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

[A. Neumaier]

the interesting situation is not when particles that are entangled (identical particles are always entangled, by my definition), but when specific attributes of the particles (spin or helicity or whatever) are entangled.

Yes, since these really live in a tensor product. Among the specific attributes there is also the momentum, which in a beam splitter becomes entangled. But not position, since photons cannot have a position.

by my definition

It is the standard (and only) definition, that you find everywhere. Misuse of the terminology not withstanding.

 

[vanhees71]

Sure, one always has to state which observables are entangled.

 

[ShayanJ]

So the more sophisticated view is that the interesting situation is not when particles that are entangled (identical particles are always entangled, by my definition), but when specific attributes of the particles (spin or helicity or whatever) are entangled.

Can you give an example of a Fermionic or Bosonic multi-particle state which has non of its observables entangled and is only entangled because of the (anti-)symmetrization?

 

[vanhees71]

The first is correct, the second only holds figuratively. For you cannot point to a single property (apart from mass 0 and spin 0, which are nondynamical) that any of the two photons whose existence you assert has. By definition you can conclude only that you have something called a 2-photon state.

Calling something (by analogy to the nonrelativistic case) a photon-number operator doesn't bring photons into existence, just as renaming the angular momentum operator ''angular particle number operator'' doesn't bring angular particles into existence.

I am talking the QFT formalism seriously as a valid description of nature, but not the talk about it, which is largely historical and to some extent inappropriate. It is a similar issue as your fight against the notion of ''second quantization''.

That's an interesting aspect, but do you say that the photon number is not an observable?

For sure, it's hard to establish that a given situation is described by a photon Fock state of determined photon number, but at least in principle the photon number should be an observable. Here's an example for a Fock-state preparation in a micromaser cavity

http://journals.aps.org/pra/abstract/10.1103/PhysRevA.36.4547

 

[A. Neumaier]

Can you give an example of a Fermionic or Bosonic multi-particle state which has none of its observables entangled and is only entangled because of the (anti-)symmetrization?

To talk about entanglement you need the tensor product structure. This depends on which (set of commuting) observables you are using to define the latter. Thus it is not all observables that count but only those observables used to define the tensor product structure under discussion.

Usually position has to be taken to be nonentangled, because it defines which particle is meant. Otherwise no discussion of small systems would make sense since an ion in an ion trap is distinguished from all the other identical ions as ''this ion'' only by its position. The problem with photon experiments (all long distance weirdness experiments are done with multi-photon states since other states decohere far too fast!) is that you cannot project to fixed position, hence talking about the position of photons is highly questionable.

 

[ShayanJ]

To talk about entanglement you need the tensor product structure. This depends on which observables you are measuring.
Thus it is not all observables that count but only those observables used to define the tensor product structure under discussion.

Usually position has to be taken to be nonentangled, because it defines which particle is meant. Otherwise no discussion of small systems would make sense since an ion in an ion trap is distinguished from all the other identical ions as ''this ion'' only by its position. The problem with photon experiments (all long distance weirdness experiments are done with multi-photon states since other states decohere far too fast!) is that you cannot project to fixed position, hence talking about the position of photons is highly questionable.

This is really hazy to me. Can you give a reference to somewhere that explains this along with the math?

 

[A. Neumaier]

do you say that the photon number is not an observable?

It is an observable in the sense that it is a self-adjoint Hermitian operator. But its name is inviting misinterpretation if taken more than figuratively, since apart from saying there are two photons you can't say anything at all about the photons involved.
At best you can say something after the photon (which of the photons?) gave up its alleged existence by exciting a photodetector.

It is also not-an-observable, in the sense that I cannot conceive of any experiment that measures photon number. Maybe one should say it is a preparable, as one can apparently prepare states with low photon number. I never looked at the techniques in detail, but maybe I should do it now with the reference you gave. In many experiments and preparations, however, the notion of a ''single photons'' does not mean ''1-photon state'' but ''photon wave packet with the energy of $\hbar\omega$''; see my slides here.

 

[A. Neumaier]

This is really hazy to me. Can you give a reference

No. It is what I read between the lines of the existing literature. There is lots of sloppiness in the terminology about quantum foundations, and once one sets one's mind on getting things really precise one notices that it is never done. If you can point to the haziness, do it here, and I'll try to explain.

 

[A. Neumaier]

I cannot conceive of any experiment that measures photon number.

instead, typical experiments change the photon number by 1 or 2. Thus if one doesn't know the number of photons from their preparation (which means almost never, since hardly ever one uses pure N-photon states as inputs to experiments) one never gets to know the photon number.

 

[ShayanJ]

No. It is what I read between the lines of the existing literature. There is lots of sloppiness in the terminology about quantum foundations, and once one sets one's mind on getting things really precise one notices that it is never done. If you can point to the haziness, do it here, and I'll try to explain.

