Graduate Bohr's solution to the EPR paradox

[PeterDonis]

So Bob (had he done X) would indeed have gotten $B^+ = +1$ with certainty.

But this only holds for the special case of Alice's and Bob's angles being exactly the same. (Or exactly opposite, in the EPR/Bell case.) There are infinitely many other possible cases, none of which your argument applies to.

The problem here appears to be that you are taking "If Bob had done X" too narrowly, possibly because that counterfactual was phrased in this thread too narrowly. It should be "If Bob had done Z", where Z is any other experimental setting that Bob could have chosen--in this case, any other angle Bob could have chosen to measure the spin at, besides the one he actually chose. A "realist" interpretation of QM, as I understand the usual interpretation of that term, would require all of those "If Bob had done Z" counterfactuals to have definite, well-defined answers, not just the particular one "If Bob had done X" in which Bob chose the exact measurement angle for which QM does happen to predict a particular result for him with certainty.

 

[Zafa Pi]

So Bob (had he done X) would indeed have gotten B+=+1B+=+1B^+ = +1 with certainty. QED per QM.

I didn't follow @PeterDonis at post#31, so I'll put in my 2cents worth.
OK, we let X be Aspect's experiment with a = b, so if Alice and Bob both do X then they will each get + or each get - with probability ½ for either case. So my 2) in post#16 is satisfied.
Alice does X and gets + while Bob times his pulse and gets 68. Now we ask what if Bob did X instead? Well that wasn't the experiment that was done. So let's do the experiment where Bob does X instead. That's a new experiment, and you say Alice will still get +. But why is that? Why not both getting -? I can tell you what a classical physicist would say (and I did), but I'm not one and don't believe in realism and neither did Bohr. I feel like I'm repeating myself; are you getting bohred?

 

[N88]

I didn't follow @PeterDonis at post#31, so I'll put in my 2cents worth.
OK, we let X be Aspect's experiment with a = b, so if Alice and Bob both do X then they will each get + or each get - with probability ½ for either case. So my 2) in post#16 is satisfied.
Alice does X and gets + while Bob times his pulse and gets 68. Now we ask what if Bob did X instead? Well that wasn't the experiment that was done. So let's do the experiment where Bob does X instead. That's a new experiment, and you say Alice will still get +. But why is that? Why not both getting -? I can tell you what a classical physicist would say (and I did), but I'm not one and don't believe in realism and neither did Bohr. I feel like I'm repeating myself; are you getting bohred?

PeterDonis' post #31 is fine with me. But there is an error in your analysis of what I wrote. I did not say: "In a new experiment Alice will still get +1."

You are right to say that they will get (with $P = \frac{1}{2}),$ +1 together of -1 together in a new experiment.

PS: Giving you a good combo of Einstein-locality and Bohr-realism, I believe in commonsense local realism (CLR), the union of local-causality (no causal influence propagates superluminally) and physical realism (some physical properties change interactively).

CLR is compatible with the principles of local realism in d’Espagnat (1979:158) and endorsed by Bell (1980:7): (i) realism = regularities in observed phenomena are caused by some physical reality whose existence is independent of human observers; (ii) locality = no influence of any kind can propagate superluminally; (iii) induction = legitimate conclusions can be drawn from consistent observations.

 

[ueit]

When someone who is not equipped with math or physics asks me what is it about QM that makes it so different from the old physics, this is what I give them:

Let us suppose that:
1) Alice and Bob are isolated from one another, so that no communication or influence can pass between them and neither knows what the other is doing.
2) If Alice and Bob both perform experiment X they will get the same result.
3) Alice performs experiment X and gets value 0, while Bob performs experiment Y and gets 1.
Then
4) If Bob had performed X instead of Y would he have necessarily gotten 0?

Classical physics says yes and quantum physics says no.

