Macro Quantum Effect witnessed: Implies time travel? Multiple Universes?

[Whack-a-Mole]

First the cool part. It seems some scientists have witnessed a macro object in a quantum (superimposed) state. The thing is still tiny but large enough to see with the eye unaided.

Here's an article on it. (http://www.nature.com/news/2010/170310/full/news.2010.130.html)

Now, turning to a much less reliable news source this article (http://www.foxnews.com/scitech/2010/04/05/freaky-physics-proves-parallel-universes/?test=faces) at FOX suggests this implies that time travel may be possible as well as many universes.

Many universes I get although I do not think this makes that theory any more likely than it was before. How does this imply time travel (even if a practical impossibility) could work? I read the article but it is not very good.

[The Hamster King]

The Fox article seems to consist of "Quantum superposition is magic! Time travel is magic! We can work magic!"

As far as I know, there's no way to interpret this finding as a step toward time travel.

[Indistinguishable]

I don't know much physics. But I do know those articles are extremely poorly and even misleadingly written. I know it because, even after reading them, I don't really have any idea what the experimenters set up, what observations they then made, and why that should be interpreted in all the various (often nonsensical) ways the articles say it should be.

[Lumpy]

The article said that the paddle was in it's quantum-mechanical ground state. Uh... wouldn't that be absolute zero, which nothing can reach?

[ZenBeam]

The article said that the paddle was in it's quantum-mechanical ground state. Uh... wouldn't that be absolute zero, which nothing can reach?It was very cold. They're looking at the vibration of the paddle, which is macroscopic, so the next energy level of the paddle from its ground state could be very large relative to the energy it would take to make a single atom vibrate. The residual energy of heat in the paddle and any surrounding gas molecules could be small enough that it couldn't excite the paddle out of its ground state, even though the temperature wasn't all the way down to absolute zero.

[RaftPeople]

After reading a few paragraphs of the object simultaneously moving and not moving, when I read this in the article "The work is simultaneously being published online today in Nature and ..."

I thought they were going to finish it with " not being published today in Nature...".

[CookingWithGas]

After reading a few paragraphs of the object simultaneously moving and not moving, when I read this in the article "The work is simultaneously being published online today in Nature and ..."

I thought they were going to finish it with " not being published today in Nature..."."Best Joke In A Thread Actually Relevant To The Thread" Award :)

[CookingWithGas]

Ok, I read both articles. IANA physicist of any kind but IMHO the Fox article is mash-up of a bona fide science experiment, a theory that is interesting but probably unfalsifiable (multiple universes), and another idea that IMHO is not quite even a theory (time travel), and draw wild conclusions about how these things are related.

"When you observe something in one state, one theory is it split the universe into two parts," Cleland told FoxNews.com, trying to explain how there can be multiple universes and we can see only one of them.

Richard Gott, a physicist at Princeton University...[says,] "With Einstein's theory of gravity, the laws of physics as we understand them today suggest that even time travel to the past is possible in principle. But to see whether time travel to the past can actually be realized we may have to learn new laws of physics that step in at the quantum level."

[and here's where that big leap occurs:] And for that, you start with a very tiny paddle in a bell jar.

The Fox reports understands just enough to be dangerous and has made some specious links from the experiment to these two theories.

[CookingWithGas]

Here is a pretty good critique of the Fox article by a physics grad student:

http://scienceblogs.com/builtonfacts/2010/04/the_worst_physics_article_ever.php

[Whack-a-Mole]

Here is a pretty good critique of the Fox article by a physics grad student:

http://scienceblogs.com/builtonfacts/2010/04/the_worst_physics_article_ever.php

Thanks...kind of what I suspected. Witnessing this thing in a superimposed state is cool because of the scale but it is not something scientists were previously unaware of. Just no one thought it could be done on a macro scale (although the article suggests some other macro quantum effects we are familiar with but personally I didn't think those were like this).

[Whack-a-Mole]

Missed the edit window and wanted to add:

I guess the most charitable thing we can say about these articles is that the scientists brought quantum weirdness into the macro world which, presumably, suggests we can harness that weirdness to do hitherto unthinkable things.

Of course this thing is still tiny so scaling it up to something we could use presents serious problems. Like one article noted this is why we do not see a bus moving and not moving at the same time. It is just suggestive that, theoretically if not practically, we could do that. The quantum weirdness is not restricted only to the super tiny world.

That blog article seems a little dismissive though of what was done. He seems to think it is cool but is a little blasé about it. Seems to me for the first time we are able to see the cat both dead and alive at the same time with our own eyes which is rather remarkable.

[GSV Consolation of Dreams]

I read the article and found it something of a tease. It didn't answer the most obvious question that just about every reader is immediately going to have: What did they SEE when it was in the superposition of states?

Did some of them see it vibrating, and some not... or what? Did they video it?

[The Hamster King]

What did they SEE when it was in the superposition of states?If you observe the superposition, it collapses. So it's impossible to see it in both states simultaneously.

[Whack-a-Mole]

If you observe the superposition, it collapses. So it's impossible to see it in both states simultaneously.

Well, what does "observing" entail? Photons are still hitting the thing and bouncing off. Why does it matter if there is an eye or camera to receive those photons? How does the quantum device "know" that a particular photon entered a camera and thus should collapse the superposition?

[Lumpy]

I read the article and found it something of a tease. It didn't answer the most obvious question that just about every reader is immediately going to have: What did they SEE when it was in the superposition of states?

Did some of them see it vibrating, and some not... or what? Did they video it?If I understand the article correctly, what they "saw" was that the current in the electric circuit the paddle was wired to was in a state of superposition between the mode it would be when the paddle was vibrating and when it was not.

[Indistinguishable]

What does it mean to see something in a state of superposition?

[MeDrewNotYou]

Here is a pretty good critique of the Fox article by a physics grad student:

http://scienceblogs.com/builtonfacts/2010/04/the_worst_physics_article_ever.php

The critique was spot on. I was screaming at the monitor reading the FOX News article and I'm a freakin' history major! The Nature article, OTOH, was a good piece.

I suggest reading the comments on the FOX article. Besides "Christian scientists have been hypothesizing for centuries that understanding quantum physics is a gateway to understanding," there was also a few great ones to the effect of, "I don't care what scientists say, something's either moving or it isn't."

[MeDrewNotYou]

What does it mean to see something in a state of superposition?

From the Nature article:
Next, the researchers put the quantum circuit into a superposition of 'push' and 'don't push', and connected it to the paddle. Through a series of careful measurements, they were able to show that the paddle was both vibrating and not vibrating simultaneously.

