# Quantum mechanics is not weird (locality and non-locality weirdness)

1. Jan 8, 2016

### zonde

Let's be fair, it's not true.
Pure states are the ones that correspond to exact physical states. And it is not intuitive that exact physical states should transform continuously. Our belief about outcome can transform continuously but belief does not correspond to pure state.

2. Jan 8, 2016

### rubi

The weirdness of QM goes away as soon as one accepts that there are quantum objects. All instances of weirdness arise from thinking classically about things that behave quantum mechanically. The existence of quantum objects is not inherently weirder than the existence of classical objects. Even if the world were purely classical, the fact that there are (classical) electrons and they seem to carry something like an electric charge for instance also needs to be accepted by us and there is no deeper explanation for it (so far). The amount of belief that is needed to accept charged classical particles is exactly as high as the amound of belief that is required to accept the existence of quantum objects. The behaviour of quantum objects is just farther from our daily experience. Finding QM weirder than CM is therefore very anthropocentric. Would anyone find QM weird, if we had grown up in a world, where quantum effects were ubiquitous in our daily experience?

3. Jan 8, 2016

### zonde

As an example say we analyze situation where two persons are in single room and we ask what is probability that one person or the other will come out first. There are two pure states: first person comes out first and second person comes out first. And we claim that there is transformation where half first/half second person comes out first.

4. Jan 8, 2016

### zonde

Weirdness of entanglement does not go away if you think only about detection records of entangled particles as classical objects. Would you claim that detection records behave quantum mechanically (a la Schrodinger's cat)?

5. Jan 8, 2016

### rubi

The detection records result from the ineraction of the detectors with quantum objects. Entangled particles just behave as they are supposed to do (because they are quantum objects) and therefore produce the observed detection records. If the detectors would interact with objects that behave sufficiently classically (like dice), the statistics of the detection records would of course be in agreement with classical statistics, but in the case of entangled particles, the detectors interact with quantum particles and therefore we observe the statistics that quantum objects are supposed to produce. Nothing is weird about the fact that (sufficiently) classical detectors interacting with quantum objects measure typtical quantum statistics. It would rather be weird if this weren't the case.

6. Jan 8, 2016

### zonde

The weirdness is in the fact that statistics calculated from classical detection records are inconsistent with locality.

7. Jan 8, 2016

### rubi

You are using a concept of locality that is only adequate for classical physics. It requires that all observables can be modeled on the same probability space, which is not true in quantum mechanics. Just like there is no probability distribution $P(x,p)$ for non-commuting $\hat x$ and $\hat p$, there isn't $P(A,\lambda)$ either, because certainly at least one observable in $\lambda$ won't commute with $A$. The fact that the world is quantum mechanical requires us to refine our classical concepts. When you say that the statistics is inconsistent with locality, you use a locality concept that is ill-defined for quantum systems. It only makes sense in a purely classical world, which we know isn't true as soon as we accept that quantum objects exist. Yes, it is true that if the world were classical, then quantum statistics were weird. But the world isn't classical.

8. Jan 8, 2016

### zonde

You assume I am referring to Bell theorem.
But I am not. Let me explain.

let me make a rather general conjecture that: For any QM prediction it is possible to have set of sequences of spacetime events (detection events) of any finite size that satisfy predicted relative frequencies exactly given we don't have to introduce rounding errors. And I will include in this conjecture assumptions of locality and no superdeterminism.

And by locality assumption here I mean that given string of factual detection events at one side it is possible to have string of detection events at other side that reproduce prediction of QM exactly (neglecting rounding errors). You see, no probabilities here. We just cover all possibilities.

And here is counterexample to that general conjecture https://www.physicsforums.com/threads/a-simple-proof-of-bells-theorem.417173/#post-2817138

Last edited: Jan 8, 2016
9. Jan 8, 2016

### rubi

Can you phrase your locality criterion in clear mathematical language and explain, why it follows from relativity? The fact that this "proof" comes from someone who writes about himself:
doesn't make me want to spend time on it. Maybe you can also refer me to a paper (not from Nick Herbert) that discusses this approach.

10. Jan 8, 2016

### zonde

Last edited by a moderator: May 7, 2017
11. Jan 8, 2016

### strangerep

I'm missing something in this argument. My understanding of "locality" (in the context of EPR, Bell's thm, etc) is that observables which have support only on mutually spacelike-separated regions necessarily commute. This seems different from the "enhanced" concept of locality you seem to be advocating above for the quantum context. Please give a precise definition of your enhanced concept of locality.