With classical physics we know that the reality facing Alice is unaffected by what Bob does, so she would still have had to get 0 if Bob did X instead, and thus yes, Bob must get 0 because of 2).
This type of reasoning leads to conclusions (e.g. something called Bell's inequality) that are contradicted by QM and experiment.Classical physics, in general, accepts local realism. If in a given experiment Alice performs experiment X and gets a value 0, then because of 1) (= locality), and due to the reality facing Alice as she does the experiment (= realism) it doesn't matter whether Bob had performed experiment X or Y. Hence if he had done X, because of 2) he must also get 0. Or, if Bob had done X he would have gotten some value b (= CFD) and because of 2) b = 0.
N.B. local realism is sufficient to derive Bell's inequality which has been shown to be experimentally false.

From the point of view of QM, the experiment that was performed was Alice doing X getting 0, while Bob doing Y getting 1. The experiment of Alice doing X while Bob doing X was not performed. Thus we cannot say that Alice would necessarily get 0 if Alice and Bob both did X.
This is exactly the objection of Bohr in the EPR "paradox" that @PeterDonis, @stevendaryl, and I were talking about in previous posts above.

In classical physics Alice's and Bob's actions as well as the properties of the particles and measurement results are uniquely determined by the initial state. If Bob does X you need an initial state, let's call it Bx. If Bob does Y you need a different initial state, say By.

In order to your above argument to hold you need to show that there exist such initial states Bx and By that only have consequences for Bob. Otherwise it might be the case that changing the initial state from Bx to By also changes what Alice is doing or the spin itself. And now it is important to say what classical theory you have in mind. If you use a theory without long-range interactions (billiard balls) finding the required initial states is trivial (you just re-position the "balls" Bob is made of). On the other hand if you use a field theory with unlimited range (classical electromagnetism for example) a change of the particles' configuration at Bob's location in the past will necessarily affect Alice as well.

Andrei

 

[DrChinese]

CLR is compatible with the principles of local realism in d’Espagnat (1979:158) and endorsed by Bell (1980:7):..

Perhaps you mean that the DEFINITION of CLR is as you describe. Certainly neither d’Espagnat nor Bell were advocates of anything like what you call CLR, since it's ruled out by Bell's Theorem.

 

[N88]

Perhaps you mean that the DEFINITION of CLR is as you describe. Certainly neither d’Espagnat nor Bell were advocates of anything like what you call CLR, since it's ruled out by Bell's Theorem.

I'd welcome a fuller explanation. For example, what part of my definition is ruled out by the three d'Espagnat/Bell definitions?
Also, is any part of my definition ruled out by QM? Thanks.

 

[bhobba]

After reading this thread I feel compelled to emphasize, as I usually do for such discussions, that all Bell, EPR etc is, is a correlation - that's it, that's all. What Bell's famous theorem simply says is you have different kinds of correlations in QM than classical physics. Want the same kind of correlations as classically - then you need FTL. How you react to that is your business, we have all sorts of positions. My position is very simple - who cares - you have a different kind of correlation in QM - big deal. But that's just me.

Thanks
Bill

 

[Zafa Pi]

In classical physics Alice's and Bob's actions as well as the properties of the particles and measurement results are uniquely determined by the initial state. If Bob does X you need an initial state, let's call it Bx. If Bob does Y you need a different initial state, say By.

In order to your above argument to hold you need to show that there exist such initial states Bx and By that only have consequences for Bob. Otherwise it might be the case that changing the initial state from Bx to By also changes what Alice is doing or the spin itself. And now it is important to say what classical theory you have in mind. If you use a theory without long-range interactions (billiard balls) finding the required initial states is trivial (you just re-position the "balls" Bob is made of). On the other hand if you use a field theory with unlimited range (classical electromagnetism for example) a change of the particles' configuration at Bob's location in the past will necessarily affect Alice as well.

Andrei

Good point. Don't you think my "1) Alice and Bob are isolated from one another, so that no communication or influence can pass between them and neither knows what the other is doing. " is adequate to take care of your concern? If not do you have a suggestion of how it might be fixed?

What I'm trying to do is somehow distinguish QM and classical in the elementary way, i.e. no technical terms.

 

[ueit]

Good point. Don't you think my "1) Alice and Bob are isolated from one another, so that no communication or influence can pass between them and neither knows what the other is doing. " is adequate to take care of your concern? If not do you have a suggestion of how it might be fixed?

