 195-200 which is taken as the second chapter in "speakable and unspeakable..." by Bell ?

Right. I have this in front of me now, and there it is written:

< sigma_1.a sigma_2.b > = -a.b

sigma_1 and sigma_2 are (the same) 3-vectors of the 3 Pauli-matrices, but which act upon the state 1 and the state 2 respectively!

They are operators over hilbert space !

And I have to say that this notation used by Bell is extremely confusing.
Bell is calculating the operator that corresponds to "the product of the outcomes of the two measurements". This is the operator over hilbertspace which corresponds to the "measurement of the correlation". He builds that operator by taking the product of the operator "Bob's measurement" and the operator "Alice's measurement".

Now, the operator "Bob's measurement" is the "measurement along the a axis" which is nothing else but the "a-component" of the three-some of angular-momentum operators on the first particle. This threesome of operators is symbolised by sigma_1, and the "a-component" is found by doing the in-product between this 3-some of operators and the real unit vector a.

Now, for spin-1/2 particles, the angular momentum operators are the three Pauli matrices, and that is what sigma_1 stands for.

However, there's a complication: we have two particles. So, for the first particle, sigma_1 acts as a matrix, while for the second particle, it acts as a real number (it commutes).

In the same way, the operator "Alices measurement" is similar, with the dot product between the 3-some of operators (the three Pauli matrices again, but this time "acting upon the second particle only"), and the b-vector, to find the operator corresponding to Alice's measurement.

Bell then takes the product of these two operators as being the operator of the product of the outcome. This is in fact a bit sloppy, but he can get away with it, because both operators acting on different spaces, they commute.

We now have the famous observable operator, which corresponds to the measurement "the correlation between the outcomes of Alice and Bob".

He next takes the quantum-mechanical expectation value of this operator over the singlet state. That's what the brackets stand for, and what they mean is the following:

First, one lets the operator act upon the quantum state vector describing the singlet state. This gives us another state in Hilbert space.
Next, one takes the Hilbert in-product of the singlet state with the result of teh previous outcome. That's the quantum mechanical technique of finding "expectation values".

After A VERY LONG CALCULATION, this comes out to be -a.b
where this time we have the simple in-product in euclidean 3-space of two unit vectors. (at least, I take it to be the correct result, I didn't verify it).

JesseM

Feb24-07, 02:47 PM

Jesse, I looked at them, and they are helpful. Thanks.

But what I am still hoping for is a derivation that starts where Bell starts <s.a*s'.b'> and ends where Bell ends -a.b'. But the derivation (http://www.mathpages.com/home/kmath521/kmath521.htm) I linked to plus the additional comments I made does show that, just using some different notation. s.a*s'.b' in Bell's notation just means the product of the two measurement results (one using angle a and the other using angle b), where each measurement yields either spin-up (+1) or spin-down (-1), so on a given trial the product will be +1 if both are measured to have the same spin on their respective measurement axes, and -1 if they are measured to have opposite spin on their respective measurement axes. What the linked page shows is that on a given trial, the probability that s.a*s'.b' is +1 will be sin^2((a-b)/2), and the probability that s.a*s'.b' is -1 will be cos^2((a-b)/2). And by the definition of "expectation value", <s.a*s'.b'> must be:

(+1)*Probability(s.a*s'.b' = +1 on each trial) + (-1)*Probability(s.a*s'.b' = -1 on each trial)

As I showed in my comments, when you work this out using the above probabilities, you conclude that <s.a*s'.b'> will be equal to -cos(a - b).

vanesch

Feb24-07, 03:33 PM

For those who are interested, I did explicitly the calculation of Bell's expression.

The expansion over the singlet state is a bit clumsy:

the singlet state is 1/sqrt(2) ( |+> |-> - |->|+>)
|+> is the (1,0) element, and |-> is the (0,1) element in the hilbert space.

So we expand the expectation value:

< singlet | O1 O2 | singlet >

as:

( < + | < - | - < - | < + |) O1 O2 ( |+> |-> - |->|+>)

= < + | O1 | +> < - |O1 |-> + < - | O1 | - > < + | O2 | + > - < + |O1 |-> < - | O2 | + > - < -| O1 | + > < + | O2 | - >

EDIT: where I forgot the front factor of 1/2, because of the double presence of the square root.
(but in the notebook, it is ok).

See attachment.

DrChinese

Feb24-07, 04:32 PM

vanesch

Feb25-07, 03:25 AM

For those who are interested, I did explicitly the calculation of Bell's expression.

BTW, it occured to me that the way Bell writes his stuff, and the mistake wm made, is a nice illustration of how quantum theory can get around doing "local" things in a way that a classical view cannot.

wm made the calculation of the correlation, thinking he was doing a kind of classical calculation, where the "sign" of (s.a) determined the outcome at Alice, and the sign of (s.b) determined the outcome at Bob. The outcomes were supposed to be +1 or -1. So the true correlation would in fact be:

< sign(a.s) . sign(b.s) >, and not < (a.s) (b.s) >

However, by some mathematical coincidence, if s is a uniformly distributed unit vector in R^3, these two expressions come out the same.

As JesseM and I demonstrated, however, they do not equate -(a.b), but rather -(a.b)/2 or -(a.b)/3, depending on whether one considers them in 2 or in 3 dimensions.

Nevertheless, the thing is that the ACTUAL RESULT OF MEASUREMENT, if it is truely "sign(a.s)" (hence, a numerical value of +1 or -1 for each trial) is then indeed "locally produced" (because only depending upon a and s)).

As we see, however, the correlation then comes out to be -(a.b)/2, which doesn't violate the Bell inequalities - as expected.

Now, quantum theory does, apparently, the same thing. So why can't we say that in its "inner workings", instead of transporting a unit vector s, this funny 3-some of Pauli matrices is transported ?

The reason is that in the expression < (sigma.a) (sigma.b) >, we write down a *quantum-mechanical* expectation value of an OPERATOR. We do not write the STATISTICAL expectation value of A PRODUCT OF TWO RESULTS.

In other words, the outcome at Bob was NOT (sigma.b) ! It was ONE OF ITS EIGENVALUES (which happens to be +1 or -1). As such, we cannot really say that "we transport the outcome at Bob, which is (sigma.b), to Alice, where her outcome is (sigma.a), and multiply the two together".

If that were true, indeed, this would have been a local mechanism. But the result at Alice is NOT (sigma.a), and the result at Bob is NOT (sigma.b). The results are of the kind +/- 1...

Or are they ?

Well, we COULD say, if we wanted to, that the outcome at Alice is not +1 or -1, but (sigma.a). And we COULD say that the outcome at Bob is (sigma.b). But that's a funny situation! It would mean that Alice didn't, after all, get a genuine numerical result such as -1 or +1, but rather a mathematical operator over hilbert space. If that were true, then wm's reasoning would be correct in a way. We take the result at Alice (again: it is not -1 or +1, but an operator over hilbert space!), which is determined purely by what happens at Alice, and similarly at Bob's, and at the point of their meeting, they multiply their outcomes (which, again, are not -1 or +1, but are now operators over hilbert space) and hurray, we get the right correlations.

But what could that possibly mean, that Alice didn't get -1 or +1 at a trial, but each time an operator ? Well, it means that Alice got BOTH results. It means that Alice and Bob now have a quantum-mechanical description, and that they are in a superposition of having -1 and +1 (the operator contains both eigenvalues). This is exactly the MWI view on things, and it illustrates how in MWI, there is indeed no problem with locality. But the price to pay is rather high: you cannot say anymore that Alice got a measurement result which was each time -1 or +1 !

Now, independently of interpretation, the reason why the quantum formalism can make predictions which defy classical theories is that in the formalism of quantum theory, there is a difference between the mathematical representation of a measurement (which is a hermitean operator), and actual individual results of a measurement (which are eigenvalues of that hermitean operator). In a classical theory, the representation of a measurement is necessarily its outcome.

Feb25-07, 05:15 AM

vanesch

Feb25-07, 05:39 AM

Hi vanesch

I see only 5 viewers so far of your welcome appended note. I wonder if others too are having trouble accessing it?

Can you tell me its format please; or where we might get the right software to read it?

I cannot open it.

Thanks, wm

Ah, sorry. It is a mathematica notebook. You can freely download a reader for them on the wolfram website http://www.wolfram.com

I guess this is the path: http://www.wolfram.com/products/mathreader/

Feb25-07, 01:02 PM

Ah, sorry. It is a mathematica notebook. You can freely download a reader for them on the wolfram website http://www.wolfram.com

I guess this is the path: http://www.wolfram.com/products/mathreader/

:!!) :!!) Thanks; got it; great. :!!) :!!)

But there are some odd looking symbols: like

o/oo. = i?

And it starts:

pauli1 = ::0, 1<, :1,0<<. = Pauli matrix in mathematica notation?

(I can work them out if you are too busy -- but what about all the new readers that will soon come to it -- see below.)

Sorry also I question but I need to be certain that there is no implication of non-locality whatsoever in the derivation. (That's why I've been requesting such a derivation here for so long.)

Because:

1. I believe that QM, correctly understood, can derive most of its results locally; and I'm pretty sure you've done that.

2. I'd like to put such LOCAL maths into my own high-school framework and understanding.

[ Being a good girl o:) ''I don't do non-local''. :devil: ]

3. Most important of all: I'd like to comment on locality in QM by saying:

The EPR-Bohm correlation can be derived wholly locally using the prescription in Bell (1964, equation (3)); refer vanesch (Physics Forums) http://www.physicsforums.com/showpost.php?p=1255185&postcount=122

Would you agree with this statement?

4. AND SO, finally: Do you need to spend a little time to polish up your notebook page? BECAUSE I think you will find it becoming very popular. :smile: It is a very nice result. Especially from my point of view as a localist: IN FACT, I suggest it would be beaucoup worth the trouble to post it in LaTeX. (Or maybe I :yuck: could learn my LaTeX on it; with some help from JesseM :!!) to whom I owe much.)

Thanks, wm

vanesch

Feb25-07, 01:11 PM

Thanks; got it; great. :!!)

But there are some odd looking symbols: like

o/oo

And it starts:

pauli1 = ::0, 1<,1 :1,0<< :confused: ???

That's very strange. I re-downloaded the notebook and it looks ok for me. (ok, I don't use the MathReader, but mathematica 4.1, but at least, the file is not corrupt)

I ask about them because I need to be sure that there is no implication of non-locality whatsoever in the derivation. (That's why I've been requesting such a derivation here for so long.)

Because:

1. I believe that QM, correctly understood, can derive most of its results locally; and I'm pretty sure you've done that.

2. I'd like to put such LOCAL maths into my own high-school framework and understanding. (''I don't do non-local''.:devil: )

3. Most important of all: I'd like to comment on locality in QM by saying:

The EPR-Bohm correlation can be derived wholly locally using the prescription in Bell (1964, equation (3)); refer vanesch (Physics Forums) http://www.physicsforums.com/showpost.php?p=1255185&postcount=122

Would you agree with this statement?

4. AND SO, finally: Do you need to spend a little time to polish up your notebook page? BECAUSE I think you will find it becoming very popular. :smile:

Thanks, wm

Unfortunately, the derivation in QM is local, or non-local, at one's interpretation. As I said, in order to be able to consider it "local", one needs to make the hypothesis that there is no genuine unique measurement result at Alice and Bob, for each particle pair. Only then is one allowed to say that the result is an operator (and not simply a real value). And in that case, one can apply the kind of reasoning you wanted to apply with the unit vector s.

The STANDARD way of looking upon things in quantum mechanics, is non-local, or undefined. Using the projection, which affects, when Alice measures, also the state at Bob, is obviously non-local as a "calculation".

Now, depending on whether one assigns any "reality" to the wavefunction, this either means 1) (wavefunction is real) that the projection is an "action-at-a-distance" or 2) (wavefunction is not real) that this is an abstract calculational procedure which has nothing to do with any local or non-local mechanism.

JesseM

Feb25-07, 01:55 PM

Unfortunately, the derivation in QM is local, or non-local, at one's interpretation. As I said, in order to be able to consider it "local", one needs to make the hypothesis that there is no genuine unique measurement result at Alice and Bob, for each particle pair. Only then is one allowed to say that the result is an operator (and not simply a real value). And in that case, one can apply the kind of reasoning you wanted to apply with the unit vector s. It's probably worth expanding on this to make sure wm understands the sense in which QM can be "local" according to mainstream physics. I'm sure vanesch would agree that Bell's theorem rules out conventional local realism in which each measurement yields a unique result; but there is a loophole in which you can regain locality if you accept something like the many-worlds interpretation (http://plato.stanford.edu/entries/qm-manyworlds/) in which there is no "collapse of the wavefunction" on measurement, instead each spin measurement simply results in a superposition of states which includes both a state where the experimenter saw a result of spin-up and a state where the experimenter saw a result of spin-down. And the key to preserving locality is that the universe doesn't have to decide how to link the versions of experimenter #1 over here with the versions of experimenter #2 over there until there has been time for a signal moving at the speed of light to pass between them.

On a previous thread I gave a simple picture which attempts to show conceptually how you can preserve locality as long as you imagine each experimenter splitting into multiple "copies" with each measurement (although this picture should be taken with a grain of salt since there are problems with using a simple frequentist notion of counting 'copies' to derive subjective probabilities of seeing different results in the many-worlds interpretation). Recall that one of the Bell inequalities says that if Alice and Bob always get opposite spins + and - when they measure along the same axis, then when they measure along different axes, conventional single-universe local realism implies the probability of getting opposite results should be greater than or equal to 1/3. But here's my conceptual picture showing how if you accept they each split into multiple copies with each measurement, you can explain how they'll get opposite results on different axes on less than 1/3 of trials: say Bob and Alice are each recieving one of an entangled pair of photons, and their decisions about which spin axis to measure are totally deterministic, so the only "splitting" necessary is in the different possible results of their measurements. Label the three spin axes a, b, and c. If they always find opposite spins when they both measure their photons along the same axis, a local hidden-variables theory would say that if they choose different axes, the probability they get opposite spins must be at least 1/3 (assuming there's no correlation between their choice of which axes to measure and the states of the photons before they make the measurement). The actual probability of opposite spins along different axes depends on the difference in their detector angles, but all that's important is that it's less than 1/3, so for the sake of the argument let's say that when Alice chooses axis c and Bob chooses axis a, they only get opposite results 1/4 of the time, a violation of Bell's inequality.

So now suppose that when Bob makes a measurement on axis a in one location and Alice makes a measurement on axis c in another, each splits into 8 parallel versions, with 4 measuring spin + and 4 measuring spin -. Label the 8 Bobs like this:

Bob 1: a+
Bob 2: a+
Bob 3: a+
Bob 4: a+
Bob 5: a-
Bob 6: a-
Bob 7: a-
Bob 8: a-

Similarly, label the 8 Alices like this:

Alice 1: c+
Alice 2: c+
Alice 3: c+
Alice 4: c+
Alice 5: c-
Alice 6: c-
Alice 7: c-
Alice 8: c-

Note that the decision of how they split is based only on the assumption that each has a 50% chance of getting + and a 50% chance of getting - on whatever axis they choose, no knowledge about what the other one was doing was needed. And again, only when a signal travelling at the speed of light or slower passes from one to the other does the universe need to decide which Alice shares the same world with which Bob...when that happens, they can be matched up like this:

Alice 1 (c+) <--> Bob 1 (a+)
Alice 2 (c+) <--> Bob 2 (a+)
Alice 3 (c+) <--> Bob 3 (a+)
Alice 4 (c+) <--> Bob 5 (a-)
Alice 5 (c-) <--> Bob 4 (a+)
Alice 6 (c-) <--> Bob 6 (a-)
Alice 7 (c-) <--> Bob 7 (a-)
Alice 8 (c-) <--> Bob 8 (a-)

This insures that each one has a 3/4 chance of finding out the other got the same spin, and a 1/4 chance that the other got the opposite spin. If Bob and Alice were two A.I.'s running on classical computers in realtime, you could simulate Bob on one computer and Alice on another, make copies of each according to purely local rules whenever each measured a quantum particle, and then use this type of matching rule to decide which of the signals from the various copies of Alice will be passed on to which copy of Bob, and you wouldn't have to make that decision until the information from the computer simulating Alice was actually transmitted to the computer simulating Bob. So using purely local rules you could insure that, after many trials like this, a randomly-selected copy of A.I. Bob or A.I. Alice would record the same type of statistics that's seen in the Aspect experiment, including the violation of Bell's inequality.

Note that you wouldn't have to simulate any hidden variables in this case--you only have to decide what the spin was along the axes each one measured, you never have to decide what the spin along the other 2 unmeasured axes of each photon was. And wm, please note that I included this loophole way back in post #133 (http://www.physicsforums.com/showpost.php?p=1215310&postcount=133) of the other thread where I stated all the conditions which must be assumed in order to prove that quantum results are incompatible with local realism: do you agree or disagree that if we have two experimenters with a spacelike separation who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like), then if they always get opposite answers when they make the same measurement on any given trial, and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement with no violations of locality allowed (and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event, as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation), then the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

Doc Al

Feb25-07, 02:24 PM

but there is a loophole in which you can regain locality if you accept something like the many-worlds interpretation (http://plato.stanford.edu/entries/qm-manyworlds/) in which there is no "collapse of the wavefunction" on measurement, instead each spin measurement simply results in a superposition of states which includes both a state where the experimenter saw a result of spin-up and a state where the experimenter saw a result of spin-down.

But there are folks--and I count myself among them--who think that the "many worlds" interpretation does such violence to the usual notion of reality--with its multiple copies of experimenters and experimental results--that preserving locality with such a model is at best a Pyrrhic victory. Given such a metaphysics, locality is the least of one's worries. With all due respect to vanesch (a gentlemen and scholar--unlike Deutsch, who belongs in the loony bin :wink:)--I just don't see how it solves anything.

Feb25-07, 03:54 PM

BTW, it occured to me that the way Bell writes his stuff, and the mistake wm made, is a nice illustration of how quantum theory can get around doing "local" things in a way that a classical view cannot.(With some minor editing and emphasis throughout.)

Comment: This post is very helpful to me. Thanks.

For the moment (for reasons to later appear): Can we have a superposition?

|Y> = V|wm wrong> + W|wm right>.

wm made the calculation of the correlation, thinking he was doing a kind of classical calculation, where the "sign" of (s.a) determined the outcome at Alice, and the sign of (s.b) determined the outcome at Bob.

Comment: wm made a classical calculation, at the level of high-school maths and logic. Maybe she went astray with her short-cut involving s and s', each a unit-vector relating to angular-momentum, and thus each a unit-axial-vector = (strictly) a bi-vector. wm believes they may transform differently.

The outcomes were supposed to be +1 or -1. So the true correlation would in fact be:

< sign(a.s) . sign(b.s) >, and not < (a.s) (b.s) >

However, by some mathematical coincidence, if s is a uniformly distributed unit vector in R^3, these two expressions come out the same.

I think this maybe not a coincidence (to be developed).

As JesseM and I demonstrated, however, they do not equate -(a.b), but rather -(a.b)/2 or -(a.b)/3, depending on whether one considers them in 2 or in 3 dimensions.

Yes; treating (standard) unit-vectors, these are both well-known relations for (standard) unit-vectors. See http://mathworld.wolfram.com/Vector.html (eg; eqn (9)).

If my comments were not clear, apologies to those who thought I was ignoring these results. My focus then was on another, equally interesting, result with bi-vectors.

SO: The question still open (for me) relates to the transformation of axial-vectors (bi-vectors; relating to angular-momentum). Maybe someone knows this answer already?

I suspect that I just need to apply Pauli matrices to classical angular momentum? Maybe some has done this already too? Would this not be possible (or not be correct) for any reason?

Nevertheless, the thing is that the ACTUAL RESULT OF MEASUREMENT, if it is truely "sign(a.s)" (hence, a numerical value of +1 or -1 for each trial) is then indeed "locally produced" (because only depending upon a and s)).

As we see, however, the correlation then comes out to be -(a.b)/2, which doesn't violate the Bell inequalities - as expected.

Yes; exactly as expected because this is the correct and well-known classical transformation for standard unit-vectors.

Now, quantum theory does, apparently, the same thing. So why can't we say that in its "inner workings", instead of transporting a unit vector s, this funny 3-some of Pauli matrices is transported ?

The reason is that in the expression < (sigma.a) (sigma.b) >, we write down a *quantum-mechanical* expectation value of an OPERATOR. We do not write the STATISTICAL expectation value of A PRODUCT OF TWO RESULTS.

In other words, the outcome at Bob was NOT (sigma.b) ! It was ONE OF ITS EIGENVALUES (which happens to be +1 or -1). As such, we cannot really say that "we transport the outcome at Bob, which is (sigma.b), to Alice, where her outcome is (sigma.a), and multiply the two together".

If that were true, indeed, this would have been a local mechanism. But the result at Alice is NOT (sigma.a), and the result at Bob is NOT (sigma.b). The results are of the kind +/- 1...

I believe it is a wholly local mechanism, relating to the conservation of angular momentum (to be developed).

Or are they ?

Yes; they certainly are. For sure! With the detectors so programmed, no one of my acquaintance has seen anything but elements from the set {+1, -1, +1', -1'}; the prime denoting Bob's results

Well, we COULD say, if we wanted to, that the outcome at Alice is not +1 or -1, but (sigma.a). And we COULD say that the outcome at Bob is (sigma.b). But that's a funny situation! It would mean that Alice didn't, after all, get a genuine numerical result such as -1 or +1, but rather a mathematical operator over hilbert space. If that were true, then wm's reasoning would be correct in a way. We take the result at Alice (again: it is not -1 or +1, but an operator over hilbert space!), which is determined purely by what happens at Alice, and similarly at Bob's, and at the point of their meeting, they multiply their outcomes (which, again, are not -1 or +1, but are now operators over hilbert space) and hurray, we get the right correlations.

vanesch, we COULD say lots of things here; BUT why would we want to say other than the truth?

1. That Alice did indeed get {+1, -1}; and while we were with her we saw these results with our own eyes and with our own friends.

2. That when we went over to Bob's lab we saw {+1', -1'}, etc.

3. And whenever we drop in at either lab (unannounced) we see the same consistent detector outputs; even when Alice and Bob were absent! Why not come with us? (For I miss your point here.)

4. Then you can join us in the correlation-checking, which we do (and can only do in this wondrously local quantum world) after we have hold of each set of related results:

Settings anti-parallel on this occasion:

Alice's paper-tape: +1, +1, -1, +1, -1, -1, ....

Bob's paper-tape: +1', +1', -1', +1', -1', -1', ....

But then; you already know the correlation from your (maybe non-local in your eyes) calculation. So all you really need is to see the tapes (and leave the correlation analysis to us). Why is not like this, please?

But what could that possibly mean, that Alice didn't get -1 or +1 at a trial, but each time an operator ? Well, it means that Alice got BOTH results. It means that Alice and Bob now have a quantum-mechanical description, and that they are in a superposition of having -1 and +1 (the operator contains both eigenvalues). This is exactly the MWI view on things, and it illustrates how in MWI, there is indeed no problem with locality. But the price to pay is rather high: you cannot say anymore that Alice got a measurement result which was each time -1 or +1 !

Seriously; save your money or give it to me.

Every Alice known to me has reported (to me, by phone or in person) the measurement results +1 and -1 only; each being a distinct printout on a permanent-record paper-tape.

Now, independently of interpretation, the reason why the quantum formalism can make predictions which defy classical theories is that in the formalism of quantum theory, there is a difference between the mathematical representation of a measurement (which is a hermitean operator), and actual individual results of a measurement (which are eigenvalues of that hermitean operator). In a classical theory, the representation of a measurement is necessarily its outcome.

I believe rather that QM is advanced probability theory and that we are quantum machines in a wholly quantum world.

That's very strange. I re-downloaded the notebook and it looks ok for me. (ok, I don't use the MathReader, but mathematica 4.1, but at least, the file is not corrupt)

OK; I'll see what I can do.

Unfortunately, the derivation in QM is local, or non-local, at one's interpretation. As I said, in order to be able to consider it "local", one needs to make the hypothesis that there is no genuine unique measurement result at Alice and Bob, for each particle pair. Only then is one allowed to say that the result is an operator (and not simply a real value). And in that case, one can apply the kind of reasoning you wanted to apply with the unit vector s.

Well I don't mind people holding non-local intepretations. BUT I'm sure glad we're living in a common-sense local and realistic quantum world.

