# Does Bell's theorem imply stronger or weaker correlation?

1. Aug 18, 2011

### lugita15

After reading so many threads here about it, I thought I would take a closer look at Bell's theorem. Consider the simple proof presented http://quantumtantra.com/bell2.html" [Broken], which I'll summarize below:

A light source produces twin state photons, and each photon of the pair goes through a polarizing filter oriented at some angle. If the two filters are oriented at 0 degrees, then it is found that the polarization of the two photons are perfectly correlated, i.e. the error rate is zero. A local realist might say that the two photons were given a definite polarization at the source. If the one of the filters is turned 30 degrees counter clockwise with respect to the other, then the error rate is found to be 25%. A local realist might say that the polarization of one out of every four photons that go through the 30-degree polarizer is changed from the initial polarization it had. So then if you turned both of the filters by 30 degrees in opposite direction, the maximum error rate you would get would be 25%+25%=50%, and the actual error rate would be even less because two simultaneous errors cancel each other out. So this is a version of Bell's inequality: the error rate at 60 degrees is less than or equal to 50%. Experimentally, this inequality has been disproven, as the error rate has been found to be 75%, so this local realist model doesn't seem to work.

My question is, what does this say about how strongly correlated entangled particles are? In the article, it is stated that local realism is disproved because of "the nature of the strong correlations observed". But that doesn't make sense to me. The local realist was expecting the particles to be correlated so strongly that even if you turned the filter by a large angle like 60 degrees, the error rate would still not exceed 50%. So isn't the experimental result that entanglement particles are more weakly correlated than allowed by local realism?

Or is it that the local realist would consider the 75% error rate at 60 degrees to be normal, but would be astounded that the error rate is so low at 30 degrees, so that the particles are more strongly correlated than local realism allows?

Any help would be greatly appreciated.

Last edited by a moderator: May 5, 2017
2. Aug 18, 2011

### xts

I recommend great, and easy to understand explanation of the strong/weak correlation and impossibility to explain them by means of local variables, presented by David Mermin:
David Mermin, Is the moon there when nobody looks?, Physics Today, Apr.1985, (google it to get free copy)

3. Aug 18, 2011

### DrChinese

Great question! The correlations are less than expected (by the local realist) at some angles, and more at other angles. For the sample settings you gave, it was less.

When they say that entanglement is strong, they really mean that the angle theta (difference between the 2 settings) is dominant to the statistics. That is consistent with a so-called contextual viewpoint. That is different than the local realist, who says there is an internal structure (hidden from us apparently) that is dominant to the outcomes.

4. Aug 18, 2011

### lugita15

How exactly can you tell whether the correlations are more or less than expected? The observed error rate is $R\left(\theta\right) = 1-cos^{2}\theta$. What requirements must a local realist's error rate satisfy? As shown above, $R\left(2\theta\right)\leq 2R\left(\theta\right)$ (at least when the angles are acute) is a necessary condition, but is it a sufficient condition? What would the graph look like?

In that case, does local realism set some upper bound for $\left|\frac{dR}{d\theta}\right|$, and does quantum mechanics violate this bound?

5. Aug 18, 2011

### xts

You must make an experiment with apparata measuring particles at at least 3 different axes to spot the Bell's inequality violation, and thus to deny local reality.

The easiest case are three axes differing by 120°.

Once again - I really recommend Mermin's article (cited above), explaining that on an example with gramophone-like machines throwing balls.

Last edited: Aug 18, 2011
6. Aug 18, 2011

### DrChinese

See the graph at Fig. 3 to see the essence of what you are describing:

http://drchinese.com/David/Bell_Theorem_Negative_Probabilities.htm

The local realist optimally needs a straight line function so that it is observer independent (although this is somewhat model dependent since there are local realistic models that are even further off from the quantum mechanical expectation values).

7. Aug 18, 2011

### lugita15

So then is there something wrong with the proof in the OP, which only involves two angles?

8. Aug 18, 2011

### DrChinese

http://www-f1.ijs.si/~ramsak/km1/mermin.moon.pdf

9. Aug 18, 2011

### xts

Nope! I was wrong, of course, posting too quick answer, having Mermin's machine in mind. You picked my post before I correct it ;)
You can't explain in terms of local reality even correlation between two sets, differing by more than 90°, you don't need the 3rd one.

10. Aug 18, 2011

### DrChinese

I think you will determine there are 3 angles in that. You need 3 to see the Bell logic emerge.

11. Aug 18, 2011

### xts

I am getting lost.... That was what I thought at quick, but it seems that two may be enough.
Just Mermin's machines, whith a lever limited to two positions. Is it explainable in terms of coloured balls? I am confused. Let me try...

12. Aug 18, 2011

### lugita15

No, the Bell inequality in the proof just compares the correlations at 30 degrees and 60 degrees.

13. Aug 18, 2011

### DrChinese

Your example is A=-30, B=0 and C=+30. Which is equivalent to A=0, B=60 and C=30 and similar.

14. Aug 18, 2011

### DrChinese

Just look at the equation and you will see 3 terms (because there are 3 angles): 25%+25%=50%

15. Aug 18, 2011

### xts

Dr.Chinese (and my first quick answer...) were right. We need at least 3 axes.
For every two-axes Mermin machine, I always make ball mixture in proportion RR, GG - $\frac{\cos^2\theta}{2}$ each, RG,GR $\frac{1-\cos^2\theta}{2}$ each

Last edited: Aug 18, 2011
16. Aug 18, 2011

### lugita15

OK, I thought the 0 degrees didn't count because that was just used to establish that the photons were in the twin state.

Now that we've got that out of the way, I would also like to know how the no-conspiracy condition enters the picture. Alice has two choices, to keep the filter straight or turn it by 30 degrees counterclockwise, and similarly Bob can either keep the filter straight or turn it by 30 degrees clockwise. If you were a god who could control the choices each person would make, how could you conspire to make Bell's inequality seem to be violated when it really wasn't?

17. Aug 18, 2011

### DrChinese

If I knew the outcomes in advance, I could have them select angle settings that would NOT yield a representative sample. If you were throwing 3 dice and only looking at the 2 I pick, I could convince you that 7 rarely occurs. Or that matches occur with unusual frequency, etc.

18. Aug 18, 2011

### DrChinese

Of course, in the seminal experiments, the selection is actually done independently by high speed selection/switching mechanisms where there is no human intervention. Computers perform the pseudo-random selection, but they are separated and otherwise independent. So one heck of a conspiracy.

19. Aug 18, 2011

### lugita15

Well, a pseudo-random algorithm is determined in some absurdly complicated way by the initial conditions, so if you were a god who knew in advance exactly what result the algorithm would reach, you could set the initial conditions such that they conspire to produce the desired result. Of course, the number of variables you would have to take into account may be ridiculous, but in principle it could be done.

But what if you didn't use pseudo-randomness, but instead the randomness of quantum mechanics? For instance, in addition to the two photons in the twin state you can bring in a third (unpolarized) photon, and put it through another polarizing filter. If it goes through, Alice keeps her filter straight, and if it doesn't go through Alice turns her filter by 30 degrees. I don't know whether an experiment like this has been done, but I'd imagine it would be easy to set up. Such an experiment would almost rule out conspiracies, but there is still one possibility: maybe the third photon is in on the conspiracy, and it's entangled with the other photons! In other words, whether it goes through or not actually depends on whether what our original photons do.

20. Aug 18, 2011

### DrChinese

Well, keep in mind that ALL scientific experiments are subject to such conspiracies! Bell tests are hardly unique in that.