Confused by nonlocal models and relativity

[kurt101]

Sure, but it is important to realize that this informal interpretation doesn't contradict relativistic QFT just as long as it ignores its relativistic content and the current consensus about non-relativistic quantum mechanics being less fundamental than relativistic QFT.

How does this informal interpretation contradict QFT by ignoring its relativistic content?

What does non-relativistic QM vs relativistic QFT have to do with this interpretation?

And the interpretation that I am discussing is what I said to PeterDonis:
In the general sense like John Bell said in his Bertlmann's socks paper "Thirdly, it may be that we have to admit that causal influences - do go faster than light.".

In a more specific way as I previously described in a previous post on this thread about the entanglement swapping paper that DrChinese mentioned. For example I said "1 is measured and non-locally gives the result of this measurement to 2".

 

[DrChinese]

As explained by PeterDonis it is only if applied to spacelike-separated operators that they commute. If you treat the particles nonrelativistically as one system there is no spacelike separation consideration possible and the operators don't commute, but then mathematically there is no "spooky action at a distance" to be puzzled about. If one insists in factoring in that distance there is no other accepted way than going to the wider scope relativistic theory that acknowledges a spacelike separation for which the operators must commute.

I don't believe relativity has anything to do with it. There are no predictive differences for a local setup versus a non-local setup for anything related to entanglement (at least AFAIK).

Conjugate operators of distant entangled particles do not commute. If they did, the EPR Paradox would not have been solved. EPR's argument was precisely that you could learn non-commuting values on a particle by reference to the entangled partner. If the partner's conjugate property DID commute, you could do that. But you can't.

Further, Bell experiments would have different outcomes. After all, the experiments have been done with strict Einsteinian non-locality a number of times. The predictions for Bob depend on the choice of measurement on distant Alice. The match statistics are dependent solely on the relative angle theta between, in a manner inconsistent with separability (Bell 1964). Bell showed there was a different relationship on theta if there were separability (i.e. and therefore they commuted). This difference does not show up in cases where the same measurement is performed on both Alice and Bob. Which makes it easy to gloss over.

 

[microsansfil]

Conjugate operators of distant entangled particles do not commute.

The commuting property of operators is independent of the states of the systems to which these operators may apply, isn't it? whether or not they're entangled.

Mathematically the two operators $Z_A \otimes I_B$ and $I_A \otimes X_B$ commute, regardless of whether Alice and Bob's particles are entangled, as shown here

/Patrick

 

[DrChinese]

The commuting property of operators is independent of the states of the systems to which these operators may apply, isn't it? whether or not they're entangled.

... as shown here

/Patrick

I am not sure where your citation is from (the image presented on the link). Could you pass that along for context? I admit that without any mention of entanglement, it is difficult to see how this relates.

On the other hand: I did mention the source for my (directly opposing) quotes in post #140, but did not provide a link. It is:

Weinberg, Lectures on Quantum Mechanics

 

[microsansfil]

I am not sure where your citation is from (the image presented on the link).

Quantum Mechanics by Claude Cohen-Tannoudji

/Patrick

 

[DrChinese]

Quantum Mechanics by Claude Cohen-Tannoudji

/Patrick

Thanks. The author is of course top notch. However, I believe this was originally written in 1973. It lacks a section on entanglement, even in the newer editions. I don't think this is a good fit for our discussion. The formulae presented do not describe entangled systems. (Or if it does, I couldn't locate it.)

 

[microsansfil]

It lacks a section on entanglement, even in the newer editions.

It is in volume 3.

/Patrick

 

[PeterDonis]

I just want to point out that εA ⊗ εB is different to εA and also different to εB.

Yes, and εA ⊗ εB is the only one of them that we need to be discussing here.

in εA ⊗ εB you can have entangled particles. i. e. which is not factorized as the tensor product, and thus can't be expressed as a single product of an "A" state and a "B" state.

Yes, this is the correct definition of an entangled state, as I think I've already said.

there exist operators in εA ⊗ εB which are not tensor products of an operator of εA and an operator of εB.

Yes, but none of them are relevant to the measurements we've been talking about.

what about the commutation of this kind of operator?

It would depend on the specific operators. But, as above, none of them are relevant to what we've been discussing. You're welcome to start a separate thread if you want to discuss operators like this, though it would be a good idea to first find a specific experiment that involves them.

 

[PeterDonis]

As explained by PeterDonis it is only if applied to spacelike-separated operators that they commute. If you treat the particles nonrelativistically as one system there is no spacelike separation consideration possible and the operators don't commute

You are misinterpreting what I said, and your statements are incorrect.

The operators I wrote down commute regardless of whether events A and B are spacelike separated or not.