I read that as their instruments, however they where set up, recorded 'movement' and 'not movement' at the same time.

[Indistinguishable]

That's not helpful to me. What kind of thing was set up to potentially record "movement"? What kind of thing was set up to potentially record "not movement"? That is, what do these labels correspond to, instrumentally?

[Chipacabra]

I can't extract a whole lot of details from either of those articles, so I don't know exactly what they were measuring for. But if I had to guess, I would say that both articles are misrepresenting what was actually measured. The whole point of quantum mechanics is that measuring collapses a superimposed state, so you can't measure both states at the same time: You get one or the other.

BUT if you repeat the experiment many many times, and nothing else changes, then sometimes you'll get one state, and sometimes you'll get the other. If you've successfully eliminated or accounted for all the variables, all that's left is quantum probability, so you can show that the state you happen to find when you measure is random, but with predictable odds.

So in this particular case, you take the paddle and reduce it to its ground state (not moving). The first excited (moving) state has some associated energy. We'll call that energy... 5. In classical physics, you could say that it takes 5 energy to bump the paddle from ground state to its excited state. Any less and it just won't make it out of the hole.

In QM, though, if you add, say, 3 energy to the paddle, then the paddle isn't exactly the ground state or the excited state. It's in a superposition of states with a total energy of 3. But since you can only observe the paddle at one quanta or the other, once you measure it, it will either be in the ground state or the excited state. (Or a higher excited state, but probably with a very low probability)

So you take the paddle, pelt it with 3 energy, and see if it starts wiggling. Then you drop it back down to ground state, pelt it with 3 energy, and see if it starts wiggling. Do this a lot. See what percentage of the experiments lead to wiggling. Compare that to expected percentages. Get misquoted by FOX. The end.


I'd really like to see the actual experiment, though. Has anyone found, say, a paper written by the actual scientists involved?

[Pasta]

I'd really like to see the actual experiment, though. Has anyone found, say, a paper written by the actual scientists involved?

The original article is published in Nature, doi:10.1038/nature08967. I just read it (or, gave it a 60%-level skim). The paper is very detailed, if terse, and the experimental approach is complex, if clever, so I'll give only a short summary.

The "drum" is piezoelectric, so it can be coupled to a circuit capacitively for readout and manipulation. That readout and manipulation is, in turn, provided by a qubit, which is a small two-state quantum system. The energy gap between the ground state and excited state of the qubit can be tuned inductively (i.e., magnetically).

The main "trick" of the experiment is that you can tune the qubit's energy gap to be either close to or far from the lowest frequency mode of the drum (energy E=hf). When these match, they exhibit the usual coupled resonator phenomenon of sharing energy back and forth. (Think: classical weakly coupled pendula. The "trading" of energy here, though, is between two systems that have only two accessible states each.)

The first half of the paper describes multiple methods used to demonstrate that they can get the coupled system into its ground state. The actual, physical, measurement performed at any given instance is What state is the qubit in? This is a quantum mechanical question, so they repeat this "~1000" times when they ask it so that they can actually determine P(e)=What is the probability that the qubit is in the excited state?

That's the crux of it. They proceed to plot P(e) under various configurations of: (1) the quality of "tune" between the qubit and the mechanical resonators, (2) how long they hold the pair in tune, (3) how long they keep them out of tune (to let the drum's state evolve a bit before reading it out resonantly), (4) how long and hard they ping the drum through a separate excitation channel (a blast of microwaves).

They look at P(e) as a function of all these things, and they extract quantities such as the energy transfer time, the resonator relaxation time, etc. These data are in excellent agreement with the quantum mechanical expectations, demonstrating, among other things, that the drum and qubit indeed seem to become quantum mechanically entangled and seem to have quantum mechanically "uncertain" states.

[Indistinguishable]

So what it amounts to is that they took a bunch of repeated measurements of some kind which didn't all come out the same, and then analyzed the statistics of this, repeating this process in various configurations as well. But it's not any nonsense like "They took a single measurement, and got two different answers at the same time, in some crazy magic that uses the word 'quantum', take our word for it...".

[Pasta]

So what it amounts to is that they took a bunch of repeated measurements of some kind which didn't all come out the same, and then analyzed the statistics of this, repeating this process in various configurations as well. But it's not any nonsense like "They took a single measurement, and got two different answers at the same time, in some crazy magic that uses the word 'quantum', take our word for it...".
That is an accurate summary. (Although, I will note that even the authors play up the mystique a bit with their opening two words: "The bizarre [...predictions of QM]." Really? For the Nature audience?)

[Mijin]

Quantum mechanics is clearly-defined until it is observed by journalists.

Then it collapses into a superposition of nonsensical states, emitting only virtual factoids.

[Whack-a-Mole]

That is an accurate summary. (Although, I will note that even the authors play up the mystique a bit with their opening two words: "The bizarre [...predictions of QM]." Really? For the Nature audience?)

Well, to be fair even the scientists who work on QM call it bizarre. I forget who said it and the exact quote but one notable scientist said something like, "Anyone who is not deeply disturbed by Quantum Mechanics does not understand it."

[Pasta]

Well, to be fair even the scientists who work on QM call it bizarre. I forget who said it and the exact quote but one notable scientist said something like, "Anyone who is not deeply disturbed by Quantum Mechanics does not understand it."
Do they? I agree it is reasonable to use such verbiage to help get the non-intuitiveness across to lay audiences, but in my experience, physicists these days do have some intuition about QM and don't find it all that bizarre. (Sort of like the "spooky" 4D spacetime concept that nowadays is pretty mundane for those that use it day to day. Or, for a more extreme example, the Copernican model of the solar system, which was about as non-intuitive as a theory could get when it was new.)

I dunno, maybe it's just me that thinks humanity (well, physicists at least) should direct their awe to some new initially-hard-to-intuit-but-after-nine-decades-really-isn't-all-that-bad topic. These are folks that presumably have been exposed to QM since their formative undergraduate days.

(BTW - The quote you mention was from Bohr, who certainly meant what he said -- in 1935 or whenever.)

[Sage Rat]

Do they? I agree it is reasonable to use such verbiage to help get the non-intuitiveness across to lay audiences, but in my experience, physicists these days do have some intuition about QM and don't find it all that bizarre. (Sort of like the "spooky" 4D spacetime concept that nowadays is pretty mundane for those that use it day to day. Or, for a more extreme example, the Copernican model of the solar system, which was about as non-intuitive as a theory could get when it was new.)

I dunno, maybe it's just me that thinks humanity (well, physicists at least) should direct their awe to some new initially-hard-to-intuit-but-after-nine-decades-really-isn't-all-that-bad topic. These are folks that presumably have been exposed to QM since their formative undergraduate days.

