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

[stevendaryl]

I had already adressed this using nothing but pure logic:

Since my criterion for "nonlocal" has nothing to do with common causes or realism, I don't see how that response is at all relevant.

 

[DrChinese]

Interesting. I'll read the articles, but I won't have time to reply before sunday evening. Maybe there is another causal explanation for these experiments that just isn't as simple as the one I proposed earlier. If not, I'll admit that I am wrong!

Take your time. One can always rationalize any position, but the fact is that photons can be entangled (via swapping) that have never interacted in any way, are from independent sources, etc. And in fact you can decide to entangle them AFTER they no longer exist. Pretty hard to make the case for a common cause in that.

And you still would not be making sense even if these experiments didn't exist. Because if you make the case for a common cause you are asserting realism (hidden variables); and we know (due to Bell) that hidden variables cannot be local. It doesn't make sense for you to assert otherwise, as this is generally accepted science (in fact see my tag line, from the Wiki page on Bell's Theorem).

In fact, reading your comments more closely: you are arguing for rejection of classical reality and acceptance of locality. Fine, that is a viable position. But in that quantum world, there is no causal explanation as you imply.

 

[jk22]

If we apply literally EPR's criterion, then the elements of reality should exist only when measuring at same angles for EPRB. But does this not already implies non locality, since how would the elements of reality "know" the state of both detectors to know if they have to exist or not ?

On this view Bell did not show that Einstein was wrong, since Bell suppose elements of reality (lambda) at any angles, whereas EPR would reply, except for same angle, there is no evidence that elements of reality exist.

 

[zonde]

Well, in the case of a Bell-test experiment, we can explain the non-local correlations if we accept that the world is quantum mechanical: The entangled quantum particles had been created locally in the past and then sent to Alice and Bob respectively at a speed lower than the speed of light. This is a perfectly causal explanation as soon as we accept that the world is quantum mechanical and quantum objects just happen to exist. If we hadn't entangled the particles in the past, before we sent them to Alice and Bob respectively, we wouldn't have seen the perfect correlations in the Bell-test experiment. A classical physicist couldn't accept such an explanation. But for a quantum physicist, who accepts that the world behaves quantum mechanically, it is not problematic. It's only the classical thinking that makes it seem weird.

The statement "if we accept that the world is quantum mechanical" is very problematic. First, it might mean a lot of different things related to different interpretations. Second predictions of a theory should be formulated in terms that are independent of theory. So if we analyze just predictions there should be no need to accept anything from the theory.
The second part about causal explanation taken alone is clearly not enough. We can see it by using reductio ad absurdum type argument:
Let's imagine that we are presented with evidence that monozygotic twins can relay messages faster than light. So if we can trace these twins back to the same zygote we can claim that there is no FTL phenomena?
Obviously we have to include into analysis relied messages and correlations between them. In case of entanglement it's measurement angles that we have to include into consideration.

 

[strangerep]

Ah, sorry, this is not factually correct. [...]

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

As well as a "like" of your post, I just wanted to give an explicit "thank you" for emphasizing that. I had not appreciated how strongly such experiments challenge (some) interpretations of physics.

 

[DrChinese]

As well as a "like" of your post, I just wanted to give an explicit "thank you" for emphasizing that. I had not appreciated how strongly such experiments challenge (some) interpretations of physics.

Thanks for your kind comment!

This class of experiments is very difficult for many QM interpretations - regardless of what one's favorites are. Because they don't fit naturally with either the Many Worlds or the Bohmian Mechanics groups. That doesn't stop those groups from claiming they are not ruled out, but again there is nothing natural about how they address this. MW says there is a splitting of worlds upon observation (its signature feature), but clearly that doesn't help much when entanglement is performed AFTER the splitting of worlds. And BM says there are non-local guide waves (its signature feature), which seemingly fails to explain why a photon that no longer exists is entangled with one that exists now - but is not entangled with anything else.

Again, not trying to turn this thread into another battle of interpretations. We can do that in a new thread if that is desired by my many MW and BM friends here ... :)

 

[jk22]

Is it that the only remaining were then the copenhagen ?

 

[atyy]

Thanks for your kind comment!

This class of experiments is very difficult for many QM interpretations - regardless of what one's favorites are. Because they don't fit naturally with either the Many Worlds or the Bohmian Mechanics groups. That doesn't stop those groups from claiming they are not ruled out, but again there is nothing natural about how they address this. MW says there is a splitting of worlds upon observation (its signature feature), but clearly that doesn't help much when entanglement is performed AFTER the splitting of worlds. And BM says there are non-local guide waves (its signature feature), which seemingly fails to explain why a photon that no longer exists is entangled with one that exists now - but is not entangled with anything else.

