Bells inequality be satisfied with equivalent local QM?

[Gerenuk]

I've just listened to an online lecture where Susskind explained Bell's inequality
()
Basically he shows that classically
[tex]
A\cap \overline{B}+B\cap \overline{C}\geq A\cap\overline{C}
[/tex]
Then he uses spin measurements with 0°, 45°, 90° to the z-axis for A, B, C to measure spins of an electron singlet. The important point is that he uses the fact that measuring a negative result A on particle 1 corresponds identically to a positive outcome on particle 2. This assumption is essential, otherwise he couldn't do all three combinations of measurement on separate particles. For me that seems to be a logical flaw if you really want to disprove hidden variable theories.

Why can't you assume that there is a theory that predicts all the same outcomes for the observables just as quantum mechanics, but is local?
The above argument is only valid if you assume
[tex]\overline{C}_1\equiv C_2[/itex]
(subscripts denote on which particle it is measured)
and a new theory might not imply that, even though it still predicts all the same observables as QM. To me it seems like cheating to modify Bell inequality to
[tex]
A_1\cap \overline{B}_2+B_2\cap \overline{C}_1\geq A_1\cap C_2
[/tex]
and be surprised why it's violated.
Or is there another version of Bell measurements where all combinations of measurements are really repeated and free from "conversions"?
Otherwise an alternative framework which predicts all the same outcomes as QM might well be local. Or is there a non-existence prove of that?

 

[DrChinese]

I've just listened to an online lecture where Susskind explained Bell's inequality
()
Basically he shows that classically
[tex]
A\cap \overline{B}+B\cap \overline{C}\geq A\cap\overline{C}
[/tex]
Then he uses spin measurements with 0°, 45°, 90° to the z-axis for A, B, C to measure spins of an electron singlet. The important point is that he uses the fact that measuring a negative result A on particle 1 corresponds identically to a positive outcome on particle 2. This assumption is essential, otherwise he couldn't do all three combinations of measurement on separate particles. For me that seems to be a logical flaw if you really want to disprove hidden variable theories.

Why can't you assume that there is a theory that predicts all the same outcomes for the observables just as quantum mechanics, but is local?
The above argument is only valid if you assume
[tex]\overline{C}_1\equiv C_2[/itex]
(subscripts denote on which particle it is measured)
and a new theory might not imply that, even though it still predicts all the same observables as QM. To me it seems like cheating to modify Bell inequality to
[tex]
A_1\cap \overline{B}_2+B_2\cap \overline{C}_1\geq A_1\cap C_2
[/tex]
and be surprised why it's violated.
Or is there another version of Bell measurements where all combinations of measurements are really repeated and free from "conversions"?

Actually there is such. Type I parametric down conversion leaves both particles in the same state in the same basis. Therefore:

[tex]C_1 \equiv C_2[/itex]

I personally don't see that it matters anyway, but hopefully you can see that issue does not play into things.

 

[Gerenuk]

I looked at link you gave in the other thread. It also mixed up measurement on particle one and particle two, so stricly speaking you cannot add or cancel probabilities.

I'm not sure what parametric down conversion is. But it is from QM theory?
It's logically incorrect to rely on the whole QM framework (rather than just the probability) predictions to disprove non-QM theories. If QM was incomplete (even though correct in predictions), then you couldn't do your argument.

I have found one way to get around it, with correctly labeling the measurements, but maybe you can explain a better way to make this logically correct?

 

[f95toli]

Why can't you assume that there is a theory that predicts all the same outcomes for the observables just as quantum mechanics, but is local?

You can but they you have to give up realism. Bell's inequality demonstrates that theories that are both local and realistic can not be correct.
Hence, non-realistic (but local) theories as well as (obviously) non-local non-realistic theories can be made to work.

There is a class of inequalities named after Tony Leggett that when tested seem -according to some- support non realistic alternatives (there was a paper about this is Nature last year).

 

[Gerenuk]

I was specifically asking about a step in Bells prove. I'd like to understand all assumptions.
But I go and also have a look at the paper you mentioned.

 

[DrChinese]

I'm not sure what parametric down conversion is. But it is from QM theory?
It's logically incorrect to rely on the whole QM framework (rather than just the probability) predictions to disprove non-QM theories. If QM was incomplete (even though correct in predictions), then you couldn't do your argument.

