Is this popular description of entanglement correct?

  • #1
entropy1
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In some popularized discussions of entanglement, you often hear that:
IF particle A is found to be spin-up, "we know that" particle B "has" spin-down.
This seems to me not necessarily the case. In this formulation, particle A is viewed through measurement outcome and particle B through ontology. If the measurement basisses of Alice and Bob are parallel, Alice's outcome practically matches (the opposite of) Bob's. If the measurement basisses of Alice and Bob have a difference of 90 degrees, Alice's outcome is uncorrelated to Bob's. In the latter case if Alice measures spin-up, Bob will measure spin-up or spin-down with 50/50 probability (ie spin-right/spin-left).

Now, if we assume like in the quote that particle B "has" spin-down, this agrees with observation. But it is still an assumption it seems to me, because what Bob measures in the second example is 50/50 spin-up and spin-down.

I am wondering if we should not confuse measurement with ontology in this case.

So I wonder if the formulation in the quote accurately reflects entanglement as a phenomenon. If not, entanglement may be more subtle to interpret.

EDIT: Likely we can describe entanglement more accurately with the wavefunction.
 
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  • #2
phinds
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IF particle A is found to be spin-up, "we know that" particle B "has" spin-down.
Would be more properly stated as "We know that for spin-entangled particles if one particle is measured as having spin up then if/when the second particle is measured for spin, it will be measured to have spin down."
 
  • #3
entropy1
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Would be more properly stated as "We know that for spin-entangled particles if one particle is measured as having spin up then if/when the second particle is measured for spin, it will be measured to have spin down."
I don't agree that it will be measured spin down. It depends on the orientation of the measurement, as I mentioned in #1.
 
  • #4
Doc Al
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I don't agree that it will be measured spin down. It depends on the orientation of the measurement, as I mentioned in #1.
Obviously. To say something is measured to be "spin down" is meaningless unless a direction is specified.

In some popularized discussions of entanglement, you often hear that:
"IF particle A is found to be spin-up, "we know that" particle B "has" spin-down."

This only makes sense if both particles have their spins measured in the same direction. Obviously, if the second particle is measured in a completely different direction you may get different results.
 
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  • #5
phinds
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I don't agree that it will be measured spin down. It depends on the orientation of the measurement, as I mentioned in #1.
Yes. I was assuming the same orientation, but assumptions ...
 
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  • #6
entropy1
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This only makes sense if both particles have their spins measured in the same direction. Obviously, if the second particle is measured in a completely different direction you may get different results.
I agree. But that still says nothing about the ontology of the spin that particle B has before measurement. So then I feel the quote is incorrect.

Just today I heard Sean Carroll say "The only probabilites for anything are 0%, 100% and 50/50." I guess this would be a case of 50/50.
 
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  • #7
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I agree. But that still says nothing about the ontology of the spin that particle B has before measurement.

If you make the assumption that physics is local (the measurement of A does not disturb B) it follows that B had that spin (DOWN in your example) not only before B is measured but even before A is measured.

Given that we have strong evidence that physics is local it follows that we have equally strong evidence that B was in a DOWN state before any measurement took place.

Obviously, we are speaking about same-direction measurements.
 
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  • #8
vanhees71
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Would be more properly stated as "We know that for spin-entangled particles if one particle is measured as having spin up then if/when the second particle is measured for spin, it will be measured to have spin down."
More precisely: If two spin-1/2 particles are prepared in a spin singlet state and A measures spin up in ##z##-direction she knows that B will find spin down with certainty when also measuring his particle's spin in ##z##-direction.
 
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  • #9
entropy1
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More precisely: If two spin-1/2 particles are prepared in a spin singlet state and A measures spin up in ##z##-direction she knows that B will find spin down with certainty when also measuring his particle's spin in ##z##-direction.
That is a good description I think. But I find that often the description is not that accurate in more popular media. Figures.
 
  • #10
Doc Al
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Just today I heard Sean Carroll say "The only probabilites for anything are 0%, 100% and 50/50." I guess this would be a case of 50/50.
By choosing the appropriate angles for measuring the spins, you can arrange for any probability, not just those three.
 
  • #11
Doc Al
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If you make the assumption that physics is local (the measurement of A does not disturb B) it follows that B had that spin (DOWN in your example) not only before B is measured but even before A is measured.

Given that we have strong evidence that physics is local it follows that we have equally strong evidence that B was in a DOWN state before any measurement took place.
You might want to review Bell's Theorem and the work done to confirm the violation of Bell's inequalities.
 
  • #12
entropy1
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By choosing the appropriate angles for measuring the spins, you can arrange for any probability, not just those three.
I think he means that any angle is not perfectly 0% or perfectly 100%. Then, any probability between those can be weighted to be equivalent to 50%.
 
  • #13
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You might want to review Bell's Theorem and the work done to confirm the violation of Bell's inequalities.
I know Bell's work. What's your point?
 
  • #14
Doc Al
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I know Bell's work. What's your point?
My point is that what you posted was incorrect.
 
  • #15
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My point is that what you posted was incorrect.
Can you be more specific about what is incorrect about my post? Just quote the relevant part and explain what's wrong!
 
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  • #16
Doc Al
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Can you be more specific about what is incorrect about my post? Just quote the relevant part and explain what's wrong!
We can start here, where you seem to assume what you wish to conclude:
Given that we have strong evidence that physics is local
But let's not do it here, as this is not relevant to the thread. If you dispute Bell and its experimental tests, perhaps you should publish a paper.
 