What I'm trying to do is somehow distinguish QM and classical in the elementary way, i.e. no technical terms.

Yes, your premise (1) is adequate. The problem is that it is not true for some classical theories (field theories like electromagnetism or general relativity). So, you either drop that premise and try to deal with all classical theories or keep it but specify that your argument only works for obsolete, old theories without long-range interactions.

You need to keep in mind that distinguishing classical from quantum mechanics might be an impossible task. There are theoretical (even if only partial) results suggesting that the two frameworks might be equivalent. In this paper:

Duality Between a Deterministic Cellular Automaton and a Bosonic Quantum Field Theory in 1+1 Dimensions
Gerard ’t Hooft, Found Phys (2013) 43:597-614

https://arxiv.org/pdf/1205.4107.pdf

't Hooft argues that:

We claim that, in spite of its quantum mechanical appearance, the Standard Model may also be a deterministic system in disguise. The author is aware of the fact that most of his readers will not be prepared to jump so such a conclusion; in particular those who have Bell’s inequaities[3][4] in mind will not be inclined to accept the idea. But then we would invite the reader to study the transformation presented in this paper. Here, we display how the mapping goes for non-interacting, massless bosons in one space- and one time dimension.

Another paper arguing for QM as being an effect of the interaction between particles and a hypothetical EM field (as it is postulated by the theory of stochastic electrodynamics) can be found here:

Pilot Wave Steerage: A Mechanism and Test

A. F. Kracklauer, Found. Phys. Lett. 12(5) 441-453 (1999)

https://arxiv.org/pdf/quant-ph/9711013.pdf

Andrei

 

[Boing3000]

Want the same kind of correlations as classically - then you need FTL.

Although I totally agree with you, I think that FLT in not a concept equivalent to non-locality.

What I'm trying to do is somehow distinguish QM and classical in the elementary way, i.e. no technical terms.

"Elementary" is indistinguishable from "classical". I still have a hard time understanding in which way QM is non-classical.
The non-technical term for non-locality is "now" and "everywhere". It clearly isn't FLT.

 

[stevendaryl]

I would say it becomes a failure to test your prediction.

I guess the way I view a "theory of physics" is that it makes statements along the lines of "in situation X, outcome Y will happen". The theory is agnostic about whether situation X ever did happen or ever will happen. So a theory of physics allows us to ask "what if?" questions. That's pretty important in science, because if the theory predicts that in situation X, we'll all die, then we know to avoid situation X. The counterfactual statement "If X then Y" is important, even when X never arises. We don't need to actually drink nitric acid for the statement "If you drink nitric acid, you'll probably die" to be meaningful.

A statement of the form "If X then Y" is not directly testable unless X actually happens, but if it is a consequence of a theory that does have a lot of empirical support, then we tend to give it credence.

 

[DrChinese]

I'd welcome a fuller explanation. For example, what part of my definition is ruled out by the three d'Espagnat/Bell definitions?
Also, is any part of my definition ruled out by QM? Thanks.

The definition of CLR is not the question. QM, Bell, etc don't have any material quibble with the definition itself. Any more than QM or Bell has anything to say about the definitions of fairies and centaurs. It is the substance of what's defined that is at issue. That is ruled out by Bell's Theorem.

 

[Elemental]

There is something that has bothered me for a long time about Bohr’s response, and 80+ years after his paper this looks like a good thread to bring it up.

In Bohr’s example two particles, whose initial states are known, pass through separate slits of known separation in a diaphragm. This entangles the states of particles A and B. If the diaphragm is free to move along the axis in the plane of the diaphragm perpendicular to the slits, one can measure the momentum the diaphragm receives due to interaction with the particles and deduce the sum of their momenta along that axis, but not their positions. Immediately after passing through the slits, the position of particle A is measured, say, by being absorbed in a photographic plate. For this to be an actual measurement, the photographic plate must be firmly attached to the rest of the laboratory apparatus to thereby establish the coordinate system in which the measurement is made.

According to the entangled state that Einstein, et. al., put forth in their EPR paper, (1) the above measurement makes particle B’s position definite but leaves its momentum indeterminate, whereas (2) if instead the momentum of particle A had been measured, it is particle B’s position that becomes indeterminate.