The STANDARD way of looking upon things in quantum mechanics, is non-local, or undefined. Using the projection, which affects, when Alice measures, also the state at Bob, is obviously non-local as a "calculation".

That's fine with me: probability theory is ideally equipped to deal with non-local calculations; WHICH, incidentally, are only ever confirmed via local communications.

Now, depending on whether one assigns any "reality" to the wavefunction, this either means 1) (wavefunction is real) that the projection is an "action-at-a-distance" or 2) (wavefunction is not real) that this is an abstract calculational procedure which has nothing to do with any local or non-local mechanism.

Thank you; I'll take 2) every time. I'm not inclined to see as real abstract-elements of abstract spaces. Moreover, there is that maths theorem from 1915 that shows that any probability distribution CAN ALWAYS be represented by the absolute square of a Fourier polynomial ( = wave-function?).

With thanks again for your expansionary comments.

Regards, wm

Feb25-07, 04:06 PM

Feb25-07, 04:15 PM

It's probably worth expanding on this to make sure wm understands the sense in which QM can be "local" according to mainstream physics. I'm sure vanesch would agree that Bell's theorem rules out conventional local realism in which each measurement yields a unique result; but there is a loophole in which you can regain locality if you accept something like the many-worlds interpretation (http://plato.stanford.edu/entries/qm-manyworlds/) in which there is no "collapse of the wavefunction" on measurement, instead each spin measurement simply results in a superposition of states which includes both a state where the experimenter saw a result of spin-up and a state where the experimenter saw a result of spin-down. And the key to preserving locality is that the universe doesn't have to decide how to link the versions of experimenter #1 over here with the versions of experimenter #2 over there until there has been time for a signal moving at the speed of light to pass between them.

On a previous thread I gave a simple picture which attempts to show conceptually how you can preserve locality as long as you imagine each experimenter splitting into multiple "copies" with each measurement (although this picture should be taken with a grain of salt since there are problems with using a simple frequentist notion of counting 'copies' to derive subjective probabilities of seeing different results in the many-worlds interpretation). Recall that one of the Bell inequalities says that if Alice and Bob always get opposite spins + and - when they measure along the same axis, then when they measure along different axes, conventional single-universe local realism implies the probability of getting opposite results should be greater than or equal to 1/3. But here's my conceptual picture showing how if you accept they each split into multiple copies with each measurement, you can explain how they'll get opposite results on different axes on less than 1/3 of trials: And wm, please note that I included this loophole way back in post #133 (http://www.physicsforums.com/showpost.php?p=1215310&postcount=133) of the other thread where I stated all the conditions which must be assumed in order to prove that quantum results are incompatible with local realism:

Jesse, it's not your fault, but I am not that good with words; I struggle.

But now that we have some maths before us; can we discuss things more in the context of specific experiments and related specific maths?

I think that it would be good to have vanesch's addendum converted out of mathematica (which I do not have) into a LaTeX post.

Is there a program that would do that? (It is only very short.)

This was something I have chased for awhile and since it is central QM, discussion of it will equally be central QM and common non-physicist mistakes.

I will still study your words, which I welcome every time. But mess-ups and time-delays are likely.

PS: To be clear, I do NOT think that QM is incompatible with common-sense local realism. NOT IN ANY WAY do I think that. Have I said that?

wm

JesseM

Feb25-07, 04:29 PM

PS: To be clear, I do NOT think that QM is incompatible with common-sense local realism. NOT IN ANY WAY do I think that. Have I said that? I never said that YOU think that. What I said is that mainstream physicists would all agree QM is incompatible with common-sense local realism, based on Bell's theorem (which need NOT include any assumption that measurements don't disturb the particle, if that's what you mean by 'Bellian realism') and that I am sure vanesch would agree, since when he talks about QM being 'local' he is only referring to a non-commonsense interpretation where measurements do not yield a single unique outcome. Do you think his derivation suggests the -cos(a-b) expectation value is compatible with common-sense local realism? If you do, then you have misunderstood something.

Also, do you still claim that you have a classical method of reproducing the -cos(a-b) expectation value based on sending two classical vectors s and -s to different experimenters, or have we managed to convince you that your math was incorrect in that case, and that the expectation value would be either -(1/2)*cos(a-b) or -(1/3)*cos(a-b) depending on whether s was a 2-vector or a 3-vector?

Doc Al

Feb25-07, 04:32 PM

Doc, you seem ''sympathetic'' to the way I see the world. And the way I see MWI. (I hope so.)
Don't jump to conclusions. :wink: While I am sympathetic to your desire for a local interpretation of QM, I just don't see it as a serious possibility given Bell's theorem and current experimental results.

Would you mind expanding on your world-view please? And pointing me to some of your papers?
I certainly have no papers on this topic. One person I admire is Bell himself.

I am not a physicist, but I see no reason to abandon my common-sense realism (which allows for measurement perturbation) and Einstein locality. (My post before this sets out some of my ideas.)
I suspect that's because you don't appreciate the import of Bell's theorem.

I do not accept what I call Bellian realism; a concept which Peres said (I believe) had nothing to do with QM. I feel that way and am truly surprised so few others see the world that way.
Not sure what you mean by "Bellian realism". As I understand it, and I'm hardly an expert, is that Bell's theorem (combined with the experimental facts of QM and the reasoning of Einstein himself in EPR) leads to the conclusion that no theory satisfying Bell locality can accurately describe the world as we know it.

Feb25-07, 04:41 PM

Although I tailored the short proofs I gave above to a particular thought-experiment, it's quite trivial to change a few words so they cover any situation where two people can measure one of three properties and they find that whenever they measure the same property they get opposite results. If you don't see how, I can do this explicitly if you'd like. I am interested in the physics of the situation, not in playing a sort of "gotcha" game where if we can show that Bell's original proof did not cover all possible local hidden variable explanations then the whole proof is declared null and void, even if it would be trivial to modify the proof to cover the new explanations we just thought up as well. I'll try reading his paper to see what modifications, if any, would be needed to cover the case where measurement is not merely revealing preexisting spins, but in the meantime let me ask you this: do you agree or disagree that if we have two experimenters with a spacelike separation who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like), then if they always get opposite answers when they make the same measurement on any given trial, and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement with no violations of locality allowed (and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event, as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation), then the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

Jesse,

1. Do you see here how long you have sentences?

2. Does not my classical model (of old) refute this Bellian-Inequality easily? Are you not giving conditions which my model meets?

3. Are you not saying (as I will let you):

a. That Alice may make a countable-inifinity of detector-settings, each delivering outcome of {+1, -1}.

b. That Bob may make a countable-infinity of detector-settings, each delivering outcomes of {+1', -1'}.

4. Anyway: Down-hill skiers, dirty-socks, books and the like will satisfy your inequality. More subtle, less wholly concrete objects will sink it for some detector combinations. Yes?

5. Is my conclusion not what vanesch has shown?

wm

JesseM

Feb25-07, 04:45 PM

But there are folks--and I count myself among them--who think that the "many worlds" interpretation does such violence to the usual notion of reality--with its multiple copies of experimenters and experimental results--that preserving locality with such a model is at best a Pyrrhic victory. Given such a metaphysics, locality is the least of one's worries. With all due respect to vanesch (a gentlemen and scholar--unlike Deutsch, who belongs in the loony bin :wink:)--I just don't see how it solves anything. Yes, I do think philosophical questions are unavoidable when discussing the many-worlds interpretation. But when discussing a possible universe where observers are constantly splitting upon measurement (which is certainly logically possible, we could simulate such a universe on a computer), I think it makes sense to relate this to the subjective probabilities experienced by each observer using the "self-sampling assumption" which the philosopher Nick Bostrom argues is implicit in all forms of anthropic reasoning. Basically, this assumption says that it makes sense in many circumstances to reason as if you were randomly sampled from the set of all observers, and to use this assumption to update your estimate of the probabilities of different events using Bayes' rule. On his website anthropic-principle.com (http://www.anthropic-principle.com/) he discusses this principle in detail, and includes many thought-experiments to show why it is plausible, such as these ones from the paper Self-Location and Observation Selection Theory (http://anthropic-principle.com/preprints/self-location.html): Dungeon

The world consists of a dungeon that has one hundred cells. In each cell there is one prisoner. Ninety of the cells are painted blue on the outside and the other ten are painted red. Each prisoner is asked to guess whether he is in a blue or a red cell. (And everybody knows all this.) You find yourself in one of these cells. What color should you think it is? – Answer: Blue, with 90% probability. In the doomsday argument FAQ (http://www.anthropic-principle.com/faq.html) he quotes a similar thought-experiment by John Leslie: firm plan was formed to rear humans in two batches: the first batch to be of three humans of one sex, the second of five thousand of the other sex. The plan called for rearing the first batch in one century. Many centuries later, the five thousand humans of the other sex would be reared. Imagine that you learn you’re one of the humans in question. You don’t know which centuries the plan specified, but you are aware of being female. You very reasonably conclude that the large batch was to be female, almost certainly. If adopted by every human in the experiment, the policy of betting that the large batch was of the same sex as oneself would yield only three failures and five thousand successes. ... [Y]ou mustn’t say: ‘My genes are female, so I have to observe myself to be female, no matter whether the female batch was to be small or large. Hence I can have no special reason for believing it was to be large.’ If we accept this sort of reasoning as valid, then it can be applied to a situation where I am constantly being copied, since if I reason as though I am randomly sampled from the set of all copies of me, I can draw probabilistic conclusions about what I am likely to see over the result of many measurements.

DrChinese

Feb25-07, 04:48 PM

PS: To be clear, I do NOT think that QM is incompatible with common-sense local realism. NOT IN ANY WAY do I think that. Have I said that?

You continue to ignore Bell's Theorem conveniently as if it does not exist. There is no substantive difference between your "common-sense" definition of realism (per your page) and Bell's. Even if there were, no one would care because Bell's maps to the debate of concern to Einstein, Bohr, and everyone who follows EPR's argument.

You should quit confusing people with your statements, and acknowledge as follows:

1. You believe in locality.
2. You believe in your version of realism, which is slightly different than Bell's but you are not sure how so mathematically.
3. You do not accept Bell's Theorem as valid.
4. You accept 3. as a matter of faith because you believe 1. and 2., and you think other folks should too.
5. You have no actual plan for developing your pet theory, but hope that those of us here at Physicsforums will help you.

I would like to point out that this discussion belongs in Theory Development, and not quantum physics.

Feb25-07, 04:52 PM

Don't jump to conclusions. :wink: While I am sympathetic to your desire for a local interpretation of QM, I just don't see it as a serious possibility given Bell's theorem and current experimental results.

I certainly have no papers on this topic. One person I admire is Bell himself.

I suspect that's because you don't appreciate the import of Bell's theorem.

Not sure what you mean by "Bellian realism". As I understand it, and I'm hardly an expert, is that Bell's theorem (combined with the experimental facts of QM and the reasoning of Einstein himself in EPR) leads to the conclusion that no theory satisfying Bell locality can accurately describe the world as we know it.

Thanks :grumpy: darn it! But while I'm here: For me I would change the last two sentences to put in my language:

"Bellian realism" is the constrained realism that we may associate with Bell (1964; for example); where A+, originally a measurement outcome, is subtly changed to a property of the particle itself; what I call the d'Espagnat move (from Sci. Am). As I understand it, and I'm hardly an expert, Bell's theorem (combined with the experimental facts of QM and the reasoning of Einstein* on locality) leads to the conclusion that no theory satisfying Bellian realism describes the world as we know it.

PS: * Einstein did not see the final draft; disagreement was later such that Einstein never again spoke to the author (Podolsky). That is why I believe many pot-shots at Einstein over EPR can be a bit misleading.

wm

JesseM

Feb25-07, 05:13 PM

Jesse,

1. Do you see here how long you have sentences? Well, if the long sentences are hard to follow, just ask for clarification, it's usually pretty easy to break them up into a list of distinct statements or assumptions...in the quote above, I could rewrite the assumptions like this:

do you agree or disagree that IF we have:

1. two experimenters with a spacelike separation
2. who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like)
3. then if they always get opposite answers when they make the same measurement on any given trial
4. and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement
5. with no violations of locality allowed
6. and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event
7. as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation)

THEN the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

2. Does not my classical model (of old) refute this Bellian-Inequality easily? Are you not giving conditions which my model meets? I would say your old model, where the source knows what detector setting Alice will use before it sends out a signal, is violating condition 6 above, "the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial". In your "yoked" experiment, the angle that Alice measures the polarization is not independent of the polarization of the light sent out (and knowing both Alice's measurement angle and the polarization of the light does predetermine the result).

Of course my condition 6 also talked about a "backwards-in-time influence", which isn't true in your yoked scenario, since you assume there is enough time between Alice choosing her measurement angle and Alice actually making a measurement for a signal to have gotten back to the source and told it what the angle would be before it sent out the polarized light. I guess in condition 1 I was implicitly assuming that each experimenter's choice of detector setting was immediately before they actually made a measurement, so that there is also a spacelike separation between these two pairs of events:

1. (Alice randomly choosing her measurement angle) AND (Bob measuring the signal/object sent to him from the source)
2. (Bob randomly choosing his measurement angle) AND (Alice measuring the signal/object sent to her from the source)

So, if you add this condition explicitly to my list, then it would be impossible for the source's choice of what signals/objects to be sent out to be correlated with Alice and Bob's choice of detector settings, unless somehow the information was travelling backwards in time (or unless there was a weird cosmic conspiracy in the initial conditions of the universe which caused the source's output to be correlated with Alice and Bob's choices on each trial even though no signal could travel between them).
3. Are you not saying (as I will let you):

a. That Alice may make a countable-inifinity of detector-settings, each delivering outcome of {+1, -1}.

b. That Bob may make a countable-infinity of detector-settings, each delivering outcomes of {+1', -1'}. Yes. 4. Anyway: Down-hill skiers, dirty-socks, books and the like will satisfy your inequality. More subtle, less wholly concrete objects will sink it for some detector combinations. Yes? Nope, as long as you obey my conditions 1-7 above (including the clarification of what I meant by condition #1...also, note that #4 is not so much a condition as a logical conclusion necessitated by the other 6), then it is impossible for the inequalities to be violated by any experiment whatsoever. Therefore, since the inequalities are empirically violated in QM, it must be that QM violates one of the assumptions 5-7...either QM allows nonlocality, or QM allows backwards-in-time signalling, or QM allows experimenters to split into multiple copies (I suppose QM could also just violate the rules of logic, but I was assuming traditional logic must be obeyed). 5. Is my conclusion not what vanesch has shown? Not at all--where did you get that idea? Vanesch just shows that when you use the conventional quantum rules to make predictions, you get the prediction that the expectation value is -cos(a-b). But the conventional quantum rules themselves do not say anything about locality or nonlocality, that's a matter for interpretation.

JesseM

Feb25-07, 05:24 PM

"Bellian realism" is the constrained realism that we may associate with Bell (1964; for example); where A+, originally a measurement outcome, is subtly changed to a property of the particle itself Again, you are free to assume that A+ is not a property of the particle itself, but that the particle just has some properties P which, when the particle is disturbed by a measurement of type A, leads deterministically to the result +. I tried to explain the logic behind the need for determinism in particle properties and measurement setting in post #56 (http://www.physicsforums.com/showpost.php?p=1252040&postcount=56), if you don't see the logic it might help if you answered my questions there. Note that the assumption here is not that the whole universe is deterministic--indeed, the particular properties P of the particle sent out on each trial may be randomly created by the source, and the experimenter's choice of measurement setting may be random too--just that, if we know the complete properties of the particle and the measurement setting on a given trial, that is enough to uniquely determine the result on that trial.

Feb25-07, 05:31 PM

You continue to ignore Bell's Theorem conveniently as if it does not exist. There is no substantive difference between your "common-sense" definition of realism (per your page) and Bell's. Even if there were, no one would care because Bell's maps to the debate of concern to Einstein, Bohr, and everyone who follows EPR's argument.

Bell's theorem (1964) is ever in my thoughts; so why do you say this?

You should quit confusing people with your statements, and acknowledge as follows:

1. You believe in locality.
2. You believe in your version of realism, which is slightly different than Bell's but you are not sure how so mathematically.
3. You do not accept Bell's Theorem as valid.
4. You accept 3. as a matter of faith because you believe 1. and 2., and you think other folks should too.
5. You have no actual plan for developing your pet theory, but hope that those of us here at Physicsforums will help you.

I would like to point out that this discussion belongs in Theory Development, and not quantum physics.

1. Yes.

2. Yes, re my realism. Then, No; I am sure now that I have seen vanesch's addendum (which I repeatedly requested of you) that mathematically I will be similar in deriving the EPR-Bohm correlation.

3. Just to be careful here: If you would define it, I would expect to give you a definite answer. Please define.

4. See 3 above. When it comes to maths, I'm not a faith-based person. So you are dead wrong again and again. (Why not follow JesseM and ask questions rather than promote lies and error (as you here do)?

5. (a) No! This is quite false and you must know that it is false!

Evidence: How would you know that it is false?

(a) I wrote to you off-PF and you replied that you were going to look at a matter and get back to me. You have not.

(b) So I am tending to read your question as an attempt to bias helpful communications from others with me. I take it you too can see where we're heading mathematically, thanks to vanesch. Put it another way: I see no theory of mine under threat.

SO why don't you ask an upfront question instead of non-locally reading my mind! Or should I rather say: Your non-locality fails again! (Many expletives deleted. But are you a PhD? By any chance, DrtC?)

5. (b) The PF communications have helped me greatly, especially JesseM, vanesch, hurkyl, DocAl +++

If I were familiar with LaTeX I would be more expansive.

PS-1. As to ''theory development'': This confirms my view that you can see where vanesch's maths (not yet mine) takes us.

PS-2. I hope the authorities will see that you are wrong (once again).

PS-3. Please provide evidence: What is the new theory that I am developing, please. (Please do not misunderstand or misrepresent the hidden-variables revealed on my website -- I should have picked you up on this before.)

PS-4. Finally: How many refereed papers do you want shoved down your fabricacious mouth?

wm

JesseM

Feb25-07, 05:42 PM

PS-1. As to ''theory development'': This confirms my view that you can see where vanesch's maths (not yet mine) takes us. If you think vanesch's math somehow shows QM is compatible with common-sense local realism, then you have totally misunderstood it (it would help if you explained why you think there is anything 'local' about his math).

And if you are not at least open to the possibility that Bell's theorem is valid and that common-sense local realism is definitively ruled out by quantum results such as the -cos(a-b) expectation value, then I agree this should go in theory development. If you are hoping to find a hole in Bell's theorem, but admit the fault may be in your understanding and are trying to improve your understanding through these discussions, then I think it's OK to continue the discussion here. So which is it?

Feb25-07, 05:42 PM

Well, if the long sentences are hard to follow, just ask for clarification, it's usually pretty easy to break them up into a list of distinct statements or assumptions...in the quote above, I could rewrite the assumptions like this:

do you agree or disagree that IF we have:

1. two experimenters with a spacelike separation
2. who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like)
3. then if they always get opposite answers when they make the same measurement on any given trial
4. and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement
5. with no violations of locality allowed
6. and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event
7. as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation)

THEN the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

I would say your old model, where the source knows what detector setting Alice will use before it sends out a signal, is violating condition 6 above, "the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial". In your "yoked" experiment, the angle that Alice measures the polarization is not independent of the polarization of the light sent out (and knowing both Alice's measurement angle and the polarization of the light does predetermine the result).

Of course my condition 6 also talked about a "backwards-in-time influence", which isn't true in your yoked scenario, since you assume there is enough time between Alice choosing her measurement angle and Alice actually making a measurement for a signal to have gotten back to the source and told it what the angle would be before it sent out the polarized light. I guess in condition 1 I was implicitly assuming that each experimenter's choice of detector setting was immediately before they actually made a measurement, so that there is also a spacelike separation between these two pairs of events:

1. (Alice randomly choosing her measurement angle) AND (Bob measuring the signal/object sent to him from the source)
2. (Bob randomly choosing his measurement angle) AND (Alice measuring the signal/object sent to her from the source)

So, if you add this condition explicitly to my list, then it would be impossible for the source's choice of what signals/objects to be sent out to be correlated with Alice and Bob's choice of detector settings, unless somehow the information was travelling backwards in time (or unless there was a weird cosmic conspiracy in the initial conditions of the universe which caused the source's output to be correlated with Alice and Bob's choices on each trial even though no signal could travel between them).
Yes. Nope, as long as you obey my conditions 1-7 above (including the clarification of what I meant by condition 1), then it is impossible for the inequalities to be violated by any experiment whatsoever. Therefore, since the inequalities are empirically violated in QM, it must be that QM violates one of the assumptions 5-7...either QM allows nonlocality, or QM allows backwards-in-time signalling, or QM allows experimenters to split into multiple copies (I suppose QM could also just violate the rules of logic, but I was assuming traditional logic must be obeyed). Not at all--where did you get that idea? Vanesch just shows that when you use the conventional quantum rules to make predictions, you get the prediction that the expectation value is -cos(a-b). But the conventional quantum rules themselves do not say anything about locality or nonlocality, that's a matter for interpretation.

Jesse, Do the following short-answers answer most of your questions:

1. vanesch has enabled me to clearly see that your last sentence is one that I can now wholeheartedly endorse!

2. You have been equally helpful in that important (for me, crucial) process; for which I thank you too.

3. Given the number of believers in non-locality, I had been searching for the origin of such a strange belief (which is totally alien to my present world-view).

4. I see now that I can happily persist with my concrete thinking-style; and bring more powerful arguments as to why non-locality is .... :rofl:

5. I'm sorry if my slow-learning style upsets you. If I'm allowed to stay here (refer DrC prior post) then I think my arguments against non-locality will be be improved; coming from a more enlightened student.

Cheers, wm

JesseM

Feb25-07, 05:55 PM

1. vanesch has enabled me to clearly see that your last sentence is one that I can now wholeheartedly endorse! But I'm sure vanesch would also agree that "the conventional quantum rules themselves" + Bell's theorem do imply that common-sense local realism is ruled out, and that the only local options involve noncommonse interpretations like the many-worlds interpretation or perhaps the transactional interpretation (in which future events can effect past events that lie in their past light cone). Obviously you do not endorse this conclusion, so if you're open to the possibility that this might be due to an error in your understanding, it would help if we discussed the reasoning behind Bell's theorem more carefully. 4. I see now that I can happily persist with my concrete thinking-style; and bring more powerful arguments as to why non-locality is .... :rofl: No you can't, not unless you wish to persist in your confused understanding of Bell's theorem.

Again, to get around this confusion, please address the following:

1. Do you agree that none of the classical experiments you've presented so far (the "yoked" polarizer experiment and the experiment sending two classical vectors) both satisfy my conditions above (including the clarifications I added) and show a violation of any Bell inequality?

2. Do you agree that in a classical universe obeying locality, if experimenters always get the same (or opposite) result when using the same setting, and there is no possibility the source can know their choice of settings in advance (see my clarification of condition 1), then the only way to explain this correlation is to assume the complete properties of the object/signal sent out by the source + the choice of detector setting -> a deterministic result for that measurement? If you don't agree, please address my questions 1-3 in post #56 (http://www.physicsforums.com/showpost.php?p=1252040&postcount=56), or provide an alternate classical explanation for the correlation.

Feb25-07, 06:01 PM

If you think vanesch's math somehow shows QM is compatible with common-sense local realism, then you have totally misunderstood it (it would help if you explained why you think there is anything 'local' about his math).

And if you are not at least open to the possibility that Bell's theorem is valid and that common-sense local realism is definitively ruled out by quantum results such as the -cos(a-b) expectation value, then I agree this should go in theory development. If you are hoping to find a hole in Bell's theorem, but admit the fault may be in your understanding and are trying to improve your understanding through these discussions, then I think it's OK to continue the discussion here. So which is it?

Jesse; more misunderstandings; is my writing that bad?

1. I had understood that there were POWERFUL arguments for NON-LOCALITY.

2. vanesch shows me (us all, surely) that there are not.

3. My world-view is common-sense local realism (CLR).

4. I would call that my interpretation of the QM and its formalisms.

5. Does my CLR interpretation cause any problems here on PF?

6. For it seems to be a fairly-mild mid-range belief compared to others and extremes that I find here on PF.

7. vanesch enables me to discuss my view in the light of QM maths; with no need to dispute the maths.

8. vanesch enables me to bring better arguments to my view in the light of QM maths.