(BTW - The quote you mention was from Bohr, who certainly meant what he said -- in 1935 or whenever.)
Actually, it wouldn't surprise me all that much that a professional scientist would consider his area of study to be wonky and cool. It's their fascination with the subject that brought them there and kept them studying it to begin with.

[enigmatic]

One thing though, doesn't verifiable macroscopic superposition potentially allow experiments into what exactly counts as an observer?

[Whack-a-Mole]

Do they? I agree it is reasonable to use such verbiage to help get the non-intuitiveness across to lay audiences, but in my experience, physicists these days do have some intuition about QM and don't find it all that bizarre. (Sort of like the "spooky" 4D spacetime concept that nowadays is pretty mundane for those that use it day to day. Or, for a more extreme example, the Copernican model of the solar system, which was about as non-intuitive as a theory could get when it was new.)


Perhaps from repeated exposure a physicist may become used to quantum weirdness and seem nonplussed but I doubt it ever really goes away.

How can anyone really find something being in two places at once, being "on"/"off" at the same time, being a wave and a particle at the same time not find it strange?

[Indistinguishable]

By not thinking of things as being a wave and a particle at the same time? We had this idea for how some things behaved, and called the concept a "particle", and we had this other idea for how some things behaved, and called the concept a "wave". But not everything has either particle-behavior or wave-behavior. Turns out, some things have a way of behaving which is different from both of those, even if there are similarities to each in certain ways. Finding this strange is like finding it strange that oranges have a thick bitter skin, like a banana, but fleshy juicy innards, like a peach. "How can something be a banana and a peach at the same time?". But, of course, it's not a banana, and it's not a peach; it's just an orange. Fretting on about some paradox of wave-particle duality is great for making things sound mystical and cool, but it's antithetical to the kind of understanding one would expect after nearly a century of familiarity.

You play Mario long enough, you get comfortable with the rules of Mario. You play Tetris long enough, you get comfortable with the rules of Tetris. You play around in the details of our world long enough, and you should get comfortable with the rules of our world.

[MeDrewNotYou]

You play Mario long enough, you get comfortable with the rules of Mario. You play Tetris long enough, you get comfortable with the rules of Tetris. You play around in the details of our world long enough, and you should get comfortable with the rules of our world.

Yes, but the real question is does understanding the quantum world let me throw fire balls or help me rescue princesses from evil turtles? ;)

I agree that studying QM should demystify the whole thing, but it still has a certain level of weirdness that, say, biology or even Newtonian physics doesn't have. Common sense from our experience in the macro world makes things like superposition counterintuitive; here a cat is either alive or dead. Sure I can study it and learn how it fits in with other laws of nature, but it's still a little weird.

To make a clumsy analogy, I can study India my whole life, but some aspects of the culture will always be a little foreign to me since all my experience is as a Westerner. Growing up in Newtonia, the customs and norms of Heisenbergistan will always be slightly foreign, no matter how long I live there.

[Indistinguishable]

Well, everyone talks about superposition like it's the ultimate weirdness of quantum mechanics, but superposition in itself is perfectly cromulent even as a classical concept. If a coin is tossed and I have not yet observed the result, I can model my uncertainty probabilistically as 1/2|Head> + 1/2|Tail>. Applying various further nondeterministic processes (e.g., a machine which kills pets with various probabilities based on the coin flip results) will cause the superposition to evolve further, in a linear manner subject to natural constraints. Perhaps I eventually reach something like 1/6|the cat is dead and the dog is alive> + 1/3|the cat is dead and the dog is dead> + 1/4|the cat is alive and the dog is alive> + 1/4|the cat is alive and the dog is dead>. Then, I actually observe that the cat is dead, and, conditioning the probability distribution representing my knowledge upon this new information, it "collapses" into 1/3|the cat is dead and the dog is alive> + 2/3|the cat is dead and the dog is dead>. And so on and so on. The mere concept of superposition, in itself, is not an illustration of quantum phenomena (this blog post puts it nicely (http://blog.sigfpe.com/2006/05/quantum-mechanics-and-probability.html)).

Rather, the interesting thing that happens in the quantum context which is different from mere classical superposition is that the weights involved aren't restricted to being nonnegative reals (as in ordinary probability), but can in fact be turned 180 degrees around from this and be negative (or even turned by intermediate angles to be complex), thus allowing for "interference" of a kind not otherwise modellable. This is something new and interesting, but here mere idea of superposition is old hat.

[Indistinguishable]

This is something new and interesting, but the mere idea of superposition is old hat.
Typo corrected in bold.

[Frylock]

Well, everyone talks about superposition like it's the ultimate weirdness of quantum mechanics, but superposition in itself is perfectly cromulent even as a classical concept. If a coin is tossed and I have not yet observed the result, I can model my uncertainty probabilistically as 1/2|Head> + 1/2|Tail>. Applying various further nondeterministic processes (e.g., a machine which kills pets with various probabilities based on the coin flip results) will cause the superposition to evolve further, in a linear manner subject to natural constraints. Perhaps I eventually reach something like 1/6|the cat is dead and the dog is alive> + 1/3|the cat is dead and the dog is dead> + 1/4|the cat is alive and the dog is alive> + 1/4|the cat is alive and the dog is dead>. Then, I actually observe that the cat is dead, and, conditioning the probability distribution representing my knowledge upon this new information, it "collapses" into 1/3|the cat is dead and the dog is alive> + 2/3|the cat is dead and the dog is dead>. And so on and so on. The mere concept of superposition, in itself, is not an illustration of quantum phenomena (this blog post puts it nicely (http://blog.sigfpe.com/2006/05/quantum-mechanics-and-probability.html)).

Rather, the interesting thing that happens in the quantum context which is different from mere classical superposition is that the weights involved aren't restricted to being nonnegative reals (as in ordinary probability), but can in fact be turned 180 degrees around from this and be negative (or even turned by intermediate angles to be complex), thus allowing for "interference" of a kind not otherwise modellable. This is something new and interesting, but here mere idea of superposition is old hat.

In the classical case, the assumption is that the system "really is" just in one state or the other, and the "superposition of states" reflects our knowledge of the system, not a fact about the system itself. But everything I've read about quantum physics (all of it firmly at the lay level of course) indicates that what's "wierd" about quantum physics is that the superposition of states reflects a fact about the system itself, and not just our knowledge about the system.

Is this wrong, though?