For now two questions come to my mind:

1) How is it that the tensor product structure depends on which observables we are measuring? We're just describing the state of a multi-particle system, why should we need any reference to our measurements?

2) The state vector is a tensor product of vectors in different Hilbert spaces each associated to an observable. The state vector as a whole should be either symmetrized or anti-symmetrized w.r.t. exchange of particles. Also it seems to me that each vector in the tensor product that gives the whole state vector should be either symmetrized or anti-symmetrized too. So I don't understand what you mean by "Usually position has to be taken to be nonentangled"!

 

[vanhees71]

It is an observable in the sense that it is a self-adjoint Hermitian operator. But its name is inviting misinterpretation if taken more than figuratively, since apart from saying there are two photons you can't say anything at all about the photons involved.
At best you can say something after the photon gave up its alleged existence by exciting a photodetector.

It is also not-an-observable, in the sense that I cannot conceive of any experiment that measures photon number. Maybe one should say it is a preparable, as one can apparently prepare states with low photon number. I never looked at the techniques in detail, but maybe I should do it now with the reference you gave. In many experiments, however, the notion of a ''single photons'' does not mean ''1-photon state'' but ''photon wave packet with the energy of $\hbar\omega$''; see the link given here.

I think this is again a clash of semantics. Of course, with "one-photon state" I mean a "wave packet" since a state must be normalizable to 1 (not "to a $\delta$ function" as used for the plane-wave states, which are generalized momentum eigenstates), i.e., something like
$$|\psi \rangle=\int_{\mathrm{d}^3 \vec{k}} \phi(\vec{k}) \hat{a}^{\dagger}(\vec{k},\lambda)|\Omega \rangle,$$
with $\phi$ a square integrable function and $\hat{a}^{\dagger}$ the usual creation operators in the plane-wave (generalized) single-particle basis, normalized such that
$$[\hat{a}(\vec{k}',\lambda'),\hat{a}^{\dagger}(\vec{k},\lambda)]=\delta^{(3)}(\vec{k}-\vec{k}') \delta_{\lambda \lambda'}.$$
There is also another interesting paper by Scully et al giving a measurement procedure to distinguish between a true single-photon state and a very-low-intensity coherent state ("dimmed laser"). Unfortunately I cannot find it. In googling, I found the following papers, which sound interesting in this context:

http://journals.aps.org/pra/abstract/10.1103/PhysRevA.71.021801
http://arxiv.org/pdf/quant-ph/0308055
http://journals.aps.org/pra/abstract/10.1103/PhysRevA.70.052308

 

[A. Neumaier]

We're just describing the state of a multi-particle system

It depends on what kind of multiparticle system you have.
The physical Hilbert space of multiparticle system consisting of distinguishable particles is a tensor product space. An example is a system of atoms in a solid (described by a lattice) at temperatures low enough that the atoms cannot change places. in this case, the position distinguishes the atoms, and tensor products of single atom states define unentangled multiatom states.
.
But the physical Hilbert space of a physical multiparticle system consisting of identical particles is not a tensor product space but a Fock space. To impose on it a tensor product structure one needs to choose a family of commuting observables whose possible values form a Cartesian product. In this case one typically [there are other possibilities], and without saying this explicitly, projects the Fock space to the (usually much smaller) space generated by the state of the system and the chosen a family of commuting observables. (In other words, one decomposes the Hilbert space into irreducible representations of the chosen family of commuting observables and only keeps the representation containing the state of the system.) The result is a tensor product space with a basis labelled by the Cartesian product. For example, each of the commuting observables projects everything to a spin up state, giving a tensor product of spins (up,down), while ignoring everything else (position, momentum, and internal degrees of freedom). To get a tensor product with helicities (left,right) you need a different family of commuting observables, and the two tensor product structures are incompatible, though the projected Hilbert space is the same . This means that what is unentangled in one of the two tensor product descriptions of the projected space is entangled in the other. This shows that the Hilbert space can carry many tensor product structures, and without saying which tensor product structure one refers to (which is often not done explicitly but silently) one cannot tell what is and what isn't entangled.

 

[A. Neumaier]

I don't understand what you mean by "Usually position has to be taken to be nonentangled"!

I mean that one projects the Hilbert space to a smaller space in which position no longer figures. One hardly ever sees an exposition of experiments involving entanglement in which position is an observable in the tensor product structure assumed silently in the discussion. Usually the state space in is finite-dimensional in the exposition. But in the interpretation of certain experiments position suddenly plays a decisive role. Weirdness introduced by sloppiness.

