# crowded quantum information

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1 hour ago, uncool said:

...if Bob could determine whether Alice has made a measurement, then Alice could send a message faster than light purely by choosing whether to perform measurements, if I'm understanding what you're saying correctly.

That sounds right. I wasn't very assertive about it, though:

On 9/10/2022 at 1:15 AM, joigus said:

He may be able to devise a clever interference experiment though, to determine that someone's been messing with the state, because coherence has been broken.

How do you devise such an interference experiment? I was thinking about this for a while, and couldn't come up with any way to do that. The question boils down to: How do you know a beam of particles that give you random spin projections is in a pure state or a mixture state? On second thought, I don't think you can, unless you gather information of all the parts. But then the superluminal character is no longer an issue.

8 minutes ago, MigL said:

I think you hould quote all of Joigus' post

Yes. Thank you for pointing that out. We almost x-posted.

I would like to think about that a bit longer...

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2 hours ago, uncool said:

...if Bob could determine whether Alice has made a measurement, then Alice could send a message faster than light purely by choosing whether to perform measurements, if I'm understanding what you're saying correctly.

OK. I know where the trouble is. Yes, Bob could in principle determine whether a measurement has been performed. Mind you, I say "in principle." Whether there is any viable way to do that is another matter. But this change in the state is subject to propagation as the interaction it actually is, and is therefore subject to delay, as any other actual interaction. IOW, if you really "touch" the system, the influence will propagate causally. But Bob would learn about it no sooner than d/c, where d is the distance between him and Alice in their inertial system. I didn't say Bob would know it instantly. This makes sense, doesn't it?

Don't forget that Einstein's criterion for what an "element of reality is" means:

(from the famous EPR paper that gave rise to the whole debate: https://cds.cern.ch/record/405662/files/PhysRev.47.777.pdf)

Now, what's troubling you, I think, is that we've heard these terms "non-locality", "superluminal action at a distance", "spooky" this or that, that we can't help but think that every time information is obtained from a quantum system, the question of non-locality is at stake. This is not the case when coherence loss is the issue. Without realising it, you're assuming this influence must be superluminal, or instantaneous. Why? It doesn't. It's an actual interaction!

Pay attention to the words "without in any way disturbing a system" in EPR's paper, because they're essential. The verification that somebody has performed a measurement on the other particle would propagate at a finite speed, because the observable, if you will, "whether someone has performed a measurement" does involve an interaction. So it is not what Einstein called an element of reality. It is only on those elements of reality (values of which information can be obtained without in any way disturbing the system) that the question is, at least, at stake.

This part is a bit more technical:

The evidence of that interaction would be reflected in what we call the density matrix, which essentially consists of statistical weights: the probabilities 1/2, 1/2 that I mentioned before (and constitute the diagonal elements of the density matrix: those would not change) plus the relative phases between the states "up" and "down" (which constitute the non-diagonal elements: those would change). But those coherences are a part of the wave function that is affected by the interaction, and they would take some time to reach Bob. Nevertheless, after enough time has passed, Bob would be able to tell (in principle, I repeat) that someone's been messing with the state. Nothing more. In fact, he wouldn't be able to tell whether it was Alice, or a gust of wind, or what the output of the experiment was unless he measured the same projection of spin they agreed to measure.

--- (end of the more technical comments).

The point is perhaps subtle. But the take-home lesson is: Not every time a leaf falls to the ground is locality at stake.

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20 minutes ago, joigus said:

But Bob would learn about it no sooner than d/c, where d is the distance between him and Alice in their inertial system. I didn't say Bob would know it instantly. This makes sense, doesn't it?

Yes, that makes sense; I had read your statement as "So he could determine in principle that Alice has performed a measurement [instantly], but he wouldn't be able to tell which outcome Alice got until he made the measurement", because the second half of the statement was qualified by timing while the first half was unqualified. Sorry about that.

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2 hours ago, uncool said:

Yes, that makes sense; I had read your statement as "So he could determine in principle that Alice has performed a measurement [instantly], but he wouldn't be able to tell which outcome Alice got until he made the measurement", because the second half of the statement was qualified by timing while the first half was unqualified. Sorry about that.