8. I think it more like that I am more here to find any holes in my world-view. (PS: While I am doing that, Am I permitted to poke holes in other views; especially using maths as my main argument now.)

9. I think I am very happy: and mostly due to about 6 lines of vanesch maths: 6 lines that you know I've been asking for for quite a while.

10. Is this above acceptable, please?

wm

JesseM

Feb25-07, 06:15 PM

Jesse; more misunderstandings; is my writing that bad? What exactly have I "misunderstood"? 1. I had understood that there were POWERFUL arguments for NON-LOCALITY.

2. vanesch shows me (us all, surely) that there are not. WHY DO YOU THINK THIS???? Vanesch just gave a recipe for calculating things in QM, the recipe itself is not based on local or nonlocal signals between events, it doesn't explain anything about why you see the correlations you do between measurements on entangled particles. However, if you take the results given to you by the recipe, and then you take Bell's theorem, it is clear that logically the results DO absolutely rule out common-sense local realism.

You seem to be confusing these two statements:

-Vanesch's calculations do not in themselves say anything either way about nonlocality vs. locality

-Vanesch's calculations are equally compatible with nonlocality and (common-sense) locality

But they are NOT equivalent--the first is true while the second is totally false! Logically the results are completely incompatible with common-sense local realism, it's just that this is not immediately obvious from looking at the calculations, you have to provide some additional logical arguments which go by the name of "Bell's theorem". 3. My world-view is common-sense local realism (CLR).

4. I would call that my interpretation of the QM and its formalisms. But it is an invalid interpretation, definitively ruled out by quantum predictions. Bell's theorem shows this. 5. Does my CLR interpretation cause any problems here on PF? Yes, no mainstream physicist would accept it as a valid "interpretation", because it does not make logical sense. You would see this if you actually made an effort to understand Bell's theorem, which is why I have been trying to walk you through it. If you're not trying to understand it, then this is equivalent to advancing the "interpretation" that perpetual motion is possible without listening to people's attempts to explain why it is ruled out by the laws of thermodynamics. 9. I think I am very happy: and mostly due to about 6 lines of vanesch maths: 6 lines that you know I've been asking for for quite a while.

10. Is this above acceptable, please? No, because you have given no explanation for why you think vanesch's math somehow supports your idea. It doesn't, all vanesch's math gives is a recipe for calculating the probabilities, and then Bell's theorem proves that these probabilities are absolutely incompatible with common-sense local realism.

Feb25-07, 06:17 PM

But I'm sure vanesch would also agree that "the conventional quantum rules themselves" + Bell's theorem do imply that common-sense local realism is ruled out, and that the only local options involve noncommonse interpretations like the many-worlds interpretation or perhaps the transactional interpretation (in which future events can effect past events that lie in their past light cone). Obviously you do not endorse this conclusion, so if you're open to the possibility that this might be due to an error in your understanding, it would help if we discussed the reasoning behind Bell's theorem more carefully. No you can't, not unless you wish to persist in your confused understanding of Bell's theorem.

Again, to get around this confusion, please address the following:

1. Do you agree that none of the classical experiments you've presented so far (the "yoked" polarizer experiment and the experiment sending two classical vectors) both satisfy my conditions above (including the clarifications I added) and show a violation of any Bell inequality?

2. Do you agree that in a classical universe obeying locality, if experimenters always get the same (or opposite) result when using the same setting, and there is no possibility the source can know their choice of settings in advance (see my clarification of condition 1), then the only way to explain this correlation is to assume the complete properties of the object/signal sent out by the source + the choice of detector setting -> a deterministic result for that measurement? If you don't agree, please address my questions 1-3 in post #56 (http://www.physicsforums.com/showpost.php?p=1252040&postcount=56), or provide an alternate classical explanation for the correlation.

I personally DO NOT NEED A CLASSICAL EXPLANATION OF ANYTHING :::: NOW THAT I HAVE VANESCH'S MATHS.

NB: THAT IS NOT me SHOUTING AT YOU. THAT IS ME SHOUTING TO THE ROOF-TOPS AND MY (sorry -- meant to turn caps off) friends that I have learned something good.

PS: I am making a big mistake in rushing all this stuff when i am so busy. I do feel that I owe you answers. But please consider them in the spirit of community dialogue and my personal learning. There is not a question I will not answer honestly; just maybe not good wording, especially when rushing.

Is this now OK please?

wm

JesseM

Feb25-07, 06:26 PM

I personally DO NOT NEED A CLASSICAL EXPLANATION OF ANYTHING :::: NOW THAT I HAVE VANESCH'S MATHS. Vanesch's math does not in any way support your conclusion that commonsense local realism (which is what I meant by the word 'classical') is compatible with QM, if you think it does, you need to explain why you think so (see my previous post #147). In fact, the probabilities vanesch calculates are absolutely incompatible with common-sense local realism, the only way for common-sense local realism to be true would be if the probabilities he calculated were incorrect. Bell's theorem shows this.

If you disagree that Bell's theorem proves that the quantum predictions derived by vanesch's math are absolutely incompatible with commonsense local realism (a conclusion I am sure vanesch and Doc Al and DrChinese would all agree with), then if you are interested in learning why everyone disagrees with you rather than just declaring everyone wrong, you need to cooperate with our attempts to try to walk you through Bell's theorem. If you're not interested in learning, but just in promoting your incorrect ideas, you should take it to theory development.

Feb25-07, 06:31 PM

What exactly have I "misunderstood"? WHY DO YOU THINK THIS???? Vanesch just gave a recipe for calculating things in QM, the recipe itself is not based on local or nonlocal signals between events, it doesn't explain anything about why you see the correlations you do between measurements on entangled particles. However, if you take the results given to you by the recipe, and then you take Bell's theorem, it is clear that logically the results DO absolutely rule out common-sense local realism.

You seem to be confusing these two statements:

-Vanesch's calculations do not in themselves say anything either way about nonlocality vs. locality

-Vanesch's calculations are equally compatible with nonlocality and (common-sense) locality

But they are NOT equivalent--the first is true while the second is totally false! Logically the results are completely incompatible with common-sense local realism, it's just that this is not immediately obvious from looking at the calculations, you have to provide some additional logical arguments which go by the name of "Bell's theorem". But it is an invalid interpretation, definitively ruled out by quantum predictions. Bell's theorem shows this. Yes, no mainstream physicist would accept it as a valid "interpretation", because it does not make logical sense. You would see this if you actually made an effort to understand Bell's theorem, which is why I have been trying to walk you through it. If you're not trying to understand it, then this is equivalent to advancing the "interpretation" that perpetual motion is possible without listening to people's attempts to explain why it is ruled out by the laws of thermodynamics. No, because you have given no explanation for why you think vanesch's math somehow supports your idea. It doesn't, all vanesch's math gives is a recipe for calculating the probabilities, and then Bell's theorem proves that these probabilities are absolutely incompatible with common-sense local realism.

vanesch gives me much personal comfort BECAUSE his QM maths I will be able to happily understand and live with.

Please, me not being rude; define Bell's theorem that you want me to swear to. I am not avoiding here but I have no idea how to answer.

I believe in LOCAL QM; shall I be thrown out for that?

I think my saints (my co-conspirators) might be Bell, Einstein, Cramer (bit re-interpreted), Peres, Rovelli, Ballentine, Griffiths, Haag, Froehner, Kracklauer, Mayants, Jaynes, Hestenes, Harrison, Gottfried, parly vanesch (because I think we might be able to agree, due his maths) +++++++++ (though I know little of them and I'm not a full-member of MWI -- see hurkyl comment early on; it seems to fit me ok but I'm not studied it much). Some gurus here seem to be ok in ideas too.

Am I fallen into a group of terrorists? Should I head voluntarily for Guantanamo Bay?

wm

Feb25-07, 06:43 PM

Vanesch's math does not in any way support your conclusion that commonsense local realism (which is what I meant by the word 'classical') is compatible with QM, if you think it does, you need to explain why you think so (see my previous post #147). In fact, the probabilities vanesch calculates are absolutely incompatible with common-sense local realism, the only way for common-sense local realism to be true would be if the probabilities he calculated were incorrect. Bell's theorem shows this.

If you disagree that Bell's theorem proves that the quantum predictions derived by vanesch's math are absolutely incompatible with commonsense local realism (a conclusion I am sure vanesch and Doc Al and DrChinese would all agree with), then if you are interested in learning why everyone disagrees with you rather than just declaring everyone wrong, you need to cooperate with our attempts to try to walk you through Bell's theorem. If you're not interested in learning, but just in promoting your incorrect ideas, you should take it to theory development.

1. This is truly getting a bit silly:

2. Are you afraid of my new learning, seriously?

3. For I will sure have some better info to discuss.

4. When I say somewhere here or all the time to friends and firmly believe: WE ARE quantum machines in a quantum world!!! Does this sound that CLR is CLASSICAL.

5. Do you seek a straw-man to destroy for fun? Otherwise how can it be so opposite to what I affirm?

6. I am less concerned with who is wrong than me having a coherent world-view that I can live with CONSISTENT with QM formalisms.

7. I guess you have no idea whatsoever what vanesch has done favourably for me in 6 lines. (Did you know how to do it?)

8. Define Bell's theorem so that I may confess and be shot? Or give you pause for thort.

9. Now that I understand QM maths a bit better; I'd be happy to be walked quietly and slowly through BT. Seems like we need that definition?

PS: Could you possibly convert vanesch maths to out of mathematica please?

wm

Feb25-07, 06:50 PM

JesseM

Feb25-07, 06:55 PM

vanesch gives me much personal comfort BECAUSE his QM maths I will be able to happily understand and live with. Even though they are logically incompatible with commonsense local realism? Please, me not being rude; define Bell's theorem that you want me to swear to. I am not avoiding here but I have no idea how to answer. I've already given this to you in post #140. Here's a slightly modified one that takes into account the clarifications I added in that post:

do you agree or disagree that IF we have:

1. two experimenters, call them "Alice" and "Bob", who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like)
2. On each trial Alice and Bob choose their measurement settings randomly, and there is a spacelike separation between the events (Alice choosing her measurement setting) and (Bob making his measurement), along with a spacelike separation between the events (Bob chooses his measurement setting) and (Alice makes her measurement)
3. they always get opposite answers when they make the same measurement on any given trial

And we make the following 3 assumptions about the laws of physics:

4. no violations of locality allowed
5. no possibility of future events affecting past ones, or "conspiracies" in the initial conditions of the universe that could create a correlation between an experimenter's choice of measurement setting and the properties of the object/signal sent to them by the source even when the source sends them out before the experimenters make their choices
6. each experiment yields a single definite result (no splitting into multiple copies with measurement)

THEN it must be true that:

7. The perfect correlation between results when they choose the same setting can only be explained by some event or events in the past light cone of both measurements (like the event of the source sending out objects/signals with correlated properties), such that knowing what happened at this past event/s + knowing what measurement an experimenter makes is enough to uniquely determine the outcome of their measurement. For example, if the past event is that of the source sending out a particle with a set of properties {P1, P2, ..., Pn}, and Alice measures the spin using angle A, then it must be true that the outcome of this measurement was completely determined by the particle's properties plus the measurement angle A.

8. Assuming conditions 1-3 are met, assumptions 4-6 about the laws of physics are valid, and you accept that 7 is logically necessary given 1-6, then the following inequalities MUST be respected:

8a. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

8b. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

So, there you have it. Presumably you disagree with either 7 or 8, since commonsense local realism requires that 4-6 be true, and yet quantum physics requires that the inequalities in 8 be violated. So, do you disagree with both 7 and 8, or just 8? I believe in LOCAL QM; shall I be thrown out for that? If you aren't willing to try to follow the argument as to why this belief is incompatible with the probabilities vanesch calculated, you might be. But if you are willing, please address my questions above.

JesseM

Feb25-07, 07:03 PM

4. When I say somewhere here or all the time to friends and firmly believe: WE ARE quantum machines in a quantum world!!! Does this sound that CLR is CLASSICAL. As I mentioned earlier, I just use the word to "classical" to mean "compatible with commonsense local realism". Sorry if this usage causes confusion, but that's why I said in the post you quoted: Vanesch's math does not in any way support your conclusion that commonsense local realism (which is what I meant by the word 'classical') is compatible with QM
7. I guess you have no idea whatsoever what vanesch has done favourably for me in 6 lines. (Did you know how to do it?) I actually haven't looked at his derivation (the first time I tried to download it my computer didn't recognize the file type, I haven't yet gone back to download the reader from the wolfram website), but I'm familiar with the type of calculation he described in the non-attachment part of post #122. 8. Define Bell's theorem so that I may confess and be shot? Or give you pause for thort.

9. Now that I understand QM maths a bit better; I'd be happy to be walked quietly and slowly through BT. Seems like we need that definition? See the last post. But Bell's theorem doesn't depend on the details of the procedure for calculating probabilities in QM, it just depends on taking those probabilities and showing they are absolutely incompatible with commonsense local realism. PS: Could you possibly convert vanesch maths to out of mathematica please? Sure, I'll give it a shot. PLEASE: What incorrect idea do I now hold? wm The idea that the probabilities predicted by QM are compatible with commonsense local realism.

Feb25-07, 07:33 PM

Even though they are logically incompatible with commonsense local realism? I've already given this to you in post #140. Here's a slightly modified one that takes into account my clarifications: do you agree or disagree that IF we have:

1. two experimenters, call them "Alice" and "Bob", who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like)
2. On each trial Alice and Bob choose their measurement settings randomly, and there is a spacelike separation between the events (Alice choosing her measurement setting) and (Bob making his measurement), along with a spacelike separation between the events (Bob chooses his measurement setting) and (Alice makes her measurement)
3. they always get opposite answers when they make the same measurement on any given trial

And we make the following 3 assumptions about the laws of physics:

4. no violations of locality allowed
5. no possibility of future events affecting past ones, or "conspiracies" in the initial conditions of the universe that could create a correlation between an experimenter's choice of measurement setting and the properties of the object/signal sent to them by the source even when the source sends them out before the experimenters make their choices
6. each experiment yields a single definite result (no splitting into multiple copies with measurement)

THEN it must be true that:

7. The perfect correlation between results when they choose the same setting can only be explained by some event or events in the past light cone of both measurements (like the event of the source sending out objects/signals with correlated properties), such that knowing what happened in this past event/s + knowing what measurement an experimenter makes is enough to uniquely determine the outcome of the measurement. For example, if the past event is that of the source sending out a particle with a set of properties {P1, P2, ..., Pn}, and Alice measures the spin using angle A, then it must be true that the outcome of this measurement was completely determined by the particle's properties plus the measurement angle A.

8. Assuming conditions 1-3 are met, assumptions 4-6 about the laws of physics are valid, and you accept that 7 is logically necessary given 1-6, then the following inequalities MUST be respected:

8a. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

8b. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

So, there you have it. Presumably you disagree with either 7 or 8, since commonsense local realism requires that 4-6 be true, and yet quantum physics requires that the inequalities in 8 be violated. So, do you disagree with both 7 and 8, or just 8? If you aren't willing to try to follow the argument as to why this belief is incompatible with the probabilities vanesch calculated, you might be. But if you are willing, please address my questions above.

I am willing to put quite some time into this; and it may take awhile; but first I would need to be comfortable with the vanesch maths.

Also, with respect: Could I be assured that the senior physicists, who know much more than me, that communicate here, have to assent to the same test.

Reason because: As a student here, I'd like some comfort that there may be others who would see some deep verbal and philosophical difficulties here. Example: I thought Peres +++ .... said some of this was not part of QM.

And I'm here to learn about QM maths; not philosophy.

Given my current stage of learning, it would be far better for me to study (with you) vanesch's maths in the light of 4-6? I said words aren't my strength; and such discussion would help me to understand the definitive QM maths better.

Would this latter be acceptable as a first step? After you have reprocessed his sheet? That way we could start discussing very soon.

You also need recall that I have yet to study those maths; to disentangle the version that comes thru my computer; to study matrices and Pauli vectors; etc. etc.

Is there any problem with the vanesch math starting point please?

wm

Feb25-07, 07:43 PM

As I mentioned earlier, I just use the word to "classical" to mean "compatible with commonsense local realism". Sorry if this usage causes confusion, but that's why I said in the post you quoted:
I actually haven't looked at his derivation (the first time I tried to download it my computer didn't recognize the file type, I haven't yet gone back to download the reader from the wolfram website), but I'm familiar with the type of calculation he described in the non-attachment part of post #122. See the last post. But Bell's theorem doesn't depend on the details of the procedure for calculating probabilities in QM, it just depends on taking those probabilities and showing they are absolutely incompatible with commonsense local realism. Sure, I'll give it a shot.

The idea that the probabilities predicted by QM are compatible with commonsense local realism.

This is good; (+ we have some common difficulties):

It will probably be a clarification of the realism for me that comes out of it all. I can't see LOCALITY going, for me; and we should agree about probability. And once I understand something; that's all I mean by common-sense.

Big PS: Know that I accept QM calculations and the Aspect-style Bell tests; I'm not into faulty weaseling outs; and hope you never think that.

I then look forward to discussing realism and VEM (new code word vanesch maths) with you (and others if they wish).

Thanks heaps, wm

JesseM

Feb25-07, 07:57 PM

I am willing to put quite some time into this; and it may take awhile; but first I would need to be comfortable with the vanesch maths. Why? I think you're misunderstanding the logic of Bell's theorem here--it has nothing to do with the methods of making calculations in QM, it only depends on the final results of those calculations, and shows that these probabilities are incompatible with commonsense local realism. Also, with respect: Could I be assured that the senior physicists, who know much more than me, that communicate here, have to assent to the same test. What test are you referring to? Reason because: As a student here, I'd like some comfort that there may be others who would see some deep verbal and philosophical difficulties here. Example: I thought Peres +++ .... said some of this was not part of QM. Some of what is not part of QM? My assumptions #4-6 were not meant to be part of QM in general, they are meant to be part of commonsense local realism. There are certainly nonlocal or non-"commonsense" interpretations of QM which would disagree with one or more of those three assumptions. And I'm here to learn about QM maths; not philosophy. Again, Bell's theorem has nothing to do with "QM maths", only with the final probabilities predicted by QM, and showing that these probabilities are incompatible with commonsense local realism. Given my current stage of learning, it would be far better for me to study (with you) vanesch's maths in the light of 4-6? No, it would not really help you in understanding Bell's theorem or why QM is incompatible with commonsense local realism. But if you're just interested in learning more about the mathematics of quantum physics independent of these issues, you could start a new thread for help with that.

Feb25-07, 08:13 PM

Why? I think you're misunderstanding the logic of Bell's theorem here--it has nothing to do with the methods of making calculations in QM, it only depends on the final results of those calculations, and shows that these probabilities are incompatible with commonsense local realism. What test are you referring to? Some of what is not part of QM? My assumptions #4-6 were not meant to be part of QM in general, they are meant to be part of commonsense local realism. There are certainly nonlocal or non-"commonsense" interpretations of QM which would disagree with one or more of those three assumptions. Again, Bell's theorem has nothing to do with "QM maths", only with the final probabilities predicted by QM, and showing that these probabilities are incompatible with commonsense local realism. No, it would not really help you in understanding Bell's theorem or why QM is incompatible with commonsense local realism. But if you're just interested in learning more about the mathematics of quantum physics independent of these issues, you could start a new thread for help with that.

Quick reply so you know where I'm coming from: I'd like to have some good hand-holding on this thread as we discuss VEM in line with OP.

PS: I will seek good book for broader QM studies: any that would be compatibel with your views please? wm

JesseM

Feb25-07, 08:29 PM

Quick reply so you know where I'm coming from: I'd like to have some good hand-holding on this thread as we discuss VEM in line with OP. Your OP was about issues of locality vs. nonlocality. Again, the techniques for calculating probabilities in QM really have nothing to do with this; it is Bell's theorem that is used to justify the claim that QM is incompatible with commonsense local realism, and Bell's theorem is just based on taking the final probabilities and showing (using arguments unrelated to the math of QM) that commonsense local realism can't possibly explain them.

If you want to just learn about the math of QM in general, without any relation to locality vs. nonlocality, you should start another thread. I'd like to discuss Bell's theorem on this thread. PS: I will seek good book for broader QM studies: any that would be compatibel with your views please? wm My introductory textbook in college was Quantum Physics (http://www.amazon.com/exec/obidos/ASIN/0471857378/) by Stephen Gasiorowicz, I think that was pretty good (although some of the reviews say it doesn't explain enough to beginners, and I was using it in a class rather than for self-study). A slightly easier introduction to the math might be What Is Quantum Mechanics?: A Physics Adventure (http://www.amazon.com/exec/obidos/ASIN/0964350416/). I haven't read Introduction to Quantum Physics (http://www.amazon.com/Introduction-Quantum-Mechanics-David-Griffiths/dp/0131118927/) by Griffiths, but his E&M textbook was good, and the amazon reviews are pretty positive and say it's very good for beginners.

But I don't remember that either of the ones I read discussed Bell's theorem, and I wouldn't be surprised if this was absent from the Griffiths textbook too, introductory textbooks usually focus on developing one's skills at making calculations, not with interpretational issues that are irrelevant to making calculations.

JesseM

Feb25-07, 11:29 PM

But there are some odd looking symbols: like

o/oo. = i?

And it starts:

pauli1 = ::0, 1<, :1,0<<. = Pauli matrix in mathematica notation? That's very strange. I re-downloaded the notebook and it looks ok for me. (ok, I don't use the MathReader, but mathematica 4.1, but at least, the file is not corrupt) Yes, I have the same problem with the file in mathreader, it seems to be in some kind of internal mathematica code rather than with the symbols displayed however they are supposed to look.

JesseM

Feb26-07, 12:40 AM

1. Hmm, but when you say "QM=correct predictions", you're talking about some subset of its predictions rather than all possible predictions made by QM, right? After all, one of QM's predictions is that the inequality will be violated in certain experiments! Are you just talking about the prediction that whenever both experimenters measure their particles at the same angle, they always get opposite results? If my guess about what you meant in the "correct predictions" step is right, then no dispute...but if it isn't, could you clarify? 1. Sure, we are talking about the situation where QM makes a prediction. In this case, the prediction is not that the Inequality is violated, it is the "cos theta" relationship. That the Inequality is violated is applicable only when local realism is also present. There is no A, B and C in QM of course, only A and B. Well, this may be the source of the confusion between us--I think we both agree that Bell showed that IF (a certain specific prediction of QM is correct) AND (local realism is correct) THEN (a certain inequality must hold for all experiments meeting certain conditions). But I understood the "certain specific prediction of QM" differently than you--I thought the QM prediction used in the Bell inequality was the one that says the experimenters always get opposite results when they choose the same detector setting, while you're saying that it's the "cos theta" relationship (though of course the opposite-result prediction is a special case of the cos theta prediction, but I thought it was the only assumption from QM that Bell used in deriving the inequalities). Now, I admit I haven't yet tried to follow Bell's paper step-by-step, I'm just going by proofs of Bell's theorem that I've seen elsewhere. But looking at the first section, I notice that after equation (2), where he shows a probability distribution for (a,b) under the assumption that the outcome of each measurement is determined by some hidden variables lamda, Bell then writes: This should equal the quantum mechanical expectation value, which for the singlet state is

<sigma_1.a sigma_2.b> = - a.b

But it will be shown that this is not possible. So if he's showing that it's "not possible" that the probability distribution based on the assumption that outcomes are determined by local hidden variables could equal -a.b = -cos(a-b), doesn't that mean that he's showing the cos theta relationship is a prediction incompatible with local realism and the inequalities, rather than using cos theta + local realism to derive the inequalities?