[CookingWithGas]

In the classical case, the assumption is that the system "really is" just in one state or the other, and the "superposition of states" reflects our knowledge of the system, not a fact about the system itself. But everything I've read about quantum physics (all of it firmly at the lay level of course) indicates that what's "wierd" about quantum physics is that the superposition of states reflects a fact about the system itself, and not just our knowledge about the system.

Is this wrong, though?I am the layest of laymen but I think this is correct. Once observed the superposition collapses to a definite state (as definite as possible; can't know velocity and momentum both).

I think the comparison of classical "superposition" to quantum is just to create an analogy, and analogies are imperfect by their nature.

[Indistinguishable]

In the classical case, the assumption is that the system "really is" just in one state or the other, and the "superposition of states" reflects our knowledge of the system, not a fact about the system itself. But everything I've read about quantum physics (all of it firmly at the lay level of course) indicates that what's "wierd" about quantum physics is that the superposition of states reflects a fact about the system itself, and not just our knowledge about the system.

Is this wrong, though?
I feign no hypotheses as to what's "really" going on; I don't want to even pretend there is any meaning to this, beyond what is ultimately observable (and perhaps as inspiration for directions in which to generalize methods by which to make predictions about what is ultimately observable). I am not proposing that even in the classical case, we should think of the system being "really" in just one state, rather than the superposition. All I am saying is that the same calculus of superpositions can be harnessed to give reliable predictions in a similar way in that context, which is, for me, the most one can say about the truth of an abstract mathematical model of the world. But, anyway, let me ask, what kind of experimental data would support for you the answer being one way as opposed to another?

Incidentally, I'm curious if you read the link from this reply of mine (http://boards.straightdope.com/sdmb/showpost.php?p=12255402&postcount=21) to a post of yours from a thread we were recently in on similar matters. For me, it was very enlightening.

[Frylock]

I feign no hypotheses as to what's "really" going on; I don't want to even pretend there is any meaning to this, beyond what is ultimately observable. But let me ask, what kind of experimental data would support for you the answer being one way as opposed to another?Heck, I don't know, I've always wondered that myself. But it seems to be what all the books I've read insist on.

Incidentally, I'm curious if you read the link from this reply of mine (http://boards.straightdope.com/sdmb/showpost.php?p=12255402&postcount=21) to a post of yours from a thread we were recently in on similar matters. For me, it was very enlightening.

Yeah, I did read it, set it aside intending to read it again, and didn't read it again.

I'll get back to you, I hope.

Repeating that conversation again: Isn't it supposed to be that the Bell's Theorem stuff shows that the observational evidence is inconsistent with the idea that systems in a superposition are in some sense "actually" just in one of the component states or the other? I haven't been able to understand how the observational evidence is supposed to be able to show that, but books by experts that I read seem to assure me that, somehow, the observational evidence does indeed accomplish that.

[Frylock]

Repeating that conversation again: Isn't it supposed to be that the Bell's Theorem stuff shows that the observational evidence is inconsistent with the idea that systems in a superposition are in some sense "actually" just in one of the component states or the other? I haven't been able to understand how the observational evidence is supposed to be able to show that, but books by experts that I read seem to assure me that, somehow, the observational evidence does indeed accomplish that.

Maybe not. It looks like (http://en.wikipedia.org/wiki/Hidden_variable_theory) I've confused "hidden variable theories" with "local hidden variable theories," and that Bell's Theorem's confirmation shows that local hidden variable theories can't work, but says nothing about hidden variable theories in general. And if the article I linked to is right (which is not clear--there are a few red flags in the article that make me unsure about the authoritativeness of some of it) then hidden variable interpretations of QM can work. I guess as long as they're not "local hidden variable" theories.

[Rigamarole]

So when will I be able to buy my matter transporter, and how much will it cost*?


*Secretly, my plan is to buy it on credit and then transport all the gold out of Fort Knox to pay for it. Muahaha!

[Whack-a-Mole]

This has been discussed up-thread some but I think it bears being clarified.

I think many people have the sense that despite the Heisenberg Uncertainty Principle a particle has a definite position and a definite momentum. We may be blocked from knowing both with perfect precision but at the bottom of it all the particle is "there" and moving "that way" with a specific speed.

A superimposed state however turns that on its head. The particle is not "there". It is, quite literally, "here" and "there" at the same time. This is most easily shown by the double-slit experiment. In a more complex fashion quantum computers have been built that exploit this feature.

I think this experiment is fascinating in that they took something we thought relegated to the ultra-tiny and brought it in to our "big" world. As mentioned up-thread what does it look like, now we can actually see it with our own eyes, to have something moving and not moving at the same time? Or, as mentioned, how does observing it collapse the wave-state so we only see one or the other? How does the object know the photon that bounced off it entered your eye and was "observed"?

Bringing this into the macro world, to me, re-opens these questions and highlights a point where science meets philosophy.

[Indistinguishable]

Just to clarify, I have no fondness for hidden variables, as such; quite the opposite, I am happiest to discuss only the ultimately predicted observations as the fundamental entities. But I don't believe an epistemological account of wavefunction collapse requires the wavefunction to be taken as merely representing uncertainty about underlying hidden variables with definite values [just as I don't believe an epistemological account of probability distribution collapse can only be carried out in the specific context of taking the probability distribution to represent uncertainty about underlying hidden variables with definite values]. While I'm at it, I think most alternative accounts of wavefunction collapse are bogged down in a morass of confusion [e.g., it certainly doesn't matter to the physics whether a photon bounces off and enters your eye or not; that's a confused reading of the "observation" terminology].

Just to clarify something else, I, er, have no fondness for hidden variables, as such, and, quite the opposite, etc., etc., am not claiming that there actually are underlying hidden variables with definite values, etc., etc. But think much popular discussion about superpositions is glib and doesn't actually emphasize any non-classical phenomena.

Anyway, all the more to stress the observational content of the manner in which quantum mechanics departs from the classical,
Repeating that conversation again: Isn't it supposed to be that the Bell's Theorem stuff shows that the observational evidence is inconsistent with the idea that systems in a superposition are in some sense "actually" just in one of the component states or the other? I haven't been able to understand how the observational evidence is supposed to be able to show that, but books by experts that I read seem to assure me that, somehow, the observational evidence does indeed accomplish that.
Well, let's walk through an illustration of Bell's theorem, and then decide what the consequences are.