Sticking with scenario (1), Bohr states that we have allowed an uncontrollable amount of momentum to be transferred to the measurement apparatus by measuring A's position, precluding any procedure which could precisely determine B’s momentum as per the uncertainty relation ΔpΔq > ħ.

Particle B’s momentum might be measured by allowing it to bounce off of another diaphragm after passing through its slit. This second diaphragm is attached to a spring and is free to move along the axis of interest (klutzy, just keeping it simple). B’s momentum is then ascertained from the recoil of the spring. It is known that in elastic collisions between unequal masses, momentum transfer is incomplete. What bothers me is, at least naively, this appears to allow a more precise measurement of particle B’s momentum than predicted by the uncertainty principle.

Doing some (non-relativistic) math, assume particle B has initial momentum p_i=mv_i (to be measured) and the measurement apparatus with which it collides has initial unknown momentum P_i=MV_i (which it acquired as “kick-back” from measuring particle A’s position; we may assume M is known from weighing the apparatus beforehand). At point of maximum recoil of the spring, their shared velocity (from conservation of momentum) is

V_s = (mv_i + MV_i)/(m + M)

At this point, part of the kinetic energy from m and M’s combined motion is stored in the spring. The amount K_s of stored energy is measured through knowledge of the spring constant and observation of the spring displacement. Also, from energy conservation, K_s is equal to the difference between the kinetic energy that M plus m had while traveling separately and their kinetic energy when traveling together with equal velocity at the point of maximum spring displacement. That is

K_s = ½(mv_i² + MV_i²) - ½ (m + M)V_s².

Combining these two and solving the resulting quadratic equation for v_i one obtains.

v_i = V_i + [2K_s(m + M)/mM]^1/2

(The other root, V_i - [2K_s(m + M)/mM]^1/2, is ruled out since no collision occurs if particle B is moving away from the measurement diaphragm.) The only unknown is V_i on the right hand side. Finally, to obtain particle B’s momentum after passing through the slit, just multiply this result by m:

p_i = mv_i = mV_i + m[2K_s(m + M)/mM]^1/2

Again, everything here is known except for V_i, but since V_i is small (because the mass M of the measurement apparatus is large so only a relatively small velocity is obtain from the “kick” of measuring particle A’s position), and it is multiplied by an also very small m (e.g., electron mass), the total uncertainty in p_i can be made arbitrarily small.

In particular, if particle A’s position is measured with an accuracy of Δq, the uncertainty relation gives the measurement apparatus a minimum momentum uncertainty of ΔP_i = MΔV_i = ħ/Δq. But the uncertainty from the forgoing calculation for particle B’s momentum is just mΔV_i, which is a factor of m/M smaller than ħ/Δq ! Thus, we beat the uncertainty relation?

Something is wrong here, and I’m not sure what it is. One possible idea is that in undertaking the measurement of B’s momentum, we are not really sure where, in momentum space, the origin of our scale is. Perhaps, since the measurement apparatus has acquired an unknown momentum whose magnitude is approximately ħ/Δq, we need to shift our entire result by that much, but we know not whether to add or subtract it?

Any comments welcome,
Elemental

 

[bhobba]

I still have a hard time understanding in which way QM is non-classical.

Its not classical probability theory - its the next simplest generalization:
https://arxiv.org/abs/1402.6562

Thanks
Bill

 

[Simon Phoenix]

I still have a hard time understanding in which way QM is non-classical.

There are, I think, quite a few ways to appreciate the 'non-classical' nature of QM. Bill has given an excellent one which may well turn out to be the most fundamental and useful way of seeing the difference.

Another way might be to consider the notion of distinguishability and how that differs within the frameworks of the two theories. In classical mechanics we can represent the state of a particle at some time $t$ as an abstract point in phase space. For a single particle we need 6 coordinates, 3 for position and 3 for momentum. We can consider another point in phase space that is very close to this - it represents a different state for our classical particle. There is nothing in the classical framework that would suggest we cannot distinguish these states from one another however close they may be. We might not be able to currently experimentally distinguish points that are very close but this would be attributed to imperfect technique or devices and the expectation is that with sufficient refinement (that is, a better experiment) we would be able to distinguish between two points in a classical phase space.