I also note that in the first page in the "Formulation" section, Bell does make use of the quantum-mechanical prediction that when both experimenters measure on the same axis, knowing the result of one measurement allows you to predict the result of the other measurement with 100% certainty: If measurement of the component sigma_1.a, where a is some unit vector, yields the value +1 then, according to quantum mechanics, measurement of sigma_2.a must yield the value -1 and vice versa. Now we make the hypothesis, and it seems at least worth considering, that if the measurements are made at places remote from one another the orientation of one magnet does not affect the result obtained with the other. Since we can predict in advance the result of measuring any chosen component of sigma_2, by previously measuring the same component of sigma_1, it follows that the result of any such measurement must actually be predetermined. Finally, I note that the cos theta relationship is itself absolutely incompatible with local realism, because it leads to a violation of an inequality based on the assumption of local realism, the CHSH inequality (http://en.wikipedia.org/wiki/CHSH_inequality). As I said in post #81: For example, look at the CHSH inequality. This inequality says that if the left detector has a choice of two arbitrary angles a and a', the right detector has a choice of two arbitrary angles b and b', then the following inequality should be satisfied under local realism:

-2 <= E(a, b) - E(a, b') + E(a', b) + E(a', b') <= 2

Now, suppose wm were correct that he had a classical experiment satisfying the conditions of Bell's theorem such that the expectation value E(a, b) would equal -cos(a - b). In this case it we could pick some specific angles a = 0 degrees, b = 0 degrees, a' = 30 degrees and b' = 90 degrees; in this case we have E(a, b) = - cos(0) = -1, E(a, b') = -cos(90) = 0, E(a', b) = -cos(30) = -0.866, and E(a', b') = -cos(60) = -0.5. So E(a, b) - E(a, b') + E(a', b) + E(a', b') would be equal to -1 - 0 - 0.866 - 0.5 = -2.366, which violates the inequality. Of course, I don't think anyone had discovered this inequality in 1964, so I suppose it's possible Bell could have used cos theta + local realism to derive his original inequality without realizing the two premises were inherently contradictory. But if you do think Bell used the cos theta relationship in deriving the inequality, as opposed to in proving that the full theory of QM violates the inequality, could you point to which step in his derivation of the inequality he uses it?
2. As I pointed out, such an attempt will not work using the path described. The logic statement I showed was equivalent to Bell's Theorem is:

IF Inequality=fails AND Local Realism=demonstrated, THEN QM=Limited Validity Again, this is not how I would understand Bell's theorem, or at least the versions I've seen derived elsewhere (look at the discussion here (http://www.upscale.utoronto.ca/GeneralInterest/Harrison/BellsTheorem/BellsTheorem.html), for example). I thought Bell's theorem said IF experimenters always get opposite results on same measurement setting AND Local realism=true, THEN Inequality must always be obeyed in experiments meeting Bell's conditions (which would also mean that QM has limited validity, since the full theory of QM predicts the inequality can be violated in these kinds of experiments). In my version, you can see that if someone produced a way of violating the inequality that was compatible with local realism and which still ensured experimenters get opposite results on the same setting, then this would demonstrate a flaw in Bell's theorem; this is what I think wm was trying to do with his example of sending vectors with definite angles to the experimenters, although he seems to have abandoned this tack now that the error in his math was pointed out.

vanesch

Feb26-07, 01:57 AM

Yes, I have the same problem with the file in mathreader, it seems to be in some kind of internal mathematica code rather than with the symbols displayed however they are supposed to look.

I don't know how to solve this. At the office, where I have also mathematica, but on a different computer and all that, the file opens correctly too.

I will try to upload it in another format (such as pdf).

EDIT: I really don't understand what's going on. I installed mathreader too (even though I have mathematica). I downloaded the file from PF, saved it on my desk, and opened it with mathreader and everything is fine...

(I had to reboot my computer after the installation of mathreader, though).

vanesch

Feb26-07, 02:30 AM

Ok, here is the pdf converted file from my notebook.
(mind you, you might need the mathematica fonts, which are normally
also installed if you have installed the mathreader - but then funny things go
on).

Feb26-07, 03:32 AM

Feb26-07, 03:40 AM

<SNIP>In my version, you can see that if someone produced a way of violating the inequality that was compatible with local realism and which still ensured experimenters get opposite results on the same setting, then this would demonstrate a flaw in Bell's theorem; this is what I think wm was trying to do with his example of sending vectors with definite angles to the experimenters, although he seems to have abandoned this tack now that the error in his math was pointed out.

Re last line, last phrase: no; not correct. Did you see my post re Pauli matrices; and s and s' being ''unit-vectors associated with angular momentum''? That is s and s' are axial-vectors (= bi-vectors).

wm

vanesch

Feb26-07, 06:19 AM

Mmm, wm not being with us anymore, I don't know if this is still of any use. But as the argument I posted has been cited and quoted in all thinkable ways, here's the thing. There's a priori no such thing as a local or a non-local "calculation". The expression taken from the Bell paper, which gives us the quantum prediction of the correlation, is only that: a calculation. The result is independent of any interpretation.

However, a calculation can be suggestive or not of a local mechanism. Now, if we had the following:

RESULT AT ALICE is given by a mathematical operation:
F(alice-settings, alice-particle,other-stuff-local-to-alice's place)

RESULT AT BOB is given by a mathematical operation:
G(bob-settings, bob-particle, other-stuff-local-to-bob's place)

and the correlation would be calculated to be < F.G >
(where the expectation is an expectation over all possible stochastic variables which occur in this business), then that would be evidence that there CAN be a local mechanism that produces the results at Alice and at Bob.

Indeed, one should then just try to make sense out of the mathematical description of F and G, and interpret it as some process that actually goes on. As they only depend on quantities local to Alice, resp. Bob, this would in principle be possible.

Now, and this is where the quantum expression (3) of Bell is both confusing and suggestive, Bell writes:

correlation = < (sigma_1.a) (sigma_2.b) >

At first sight, this looks exactly like our < F.G >. Indeed, (sigma_1.a) seems to be a mathematical expression local to Alice, and (sigma_2.b) seems to be a mathematical expression at Bob.

BUT! Let us not forget that in < F.G >, F had to be the OUTCOME at Alice, and G had to be an outcome at Bob. Moreover, < > was supposed to be a statistical average.
This is where the superficial comparison goes wrong. (sigma_1.a) is NOT the outcome at Alice, if we require that outcome to be +1 or -1. sigma_1.a is an OPERATOR. Same at Bob. Moreover, < > is a hilbert space operation, not the usual "integration over stochastical variables" operation.

As such, the superficial formal equivalence between < F.G > and < (sigma_1.a)(sigma_2.b) > is misleading and confusing. The quantum calculation is hence not an indication that a local mechanism is at work.

At least, if we require that the outcomes at Alice and Bob are genuine, objective (observer-independent) physical outcomes (which is a tacit but obvious assumption in the derivation of Bell's theorem).

The only way to interpret the quantum-mechanical computation < (sigma_1.a)(sigma_2.b) > as a suggestion for a local mechanism, would be when we interpret (sigma_1.a) as the OUTCOME at Alice, and we interpret (sigma_2.b) as the outcome at Bob. But that's almost too crazy to consider. And some people leave out the "too" :smile:
It would mean that there is no objective result at Bob, and no objective result at Alice, which would be a list of {+1,-1,-1,...} The result would be "an operator" and NOT a +1 or a -1.

If you are mentally capable of stretching your imagination so far as to claim that there IS no outcome at Alice, that looks like a -1 or a +1, but that it is an operator, AND ONLY IN THAT CASE, then you CAN interpret the quantum expression as being of the kind < F.G >. This is the MWI view on things, and the only way to keep a local mechanism compatible with the quantum-mechanical predictions. Moreover, the local mechanism is then given exactly by the formal expression that was thought not to stand for any mechanism (but "just a calculation"). But one should really realise the stretch of imagination that is needed for that case: there is no objective outcome at Alice which takes on a +1 or -1 value :bugeye:
Nevertheless, to be able to make this crazy view compatible with the obvious fact that Alice HAS SEEN a +1 or a -1, the trick is to consider that there are now TWO ALICEs, one who has seen +1 and another who has seen -1. The "overall state" of Alice is then described not by a +1 or a -1, but by, exactly, an operator which is sigma_1.a.
When this "superposition of Alices" meets the "superposition of Bobs" (he will suffer a similar fate), then upon meeting, they will get together in SEVERAL DISTINCT COUPLES Alice/Bob (this time described by (sigma_1.a)(sigma_2.b) ). The statistics of this set of couples is then given by the quantum-mechanical expectation value < >, and gives us the correct correlations.

This is what the formalism suggests. It is also the IMO only way in which a local mechanism can be preserved. But it is of course totally crazy. That's MWI.

vanesch

Feb26-07, 08:32 AM

BTW, ttn, I moved your post, and my answer, to the thread of "what we see is bogus" in MWI.

DrChinese

Feb26-07, 08:39 AM

1. I had understood that there were POWERFUL arguments for NON-LOCALITY.

2. vanesch shows me (us all, surely) that there are not.

WHOA, you are again throwing the baby out with the bath water!!

There ARE powerful arguments for non-locality. ttn is our standard bearer on this, but he has chosen not to chime in. So I will point out that the standard interpretation of Bell + Aspect is that either locality or realism must be rejected. ttn makes a very strong argument that is is locality that must be rejected.

Vanesch, on the other hand, takes a different approach. Both of their views are interpretations which are consistent with Bell. Your interpretation is not, and requires you to change something if you want to be consistent with the facts.

vanesch

Feb26-07, 09:25 AM

There ARE powerful arguments for non-locality. ttn is our standard bearer on this, but he has chosen not to chime in. So I will point out that the standard interpretation of Bell + Aspect is that either locality or realism must be rejected. ttn makes a very strong argument that is is locality that must be rejected.

Vanesch, on the other hand, takes a different approach. Both of their views are interpretations which are consistent with Bell. Your interpretation is not, and requires you to change something if you want to be consistent with the facts.

Indeed. I think these are about the two "ontology" positions one can take: non-local mechanism (Bohmian), or a MWI-type mechanism.

Next to that, there are still a few possibilities:

- superdeterminism (the settings at Alice and Bob are pre-determined by a common origin in the (far) past, and you don't really have any statistically independent choice)

- shut-up-and-calculate

- I don't know where to put signaling from the future.

But in all these cases, some explicit or implicit assumption by Bell has been violated.

In Bohm, that's clear (locality). In MWI, "realism" (although there is a kind of realism, but not one in which there are objectively real and unique outcomes, which is what Bell needed).

In superdeterminism, the independence of choice was a necessary (though implicitly assumed) condition in Bell's derivation.

In the "shut up and calculate" approach, given that one doesn't consider any reality, or any mechanism, Bell's hypotheses don't hold necessarily.

Signaling from the future was also not considered, because by a front-and-back loop, the outcome at Alice can be influenced by the result at Bob's. I tend to think that "signaling from the future" should be some kind of non-locality, though in principle potentially compatible with relativity.

BTW, ttn DID chime in of course, but as his discussion was not on the topic of Bell's theorem, but just another attack on MWI, I moved it to the thread where his previous attack on MWI is housed.

DrChinese

Feb26-07, 10:16 AM

1. Well, I think may be the source of the confusion between us--we both agree that Bell showed that IF (a certain specific prediction of QM is correct) AND (local realism is correct) THEN (a certain inequality must hold for all experiments meeting certain conditions). But I understood the "certain specific prediction of QM" differently than you--I thought the QM prediction used in the Bell inequality was the one that says the experimenters always get opposite results when they choose the same detector setting, while you're saying that it's the "cos theta" relationship. Now, I admit I haven't yet tried to follow Bell's paper step-by-step, I'm just going by proofs of Bell's theorem that I've seen elsewhere. But looking at the first section, I notice that after equation (2), where he shows a probability distribution for (a,b) under the assumption that the outcome of each measurement is determined by some hidden variables lamda, Bell then writes:

2. So if he's showing that it's "not possible" that the probability distribution based on the assumption that outcomes are determined by local hidden variables could equal -a.b = -cos(a-b), doesn't that mean that he's showing the cos theta relationship is a prediction incompatible with local realism and the inequalities, rather than using cos theta + local realism to derive the inequalities?

3. I also note that in the first page in the "Formulation" section, Bell does make use of the quantum-mechanical prediction that when both experimenters measure on the same axis, knowing the result of one measurement allows you to predict the result of the other measurement with 100% certainty: Finally, I note that the cos theta relationship is itself absolutely incompatible with local realism, because it leads to a violation of an inequality based on the assumption of local realism, the CHSH inequality (http://en.wikipedia.org/wiki/CHSH_inequality). As I said in post #81: Of course, I don't think anyone had discovered this inequality in 1964, so I suppose it's possible Bell could have used cos theta + local realism to derive his original inequality without realizing the two premises were inherently contradictory. But if you do think Bell used the cos theta relationship in deriving the inequality, as opposed to in proving that the full theory of QM violates the inequality, could you point to which step in his derivation of the inequality he uses it?

4. Again, this is not how I would understand Bell's theorem, or at least the versions I've seen derived elsewhere (look at the discussion here (http://www.upscale.utoronto.ca/GeneralInterest/Harrison/BellsTheorem/BellsTheorem.html), for example). I thought Bell's theorem said IF experimenters always get opposite results on same measurement setting AND Local realism=true, THEN Inequality must always be obeyed in experiments meeting Bell's conditions (which would also mean that QM has limited validity, since the full theory of QM predicts the inequality can be violated in these kinds of experiments). In my version, you can see that if someone produced a way of violating the inequality that was compatible with local realism and which still ensured experimenters get opposite results on the same setting, then this would demonstrate a flaw in Bell's theorem; this is what I think wm was trying to do with his example of sending vectors with definite angles to the experimenters, although he seems to have abandoned this tack now that the error in his math was pointed out.

A bit of good ground to cover here, so let's see what we get:

1. Bell actually shows both:

a) You get opposite results when the detector settings are the same - see his (8) for the various cases of 0, 90, 180 degrees. This is a subset of the general case b).
b) The -cos(\theta) relationship, which is "net correlation" basis (ranges from -1 to 1, matches less mismatches), or: [tex]sin^2(\theta/2) which is the "gross correlation" basis (ranges from 0 to 1, matches). I realize that these bases (gross and net) can be very confusing and I probably shouldn't mention them, but you often see them interchanged without being labelled (and I am often guilty of this too).

In the net basis:

1=completely correlated.
-1=completely anti-correlated
0=no correlation (due completely to chance, in other words)

This matches Bell's (8) exactly. You can easily see this because sin^2(0 degrees/2)=0 (gross basis) or -cos(0) degrees=-1 (net basis). So what I am saying is this: historically, after EPR, it was generally accepted that the a) case worked for both classical explanations and was consistent with QM as well. The inequality had NOT yet been discovered of course.

2. Well, yes and no. You can read it a couple of different ways, but there is really only one meaning. The inequality comes from the following:

a) Assume the QM predictions for the singlet state of an entangled spin 1/2 pair must hold.
b) Apply those same predictions to 3 angle settings and require that they be internally consistent as well, so you expand the relationships that work OK for unit vectors a and b to a, b and c.

So the Inequality does hold if QM and local realism are both valid. Keep in mind that you can derive many different forms of the Inequality using particle spin/polatization attributes, and all essentially lead down the same path - incompatibility with experimental results at some settings, but not at others.

3. The cos theta relationship is not incompatible with QM if you only look at 2 settings (a and b) as EPR did. And as you say, Bell discovered the Inequality as that is the core of his paper. The Inequality uses a, b and c, and yes Bell absolutely knows and uses the cos theta relationship in his paper. Unfortunately, he did not see that as very important to point out the step but I can show it to you (it is indirect):

Look at (22) and the next line: a.c=0, a.b=b.c=1/sqrt(2)

You must make the substitution Bell makes: a.c=0 because a and c are crossed, i.e. 90 degrees apart. b is midway between a and c so ab=bc=135 degrees. And of course -cos(ab)=-cos(bc)=1/sqrt(2)=.707.

He is saying that the realistic assumption at these angles is violated but he is using (22) to show it. I always use (15) to get to the same point, because I get confused trying to manipulate the signs.

So I will do a separate post to more readily show the angle setting violations.

4. Again, yes and no. The showing of the opposite results is needed for calibration and to show that you have entangled pairs. It alone does not violate any inequality nor does it support or refute local realism in any way. This was assumed to be the case in 1935 and did not pose a problem at that time. As I mention, the opposite results is a simple extension of the general QM correlation function, and there historically was never any worry about the fact that the same function might apply at any 2 angle settings. Of course, that ignores the realism assumption which changes everything. As I have said many times, a and b alone don't lead to inequalities. It is adding c into the picture that creates the Inequalities, and QM does NOT postulate a, b and c exist. Only realistic theories add this.

DrChinese

Feb26-07, 11:03 AM

...Signaling from the future was also not considered, because by a front-and-back loop, the outcome at Alice can be influenced by the result at Bob's. I tend to think that "signaling from the future" should be some kind of non-locality, though in principle potentially compatible with relativity.

BTW, ttn DID chime in of course, but as his discussion was not on the topic of Bell's theorem...

Yes, effects from the future - or otherwise traversing the direction of time in feedback loops, is a possibility. But I can't figure out if they tend to be "non-realistic" or "non-local". Suppose the effect propagates at c (so relativity is respected) but the direction of movement through time was reversed. Now the definition of locality gets blurred as we normally assume a single direction of movement.

Yes, I had seen that and felt you guys were better off without me in the mix. Of course ttn supports Bell strongly, and I was dropping his name... :biggrin:

enotstrebor

Mar12-07, 02:33 AM

My apologies for not getting back.

You can't give a classical example which satisfies all the conditions laid out in Bell's theorem .... and still violates an inequality. If you think you have one, please present it.

1) The reason that a classical example can violate Bell's is related to Bayes formula and as I pointed out cases two and seven are not classically valid in the case of correlated events. When one introduces the new "c" condition one needs to take into account that P(c,b) is not independent of P(a,b) when subtracting P(c,b). That is if P(a,b) is correlation dependent, then if one could also measure "c" at the same time as "a" the two probabilities P(a,b) and P(c,b) are not independent. One should be subtracting P(c,b) given "a" from P(a,b) given "a". Thus when one uses the average P(c,b) subtracting it from the average P(a,b) one gets the violation.

Note that taking the average "violation" over all angles (some above/positive Bells expected linear (versus angle) result and some below/negative) the average does not violate the inequality while at all specific angles, except 0,90,180, there is a violation. Thus one must be careful because the average probability is not the same as the individual event (or individual angle) probability.

For correlated events one can not deal with averages ( bar-P(c,b) ) but must subtract the specific P(c,b) given "a" from the specific P(a,b) given "a".

Thus although Bell appears to include correlated events, the P(c,b) is used as if (effectively assumes) there is no correlation in the physics process (Bayes chain rule is only specifically valid for independent events when averages are use or when bar-P(c,b)= P(c,b))

2) But there is also another error often used in "classical" approaches which use Malus Law probabilities (associate with integrals of some "probability of interaction", e.g. Malus Law cos^2(a-\phi), over all angles). That is that Malus Law is true for all potential causes of photon correlations. But this assumes! And there are reasons to believe that the photon, a bi-vectored object, has at least two (potentially three) "phase" type variables to describe its behavior, not the single phase of Malus Law (average behavior over the other phase variables). If correlated in the "hidden phase" then for example the actual probability for correlated photons could be some other relationship (.e.g cos(a-\phi)), rather than cos^2(a-\phi) which gives an entirely different answer.

If you're implying each particle could be a classical quadrupole, then no, this could not possibly explain quantum experiments which violate Bell inequalities.

This is related to item 2) above when using classical probabilities (associate with integrals of some "probability of interaction" over all angles to calculate probabilities) to show the violation of Bell type inequalities using the bi-state SM view rather than the quadrapole view. It again changes the probabilites traditionally associated with the SM single spin up/down state.

If you're implying each particle could be a classical quadrupole, .

Also note that Stern-Gerlach experiments make physics sense if the electron is magnetic quadrapole at 90 degrees.

JesseM

Mar12-07, 12:52 PM

1) The reason that a classical example can violate Bell's No, it really can't. If you think it can, you are either misunderstanding the conditions of Bell's theorem, or failing to understand the proof. But if you think it can, then please provide a specific classical example. is related to Bayes formula and as I pointed out cases two and seven are not classically valid in the case of correlated events. Cases 2 and 7 of what? If you're talking about the eight possible hidden states, i.e.

1. a+ b+ c+
2. a+ b+ c-
3. a+ b- c+
4. a+ b- c-
5. a- b+ c+
6. a- b+ c-
7. a- b- c+
8. a- b- c-

...then there is no assumption that the source must emit hidden states in such a way that each has a nonzero probability. It is quite possible that the probability of a+ b+ c- could be 0, and that the probability of a- b- c+ could be 0 as well; the only assumption is that every pair emitted is in one of these states on every trial. When one introduces the new "c" condition one needs to take into account that P(c,b) is not independent of P(a,b) when subtracting P(c,b). That is if P(a,b) is correlation dependent, then if one could also measure "c" at the same time as "a" the two probabilities P(a,b) and P(c,b) are not independent. What does P(a,b) represent in your notation, precisely? And can you be specific about which of the various Bell inequalities you think can be violated classically? For example, if you're talking about this Bell inequality:

P(a+, b-) + P(b+, c-) >= P(a+, c-)

then in this case P(a+, b-) means "the probability that experimenter 1 chooses detector setting a and gets result +, while experimenter 2 chooses detector setting b and gets -". Do you think this inequality can be violated classically, given all the conditions assumed in Bell's theorem? (and note that I'm assuming here that whenever both experimenters choose the same setting, they always get the same result, so P(a+, a-) and P(a-, a+) = 0) Or is it some other inequality you think can be violated? Can you explain in words what the notation P(a,b) represents, as I did for P(a+, b-)? One should be subtracting P(c,b) given "a" from P(a,b) given "a". Why "should" one be doing that? Note that taking the average "violation" over all angles (some above/positive Bells expected linear (versus angle) result and some below/negative) the average does not violate the inequality What specific inequality are you talking about? All the Bell inequalities I've seen involve some finite number of detector angles, I haven't seen any that involve taking an average over all angles. If you are going to claim Bell's theorem is wrong, you'd better show that some specific Bellian inequality which Bell's theorem says can never be violated classically actually is violated classically, you can't just make up some new inequality that Bell's theorem doesn't even address. while at all specific angles, except 0,90,180, there is a violation. What inequality is violated at angles 0, 90, 180? Can you give in detail a specific classical setup where a specific inequality will be violated using these angles? For correlated events one can not deal with averages ( bar-P(c,b) ) but must subtract the specific P(c,b) given "a" from the specific P(a,b) given "a". Again, why "must" one do this? If you understand what Bell's theorem is actually saying, you must understand that an expression like P(a+, b-) refers to the probability of experimenter 1 getting + and experimenter 2 getting - over many trials where experimenter 1 used setting a and experimenter 2 used setting b. This probability would be perfectly well-defined, and would depend on the probability of the source emitting different possible "hidden states" on each trial. For example, suppose the source emits identical particles in 3 possible hidden states (so on a given trial, both particles will always have the same hidden state) with the following probabilities:

A. a+ b- c+ (P=20%)
B. a- b+ c- (P=60%)
C. a+ b- c- (P=20%)

In this case, if experimenter 1 uses setting a and experimenter 2 uses setting b, then experimenter 1 is guaranteed to get + and experimenter 2 is guaranteed to get - if the hidden state is A or C, while experimenter 1 will get - and experimenter 2 will get + if the hidden state is B. So, since there is a 20% chance of A and a 20% chance of C, P(a+, b-) would be 40%. Do you disagree? Thus although Bell appears to include correlated events, the P(c,b) is used as if (effectively assumes) there is no correlation in the physics process (Bayes chain rule is only specifically valid for independent events when averages are use or when bar-P(c,b)= P(c,b)) Again, I don't even know what P(c,b) means in your notation, so I don't know what you mean by "P(c,b) is used as if there is no correlation in the physics process". Correlation between what and what? There can certainly be a correlation between the variables of the possible hidden states...for example, in my above example if the hidden state includes a+ there's a 100% chance it also includes b- (cases A and C), and if the hidden state includes a- there's a 100% chance it also includes b+ (case B). 2) But there is also another error often used in "classical" approaches which use Malus Law probabilities (associate with integrals of some "probability of interaction", e.g. Malus Law cos^2(a-\phi), over all angles). "Probability of interaction"? Malus' law gives the intensity of light polarized at a certain angle after it passes through a polarizer at a different angle, which is the same as the probability that a given photon makes it through the polarizer in the case of a beam where all the photons have the same frequency (since the intensity of a single-frequency beam is just proportional to the number of photons). You can test this experimentally to see that it does hold for any combination of angles. Also note that Stern-Gerlach experiments make physics sense if the electron is magnetic quadrapole at 90 degrees. Not sure what you mean by this, can you provide a calculation showing how this works? With a classical quadrupole, can you explain why no matter how you orient your Stern-Gerlach apparatus, you always get only two possible deflection angles (spin-up or spin-down) rather than a continuous range of them? (see this page (http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SternGerlach/SternGerlach.html) for an explanation of how classical charged spinning objects would behave differently than electrons when passing through a Stern-Gerlach apparatus) Also, keep in mind that the Bell inequalities deal with pairs of entangled electrons which have the property that when the Stern-Gerlach apparatus of second experimenter is at a 180-degree angle from the Stern-Gerlach apparatus of the first experimenter, they always measure the same spin, regardless of what specific angle the first experimenter chose. In order to explain this in terms of quadrupoles, presumably you'd have to say that the source emits pairs of electrons in such a way that the quadrupole moment of the first is correlated to the quadrupole moment of the second, in such a way that you're guaranteed to get the same deflection when the two Stern-Gerlach apparatus are at 180 degree angles. But in this case you will not be able to violate any of the Bell inequalities when you choose any three angles for the first experimenter a, b, and c, and also label the corresponding 180-degree shifted angles for the second experimenter a, b, and c. For example, you will not find a violation of this inequality:

P(a+, b-) + P(b+, c-) >= P(a+, c-)

Do you disagree, and think you can violate this inequality using classical quadrupoles, given the correct understanding of what the symbols mean, and given the condition that experimenters always get the same spin when they choose the same setting, along with the condition that the source has no foreknowledge of what settings they'll choose on a given trial? If so, then once again I must ask you to provide some kind of detailed numerical example showing how it would work.

enotstrebor

Apr5-07, 11:49 PM

I again apologize for not getting back to you in a timely manor.