12. Jan 8, 2016

### zonde

You mean something like this?
1) given set A there exists set B' such that certain relationship between sets A and B' equals QM predicted value
2) given set B there exists set A' such that certain relationship between sets B and A' equals QM predicted value
Theory is local if within all sets B' that satisfy point 1) and all sets A' that satisfy point 2) we can find pair of sets A' and B' such that certain relationship between these sets equals QM predicted value.
Hmm, I have to think about that.
Can you give as an example explanation how "no FTL" follows from relativity?

13. Jan 8, 2016

### rubi

It's true that spacelike separated observables commute, but there are also observables that are not spacelike separated. For example the $S_x$, $S_y$ and $S_z$ observables are not spacelike separated for Alice and Bobs individual particles. If you measure $S_z$ in a Bell test, the $\lambda$ observables would have to include $S_x$ and $S_y$ (and possibly many other observables) and there can't be a probability distribution $P(S_x, S_y, S_z)$ in quantum mechanics. Such combined probability distributions only exist for commuting observables. Whatever $\lambda$ is, in a quantum context, it certainly includes non-commuting observables that are localized in the same region as the one you are considering. That fact that Bobs variables are not among the ones that don't commute with Alices (and the other way around) doesn't help.

I don't have an enhanced concept of locality. I'm just saying that the classical one doesn't apply anymore. It might as well be that in a quantum world, statistics just can't be used anymore to make conclusions about locality.

14. Jan 8, 2016

### rubi

Last edited by a moderator: May 7, 2017
15. Jan 8, 2016

### strangerep

Imho, this sounds more like an argument against local realism (meaning the notion that all possible observables have a value before being measured). It's not really an argument against (the standard notion of) "locality" per se.

16. Jan 8, 2016

### zonde

So, do you need explanation why Bell's or Eberhard's locality criterion follows from relativity?
You have misunderstood his derivation. He derives his inequality in presence of non-detections. But non-detections can be set to zero (there are no unpaired single detections) and of course derivation will remain valid. So his derivation is more general.

17. Jan 8, 2016

### rubi

A realistic theory is non-local if Bell's inequality is violated. But QM is not a realistic theory, so the violation of Bell's inequality doesn't imply that QM is non-local. Thus, Bell's criterion doesn't capture locality adequately for theories that are not realistic. Bell's factorization criterion can't even be formulated in the context of QM, since QM doesn't have the combined probability distributions, which are needed to write it down.

18. Jan 8, 2016

### rubi

I know it for Bell. I basically just want to know, why your definition of "locality" is appropriate, i.e. why it really captures the concept of locality appropriately.

I'm aware of that. One can derive Bell type inequalities for more realistic situations, where detectors don't have 100% efficiency. Nevertheless, the derivations always assume Bell's factorization property somewhere and this is the one that supposedly encodes locality.

19. Jan 8, 2016

### stevendaryl

Staff Emeritus
I think that Bell explained his notion of locality in his essay about "Local Be-ables". Let me try to paraphrase, rather than repeat him.

Let's focus on a measurement event that takes place in a small region of spacetime. For example, we set up a Stern-Gerlach device for measuring the spin of an electron. Before the measurement takes place, we can figure out a set of possible outcomes for the measurement. Now, if we know more about the history of the universe prior to that measurement event, we may use that knowledge to narrow down the set of possibilities. For example, if we know that the electrons were produced in the spin-up state along some axis (and that there are no fields present between the source and the device), then we know that the device will definitely not measure spin-down along that axis. Bell's locality criterion is that the set of possibilities should be determined by local conditions. If information about conditions in distant regions of space allow you to narrow down the set of possibilities beyond what can be deduced from local conditions, then the theory is nonlocal in Bell's sense. The sense of "nonlocal" is pretty straight-forward: the best (most accurate) prediction about what happens in one region of space requires information about distant regions of spacetime.

And the EPR experiment is definitely nonlocal in this sense. A pair of anti-correlated spin-1/2 particles are produced. Alice measures the spin of one particle along axis $\vec{A}$ and finds it is spin-up along that axis. Far, far, away, Bob measures the spin of the other particle along the same axis. Then it is definitely the case that Bob will measure spin-down, but there is no local information that would allow him to predict this. But there is nonlocal information: the fact that Alice already measured spin-up along that axis.

This notion of nonlocality is about nonlocality of inference, rather than causality. The mere fact that Alice can predict Bob's result doesn't say anything about whether there is a causal influence between the two measurements. But it does say that there is nonlocal information. Any correlation between distant events that doesn't factor into facts about local regions is a nonlocal correlation in Bell's sense. As to whether that's the "appropriate" notion of locality, it depends on what you mean: appropriate for what?