No problem. Part of the reason why this is all very confusing is because so much nonsense has been spread for so long that we think these theorems, as well as the experiments, say what they actually don't. The abstract nature of quantum mechanics makes it quite difficult --if not impossible-- to get any intuitive picture of the part of it that's really bizarre. So we forget that another part of it is just propagation of waves.

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12 hours ago, joigus said:

Then you absolutely have missed the point of what's going on, and are still living in the confusion. This is what Murray Gell-Mann calls the "widespread foolishness associated with the EPR effect."

The EPR effect was first demonstrated to be invalid in the early sixties by the experiments performed by Bell and Aspect. I think I understand what you and Gell-Mann are saying but I can’t say I agree with either. I will try to point out items where I find we disagree.

12 hours ago, joigus said:

Go back to the example of the gloves that MigL was talking about. One glove goes to Australia and the other stays with me. I open the box and find out that it's LH. I thereby know immediately that hoola got the RH one. Would you think for a moment that one glove corresponds to the right hand and the other to the left hand because some "spooky action at a distance" has taken place between them? That's what's foolish to say. The gloves are perfectly anti-correlated just because the correlation was there from the beginning.

I can repeat the point over and over if you wish, but I can't make it any more clear.

That is perfectly clear so no need to repeat it.

The correlation with the gloves was there from the beginning, but with entangled particles, their quantum identities are indeterminate until the first measurement is made. Some say they are in a state of superposition prior to the first measurement so the glove analogy does not apply. I thought we had agreed on this.

12 hours ago, joigus said:

The story of the argument is somewhat contorted, because Einstein thought of a different example, with position and momentum, and then David Bohm proposed one with spins. But you got the story wrong too.

That's for another post, though.

As I recall, Niels Bohr claimed there were no hidden variables while Einstein claimed there were and Bell’s inequality was a test to decide between the two and Bell’s results favored Bohr’s view.

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14 minutes ago, bangstrom said:

The correlation with the gloves was there from the beginning, but with entangled particles, their quantum identities are indeterminate until the first measurement is made. Some say they are in a state of superposition prior to the first measurement so the glove analogy does not apply. I thought we had agreed on this.

The individual states are indeterminate, but the correlation is there.

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13 hours ago, joigus said:

4 hours ago, joigus said:

The evidence of that interaction would be reflected in what we call the density matrix, which essentially consists of statistical weights: the probabilities 1/2, 1/2 that I mentioned before (and constitute the diagonal elements of the density matrix: those would not change) plus the relative phases between the states "up" and "down" (which constitute the non-diagonal elements: those would change). But those coherences are a part of the wave function that is affected by the interaction, and they would take some time to reach Bob. Nevertheless, after enough time has passed, Bob would be able to tell (in principle, I repeat) that someone's been messing with the state. Nothing more. In fact, he wouldn't be able to tell whether it was Alice, or a gust of wind, or what the output of the experiment was unless he measured the same projection of spin they agreed to measure.

This makes no sense. QM does not care if the message is understood by Bob or when it arrives at his level. The loss of entanglement is a signal from one particle to its entangled partner and the partner receives the interaction instantly and non-locally and responds accordingly. Alice and Bobs perception of the timing may be delayed but it is irrelevant to what is happening at the particle level. The message is for the paired particle and not for Bob.

Edited by bangstrom
duplicate post I thought the other did not go through.
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30 minutes ago, bangstrom said:

As I recall, Niels Bohr claimed there were no hidden variables while Einstein claimed there were and Bell’s inequality was a test to decide between the two and Bell’s results favored Bohr’s view.

Kind of... What Bohr argued was that the very nature of the measurement process somehow produces the results themselves, in such a way that the statistical constrictions on the outcomes and the measurement process itself were inseparable. Analysis of measurement in later decades clarified that even if you have filtering measurements, preparations... (no intervention with an interacting piece of equipment) quantum correlations are there. So they're not due to intervention during the measurement.

The question of hidden variables really started with David Bohm (pro) and John Von Neumann (against).

It was David Bohm who, to the best of my knowledge, started to work on "hidden variables", and Von Neumann who started thinking deeply about this problem, and formulated a first version of a theorem of impossibility. Then came Gleason with a theorem about the impossibility of assigning a binary function (true/false) which was continuous on the Bloch sphere, and from there, the last --more powerful for some-- version of that is the Kochen-Specker theorem.

Bell's inequalities are a further refinement of an argument by Clauser, Horne, Shimony and Holt.