PDC is a method of generating entangled photon pairs. A single input photon passes through a special crystal and 2 entangled photons come out (this is a simplified description). Bell tests do not rely on the QM predictions per se, although they do yield results consistent with the predictions of QM.

Now that you know that the labeling is not an issue, do you still have questions about Bell's Theorem?

 

[Gerenuk]

PDC is a method of generating entangled photon pairs. A single input photon passes through a special crystal and 2 entangled photons come out (this is a simplified description).

I think I wasn't clear enough. How would you go on to prove [itex]\overline{C}_1\equiv C_2[/itex] by a mathematically rigorous way? Remember, you are not allowed to use any QM formalism, because you have not proven the theorem, that QM is the complete underlying theory. You know that after crunching the numbers every time [itex]|\psi|^2[/itex] is consistent with experiment, but you do not know if crunching other number another way wouldn't be just as good.
And an experimental test for [itex]\overline{C}_1\equiv C_2[/itex] doesn't exist either. since just observing the same probabilities doesn't rule out a difference in some "hidden" variables" for these states. I mean you are only proving [itex]P(\overline{C}_1)=P(C_2)[/itex] this way.

Now that you know that the labeling is not an issue, do you still have questions about Bell's Theorem?

I understand the derivation on your site, but I do not see a way to justify mathematically correct the assumption [itex]{X}_1\equiv {X}_2[/itex]

Later I want to check if finite size detector errors and detector hidden variables can are an issue.

 

[DrChinese]

I think I wasn't clear enough. How would you go on to prove [itex]\overline{C}_1\equiv C_2[/itex] by a mathematically rigorous way? Remember, you are not allowed to use any QM formalism, because you have not proven the theorem, that QM is the complete underlying theory. You know that after crunching the numbers every time [itex]|\psi|^2[/itex] is consistent with experiment, but you do not know if crunching other number another way wouldn't be just as good.
And an experimental test for [itex]\overline{C}_1\equiv C_2[/itex] doesn't exist either. since just observing the same probabilities doesn't rule out a difference in some "hidden" variables" for these states. I mean you are only proving [itex]P(\overline{C}_1)=P(C_2)[/itex] this way.

I am confunsed by your question. You don't need to prove a QM formalism in Bell. You only need to know what QM predicts.

As to [itex]\overline{C}_1\equiv C_2[/itex], I am similarly confused. This is part of the experiment. With Type I PDC, each trial (1 entangled pair) produces identical results for Alice and Bob for identical settings (within experimental limits).

You justify by experiment. The question is whether the identical results are because a) the system is entangled as predicted by QM; or b) there are hidden variables as predicted by local realism. To start with, we check to see if both a) and b) are possibilities. If we see perfect correlations, then they are both possibilities. THEN, we do the Bell test.

 

[Gerenuk]

As to [itex]\overline{C}_1\equiv C_2[/itex], I am similarly confused. This is part of the experiment. With Type I PDC, each trial (1 entangled pair) produces identical results for Alice and Bob for identical settings (within experimental limits).

I think it is better to write it out in mathematical notation, because logic might be deceiving.
Identical results do not imply that the states (or hidden variables) were the same!
What you are actually proving by experiment is that [itex]P(\overline{C}_1)=P(C_2)[/itex], but not that these states are equivalent, i.e. [itex]\overline{C}_1=C_2[/itex]. You see the difference and agree?
A simple example, which is not the most general that one can think of, is where hidden variables are disregarded for the measurement but still existent. Thus the states differ in their hidden variables and are stricly speaking not identical.

Without being able to prove the equivalence you cannot do the Bell test, which as written out mathematically, only applies when you really use the same states and not some additional conversion to "equivalent states".

 

[DrChinese]

I think it is better to write it out in mathematical notation, because logic might be deceiving.
Identical results do not imply that the states (or hidden variables) were the same!
What you are actually proving by experiment is that [itex]P(\overline{C}_1)=P(C_2)[/itex], but not that these states are equivalent, i.e. [itex]\overline{C}_1=C_2[/itex]. You see the difference and agree?
A simple example, which is not the most general that one can think of, is where hidden variables are disregarded for the measurement but still existent. Thus the states differ in their hidden variables and are stricly speaking not identical.

Without being able to prove the equivalence you cannot do the Bell test, which as written out mathematically, only applies when you really use the same states and not some additional conversion to "equivalent states".