20. Jan 8, 2016

### stevendaryl

Staff Emeritus
It seems clear to me that QM is nonlocal, in the sense that it makes nonlocal predictions. In a spin-1/2 EPR experiment, QM predicts that if Alice and Bob both measure spins of their particles along the same axis, they will get opposite results. That's a nonlocal fact about those results. The point about hidden variables is that such nonlocal facts can arise in classical probability through ignorance about local hidden (unknown) variables. So classically, nonlocal facts can be accounted for by local facts, and so the nonlocality is seen to be an artifact of our ignorance, rather than an objective fact about the world. But in contrast, QM appears to involve nonlocal facts that can't be accounted for by local facts. So it's nonlocal, in Bell's sense. I don't see how realistic versus nonrealistic is relevant to the conclusion.

21. Jan 8, 2016

### zonde

As I understand concept of locality, spacelike separated regions are independent as tested by "external" parameters ("external" parameters from one region can not influence what happens in the other region).

22. Jan 8, 2016

### rubi

I understand this concept of locality, but zonde claims to be using a different one.

Non-local correlations are not problematic though, as the example of Bertlmann's socks shows. People claim that Bell violations are a sign for non-local causation, which doesn't follow. It is true that non-local correlations between quantum objects can be more severe than non-local correlations between classical objects. But concluding that QM is non-local (i.e. it features non-local causation rather than just non-local correlations) is not possible, since Bell's argument for non-local causation in the case of inequality violation makes crucial use of the realism of the underlying theory and therefore doesn't apply to QM. In other words: If the world were classical, then a Bell violation excludes a common cause for the non-local correlations. Since, the world is quantum mechanical, a common cause cannot be excluded, so the non-local correlations needn't be problematic.

As I said, non-local correlations are not problematic even in a local theory. It would only be problematic if there were no common cause that could explain them, because that would hint at non-local causation. I'm only arguing that Bell violations can't be used to exclude a common cause in the past in the case of a non-realistic theory.

Edit: I'll explain it differently: Assume Alice has a red sock. Then she immediately knows that Bob has a Blue sock. Assume Alice has a spin-up partricle. Then she immediately knows that Bob has a spin-down particle. What is the difference? The difference is that the first case can be modeled by a local classical theory, in which there was a common cause of Alice and Bobs measurements. Someone just sent a red sock to Alice and a blue sock to Bob. The statistics of such experiments satisfy Bell's inequality. However, the statistics of the second case violates Bell's inequality. Does it mean that there is no common cause? No! What happened is the following: The world is quantum mechanical. Someone in the past generated a pair of entangled quantum objects and sent one to Alice and the other one to Bob. This is the common cause that explains the non-local correlations. This is only mysterious to someone who says: "I don't believe this mumbo-jumbo about superpositions. The spins clearly had definite values throughout the experiment. How can it be different? Therefore the world must be non-local." However, if you are comfortable with the fact that the world is quantum mechanical, the idea that there was a common cause should not trouble you.

Right, but how does this imply the locality criterion of Nick Herbert (for instance), which says something about errors rather than spacelike separated regions? Why is the statement about errors the same as a statement about spacelike separated regions?

Last edited: Jan 8, 2016
23. Jan 8, 2016

### zonde

I reread your post #123 and I think it's worth returning to it.
Let's say we have polarization entangled photons and two non-commuting probability spaces for H/V detections and +45/-45 detections. Let's say Alice and Bob both measure photons in H/V basis so they use H/V probability space and observe perfect correlations. If they would measure photons in +45/-45 basis they would use +45/-45 probability space and would observe perfect correlations. Is this approximately your idea of non-commuting probability spaces?

24. Jan 8, 2016

### rubi

It's not my idea, it's Andrei Khrennikov's or Fine's I think. You have the following spaces: $\Omega_1 = \{\uparrow,\downarrow\}$ and $\Omega_2 = \{\leftarrow,\rightarrow\}$, and probabilities on them, but you don't have $\Omega_1\times\Omega_2$, since the events in that space cannot occur due to the non-commutativity of observables. In a quantum system, it is just impossible in principle to measure the system in the state $(\uparrow,\leftarrow)$ for example, because an eigenstate of $S_z$ is never an eigenstate of $S_x$ at the same time. There is a probability $P_1(\uparrow)$ and a probability $P_2(\leftarrow)$, but no combined probability $P_{12}(\uparrow,\leftarrow)$.

25. Jan 8, 2016

### zonde

Are you trying to guess my next question?
Because I was asking about both parties measuring their entangled photons in the same commuting basis: H/V or +45/-45 that re analog to your $\Omega_1 = \{\uparrow,\downarrow\}$ and $\Omega_2 = \{\leftarrow,\rightarrow\}$