Also I did make some statements that, after reflection, would not effect this specific situation.

No, it really can't. If you think it can, you are either misunderstanding the conditions of Bell's theorem, or failing to understand the proof. But if you think it can, then please provide a specific classical example.

Do you think this inequality can be violated classically, given all the conditions assumed in Bell's theorem?
.

Violation can occur when one does not have the full facts, correct model or one makes incorrect assumptions.

It has always been assumed that the "entangled'' photon is no different EM-wise than any other photon.

If one had a physical model of the photon one might see that there are actually two types of linear polarized photons. Regular and "entangled". The entangled photon has a electric vector which at maximum is twice (mag. 2) that of the normal photon (mag 1). Thus in fact rather than the correlation being <=2 (1+1) the result actually can be <= 4 (2+2).

What is actually occurring is that the larger maximum of the "entangled" photon E-vector produces a higher rate (than the cos^2) of passing through the polarizer between 0 and 45 and a higher rate of no-pass (sin^2) between 45-90. Basically, a sin(\theta/2) like modulation of the normal cos^2 curve. The observe correlation curve (cos(\theta/2) from +1 to -1) from 0 to 90 is the resultant.

Phenomenologically one can perform the ``classical'' local and realistic calculation of this probability using the cos(A-x)^2 and sin(B-x)^2 integrations (x is the angle of the photon polarization) for the product of A and B polarization angles. One can compute the ++ coincidence with the factor of 2 for the amplitude (2*int(cos(A-x)^2*cos(B-x)^2)dx) and -- coincidence (2*int(sin(A-x)^2*sin(B-x)^2)dx) where now the coincidence minus anti-coincidence is calculation based on the amplitude of 2 rather than 1, i.e. correlation= 2*coincidence-2 (rather than 2*(++ plus --) - 1 ).

Note that one can get photon/polarizer "probabilities" >1 which is why one gets negative probabilities also (though I prefer in this case and in QM not to use the term probabilities, at best relative probabilities).

This local realistic calculation produces the correct (experimental) results.

If this model is correct, although the average over 360 is the same the model predicts that if one put a second polarizer after the first (one only needs one side of the EPR experiment) that one will see a non-malus law function between the two polarizers, i.e. the modulation of cos(theta/2). I also have yet to find a way to circularly polarize this "entangled" photon.

To my knowledge neither of these aspects have been looked at. I believe it has always been assumed that the "entangled'' photon is no different EM-wise.

If you know of published or unpublished (but documented) sources on these two "predicted" aspects of the "entangled" photon. Please quote me the sources.

With a classical quadrupole, can you explain why no matter how you orient your Stern-Gerlach apparatus, you always get only two possible deflection angles (spin-up or spin-down) rather than a continuous range of them?
.

First don't confuse the ``spinning charge'' concepts with anything that follows. It does not apply. The particle does not have a spinning charge. It does have a spinning vector potential (like) force which has nothing to do with charge but results in a magnetic (like) interaction.

Having two spin planes at 90 degrees it also has two magnetic interactions orientated at 90 degrees. The magnetic field of the Stern-Gerlach orients one of the two spin planes (50/50 chance where the other being normal is not effected - this is not a bar magnet type interaction, i.e. no true north south, just orientational to the magnetic field) is oriented field normal (field vertically through the spin plane).

This oriented plane can be spin spin up or spin down with respect to the magnetic field. The spin plane at 90 degrees is un-oriented.

As long as the magnetic field is kept normal to this oriented plane the particle stays oriented. When passing onto a second magnetic orientation, this new magnetic orientation, not being normal to the oriented plane interacts with both spin planes (unless of course it is normal to the first magnetic field).

The case, the second SG-magnet being normal to the first is more straight forward so I will deal with this only. In this case this second SG-magnet does not effect the originally oriented plane (the magnetic field lines are parallel to the plane) but orients the second plane which can be rotated by the magnetic interaction to be normal to the second SG-magnets field lines. Again, depending on the specific relative rotational angle of this plane's spin with respect to the field the plane, this second spin plane is now either oriented spin up or spin down with respect to this plane (while the first spin plane is now un-oriented). As this second plane was rotationally unoriented by the first magnet then the result is again 50/50 mix of spin up or spin down. The first SG-magnet is irrelevant to the probabilities of the second 90 degree SG-magnet.

Note that calculations of probabilities for a quadrapole also introduces a "hidden" factor of two.

There ARE powerful arguments for non-locality.

In deed there are, but only if one assumes that one can have a complete understanding of the physics of the particle by modelling its behavior and without having a model of the particle, the source and cause of the behavior.

"The map is not the territory." and "The behavior is not the particle."

JesseM

Apr6-07, 12:44 AM

Violation can occur when one does not have the full facts, correct model or one makes incorrect assumptions. No, it is impossible to violate Bellian inequalities classically, provided you respect all the conditions in the proof of Bell's theorem. If you disagree, please provide a specific numerical example which you think shows a violation of one of the inequalities in a classical context, but still respects all the conditions of the proof. It has always been assumed that the "entangled'' photon is no different EM-wise than any other photon. There is no such assumption made in the proof of Bell's theorem, every photon can be in a completely different state for its "hidden variables". Where did you get this idea? Have you actually studied the proof, and if so, what was your source? If one had a physical model of the photon one might see that there are actually two types of linear polarized photons. Regular and "entangled". The entangled photon has a electric vector which at maximum is twice (mag. 2) that of the normal photon (mag 1). Thus in fact rather than the correlation being <=2 (1+1) the result actually can be <= 4 (2+2). What correlation are you talking about? Are you referring to the correlation in the CHSH inequality (http://en.wikipedia.org/wiki/CHSH_Bell_test)? If so, do you understand that when the inequality says that S <=2, S refers to the sum of E(a, b) - E(a, b') + E(a', b) + E(a', b'), where each E is the expected value of the product of the two measurements given the stated settings (for example, if I use setting a and you use setting b', and I get the result +1 and you get -1 on that trial, (a,b') for that trial is 1*-1, and E(a,b') is the average of this product (a,b') over a large number of trials on which I use setting a and you use setting b').

If you are interpretating the correlation in the correct way, then are you saying you have a specific example in mind where E(a, b) - E(a, b') + E(a', b) + E(a', b') is larger than 2? What are the individual values of E(a,b) and E(a, b') and E(a',b) and E(a', b') in your example? What angles have you chosen for a, a', b and b'? What is actually occurring is that the larger maximum of the "entangled" photon E-vector produces a higher rate (than the cos^2) of passing through the polarizer between 0 and 45 and a higher rate of no-pass (sin^2) between 45-90. "Larger maximum" of what quantity, exactly? Please give more specifics...what are the exact characteristics of these classical versions of "entangled" photons which determine the probability they pass through a given filter angle? You said something about them having a larger "electric vector" but I'm not clear what you mean. So again, it would help if you describe the specific variables which characterize each photon (and what range of values these variables can take, and the probability distribution for different possible values over multiple trials), and give an equation for the probability they pass through the filter as a function of these variables. If it helps, imagine that we were just trying to simulate the situation you're imagining, by sending a data packet representing a particular entangled photon to a computer, and the computer using some algorithm to determine the probability the simulated photon makes it through its simulated filter, based on the filter angle and on the properties of the simulated photon given in the data packet...what would need to be in the data packet, and what algorithm should the computer use?

DrChinese

Apr6-07, 10:17 AM

It has always been assumed that the "entangled'' photon is no different EM-wise than any other photon.

If one had a physical model of the photon one might see that there are actually two types of linear polarized photons. Regular and "entangled". The entangled photon has a electric vector which at maximum is twice (mag. 2) that of the normal photon (mag 1). Thus in fact rather than the correlation being <=2 (1+1) the result actually can be <= 4 (2+2).

etc.

OK, let's test your hypothesis against Bell's Theorem. For the 8 cases
below, please give your expectation probabilities for a specific a, b and c:

1. a+ b+ c+ : ?
2. a+ b+ c- : ?
3. a+ b- c+ : ?
4. a+ b- c- : ?
5. a- b+ c+ : ?
6. a- b+ c- : ?
7. a- b- c+ : ?
8. a- b- c- : ?

If they add to 100% and none are less than 0%, then your hypothesis is realistic (as this is the precise definition of realism). Try settings a=0, b=67.5, c=45 for your entangled photons.

You see, a model does not become realistic simply because you say it is. It must meet a very specific condition, one that Bell found was too difficult for local realistic theories to achieve.

Unless you are willing to share your predictions, I don't think we will be able to evaluate your idea.

Mentz114

Apr8-07, 08:32 AM

QuantunEnigma

Apr15-07, 05:41 PM

Robert, can you point me to any reference where this is explicated, I'm very interested to find out more.

Interesting thread. Has anyone read this ?

Attached Files: Clifford alg values and Bell quant-ph. 0703179.pdf (131.3 KB, 1 views)

Please. The attached files, are they accessible by clicking, or by how?

Mentz114

Apr15-07, 05:43 PM

Yes. I can click on that link and download from arXiv.

It does not work for you, go to the xxx.lanl.gov and download it.

QuantunEnigma

Apr15-07, 08:14 PM

Yes. I can click on that link and download from arXiv.

It does not work for you, go to the xxx.lanl.gov and download it.

Is this Bell's theorem refuted? Expert comment please?

From abstract quant-ph/0703179

Disproof of Bell’s Theorem by Clifford Algebra Valued Local Variables

Joy Christian, Perimeter Institute, 31 Caroline Street North, Waterloo, Ontario N2L 2Y5, Canada, and Department of Physics, University of Oxford, Parks Road, Oxford OX1 3PU, England

It is shown that Bell’s theorem fails for the Clifford algebra valued local realistic variables. This is made evident by exactly reproducing quantum mechanical expectation value for the EPR-Bohm type spin correlations observable by means of a local, deterministic, Clifford algebra valued variable, without necessitating either remote contextuality or backward causation. Since Clifford product of multivector variables is non-commutative in general, the spin correlations derived within our locally causal model violate the CHSH inequality just as strongly as their quantum mechanical counterparts.

JesseM

Apr15-07, 09:02 PM

Is this Bell's theorem refuted? Expert comment please?

From abstract quant-ph/0703179

Disproof of Bell’s Theorem by Clifford Algebra Valued Local Variables I don't know anything about Clifford Algebra so I can't follow it myself, but I came across a short critical response here:

http://www.arxiv.org/abs/quant-ph/0703218

Christian also has a "reply to critics" here:

http://www.arxiv.org/abs/quant-ph/0703244

DrChinese

Apr15-07, 10:28 PM

Is this Bell's theorem refuted? Expert comment please?

From abstract quant-ph/0703179

Disproof of Bell’s Theorem by Clifford Algebra Valued Local Variables

Joy Christian, Perimeter Institute, 31 Caroline Street North, Waterloo, Ontario N2L 2Y5, Canada, and Department of Physics, University of Oxford, Parks Road, Oxford OX1 3PU, England

It is shown that Bell’s theorem fails for the Clifford algebra valued local realistic variables. This is made evident by exactly reproducing quantum mechanical expectation value for the EPR-Bohm type spin correlations observable by means of a local, deterministic, Clifford algebra valued variable, without necessitating either remote contextuality or backward causation. Since Clifford product of multivector variables is non-commutative in general, the spin correlations derived within our locally causal model violate the CHSH inequality just as strongly as their quantum mechanical counterparts.

I have previously read this. In my opinion, it will never be accepted as being of substance sufficient to change minds about Bell's Theorem. It is highly technical, and I don't believe it addresses any of the elements Bell has laid out.

There are new "disproofs" of Bell being published in the archives every month or so. These are rejected regularly for publication in peer reviewed journals.

I have a simple test for any disproof - I have posted it here many times. So far, I have no takers. (All I ask is that someone provide their predictions for a pair of entangled particles when at observation settings a, b and c - where the outcomes are independent of which a/b/c are to be observed. The predictions need only add to 100% and none should be less than zero, which is of course the requirement of Bell Realism.)

Because most of these disproofs hinge on technical issues, they miss the entire point Bell made. Bell used the definition of reality associated with Einstein, that particle observables must have values independent of the act of observation. This is directly opposed to the Heisenberg Uncertainty Principle, as Einstein was acutely aware. Disproofs provide different definitions, so it becomes a "straw man" argument to tear it down.

This paper's argument attempts to show that a local realistic theory will also violate a Bell Inequality - something that presumably can only be accomplished by a non-local or non-realistic theory such as Quantum Mechanics. This is a flawed logic model, as the issue is to demonstrate that a local realistic theory can both match experiment AND meet (not violate) the standard set by a Bell Inequality. If it were capable of this, it could pass my little test. Note that Quantum Mechanics does NOT need to address my test, since it does not claim to be realistic (and local).

In summary: it is NOT true that a purported classical (local & realistic) theory which violates a Bell Inequality will render Bell's Theorem invalid. Therefore, Christian's paper ultimately fails. Bell provided a specific set of settings for a/b/c to consider for any local realistic theory, and I note that these were not addressed.

ueit

Apr16-07, 03:31 PM

This paper's argument attempts to show that a local realistic theory will also violate a Bell Inequality - something that presumably can only be accomplished by a non-local or non-realistic theory such as Quantum Mechanics. This is a flawed logic model, as the issue is to demonstrate that a local realistic theory can both match experiment AND meet (not violate) the standard set by a Bell Inequality. If it were capable of this, it could pass my little test. Note that Quantum Mechanics does NOT need to address my test, since it does not claim to be realistic (and local).

A local realistic theory, in order to predict the observed correlations, must be deterministic. If it is stochastic it needs non-locality.

Bell's theorem rejects from the start any deterministic theory because of its "free-choice" assumption.

What you are asking with your test is a logical impossibility. Any local theory that tries to "beat" Bell's theorem and does not deny the "free choice" assumption is doomed because it is logically contradictory (deterministic and non-deterministic in the same time).

JesseM

Apr16-07, 03:59 PM

A local realistic theory, in order to predict the observed correlations, must be deterministic. If it is stochastic it needs non-locality.

Bell's theorem rejects from the start any deterministic theory because of its "free-choice" assumption. It doesn't reject deterministic theories, it just rejects bizarre "conspiracy" theories where somehow the initial conditions of the universe determine both the state of the particles emitted by the source on a given trial and the brain state of the experimenter on the same trial in just the right way to give the required correlations. As long as you assume the brain state of the experimenter before making a choice on a given trial is statistically independent of the hidden states of the particle emitted by the source on the same trial, then Bell's theorem can rule out local realism, it doesn't matter whether the universe is fundamentally deterministic or not.

In any case, although I don't understand the details of what Christian is proposing in his Clifford Algebra paper, I didn't get the impression he was proposing this sort of "conspiracy" explanation.

DrChinese

Apr16-07, 06:21 PM

A local realistic theory, in order to predict the observed correlations, must be deterministic. If it is stochastic it needs non-locality.

Bell's theorem rejects from the start any deterministic theory because of its "free-choice" assumption.

What you are asking with your test is a logical impossibility. Any local theory that tries to "beat" Bell's theorem and does not deny the "free choice" assumption is doomed because it is logically contradictory (deterministic and non-deterministic in the same time).

You imply that Bell's requirements are too strict, and possibly unnecessary as well.

IF...*you* postulate a local realistic theory, THEN Bell applies. If you don't like the results, that is your problem, and I cannot help you. Bell does not apply to a non-realistic theory such as QM, nor does it apply to non-local theories such as BM.

I mean, it is not like Bell randomly came up with his theory just to confound you! :tongue:

ueit

Apr18-07, 03:48 PM

It doesn't reject deterministic theories, it just rejects bizarre "conspiracy" theories where somehow the initial conditions of the universe determine both the state of the particles emitted by the source on a given trial and the brain state of the experimenter on the same trial in just the right way to give the required correlations. As long as you assume the brain state of the experimenter before making a choice on a given trial is statistically independent of the hidden states of the particle emitted by the source on the same trial, then Bell's theorem can rule out local realism, it doesn't matter whether the universe is fundamentally deterministic or not.

I think there are two types of deterministic theories:

1. theories that lack long-range forces (Newtonian billiard balls that interact only when collisions take place)

2. GR type theories, where each particle (in the case of GR-each massive body) interacts with every other particle.

I agree with you that "billiard balls" theories are pretty much rejected by Bell's theorem, not because they cannot possibly work, but because, in order for them to work, "bizarre conspiracies" must be postulated about the initial state of the universe.

However, the second type theories need not to posit such conspiracies. The assumption of statistical independence between distant parts of a system is highly questionable. A change in one part of the system requires a change of the whole. You cannot, say, move Mars on a different orbit while keeping the other bodies in place. Of course, with a more complex system, like two distant galaxies, it is not so obvious that a change in one is not possible without a corresponding adjustment of the second so one might be fooled to think that there are two statistically independent systems.

The bottom line is that local-deterministic theories are not ruled out by the experimental evidence but by the assumption of statistical independence used for the derivation of Bell's theorem.

ueit

Apr18-07, 03:55 PM

IF...*you* postulate a local realistic theory, THEN Bell applies.

Not if I deny statistical independence between the source and detectors.

JesseM

Apr18-07, 04:50 PM

I think there are two types of deterministic theories:

1. theories that lack long-range forces (Newtonian billiard balls that interact only when collisions take place)

2. GR type theories, where each particle (in the case of GR-each massive body) interacts with every other particle. But objects don't interact instantaneously--GR still has a light cone structure, so if you foliate spacetime into a stack of spacelike surfaces, everything going on in one region of space in a given surface should be determined by what was going on in the complete set of points in space in an earlier surface that lie in the later region's past light cone, and nothing outside that region of the earlier surface should have an effect on the chosen region of the later surface (There are weird spacetimes that apparently can't be foliated in this way, like ones containing closed timelike curves, but I think this is true as long as you assume a globally hyperbolic spacetime (http://en.wikipedia.org/wiki/Globally_hyperbolic)).

So, it seems to me the situation is no different with GR than with billiard balls--if the event of the experimenter choosing what measurement setting to use and the event of the source generating the two particles are each outside the other's future and past light cone (a spacelike separation), the only way they could fail to be statistically independent is if you assume that some event or events in their mutual past light cone determined these two events in just the right way to create the correlations--the "conspiracy" assumption. However, the second type theories need not to posit such conspiracies. The assumption of statistical independence between distant parts of a system is highly questionable. A change in one part of the system requires a change of the whole. You cannot, say, move Mars on a different orbit while keeping the other bodies in place. If you set off a bunch of nuclear bombs or something on Mars to shift its orbit, we wouldn't feel any gravitational effects of this event any sooner than we'd receive light waves from the event--gravitational waves travel at c just like electromagnetic waves. The bottom line is that local-deterministic theories are not ruled out by the experimental evidence but by the assumption of statistical independence used for the derivation of Bell's theorem. As I understand it, "local" means "having a light cone structure", and you can use the type of argument I made above to show that no local theory where each event has a single definite outcome (as opposed to a many-worlds type theory) can explain the violation of Bell inequalities without positing a "conspiracy" where events in the past light cone of both the source's particle emission and the experimenter's choosing of setting always causes them to be correlated in just the right way to give the observed results. If you don't posit such a conspiracy, how can you explain a correlation between two events with a spacelike separation?

DrChinese

Apr18-07, 06:33 PM

Not if I deny statistical independence between the source and detectors.

Again, you speak in generalities when you imply such a connection. The source is a polarized laser beam. The detectors are polarized as well. Exactly how do you propose that the results depend on the source? The usual formula, Cos^2(a-b), relates the results of the detector settings but lacks a term for the source setting. This formula has substantial experimental validation.

In my scientific opinion: unless you can predict the results of specific cases in advance or otherwise improve the accuracy of the usual formula by adding a term for a source setting, you may as well be asserting that the results are a function of the phase of the moon.

Or perhaps - gasp - it is a hidden variable. But haven't we been down that road before? Isn't that exactly what Bell started with? :tongue:

ueit

Apr20-07, 03:46 PM

But objects don't interact instantaneously--GR still has a light cone structure, so if you foliate spacetime into a stack of spacelike surfaces, everything going on in one region of space in a given surface should be determined by what was going on in the complete set of points in space in an earlier surface that lie in the later region's past light cone, and nothing outside that region of the earlier surface should have an effect on the chosen region of the later surface (There are weird spacetimes that apparently can't be foliated in this way, like ones containing closed timelike curves, but I think this is true as long as you assume a globally hyperbolic spacetime (http://en.wikipedia.org/wiki/Globally_hyperbolic)).

I agree.

So, it seems to me the situation is no different with GR than with billiard balls--if the event of the experimenter choosing what measurement setting to use and the event of the source generating the two particles are each outside the other's future and past light cone (a spacelike separation), the only way they could fail to be statistically independent is if you assume that some event or events in their mutual past light cone determined these two events in just the right way to create the correlations--the "conspiracy" assumption. If you set off a bunch of nuclear bombs or something on Mars to shift its orbit, we wouldn't feel any gravitational effects of this event any sooner than we'd receive light waves from the event--gravitational waves travel at c just like electromagnetic waves.

There is a subtle error in your above line of reasoning. It is true that if we change Mars orbit by using a nuke, the effect will manifest on Earth at the same moment we see the explosion. However, in this case we do not deal with a deterministic theory anymore. While GR is deterministic, the nuclear explosion is not governed by GR and therefore, from GR's "point of view" it is a true unpredictable event. We have a mixture of a deterministic theory with random events and this is not what I propose as a local realistic explanation of EPR.

Now, let's make a correct analogy, by letting a stray planet, coming from a distant galaxy, to alter the orbit of Mars. In this case we deal with a true deterministic system and the effect will be felt instantaneously on Earth, as Newton's law of gravity (which is a good approximation for this case) predicts, before the light from Mars will reach us. This is because the stray planet does not interact only with Mars, but with Earth, and Jupiter, and all other bodies at once.

If we return to Bell's theorem we see that we are in the "stray planet" case and not in the "nuke" case. The mechanism behind the decision to move the detector on a different axis is entirely covered by QM (unless you do not propose a mind/body dualism) so it does not and cannot "inject" randomness into the quantum system in the way the nuke does for the gravitational system.

The difference between the "billiard ball" theory and GR is that in the former the particles are not aware of each other (a particle in a distant place has no effect on another) while in the later such awareness exists even beyond the light cone (if no non-deterministic events like nukes are allowed). For example, in the solar system Earth accelerates towards the future position of the Sun, and not towards its retarded position (the place we see the Sun).

As I understand it, "local" means "having a light cone structure", and you can use the type of argument I made above to show that no local theory where each event has a single definite outcome (as opposed to a many-worlds type theory) can explain the violation of Bell inequalities without positing a "conspiracy" where events in the past light cone of both the source's particle emission and the experimenter's choosing of setting always causes them to be correlated in just the right way to give the observed results. If you don't posit such a conspiracy, how can you explain a correlation between two events with a spacelike separation?

I think I've provided an explanation why your above argument does not apply to EPR. If QM is deterministic there is no source of randomness that can be used to "fool" the PDC about the future detector orientation. The light cone structure is not a problem because the information about the past is enough to perfectly predict the future.