We can set up a situation where there are two machines, located at different points in spacetime, outside of each other's lightcones (basically, you can think of the setup as two machines at different locations in space, but with everything happening at the same time), with the following property: each machine can be fed an angle as input, and will then produce one bit of output (let's call this "red" or "blue"). The interesting thing, observed empirically over many runs of this experiment, is that, if both machines are given the same angle as input, they always produce the same output. More generally, again as observed empirically over many runs of this experiment, out of all the situations where the difference between the two machine's input angles is theta, the proportion of times where both machines output "red" is about cos^2(theta)/2, the proportion where both output "blue" is also about cos^2(theta)/2, the proportion where Machine 1 outputs "red" and Machine 2 outputs "blue" is about sin^2(theta)/2, and the proportion where Machine 1 outputs "blue" and Machine 2 outputs "red" is about sin^2(theta)/2. That is, about cos^2(theta) of the time, the machines produce the same output.

In particular, let us simplify things even further, and only concern ourselves with three angles, each separated by 120 degrees. Then the setup is that each machine has three buttons (let's call them A, B, and C) out of which one can be pressed, after which a single bit of output is produced; if the same button is pressed on both machines, then they always produce the same output. However, if different buttons are pressed, then only about a quarter of the time do they produce the same output.

So what? Well, now let's think about modelling this with the particular mathematical construct of a probability distribution (on the space of all 30 possible complete input-and output-configurations; i.e., an assignment of a number >= 0 to each of these, with the total summing to 1). Which distribution? Well, let's discuss various properties such a distribution could have. It could be the case that each of the nine particular input-configurations is equiprobable (i.e., the machines' inputs are chosen independently at uniform random); we'll say that a distribution with this property "doesn't predict the inputs". It could be the case that the events at Machine 1 are independent of those at Machine 2, in the sense that the probability of each complete input-and-output configuration is simply the product of the probabilities of the corresponding Machine 1 and Machine 2 configurations; we'll say that a distribution with this property "fixes the common causes". Finally, it could be the case that the distribution "satisfies the empirical condition", in the sense that the probability of any particular complete input-and-output configuration with mismatched inputs is a quarter of the probability of the corresponding input-configuration.

Again, so what? Well, the interesting thing now is that we can now prove an interesting fact. We'll say that a probability distribution is "local" if it is the weighted average of distributions which fix the common causes and don't predict the inputs (in mathematical jargon, it arises as a "marginal" on such distributions). The interesting fact is that no local distribution satisfies the empirical condition.

Proof: Let P be an arbitrary distribution which fixes the common causes and doesn't predict the inputs. First, we will show that, under P, each input (A, B, or C) must make either an output of blue almost certain or an output of red almost certain (in the sense of the conditional probability being 1). This is because 0 = P(Machine 1 outputs red and Machine 2 outputs blue and Machine 1 has input A and Machine 2 has input A) = P(Machine 1 outputs red and has input A) * P(Machine 2 outputs blue and has input A) [this last step following from the independence defining fixing the common cause]. Thus, either P(Machine 1 outputs red and has input A) or P(Machine 2 outputs blue and has input A) is equal to 0. Symmetrically, either P(Machine 1 outputs blue and has input A) or P(Machine 2 outputs red and has input A) is equal to 0. But P(Machine 1 outputs red and has input A) + P(Machine 1 outputs blue and has input A) = 1/3, and similarly for Machine 2; it follows that either P(Machine 1 outputs red and has input A) = P(Machine 2 outputs red and has input A) = 0, or the same with blue substituted in for red. And symmetrically for inputs B and C. Accordingly, each input has a corresponding color which it makes almost certain.

Next, we will show that this means P(the two machines give the same output) is at least 5/9. For each of A, B, and C there is one color which it makes almost certain as output; this color has to be the same for at least two of the inputs. Accordingly, P(the two machines give the same output) = P(the two machines give the same output and have the same input) + P(the two machines give the same output and have different inputs) = P(the two machines have the same input) + P(the two machines have differing inputs with matching assigned colors) = 1/3 + P(the two machines have differing inputs with matching assigned colors) >= 1/3 + 2/3 * (2 - 1)/3 = 5/9.

Finally, as the above held for an arbitrary distribution fixing the common causes and not predicting the inputs, we can conclude that, for any local distribution P, the probability that the two machines give the same output is a weighted average of values which are at least as large as 5/9, and thus is at least as large as 5/9 itself. And also, for any local distribution, the probability that both machines have the same input = 1/3, just by taking the weighted average of values which are all equal to 1/3. But then this cannot satisfy the empirical condition, for the empirical condition tells us that the probability that the two machines give the same output = the probability that both machines have the same input + 1/4 * (1 - the probability that both machines give the same input), which would equal 1/3 + 1/4 * 2/3 = 1/2, which is less than 5/9. This completes the theorem.

Discussion of its implications can now follow.

[Indistinguishable]

Like any long post, there are various typos in the above which I am now too late to fix. Persevere through, valiant reader...

[Indistinguishable]

More generally, again as observed empirically over many runs of this experiment, out of all the situations where the difference between the two machine's input angles is theta, the proportion of times where both machines output "red" is about cos^2(theta/2)/2, the proportion where both output "blue" is also about cos^2(theta/2)/2, the proportion where Machine 1 outputs "red" and Machine 2 outputs "blue" is about sin^2(theta/2)/2, and the proportion where Machine 1 outputs "blue" and Machine 2 outputs "red" is about sin^2(theta/2)/2. That is, about cos^2(theta) of the time, the machines produce the same output.
Oops, one important typo. I forgot some halvings, as reinstated in bold. Luckily, it doesn't change anything in the rest of the example (cos^2(120 degrees/2) is still 1/4).

I should also perhaps clarify that by "angle", all I really mean is "direction in (in fact, three-dimensional) space". (And if any physicists are reading, yes, I know that the most direct setup has agreement and disagreement switched around from the way I did it, but my way seemed a little easier to write up concisely, so I went ahead and had one detector "flipped", so to speak.)

[Frylock]

Since the empirical condition is satisfied, the probability distribution either predicts the inputs or doesn't fix the common causes.

Having only a basic understanding of probability, I would have thought that all distributions on the inputs would fix the common causes--just relying here on the grade school notion that the probability of A AND B is the probability of A multiplied by the probability of B. This would imply that the probability distribution discussed in the theorem must predict the inputs. But I'll stop there and ask what I've misunderstood already. (For example, am I right that the grade school notion of the probability of A AND B applies here and implies that all probability distributions on the inputs must fix the common causes?)

[Indistinguishable]

Remember, a probability distribution (at least, so far as our finitary purposes go) is just some way of assigning numbers >= 0 to each possibility in some space, with them all adding up to 1. The probability of any subset of all the possibilities is then just the sum of the encapsulated numbers. So nothing's forcing P(A AND B) to equal P(A) * P(B) for arbitrary subsets A and B of possibilities.