  • #17
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If you dispute Bell and its experimental tests, perhaps you should publish a paper.
So, you disagree that we have strong evidence that physics is local? I also don't get what is circular about that. I assume locality to prove that the particle was in a spin DOWN state before the measurement. I'm not assuming locality to prove locality, which would be circular, indeed.

If you dispute Bell and its experimental tests, perhaps you should publish a paper.
I do not dispute any experimental test. I also do not dispute the validity of Bell's theorem. But since you didn't tell me what it is you think Bell proved I can't say if I dispute "Bell" or not.
 
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  • #18
vanhees71
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Bell assumed a general type of local deterministic hidden-variable theories and proved an inequality for certain correlation measures. This inequality is predicted to be violated by QT, including local relativistic QFTs, and all experimental tests disfavored with overwhelming significance the local deterministic HV theories and confirmed local relativistic QFT. For me the conclusion is clear.
 
  • #19
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Bell assumed a general type of local deterministic hidden-variable theories...
He assumed a type of local deterministic hidden-variable theories in which the hidden variables are independent of measurements' settings. Those are indeed ruled out. The others still remain on the table.
 
  • #20
PeterDonis
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He assumed a type of local deterministic hidden-variable theories in which the hidden variables are independent of measurements' settings. Those are indeed ruled out. The others still remain on the table.
The type of model in which the hidden variables are not independent of the measurement settings is called "superdeterminism". In such a model, you have to accept that, for example, in all of the experiments we have run on entangled particles that have shown violations of the Bell inequalities, the measurement settings for each individual particle were not independent of the process that produced the particles themselves. Those physical processes had some hidden connection between them that ensured that only certain combinations of particle states and measurement settings occurred, in order to make it seem like the world works according to quantum mechanics even though the actual underlying model is very different.

To say that such a model seems implausible is a vast understatement. But if you are going to take the position you are taking in what I quoted above, that is the kind of model you are committed to.
 
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  • #21
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The type of model in which the hidden variables are not independent of the measurement settings is called "superdeterminism".
Indeed.
In such a model, you have to accept that, for example, in all of the experiments we have run on entangled particles that have shown violations of the Bell inequalities, the measurement settings for each individual particle were not independent of the process that produced the particles themselves.
True.
Those physical processes had some hidden connection between them that ensured that only certain combinations of particle states and measurement settings occurred, in order to make it seem like the world works according to quantum mechanics even though the actual underlying model is very different.
The connection is not necessary hidden. What you need is an interaction (with infinite range) and we know for a fact that the source and detectors interact. They interact gravitationally (which is probably not that important) but they also interact electromagnetically (since they are composed of charge particles). Bell's statistical independence assumption is in fact a non-interaction assumption. In the general case, if N bodies interact their state would be described by a solution to the N-body problem, not by N solutions of the 1-body-problem, so their states are not independent. A Bell test is a particular case of an N body electromagnetical problem, where N is the number of charged particles (electrons and nuclei) taking part in the experiment, so, I would say, the statistical independence assumption is (almost) clearly false. I say almost, because it might be the case that independence is restored at the statistical level. At the level of single states there is no independence since the state of each object clearly depends on all N objects, that's a mathematical fact.

To say that such a model seems implausible is a vast understatement.
1. How do you estimate the probability of such a model to be true? Based on what?
2. Even if you can prove that this type of model is unlikely it doesn't mean much if the alternative (non-locality) isn't more likely. Can you tell me why would you ascribe a greater probability for the non-locality to be true? I would say the opposite is more reasonable, since superdeterminism does not violate any known physical principle, whyle non-locality does.

But if you are going to take the position you are taking in what I quoted above, that is the kind of model you are committed to.
Sure.
 
  • #22
PeterDonis
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What you need is an interaction (with infinite range) and we know for a fact that the source and detectors interact.
An interaction by itself is not enough. You need an interaction whose effects are very precisely tailored to make it look like quantum mechanics is correct, when in fact the underlying laws are very different. Ordinary gravitational or electromagnetic interactions will not do that.

How do you estimate the probability of such a model to be true?
"Implausible" is a judgment, not an estimate of probability. You are free to disagree with such a judgment; I am simply pointing out what such a disagreement commits you to. You appear to be fine with that.

superdeterminism does not violate any known physical principle, whyle non-locality does.
What "known physical principle" does non-locality (in the sense of violating the Bell inequalities) violate? Note that quantum mechanics, including quantum field theory, predicts Bell inequality violations, so you are claiming that QM/QFT violates some "known physical principle", which is a very, very strong claim.
 
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  • #23
vanhees71
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Again we have the problem that you and @AndreiB did not specify what you mean with "non-locality". The standard relativistic local QFTs are local, as the name local QFT says, and here locality means that local observables fulfill the microcausality constraint. It's also a realization of what Bell means by "locality" at least in his original papers containing his inequalities based on local deterministic hidden-variable models.

Of course local relativistic QFTs don't violate any known physical principles but are by construction a realization of those principles as far as causality in accordance with (special) relativity and the fundamental spacetime symmetries following from the spacetime model (proper orthochonous) Poincare invariance as well as the unitarity of the quantum-theoretical time-evolution is concerned.
 
  • #24
PeterDonis
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Again we have the problem that you and @AndreiB did not specify what you mean with "non-locality".
If you are referring to me, I did specify what I meant by it in post #22.
 
  • #25
vanhees71
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There you just claim the violation of Bell's inequality implies non-locality. Relativistic local QFT violates Bell's inequality too and it's local (in the sense of no FTL propagating causal effects).
 

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