# Perceived contradiction in non-locality principle

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1. Dec 3, 2015

### entropy1

I have had some criticism on a post of mine in another topic. Since I don't want to pollute that thread with my own discussion, and since I am a layman and am really curious about the answer, I'll pose my question here.

Consider two polarisation-entangled photons A and B measured by Alice and Bob by means of polarisation filters, which, for this purpose, are at a different spacial angle. Now, if a photon passes Alice's filter, it gets polarisation pol(A). Now B behaves as if it has polarisation -pol(A), and either passes Bob's filter or not. Anyway there is a correlation between the two measurements.

However, the same line of reasoning holds for Bob: if a photon passes Bob's filter, it gets polarisation pol(B). Now A behaves as if it has polarisation -pol(B), and either passes Alice's filter or not.

Since there is a correlation between the measurements, at some occasions both photon A passes Alice's filter and photon B passes Bob's filter. However, following the above reasoning, Bob's photon has polarisation pol(B) and polarisation -pol(A). Similarly, Alice's photon has polarisation pol(A) and polarisation -pol(B). So we have pol(B)=-pol(A). But to which filter is this polarisation aligned?? Alice's or Bob's?

It seems to me a contradiction, unless you pose that the latter photon doesn't take on any defined values as a result of measurement of the first, but instead, for instance, both rather only exhibit a measurement correlation! Moreover, if you take relativity in consideration, you can't tell who causes the the value of the other's photon to get a definite value, for it depends on which reference frame you are in.

So who can explain what I am overlooking??

Last edited: Dec 3, 2015
2. Dec 3, 2015

### DrChinese

First, A is always measured at Alice's filter angle, and B is always at Bob's. In other words, it is not correct to state that A is at some angle per Bob so much as instead saying the the probability of a particular outcome is some value X. It is speculative to make inferences about unmeasured properties.

Second, you cannot - at the end of the experiment - make a statement about what is the cause and what is the effect (ie is it due to Alice *or* Bob). Relativity has nothing to do with it.

3. Dec 3, 2015

### entropy1

Let me check if I understand. If photon A passes Alice's filter, photon B is supposedly (I understand) supposed to take on the polarisation photon A has gotten by passing Alice's filter. This also holds from the viewpoint of Bob, but vice-versa. 'Supposedly', because I don't endorse this point of view, but I've seen it come along. So, in the case photon A passed Alice's filter, photon B can now pass Bob's filter or not. In the case it does, my example applies.

Secondly: why does my relativity example not apply?

4. Dec 3, 2015

### DrChinese

1. Precisely because the case [A causes B] is not distinguishable from the case [B causes A]. Ordering does not matter. Saying one happens before the other is a convenience, but not otherwise meaningful or accurate.

2. Because relative motion (or accelerated motion) does not change any of the predictions. Consider that nothing is different in the following situations:
a. You observe Alice and Bob in the same frame of motion.
b. You observe Alice and Bob in different frames.
c. The entangled particles don't even need to exist at the same time (they never coexisted in a common area of spacetime).

5. Dec 3, 2015

### entropy1

Then we agree.

6. Dec 3, 2015

### entropy1

So does that also mean that a polarisation filter does not 'cause' a wave-function collapse? That is, measuring A does not cause B's? And if that is the case, the filter also doesn't cause a wave-function collapse of A itself? (since B's wavefunction is entangled with A's?) Or does that just mean that we can't say anything about B's state immediately after measuring A?

7. Dec 3, 2015

### StevieTNZ

I don't believe a filter will cause "wave function collapse" but its subsequent detection by the detector will.

8. Dec 3, 2015

### Staff: Mentor

As long as we're going to go with a collapse interpretation then the polarizing filter causes a collapse.

In any multiparticle system (not just entangled pairs) there is no "A's wave function" and "B's wave function". There is only one wave function for the entire quantum system, and it is collapsed by any measurement on the system. The wave function of the system of two entangled particles might be $(|UD\rangle+|DU\rangle)/\sqrt{2}$ (here $|UD\rangle>$ is the state "Alice's particle is up and Bob's particle is down" and $|DU\rangle>$ is "Alice's particle is down and Bob's particle is up"). A measurement by either Alice or Bob will collapse the wave function to either $|UD\rangle>$ or $|DU\rangle>$; if it collapses to $|UD\rangle>$ there is no way of saying whether Alice's up measurement or Bob's down measurement caused the collapse (and vice versa for $|DU\rangle>$).

9. Dec 3, 2015

### Staff: Mentor

In Copenhagen-like collapse interpretations, the collapse happens when the quantum system interacts with some macroscopic object (and here "macroscopic" is defined with annoying circularity as "causes collapse"). Most physically realizable polarizing filters count as macroscopic objects so we'd say they collapse the wave function.