There is another argument of impossibility by Greenberger, Horne and Zeilinger, that Mermin simplified, with just one observable.

There is another argument by Conway and Kochen...

To this day, there's so much literature about the subject that it's possible to spend a lifetime's worth of study learning it.

28 minutes ago, bangstrom said:

The correlation with the gloves was there from the beginning, but with entangled particles, their quantum identities are indeterminate until the first measurement is made. Some say they are in a state of superposition prior to the first measurement so the glove analogy does not apply. Unlike with the gloves, the quantum identities are not fixed from the start. I thought we had agreed on this.

As Swansont said:

32 minutes ago, swansont said:

The individual states are indeterminate, but the correlation is there.

If you know some Pauli-matrix and angular momentum algebra, it's an interesting exercise to write down the singlet state and rotate it. People normally write it as

|up, down> - |down, up>

with a normalisation factor, and referred to the z direction. As if the z-direction played some kind of role in it. I doesn't. The states really are indeterminate. You can use any axis you want and it has the same form:

|upx, downx> - |upx, downx> = |upy, downy>- |downy, upy> = |upn, downn>- |downn, upn> = etc.

It's all a whole quantum state with like "no parts in it", "no internal arrows", so to speak. That's entanglement for you.

Try it, it's very illuminating.

The only thing that's physical is the whole vector. Then you obtain the expected value of spin along any direction you want and it always gives you zero for the sum.

Then you do some further quantum mechanical calculations and consider the evolution operator from, say, t=0 (when the singlet is prepared and the particles are next to each other) and a time T when the particles have come apart, and you will see that no expected value depends on the fact that the spatial factor of the states has taken them apart. It's all in the maths of QM.

As I said: The correlations are there when the singlet is prepared, they're there a minute later, they're there until you perform another measurement. And no experiment that I know of contradicts this.

So yes, it's like the gloves in the sense that the correlations are initial. No superluminal action at a distance. Period.

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3 hours ago, joigus said:

If you know some Pauli-matrix and angular momentum algebra, it's an interesting exercise to write down the singlet state and rotate it. People normally write it as

|up, down> - |down, up>

with a normalisation factor, and referred to the z direction. As if the z-direction played some kind of role in it. I doesn't. The states really are indeterminate. You can use any axis you want and it has the same form:

|upx, downx> - |upx, downx> = |upy, downy>- |downy, upy> = |upn, downn>- |downn, upn> = etc.

It's all a whole quantum state with like "no parts in it", "no internal arrows", so to speak. That's entanglement for you.

Try it, it's very illuminating.

I have read (not that I follow it) a paper by A.F. Kracklauer where expresses something similar where he claims entanglement can be explained as rotations through Q-bit space. I think rotations are looking at events in reverse. The initial orientation of the detectors is all that matters.

There is a speculation I find favorable that any electron can spontaneously entangle with any, and likely several, other electrons existing on the same light cone.

There is a precise orientation where one electron, and likely under the influence of other electrons nearby, can share a common Schroedinger wave function with another electrons such that strong interactions can occur with the result that remote electrons can function as if side by side no matter what the distance between them. That is entanglement.

Because the interaction is instant and non-local, there is no time between the signal and its reception to alter the orientation of a detected result so that it is anything other than anti-correlated.

Because the entangled electrons are found at different locations on the light cone, we observe a light related time delay. Our observations have a “space like” time delay because the observed events are simultaneous but separated by space rather than happening at different times, in which case the delay would be “time like.”

4 hours ago, joigus said:

As I said: The correlations are there when the singlet is prepared, they're there a minute later, they're there until you perform another measurement. And no experiment that I know of contradicts this.

So yes, it's like the gloves in the sense that the correlations are initial. No superluminal action at a distance. Period.

It is my understanding that the general consensus is that entangled particles are in a state of superposition prior to their first measurement, in which case, the correlations are not necessarily fixed at the moment of preparation. Rather, they are fixed a the the instant of first measurement.

I don”t favor the idea of superposition but I think it can be better explained without the magic.

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4 hours ago, bangstrom said:

I have read (not that I follow it) a paper by A.F. Kracklauer where expresses something similar where he claims entanglement can be explained as rotations through Q-bit space. I think rotations are looking at events in reverse. The initial orientation of the detectors is all that matters.