In fact, the equality between the two is characteristic of A and B being probabilistically independent. Consider a coin flip (with heads and tails equiprobable). What's the probability that it comes up heads? What's the probability that it comes up heads AND it comes up heads? Is the latter the square of the former? Less trivially, consider the probability that a digit from 0 - 9 (each equiprobable) is even, the probability that it is prime, and the probability that it is even AND a prime. Is the third the product of the first two?

[ivan astikov]

< SNIP --- a lot of incomprehensible stuff >
Discussion of its implications can now follow.

Seriously, is there no way of simplifying that?

What's the least I'd need to learn to understand anything you said after the bit about Bell's Theorem and deciding what the consequences are?

[Indistinguishable]

What's the least I'd need to learn to understand anything you said after the bit about Bell's Theorem and deciding what the consequences are?
Well, you can start reading only from "In particular, let us simplify things further...", and then skip the proof at the end, if you like. That's just three paragraphs, with only grade school math. If that's not suitably simple, then let me know what the stumbling block is.

[Frylock]

Remember, a probability distribution (at least, so far as our finitary purposes go) is just some way of assigning numbers >= 0 to each possibility in some space, with them all adding up to 1. The probability of any subset of all the possibilities is then just the sum of the encapsulated numbers. So nothing's forcing P(A AND B) to equal P(A) * P(B) for arbitrary subsets A and B of possibilities.

In fact, the equality between the two is characteristic of A and B being probabilistically independent. Consider a coin flip (with heads and tails equiprobable). What's the probability that it comes up heads? What's the probability that it comes up heads AND it comes up heads? Is the latter the square of the former? Less trivially, consider the probability that a digit from 0 - 9 (each equiprobable) is even, the probability that it is prime, and the probability that it is even AND a prime. Is the third the product of the first two?

:smack:Yes yes, I remember now, this is also from grade school.

Sorry!

I'll give this some more thought before responding again.

[Indistinguishable]

Since the empirical condition is satisfied, the probability distribution either predicts the inputs or doesn't fix the common causes.
Well, it doesn't make sense to say "the empirical condition is satisfied" in itself. But, of course, it's clear what you were saying:

A particular probability distribution on our sample space of 30 possibilities may or may not satisfy the empirical condition. I gave a very weak empirical condition which doesn't uniquely specify a distribution since it was all that was needed for my theorem, but there is one particular distribution which concerns us more than others: the one in which the probability of "Machine 1 has input X, Machine 2 has input Y, Machine 1 has output Z, and Machine 2 has output W" is 1/18 when X = Y and Z = W, 1/72 when X is not equal to Y but Z = W, and 1/24 when X is not equal to Y and Z is not equal to W. We may as well call this "the empirical distribution". This is the distribution which matches all the experimentally observed frequency proportions when we try to choose the machine inputs at uniform random. It's the unique distribution which doesn't predict the inputs, satisfies what I called "the empirical condition", and is also symmetric under interchange of the two possible output values.

You can read right off that the empirical distribution does not "fix the common causes". Remember, this just means that it does not make the goings-on at Machine 1 probabilistically independent from those at Machine 2. That's clear enough; originally, the probability that Machine 2 has input A and output blue, for example, is 1/6. However, conditioned on the event that Machine 1 has input A and output red, the probability that Machine 2 has input A and output blue switches to 0. Learning information about the goings-on at Machine 1 can tell you something about the goings-on at Machine 2.

In itself, this is not terribly surprising; why shouldn't there be correlation? But we might expect that any correlation arises from some property of a common cause ("If the winning numbers printed in New York are correlated with the winning numbers printed in LA, it's no surprise; they're both taken from readings originally made in Chicago, and then sent out to the coasts (with some small chance of printing errors at the end)"), and that, therefore, a probability distribution which has already incorporated all the information there is to know about all the properties of all the common causes would make the goings-on independent ("Once I know the readings in Chicago, I can't gain any more information about the numbers in New York by asking about the numbers in LA; learning about printing errors at one end tells me nothing about printing errors at the other end"). Hence, the name I gave to that condition.

Furthermore, we would then expect the empirical distribution itself is just a probabilistically weighted average of the various distributions conditioned on further information about the common causes. (If Y_1, Y_2, ..., are such that one and only one of them will occur, then the probability of any arbitrary X can always be taken as the weighted average of the probabilities of X conditioned on each Y_i, each weighted by the probability of that Y_i).

Now, remember, the interesting theorem isn't that the empirical distribution happens not to fix the common causes. The interesting theorem is that the empirical distribution can't even arise as a weighted average of distributions which fix the common causes and don't predict the input.

One reading of this: it's not possible to have extra information which both "accounts for" the correlation between Machine 1 and Machine 2 and is not itself correlated with the inputs. (Of course, the whole point of writing out the example and the (very simple) math is so you can decide for yourself what kind of reading you would like to give it.)

[Frylock]

I don't understand why people have takn this to mean local hidden variable theories can't work. By the theorem together with the fact that the empirical condition is satisfied, it seems like the most natural thing to say is that either the two inputs aren't independent of each other, or the 9 combinations aren't equiprobable. The most natural way to understand that, in turn, would seem to me to be to hypothesize there is some kind of "common cause" to the two events--a hidden variable, in other words. And why not a local one?

So--if you're explaining Bell's theorem (or anyway, if I am understanding you correctly, a special but representative particular case of it) and if I've understood it right, I do not understand why mathematicians and phycisists have taken the empirical confirmation of it to mean local hidden variable theories are false. What you've explained seems to me rather to suggest the truth of some local hidden variable theory or other.

(The above was written before your last post)

[Frylock]

Well, it doesn't make sense to say "the empirical condition is satisfied" in itself.

Wait, why not? Doesn't this just mean that observations confirm the probability is 1/4?

[Indistinguishable]

Well, when you run the experiments, you're free to choose the inputs to the machine however you want. You can roll dies at each end, if you like. If you go down the road of denying that the inputs are independent, you're saying that die rolls at the two ends are correlated; if this arises locally, it must mean that the separate die rolls' values are actually influenced by some event way back which communicated with both dice. Seems implausible, doesn't it?

ETA: Er, this was in response to your penultimate post, not your ultimate post.

[Indistinguishable]

Wait, why not? Doesn't this just mean that observations confirm the probability is 1/4?
All I'm pointing out is that I defined "the empirical condition" as a predicate upon probability distributions. Some satisfy it, some don't. There is a particular probability distribution which satisfies it which we are concerned with, it is true (what I called "the empirical distribution" in that post), but there are also others which don't. I'm just highlighting this conceptual organization; we can talk about any probability distribution we like, even ones which aren't the empirical distribution. Indeed, we have to to even begin to discuss the theorem.