A local, realistic, non-conspiracy type mechanism for EPR would be as follows:

1. every particle in the experimental setup sends a signal, at light speed towards the PDC source. (this is in fact true for every particle in the visible universe)

2. The calcium atom "reads" from those signals the position/momentum for each particle and "computes" their future evolution (including how the detector will be oriented at the time of detection). This might resemble the way Earth "reads" from the space curvature around it the position/momentum of other massive bodies and then "decides" how to accelerate.

3. When a suitable future detector orientation is detected (suitable in the sense that it must conform with Malus's law and conservation laws) a pair of "entangled" particles is emitted, "laughing" about the futile attempts of the experimenter to "beat the system".

ueit

Apr20-07, 04:01 PM

Again, you speak in generalities when you imply such a connection. The source is a polarized laser beam. The detectors are polarized as well. Exactly how do you propose that the results depend on the source? The usual formula, Cos^2(a-b), relates the results of the detector settings but lacks a term for the source setting. This formula has substantial experimental validation.

Please take a look at my above post to JesseM

In my scientific opinion: unless you can predict the results of specific cases in advance or otherwise improve the accuracy of the usual formula by adding a term for a source setting, you may as well be asserting that the results are a function of the phase of the moon.

I do not need to provide a physically plausible local-realistic mechanism for EPR, only a logically consistent one (without appealing to conspiracies as these are extremely non-parsimonious). That is enough to prove that your assertion regarding the applicability of Bell's theorem (in spite of what Bell himself said) is false.

JesseM

Apr20-07, 05:05 PM

There is a subtle error in your above line of reasoning. It is true that if we change Mars orbit by using a nuke, the effect will manifest on Earth at the same moment we see the explosion. However, in this case we do not deal with a deterministic theory anymore. While GR is deterministic, the nuclear explosion is not governed by GR and therefore, from GR's "point of view" it is a true unpredictable event. We have a mixture of a deterministic theory with random events and this is not what I propose as a local realistic explanation of EPR. I didn't assume the nuclear explosion was random, though. You are free to assume that whatever non-gravitational forces are involved are also governed by deterministic laws, like classical electronmagnetism, which can certainly be incorporated into GR.

The point is just that no matter what your complete set of fundamental laws are, as long as they have a light cone structure, then there should be no statistical correlation between events A and B with a spacelike separation unless there's some event or events in the past light cone of A and B which predetermines them in the right way to create the correlation. And if A is the event of the source emitting particles in a certain state, and B is the event of an experimenter's brain making a choice of which setting to use on a given trial, then explaining the violation of Bell inequalities in terms of such a predetermining event in A and B's past light cone amounts to the "conspiracy" assumption discussed earlier. Do you disagree with any of this? If so, what specifically do you disagree with? Now, let's make a correct analogy, by letting a stray planet, coming from a distant galaxy, to alter the orbit of Mars. In this case we deal with a true deterministic system and the effect will be felt instantaneously on Earth, as Newton's law of gravity (which is a good approximation for this case) predicts, before the light from Mars will reach us. This is because the stray planet does not interact only with Mars, but with Earth, and Jupiter, and all other bodies at once. Are you sure about that? I believe it is true that if a body is moving at a constant velocity then we'll feel the pull from its current position. This is analogous to the situation in classical electromagnetism, where if you have a charge moving at constant velocity, other charges will be attracted to its current position rather than its retarded position; but this is in effect because the electromagnetic field has a built-in ability to "extrapolate" linear movement, there's no actual signals moving faster than light, and if the charge were to accelerate other charges would continue to be attracted to where the original charge would have been had it continued to move in a straight line, until they receive an "update" on its position in the form of electromagnetic waves. Because electromagnetic waves depend on a dipole moment while gravitational waves depend on a quadrupole moment, the gravitational field can "extrapolate" some more general types of movement than the electromagnetic field, like a spherically symmetric collapsing star, but in any situation where gravitational waves are generated, other objects do not anticipate all the motions, and continue to be attracted to the "wrong" positions until the gravitational waves reach them. And wouldn't one planet smashing into another and knocking it off course generate gravitational waves? See Sources of gravitational waves (http://en.wikipedia.org/wiki/Gravitational_wave#Sources_of_gravitational_waves) on wikipedia.

In any case, it seems to me the argument about the light cone structure is pretty airtight. In electromagnetism there is a correlation between the direction one charge A is being pulled at a given moment and the current position of another charge B moving at constant velocity, and these events have a spacelike separation, but this could be explained in terms of the position of the charge B at some previous time, an event in the past light cone of charge A, plus the electromagnetic field's ability to naturally "extrapolate" the position of charge B as long as it keeps moving at constant velocity. But any such dependence on events in the past light cone for Bell experiments would either involve a "conspiracy" in the initial conditions, or it would involve ridiculously complex laws of nature that were somehow "extrapolating" the precise future brain state of the experimenter at the moment of choice using only events in the past light cone of the event of the source emitting particles (and even if you are willing to allow such ridiculously complex laws of nature, this probably doesn't make sense anyway since the brain is a chaotic system and the choice would probably depend on everything in the past light cone of the experimenter's choice at a given time, but at any given time some of the events which lie in the past light cone of the choice-event are outside the past light cone of the event of the source emitting the particles, so even Laplace's demon couldn't predict the choice using only the set of events in the past light cone of the emission-event). The difference between the "billiard ball" theory and GR is that in the former the particles are not aware of each other (a particle in a distant place has no effect on another) while in the later such awareness exists even beyond the light cone (if no non-deterministic events like nukes are allowed). For example, in the solar system Earth accelerates towards the future position of the Sun, and not towards its retarded position (the place we see the Sun). But again, the type of motions that GR can "extrapolate" in this way are pretty limited, I think it may just be constant-velocity motion and spherically or cylindrically symmetric acceleration; any type of motion complicated enough to result in gravitational waves cannot be extrapolated in this way, so objects will not be pulled in exactly the direction of other object's current position in these circumstances. A local, realistic, non-conspiracy type mechanism for EPR would be as follows:

1. every particle in the experimental setup sends a signal, at light speed towards the PDC source. (this is in fact true for every particle in the visible universe)

2. The calcium atom "reads" from those signals the position/momentum for each particle and "computes" their future evolution (including how the detector will be oriented at the time of detection). This might resemble the way Earth "reads" from the space curvature around it the position/momentum of other massive bodies and then "decides" how to accelerate. But like I said, the more complicated the types of motion you want objects to be able to "extrapolate", the more complicated your fundamental laws have to be; and I think my parenthetical comment about how even Laplace's demon probably couldn't predict the experimenter's choice using only information about events in the past light cone of the source's emission event suggests that even ridiculously complicated laws couldn't do what you're suggesting without "conspiracy-like" restrictions on the initial conditions of the universe.

ueit

Apr24-07, 04:50 AM

I didn't assume the nuclear explosion was random, though. You are free to assume that whatever non-gravitational forces are involved are also governed by deterministic laws, like classical electromagnetism, which can certainly be incorporated into GR.

Yeah, but introducing additional complexity in my analogy does no good.

The point is just that no matter what your complete set of fundamental laws are, as long as they have a light cone structure, then there should be no statistical correlation between events A and B with a spacelike separation unless there's some event or events in the past light cone of A and B which predetermines them in the right way to create the correlation.

I agree with this. More, I think there is good evidence (the uniformity of microwave background radiation) that all the visible universe passed a period when all its particles were able to "make contact" with each other:
http://en.wikipedia.org/wiki/Inflationary_theory

In physical cosmology, cosmic inflation is the idea that the nascent universe passed through a phase of exponential expansion that was driven by a negative-pressure vacuum energy density.[1] As a direct consequence of this expansion, all of the observable universe originated in a small causally-connected region. Inflation answers the classic conundrums of the big bang cosmology: why does the universe appear flat, homogeneous and isotropic in accordance with the cosmological principle when one would expect, on the basis of the physics of the big bang, a highly curved, inhomogeneous universe. (emphasis mine)

And if A is the event of the source emitting particles in a certain state, and B is the event of an experimenter's brain making a choice of which setting to use on a given trial, then explaining the violation of Bell inequalities in terms of such a predetermining event in A and B's past light cone amounts to the "conspiracy" assumption discussed earlier. Do you disagree with any of this?

I disagree with "predetermining event in A and B's past light cone" formulation. All the particles in the universe are correlated with each other from the time of big-bang. Even if those particles are now far from each other, the correlation between their motion remains.

Are you sure about that?

Of course I'm not. It's just one of many possible scenarios.

I believe it is true that if a body is moving at a constant velocity then we'll feel the pull from its current position. This is analogous to the situation in classical electromagnetism, where if you have a charge moving at constant velocity, other charges will be attracted to its current position rather than its retarded position; but this is in effect because the electromagnetic field has a built-in ability to "extrapolate" linear movement, there's no actual signals moving faster than light, and if the charge were to accelerate other charges would continue to be attracted to where the original charge would have been had it continued to move in a straight line, until they receive an "update" on its position in the form of electromagnetic waves. Because electromagnetic waves depend on a dipole moment while gravitational waves depend on a quadrupole moment, the gravitational field can "extrapolate" some more general types of movement than the electromagnetic field, like a spherically symmetric collapsing star, but in any situation where gravitational waves are generated, other objects do not anticipate all the motions, and continue to be attracted to the "wrong" positions until the gravitational waves reach them. And wouldn't one planet smashing into another and knocking it off course generate gravitational waves? See Sources of gravitational waves (http://en.wikipedia.org/wiki/Gravitational_wave#Sources_of_gravitational_waves) on wikipedia.

I didn't think about the planet "smashing into" Mars, only passing close enough to significantly alter its orbit. In a collision, a lot of energy is lost as heat, and the analogy wouldn't work (the error introduced by gravity waves is negligible though). Now, GR is only an analogy. I do not claim that the mechanism behind EPR is exactly like GR. In any case, the accelerated motion is "extrapolated" very well by GR so that a non-local mechanism as the one proposed by Newtonian gravity works very well for all but extreme situations like the merging of black holes or neutron stars. I know that it doesn't work perfectly and that's why I specified that for the case I gave you, Newtonian theory is a good approximation. I see no reason to assume that a better or even perfect "extrapolation" of accelerated motion (which is the only possible motion of a point particle except the uniform one) cannot be accomplished by a theory. At least, I know of no proof of that.

In any case, it seems to me the argument about the light cone structure is pretty airtight. In electromagnetism there is a correlation between the direction one charge A is being pulled at a given moment and the current position of another charge B moving at constant velocity, and these events have a spacelike separation, but this could be explained in terms of the position of the charge B at some previous time, an event in the past light cone of charge A, plus the electromagnetic field's ability to naturally "extrapolate" the position of charge B as long as it keeps moving at constant velocity. But any such dependence on events in the past light cone for Bell experiments would either involve a "conspiracy" in the initial conditions, or it would involve ridiculously complex laws of nature that were somehow "extrapolating" the precise future brain state of the experimenter at the moment of choice using only events in the past light cone of the event of the source emitting particles (and even if you are willing to allow such ridiculously complex laws of nature, this probably doesn't make sense anyway since the brain is a chaotic system and the choice would probably depend on everything in the past light cone of the experimenter's choice at a given time, but at any given time some of the events which lie in the past light cone of the choice-event are outside the past light cone of the event of the source emitting the particles, so even Laplace's demon couldn't predict the choice using only the set of events in the past light cone of the emission-event).

When you are speaking about “ridiculously complex laws of nature that were somehow "extrapolating" the precise future brain state of the experimenter at the moment of choice” you are referring to a high level description of facts. The mechanism I propose works at the lowest level. A calcium atom doesn’t “know” anything about brains, computers or experiments; it only “looks” for two suitable absorbers (other atoms) for the entangled photons. When such absorbers are found, a pair of photons is send towards their extrapolated position. That’s all. The chain of events by which those absorbers arrive at their position is irrelevant. You may have a human pushing a button that hits a monkey; then the monkey starts a computer running a random number generator that in turn commands an engine to change the polarizer’s position. If, at low level, the “extrapolation” mechanism works perfectly, or at least with a good enough accuracy the calcium atom would not be “fooled” and Bell’s inequality would be violated.

But again, the type of motions that GR can "extrapolate" in this way are pretty limited, I think it may just be constant-velocity motion and spherically or cylindrically symmetric acceleration; any type of motion complicated enough to result in gravitational waves cannot be extrapolated in this way, so objects will not be pulled in exactly the direction of other object's current position in these circumstances. But like I said, the more complicated the types of motion you want objects to be able to "extrapolate", the more complicated your fundamental laws have to be; and I think my parenthetical comment about how even Laplace's demon probably couldn't predict the experimenter's choice using only information about events in the past light cone of the source's emission event suggests that even ridiculously complicated laws couldn't do what you're suggesting without "conspiracy-like" restrictions on the initial conditions of the universe.

From my answer above I conclude:

1. It should be enough to extrapolate accelerated motion. Other types are not possible for a point particle. Probably even the imperfect extrapolation of GR is enough to explain all experiments to date.
2. “all of the observable universe originated in a small causally-connected region”. This is the event that “links” the whole experimental setup. Since then all particles are correlated with each other because of the “extrapolation” effect.
3. No complicated laws must be postulated to deal with brains or different experimental tricks. Conservation laws, the microscopic equivalent of Mallus’ law plus the extrapolation mechanism will do.

Demystifier

Apr24-07, 07:01 AM

Is this Bell's theorem refuted? Expert comment please?

From abstract quant-ph/0703179

Disproof of Bell’s Theorem by Clifford Algebra Valued Local Variables

Joy Christian, Perimeter Institute, 31 Caroline Street North, Waterloo, Ontario N2L 2Y5, Canada, and Department of Physics, University of Oxford, Parks Road, Oxford OX1 3PU, England

It is shown that Bell’s theorem fails for the Clifford algebra valued local realistic variables. This is made evident by exactly reproducing quantum mechanical expectation value for the EPR-Bohm type spin correlations observable by means of a local, deterministic, Clifford algebra valued variable, without necessitating either remote contextuality or backward causation. Since Clifford product of multivector variables is non-commutative in general, the spin correlations derived within our locally causal model violate the CHSH inequality just as strongly as their quantum mechanical counterparts.
Let me make a comment on the Clifford-valued local realistic variables.
Although I have not completely understood the paper, it is not a surprise to me that local Clifford-valued realistic variables may simulate QM. This is because, in a sense, non-commuting variables are never truly local, even if they are local formally. Let me explain what I mean by this:
A formally local quantity is a quantity of the form A(x) or B(y), where x and y are positions of the first and the second particle, respectively. Now, if they are not commuting, then
A(x)B(y) \neq B(y)A(x)
But how two quantities A and B know that they should not commute if x is very far from y? This knowledge is a sort of nonlocality as well.

My opinion is that realistic variables (local or not) must be not only commuting, but represented by real numbers. This is because they are supposed to be measurable, while a measurable quantity must be a real number. Therefore, I believe that the Clifford-valued realistic variables are physically meaningless.

In fact, the claim that physical variables could be noncommuting numbers does not differ much from the claim that physical variables could be noncommuting operators or noncommuting matrices. But this is exactly what the realistic physical variables in QM are NOT supposed to be, because otherwise we deal with QM in the usual matrix/operator form.

DrChinese

Apr24-07, 08:13 AM

I do not need to provide a physically plausible local-realistic mechanism for EPR, only a logically consistent one (without appealing to conspiracies as these are extremely non-parsimonious). That is enough to prove that your assertion regarding the applicability of Bell's theorem (in spite of what Bell himself said) is false.

Well, I'm not too sure how "logically consistent" your mechanism is if it can't meet my simple test (in which each possible outcome probability is non-negative).

I place your hypothesis in the bin with the other "proofs" that Bell is not applicable.

DrChinese

Apr24-07, 08:21 AM

My opinion is that realistic variables (local or not) must be not only commuting, but represented by real numbers. This is because they are supposed to be measurable, while a measurable quantity must be a real number. Therefore, I believe that the Clifford-valued realistic variables are physically meaningless.

In fact, the claim that physical variables could be noncommuting numbers does not differ much from the claim that physical variables could be noncommuting operators or noncommuting matrices. But this is exactly what the realistic physical variables in QM are NOT supposed to be, because otherwise we deal with QM in the usual matrix/operator form.

Well said.

I am always amazed at new hypotheses (such as Christian's) which purport to show a local realistic scenario which agree with the predictions of QM - yet do not discuss the negative probabilities which result when the observer freely chooses between measurement settings. The entire realistic argument IS that the observer could do this! That is what everyone cares about - whether the results are observer dependent or not. So when the observer independence issue is magically dropped (in this case by having non-commutivity), it is no big surprise that Bell is bypassed in the results. Of course, that is why Bell's Theorem is so important. His assumptions are very straightforward and easy to agree with.

JesseM

Apr24-07, 01:44 PM

I agree with this. More, I think there is good evidence (the uniformity of microwave background radiation) that all the visible universe passed a period when all its particles were able to "make contact" with each other Unless I'm misunderstanding something, that doesn't mean that there was any time when all the events in the past light cone of the event of the experimenter making a choice of what to measure were also in the past light cone of the event of the the source sending out the particles. Again, if you don't place any special constraints on initial conditions, then even in a deterministic universe, a Laplacian demon with knowledge of everything in the past light cone of the source sending out the particles would not necessarily be able to predict the brain state of the experimenter at the time he made his choice of what to measure. Do you disagree? I disagree with "predetermining event in A and B's past light cone" formulation. All the particles in the universe are correlated with each other from the time of big-bang. Even if those particles are now far from each other, the correlation between their motion remains. "Correlated" is too vague. I think that inflationary theory would say that the past light-cones of the most widely-separated events we can see will partially overlap, so that the similarity of the CMBR in different regions can have a common past cause. But again, it doesn't mean that knowing the past light cone of one event would allow you to predict every other event, even in a perfectly deterministic universe, because any pair of spacelike separated events would have parts of their past light cones that are outside the past light cone of the other event. (This is assuming you don't try to define the past light cone of each event at the exact time of the initial singularity itself, since the singularity doesn't seem to have a state that could allow you to extrapolate later events by knowing it...for every time slice after the singularity, though, knowing the complete physical state of a region of space would allow you to predict any future event whose past light cone lies entirely in that region, in a deterministic universe.) Are you sure about that? Of course I'm not. It's just one of many possible scenarios. I was asking if you were sure about your claim that in the situation where Mars was deflected by a passing body, the Earth would continue to feel a gravitational pull towards Mars' present position rather than its retarded position, throughout the process. This is a question about GR that would presumably have a single correct answer, so I'm not sure what you mean by "many possible scenarios"--perhaps you misunderstood what I was asking. I didn't think about the planet "smashing into" Mars, only passing close enough to significantly alter its orbit. That's fine, but like I said, my understanding is that GR can only "extrapolate" constant-velocity motion or situations involving acceleration which are spherically or cylindrically symmetric. I don't see how the situation of Mars being deflected from its orbit by a passing body could exhibit this kind of symmetry, so I'm pretty sure the Earth would not continue to be pulled towards Mars' present position throughout the process. In any case, the accelerated motion is "extrapolated" very well by GR so that a non-local mechanism as the one proposed by Newtonian gravity works very well for all but extreme situations like the merging of black holes or neutron stars. It only works as an approximation. If you're claiming that it works in the specific sense of objects continuing to be pulled towards other object's present positions rather than retarded positions, I believe you're wrong about that--again, the "extrapolation" only happens in the case of constant velocity or spherically/cylindrically symmetric motion AFAIK. When you are speaking about “ridiculously complex laws of nature that were somehow "extrapolating" the precise future brain state of the experimenter at the moment of choice” you are referring to a high level description of facts. The mechanism I propose works at the lowest level. A calcium atom doesn’t “know” anything about brains, computers or experiments; it only “looks” for two suitable absorbers (other atoms) for the entangled photons. When such absorbers are found, a pair of photons is send towards their extrapolated position. That’s all. By "complexity" I was referring to the mathematical complexity of the laws involved. We could say that in electromagnetism a charged particle "knows" where another particle would be now if it kept moving at constant velocity, and in GR a test particle "knows" where the surface of a collapsing shell would be if it maintains spherical symmetry; there isn't a literal calculation of this of course, but the laws are such that the particles act as if they know in terms of what direction they are pulled. In order for the source to act as though it knows the orientation of a distant polarizer which was fixed by the brain of a human experimenter, then even if we ignore the issue of some events in the past light cone of the experimenter's choice being outside the past light cone of the source emitting the particles, the "extrapolation" here would be far more complicated because of the extremely complicated and non-symmetrical motions of all the mutually interacting particles in the experimenter's brain which must be extrapolated from some past state, and presumably the laws that would make the source act this way would not have anything like the simplicity of electromagnetism or GR. We could think in terms of algorithmic complexity, for example--the local rules in a cellular-automata program simulating EM or GR would not require a hugely long program (although the actual calculations for a large number of 'cells' might require a lot of computing power), while it seems to me that the sort of rules you're imagining would involve a much, much longer program just to state the fundamental local rules. 1. It should be enough to extrapolate accelerated motion. Other types are not possible for a point particle. Probably even the imperfect extrapolation of GR is enough to explain all experiments to date. You refer to "imperfect" extrapolation, but I'm pretty sure it's not as if GR can kinda-sorta extrapolate accelerations that aren't perfectly spherically or cylindrically symmetric, it's an all-or-nothing deal, just like with EM where the extrapolation is to where the other particle would be if it kept moving at an exactly constant velocity, not somewhere between a constant velocity and its true acceleration. GR wouldn't in any way begin to extrapolate the current positions of particles which are accelerating in all sorts of different directions in a non-symmetric way, with the direction and magnitude of each particle's acceleration always changing due to interactions with other particles (like all the different molecules and electrons in your brain).

And of course, even if you set things up so the detector angle was determined by some simple mechanism which GR could extrapolate, like the radius of a collapsing star at the moment the source emits its particles, the "extrapolation" just refers to where other objects will experience a gravitational pull, what sort of laws do you propose that would allow the source to "know" that the detector angle depends on this variable, and to modify the hidden variables based on the detector angles? Obviously there's nothing in GR itself that could do this. 2. “all of the observable universe originated in a small causally-connected region”. This is the event that “links” the whole experimental setup. Since then all particles are correlated with each other because of the “extrapolation” effect. See above--like I said, this doesn't mean that knowing the past light cone of one event would allow you to automatically predict the outcome of another event with a spacelike separation from the first. The regions of the two past light cones will overlap in the very early universe, but there will be no finite moment after the singularity where the regions encompassed by the two past light cones at that moment are identical, there will always be some points in the past light cone of one that are outside the past light cone of the other. If the event we're talking about is the product of a nonlinear system exhibiting sensitive dependence on initial conditions like the brain, then it seems to me that even in a deterministic universe you'd need to know the complete state of the region of space inside the past light cone at an earlier time in order to predict the event. This is why I think that even Laplace's demon could not predict what the detector setting would be if he only knew about events in the past light cone of the source emitting the entangled particles. Do you disagree, and if so, why?

Hurkyl

Apr24-07, 09:00 PM

Let me make a comment on the Clifford-valued local realistic variables.
Although I have not completely understood the paper, it is not a surprise to me that local Clifford-valued realistic variables may simulate QM. This is because, in a sense, non-commuting variables are never truly local, even if they are local formally. Let me explain what I mean by this:
A formally local quantity is a quantity of the form A(x) or B(y), where x and y are positions of the first and the second particle, respectively. Now, if they are not commuting, then
A(x)B(y) \neq B(y)A(x)
But how two quantities A and B know that they should not commute if x is very far from y? This knowledge is a sort of nonlocality as well.
Well, there are some problems with this. First, a RAA: let's suppose A and B are real-valued functions. How can A and B know that they should commute if x is very far from y?

Anyways, this is very simple. If A and B are Clifford-valued functions, (or if they are real-valued functions), then A(x) and B(y) are numbers. I repeat, they are not numbers located someplace in space-time: they are simply numbers.

OTOH, if A and B took values in a line bundle, so that A(x) is a number located someplace in space-time, then A(x)B(y) is nonsensical: we need a connection (and a path from x to y) to transport a value at x to the fiber at y before we can do any arithmetic with them. (This is true, even if our line bundle is of real numbers)

My opinion is that realistic variables (local or not) must be not only commuting, but represented by real numbers. This is because they are supposed to be measurable, while a measurable quantity must be a real number.
Just to be clear -- is "a measurable quantity must be a real number" your opinion, or are you claiming that as fact?