 

[A. Neumaier]

Of course, with "one-photon state" I mean a "wave packet" since a state must be normalizable to 1

Your wave packets can have an arbitrary energy not necessarily related to the frequency. But I said:

In many experiments and preparations, however, the notion of a ''single photons'' does not mean ''1-photon state'' but ''photon wave packet with the energy of $\hbar\omega$''; see the link given here.

Which means that a coherent state whose mode (= normalizable solution of the Maxwell equation) consists of a sequence of N pulses each with the energy of $\hbar\omega$ is considered to contain N photons. (In contrast to the most orthodox view, where a coherent state is a superposition of N-photon states of all N, independent of its mode.)

 

[stevendaryl]

Can you give an example of a Fermionic or Bosonic multi-particle state which has non of its observables entangled and is only entangled because of the (anti-)symmetrization?

Well, suppose we have a two-particle state that looks like this:

$\left( \begin{array} \\ \psi(x_1) \\ 0 \end{array} \right) \otimes \left( \begin{array} \\ 0 \\ \phi(x_2) \end{array} \right) - \left( \begin{array} \\ 0 \\ \phi(x_1)\end{array} \right) \otimes \left( \begin{array} \\ \psi(x_2) \\ 0 \end{array} \right)$

where $\psi(x_1)$ is basically zero everywhere except when $x_1$ is in region $A$ (near Alice's detector), and $\phi(x_2)$ is basically zero everywhere except when $x_2$ is in region $B$ (near Bob's detector).

Then it's true that particle 1 is entangled with particle 2. However, if instead of referring to "particle 1" and "particle 2", we refer to "the particle in region A" and "the particle in region B", then the particle in region A does not have its spin entangled with the particle in region B. The first particle (whichever one is in region A) has definite spin-up, and the second particle (whichever one is in region B) has definite spin-down.

For practical purposes, particles that are far apart, spatially, can be thought of as distinguishable: you distinguish them by their location.

 

[zonde]

As there is some discussion about QFT treatment of entangled states I would like to ask question related to that.
When entangled two-particle state at one side is subject to PBS that is in different base as the one in we which we have expressed two-particle state, how is calculations of output probability amplitudes done and how this is reflected at remote side?
As I understand we write annihilation operators for input states and creation operators for output states. As there are two different modes (H and V) present in each output we sum probability amplitudes for these two modes. So we get new probability amplitudes for two (output) states in this new basis.
And the question is: How do we fulfill symmetrization requirement with remote side? Do we automatically view the remote side in this new basis with the same amplitudes as in the local side (does not sound quite right to me)?

 

[rubi]

Ah, sorry, this is not factually correct. I say this not even considering the general Bell Theorem issues that others have pointed out.

You can entangle, and get perfect correlations, from particles that have never existed in any common light cone. There is no common cause.

http://arxiv.org/abs/1209.4191

"The role of the timing and order of quantum measurements is not just a fundamental question of quantum mechanics, but also a puzzling one. Any part of a quantum system that has finished evolving, can be measured immediately or saved for later, without affecting the final results, regardless of the continued evolution of the rest of the system. In addition, the non-locality of quantum mechanics, as manifested by entanglement, does not apply only to particles with spatial separation, but also with temporal separation. Here we demonstrate these principles by generating and fully characterizing an entangled pair of photons that never coexisted. Using entanglement swapping between two temporally separated photon pairs we entangle one photon from the first pair with another photon from the second pair. The first photon was detected even before the other was created. The observed quantum correlations manifest the non-locality of quantum mechanics in spacetime."

And

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.80.3891

We experimentally entangle freely propagating particles that never physically interacted with one another or which have never been dynamically coupled by any other means. This demonstrates that quantum entanglement requires the entangled particles neither to come from a common source nor to have interacted in the past. In our experiment we take two pairs of polarization entangled photons and subject one photon from each pair to a Bell-state measurement. This results in projecting the other two outgoing photons into an entangled state.

Your paper (let's talk about the first one) was very challenging, but I think I sorted it out now. Let's say Alice measures photon 1 and Bob measures photon 4, like in a usual Bell-test experiment, and they can freely choose their detector angles. The statistics they measure will not show the typical non-local Bell correlations. In order to find the non-local correlations you were talking about, Alice and Bob need access to observables outside their region of spacetime (i.e. they need to measure photons 2 and 3) and postselect the events. So they need to make their local probability distributions dependent on non-local beables, contrary to what is required in the proof of Bell's inequality.

 

[A. Neumaier]

How do we fulfill symmetrization requirement with remote side?

In quantum field theory, there are only symmetrized (or antisymmetrized) multiparticle states. One cannot create any others using creation operators - they are unphysical.