Again, not trying to turn this thread into another battle of interpretations. We can do that in a new thread if that is desired by my many MW and BM friends here ... :)

I'm not sure the word "entanglement" is used in a standard way in those papers.

I labelled the following statements from http://arxiv.org/abs/1102.1490 as "A" and "B".

"A: We also demonstrate that this two-photon signal may violate Bell's inequalities in the Clauser and Horne (CH74) formulation [12].

B: In view of the above statement this implies that entanglement among two photons may exist even though the two photons do not overlap in time."

In fact, B does not follow from A, if one uses a standard meaning of entanglement. A is what is observed, so as long as an interpretation can explain A, it is fine.

 

[stevendaryl]

I'm not sure the word "entanglement" is used in a standard way in those papers.

I always had a simplistic view of entanglement: A two-particle state is entangled if its state cannot be written as a product. But that doesn't actually make sense, because the Fermi and Bose statistics forces the state to be symmetrized in a way that it can't be a simple product. So I'm not sure what the mathematical definition of entangled state ought to be.

 

[kith]

I'm not sure the word "entanglement" is used in a standard way in those papers.

That's also my sentiment. I didn't have a look at the papers yet but I don't see how the standard definition of entanglement (inseparable states of a tensor product space) can be applied to photons which didn't coexist. On the other hand, violations of Bell-type inequalities without entanglement are maybe even more interesting.

 

[A. Neumaier]

So I'm not sure what the mathematical definition of entangled state ought to be.

It is precisely what you say. Entanglement is a mathematical property that makes only sense between distinguishable particles. They are typically distinguished by their preparation (label the particles by the beam in which they are at the beginning) before they get entangled.

Indistinguishable particles in a multiparticle state have no identity - they don't have a true particle existence since the physical Hilbert space for them has no position operator for one particle! This is why it is much more natural to describe them by fields, which give naturally rise to indistinguishable particle states as anonymous excitations.

If you want to treat indistinguishable particles in a 2-particle state them as two particles with an identity you need to describe them in an unphysical bigger Hilbert space of distinguished particles. There they will be automatically entangled, and remain so if the interaction is physical, since they will remain indistinguishable.

Thus forcing realistic quantum physics into a particle picture creates weirdness almost from the start.

 

[zonde]

On the other hand, violations of Bell-type inequalities without entanglement are maybe even more interesting.

I find it interesting that experiments that test Bell inequalities with efficient detection (with fair sampling loophole closed) actually use non-maximally entangled states as they allow violation of Bell inequalities at lower efficiency.
By itself it does not mean anything but it sort of suggests that entanglement might be just special case of some other more fundamental phenomena (say interference).

 

[DrChinese]

That's also my sentiment. I didn't have a look at the papers yet but I don't see how the standard definition of entanglement (inseparable states of a tensor product space) can be applied to photons which didn't coexist.

Sure they are inseparable, no different in any way mathematically than any other entangled system if drawn up appropriately. The question is: what is the physical meaning of an entangled system with components that do not co-exist? That is certainly no "weirder" (see thread name) than when the entangled components are not co-located (i.e. not local to each other). Keep in mind that standard QM does not favor one over the other (non-local vs. non-contemporaneous entanglement) in any manner. It does not favor entanglement before detection over entanglement after detection either. All of these are equivalently entangled, and you cannot signal with any of the variations.

 

[atyy]

That's also my sentiment. I didn't have a look at the papers yet but I don't see how the standard definition of entanglement (inseparable states of a tensor product space) can be applied to photons which didn't coexist. On the other hand, violations of Bell-type inequalities without entanglement are maybe even more interesting.

http://arxiv.org/abs/1209.4191 Eq 2 and 3.

It's just entanglement swapping. Nothing very mysterious. What is observed is "Entanglement swapping creates correlations between the first and last photons non-locally not only in space, but also in time. Quantum correlations are only observed a posteriori, after the measurement of all photons is completed."

So as long as one can explain the correlations, that's fine. I'm sure there should be a notation (probably using second quantization to fully allow creation and destruction of photons) that will allow even the quantum mechanics to be put into non-mysterious English.

 

[atyy]

I always had a simplistic view of entanglement: A two-particle state is entangled if its state cannot be written as a product. But that doesn't actually make sense, because the Fermi and Bose statistics forces the state to be symmetrized in a way that it can't be a simple product. So I'm not sure what the mathematical definition of entangled state ought to be.

But in this case, one can use the simplistic view, since identical particles need not be involved.

 

[vanhees71]

It is precisely what you say. Entanglement is a mathematical property that makes only sense between distinguishable particles. They are typically distinguished by their preparation (label the particles by the beam in which they are at the beginning) before they get entangled.