Anyway, if you like, we can forget "the empirical condition", and only talk about "the empirical distribution"; the theorem is then that "the empirical distribution is not a weighted average of distributions which fix the common causes and don't predict the inputs". The empirical condition was just one particular property which the empirical distribution satisfied, which happened to be sufficient for the theorem, but, you know, the further generality doesn't really concern us, so, whatever.

[Indistinguishable]

(I'll explain why it is that I used the word "local" for being a weighted average of distributions which fix the common causes and don't predict the inputs in a bit, but the idea is basically all in those parentheticals about why we wouldn't be surprised by correlations in NY and LA printings of some Chicago lottery.)

[Frylock]

Seems implausible, doesn't it?


I wouldn't think it implausible, if in fact the outcomes are conforming to the empirical distribution.

Now, if I actually did this with two machines by rolling two dice which never had anything to do with each other, then I wouldn't come up with the empirical distribution, and there'd be no reason to think the two dice have some single influence causing both their results.

But if I do get the empirical distrbution, then don't I have a reason to think the two dice have somehow been coordinated by some common cause?

[Indistinguishable]

I think you may be confusing the inputs to the machines and the outputs to the machines.

The inputs to the machines (one of three possibilities at each) are observed to come up in each of the nine possibilities with equal frequencies. No correlation is observed there. These are the dice, so to speak.

The correlation observed is that the outputs of the machines (one of two possibilities at each) are never different if their inputs are the same, while the outputs of the machines are the same only about a quarter of the time when the inputs are different.

(Our whole setup has 30 complete input-output possibilities: 6 where both machines have the same input and output, and 24 where both machines have different inputs and some (maybe different, maybe the same) outputs.)

[Frylock]

I think you may be confusing the inputs to the machines and the outputs to the machines.

You're right...

ETA I'll come back to the thread later, re-reading paying special care not to get confused about that.

[Indistinguishable]

I should also say, the usual presentation of the theorem is a priori much weaker than the way I've put it, the usual presentation only demonstrating that the empirical distribution cannot arise as a weighted average of distributions which don't predict the inputs but completely specify everything else in the sense that "Each of the three inputs is assigned a corresponding output which is guaranteed for any machine by that input". However, as demonstrated by the beginning of my proof, we can actually generalize, since it turns out fixing the common causes, even though a priori a much weaker property, will entail the quoted property, which I think is well worth observing.

[MeDrewNotYou]

It really belongs in GD, but here's a quick finish to my earlier side-track.
I think I'll revise my point a little bit re: QM is 'weird.' I still think it is to the lay audience, but only because of its relative newness. ~90yrs is certainly old, but compare that to the 400yrs we've had to get used to Newton. Also, while it does wonders for fleshing out our understanding of the universe, QM hasn't contributed any commonly used technology that would lead to more widespread familiarity. In 50-100yrs, people may very well take QM for granted while the newest big thing (brane theory?) will seem bizarre even though it accurately describes things.

[Indistinguishable]

I think perhaps better evocative terminology for "fixes the common causes" would be "is entanglement-free". Or, rather, I think it's useful to see the connection between both perspectives on what it means (for a probability distribution to decompose into an independent product). The point being that entanglement merely amounts to correlation. And that, in the relevant situation, the interesting thing isn't the correlation itself, but the fact that any variable which accounts for the correlation (in the sense that the correlation disappears after conditioning on this variable) must potentially carry information about the outcome of the dice rolls.

[Frylock]

I forgot this thread existed. Partial zombie, I guess, but I intended to reply all along so I hope it's okay.

Going back to a previous post:


We can set up a situation where there are two machines, located at different points in spacetime, outside of each other's lightcones (basically, you can think of the setup as two machines at different locations in space, but with everything happening at the same time), with the following property: each machine can be fed an angle as input, and will then produce one bit of output (let's call this "red" or "blue"). The interesting thing, observed empirically over many runs of this experiment, is that, if both machines are given the same angle as input, they always produce the same output. More generally, again as observed empirically over many runs of this experiment, out of all the situations where the difference between the two machine's input angles is theta, the proportion of times where both machines output "red" is about cos^2(theta)/2, the proportion where both output "blue" is also about cos^2(theta)/2, the proportion where Machine 1 outputs "red" and Machine 2 outputs "blue" is about sin^2(theta)/2, and the proportion where Machine 1 outputs "blue" and Machine 2 outputs "red" is about sin^2(theta)/2. That is, about cos^2(theta) of the time, the machines produce the same output.

In particular, let us simplify things even further, and only concern ourselves with three angles, each separated by 120 degrees. Then the setup is that each machine has three buttons (let's call them A, B, and C) out of which one can be pressed, after which a single bit of output is produced; if the same button is pressed on both machines, then they always produce the same output. However, if different buttons are pressed, then only about a quarter of the time do they produce the same output.

So what? Well, now let's think about modelling this with the particular mathematical construct of a probability distribution (on the space of all 30 possible complete input-and output-configurations; i.e., an assignment of a number >= 0 to each of these, with the total summing to 1). Which distribution? Well, let's discuss various properties such a distribution could have. It could be the case that each of the nine particular input-configurations is equiprobable (i.e., the machines' inputs are chosen independently at uniform random); we'll say that a distribution with this property "doesn't predict the inputs". It could be the case that the events at Machine 1 are independent of those at Machine 2, in the sense that the probability of each complete input-and-output configuration is simply the product of the probabilities of the corresponding Machine 1 and Machine 2 configurations; we'll say that a distribution with this property "fixes the common causes".

(Interrupting here)

Something I don't understand about the bolded part at the end. Here's a set of "Machine 1 and Machine 2 configuration":

{A, B}.

Here are four "complete input-and-output configurations":

{<A, red> <B, red>},
{<A, blue> <B, red>},
{<A, red> <B, blue>}, and
{<A, blue> <B, blue>}.

Each of those four "complete input-and-output configurations" has as its set of "corresponding Machine 1 and Machine 2 congfigurations," I take it, the very same set: {A, B}.

But how can the probability of each of those complete input-and-output configurations be the product of the probabilities of the corresponding Machine 1 configuration (in this case, "A") and Machine 2 configuration (in this case, "B")? Suppose the probability of Machine 1 being in configuration "A" is 1/3. Suppose the probability of Machine 2 bing in configuration "B" is 1/3. This should mean that each of the four complete input-and-output configurations listed above should have a probability of 1/9. But that causes the four probabilities to add up to 4/9. Shouldn't they add up to 1/9? (Since they are four exclusive possibilities given the particular set of Machine 1 and Machine 2 configurations, shouldn't their probabilities add up to the probability of that particular Machine 1 and Machine 2 configuration?