 

[DrChinese]

Let's say Alice measures photon 1 and Bob measures photon 4, like in a usual Bell-test experiment, and they can freely choose their detector angles. The statistics they measure will not show the typical non-local Bell correlations. In order to find the non-local correlations you were talking about, Alice and Bob need access to observables outside their region of spacetime (i.e. they need to measure photons 2 and 3) and postselect the events. So they need to make their local probability distributions dependent on non-local beables, contrary to what is required in the proof of Bell's inequality.

I am not sure what you mean about something being "contrary" to the requirements of a Bell proof. Each experiment is different. In this case, Alice and Bob are outside each others' light cone, but their respective photon source light cones overlap exactly where the Bell state measurement (BSM) is performed.

You are welcome to reject post-selection, just as you are welcome to reject the results of any scientific paper. Not sure why that would be a reason to dismiss this incredible paper. It neatly demonstrates a very different reality than the one you describe.

The statistics from each of the projective BSMs are accumulated separately, true, but there is still no "simple" physical description of the experiment possible. (Unless you allow the BSM to have a retrocausal impact which can be seen and measured by Alice and Bob. Which violates your original premise of a preceding common cause.)

 

[rubi]

I am not sure what you mean about something being "contrary" to the requirements of a Bell proof. Each experiment is different. In this case, Alice and Bob are outside each others' light cone, but their respective photon source light cones overlap exactly where the Bell state measurement (BSM) is performed.

You are welcome to reject post-selection, just as you are welcome to reject the results of any scientific paper. Not sure why that would be a reason to dismiss this incredible paper. It neatly demonstrates a very different reality than the one you describe.

The statistics from each of the projective BSMs are accumulated separately, true, but there is still no "simple" physical description of the experiment possible. (Unless you allow the BSM to have a retrocausal impact which can be seen and measured by Alice and Bob. Which violates your original premise of a preceding common cause.)

There is no common cause, but there needn't be one in this case. There only needs to be a common cause if the Bell-violation comes from probability distributions that depend only on local beables. I don't reject postselection and I also don't reject the paper. The paper is great and completely right and also everybody is free to perform postselection whenever they want to. It's just that locality doesn't require a common cause in this situation. The paper doesn't make any claims that are in contradiction to what I've said.

In other words: The probability distributions of Alice and Bob that only depend on local beables (as Bell requires, check out the proof again) don't feature the non-locality and correlations computed from them won't violate Bell's inequality. Only the postselected probability distributions feature non-local correlations, but this is fine, because they don't depend only on local beables.

 

[zonde]

In quantum field theory, there are only symmetrized (or antisymmetrized) multiparticle states. One cannot create any others using creation operators - they are unphysical.

This happens because any state in Fock space is symmetrized (or antisymmetrized) so that output modes are symmetrized simply because they are components of state in Fock space (that describes remote side too), right?

Then the answer to my question:
"Do we automatically view the remote side in this new basis with the same amplitudes as in the local side?"
is yes, right? Because amplitudes at remote side are represented by the same mathematical object (state in a Fock space).

 

[A. Neumaier]

Do we automatically view the remote side in this new basis with the same amplitudes as in the local side?

There is only a single amplitude, namely that for the complete symmetrized state $\psi$. One cannot ascribe the amplitude to a particular region in space. Thus your questions doesn't make sense.

If you want to consider local pieces of the state you need to create them by projection. Thus you need to define projection operators $P_L$ and $P_R$ that project out the local part $P_L\psi$ and the remote part $P_R\psi$. Typically, these do not sum up to $\psi$.

 

[DrChinese]

1. There is no common cause, but there needn't be one in this case.

2. In other words: The probability distributions of Alice and Bob that only depend on local beables (as Bell requires, check out the proof again) don't feature the non-locality and correlations computed from them won't violate Bell's inequality. Only the postselected probability distributions feature non-local correlations, but this is fine, because they don't depend only on local beables.

1. There are perfect correlations, and no opportunity for a common cause. QED.

2. "Local" is an assumption of the usual Bell inequalities. In cases in which this is strictly enforced (for the usual Bell tests), Alice and Bob are non-local to each other (as in the referenced experiment) but they share a common past (unlike the referenced experiment). So I really don't follow your point. When Bell inequalities are violated, under either scenario, no local realistic explanation is possible. This is true whether you accept QM or not.

So to repeat my earlier statement: you are arguing for rejection of classical reality and acceptance of locality. Fine, that is a viable position. But in that quantum ("non-classical) world, there is (still) no causal explanation as you imply. If there were, the time ordering would be different; and it would require Alice and Bob to measure their perfect correlations (and violations of Bell inequalities) within a common light cone (presumably from a common source of entangled pairs) - which they don't.

 

[rubi]

There are perfect correlations, and no opportunity for a common cause.

This is both completely true and completely irrelevant at the same time. Perfect correlations are not problematic if the underlying probability distributions depend on non-local beables. It is only problematic for locality if the local probability distributions of Alice and Bob would satisfy Bell's factorization criterion.