I really have no clue as to what you are talking about. I can see you have something you are trying to "prove", but you might want to start at the beginning rather than the end. You might also want to consider that others won't agree with you, even though you are convinced you are "right".

The EPR argument was that 2 paticles that interacted in the past would have certain predictable future elements - they were called elements of reality" - based on conservation rules. They said a more complete specification of state was possible. Bell showed us that the predictions of QM - let's say for polarizations of a pair of entangled photons - would not be consistent with the reasonable assumptions of local realism (a la EPR). It doesn't matter whether you do the test or not, the Bell conclusion is the same.

For the tests themselves: If you choose to nitpick over the difference between "same states" versus "equivalent states", I think you are completely missing the point. It doesn't matter at all as long as there are elements of reality.

 

[DrChinese]

Why can't you assume that there is a theory that predicts all the same outcomes for the observables just as quantum mechanics, but is local?

...

Or is there another version of Bell measurements where all combinations of measurements are really repeated and free from "conversions"?
Otherwise an alternative framework which predicts all the same outcomes as QM might well be local. Or is there a non-existence prove of that?

After re-reading your first post: Are you simply asking if QM is local? The answer to this depends on interpretation.

Or are you specifically asking if there is a local realistic theory possible which makes the same predictions as QM? The answer to this is NO, per Bell.

 

[Gerenuk]

I think you are not reading all of what I write.

You might also want to consider that others won't agree with you, even though you are convinced you are "right".

I suppose someone out there has a correct prove and I'd like to read that. But your prove on the website is wrong because it makes an assumption which doesn't have to be true.
I summarize:
-I said it is important to prove the assumption C1=C2
-You said experiments do that.
-I said, no, experiments only prove P(C1)=P(C2)
I gave you a simple example which illustrates that:
Imagine a function P being defined as 1 if the first letter in a two-letter word is A.
Let's say the possible outcomes in that model world are (AX,AX), (AX,AY) and (BX,BY)
Now for a states (S1,S2) of these three, P will give you P(S1)=1 and P(S2)=1 in about 2/3s of the cases. And yet for half of these "positive results" S1 is not equal to S2. This difference in the second letter might be crucial for other experiments however!

For the tests themselves: If you choose to nitpick over the difference between "same states" versus "equivalent states", I think you are completely missing the point.

C1 and C2 are neither the same nor equivalent. They have merely the same probabilities.

 

[Gerenuk]

After re-reading your first post: Are you simply asking if QM is local? The answer to this depends on interpretation.
Or are you specifically asking if there is a local realistic theory possible which makes the same predictions as QM? The answer to this is NO, per Bell.

Actually I only want to know the very precise assumptions of Bell's prove. For me it's important to know a mathematically rigorous prove.
I know people who think "Mermin said in 2D there is no ordering". These people haven't looked at the prove and thus don't know the assumptions. That's why they are surprised when their "law" fails, whereas other can predict when the law is applicable.

 

[DrChinese]

Actually I only want to know the very precise assumptions of Bell's prove. For me it's important to know a mathematically rigorous prove.
I know people who think "Mermin said in 2D there is no ordering". These people haven't looked at the prove and thus don't know the assumptions. That's why they are surprised when their "law" fails, whereas other can predict when the law is applicable.

Actually plenty of people have looked over the Bell proofs. You are the only person I have ever heard mention state identity as an issue. The question is usually put in terms of "elements of reality", which is defined in EPR.

 

[DrChinese]

I think you are not reading all of what I write.

-I said it is important to prove the assumption C1=C2
-I said, no, experiments only prove P(C1)=P(C2)
I gave you a simple example which illustrates that:
Imagine a function P being defined as 1 if the first letter in a two-letter word is A.
Let's say the possible outcomes in that model world are (AX,AX), (AX,AY) and (BX,BY)
Now for a states (S1,S2) of these three, P will give you P(S1)=1 and P(S2)=1 in about 2/3s of the cases. And yet for half of these "positive results" S1 is not equal to S2. This difference in the second letter might be crucial for other experiments however!

Can you put your example in physical terms? It makes little sense as written. Are you trying to somehow identify hidden variables as X and Y?

In a traditional example, Alice and Bob measure pairs of entangled photons with identical polarizer settings.

They get results such as the following, where the first element is Alice's result and the second is Bob's:

(1,1), (0,0), (0,0), (1,1), (0,0), (1,1), (1,1), (0,0), etc.

