# Hidden variables and Bell's inequality

1. Nov 23, 2015

### morrobay

[Mentor's note: Moved from a thread about field theories as this is just about the basic meaning of the theorem for ordinary entangled particles]]
Suppose that there are photon spin outcomes that are pre existing from entanglement or from local hidden variables.
For every detector angle Alice selects there is a result that is correlated with every detector angle Bob selects.
With 10 units that would be 360 x 360 mutual outcomes.
* In order to have the probabilities that produce correlations that match quantum mechanical predictions;
opposite results cos2 (β - α ) or same sin2 (β - α ) you allow for the probability that Alice gets a given result on her particle (for any detector angle)
to vary with the direction that Bob chooses to measure on.
This could be a local hidden variable model for relative angles between detectors
*In part from post #12 comparison between quantum entanglement and a classical version.

Last edited by a moderator: Nov 28, 2015
2. Nov 23, 2015

### gill1109

In a good experiment, Alice and Bob choose their measurement angles freely. Moreover, the time elapsed between choice of angle and registration of measurement outcome (on each side of the experiment) is so short (relative to the distance between the two measurement locations), that there is no way that Alice's angle could be known at Bob's location before Bob's measurement outcome is fixed.

3. Nov 24, 2015

### Truecrimson

I totally agree with you. To clarify, "Everything is determined ahead of time" is what I call superdeterministic. Morrobay's hidden variable theory seems to fall into that type.

4. Nov 25, 2015

### morrobay

Where do you draw the line between a (super)deterministic hidden variable theory where probability outcomes vary
with β - α and match QM predictions
And experiments that show maximum inequality violations, Sqm = 2√2 When a = -450 , a' = 00 b = 22.50 , b' =22.50
The outcome probabilities for above relative θ also vary in a determined, "ahead of time" and pre existing way ?

5. Nov 25, 2015

### Truecrimson

I don't know what you are talking about. Experiment results are experiment results. They are not theory. There is a theory (QM) that predicts those results without positing pre-existing values of the observables.

6. Nov 25, 2015

### morrobay

Let me restate the question (originally from post # 45) with emphasis on local hidden variables
rather than pre existing values that are associated with superdeterminism.
This local model allows for probability outcomes at A & B to vary with β - α based on physical mechanisms (interactions with particle , hidden variables and detector) and in agreement with QM predictions : S = 2√2
Then how can this LHV model be distinguished from QM theory when they both produce the same results ?
Ie: Could the LHV model produce the same curve that QM predicts
With the former the result of physical mechanisms that are not completely understood.

Last edited: Nov 25, 2015
7. Nov 25, 2015

### zonde

It can't do that. If LHV is not exploiting loophole it can't violate Bell inequality.

Edit:
The whole point is that Alice's particle does not know Bob's detector angle so it can't select result based on that information.

Last edited: Nov 25, 2015
8. Nov 27, 2015

### morrobay

This is what I cannot understand : " The hidden variables may be anything " So if hidden variable is such a flexible term ,
then why cannot hidden variables account for inequality violations ? A local hidden variable theory that matches QM predictions.

Last edited by a moderator: Nov 28, 2015
9. Nov 28, 2015

### Staff: Mentor

Have you tried working through the mathematical logic in Bell's original proof of his theorem? That's really the best answer to your question. The next answer would be "No matter what they are, as long as they are local they cannot explain the correlations".

Basically, Bell shows that no matter what the hidden variables are, you need to use the hidden variables at both detectors (that is, non-local) to correctly calculate the correlations.

10. Nov 28, 2015

### DrChinese

This is the crux of Bell's Theorem: there are no such theories. That this is true, you can determine for yourself with some simple trial and error with photons (Type I produces matches at the same angles, mismatches at theta=90 degrees). Pick any 3 angles (not the same or different by multiples of 45 degrees). You cannot hand pick a dataset which will yield the proper QM average predictions - cos^2(theta) - for all 3 angles. If you can't even hand pick them, obviously there is no theory which will work either.

Example angles:
0, 30, 60
0, 22.5, 67.5
0, 10, 20

11. Nov 29, 2015

### entropy1

Isn't the sheer fact that Alice can in principle change her measurement base at any time she wishes before measuring, a certain clue to that her measurement result, if correlated with Bob's, is communicated FTL (or non-local) to Bob? All terminology depending on the interpretation of QM of course.

12. Nov 29, 2015

### DrChinese

No, you cannot say what is communicated or where. Or even whether it is from Alice to Bob or Bob to Alice. We would all be speculating at that point.

13. Dec 2, 2015

### eloheim

Here's the point: Imagine your little algorithm for Bob's result. It says, "if Bob measures at angle x, consult Alice's angle (y) for that same measurement, to get Bob's result (z)." Simple enough.

Now imagine the particles have been sent in opposite directions for a very long period of time, so they're very far apart. Bob sets his angle and makes his measurement. But to calculate his result, you need to know what angle Alice is picking for her (simultaneous) measurement halfway around the globe.

Bob gets his result as SOON as he makes his measurement (obviously), but how could the little algorithm give him that instant result when it needs to know what angle Alice is picking thousands of miles away? Of course, it can't.

Yes, it can give you a list of what the results WOULD be, given that Alice chooses whichever angle such-and-such for her particle measurement, but without actually knowing what Alice is up to at that moment, you can't calculate a specific, actual result when Bob needs it. Which is as soon he makes the measurement.

I hope this helps, perhaps? :-)