Contrast this with the situation in QM where distinguishability is characterized by orthogonality. In QM (pure) states are represented as rays in an abstract complex Hilbert space. Two states that are 'close' together cannot, in principle, be distinguished, except with a certain probability dependent on their overlap. Two orthogonal states have zero overlap and can be distinguished with unit probability. There is no experimental 'refinement' we can make to overcome our inability to perfectly distinguish two non-orthogonal states in QM; it's a limit imposed by nature, not by our technology or ingenuity (or lack of).

There's a danger with this picture though. We might, naively, then think something along the lines of "Oh well, if that's the case, all QM is just classical physics where we build in this limitation using a stochastic approach". So we might think of a trajectory in (classical) phase space as being a tube rather than a curve, to account for this limitation on distinguishability. Whilst there is some merit in these kinds of approaches they don't really properly work - for reasons that are outlined in the article Bill linked to. QM isn't just classical physics with some probability stuff tacked on.

In some sense it all boils down to the difference between how the composition of conditional probabilities is handled in QM and classical physics. Feynman gave a really nice discussion of this in the introduction to his classic path integral paper. If we imagine some system to start off in some state $A$ and finish up in some state $C$ and we further imagine that it can get to $C$ by some collection of intermediate states $B_k$ then the usual (classical) law for the composition of probabilities would say that the probability of being in state $C$, given that we started in some state $A$ is given by the composition rule $$P(C | A) = \sum_k P(C | B_k ) P (B_k | A)$$However, in QM this composition rule is applied to amplitudes and not to probabilities so that we have $$ \langle C | A \rangle =\sum_k \langle C | B_k \rangle \langle B_k | A \rangle $$ and to get the conditional probability $P(C | A)$ we must form $| \langle C | A \rangle |^2$

Now we can see that in the QM case we're going to have the possibility of interference - interference between the different paths on the probability tree, if you like. If we try to answer the question (by experiment) "which path on the probability tree did it go on?" then we recover the classical composition rule. So we can see that QM 'contains' the classical case - but is much more general.

So personally I see a big difference between QM and classical approaches. Maybe others do not and certainly one can formalize the hell out of things and say that QM and CM have the same mathematical algebraic structure but it's just that QM has non-commuting variables - but I have never found much useful insight in that. For me the real mystery is in why nature is like this - but again many would argue that this is not the job of physics which is simply to cook up some formalism that accurately predicts observations. I find this latter position a little too prosaic for my tastes

 

[bhobba]

For me the real mystery is in why nature is like this - but again many would argue that this is not the job of physics which is simply to cook up some formalism that accurately predicts observations. I find this latter position a little too prosaic for my tastes

For fundamental theories its ultimately impossible to answer. Why is nature like that? You rack your brains like Einstein did about gravity and you get a more fundamental answer - its not Newtons law - it's because nature has no prior geometry at the space-time level. Its very intuitive - it would be rather strange if nature did single out a particular geometry - but a priori there is no reason for it to be true. Nature just is like that. And that is the sad truth about why questions in science - it doesn't matter how much you rack your brain to answer it, and you may come up with one so profound you get a Nobel, it always has assumptions where you have to say - nature is just like that - and we know its like that from observation.

I always post what Feynman said because, while short and pithy it simply can't be said better:


The most that can be done is express it in such a way it, like no prior geometry, seems intuitive and obvious. So in QM what is the equivalent? Its simply this - in the framework of generalized probability theories physically you would want the so called pure states to be continuous. That is if a system is represented by a pure state you would expect if it was in some state at time t1 and another at time t2 it would go through some other state. Seems sort of obvious - but just like no prior geometry you are making all sorts of assumptions - eg why should nature be a generalized probability model rather than deterministic? We don't know - its just how nature is. We have all sorts of conjectures but no experimental way to decide. So again in a certain sense it is a self defeating question.

I don't generally like quoting philosophers but many would side with Wittgenstein - Whereof one cannot speak, thereof one must be silent.