So the states are identical as measured. This is verified experimentally, and does not imply that the particles are identical or that other "unmeasured" observables are or are not identical. But the measured observable is an element of reality. Now, what are you asking?

 

[friend]

Sorry for my ignorance, but does Bell's inequality have anything to do with the bell curve, the normal distribution of completely random processes? It occurs to me that a bell curve in physics would indicate that no other determinative, mathematical structure can be involved, or no "hidden variable" can be used to describe a completely random process.

 

[jtbell]

does Bell's inequality have anything to do with the bell curve, the normal distribution of completely random processes?

No, it doesn't.

 

[Gerenuk]

It makes little sense as written. Are you trying to somehow identify hidden variables as X and Y?

Yes.

But actually as I think more about it, I think you are right that in equations one can replace [itex]\overline{C}_1=C_2[/itex]. Sorry about that.
I prefer another solution to what I saw as a problem, but that's similar.

I'd like to check another point and maybe you can help me with a reference about it. I want to check if it's possible that entangled states do not very perfectly obey quantum mechanical correlation predictions. So I'd need a reference where they find an upper bound for the deviation from perfect correlation. Experiments are never perfect and so there must be a finite limit that experimenters give?

 

[DrChinese]

I'd like to check another point and maybe you can help me with a reference about it. I want to check if it's possible that entangled states do not very perfectly obey quantum mechanical correlation predictions. So I'd need a reference where they find an upper bound for the deviation from perfect correlation. Experiments are never perfect and so there must be a finite limit that experimenters give?

Yes, there is a small amount of experimental noise in experiments. I have a couple of references which talk about this a bit:

http://arxiv.org/abs/quant-ph/9806043

http://arxiv.org/PS_cache/quant-ph/pdf/0205/0205171v1.pdf

The second reference especially talks about the setup and Type I PDC in good detail, I often refer to this. Also, another good reference - which is not about Bell but uses a similar setup and demonstrates the quantum nature of light:

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

 

[Gerenuk]

Thanks for the reference! I try to think about them in detail.

Another question: Is this non-locality observed for spins only, or can you construct similar experiments for particle position and momentum?

 

[DrChinese]

Another question: Is this non-locality observed for spins only, or can you construct similar experiments for particle position and momentum?

Conceptually, you can do it for any pair of non-commuting observables. A number of experiments have been done with time-energy for example. And there can be some other observables generated that show entanglement too, going beyond the usual quantum observables. So the answer is that spin does not occupy a special place. It is simply easier to setup (usually) and easier to talk about.

 

[Gerenuk]

OK, I was just wondering if there were enough way to measure non-commuting operators to set up a Bell experiment.

You seem to be well informed. Maybe you know the answer to this question:
Most people try to measure the states of the two particle as simultaneously as possible to avoid "communication". I'm rather interested in whether that state of the second particle is really constant and would be the same if the second particle is measured much later than the first. This is to avoid temporal correlations. Do you know if there are experiments performed to detect this? Obviously light isn't a good candidate for this.

 

[zonde]

I'm rather interested in whether that state of the second particle is really constant and would be the same if the second particle is measured much later than the first. This is to avoid temporal correlations. Do you know if there are experiments performed to detect this? Obviously light isn't a good candidate for this.

If entanglement is statistical correlation of two measurements then the limitation is how well you can preserve the sate of particle and there should be no other limitation regarding time till measurement.
I do not know about Bell inequality tests that try to test large asymmetry in time till measurement but you might consider http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser" as an experimental test for time independence of entanglement.

 

[Gerenuk]

If entanglement is statistical correlation of two measurements then the limitation is how well you can preserve the sate of particle and there should be no other limitation regarding time till measurement.

Yes, and I'd like to see an experiment being able to conserve the state of a particle an appreciable time.

I do not know about Bell inequality tests that try to test large asymmetry in time till measurement but you might consider http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser" as an experimental test for time independence of entanglement.

I quickly read through it and might have missed a point. In the eraser experiments some causility dependences are checked, but still the time asymmetry isn't really large?!
I'd like to see an experiment detecting a predicted state arising from a split up in entanglement seconds after it has been created.

 

[zonde]

I'd like to see an experiment detecting a predicted state arising from a split up in entanglement seconds after it has been created.

I found this one with around 0.5 ms delay:
"High-fidelity transmission of polarization encoded qubits from an entangled source over 100 km of fiber" - http://arxiv.org/abs/0801.3620"
There one photon is measured locally but the other is sent over 100km fiber. That makes delay around 0.5 ms.
Delay of 1s will require 200 000 km fiber - seems like very far from realistic experiment.

