# Bell inequality experiments and photon polarisation issue

1. Dec 27, 2011

### hollowsolid

I am confused by the role of photon polarisation in Bell inequality experiments.

The original logic of EPR as I understand it is based on the HUP such that QM predicts that
measurement of momentum on one particle should affect the measurement of position
of the other particle. Yet across large distances this would required superluminal communication. EPR

I can see that measurement of spin is a form of angular momentum but AFAIK bell inequality
experiments do not measure corresponding position, they measure correlations between
spin measurements.

So I don't understand the involvement of the HUP in entangled pairs. Even worse, I am unsure whether photon polarisation has anything to do with angular momentum at all and hence is even less related to the HUP.

In short, how does the HUP relate to measurement of singlet states and what is the logic behind redesigning EPR experiments from position/momentum uncertainty to correlations between spin or polarisation?

Thanks for any help

2. Dec 27, 2011

### DrChinese

The HUP applies to spin and polarization just as it does to position and momentum. For a photon, its polarization at 0 and 45 degrees (relative) do not commute. For an electron, that would be 90 degrees instead of 45.

So if you know photon polarization at 0 degrees, the polarization at 45 degrees is completely uncertain (and vice versa). And this then becomes the starting point for the analogy with EPR.

3. Dec 28, 2011

### hollowsolid

Thank you for this clarification Dr. Chinese.

However ir confuses me further. The HUP is specifically about the concommutation of
momentum and position.

I don't understand the extension of the HUP to polarisation angles which on the face of it are unrelated to momentum and position.

If I could understand the generalisation of the HUP to polarisation the rest of EPR/Bell is fine
but it's this step that eludes me.

Please provide the clue i'm missing.

cheers

4. Dec 28, 2011

### Edgardo

Run this simulation of the Stern-Gerlach experiment:

Situation 1:
- In the simulation set the initial spin of the atom to +z (see left green box).
- Click on Auto-Fire (see left side) and set the fire rate to fast.
- You'll notice that if your SG-apparatus measures the spin along the z-direction (magnet at angle=0°, see top green box)
then you'll get 100% z+, i.e. spin-up atoms (see red circle at the bottom).

Situation 2a:
- In the simulation set the initial spin of the atom to +z (see left green box).
- Change the number of magnets to 2. Leave both magnets at angle 0° (see top green box).
- Click on Auto-Fire and set the fire rate to fast.
- You still get 100% z+.

Situation 2b:
- In the simulation set the initial spin of the atom to +z (see left green box).
- Now, consider again 2 magnets.
- Set the first magnet to 90° and the second to 0° (see top green box: angle=90° and angle2=0°)
- Now, what do you expect is the result after magnet 2?
(Don't click on Auto-Fire yet)

We know from Situation 1) that the atoms coming out of the source all have spin-up.
So we would expect that if we measure the z-component in Situation 2b) with magnet 2, all the atoms have spin-up.

Now, click on Auto-Fire. The result is not what we would have expected. Suddenly, only half of the atoms have spin-up.

It must have something to do with magnet 1. Magnet 1 measures the spin in x-direction, and this x-measurement changes the z-component of the atom.

This is similar to position and momentum. A position measurement changes momentum.
Here, a Sx measurement changes the result of the Sz measurement.
(where Sz stands for a measurement of spin in z-direction)

---

You probably know the uncertainty relation for position and momentum: $\Delta x \Delta p \geq \hbar/2$

For spin there is also a uncertainty relation: $\Delta S_x \Delta S_y \geq |\langle S_z \rangle|$
(see page 95 here)

which follows from the general uncertainty relation.

---

Here is more material to read:
1. Angular Momentum in Quantum Mechanics
2. Angular Momentum in Quantum Mechanics: Spin Indepth

5. Dec 28, 2011

### Edgardo

For photon polarization I recommend looking at an experiment with three polarizers.
It's similar to the Stern-Gerlach experiment with two magnets (actually we have three magnets too, where the third is the one preparing the atoms in a certain state, e.g. z+):

1. Polarisation - Sixty Symbols (youtube video)
2. Polarization of Light and Microwaves (Quantum Physics) (youtube video)
3. Three Polarizers (PF Thread)
4. Three Polarizer “Paradox” (an article by Prof. Frank Rioux)

6. Dec 28, 2011

### Ken G

I think you should forget about the HUP altogether, because what you already know about the HUP is not triggering anything that is helping you understand polarization. Basically, the HUP, and the polarization results you are trying to understand, both stem from the same source-- so understanding that source is what matters. The source is the principle of superposition, which says that a given state can be written as a linear combination of basis vectors, and the combination depends on the choice of basis vectors but the state doesn't. So in one basis, you might have a polarization "eigenstate" (a state with a definite outcome for some particular polarization measurement), but in another basis, the state might be a superposition of eigenstates (so no definite outcome for the polarization measurements corresponding to the new basis). That's really what "noncommuting" means, it means that the eigenvectors of one measurement are not eigenvectors of a noncommuting measurement.

This also means that the order you do the measurements matters, because you always end up in an eigenstate of the second measurement, which is not an eigenstate of the first measurement if the two are complementary (i.e., completely noncommuting). All of this holds either for position/momentum measurements, or for polarization measurements, and you can think of it as being the reason for the HUP but not quite the same thing as how you think of the HUP now.

7. Dec 29, 2011

### hollowsolid

Thank you Edgardo and KenG.

Excellent help. You have a real talent for explaining the complex in an understandable way.

I now think i understand how QM polarisation probabilities are calculated and how these mismatch those from the Bell inequality.

I am still murky on the significance of 3 polarisation measures in terms of hidden variables- apart from a vague inkling that any measure of A and B polarisation leaves
C as a hidden variable- and so on for all combinations.

I'm therefore concluding that if correlations between these pairings do not match
those of QM- which are experimentally supported- then hidden variables are insufficient
to explain what actually happens. Indeed QM correctly predicts what happens while
hidden variables do not.

I still need to read much more deeply into the HUP since the basic inequalities
of momentum, position, energy and time do not provide me with much insight
into what observables fail to commute and when.

I may have to learn bra-ket notation to achieve that since there are subtleties about
it that seem to depend upon the mathematics.

Once again thank you so much for the time and effort to help others.

It makes this site so very valuable.

8. Dec 30, 2011

### Ken G

That's right, though be sure to attach the term "local" to the hidden variables. It was originally hoped, by people like Einstein, that the hidden variables could in a very real sense 'belong to' the particles and be 'carried with' them, but that's what Bell's theorem rules out. However, it is still possible to have hidden variables theories with nonlocal connections, so one can still interpret quantum mechanics as the theory you get when you don't know what the hidden variables are doing. Still, if the hidden variables aren't local to the particles, it's not clear what they mean any more, they seem magical in the absence of some actual way to manipulate and test them.
It all traces back to the fundamental breakthrough of quantum mechanics, that "action" is quantized. Action has the units of angular momentum, which is the units of distance times momentum, or energy times time. That's basically where those combinations come from, but you're right there's a lot more to it.