Demystifier

Apr25-07, 04:30 AM

Just to be clear -- is "a measurable quantity must be a real number" your opinion, or are you claiming that as fact?
It is my opinion. But I am quite certain about it, so I would even dare to claim that it is a fact.

ueit

Apr25-07, 04:49 AM

Well, I'm not too sure how "logically consistent" your mechanism is if it can't meet my simple test (in which each possible outcome probability is non-negative).

My mechanism is as follows:

The PDC source generates photon pairs that obey Malus’ law (n = cos^2(alpha)), where:

n = the probability that the two photons have the same spin on the two measurement axes.

alpha = angle between the polarizers.

The detectors' settings are not communicated non-locally but are "extrapolated" from the past state of the system.

This mechanism is therefore local, realistic (the photons had the measured spin all along) and gives the same predictions as QM, but would not pass your test. This is because your constraints are irrelevant as locality and realism are concerned. It is the statistical independence assumption that requires the probabilities to add to 100% and my mechanism denies this.

I place your hypothesis in the bin with the other "proofs" that Bell is not applicable.

There is nothing to prove. Bell himself clearly stated that the theorem depends of the assumption of statistical independence. You seem not to be able to accept this, for a reason I can't understand.

Bell J., Speakable And Unspeakable In Quantum Mechanics, p. 100:

It has been argued the QM is not locally causal and cannot be embedded in a local causal theory. That conclusion depends on treating certain experimental parameters, typically the orientations of polarization filters, as free variables (emphasis mine)

Please read carefully the above quote and try to understand the irrelevance of your "test" in my case.

NateTG

Apr25-07, 12:16 PM

There is nothing to prove. Bell himself clearly stated that the theorem depends of the assumption of statistical independence. [DrChinese seems] not to be able to accept this, for a reason I can't understand.

Among other reasons, people find strong determinism unpalatable because it is not useful for producing testable theories. Of course, the same is true for MWI, which people seem to have much less trouble with.

Although it's not part of Bell's Theorem, the assumption that pairs of entangled particles can be space-like separated is untested (and possibly untestable), but necessary for valid Bell experiments.

JesseM

Apr25-07, 12:36 PM

Among other reasons, people find strong determinism unpalatable because it is not useful for producing testable theories. Of course, the same is true for MWI, which people seem to have much less trouble with. What does the assumption of statistical independence between spacelike separated events have to do with strong determinism? This statistical independence would be expected even in a completely deterministic universe, unless some past event that influenced both later events caused the correlation (but I explained in my last post to ueit why this doesn't seem to work if one event is that of a human brain making a choice, or any other event with sensitive dependence on initial conditions). Although it's not part of Bell's Theorem, the assumption that pairs of entangled particles can be space-like separated is untested (and possibly untestable), but necessary for valid Bell experiments. What do you mean by "the assumption that pairs of entangled partices can be space-like separated"? Spacelike separation only applies to events, not particles with extended worldlines. And there's no disputing that the event of the experimenter choosing a detector setting and the source emitting the particles can be spacelike separated, all you have to do is find the coordinates of each event and verify that the spatial distance between them is greater than c^2 times the time interval between them, that's all that "spacelike separated" means.

DrChinese

Apr25-07, 02:04 PM

My mechanism is as follows:

The PDC source generates photon pairs that obey Malus’ law (n = cos^2(alpha)), where:

n = the probability that the two photons have the same spin on the two measurement axes.

alpha = angle between the polarizers.

The detectors' settings are not communicated non-locally but are "extrapolated" from the past state of the system.

This is absurd, and has been already ruled out by experiment (Aspect and many subsequent variations).

Detector orientations were changed mid-flight so it is too late for them to in any way be related to the state of the system at the time the entangled photons were created. You are certainly free to reject generally accepted science, but you should not expect others to follow suit without a better argument than that. Your hypothesis is akin to "intelligent design" arguments: there is no evidence to support your viewpoint - and all evidence that should be there is completely missing.

JesseM

Apr25-07, 02:22 PM

This is absurd, and has been already ruled out by experiment (Aspect and many subsequent variations).

Detector orientations were changed mid-flight so it is too late for them to in any way be related to the state of the system at the time the entangled photons were created. You are certainly free to reject generally accepted science, but you should not expect others to follow suit without a better argument than that. Your hypothesis is akin to "intelligent design" arguments: there is no evidence to support your viewpoint - and all evidence that should be there is completely missing. What ueit is proposing is that the source is able to predict the actions of the experimenters in advance, including anything they do while the photons are in mid-flight--this is basically similar to the retrocausality loophole in Bell's theorem (ueit bases his idea on a half-baked analogy with the way the electromagnetic and gravitational forces allow objects to "extrapolate" certain limited kinds of motions of other objects--see this article (http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html) from John Baez's site). But aside from this requiring absurdly complicated laws of physics, I've already pointed out to ueit that it won't work without "conspiracies" in initial conditions, because there are events in the past light cone of the experimenter's choice of detector setting that are outside the past light cone of the source's emitting the particles, so that even assuming perfect determinism, a Laplacian demon sitting at the source would not be able to predict the experimenter's choice given knowledge of everything in the past light cone of the emission-event, all the way back to the Big Bang.

DrChinese

Apr25-07, 03:21 PM

What ueit is proposing is that the source is able to predict the actions of the experimenters in advance, including anything they do while the photons are in mid-flight--this is basically similar to the retrocausality loophole in Bell's theorem (ueit bases his idea on a half-baked analogy with the way the electromagnetic and gravitational forces allow objects to "extrapolate" certain limited kinds of motions of other objects--see this article (http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html) from John Baez's site). But aside from this requiring absurdly complicated laws of physics, I've already pointed out to ueit that it won't work without "conspiracies" in initial conditions, because there are events in the past light cone of the experimenter's choice of detector setting that are outside the past light cone of the source's emitting the particles, so that even assuming perfect determinism, a Laplacian demon sitting at the source would not be able to predict the experimenter's choice given knowledge of everything in the past light cone of the emission-event, all the way back to the Big Bang.

Yes, I quite agree with you. I thought you raised several good points about the entire concept (for instance, the idea that there are many light cones which start to come into play, not just the common one).

But ueit's idea is STILL absurd because it is not really science. There is not the slightest bit of evidence that the polarizer settings have any causal connection to the creation of the entangled particles - nor does that of any other prior event (or set of events) whatsoever. You may as well say that God wants to trick us about the cos^2 theta relationship (which is your "conspiracy" concept) as this relationship evolves regardless of whether the polarizers are set randomly by computer or by human hand.

The fact is, QM specifies the cos^2 theta relationship while ueit's "hypothesis" actually does not. The same hypothesis also fails to accurately predict a single physical law or any known phenomena whatsoever. Additionally, there is no known mechanism by which such causality can be transmitted. Ergo, I personally do not believe it qualifies for discussion on this board.

NateTG

Apr25-07, 10:53 PM

What do you mean by "the assumption that pairs of entangled partices can be space-like separated"? Spacelike separation only applies to events, not particles with extended worldlines. And there's no disputing that the event of the experimenter choosing a detector setting and the source emitting the particles can be spacelike separated.

It's a bit silly, but suppose, for a moment, that whenever two particles are entangled, there is a very tiny wormhole connecting them so that although these two particles appear to be seperated, they are both really aspects of a single particle and local . Of course, these wormholes would have to have some rather odd properties, but, at that point it's impossible to have spacelike seperation between the measurements.

RandallB

Apr26-07, 08:45 AM

It's a bit silly, but suppose, .......tiny wormhole connecting them so that although these two particles appear to be seperated, they are both really aspects of a single particle and local . If true Bell is still correct, the "and local" you refer to is not "Bell Local". Your discribing a reality that is still not local and realistic, just as a theory like BM can use guide waves or MWI can use extra demensions to create a their own version of "local". But not "local and realistic" in the Classical meaning. Which is all Bell is disigned to test for. The definition of QM is Bell non-local.

NateTG

Apr26-07, 01:22 PM

If true Bell is still correct, the "and local" you refer to is not "Bell Local". Your discribing a reality that is still not local and realistic, just as a theory like BM can use guide waves or MWI can use extra demensions to create a their own version of "local". But not "local and realistic" in the Classical meaning. Which is all Bell is disigned to test for.

Just to be clear, this an attempt to illustrate an 'unstated assumption' of Bell's Theorem and not an attempt to refute it, QM, or any experimental results.

Although the presence of wormholes can cause some problems with causality, even Einstein knew that they were consistent with GR (and are thus local). AFAICT It's a bit ambiguous whether this qualifies as Bell-local because people generally assume that pairs of entangled particles can be space-like separated.

Now, if for every pair of particles A and B there is a zero length wormhole W between them, then the order of measurements on A or B is no longer dependent on the observer's frame of reference. Instead the measurement of A and the measurement of B can be considered to occur twice for any particular frame of reference (although each is only observable once). Since the 'first measurement' is well-defined, it's trivial to assign values to the particles at that point.

DrChinese

Apr26-07, 01:36 PM

Let me make a comment on the Clifford-valued local realistic variables.

Although I have not completely understood the paper, it is not a surprise to me that local Clifford-valued realistic variables may simulate QM. This is because, in a sense, non-commuting variables are never truly local, even if they are local formally. Let me explain what I mean by this:

A formally local quantity is a quantity of the form A(x) or B(y), where x and y are positions of the first and the second particle, respectively. Now, if they are not commuting, then
A(x)B(y) \neq B(y)A(x)

But how two quantities A and B know that they should not commute if x is very far from y? This knowledge is a sort of nonlocality as well.

My opinion is that realistic variables (local or not) must be not only commuting, but represented by real numbers. This is because they are supposed to be measurable, while a measurable quantity must be a real number. Therefore, I believe that the Clifford-valued realistic variables are physically meaningless.

In fact, the claim that physical variables could be noncommuting numbers does not differ much from the claim that physical variables could be noncommuting operators or noncommuting matrices. But this is exactly what the realistic physical variables in QM are NOT supposed to be, because otherwise we deal with QM in the usual matrix/operator form.

A paper has been added to the archive that discusses some of this in greater depth and may be of interest:

Title: Non-Viability of a Counter-Argument to Bell's Theorem (http://arxiv.org/abs/0704.2038)

ueit

Apr27-07, 06:10 AM

Unless I'm misunderstanding something, that doesn't mean that there was any time when all the events in the past light cone of the event of the experimenter making a choice of what to measure were also in the past light cone of the event of the the source sending out the particles. Again, if you don't place any special constraints on initial conditions, then even in a deterministic universe, a Laplacian demon with knowledge of everything in the past light cone of the source sending out the particles would not necessarily be able to predict the brain state of the experimenter at the time he made his choice of what to measure. Do you disagree?

I think that inflationary theory would say that the past light-cones of the most widely-separated events we can see will partially overlap, so that the similarity of the CMBR in different regions can have a common past cause. But again, it doesn't mean that knowing the past light cone of one event would allow you to predict every other event, even in a perfectly deterministic universe, because any pair of spacelike separated events would have parts of their past light cones that are outside the past light cone of the other event. (This is assuming you don't try to define the past light cone of each event at the exact time of the initial singularity itself, since the singularity doesn't seem to have a state that could allow you to extrapolate later events by knowing it...for every time slice after the singularity, though, knowing the complete physical state of a region of space would allow you to predict any future event whose past light cone lies entirely in that region, in a deterministic universe.)

Juao Magueijo’s article “Plan B for the cosmos” (Scientific American, Jan. 2001, p.47) reads:

Inflationary theory postulates that the early universe expanded so fast that the range of light was phenomenally large. Seemingly disjointed regions could thus have communicated with one another and reached a common temperature and density. When the inflationary expansion ended, these regions began to fall out of touch.

It does not take much thought to realize that the same thing could have been achieved if light simply had traveled faster in the early universe than it does today. Fast light could have stitched together a patchwork of otherwise disconnected regions. These regions could have homogenized themselves. As the speed of light slowed, those regions would have fallen out of contact

It is clear from the above quote that the early universe was in thermal equilibrium. That means that there was enough time for the EM field of each particle to reach all other particles (it only takes light one second to travel between two opposite points on a sphere with a diameter of 3 x 10^8 m but this time is hardly enough to bring such a sphere of gas at an almost perfect thermal equilibrium). A Laplacian demon “riding on a particle” could infer the position/momentum of every other particle in that early universe by looking at the field around him. This is still true today because of the extrapolation mechanism.

I also disagree that “the singularity doesn't seem to have a state that could allow you to extrapolate later events by knowing it”. We don’t have a theory to describe the big-bang so I don’t see why we should assume that it was a non-deterministic phenomena rather than a deterministic one. If QM is deterministic after all I don’t see where a stochastic big-bang could come from.

To summarize, my questions are:

1. Do we have compelling evidence that the big-bang was a non-deterministic process?
2. Does the present evidence exclude the possibility that “past light-cones of the most widely-separated events” overlap completely?

If the answer to any of the these questions is “no” my hypothesis stands.

I was asking if you were sure about your claim that in the situation where Mars was deflected by a passing body, the Earth would continue to feel a gravitational pull towards Mars' present position rather than its retarded position, throughout the process.

Yes, because this is a case where Newtonian theory applies well (small mass density). I’m not accustomed with GR formalism but I bet that the difference between the predictions of the two theories is very small.

This is a question about GR that would presumably have a single correct answer, so I'm not sure what you mean by "many possible scenarios"--perhaps you misunderstood what I was asking.

Yes, I misunderstood you.

It only works as an approximation. If you're claiming that it works in the specific sense of objects continuing to be pulled towards other object's present positions rather than retarded positions, I believe you're wrong about that--again, the "extrapolation" only happens in the case of constant velocity or spherically/cylindrically symmetric motion AFAIK.

In Newtonian gravity the force is instantaneous. So, yes, in any system for which Newtonian gravity is a good approximation the objects are “pulled towards other object's present positions”. The article you linked from John Baez’s site claims that uniform accelerated motion is extrapolated by GR as well.

By "complexity" I was referring to the mathematical complexity of the laws involved. We could say that in electromagnetism a charged particle "knows" where another particle would be now if it kept moving at constant velocity, and in GR a test particle "knows" where the surface of a collapsing shell would be if it maintains spherical symmetry; there isn't a literal calculation of this of course, but the laws are such that the particles act as if they know in terms of what direction they are pulled. In order for the source to act as though it knows the orientation of a distant polarizer which was fixed by the brain of a human experimenter, then even if we ignore the issue of some events in the past light cone of the experimenter's choice being outside the past light cone of the source emitting the particles, the "extrapolation" here would be far more complicated because of the extremely complicated and non-symmetrical motions of all the mutually interacting particles in the experimenter's brain which must be extrapolated from some past state, and presumably the laws that would make the source act this way would not have anything like the simplicity of electromagnetism or GR. We could think in terms of algorithmic complexity, for example--the local rules in a cellular-automata program simulating EM or GR would not require a hugely long program (although the actual calculations for a large number of 'cells' might require a lot of computing power), while it seems to me that the sort of rules you're imagining would involve a much, much longer program just to state the fundamental local rules.

EM extrapolates uniform motion, GR uniform accelerated motion. I’m not a mathematician so I have no idea if a mechanism able to extrapolate a generic accelerated motion should necessarily be as complex or so difficult to simulate on a computer as you imply. You are, of course, free to express an opinion but at this point I don’t think you’ve put forward a compelling argument.

You refer to "imperfect" extrapolation, but I'm pretty sure it's not as if GR can kinda-sorta extrapolate accelerations that aren't perfectly spherically or cylindrically symmetric, it's an all-or-nothing deal, just like with EM where the extrapolation is to where the other particle would be if it kept moving at an exactly constant velocity, not somewhere between a constant velocity and its true acceleration. GR wouldn't in any way begin to extrapolate the current positions of particles which are accelerating in all sorts of different directions in a non-symmetric way, with the direction and magnitude of each particle's acceleration always changing due to interactions with other particles (like all the different molecules and electrons in your brain).

If what you are saying is true then we should expect Newtonian gravity to miserably fail when dealing with a non-uniform accelerated motion, like a planet in an elliptical orbit, right? Anyway, probably you are right that an imperfect extrapolation would be useless because of chaos, so a mechanism able to perfectly extrapolate accelerated motion is required.

And of course, even if you set things up so the detector angle was determined by some simple mechanism which GR could extrapolate, like the radius of a collapsing star at the moment the source emits its particles, the "extrapolation" just refers to where other objects will experience a gravitational pull, what sort of laws do you propose that would allow the source to "know" that the detector angle depends on this variable, and to modify the hidden variables based on the detector angles? Obviously there's nothing in GR itself that could do this.

I have no idea of how the mathematical implementation of such a mechanism would look like. Probably one could start with the Cramer’s transactional interpretation and replace the advanced wave that is send by the absorber back in time towards the emitter with a “normal”, retarded wave coming from the detector prior to emission and make the emission event depend on the “extrapolated” position of the absorber.

See above--like I said, this doesn't mean that knowing the past light cone of one event would allow you to automatically predict the outcome of another event with a spacelike separation from the first. The regions of the two past light cones will overlap in the very early universe, but there will be no finite moment after the singularity where the regions encompassed by the two past light cones at that moment are identical, there will always be some points in the past light cone of one that are outside the past light cone of the other. If the event we're talking about is the product of a nonlinear system exhibiting sensitive dependence on initial conditions like the brain, then it seems to me that even in a deterministic universe you'd need to know the complete state of the region of space inside the past light cone at an earlier time in order to predict the event. This is why I think that even Laplace's demon could not predict what the detector setting would be if he only knew about events in the past light cone of the source emitting the entangled particles. Do you disagree, and if so, why?

I disagree, see my two questions above.

ueit

Apr27-07, 06:37 AM

But ueit's idea is STILL absurd because it is not really science. There is not the slightest bit of evidence that the polarizer settings have any causal connection to the creation of the entangled particles - nor does that of any other prior event (or set of events) whatsoever.

Please specify what evidence supports the idea of a non-realistic universe.
Please specify what evidence supports the idea of a non-local universe.

The fact is, QM specifies the cos^2 theta relationship while ueit's "hypothesis" actually does not. The same hypothesis also fails to accurately predict a single physical law or any known phenomena whatsoever. Additionally, there is no known mechanism by which such causality can be transmitted. Ergo, I personally do not believe it qualifies for discussion on this board.

You shift the burden of proof. Again.

DrChinese

Apr27-07, 09:02 AM

A Laplacian demon “riding on a particle” could infer the position/momentum of every other particle in that early universe by looking at the field around him. This is still true today because of the extrapolation mechanism.

Another exaggeration. This is not science! If you are such a demon with amazing god-like powers, please show how you might do this.

Otherwise, you should stick to accepted theory & experiment. Obviously, your hypothesis is FAR from accepted as there are no known field effects capable of providing this amount of information. And the Heisenberg Uncertainty Principle flat out excludes this hypothesis for even ONE particle.

DrChinese

Apr27-07, 09:09 AM

jtbell

Apr27-07, 09:33 AM

Just as a reminder, the old "Theory Development" forum was superseded some time ago by the "Independent Research" forum.

DrChinese

Apr27-07, 09:41 AM

Just as a reminder, the old "Theory Development" forum was superseded some time ago by the "Independent Research" forum.

I stand corrected. :smile:

I would say that ueit has had a run with this idea; it has been fairly discussed; forum members (besides myself) have addressed the substantial weaknesses in the concept; and ueit has failed to provide any substantiation for claims that are getting bolder and bolder.

Given PF guidelines, I think it is time for ueit to move on regarding this ad hoc and untestable hypothesis.

Mentz114

Apr27-07, 09:47 AM

Please quit telling readers here that entangled particle wave functions were predetermined from the initial pre-inflationary state of the universe unless you have some specific evidence of that.

We are not so easily influenced. But I believe, if wavefunctions exist, and the big bang happened ( and other ifs too) that ALL wavefunctions are entangled - but not in a conspiratorial way. So that kind of entanglement is irrelevant to the Bell scenario.

NateTG

Apr27-07, 10:20 AM

You shift the burden of proof. Again.

There is a big difference between your position and mine, and that allows me to do this successfully.

Your ad hoc opinion belongs in Theory Development, where you can attempt to develop it into a testable scientific hypothesis. On the other hand, my position is orthodox science with the backing of both theory and experiment.

Please quit telling readers here that entangled particle wave functions were predetermined from the initial pre-inflationary state of the universe unless you have some specific evidence of that.

This is a discussion about interpretations of QM. They're called interpretations, rather than theories, because they do not make testable predictions.

It's unreasonable to require experimental evidence to validate an interpretation. As a theory, QM is agnostic regarding determinism, so QM makes no predictions that support or contradict the notion that the current state of the universe is completely determined by a prior state. (Technically, one could stay that QM is a deterministic theory in the sense that it can model a deterministic reality.)

There are prediction-identical strongly deterministic interpretations of QM. Many worlds and Bohmian mechanics are two well-known examples - neither of which is at odds with orthodox science. Any experimental result that falsifies the notion of strong determinism would also invalidate both of those interpretations.

There are (as previously mentioned) philosophical and aesthetic reasons for refusing to accept arbitrary strong determinism. Arbitrary strong determinism is not predictive, and could easily be described as a 'conspiring universe'. However, since since strong determinism is not falsifiable, it cannot be contradicted by scientific experiments.

To be clear, it is categorically impossible to experimentally contradict the notion that wave functions are predetermined.

N.B.: Talking (or posting) about things in terms of 'your position' or 'my position' can be useful and convenient, but can also foster confusion or lead to people taking things personally.

NateTG

Apr27-07, 10:44 AM

Among other reasons, people find strong determinism unpalatable because it is not useful for producing testable theories. Of course, the same is true for MWI, which people seem to have much less trouble with.

What does the assumption of statistical independence between spacelike separated events have to do with strong determinism? This statistical independence would be expected even in a completely deterministic universe, unless some past event that influenced both later events caused the correlation (but I explained in my last post to ueit why this doesn't seem to work if one event is that of a human brain making a choice, or any other event with sensitive dependence on initial conditions).

As far as I can tell, ueit is basically arguing an interpretation of arbitrary strong determinism, which is then made local by assuming that each particle consults it's own model of the entire universe. In effect, each particle carries some hidden state \vec{h} which corresponds to a complete list of the results of 'wavefunction collapses'.

What strong determinism has to do with statistical independence is that statistical independence may not be possible in a deterministic universe. As you have alluded to, by your reference to the human brain, the notion of statistical independence has also been called 'free will'.

DrChinese

Apr27-07, 11:03 AM

A Laplacian demon “riding on a particle” could infer the position/momentum of every other particle in that early universe by looking at the field around him. This is still true today because of the extrapolation mechanism.

As far as I can tell, ueit is basically arguing an interpretation of arbitrary strong determinism, which is then made local by assuming that each particle consults it's own model of the entire universe. In effect, each particle carries some hidden state \vec{h} which corresponds to a complete list of the results of 'wavefunction collapses'.

Listen to what is being said! It is absolutely as extreme as arguing that "Jesus said so" is an interpretation of physics that we need to discuss. Yes, there would need to be a local copy of the history (past and present, and presumably future as well) of the entire universe present in every particle. Yet ueit even acknowledged this as absurd earlier (but has apparently returned to it).

I like my interpretations with at least a modicum of science included. :rofl: As you can tell, I am no fan of ad hoc theories that make NO specific predictions (testable or otherwise). Clearly there is not one IOTA of connection between this "interpretation" and the results we actually observe since "superdeterminism" makes no predictive results whatsoever other than "anything goes".

As regards this thread specifically: Where does the cos^2 theta relationship come from? Bohmian Mechanics has a framework, as does QM. So please, don't elevate ueit's ideas from ad hoc personal philosophy to the level of legitimate science.

DrChinese

Apr27-07, 11:22 AM

It's unreasonable to require experimental evidence to validate an interpretation.

NateTG, I'll disagree with you on this one. We should expect an physical interpretation to pass muster with an existing substantial body evidence.

1. If there is a map of the entire history of the universe embedded in every particle (as ueit says), why have we never noticed this in any experiment? There is not a single shed of evidence, direct or indirect, that this is so.

2. And there is substantial evidence - in the form of random results to any desired level - that there is NO connection between quantum states of many collections of particles that are in extremely close causal contact. An example being a radioactive sample of uranium, or polarization of photons emitted from ordinary light bulbs. It is amazing that ONLY the entangled photons from a PDC source (or similar) display these fascinating correlations, while all other particles are purely random. Yet ueit says all share a "common" knowledge of the evolution of the universe.