 

[Gerenuk]

Oh OK. Thanks for your research.

That's why I thought light might not be the ideal candidate to check to time-independence. With a particle it might be easy to trap it for a while and read it out later.

Are most Bell experiments performed with light then?

 

[Hootenanny]

Are most Bell experiments performed with light then?

Yes. In fact, I am not aware of any Bell test experiments that do not use photons.

 

[zonde]

That's why I thought light might not be the ideal candidate to check to time-independence. With a particle it might be easy to trap it for a while and read it out later.

But how many coincidences do you want in that delay interval? In that reference I gave there was 100coinc./s rate reported and it makes 1 coincidence in 10ms on average.

As well maybe that can be interesting for you:
Hnilo et al "Low Dimension Dynamics in the EPRB Experiment with Random Variable Analyzers"
http://www.springerlink.com/content/0136072643546224/"
There is some time dependence analysis performed.

Are most Bell experiments performed with light then?

Yes

As an example of non photon experiment:
http://arxiv.org/abs/quant-ph/0310112"

And you can look in this paper for references to non photon experiments:
http://arxiv.org/abs/quant-ph/0701071"

 

[Hootenanny]

And you can look in this paper for references to non photon experiments:
http://arxiv.org/abs/quant-ph/0701071"

Very interesting! Thanks for posting this, zonde. I'll have to read through that this weekend.

 

[zonde]

Very interesting! Thanks for posting this, zonde. I'll have to read through that this weekend.

You are welcome. Probably you should thank DrChinese as he have posted this paper in different threads.
Quite some number of references in this paper. But do not be cheated in thinking that there are references to all major types of experiments. For example, I think there wasn't references to proton spin correlation experiments like the one I posted (I hope I'm not wrong ).

 

[Galteeth]

How do you know in this experiment that the correlations are from specific particle pairs as oppossed to a general statistical interaction (like in some hypothetical casino where a roulette wheel that turns up black will make some other whell in the casino turn up red)?

 

[DrChinese]

How do you know in this experiment that the correlations are from specific particle pairs as oppossed to a general statistical interaction (like in some hypothetical casino where a roulette wheel that turns up black will make some other whell in the casino turn up red)?

When you look at red and black (which correspond in the analogy to settings of 0 or 90 degrees) alone, you may not notice the entanglement. I.e. it may not appear very persuasive. But when you look at a variety of angles other than 0 and 90 degrees, you can see that the particles act as a system and not as independent particles.

 

[Galteeth]

When you look at red and black (which correspond in the analogy to settings of 0 or 90 degrees) alone, you may not notice the entanglement. I.e. it may not appear very persuasive. But when you look at a variety of angles other than 0 and 90 degrees, you can see that the particles act as a system and not as independent particles.

No, I understand they act as a system. What I meant was, how do you know that specific particle pairs are correlated as oppossed to a more general systemic co-ordination?

 

[DrChinese]

No, I understand they act as a system. What I meant was, how do you know that specific particle pairs are correlated as oppossed to a more general systemic co-ordination?

Not sure I follow. Experimental correlation IS evidence of entanglement. Entanglement means the pair is acting as a system. What other option do you propose as a "general systemic coordination" ? Unentangled pairs do not exhibit cos^2 correlations.

But not all correlations are equal indicators of entanglement. For example, "perfect" correlations (0 degrees) indicate entanglement and this is the easiest way to calibrate the apparatus. But this alone does not violate a Bell inequality.

 

[Galteeth]

Not sure I follow. Experimental correlation IS evidence of entanglement. Entanglement means the pair is acting as a system. What other option do you propose as a "general systemic coordination" ? Unentangled pairs do not exhibit cos^2 correlations.

But not all correlations are equal indicators of entanglement. For example, "perfect" correlations (0 degrees) indicate entanglement and this is the easiest way to calibrate the apparatus. But this alone does not violate a Bell inequality.

I wasn't saying that what I was asking was likely per ce, just wanted to see if there is a way to know for sure. What I mean is, hypothetically, photon pair 1 and 2 could give uncorrelated results, and photon pair 3 and 4 could give uncorrelated results, but taken together as a system there could be a perfect systematic correlation. I know that's unlikely, but is there a way to know that isn't the case?