No. The singlet state is totally trivial under rotations. Rotations act on it trivially: they don't change it at all. That's why it looks like (up)x(down)-(down)(up) in any representation you choose. It does not code any orientation in it. The matrix that rotates it is the identity matrix. It is blind to rotations. I can try to rephrase this over and over... That's all I can do, I'm afraid.

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13 hours ago, joigus said:

No. The singlet state is totally trivial under rotations. Rotations act on it trivially: they don't change it at all. That's why it looks like (up)x(down)-(down)(up) in any representation you choose. It does not code any orientation in it. The matrix that rotates it is the identity matrix. It is blind to rotations. I can try to rephrase this over and over... That's all I can do, I'm afraid.

I think we are largely in agreement about the impossibility of rotating a signet state.

Swansont mentioned something about contemplating a rotation of the signet state so I mentioned Kracklauer’s paper where he described entanglement as a rotation through Q-bit space just to indicate that I had given rotation some thought but I also thought I made it clear that I did not share Kracklauer’s view.

On other matters, if I understand correctly, your view is that that there is nothing superluminal about entanglement? Could you explain?

Also, I mentioned that the general consensus about entanglement is that the particles involved are in a state of superposition prior to their first measurement. I qualified this statement with my view that it could be better explained.

I got some contrary opinions about this but a quick Google search should indicate that the main stream, hyper peer reviewed, everybody knows it, consensus of opinion is that entangled particles are in a state of superposition prior to their first measurement and this has been the case for many years.

I don’t always agree with Don Lincoln but his video below is one I can agree with from start to finish.

He explains why entanglement is superluminal, why the gloves/ balls in a box analogy is not valid, and rather than saying the particles are in superposition, he says their spin directions can be in any possible direction governed by their common wavefunction. This is essentially the same thing but better than saying they are in a superposition of both spin directions at the same tine.

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54 minutes ago, bangstrom said:

Swansont mentioned something about contemplating a rotation of the signet state

No, not me, and it’s singlet (as opposed to triplet)

Quote

This is essentially the same thing but better than saying they are in a superposition of both spin directions at the same tine.

Better is in the eye of the beholder. If you look at the wave function, it’s a superposition of the two states (|ud>-|du>)/sqrt(2)) so there’s nothing wrong with saying that.

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1 hour ago, bangstrom said:

I don’t always agree with Don Lincoln but his video below is one I can agree with from start to finish.

He explains why entanglement is superluminal, why the gloves/ balls in a box analogy is not valid, and rather than saying the particles are in superposition, he says their spin directions can be in any possible direction governed by their common wavefunction. This is essentially the same thing but better than saying they are in a superposition of both spin directions at the same tine.

On 9/13/2022 at 2:05 AM, joigus said:

Then you do some further quantum mechanical calculations and consider the evolution operator from, say, t=0 (when the singlet is prepared and the particles are next to each other) and a time T when the particles have come apart, and you will see that no expected value depends on the fact that the spatial factor of the states has taken them apart. It's all in the maths of QM.

And do QM calculations for ten years as a punishment for making the widespread foolishness that Gell-Mann talked about, even more widespread.

The gloves example serves the purpose of showing that initial correlations don't require spooky action at a distance. Nothing more. They differ --very importantly-- in that left or right-handedness, colour, material, etc., are well-defined at all times. Contrary to quantum mechanical systems, for which the mere assumption that these properties have a definite value would lead you to untenable assumptions like non-locality or existence of negative probabilities, or both.

Because we think quantum mechanics is correct, we don't need to assume such foolish things. Do you see my point?

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3 hours ago, joigus said:

And do QM calculations for ten years as a punishment for making the widespread foolishness that Gell-Mann talked about, even more widespread.

I have a lot of respect for Gell-Mann but two things made me suspect his views. One was that they appeared to be based on the long discredited EPR paper and mainly because I find non-locality to be a long established reality.

4 hours ago, joigus said:

The gloves example serves the purpose of showing that initial correlations don't require spooky action at a distance. Nothing more.

I can agree with that.

4 hours ago, joigus said:

They differ --very importantly-- in that left or right-handedness, colour, material, etc., are well-defined at all times. Contrary to quantum mechanical systems, for which the mere assumption that these properties have a definite value would lead you to untenable assumptions like non-locality or existence of negative probabilities, or both.