You can never get ultimate answers to why questions - all you can get is more 'intuitive' ones.

Thanks
Bill

 

[Simon Phoenix]

I always post what Feynman said because, while short and pithy it simply can't be said better:

Yes that's a marvellous lecture, and one of my favourites. Worth watching in its entirety. I love the way Feynman just seems to effortlessly cut through the BS. I totally agree with everything you say above. There is, ultimately, no ultimate 'why'

The fact that searching for a deeper 'why' is (eventually) a futile endeavour should not deter one from the quest I feel - but I guess it would depend on whether we think we've hit the brick wall beyond which all we can say is "it just is".

In the case of QM I feel we've still got a way to go - at least I hope so, because despite the prevalent view on these forums that QM isn't 'weird' and that this is just the product of those naughty popularisers, I think there's something delightfully bonkers going on that I want to understand a bit better.

Trying to pin down QM reminds me of the (probably) apocryphal story of the professor who, in the middle of a complex and difficult lecture, uttered the dreaded phrase "and it is obvious that . . ." and proceeded to write down the result. One brave soul put their hand up and asked "why is it obvious?" The professor, taken aback, spent the next 15 minutes furiously scribbling equations on a corner of the blackboard and muttering to himself. Eventually he turned round and triumphantly stated "Ah yes, it is obvious!" and then continued with his lecture.

 

[bhobba]

The fact that searching for a deeper 'why' is (eventually) a futile endeavour should not deter one from the quest I feel - but I guess it would depend on whether we think we've hit the brick wall beyond which all we can say is "it just is".

Most definitely not. I will repeat it again for emphasis - you should never cease asking why. Like with Bell your answer may be a big advance. What is important is to realize what you are actually doing. You can never answer why - but what was it Murray Gell-Mann said about his 'augments' with Feynman about physics - we didn't resolve anything but we tweaked it around a bit .

Thanks
Bill

 

[Simon Phoenix]

but what was it Murray Gell-Mann said about his 'augments' with Feynman about physics - we didn't resolve anything but we tweaked it around a bit

Newton : If I have seen further than others it's because I have stood on the shoulders of giants
Gell-Mann : If I have seen further than others it's because everyone around me is a dwarf

 

[Elemental]

Bohr states that we have allowed an uncontrollable amount of momentum to be transferred to the measurement apparatus by measuring A's position, precluding any procedure which could precisely determine B’s momentum as per the uncertainty relation ΔpΔq > ħ.

Typo, I meant ΔpΔq = ħ/2, and carrying though the correction, the sentence toward the bottom should read, "But the uncertainty from the forgoing calculation for particle B’s momentum is just mΔVi, which is a factor of m/M smaller than ħ/2Δq!"

Anyway, the point I tried to make is that, sticking strictly to Bohr's argument that, "by allowing an essentially uncontrollable momentum to pass from the first particle into the mentioned support, however, we have by this procedure cut ourselves off from any future possibility of applying the law of conservation of momentum to the system consisting of the diaphragm and the two particles and therefore have lost our only basis for an unambiguous application of the idea of momentum in predictions regarding the behavior of the second particle," I don't obtain the requisite uncertainty in the second particle's momentum.

Most of the unknown momentum the apparatus receives from measuring the first particle's position stays in the apparatus, is not transferred to the second particle, and so seemingly should not interfere substantially with measuring the second particle's momentum. A similar objection could be raised using angular momentum for versions of the experiment involving spin correlation (successfully carried out by Aspect and many others and confirming Bell's inequality). I'm not disputing Bell's inequality, just questioning the argument that it can be explained solely on the basis of the disturbance of the measurement apparatus from the initial measurement on one of the particles whose quantum correlations are being investigated.

 

[Zafa Pi]

You can never get ultimate answers to why questions - all you can get is more 'intuitive' ones.

Why is that?

 

[Zafa Pi]

Newton : If I have seen further than others it's because I have stood on the shoulders of giants
Gell-Mann : If I have seen further than others it's because everyone around me is a dwarf

Dwarf : Hey guys I found the car keys here on the ground, we can go now.

 

[bhobba]

Why is that?