"Local" is an assumption of the usual Bell inequalities. In cases in which this is strictly enforced (for the usual Bell tests), Alice and Bob are non-local to each other (as in the referenced experiment) but they share a common past (unlike the referenced experiment). So I really don't follow your point. When Bell inequalities are violated, under either scenario, no local realistic explanation is possible. This is true whether you accept QM or not.

The postselected probability distributions violate Bell's factorization criterion (because the local probability distributions depend on the beables of photon 2 and 3, which are localized outside the past lightcone of both photon 1 and 4), so a violation of Bell's inequality is not problematic (i.e. the correlations don't require a causal explanation, at least not according to Bell). A violation of Bell's inequality would only be surprising if Bell's factorization criterion did hold. The fact that correlations derived from non-local probability distributions appear to be non-local is not problematic! You might even be able to violate Bell's inequality in a local realistic theory if you use non-local probability distributions, since even in local realistic theories, Bell's inequalities must only hold if Bell's factorization criterion is satisfied.

Bell's factorization criterion holds $\Rightarrow$ Bell's inequality holds
Bell's factorization criterion doesn't hold (as is the case for the postselected probability distributions) $\Rightarrow$ anything can happen

I demand a proof that Bell's inequality is supposed to hold even if the factorization criterion is violated. Otherwise, I'm not going to accept the necessity of a causal explanation.

So to repeat my earlier statement: you are arguing for rejection of classical reality and acceptance of locality. Fine, that is a viable position. But in that quantum ("non-classical) world, there is (still) no causal explanation as you imply.

There is a causal explanation for everything that requires a causal explanation. Your experiment just doesn't require one. I fully agree that there is no causal explanation for your experiment!

There is still non-classical non-locality in the correlations between post-selected photons 1&4, but not of the kind that requires a causal explanation. There is also non-classical non-locality which requires a causal explanation (the correlations between photons 1&2 and 3&4), but there is a causal explanation for them according to quantum reasoning.

If you take non-local correlations that can be explained and combine them in a non-local way, you get non-local correlations again, but these needn't necessarily require an explanation as well. Neither in the case of quantum mechanics nor in the case of classical mechanics.

 

[atyy]

Bell's factorization criterion holds $\Rightarrow$ Bell's inequality holds
Bell's factorization criterion doesn't hold (as is the case for the postselected probability distributions) $\Rightarrow$ anything can happen

I demand a proof that Bell's inequality is supposed to hold even if the factorization criterion is violated. Otherwise, I'm not going to accept the necessity of a causal explanation.

I'm not entirely sure if this is what you had in mind, but the Bell inequalities for entanglement swapping require a slightly different argument than the usual Bell scenario.
http://arxiv.org/abs/0911.1314

 

[DrChinese]

This is both completely true and completely irrelevant at the same time. ...

I demand a proof that Bell's inequality is supposed to hold even if the factorization criterion is violated. Otherwise, I'm not going to accept the necessity of a causal explanation.

There is a causal explanation for everything that requires a causal explanation. Your experiment just doesn't require one. I fully agree that there is no causal explanation for your experiment!
...

Your comments do not make sense to me. In the Bell scheme, all separated observations should be able to factorize. When the results are inconsistent with such factoring, a Bell inequality is violated. Any entangled system does not factorize, according to QM. That includes the usual Bell tests, as well as ones per the supplied references in post #40. Entanglement is entanglement (from swapping) is entanglement (post selected), and that's what you get from entanglement swapping experiments.

According to your argument that classical realism should be rejected when considering quantum systems, there should still be common cause. There isn't, and this is equally true in the usual Bell test regimen. The only difference is that in one group, there is a common source; and in the other group, there isn't. Your distinctions don't hold water. It is good that you agree that "there is no causal explanation for" the referenced experiments; there is no such physical explanation for ordinary Bell tests either. The accepted explanation is that we apply QM and get the correct answer in all cases. And that is as much of which anyone can reasonably be sure.

 

[rubi]

Your comments do not make sense to me. In the Bell scheme, all separated observations should be able to factorize. When the results are inconsistent with such factoring, a Bell inequality is violated. Any entangled system does not factorize, according to QM.

Yes, but in the proof of Bell's inequality, the local probability distributions must only depend on beables in a region that shields the relevant beable from the overlap of the light cones. This is how locality shows up in the proof. The beables of photons 2&3 are not localized in this region, but the postselected probability distribution depends on them nevertheless.

A full specification of the beables in region 3 must determine the probability distribution for region 1 completely. This is not fulfilled for the postselected probability distributions, since they depend on the beables of photons 2&3, which are not even localized within the past lightcone of region 1, let alone region 3.

(@atyy: I hope this clarifies it.)