I don’t find that “well-defined at all times” applies to quantum identities which are an entirely different matter.

4 hours ago, joigus said:

Because we think quantum mechanics is correct, we don't need to assume such foolish things. Do you see my point?

I don’t see your point. What principles of QM make quantum identities constant at all times or rule out the possibility of non-locality? Have you heard of “identity swapping” where entangled particles non-locally swap identities? Quantum teleportation is an example.

Consider this possibility. One can generate two entangled particles A and B. Later they can generate two more entangled particles C and B. Then if they entangle particles B and C, the result is a four-way entanglement ABCD. As with single entanglements, the pairs are all anti-correlated.

A multiple ABCD entanglement is called a GHZ state named after the first experimenters to study multiple entanglements Greenberger, Horne, and Zeilinger. If one measures just the identity of particle D as spin-up, they can predict the identity of particle A as spin-down.

Since particles A and D are always anti-correlated, this means that particle A must change its identity half the time to conform to the later ABCD pairing.

This and several other experiments suggest that quantum identities are not fixed at the moment of their origin but they only become fixed at the first measurement. This is why entangled particles are considered to be in a state of superposition prior to observation. In QM, the cat is neither dead nor alive until examined.

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5 hours ago, bangstrom said:

Since particles A and D are always anti-correlated, this means that particle A must change its identity half the time to conform to the later ABCD pairing.

A doesn’t “change” its identity, since it doesn’t have one in the first place. Its spin is not determined until the measurement

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6 hours ago, bangstrom said:

I have a lot of respect for Gell-Mann but two things made me suspect his views. One was that they appeared to be based on the long discredited EPR paper and mainly because I find non-locality to be a long established reality.

No, they're not based on EPR. EPR published their paper hoping it would settle the question. They thought quantum mechanics is incomplete. Murray Gell-Mann thought otherwise.

It's not discredited. EPR was conceived to coin a concept that would be able to discern if quantum mechanics was right or he --and other critics-- were right.

Non-locality is not a long-established reality. It's sometimes actually used --wrongly, because many people do not understand what it means-- to discredit new ideas on the grounds that they would be non-local.

6 hours ago, bangstrom said:

I don’t find that “well-defined at all times” applies to quantum identities which are an entirely different matter.

Then you certainly don't understand the question. $$\frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\right\rangle -\left|\downarrow\uparrow\right\rangle \right)$$ independently of the space-time factor of the state. In fact, the space-time factor of the state is completely omitted. Don't you find that peculiar?

6 hours ago, bangstrom said:

I don’t see your point. What principles of QM make quantum identities constant at all times or rule out the possibility of non-locality? Have you heard of “identity swapping” where entangled particles non-locally swap identities? Quantum teleportation is an example.

OK. Let me stop you right there, because it is plainly obvious you don't understand quantum mechanics here. Quantum particles have no identity. They are indistinguishable, and they are in a way much more profound than macroscopic objects can be made extremely difficult to tell apart. Not even Nature "knows" which electron is which. There is no "which electron." They're just instantiations of a quantum field. It's actually more profound than instances of a computer program, for example, which have a process tag and a time stamp. Electrons have no tags.

GHZ in its original form is about three particles, and the GHZ state is $$\frac{1}{\sqrt{2}}\left(\left|\uparrow\uparrow\uparrow\right\rangle -\left|\downarrow\downarrow\downarrow\right\rangle \right)$$. The observable to measure here is $$\sigma_{x}\left(1\right)\sigma_{x}\left(2\right)\sigma_{x}\left(3\right)$$. You can extend that to more than three particles, and I'm sure people have been busying themselves doing that.

But again, there is no mystery but the mystery of quantum mechanical correlations. It all comes from a local conservation law, which is conservation of angular momentum. In this case, spin angular momentum. The GHZ observable is a diagonal (eigenvale-to-eigenvalue) function of total x-component of angular momentum, which is locally conserved.

Edited by joigus
minor correction
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10 hours ago, swansont said:
16 hours ago, bangstrom said:

Since particles A and D are always anti-correlated, this means that particle A must change its identity half the time to conform to the later ABCD pairing.

A doesn’t “change” its identity, since it doesn’t have one in the first place. Its spin is not determined until the measurement

That is a valid point but when A was entangled with AB its spin could have been either up or down.