Because it doesn't matter what answer you come up with it assumes some things exactly as post 47 said.

Thanks
Bill

 

[Zafa Pi]

Because it doesn't matter what answer you come up with it assumes some things exactly as post 47 said.

Thanks
Bill

Thanks Bill, you just answered my why question.

 

[bhobba]

Thanks Bill, you just answered my why question.

Good one

Thanks
Bill

 

[Zafa Pi]

Yes, your premise (1) is adequate. The problem is that it is not true for some classical theories (field theories like electromagnetism or general relativity). So, you either drop that premise and try to deal with all classical theories or keep it but specify that your argument only works for obsolete, old theories without long-range interactions.

In the usual manner that Bell's inequality is explicated, Alice and Bob are 2 light minutes apart with Eve half way between them and simultaneously send a light signal to each of them. Then Alice and Bob do their experiments within 30 seconds. Thus my 1) is satisfied during their experiments even for classical field theories since for those theories any communication takes place at less than or equal to light speed.

What I was trying to capture in my post #29 was a nontechnical way of describing which assumption QM refutes in the classically derived Bell inequality.

 

[ueit]

In the usual manner that Bell's inequality is explicated, Alice and Bob are 2 light minutes apart with Eve half way between them and simultaneously send a light signal to each of them. Then Alice and Bob do their experiments within 30 seconds. Thus my 1) is satisfied during their experiments even for classical field theories since for those theories any communication takes place at less than or equal to light speed.

What I was trying to capture in my post #29 was a nontechnical way of describing which assumption QM refutes in the classically derived Bell inequality.

The speed of communication is not the issue here. Eve, just like Alice Alice and Bob is a collection of field sources (electrons and nuclei are field sources in classical EM). Therefore what Eve does is a function of position momenta of all particles (including those in Bob and Alice), what Bob does is a function of Alice and Eve and what Alice does is a function of Bob and Eve. There is no way you can make them behave independently while still obeying the equations of classical EM.

 

[Zafa Pi]

The speed of communication is not the issue here. Eve, just like Alice Alice and Bob is a collection of field sources (electrons and nuclei are field sources in classical EM). Therefore what Eve does is a function of position momenta of all particles (including those in Bob and Alice), what Bob does is a function of Alice and Eve and what Alice does is a function of Bob and Eve. There is no way you can make them behave independently while still obeying the equations of classical EM.

Let me get this straight. You are saying that given the setup I described in post #56 what Alice does in her experiment can affect Bob's results in his experiment, under the assumption of classical EM theory. That doesn't violate locality for classical theory?

 

[Denis]

Its not classical probability theory - its the next simplest generalization:
https://arxiv.org/abs/1402.6562

Probability theory is simply logic - the logic of plausible reasoning. Read Jaynes about this. There is only one logic. It makes no sense to generalize logic.

Some "generalized logic" as well as "generalized probability theory" may be some nice mathematical exercise. But it does not change classical logic, or classical probability theory. Usually, behind such "generalizations" is simply a bad, misleading use of words. If one reintroduces appropriate words, the conflict with logic resp. probability theory disappears.

The classical example is the reintroduction of the "exists" of classical logic into intuitionism by defining it as "not for all not". It appeared that the intuitionistic "exists" was simply a different thing than the classical "exists".

Some similar bad naming is probably the cause of "quantum logic" and "quantum probability theory". One clear candidate of confusing naming is "measurement" for something which is the outcome of some interaction.

If one, instead, really "generalizes" logic or probability theory, one simply ends in logical contradictions (of course, contradictions of classical logic). Some sort of "generalization" of the scientific method which covers also astrology.

 

[Zafa Pi]

Probability theory is simply logic - the logic of plausible reasoning. Read Jaynes about this. There is only one logic. It makes no sense to generalize logic.

This is amazing, since I was just about to comment on the same post by @bhobba.
I have seen many comments that say QM requires a new or different kind of probability theory. This is not true and perhaps not what they mean. For example, urn models with or without replacement, or Bose-Einstein v Fermi-Dirac, may require different sample spaces with different probabilities, but all of them are using the usual classical probability theory, as say given in Feller.