 

[DrChinese]

Yes, but in the proof of Bell's inequality, the local probability distributions must only depend on beables in a region that shields the relevant beable from the overlap of the light cones. This is how locality shows up in the proof. The beables of photons 2&3 are not localized in this region, but the postselected probability distribution depends on them nevertheless.

A full specification of the beables in region 3 must determine the probability distribution for region 1 completely. This is not fulfilled for the postselected probability distributions, since they depend on the beables of photons 2&3, which are not even localized within the past lightcone of region 1, let alone region 3.

Your diagram and explanation, per your perspective, would not even apply to a normal Bell test. That is because Alice and Bob would both occupy your region 3 in a Bell test. And in strict Bell tests, Alice and Bob are unable to communicate their observation decisions to each other. So no, I disagree with your characterization. Bell actually says that the decision of Alice cannot affect the outcome Bob sees. This is the criteria. This applies to entanglement swapping experiments equally as well.

Basically, you are saying that local realism - where there is a common past - is ruled out by normal Bell tests using entanglement. That is true enough. But it is a special case (there exists a common source) of a more general entanglement scenario in which there is no common source. This is a more far reaching statement, and is a deduction from standard QM. If there *is* something about the requirement of a common source in Bell tests (to achieve its conclusion ruling out local realism), clearly that requirement can be dispensed with. That is what the references I provided tell us. Quantum non-locality (between Alice and Bob) is not dependent on the existence of local beables in a common prior light cone, as you have implied. Because those beables would in the end be classical if there is to be a common cause. Obviously, the experimental results do not change when your hypothetical "common cause" is eliminated. There is no "quantum" (non-classical) local common cause. Drop the common cause entirely, or drop the local part entirely, or both.

 

[rubi]

Your diagram and explanation, per your perspective, would not even apply to a normal Bell test. That is because Alice and Bob would both occupy your region 3 in a Bell test.

This is not true. Alice is localized in region 1 during her measurement and Bob is localized in region 2 during his measurement. Bell requires that a full specification of the beables in region 3 already fully determine Alice's probability distribution. An analogous requirement must hold for Bob.
http://www.scholarpedia.org/article/Bell's_theorem#Bell.27s_definition_of_locality

Obviously, the experimental results do not change when your hypothetical "common cause" is eliminated.

If I choose not to produce entangled particles in the past, I will not see non-local correlations in the future and the other way around. From this I conclude that the cause of the correlations is my choice in the past. We can agree to disagree that this is a valid way of reasoning.

 

[DrChinese]

This is not true. Alice is localized in region 1 during her measurement and Bob is localized in region 2 during his measurement. Bell requires that a full specification of the beables in region 3 already fully determine Alice's probability distribution. An analogous requirement must hold for Bob.
http://www.scholarpedia.org/article/Bell's_theorem#Bell.27s_definition_of_locality

If I choose not to produce entangled particles in the past, I will not see non-local correlations in the future and the other way around. From this I conclude that the cause of the correlations is my choice in the past. We can agree to disagree that this is a valid way of reasoning.

You are mixing a lot of different things here.

First, you are quoting from an article by Norsen (and yes I can read the full author list - but this is mostly Norsen talking). This is a skewed article. Although it is technically correct in most particulars, I don't consider his/their use of terminology to be very good. It causes just the kind of problems in communication that we are having. For example: the use of the word "beable" is most often associated with Bohmians and this article clearly reflects that (see the first sentence if you are not sure). This word causes all kinds of problems. (FYI: A lot of folks do not support Norsen on his interpretations of Bell, so use this article as a source at your own risk. He gets a lot of opposing comments on his articles on the subject from top physicists.)

Second, the diagram (as originally supplied) meant something completely different to me that how it is used in the context of the article (and by you I now presume). So by supplying that context, we can get on the same page on that - so thanks. Norsen uses it to say that to a local realist, observations in area 3 by Alice and Bob cannot be affected by events in area 2. Please, this has little or nothing to do with the usual Bell test. As I said previously, Bell instead says that Alice's outcome should not be influenced by Bob's choice of measurement basis, and vice versa. This is a generally accepted assumption of Bell, and is directly connected to the EPR paper it is addressing.

Last: In reality, your diagram is a better description of a more general conclusion on entanglement described in the references I supplied. That being that local realists assert there cannot be entanglement of photons from sources 1 and 2 in regions that do not overlap (reading the diagram a different way). Obviously, that is wrong (as experiment plainly shows). I would conclude from the experiment that there are no non-local hidden variables either. However, technically such conclusion is still interpretation dependent and is not strictly justified.

 

[stevendaryl]

Any entangled system does not factorize, according to QM.

Bell's factorization criterion is not exactly the same as the criterion that the wave function (or density matrix) factors, because is talking about whether the probabilities for outcomes of measurements factors, rather than whether the wave function factors. Those aren't exactly the same criteria.