The alternative would be to say the the filter and the particle enter an entangled superposition of "passed" and "didn't pass", and the collapse happens when the filter+particle system encounters the detector. This gets awkward when the particle is absorbed by the filter so we're asking the detector to detect the non-arrival of a particle, but it's still a valid interpretation of what the mathematical formalism of QM is telling us.

10. Dec 3, 2015

### StevieTNZ

Hi Nugatory

What about in the case of a birefringent crystal, with a setup like that on page 86 of "Sneaking a Look at God's Cards" where by a 45 degree polarized photon goes through a (45 degree) filter [passes], then hits a birefringent crystal (where the extraordinary ray output is vertical polarization, and the ordinary ray is horizontal polarization [one expects with 0.5 probability the photon to come out the extraordinary ray, the other 0.5 the ordinary ray, over many runs]).

Then those ray(s) hit another birefringent crystal reversed, go through another 45 degree filter towards a detector but all photons inputted at the start are detected? (i.e. implying that at the first crystal the photon was in a superposition of exiting both the extraordinary, and ordinary, ray) Wouldn't "collapse" happen at the first birefringent crystal (and therefore when the photon hits the 2nd filter, it has 1/2 probability of going through)?

Unless I have overlooked some detail.

Stevie

EDIT: tidied up some details.

Last edited: Dec 3, 2015
11. Dec 3, 2015

### Weddgyr

To add two cents here, this is essentially my current understanding of Bell's theorem and entanglement measurements. While a non-local model explaining the correlations can be cobbled together if you ignore relativity, once you introduce special relativity and in particular the relativity of simultaneity, non-local explanations of entanglement also fall apart. As far as I understand it, it is impossible to model an entanglement experiments in a relativistic framework on a computer. Effectively entanglement correlations behave as though the two detectors were preprogrammed to take on a predefined set of angles and produce detector beeps accordingly. There is no common cause explanation, and in particular no model of elements of physical reality explaining why anything from the source would cause the detectors to beep.

And having said that, to add a whole nickel, I don't understand how any of the usual interpretations of QM work when entanglement and relativity are combined. I have read comments to the effect that QM can be explained by local but non-realist theories, but as far as I can see all local physical models /hidden variable of detection will fail to produce the correlations. On weekends, I'm actually tempted to view the QM formalism itself as a kind of non-local hidden variable model given that it refers the correlations back to something, and I wonder if we shouldn't simply declare the correlations by fiat and have done.

But I think we should only decide this question every second Tuesday.

12. Dec 3, 2015

### zonde

Measurement correlation is statement of a bare physical fact. It is no replacement for explanation, it is basis for explanation.
Idea that all inertial reference frames are fundamentally equivalent is interpretation of relativity, very common interpretation but still just an interpretation.

13. Dec 3, 2015

### zonde

In special relativity simultaneity is a convention. There are no physical consequences specific to simultaneity. So one can use different convention without breaking (physical) predictions of relativity.

14. Dec 3, 2015

### Weddgyr

Don't we need simultaneity to avoid the various paradoxes of special relativity? Are there conventions that could be employed to avoid both these and which would still accommodate entanglement?

15. Dec 3, 2015

### zonde

No, you can construct valid space-time diagram without using simultaneity lines. They have only illustrative function in space-time diagram.
If by entanglement you mean FTL type model (explanation) of entanglement then it would require preferred reference frame convention.

16. Dec 4, 2015

### DrChinese

The only meaningful thing is the full context of the setup (source, 2 polarizers, detectors, etc). It is not possible to point to a single component of the setup (for example, a single polarizer) and say "this is the cause".

17. Dec 4, 2015

### DrChinese

Welcome to PhysicsForums, Weddgyr! On some days, I would give you a dime for the above.

Even if you try to fit relativity into the equation, I think you will see there is no impact. The entangled photon correlation ratio is a function of cos^2(Alice - Bob). That never changes, even in non-inertial reference frames.

18. Dec 5, 2015

### entropy1

So, am I right when I say that if Alice measures before Bob does, that A collapses and B takes on A's polarisation, and vice-versa: if Bob measures before Alice does, that B collapses and A takes on B's polarisation? Then how can you establish who measures first (Alice or Bob), and which photon of the pair takes on the other's polarisation? Is it even defined?

19. Dec 5, 2015

### entropy1

So, if you are not using the FTL explanation, a preferred frame is not required?

20. Dec 5, 2015

### entropy1

If I want to establish who measures first, don't I need a definition of temporal order, and in that case wouldn't I need the context of spacetime and thus relativity?