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

[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.