I gave an example earlier in this thread, but I'll repeat it: Let $|\psi, u\rangle$ be the one-electron state in which the electron is definitely spin-up along the z-axis, and has probability amplitude $\psi(\vec{r})$ of being found in position $\vec{r}$. Let $|\phi, d\rangle$ be the one-electron state in which the electron is definitely spin-down along the z-axis, and has probability amplitude $\phi(\vec{r})$ of being found in position $\vec{r}$. Then we can form the two-electron state:

$|\Psi\rangle = |\psi, u\rangle \otimes |\phi, d\rangle - |\phi, d\rangle \otimes |\psi, u\rangle$

That is an entangled state. But if the two spatial dependencies $\phi(\vec{r})$ and $\psi(\vec{r})$ have non-overlapping support (there is no place where both are nonzero), then the corresponding probabilities for spin measurements at two distant locations $\vec{r_1}$ and $\vec{r_2}$, where $\psi(\vec{r_1})$ and $\phi(\vec{r_2})$ are both nonzero, factor:

$P(A \& B | \vec{\alpha}, \vec{\beta}) = |\psi(\vec{r_1})|^2 cos^2(\theta_1/2) |\phi(\vec{r_1})|^2 sin^2(\theta_2/2)$

(where $A$ is true if an electron is found to have spin-up along $\vec{\alpha}$ at location $\vec{r_1}$, and $B$ is true if an electron is found to have spin-up along $\vec{\beta}$ at location $\vec{r_1}$, and $\theta_1$ is the angle between $\vec{\alpha}$ and the z-axis, and $\theta_2$ is the angle between $\vec{\beta}$ and the z-axis.)

 

[DrChinese]

Bell's factorization criterion is not exactly the same as the criterion that the wave function (or density matrix) factors, because is talking about whether the probabilities for outcomes of measurements factors, rather than whether the wave function factors. Those aren't exactly the same criteria. ...

OK. The linear polarization observables for any 2 photons entangled on that basis will either be something like H>H>+V>V> or H>V>+V>H> depending on whether they are in the + or - Bell state. Neither of these factor. The photons do not need to be co-existing, may be entangled before/after observation, and can be from the same or different sources.

If you ask: is Alice's outcome independent of Bob's choice of measurement basis, and vice versa? I would answer that under any "reasonable" interpretation of QM the answer is NO. On the other hand, I have no explanation of what influences what, or by what mechanism any of this occurs. Further, given that there is no requirement that Alice and Bob's photons share any prior causal contact (or common contact with any other object), I would be hard pressed to say there is any common preceding cause. And finally, I would say that the only apparent variable which explains the correlations of Alice and Bob is the relationship of their measurement bases, and nothing else.

So in total, the evidence leads us to reject the idea that there is anything objectively real other than the relationship of the observation basis. Everything else (local and nonlocal) appears to reduce to a single random value, if it reduces to anything. Bohmians associate that with a "pilot wave" and MWIers relate that to the world they inhabit.

 

[rubi]

First, you are quoting from an article by Norsen (and yes I can read the full author list - but this is mostly Norsen talking). This is a skewed article. Although it is technically correct in most particulars, I don't consider his/their use of terminology to be very good. It causes just the kind of problems in communication that we are having. For example: the use of the word "beable" is most often associated with Bohmians and this article clearly reflects that (see the first sentence if you are not sure). This word causes all kinds of problems. (FYI: A lot of folks do not support Norsen on his interpretations of Bell, so use this article as a source at your own risk. He gets a lot of opposing comments on his articles on the subject from top physicists.)

I don't agree with Bohmians either (BM clearly is a proof-of-concept conspiracy theory that shows that one can get a realistic theory by dropping locality, but it is not a viable theory of physics) and I don't agree with everything in that article, but the presentation of the locality concept is correct, as it does indeed capture precisely (for a realistic theory) what relativists mean when they talk about locality. The past lightcone around region 1 (or 2 respectively) is called the domain of dependence of region 1 (or 2 respectively). In a local theory, everything that can be known about region 1 (or 2) can only depend on data from the domain of dependence and in fact it even depends only on the data in region 3. Thus, a full specification of all data in region 3 fully determines the data in region 1 (or 2) in every realistic local theory. If you use probability distributions that explicitely depend on non-local beables (I'm using this word only to be consistent with Bell's writing, not because I'm a Bohmian. It refers to the clicks of the detectors, nothing more.), then you shouldn't be surprised about Bell violations.

Assume I gave dice to Alice and Bob and each of them threw them a thousand times. Obviously, their statistics will be completely independent. However, if I know the list of records of both Alice and Bob (this is non-local information) and I remove all entries with unequal results, i.e. I postselect the probability distributions, would you be surprised that the postselected probability distributions feature perfect correlations, although Alice and Bob were spacelike separated when they performed the experiment? Would you require me to find a common cause for that in the intersection of their past lightcones?