So while these blatant gaps may not bother you, they point out to me that this is a purely ad hoc personal theory and one which does not qualify as a legitimate interpretation of particle physics. Or perhaps there are papers that could be cited to at least add some air of science to this?

I think that discussions of interpretations are themselves completely legitimate, but ueit is well beyond that point. I recognize that you may have a different opinion of where that point is.

NateTG

Apr27-07, 12:12 PM

Listen to what is being said! It is absolutely as extreme as arguing that "Jesus said so" is an interpretation of physics that we need to discuss.

What I'm arguing is that it's incorrect to say that 'Jesus said so' can be (scientifically) falsified. 'Jesus said so' is an awful interpretation of QM, but the salient argument for that is philosophical and not scientific in nature. However this claim:
On the other hand, my position is orthodox science with the backing of both theory and experiment.
represents itself as a scientific one.

Yes, there would need to be a local copy of the history (past and present, and presumably future as well) of the entire universe present in every particle. Yet ueit even acknowledged this as absurd earlier (but has apparently returned to it).

Philosophically speaking, nobody seems to have any trouble with every particle having a local copy of its own past. It's not difficult to conceive of a universe where every particle's past includes sufficient information to predict the entire universe's space-time.

As regards this thread specifically: Where does the cos^2 theta relationship come from? Bohmian Mechanics has a framework, as does QM. So please, don't elevate ueit's ideas from ad hoc personal philosophy to the level of legitimate science.

I don't want to put words into ueit's mouth, but what he seems to be trying to describe is very similar to Bohmian Mechanics in a universe with a singular origin, which is local, realistic, and prediction equivalent to QM.

NateTG

Apr27-07, 12:36 PM

1. If there is a map of the entire history of the universe embedded in every particle (as ueit says), why have we never noticed this in any experiment? There is not a single shed of evidence, direct or indirect, that this is so.

If there is a wavefunction and it collapses why have we never noticed wavefunction collapse in an experiment? There is not a single shred of evidence, direct, or indirect, that a wavefunction physically exists or wavefunction collapse occurs.

2. And there is substantial evidence - in the form of random results to any desired level - that there is NO connection between quantum states of many collections of particles that are in extremely close causal contact. An example being a radioactive sample of uranium, or polarization of photons emitted from ordinary light bulbs. It is amazing that ONLY the entangled photons from a PDC source (or similar) display these fascinating correlations, while all other particles are purely random. Yet ueit says all share a "common" knowledge of the evolution of the universe.

This is a fallacious argument. Just because something something is unpredictable does not make it non-deterministic. This is true even in the classical example of a ball at the top of a hill.

RandallB

Apr27-07, 01:29 PM

Just to be clear, this an attempt to illustrate an 'unstated assumption' of Bell's Theorem and not an attempt to refute it, QM, or any experimental results.

Although the presence of wormholes can cause some problems with causality, even Einstein knew that they were consistent with GR (and are thus local). AFAICT It's a bit ambiguous whether this qualifies as Bell-local because people generally assume that pairs of entangled particles can be space-like separated. [emphasis added]

Sorry I’ve totally missed any 'unstated assumption' by Bell that should be included in his theorem. Do you have a clear definition of what you’re referring to here?

Also I don’t see what you think is “a bit ambiguous” - - - are you trying to say it could be fair to call a pair of wormholes Bell-local?? If you understand Bell-local there is nothing ambiguous at all; the idea is just not Bell Local any more than QM is Bell Local!

There is nothing wrong with being non Bell-local that is how QM defines itself as a “complete” theory (no other can improve on) and a Local Realist cannot be complete. Remember Bell-local means BOTH Einstein local and Einstein realistic i.e. Einstein local and realistic. That means local variables created as part of the two separate photons when they were created including hidden variables. But that is not all; these variables do not change with any “Weird Action at’a Distance” (WAAD) of any kind! Any such requirement is NOT Bell-local.

All the Bell experiments have said is “Sorry Bell we still see weird action at a distance that cannot be answered by any Local Realistic Variable, not even a hidden but unknown one.”

Wormholes theories and "common histories able to manipulate and correlate outcomes" theories can solve Bell is because they are non Bell-local.

Just like MWI and Bohmian Mechanics are not Bell-local, they all allow WAAD to appear, they just have different ways to account for it - but it is still WAAD that a Local Realist cannot understand without extra demensions of invisable guide waves. They provide predictions equivalent to QM because they are using non Bell-local solutions,

If they are Bell-local theories they could be provide a definition of Einstein’s unknown hidden variable; not a solution to WAAD equivalent to QM non-local solution.

A theory can not just change the rules for what Bell-local means and expect to gain more respect than QM; if it wants to claim being local it must define and describe the variable that replaces the uncertainty principle.

ueit

Apr27-07, 02:48 PM

Among other reasons, people find strong determinism unpalatable because it is not useful for producing testable theories. Of course, the same is true for MWI, which people seem to have much less trouble with.

Strong determinism is nothing but plain old determinism followed to its logical conclusion. If one doesn't like the conclusion, too bad for him. However, it should be clearly stated that the reason for rejecting such theories has nothing to do with Bell.

ueit

Apr27-07, 02:55 PM

DrChinese, please answer my two questions:

Please specify what evidence supports the idea of a non-realistic universe.
Please specify what evidence supports the idea of a non-local universe.

Thanks!

paw

Apr27-07, 03:05 PM

....There is not a single shred of evidence, direct, or indirect, that a wavefunction physically exists....

Have I missed something here? Aren't there solutions of the Schroedinger equation that predict accurately the orbitals of the hydrogen atom, the electron density of H2? Didn't solutions of the SE allow development of the scanning tunnelling EM? Aren't DeBroglie waves wavefunctions? Don't they accurately predict the various interferance patterns of electrons? I think the evidence is almost overwhelming.

Of course you could argue that all these phenomena are caused by some property that behaves exactly like the wavefunction, but a gentleman by the name of Occam had something to say about that.:biggrin:If it walks like a duck and quacks like a duck.......

JesseM

Apr27-07, 03:30 PM

Juao Magueijo’s article “Plan B for the cosmos” (Scientific American, Jan. 2001, p.47) reads:
Inflationary theory postulates that the early universe expanded so fast that the range of light was phenomenally large. Seemingly disjointed regions could thus have communicated with one another and reached a common temperature and density. When the inflationary expansion ended, these regions began to fall out of touch.

It does not take much thought to realize that the same thing could have been achieved if light simply had traveled faster in the early universe than it does today. Fast light could have stitched together a patchwork of otherwise disconnected regions. These regions could have homogenized themselves. As the speed of light slowed, those regions would have fallen out of contact
It is clear from the above quote that the early universe was in thermal equilibrium. That means that there was enough time for the EM field of each particle to reach all other particles (it only takes light one second to travel between two opposite points on a sphere with a diameter of 3 x 10^8 m but this time is hardly enough to bring such a sphere of gas at an almost perfect thermal equilibrium). A Laplacian demon “riding on a particle” could infer the position/momentum of every other particle in that early universe by looking at the field around him. This is still true today because of the extrapolation mechanism. Your logic here is faulty--even if the observable universe had reached thermal equilibrium, that definitely doesn't mean that each particle's past light cone would become identical at some early time. This is easier to see if we consider a situation of a local region reaching equilibrium in SR. Suppose at some time t0 we fill a box many light-years long with an inhomogenous distribution of gas, and immediately seal the box. We pick a particular region which is small compared to the entire box--say, a region 1 light-second wide--and wait just long enough for this region to get very close to thermal equilibrium. The box is much larger than the region so this will not have been long enough for the whole thing to reach equilibrium, so perhaps there will be large-scale gradients in density/pressure/temperature etc., even if any given region 1 light-second wide is very close to homogenous.

So, does this mean that if we take two spacelike-separated events inside the region which happen after it has reached equilibrium, we can predict one by knowing the complete light cone of the other? Of course not--this scenario is based entirely on the flat spacetime of SR, so it's easy to see that for any spacelike-separated events in SR, there must be events in the past light cone of one which lie outside the past light cone of the other, no matter how far back in time you go. In fact, as measured in the inertial frame where the events are simultaneous, the distance between the two events must be identical to the distance between the edges of the two past light cones at all earlier times. Also, if we've left enough time for the 1 light-second region to reach equilibrium, this will probably be a lot longer than 1 second, meaning the size of each event's past light cone at t0 will be much larger than the 1 light-second region itself.

The situation is a little more complicated in GR due to curved spacetime distorting the light cones (look at some of the diagrams on Ned Wright's Cosmology Tutorial (http://www.astro.ucla.edu/~wright/cosmoall.htm), for example), but I'm confident you wouldn't see two light cones smoothly join up and encompass identical regions at earlier times--it seems to me this would imply at at the event of the joining-up, this would mean photons at the same position and moving in the same direction would have more than one possible geodesic path (leading either to the first event or the second event), which isn't supposed to be possible. In any case, your argument didn't depend specifically on any features of GR, it just suggested that if the universe had reached equilibrium this would mean that knowing the past light cone of one event in the region would allow a Laplacian demon to predict the outcome of another spacelike-separated event, but my SR example shows this doesn't make sense. I also disagree that “the singularity doesn't seem to have a state that could allow you to extrapolate later events by knowing it”. We don’t have a theory to describe the big-bang so I don’t see why we should assume that it was a non-deterministic phenomena rather than a deterministic one. If QM is deterministic after all I don’t see where a stochastic big-bang could come from. I wasn't saying anything about the big bang being stochastic, just about the initial singularity in GR being fairly "featurless", you can't extrapolate the later state of the universe from some sort of description of the singularity itself--this doesn't really mean GR is non-deterministic, you could just consider the singularity to not be a part of the spacetime manifold, but more like a point-sized "hole" in it. Of course GR's prediction of a "singularity" may be wrong, but in that case the past light cones of different events wouldn't converge on a single point of zero volume in the same way, so as long as we assume the new theory still has a light cone structure, we're back to my old argument about the past light cones of spacelike-separated events never becoming identical. I was asking if you were sure about your claim that in the situation where Mars was deflected by a passing body, the Earth would continue to feel a gravitational pull towards Mars' present position rather than its retarded position, throughout the process. Yes, because this is a case where Newtonian theory applies well (small mass density). I’m not accustomed with GR formalism but I bet that the difference between the predictions of the two theories is very small. Probably, but that doesn't imply that GR predicts that the Earth will be attracted to Mars' current position, since after all one can ignore Mars altogether in Newtonian gravity and still get a very good prediction of the movement of the Earth. If you really think it's plausible that GR predicts Earth can "extrapolate" the motions of Mars in this situation which obviously departs significantly from spherical/cylindrical symmetry, perhaps we should start a thread on the relativity forum to get confirmation from GR experts over there? In Newtonian gravity the force is instantaneous. So, yes, in any system for which Newtonian gravity is a good approximation the objects are “pulled towards other object's present positions”. You're talking as though the only reason Newtonian gravity could fail to be a good approximation is because of the retarded vs. current position issue! But there are all kinds of ways in which GR departs wildly from Newtonian gravity which have nothing to do with this issue, like the prediction that sufficiently massive objects can form black holes, or the prediction of gravitational time dilation. And the fact is that the orbit of a given planet can be approximated well by ignoring the other planets altogether (or only including Jupiter), so obviously the issue of the Earth being attracted to the current vs. retarded position of Mars is going to have little effect on our predictions. The article you linked from John Baez’s site claims that uniform accelerated motion is extrapolated by GR as well. Well, the wikipedia article (http://en.wikipedia.org/wiki/Gravitational_wave#Sources_of_gravitational_waves) says: In general terms, gravitational waves are radiated by objects whose motion involves acceleration, provided that the motion is not perfectly spherically symmetric (like a spinning, expanding or contracting sphere) or cylindrically symmetric (like a spinning disk). So either one is wrong or we're misunderstanding what "uniform acceleration" means...is it possible that Baez was only talking about uniform acceleration caused by gravity as opposed to other forces, and that gravity only causes uniform acceleration in an orbit situation which also has spherical/cylindrical symmetry? I don't know the answer, this might be another question to ask on the relativity forum...in any case, I'm pretty sure that the situation you envisioned where Mars is deflected from its orbit by a passing body would not qualify as either "uniform acceleration" or "spherically/cylindrically symmetric". EM extrapolates uniform motion, GR uniform accelerated motion. I’m not a mathematician so I have no idea if a mechanism able to extrapolate a generic accelerated motion should necessarily be as complex or so difficult to simulate on a computer as you imply. You are, of course, free to express an opinion but at this point I don’t think you’ve put forward a compelling argument. You're right that I don't have a rigorous argument, but I'm just using the following intuition--if you know the current position of an object moving at constant velocity, how much calculation would it take to predict its future position under the assumption it continued to move at this velocity? How much calculation would it take to predict the future position of an object which we assume is undergoing uniform acceleration? And given a system involving many components with constantly-changing accelerations due to constant interactions with each other, like water molecules in a jar or molecules in a brain, how much calculation would it take to predict the future position of one of these parts given knowledge of the system's state in the past. Obviously the amount of calculation needed in the third situation is many orders of magnitude greater than in the first two. If what you are saying is true then we should expect Newtonian gravity to miserably fail when dealing with a non-uniform accelerated motion, like a planet in an elliptical orbit, right? No. If our predictions don't "miserably fail" when we ignore Mars altogether, they aren't going to miserably fail if we predict the Earth is attracted to Mars' current position as opposed to where GR says it should be attracted to, which is not going to be very different anyway since a signal from Mars moving at the speed of light takes a maximum of 22 minutes to reach Earth according to this page (http://www.aerospaceweb.org/question/astronomy/q0254.shtml). Again, in the situations where GR and Newtonian gravity give very different predictions, this is not mainly because of the retarded vs. current position issue.

JesseM

Apr27-07, 03:37 PM

As far as I can tell, ueit is basically arguing an interpretation of arbitrary strong determinism, which is then made local by assuming that each particle consults it's own model of the entire universe. But if it's really local, each particle should only be allowed to consult its own model of everything in its past light cone back to the Big Bang--for the particle to have a model of anything outside that would require either nonlocality or a "conspiracy" in initial conditions. As I argued in my last post, ueit's argument about thermal equilibrium in the early universe establishing that all past light cones merge and become identical at some point doesn't make sense.

ueit

Apr27-07, 04:10 PM

As far as I can tell, ueit is basically arguing an interpretation of arbitrary strong determinism, which is then made local by assuming that each particle consults it's own model of the entire universe. In effect, each particle carries some hidden state \vec{h} which corresponds to a complete list of the results of 'wavefunction collapses'.

It's not exactly what I propose. Take the case of gravity in a Newtonian framework. Each object "knows" where all other objects are, instantaneously. It then acts as if it's doing all the calculations, applying the inverse square law. General relativity explains this apparently non-local behavior through a local mechanism where the instantaneous position of each body in the system is extrapolated from the past state. That past state is "inferred" from the space curvature around the object.
By analogy, we might think that the EPR source "infers" the past state of the detectors from the EM field around it, extrapolates the future detector orientation and generates a pair of entangled particles with a suitable spin.

DrChinese

Apr27-07, 08:13 PM

DrC said: If there is a wavefunction and it collapses why have we never noticed wavefunction collapse in an experiment? There is not a single shred of evidence, direct, or indirect, that a wavefunction physically exists or wavefunction collapse occurs.

This is a fallacious argument. Just because something something is unpredictable does not make it non-deterministic. This is true even in the classical example of a ball at the top of a hill.

Not true. If entangled particles are spin correlated due to a prior state of the system, why aren't ALL particles similarly correlated? We should see correlations of spin observables everywhere we look! But we don't, we see randomness everywhere else. So the ONLY time we see these are with entangled particles. Hmmm. Gee, is this a strained explanation or what? And gosh, the actual experimental correlation just happens to match QM, while there is absolutely no reason (with this hypothesis) it couldn't have ANY value (how about sin^2 theta, for example). Why is that?

It is sorta like invoking the phases of the moon to explain why there are more murders during a full moon, and not being willing to accept that there are no fewer murders at other times. Or do we use this as an explanation only when it suits us?

If this isn't ad hoc science, what is?

ueit

Apr28-07, 05:38 AM

Not true. If entangled particles are spin correlated due to a prior state of the system, why aren't ALL particles similarly correlated? We should see correlations of spin observables everywhere we look! But we don't, we see randomness everywhere else. So the ONLY time we see these are with entangled particles. Hmmm. Gee, is this a strained explanation or what?

That's simple. Even if each particle "looks" for a suitable detector orientation before emission, only for entangled particles we have a set of supplementary conditions (conservation of angular momentum, same emission time) that enable us to observe the correlations. In order to release a pair of entangled particles both detectors must be in a suitable state, that's not the case for a "normal" particle.

And gosh, the actual experimental correlation just happens to match QM, while there is absolutely no reason (with this hypothesis) it couldn't have ANY value (how about sin^2 theta, for example). Why is that?

I put Malus' law by hand without any particular reason other than reproduce QM's prediction. I'm getting tired of pointing out that the burden of proof is on you. You make the strong claim that no local-realistic mechanism can reproduce QM's prediction. On the other hand I don't claim that my hypothesis is true or even likely. I only claim that it is possible.

To give you an example, von Newman's proof against the existence of hidden-variable theories is wrong even if no such theory is provided. It was wrong even before Bohm published his interpretation and will remain wrong even if BM is falsified. So, asking me to provide evidence for the local-realistic mechanism I propose is a red-herring.

If this isn't ad hoc science, what is?

It certainly is ad-hoc but so what? Your bold claim regarding Bell's theorem is still proven false.

DrChinese

Apr28-07, 10:24 AM

1. In order to release a pair of entangled particles both detectors must be in a suitable state, that's not the case for a "normal" particle.

2. I put Malus' law by hand without any particular reason other than reproduce QM's prediction.

3. Your bold claim regarding Bell's theorem is still proven false.

1. :rofl:

2. :rofl:

3. Still agreed to by virtually every scientist in the field.

JesseM

Apr28-07, 11:59 AM

The question of what is or is not a valid loophole in Bell's theorem should not be a matter of opinion, and it also should not be affected by how ridiculous or implausible a theory based on the loophole would have to be. For example, everyone agrees the "conspiracy in initial conditions" is a logically valid loophole, even though virtually everyone also agrees that it's not worth taking seriously as a real possibility. If it wasn't for the light cone objection, I'd say that ueit had pointed out another valid loophole, even though I personally wouldn't take it seriously because of the separate objection of the need for ridiculously complicated laws of physics to "extrapolate" the future states of nonlinear systems with a huge number of interacting parts like the human brain. But I do think the light cone objection shows that ueit's idea doesn't work even as a logical possibility. If he wanted to argue that each particle has, in effect, not just a record of everything in its past light cone, but a record of the state of the entire universe immediately after the Big Bang (or at the singularity, if you imagine the singularity itself has 'hidden variables' which determine future states of the universe), then this would be a logically valid loophole, although I would see it as just a version of the "conspiracy" loophole (since each particle's 'record' of the entire universe's past state can't really be explained dynamically, it would seem to be part of the initial conditions).

ueit

Apr29-07, 04:09 PM

Your logic here is faulty--even if the observable universe had reached thermal equilibrium, that definitely doesn't mean that each particle's past light cone would become identical at some early time. This is easier to see if we consider a situation of a local region reaching equilibrium in SR. Suppose at some time t0 we fill a box many light-years long with an inhomogenous distribution of gas, and immediately seal the box. We pick a particular region which is small compared to the entire box--say, a region 1 light-second wide--and wait just long enough for this region to get very close to thermal equilibrium. The box is much larger than the region so this will not have been long enough for the whole thing to reach equilibrium, so perhaps there will be large-scale gradients in density/pressure/temperature etc., even if any given region 1 light-second wide is very close to homogenous.

So, does this mean that if we take two spacelike-separated events inside the region which happen after it has reached equilibrium, we can predict one by knowing the complete light cone of the other? Of course not--this scenario is based entirely on the flat spacetime of SR, so it's easy to see that for any spacelike-separated events in SR, there must be events in the past light cone of one which lie outside the past light cone of the other, no matter how far back in time you go. In fact, as measured in the inertial frame where the events are simultaneous, the distance between the two events must be identical to the distance between the edges of the two past light cones at all earlier times. Also, if we've left enough time for the 1 light-second region to reach equilibrium, this will probably be a lot longer than 1 second, meaning the size of each event's past light cone at t0 will be much larger than the 1 light-second region itself.

The situation is a little more complicated in GR due to curved spacetime distorting the light cones (look at some of the diagrams on Ned Wright's Cosmology Tutorial (http://www.astro.ucla.edu/~wright/cosmoall.htm), for example), but I'm confident you wouldn't see two light cones smoothly join up and encompass identical regions at earlier times--it seems to me this would imply at at the event of the joining-up, this would mean photons at the same position and moving in the same direction would have more than one possible geodesic path (leading either to the first event or the second event), which isn't supposed to be possible. In any case, your argument didn't depend specifically on any features of GR, it just suggested that if the universe had reached equilibrium this would mean that knowing the past light cone of one event in the region would allow a Laplacian demon to predict the outcome of another spacelike-separated event, but my SR example shows this doesn't make sense.

OK, I think I understand your point. The CMB isotropy does not require the whole early universe to be in thermal equilibrium. But, does the evidence we have require the opposite, that the whole universe was not in equilibrium? If not, my hypothesis is still consistent with extant data.

I wasn't saying anything about the big bang being stochastic, just about the initial singularity in GR being fairly "featurless", you can't extrapolate the later state of the universe from some sort of description of the singularity itself--this doesn't really mean GR is non-deterministic, you could just consider the singularity to not be a part of the spacetime manifold, but more like a point-sized "hole" in it. Of course GR's prediction of a "singularity" may be wrong, but in that case the past light cones of different events wouldn't converge on a single point of zero volume in the same way, so as long as we assume the new theory still has a light cone structure, we're back to my old argument about the past light cones of spacelike-separated events never becoming identical.

I don't think your argument applies in this case. For example, the pre-big bang universe might have been a Planck-sized "molecule" of an exotic type, that produced all particles in a deterministic manner.

Probably, but that doesn't imply that GR predicts that the Earth will be attracted to Mars' current position, since after all one can ignore Mars altogether in Newtonian gravity and still get a very good prediction of the movement of the Earth.

Forget about that example. Take Pluto's orbit or the Earth-Moon-Sun system. In both cases the acceleration felt by each object is non-uniform (the distance between Pluto and Sun ranges from 4.3 to 7.3 billion km, during a Sun eclipse the force acting on the Moon differs significantly from the case of a Moon eclipse). However, both systems are well described by Newtonian gravity hence the retardation effect is almost null. I think the main reason is that, quantitatively, the gravitational radiation is extremely small. The Wikipedia article you've linked says that Earth loses about 300 joules as gravitational radiation from a total of 2.7 x 10^33 joules.

If you really think it's plausible that GR predicts Earth can "extrapolate" the motions of Mars in this situation which obviously departs significantly from spherical/cylindrical symmetry, perhaps we should start a thread on the relativity forum to get confirmation from GR experts over there?

I'll do that.

You're right that I don't have a rigorous argument, but I'm just using the following intuition--if you know the current position of an object moving at constant velocity, how much calculation would it take to predict its future position under the assumption it continued to move at this velocity? How much calculation would it take to predict the future position of an object which we assume is undergoing uniform acceleration? And given a system involving many components with constantly-changing accelerations due to constant interactions with each other, like water molecules in a jar or molecules in a brain, how much calculation would it take to predict the future position of one of these parts given knowledge of the system's state in the past. Obviously the amount of calculation needed in the third situation is many orders of magnitude greater than in the first two.

If my analogy with gravity stands (all kinds of motions are well extrapolated in the small mass density regime), the difference in complexity should be about the same as between the Newtonian inverse square law and GR.

No. If our predictions don't "miserably fail" when we ignore Mars altogether, they aren't going to miserably fail if we predict the Earth is attracted to Mars' current position as opposed to where GR says it should be attracted to, which is not going to be very different anyway since a signal from Mars moving at the speed of light takes a maximum of 22 minutes to reach Earth according to this page (http://www.aerospaceweb.org/question/astronomy/q0254.shtml). Again, in the situations where GR and Newtonian gravity give very different predictions, this is not mainly because of the retarded vs. current position issue.

See my other examples above.