But when the particles are in a three way entanglement D-CB-A and particle D is observed to have a spin-down, that collapses the wave function for all the particles which fixes the observation of A to a predictable spin-up.

This is why your gloves in boxes model does not work for entangled particles. Gloves have a fixed identity before observation but entangled particles do not.

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10 minutes ago, bangstrom said:

That is a valid point but when A was entangled with AB its spin could have been either up or down.

Yes, and we don’t know which state it’s in once entangled

10 minutes ago, bangstrom said:

But when the particles are in a three way entanglement D-CB-A and particle D is observed to have a spin-down, that collapses the wave function for all the particles which fixes the observation of A to a predictable spin-up.

That’s not the point in dispute. Before the measurement the spin is undetermined, so to say it changed (“particle A must change its identity”) has no meaning.

10 minutes ago, bangstrom said:

This is why your gloves in boxes model does not work for entangled particles. Gloves have a fixed identity before observation but entangled particles do not.

It’s not a model, it’s an analogy, and one that (when properly presented) acknowledges that the indeterminate state aspect is not covered by it.

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9 hours ago, joigus said:
16 hours ago, bangstrom said:

I don’t find that “well-defined at all times” applies to quantum identities which are an entirely different matter.

Then you certainly don't understand the question. 12(||) independently of the space-time factor of the state. In fact, the space-time factor of the state is completely omitted. Don't you find that peculiar?

What you are saying is largely what I am saying so I do find it peculiar because what you are saying sounds like non-locality.

"Independently of the space-time factor of the state. In fact, the space-time factor of the state is completely omitted".

That sounds like what I would call non-locality.

If two entangled particles are moving in opposite directions and beyond the range of light speed communication, Observing the identity of particle A as spin-up instantly fixes the identity of its partner B as a predictable spin-down. How does particle B “know” what has happened to particle A?

Or, how would you explain it without non-locality.

9 hours ago, joigus said:

OK. Let me stop you right there, because it is plainly obvious you don't understand quantum mechanics here. Quantum particles have no identity. They are indistinguishable, and they are in a way much more profound than macroscopic objects can be made extremely difficult to tell apart. Not even Nature "knows" which electron is which. There is no "which electron." They're just instantiations of a quantum field. It's actually more profound than instances of a computer program, for example, which have a process tag and a time stamp. Electrons have no tags.

This appears to be an overly broad a generalization that may apply to entangled particles or entangled particles in quantum computers but normally particles have a largely predictable nature otherwise there would be nothing but chaos. All electrons may look alike but Nature at least "knows" where they are.

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33 minutes ago, bangstrom said:

What you are saying is largely what I am saying so I do find it peculiar because what you are saying sounds like non-locality.

"Independently of the space-time factor of the state. In fact, the space-time factor of the state is completely omitted".

That sounds like what I would call non-locality.

If two entangled particles are moving in opposite directions and beyond the range of light speed communication, Observing the identity of particle A as spin-up instantly fixes the identity of its partner B as a predictable spin-down. How does particle B “know” what has happened to particle A?

Or, how would you explain it without non-locality.

This reminds me of an old debate I had with someone who asked me if/why a cyclist who is more massive than another has an advantage when going downhill. I said he does.

"Is it not true that the acceleration of the cyclist does not depend on how massive he is? So both the massive, and the less massive cyclists would experience the same acceleration."

"Yes", I said. "For cyclists falling downhill on a frictionless slope, that's true. But you're forgetting friction."

"But is it not true that the friction coefficient does not depend on the cyclist's mass either?

"Errr... sure." I said.

If you listen to the argument like that, in words, it sounds like he was right. But,

Because he couln't be bothered with writing a simple equation, he was incapable of understanding why I was right. In the first case, the acceleration is the same because the force is proportional to the mass. In the second case --with friction-- the force of gravity is still proportional to the mass --necessary for acceleration to be the same in the 1st case--, while the force of friction is not. That's why the acceleration is different. The things that are independent of mass are different things.

The problem is, if you just follow the words, you're incapable of understanding the reasoning. Words in physics, by themselves, are very deceptive.

If the probability amplitude, with its correlations born in it, can be written with the spatial terms factored out, it's precisely because it doesn't matter where they are as long as we don't measure spin. The spin structure of the state was the same when the particles were together.