Second, the diagram (as originally supplied) meant something completely different to me that how it is used in the context of the article (and by you I now presume). So by supplying that context, we can get on the same page on that - so thanks. Norsen uses it to say that to a local realist, observations in area 3 by Alice and Bob cannot be affected by events in area 2. Please, this has little or nothing to do with the usual Bell test. As I said previously, Bell instead says that Alice's outcome should not be influenced by Bob's choice of measurement basis, and vice versa. This is a generally accepted assumption of Bell, and is directly connected to the EPR paper it is addressing.

Last: In reality, your diagram is a better description of a more general conclusion on entanglement described in the references I supplied. That being that local realists assert there cannot be entanglement of photons from sources 1 and 2 in regions that do not overlap (reading the diagram a different way). Obviously, that is wrong (as experiment plainly shows). I would conclude from the experiment that there are no non-local hidden variables either. However, technically such conclusion is still interpretation dependent and is not strictly justified.

I don't want to defend Norsen's arguments. I only referred to parts of his writings which I think are uncontroversial. If you don't agree that the above description captures locality, can you please point me to the mistake?

 

[DrChinese]

1. The past lightcone around region 1 (or 2 respectively) is called the domain of dependence of region 1 (or 2 respectively). In a local theory, everything that can be known about region 1 (or 2) can only depend on data from the domain of dependence and in fact it even depends only on the data in region 3. Thus, a full specification of all data in region 3 fully determines the data in region 1 (or 2) in every realistic local theory.

2. Assume I gave dice to Alice and Bob and each of them threw them a thousand times. Obviously, their statistics will be completely independent. However, if I know the list of records of both Alice and Bob (this is non-local information) and I remove all entries with unequal results, i.e. I postselect the probability distributions, would you be surprised that the postselected probability distributions feature perfect correlations, although Alice and Bob were spacelike separated when they performed the experiment?

3. Would you require me to find a common cause for that in the intersection of their past lightcones?

1. I follow your notation in this example, and agree that this is what a local realist would assert. Region 3 would receive information (potentially) from an area that is common to both 1 and 2 (which I guess is Alice and Bob in our example).

2. This bears no resemblance to the post-selection done in any Bell tests, including the references I provided. Post-selection criteria is independent of the values that Alice and Bob record. The criteria relates to having pairs arrive at Alice and Bob that can be matched as to time window and Bell State entanglement type (+ or -). Because the criteria is independent of the values Alice and Bob observe, it is a fair test of entanglement. In the swapping-type experiments, a large portion of the photons that Alice and Bob see are not entangled at all. Only some pairs meet the qualifications, because only some pairs are cast into a Bell State. But Alice and Bob have no way to know which photons they see will be a part of a pair. The entanglement can be performed before or after Alice and Bob record anything. Again, these are independently observed and recorded.

3. I would require it if you continue to insist that there is some common cause to what Alice and Bob see. You are the one asserting that you can reject classical realism, maintain locality, and still have a common cause in the past. You can't, that is what all of these experiments show. That, to me, is the point of our discussion.

 

[rubi]

I follow your notation in this example, and agree that this is what a local realist would assert. Region 3 would receive information (potentially) from an area that is common to both 1 and 2 (which I guess is Alice and Bob in our example).

So if a local realist would assert this, then a violation of Bell's inequality, derived from this local realism assumption, let's us discard local realism. However, a violation of Bell's inequality that isn't derived from the local realism assumption tells us nothing, neither about realism nor locality.

This bears no resemblance to the post-selection done in any Bell tests, including the references I provided.

It bears resemblance insofar as it also uses non-local information to make the post-selection.

Post-selection criteria is independent of the values that Alice and Bob record. The criteria relates to having pairs arrive at Alice and Bob that can be matched as to time window and Bell State entanglement type (+ or -). Because the criteria is independent of the values Alice and Bob observe, it is a fair test of entanglement.

In order to be a test for locality, it must also be independent of non-local beables.

I would require it if you continue to insist that there is some common cause to what Alice and Bob see. You are the one asserting that you can reject classical realism, maintain locality, and still have a common cause in the past. You can't, that is what all of these experiments show. That, to me, is the point of our discussion.

I thought I made it perfectly clear that I don't insist on a common cause for an experiment that uses entanglement swapping, precisely because the postselected probability distributions depend on non-local information. Non-local correlations are only weird, if they are obtained from data that is collected locally. Otherwise, it is perfectly fine to have non-local correlations without common cause. Not even a local realist would insist on a common cause for non-local correlations that have been postselected using non-local data.