Now it's indeed very confusing (not to say weird). Usually the experiments on entanglement are done with photons, which are indistinguishable bosons. Using parametric down conversion they prepare, e.g., the singlet state
$$|\Psi \rangle =\frac{1}{\sqrt{2}} [|\phi_A, \phi_B \rangle -|\phi_B,\phi_A \rangle] \otimes [|1,-1 \rangle-|-1,1 \rangle],$$
where I've factorized the states in a spatial and a helicity (in one arbitrarily given direction) part. It's a symmetrized state as it must be; $|\phi_A \rangle$ denotes a state that refers to a single-photon "wave packet" moving in A's direction. The photons are indistinguishable as it must be. What's entangled are the polarizations, i.e., if A finds $+1$, B finds $-1$ and vice versa. You can't say who measures which individual photon. You can only say that A measures a photon and its polarization state as well as B at the location of their experimental setups (polarizer+photon detector).

I think it's very clear, if you write the state in this complete way, including the spatial (or momentum) part of the states, that the photons are indistinguishable, particularly in this case. You can't say, which individual photon has which helicity. It doesn't even make any sense to try so, because of the very preparation discussed here.

Also you don't need many-body states to have entanglement. A nice example is the Stern-Gerlach experiment which can be seen as an apparatus preparing single-particle states, where position and spin are entangled. In the above notation this single-particle state would read as follows
$$|\psi \rangle=c_1 |\phi_1 \rangle \otimes |+1/2 \rangle + c_2 |\phi_2 \rangle \otimes |-1/2 \rangle,$$
where $|\phi_j \rangle$ refers to wave packets that peak in FAPP well separated regions of space. Then the particle has a spin component +1/2 if found at location 1 and -1/2 if found in region 2.

Indistinguishable particles in a multiparticle state have no identity - they don't have a true particle existence since the physical Hilbert space for them has no position operator for one particle! This is why it is much more natural to describe them by fields, which give naturally rise to indistinguishable particle states as anonymous excitations.

If you want to treat indistinguishable particles in a 2-particle state them as two particles with an identity you need to describe them in an unphysical bigger Hilbert space of distinguished particles. There they will be automatically entangled, and remain so if the interaction is physical, since they will remain indistinguishable.

Thus forcing realistic quantum physics into a particle picture creates weirdness almost from the start.

Well, in non-relativistic QT, where you have a fixed number of particles you can describe everything in terms of appropriate symmetrized or antisymmetrized wave functions. There's no need for QFT, although of course you can use QFT in this case either, and creation and annihilation operators are just more convenient to handle than the cumbersome (anti)symmetrized wave functions of the "1st-quantization formalism".

 

[A. Neumaier]

it's indeed very confusing (not to say weird)

Indeed, it is,. Many experiments about quantum foundations are confusing and hence weird by choice if their language.
Much of it t makes sense only by being sloppy enough. This sloppiness is enough to widely open the gate for all sorts
of weirdness to enter, and for rationality to leave.

Entanglement is something very useful for discussing quantum information theory, where one deals from the start with true tensor products, and exploits superpositions to get computational advantages for cryptographic security or faster algorithms.

it is misplaced for the description of 2-photon states. The two photons in a 2-photon state exist only figuratively. In no sense covered by the formal side of QED, a 2-photon state contains two single photons since photons are intrinsic relativistic objects and there are no associated operators in photon Fock space. The fact that the photon number operator has a discrete spectrum doesn't make single photons existent in a 2-photon state. If it did, we'd also have angular particles and angular antiparticles describing quantum states of high angular momentum.

 

[vanhees71]

This I don't understand either. The usual definition of an $N$-photon state is that it is an eigenstate of the total-photon-number operator of eigenvalue $N$. Then you have two photons by definition. All this, of course, refers to non-interacting photons, because there is no clear definition of a photon number for interacting quantum fields.

 

[stevendaryl]

Now it's indeed very confusing (not to say weird). Usually the experiments on entanglement are done with photons, which are indistinguishable bosons. Using parametric down conversion they prepare, e.g., the singlet state
$$|\Psi \rangle =\frac{1}{\sqrt{2}} [|\phi_A, \phi_B \rangle -|\phi_B,\phi_A \rangle] \otimes [|1,-1 \rangle-|-1,1 \rangle],$$
where I've factorized the states in a spatial and a helicity (in one arbitrarily given direction) part.

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.

 

[A. Neumaier]

The usual definition of an N-photon state is that it is an eigenstate of the total-photon-number operator of eigenvalue N. Then you have two photons by definition.

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 taking 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''.

 

[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)<br /> - \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.