It's like the person in my story. He was puzzled. We're saying the same thing!

Yes, but you're interpreting incorrectly, and drawing wrong conclusions!

Is that any better?

Edited by joigus
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10 hours ago, joigus said:

Is that any better?

Not at all and you side-stepped my question.

10 hours ago, joigus said:

Yes, but you're interpreting incorrectly, and drawing wrong conclusions!

I don't find your explanation of the bicycle story to be an example of of either interpreting correctly or reaching the right conclusions. Here is your explanation.

10 hours ago, joigus said:

If you listen to the argument like that, in words, it sounds like he was right. But,

Because he couln't be bothered with writing a simple equation, he was incapable of understanding why I was right. In the first case, the acceleration is the same because the force is proportional to the mass. In the second case --with friction-- the force of gravity is still proportional to the mass --necessary for acceleration to be the same in the 1st case--, while the force of friction is not. That's why the acceleration is different. The things that are independent of mass are different things.

I find the bicycle story to be intuitively simple even without any math.

A heavy rider is at a big disadvantage when climbing a hill because he has to input so much more energy to get to the top. When he is at the top, he has more potential energy with which to overcome the friction when going down. Therefore he can go faster.

The acceleration is not that different because of the force of friction. The heavier rider has even more wind friction because of his speed. The potential energy of the riders is the dominating factor.

As for the rest of our discussion:

I don’t see where you have been addressing the problems at hand with much more than unsupported personal opinions.

Why do the same math examples you give not work with non-locality as well as with locality?

How do you define non-locality?

I think a quick view of the literature will show your views to be to be long out of date.

Things like the non-locality of entangled particles, and the super position state of entangled particles prior to their measurement are the commonly accepted, mainstream views. Not that that makes it right.

And, can you support your views with anything more than just personal opinions?

Edited by bangstrom
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16 minutes ago, bangstrom said:

How do you define non-locality?

No, no. You go. What is non-locality to you?

So far, I'm the only one of us that's shown you the calculations and basic principles (conservation laws, observables, quantum evolution, maximally entangled systems) at play, in a way that seems to be to, at least to a certain extent, to the satisfaction/agreement of everybody else but you.

You've shown nothing but your unconditional adhesion to a well-known silly and incorrect interpretation that's been running around for decades to the desperation of many renowned physicists.

Then I give you a simple problem in classical physics to illustrate how if you try to solve a "paradox" by using words instead of writing down the maths, you can be easily mislead.

You don't even understand that simple problem (the cyclists are not pedalling downhill; rather, they're falling downhill, etc.) You misinterpret every single thing I say. That was designed as a test for your attention span. And the key to why you misinterpret the physics is in your own words:

33 minutes ago, bangstrom said:

I think a quick view of the literature will show your views to be to be long out of date.

(My emphasis.)

It takes a lot more than a quick view of the literature to understand physics.

So now I think it's your turn. What is non-locality?

Be careful, because physicists use this term in loosely overlapping senses sometimes, and it's necessary to tread carefully. This is not something that you will be able to sort out by googling it up in a couple of minutes.

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1 hour ago, joigus said:

(the cyclists are not pedalling uphill downhill; rather, they're falling downhill, etc.)

Sorry, mistake there.

They're not pedalling downhill either. But anyway. Feel free to ignore the cyclist's story from here on out.

To the point. What exactly is non-locality?

Get ready, for my next question is going to be: Which one of these (2-3) mathematical models of evolution is non-local, and why?

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11 hours ago, joigus said:

So far, I'm the only one of us that's shown you the calculations and basic principles (conservation laws, observables, quantum evolution, maximally entangled systems) at play, in a way that seems to be to, at least to a certain extent, to the satisfaction/agreement of everybody else but you.

The main calculation you gave me is the same calculation physicists give to explain non-locality. The same maths that work for non-locality also work for locality.

11 hours ago, joigus said:

You've shown nothing but your unconditional adhesion to a well-known silly and incorrect interpretation that's been running around for decades to the desperation of many renowned physicists.

That means the many renowned physicists may be right.

11 hours ago, joigus said:

Then I give you a simple problem in classical physics to illustrate how if you try to solve a "paradox" by using words instead of writing down the maths, you can be easily mislead.

