# Bells inequality be satisfied with equivalent local QM?

1. Nov 9, 2009

### Gerenuk

I've just listened to an online lecture where Susskind explained Bell's inequality
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Basically he shows that classically
$$A\cap \overline{B}+B\cap \overline{C}\geq A\cap\overline{C}$$
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
$$\overline{C}_1\equiv C_2[/itex] (subscripts denote on which particle it is measured) and a new theory might not imply that, even though it still predicts all the same observables as QM. To me it seems like cheating to modify Bell inequality to [tex] A_1\cap \overline{B}_2+B_2\cap \overline{C}_1\geq A_1\cap C_2$$
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?

Last edited by a moderator: Sep 25, 2014
2. Nov 17, 2009

### DrChinese

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.

Last edited by a moderator: Sep 25, 2014
3. Nov 19, 2009

### 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?

4. Nov 19, 2009

### f95toli

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).

5. Nov 19, 2009

### 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.

6. Nov 20, 2009

### DrChinese

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?

7. Nov 20, 2009

### Gerenuk

I think I wasn't clear enough. How would you go on to prove $\overline{C}_1\equiv C_2$ 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 $|\psi|^2$ 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 $\overline{C}_1\equiv C_2$ 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 $P(\overline{C}_1)=P(C_2)$ this way.
I understand the derivation on your site, but I do not see a way to justify mathematically correct the assumption ${X}_1\equiv {X}_2$

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

8. Nov 21, 2009

### DrChinese

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 $\overline{C}_1\equiv C_2$, 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.

9. Nov 21, 2009

### Gerenuk

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 $P(\overline{C}_1)=P(C_2)$, but not that these states are equivalent, i.e. $\overline{C}_1=C_2$. 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".

10. Nov 21, 2009

### DrChinese

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.

11. Nov 21, 2009

### DrChinese

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.

12. Nov 21, 2009

### Gerenuk

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

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

13. Nov 21, 2009

### Gerenuk

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.

14. Nov 21, 2009

### DrChinese

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.

15. Nov 21, 2009

### DrChinese

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?

16. Nov 21, 2009

### 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.

17. Nov 21, 2009

### Staff: Mentor

No, it doesn't.

18. Nov 21, 2009

### Gerenuk

Yes.

But actually as I think more about it, I think you are right that in equations one can replace $\overline{C}_1=C_2$. 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?

19. Nov 22, 2009

### DrChinese

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

20. Nov 24, 2009

### 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?