Considering two entangled particles nonrelativistically as one system certainly does not rule out spacelike separation. It just means you don't make any use of it in your analysis. And the operators I wrote down commute regardless of whether you do a relativistic or a non-relativistic analysis.

 

[PeterDonis]

The QM math does not describe what happens when you measure a single instance of entangled particles.

That depends on which interpretation you adopt.

If the universe can be simulated (in a theoretical sense) and has hidden variables that describe its state and you can pause the simulation then you can say what measurement happened first in the simulation.

This is way off topic for this thread, and indeed this forum.

What we observe in the experiment is also a basis for talking about counterfactuals.

I don't see how, since we only observe what actually happened, not what might have happened counterfactually.

In the general sense like John Bell said in his Bertlmann's socks paper "Thirdly, it may be that we have to admit that causal influences - do go faster than light.".

Then I think you need to pick a specific interpretation that is based on this premise if you want to discuss it. Bell, IIRC, had Bohmian mechanics in mind as falling into this category.

 

[PeterDonis]

this informal interpretation doesn't contradict relativistic QFT just as long as it ignores its relativistic content and the current consensus about non-relativistic quantum mechanics being less fundamental than relativistic QFT.

I have no idea what you're talking about here. Do you have a reference that you are basing your claims on?

 

[PeterDonis]

The question is whether entangled Alice(p) commutes (or not) with Bob(q).

"Alice" and "Bob" aren't operators, so asking whether "Alice" and "Bob" commute makes no sense to me.

For the operators I wrote down (and I'll write them again below), they obviously commute, for the reason I gave. And those operators are the obvious operators that represent "Alice measures her particle" and "Bob measures his particle" in directions $Z$ or $X$. So I'm really, really confused at this point about what you're trying to say.

let's discuss entangled spin-1 particle spins x and z (singlet). These are not separable, do NOT commute, and the uncertainty principle should be applied.

Please write down the explicit math behind this statement, since I have no idea what it is at this point. The math I would write down is that we have an entangled state (it looks the same in either the $Z$ or the $X$ basis)

$$
| \psi \rangle = |\uparrow \rangle_A \otimes |\downarrow \rangle_B - |\downarrow \rangle_A \otimes |\uparrow \rangle_B
$$

and the operators applied to it are, by Alice, either $Z_A \otimes I_B$ or $X_A \otimes I_B$, and, by Bob, either $I_A \otimes Z_B$ or $I_A \otimes X_B$. Both "Alice" operators obviously commute with both "Bob" operators, so regardless of which choice of measurement Alice and Bob make, their measurements will commute.

It seems like you must have some different math in mind since you are claiming that the measurements don't commute, but I have no idea what it is. And I can't even respond to the rest of what you're saying at this point because I don't know what mathematical description you have in mind; the only mathematical description I can come up with is the one I wrote down above, which makes what you are saying seem obviously wrong. So, again, I'm really, really confused at this point.

 

[kurt101]

From the book "Quantum Field Theory for the Gifted Amateur" Link to preview page it says:
"If we say that the operator $\hat O_1$ corresponds to an observable measured at spacetime point x and $\hat O_2$ corresponds to an observable measured at y, then if x and y have a space-like separation $[(x - y)^2 < 0]$ then the operators must commute."

This is similar to what @PeterDonis has said. However, isn't the space-time distance between entangled photons 0? So does this statement not apply to entangled photons?

 

[Mordred]

Don't all spin 1 particles commute, while fermionic spins anticommute ?

Never mind your looking at different operators

 

[PeterDonis]

isn't the space-time distance between entangled photons 0?

No. The spacetime interval along a single photon's worldline is 0. But the spacetime interval between events on the worldlines of different photons will not, in general, be 0.

 

[PeterDonis]

Don't all spin 1 particles commute, while fermionic spins anticommute ?

No, "commute" and "anticommute" are the wrong terms here. Wave functions of bosons are symmetric, while wave functions of fermions are antisymmetric.

 

[Mordred]

Do'h how did I confuse those terms grr.. never taking a year break from physics again grr

 

[Mordred]

Ah that's where I messed up the boson creation annihilation operators commute while the fermionic creation annihilation operators anticommute.

Still my bad lol

 

[Tendex]

You are misinterpreting what I said, and your statements are incorrect.

The operators I wrote down commute regardless of whether events A and B are spacelike separated or not.

Considering two entangled particles nonrelativistically as one system certainly does not rule out spacelike separation. It just means you don't make any use of it in your analysis. And the operators I wrote down commute regardless of whether you do a relativistic or a non-relativistic analysis.