You don't even understand that simple problem (the cyclists are not pedalling downhill; rather, they're falling downhill, etc.) You misinterpret every single thing I say. That was designed as a test for your attention span. And the key to why you misinterpret the physics is in your own words:

12 hours ago, bangstrom said:

The quote below is what I said.

12 hours ago, bangstrom said:

I find the bicycle story to be intuitively simple even without any math.

A heavy rider is at a big disadvantage when climbing a hill because he has to input so much more energy to get to the top. When he is at the top, he has more potential energy with which to overcome the friction when going down. Therefore he can go faster.

The acceleration is not that different because of the force of friction. The heavier rider has even more wind friction because of his speed. The potential energy of the riders is the dominating factor.

And here is what you said.

22 hours ago, joigus said:

If you listen to the argument like that, in words, it sounds like he was right. But,

Because he couln't be bothered with writing a simple equation, he was incapable of understanding why I was right. In the first case, the acceleration is the same because the force is proportional to the mass. In the second case --with friction-- the force of gravity is still proportional to the mass --necessary for acceleration to be the same in the 1st case--, while the force of friction is not. That's why the acceleration is different. The things that are independent of mass are different things.

The problem is, if you just follow the words, you're incapable of understanding the reasoning. Words in physics, by themselves, are very deceptive.

Show me where I failed to understand the problem, misinterpreted your words, or came to the wrong conclusion.

11 hours ago, joigus said:

11 hours ago, joigus said:

It takes a lot more than a quick view of the literature to understand physics.

It doesn’t take long to look up what the current authorities are writing or saying about commonly discussed topics such as non-locality or entanglement. Naturally, understanding takes longer.

11 hours ago, joigus said:

So now I think it's your turn. What is non-locality?

You never responded when I asked you the question… so now its my turn?

Einstein’s view of non-locality works for me. “Spooky action at a distance.”

Non-locality is a non-observable time interval between an action and a reaction because the timing is either instant or far too fast to measure. It is instant action at a distance.

Edited by bangstrom
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2 hours ago, bangstrom said:

The same maths that work for non-locality also work for locality.

OK. This kind of says it all. The same maths work both for locality or non-locality depending on what words you use to describe what you see in those maths.

That is what you're saying. So it's just a word that you put on top of the maths that says whether some pattern of evolution is local or non-local. (!!!)

2 hours ago, bangstrom said:

Show me where I failed to understand the problem, misinterpreted your words, or came to the wrong conclusion.

Later, as it was just meant to illustrate 1) how words mislead you easily in physics, and 2) your misunderstanding of basic physical concepts.

2 hours ago, bangstrom said:

It doesn’t take long to look up what the current authorities are writing or saying about commonly discussed topics such as non-locality or entanglement. Naturally, understanding takes longer.

For better or worse, sanity checks take a lot less out of one than insanity checks. And I'm afraid your "current authorities" have a lot much of the first attribute than of the second one.

2 hours ago, bangstrom said:

Einstein’s view of non-locality works for me. “Spooky action at a distance.”

Funny. You said the EPR was "discredited." Now Einstein is summoned here to save the day for the believers of spooky action at a distance.

2 hours ago, bangstrom said:

Non-locality is a non-observable time interval between an action and a reaction because the timing is either instant or far too fast to measure. It is instant action at a distance.

Quite honestly, I don't know what to do with that. Non-locality is a time? Rather, non-locality (or its negation) is an attribute of the evolution law, the law that tells you how the state of a system is updated with time.

Before I spend --presumably waste-- more time talking about the meaning of words, let's go for my sanity check:

If $$\varphi\left(x,t\right)$$ is a scalar (number-valued) field on the real line, $$\dot{\varphi}$$ is its time derivative, $$\frac{d\varphi}{dx}$$ its spatial derivative, and $$f$$ is an arbitrary numeric function, which one of these models is non-local, which one has propagation at finite speed, which one has no propagation at all, and for which one it depends on some non-specified conditions?

$\dot{\varphi}=f\left(\frac{d}{dx}\right)\varphi$

$\dot{\varphi}=f\left(x,t\right)$

$\dot{\varphi}=f\left(\varphi\right)$

$\dot{\varphi}=\frac{d\varphi}{dx}+f\left(x,t\right)$

You can give that to your authorities if you want.

Edited by joigus
minor deletion

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