Now that I asked that, I see I didn't have to go through all the rigmarole. What I didn't understand was simply this: You said that there could be a probability distribution where the probability of (sorry for not knowing the notation here) "X and Y and Z" occuring is the same as the probability of "X and Y and W" occuring, both of which are the same as the probability of "X and Y" occuring. This could happen if Z and W both have a probability of 1. But otherwise, this doesn't seem possible to me.

What did I misunderstand?

Am I wrong to think that if A and B are exclusive and exhaustive possibilities given C, then the probabilities of A and B add up to the probability of C?

ETA: Oh. I bet I'm wrong to think that the product of the probabilities of the two machine inputs is the same as the probability of both inputs occuring. Something about independent and dependent probabilities here. Curse you, my ignorance of probability.

[CalMeacham]

The article said that the paddle was in it's quantum-mechanical ground state. Uh... wouldn't that be absolute zero, which nothing can reach?

No -- the ground state of an atom, a molecule, a harmonic oscillator, or a particle in a box (even one with walls at infinite potential) is not at absolute zero -- the particle wavefunction has a finite spread in space and a corresponding spread in momentum, neither of which is infinite or infintesimal in extent. The ground state is the lowest lying energy state, which isn't the state with no momentum.


see a book on quantum mechanics, or look up subjects like "harmonic oscillator" and look at the quantum mechanical version.

[Indistinguishable]

I forgot this thread existed. Partial zombie, I guess, but I intended to reply all along so I hope it's okay.
Well, it's certainly fine by me. (Incidentally, I have a partially typed up resurrection still intended in reply to your "Does physics support realism about transcendental numbers?" thread, which I keep putting off completing in the vain hopes that I can get the work I actually have to do done first. But when it comes, it will be glorious...)

Something I don't understand about the bolded part at the end. Here's a set of "Machine 1 and Machine 2 configuration":

{A, B}
By a "Machine 1 configuration", I mean a specification of both the input and the output at Machine 1, and likewise for a "Machine 2 configuration". So a machine 1 configuration might look like "Machine 1: A red". What you've given is not a set of Machine 1 and Machine 2 configurations; it's just a set of possible input values.

By a "complete input-and-output configuration", I mean a pair consisting of a Machine 1 configuration and a Machine 2 configuration. So, for example, a complete input-and-output configuration might look like this: (Machine 1: A red, Machine 2: B blue). Its component Machine 1 configuration is "Machine 1: A red" and its component Machine 2 configuration is "Machine 2: B blue". If the distribution fixes the common causes, then it needs to be the case that P(Machine 1: A red, Machine 2: B blue) = P(Machine 1: A red) * P(Machine 2: B blue), and so on for every other complete input-and-output configuration.

That I think was the most major confusion, but I'll address some of the other questions as well:

Now that I asked that, I see I didn't have to go through all the rigmarole. What I didn't understand was simply this: You said that there could be a probability distribution where the probability of (sorry for not knowing the notation here) "X and Y and Z" occuring is the same as the probability of "X and Y and W" occuring, both of which are the same as the probability of "X and Y" occuring. This could happen if Z and W both have a probability of 1. But otherwise, this doesn't seem possible to me.

What did I misunderstand?
Certainly, this would occur just in case, conditioned on X and Y, it is the case that W and Z has probability 1. But this needn't mean either W or Z has an unconditioned probability of 1. For example, take X, Y, Z, and W to all be "The (sole) coin I flip comes up heads".

Am I wrong to think that if A and B are exclusive and exhaustive possibilities given C, then the probabilities of A and B add up to the probability of C?
No, you are not wrong about that; this is (basically) correct. [That is, it's correct so long as A is entirely contained within C and likewise B is entirely contained within C]

ETA: Oh. I bet I'm wrong to think that the product of the probabilities of the two machine inputs is the same as the probability of both inputs occuring. Something about independent and dependent probabilities here. Curse you, my ignorance of probability.
Yes, P(X and Y) = P(X) * P(Y) if and only if X and Y are probabilistically independent.

[Indistinguishable]

(Thus, to say that a distribution fixes the common causes is to say that it makes machine 1 configurations probabilistically independent from machine 2 configurations.)

[TriPolar]

Hi Indistinquishable, crossing over from the other thread.

The math isn't that hard to deal with conceptually. I need to go over this some more to get all the terminology straight, and try to work out some of the details.
But in general you seem to be saying a local variable would add a factor that would change the statistical results, I guess because a degree of randomness has been removed. Am I at least looking in the right direction?

[Indistinguishable]

But in general you seem to be saying a local variable would add a factor that would change the statistical results, I guess because a degree of randomness has been removed. Am I at least looking in the right direction?
Unfortunately, I'm not sure what you mean; could you clarify?

[TriPolar]

Actually I'm figuring things out a little more. I guess probability is a better term than statitistics. So I'm hearing that the probabilities don't match up with conditions that would result from a local variable being present. I'm going to look things over more so I can form better questions. But I'm trying to discern how the local variable changes those conditions. I'm sensing that the definition of 'local variable' is based on a determination of the spin at the time of entanglement, not a choice that can appear to be random later at the time of the measurement.
If that doesn't make sense at all, I just need more time to look at this. Then I can ask 'less dumb' questions.

[TriPolar]

That last statement used words that don't seem to correspond in context. A better way to say it may be that if there was a 'hidden local variable', it couldn't have any affect on the probability of an outcome. The nature of a hidden variable would be a deterministic result in the spins. If the spin were determined at the time of the entanglement, the measured results wouldn't match the expected probability.
I am getting the sense that comparison of results from both particles, and the introduction of other factors in the definition of the probability can highlight that. And that sounds like the area I have to study up more on.
Veering off, I think I'm distracted by the simplistic idea that the hidden variable could cause the measured spin to be just as apparently random as without it.
Other explanations have presented a case that the there would be an appearance of information sharing that doesn't conform to the probability of the outcome if the resulting spin was truly random instead of predetermined. I understand that concept, but I'm not sure how the hidden variable would have that effect.
I'll let this stuff gestate in my brain for a while. I'm probably still finding the slots to stick each term and definition into, and the connections and conclusions won't come until that basic organization is established. Sort of like walking into a crowded room full of people you don't know in mid-conversation. After you identify the people, and who's talking to who, in a particular order, you can start to make sense of the conversations.