My mistake. I did misinterpret DrChinese posts that I quoted there as still referring to the microcausality issue since in fact when mentioning you I was thinking about your remark in post 97 in connection to spacelike separation of operators and its effect on commutation as prescribed by microcausality, no relation to any other operators you might have written down:

"Operators that don't commute when applied to the same individual particle. But here we are talking about two spacelike separated operators applied to two different particles. Not the same thing."

Reference https://www.physicsforums.com/threads/confused-by-nonlocal-models-and-relativity.973876/page-4

 

[Tendex]

I have no idea what you're talking about here. Do you have a reference that you are basing your claims on?

It's just the point bobob made in the first few posts of this thread with different words. I agree with him that the origin of all these disputes about local vs nonlocal models lies in (either intentionally or not) ignoring relativity of simultaneity.

 

[vanhees71]

The question is whether entangled Alice(p) commutes (or not) with Bob(q). Or substitute any conjugate pair for p and q (such as spin at various angles). To be specific, let's discuss entangled spin-1 particle spins x and z (singlet). These are not separable, do NOT commute, and the uncertainty principle should be applied.

a. There is no question that if Alice(x) and Bob(x) are measured, they will be opposite. Similarly, if Alice(z) and Bob(z) are measured, they too will be opposite. The uncertainty principle is not a constraint.

b. In the EPR-B case, as summarized by Weinberg, Lectures on Quantum Mechanics, 12.1 Paradoxes of Entanglement: "... the observer could have measured the x-component of the spin of particle 1 instead of its z-component, and by the same reasoning, if a value h/2 or −h/2 were found for the x-component of the spin of particle 1 then also the x-component of the spin of particle 2 must have been −h/2 or h/2 all along. Likewise for the y-components. So according to this reasoning, all three components of the spin of particle 2 have definite values, which is impossible since these spin components do not commute..."

Your recent comment (I know, this is cheating!) on the above quote was: "Weinberg is not talking about successive measurements. He's talking about alternative possibilities for single measurements (one on each particle). He's simply pointing out that, since measuring the same spin component of both particles will always give opposite results, no matter which component is measured (x or y or z), any hidden variable model, i.e., any model that attributes the correlations between these measurements to pre-existing properties of the particles, would have to attribute definite spin components (+1 or -1) in all three directions (x and y and z) to each particle, i.e., the pre-measurement state of each particle would have to have definite values for spin-x, spin-y, and spin-z. But that is not possible because no quantum state can have definite values for multiple non-commuting operators. "

I think you agree with Weinberg that the Uncertainty Principle applies to non-commuting components of entangled pairs. And again, any analysis of EPR-B is going to say virtually the same as Weinberg. Weinberg further comments: "There is a troubling weirdness about quantum mechanics. Perhaps its weirdest feature is entanglement, the need to describe even systems that extend over macroscopic distances in ways that are inconsistent with classical ideas." . One of the classical ideas being local causality.

c. And in fact, Weinberg goes on to say as follows: "Of course, according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem - it just doesn't change the density matrix." Which is what I assert: A measurement on Alice's particle changes the physical state of Bob's remote entangled particle (what is observed). Although if you and vanhees71 are instead referring to the density matrix, you would be right about that. (Weinberg personally believes in the reality of the reduced density matrix rather than the state vector, I believe, but that does not seem a common interpretation.)

d. This from Wikipedia, which is not cited as a source but rather to indicate the generally accepted viewpoint and related derivation:

EPR Paradox
"...how does Bob's [entangled] positron know which way to point if Alice decides (based on information unavailable to Bob) to measure x (i.e., to be the opposite of Alice's electron's spin about the x-axis) and also how to point if Alice measures z, since it is only supposed to know one thing at a time? The Copenhagen interpretation* rules that the wave function "collapses" at the time of measurement, so there must be action at a distance..." Of course, the mathematical presentation shows that Alice's x and Bob's z are constrained by the Uncertainty Principle and clearly do not commute. That portion is identical to nearly any presentation of this problem.*Copenhagen simply being one viable interpretation, there are many others too; such would not change the sense of this passage.

In the experiment we discuss Alice's projection operator acting on her two-photon state is NOT acting on Bob's two-photon state by construction. To see this you have to write the initial state in form of Eq. (3), which is written as a sum over products of Alice's and Bob's two photons. Then Alice's projector in the four-photon space is
$$\hat{P}_A \otimes \hat{1}_B=|\Psi_{-}^{12} \rangle \langle \Psi_-^{12} \rangle \otimes hat{1}^{03}.$$
As is immediately seen from Eq. (3) of the paper the subensemble projected out by Alice is nevertheless given by
$$\hat{P}_A \otimes \hat{1}_B|\Psi \rangle=\frac{1}{2} |\psi_{-}^{12} \rangle \otimes |\psi_{-}^{03} \rangle.$$
Nothing interacted with Bob's photons due to Alice's manipulations of her photons.

