# I Nature of hidden variables

1. Sep 21, 2016

### Jilang

In the Bell type experiment is the hidden variable just a function of the state or a function of the state and the detector?

2. Sep 21, 2016

### DrChinese

It could be either in the Bell formulation.

3. Sep 22, 2016

### Jilang

So are they said to be pre-existing (before the particle meets the detector)?

4. Sep 22, 2016

### zonde

Yes, they are pre-existing, meaning they can be inherited from common past of both measurements.

I would like to throw in some links that make the same point as Bell theorem but without hidden variables. There locality is expressed as assumption that measurements at different angles at one side should not affect possible measurement results at other side.
One is Eberhard's paper that contains version of Bell theorem http://journals.aps.org/pra/abstract/10.1103/PhysRevA.47.R747
as it is behind paywall I tried to give a bit simplified version here.
The other one is very simple counter example type argument that is using exact angles and assumption that different hypothetical measurements at one side leave measurement results at other side exactly the same.

5. Sep 22, 2016

### Jilang

Thanks for the links. I read them and followed them I think. I don't see how the remote measurement could change anything. but what assumption is the idea that mismatches must be some kind of linear thing based on? The angle between the detectors is in a plane, but the original angle between the spin and detector might not lie in that plane?

6. Sep 23, 2016

### zonde

This is really not an assumptions (if I understand correctly what you are talking about) but conditions of experimental setup for which we take QM predictions. We measure photon polarization by placing polarizer orthogonally to direction of photon beam and the numbers we use are predictions for such setup.

7. Sep 23, 2016

### Jilang

I am thinking about great circles here. Doubling the longitude difference between two points on a circle doesn't double the angle between them. This would only be the case if they lay in the same latitude. So I am confused about the linear argument at least for spin.

8. Sep 23, 2016

### forcefield

The doubling is spacelike separated so it shouldn't affect measurement results but it does. Or something like that.

9. Sep 23, 2016

### secur

That example comes from Nick Herbert's "Quantum Reality" (1985). Overall it's a decent book, a bit outdated now, but this explanation of Bell's basic point (not his inequality) is great. Everyone should be familiar with it. When someone incorrectly thinks Bell situation is analogous to Bertleman's socks, you can be certain they never read this book. Unfortunately they won't even glance at this example (as I've found from experience) - it's too simple.

@Jilang, the linearity involved is by design (as @zonde indicated). Herbert is measuring angles only in a single plane, normally thought of as the x-y plane. On Earth, you could picture it as the equatorial line. So the 30 degrees in one direction (East, say), and the -30 in the other direction (West), add to give total separation of 60. You're right, it doesn't apply generally: for instance, if one angle were North and the other East.

10. Sep 23, 2016

### Jilang

@secur, I think I haven't explained my question well enough. The two measurement angles always form a plane (since there are only two of them). I am wondering why mismatches (related to projections) into the said plane are assumed to be linear with respect to the angle round that plane. The square root of two seems to be the maximum violation of the Bell inequality, but isn't that just a consequence of the geometry?

11. Sep 23, 2016

### zonde

But mismatches are not linear. They follow $cos^2(\phi)$ rule.

And two measurement angles do not form a plane as they are measured at two remote places. There is just "zero angle" direction in respect to which you measure polarization angle and that is taken freely at one side and determined by the setup at the other side (photon polarization might be rotated along the way).

12. Sep 23, 2016

### secur

@Jilang, there are three angles involved, not two. Originally Bob and Alice each set detectors at angle 0, which can be arbitrarily chosen. In the Earth analogy let's suppose it's at longitude 0, at the equator. Then Alice uses 30 degrees East of that, and Bob 30 degrees West (let's say). All three angles must be in the same plane or the example doesn't work. The max violation of Bell inequality is, indeed, a consequence of "the" geometry - but what geometry? In "our" space - be it the Earth, or the detector angles - it's SO(3), rotations in Euclidean space. (Minkowski space can be ignored, has nothing to do with this simple but 100% valid example.) But in spin space it's SU(2), which is most easily pictured as unit quaternion space. The inequality comes from comparing the correlation that can be achieved in real space, SO(3), to that of SU(2). The probability of "success" there is, as @zonde says, cos^2; the correlation therefore goes by the cos (skipping some detail). Whereas in SO(3) it's linear. SU(2) correlation is too strong near zero to achieve in real space, without FTL influence, or something at least as weird, based on denying Bell's realism assumption.

13. Sep 24, 2016

### Jilang

Indeed, and at larger angles the correlation is less. I am thinking that this factor of the square root of two suggests something beyond our SO(3) that is related to the geometry, it will be interesting to see how it pans out.

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