Spooky action at a distance

[Umbran]

A little more precisely, the entire system (ie, both electrons in this case), is in a definite state in quantum terms, but that state does not correspond to a definite value of any measurement for the individual electrons.

Yah. I didn't want to try to explain being in a state that *doesn't* correspond to a measurement. It isn't exactly intuitive.

 

[freyar]

Yah. I didn't want to try to explain being in a state that *doesn't* correspond to a measurement. It isn't exactly intuitive.

No arguments there. I just don't like using technical words that have pretty clear precise definitions in ways that aren't technically correct. That said, it's really hard to come up with a non-technical word to replace "state" in this context, so I sympathize. Just wanted to be as clear as possible, since no state corresponds to measurement of ALL observables (except in a trivial system).

 

[Scott DeWar]

I [THINK I am understanding what the general statement is here.I say this as a hypothetical statement:

we have two electrons in a (And I use this only loosely) Schrodinger's quantum state where electron A and Electron B [here to fore to be referred as A and B] spinning in both clockwise and counterclockwise directions. How one is vied, the other, at the speed of C, influences the other to spin in the opposite direction.

If this was able to be used as a communication of data, you could have a VIRTUALLY unbreakable security as the source would, without any physical connection have a transfer of data stream from A to B with no way to intercept it. I see it would be a parity bit of sorts, but that could easily be countered with a Schmidt trigger, which inverts and sharpens a square wave.

Do I have this little bit of thought correct?

 

[Umbran]

we have two electrons in a (And I use this only loosely) Schrodinger's quantum state where electron A and Electron B [here to fore to be referred as A and B] spinning in both clockwise and counterclockwise directions. How one is vied, the other, at the speed of C, influences the other to spin in the opposite direction.

Nope. There is no influence that travels from one to the other. That's the difficult point. When you view one, the other is also in its corresponding state instantaneously.

One way to think of it is that, once entangled, they are *not* really separate objects. They are both now part of one entangled thing that is not directly analogous to one single physical object or two separate physical objects.

If this was able to be used as a communication of data, you could have a VIRTUALLY unbreakable security as the source would, without any physical connection have a transfer of data stream from A to B with no way to intercept it. I see it would be a parity bit of sorts, but that could easily be countered with a Schmidt trigger, which inverts and sharpens a square wave.

No, because there's no communication between the two.

When used as a communication of data, what this allows is a cryptographic key that can only be known by the two ends, because eavesdropping on it breaks the entanglement.

 

[Scott DeWar]

**sigh**

 

[freyar]

**sigh**

Don't feel too bad --- we don't teach this until people have had 3 to 4 years of specialized university courses in physics already. But the point is that it's weird and totally counterintuitive. It's kind of amazing people were able to figure this stuff out in the first place.

 

[freyar]

As I envision, entangled particles become a system. We can know the internal properties of the system "the particles will assume opposite spin" -- assuming no interference to destabilise it -- but we can't predict external specifics of a part of the system "this particle will spin up" until we measure at least one part. Once we have a measurement, we can use it and the expected properties of the system to predict how other parts of the system behaved.

To some extent --- if I make one measurement on electron A and you make a different measurement on electron B, then neither of us can predict with certainty how the other's measurement will go (though we could work out probabilities).

The only part that "feels weird" to me is Bell's Theorem. I would be more at ease if we could say "there are hidden variables that underlie the system, but we can't get to them". I guess we're just getting too close to the underlying limits of the universe simulation...

Except the point of Bell's theorem is that there isn't a (deterministic) simulation of the universe. That would be a hidden variable.

 

[Nagol]

To some extent --- if I make one measurement on electron A and you make a different measurement on electron B, then neither of us can predict with certainty how the other's measurement will go (though we could work out probabilities).



Except the point of Bell's theorem is that there isn't a (deterministic) simulation of the universe. That would be a hidden variable.

Different measurements would destroy entanglement wouldn't they? That's what I meant by something interfering.

Well, there can still be hidden variables, just not local ones (I am going on relatively old and unused memories here so am probably wrong). and the simulation would definitely not be local!

 

[Umbran]

Different measurements would destroy entanglement wouldn't they? That's what I meant by something interfering.

Yes. But, until you measure it, it doesn't have external specifics to predict! Any interaction in which it shows that it was in a specific measurable state *is* a measurement.

 

[freyar]

Different measurements would destroy entanglement wouldn't they? That's what I meant by something interfering.

Oh, I didn't mean another measurement or interference outside of the experiment that ruins the entanglement. What I mean is that the person at A could measure whether the electron in spinning up or down while the person at B measures whether that electron is spinning left or right. From their result, the person at A only knows that the electron at B has a 50% chance of spinning left vs 50% chance of spinning right.

Well, there can still be hidden variables, just not local ones (I am going on relatively old and unused memories here so am probably wrong). and the simulation would definitely not be local!

Yes, I do believe that's correct. But now you run into Occam's razor.