This is insured in the delayed-choice version of the experiment. There Alice performs her experiment after Bob's photons are measured and thus absorbed in his photodetectors since there cannot be any interactions from the future to the past by construction in relativistic QFT!

 

[vanhees71]

"Alice" and "Bob" aren't operators, so asking whether "Alice" and "Bob" commute makes no sense to me.

For the operators I wrote down (and I'll write them again below), they obviously commute, for the reason I gave. And those operators are the obvious operators that represent "Alice measures her particle" and "Bob measures his particle" in directions $Z$ or $X$. So I'm really, really confused at this point about what you're trying to say.
Please write down the explicit math behind this statement, since I have no idea what it is at this point. The math I would write down is that we have an entangled state (it looks the same in either the $Z$ or the $X$ basis)

$$
| \psi \rangle = |\uparrow \rangle_A \otimes |\downarrow \rangle_B - |\downarrow \rangle_A \otimes |\uparrow \rangle_B
$$

and the operators applied to it are, by Alice, either $Z_A \otimes I_B$ or $X_A \otimes I_B$, and, by Bob, either $I_A \otimes Z_B$ or $I_A \otimes X_B$. Both "Alice" operators obviously commute with both "Bob" operators, so regardless of which choice of measurement Alice and Bob make, their measurements will commute.

It seems like you must have some different math in mind since you are claiming that the measurements don't commute, but I have no idea what it is. And I can't even respond to the rest of what you're saying at this point because I don't know what mathematical description you have in mind; the only mathematical description I can come up with is the one I wrote down above, which makes what you are saying seem obviously wrong. So, again, I'm really, really confused at this point.

What we have in the here discussed experiment before any manipulations by A or B is the four-photon state given in the paper in Eq. (2) which can be written in form of products $|\psi_{\text{Alice}} \otimes |\psi_{\text{Bob}} \rangle$ as given in Eq. (3):

where
$$\langle \Psi^{\pm}_{ij} \rangle = \frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_i,H) \hat{a}^{\dagger}(\vec{p}_j,V) \pm \hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_i,V) \right]|\Omega \rangle,\\
\langle \Psi^{\pm}_{ij} \rangle = \frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_i,H) \hat{a}^{\dagger}(\vec{p}_j,H) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_i,V) \right] |\Omega \rangle.$$

 

[Mentz114]

1. Yay!

2. Adding delay to change the ordering (sequence) does not change the statistics. The following can occur in any order:

a. Detection of photon 1.
b. Detection of photon 4.
c. Projection of the photons 2 & 3 into a Bell state via co-arrival at the beam splitter.
d. Creation of photons 1 & 2 (must precede a. and c. though).
e. Creation of photons 3 & 4 (must precede b. and c. though).

3. I don't know if it makes sense to refer to "everything happening at once" in a normal temporal sense. Precisely because there is no required order other than that photons must be created before they are detected, and photons 2 & 3 must be created prior to projection.

My "narrative" to describe entanglement swapping is as follows:

When photons 1 & 2 are created, they form an entangled system "X" which grows to have spatio-temporal extent. When photons 3 & 4 are created, they too form an entangled system "Y" which grows to have spatio-temporal extent. As elements of quantum systems X and Y intersect at the beam splitter, they split into 2 new systems that are likewise entangled, but consisting of different pairing of the photons. After the *beamsplitter* portion of the BSA, ALL 4 PHOTONS ARE STILL ENTANGLED: 1 & 4, and 2 & 3. And in some experimental versions, the 2 & 3 pair is in the singlet state and therefore otherwise has the same characteristics as the 1 & 4 pair. Both sets now exhibit perfect correlations.

a. What can't be described in this narrative is the nature of how systems with spatio-temporal extent "collapse", if indeed there is something that can be called collapse. Because of Bell: this implies that "something" changes non-locally, and it certainly appears that it is NOT the revealing of quantum properties that had preexisting values. Because entangled particles lack well-defined preexisting values until observation (again per Bell, and this particular characteristic appears one way or another in all interpretations).

b. When can it be said that 1 & 4 become entangled? They need not ever have been in causal contact, don't need to exist at the same time, don't need to even exist when they became entangled. And because of entanglement monogamy, they cannot remain entangled (as they were previously) with their birth twins.

c. Returning to the OP: special relativity does NOT in any way figure in, constrain, or otherwise involve itself in the quantum description. In fact, SR can be even considered time symmetric (just to add to the confusion).

For some reason I missed the alert when you posted the above and I first saw it yesterday while looking for a link. It must seem impolite that I did not reply but it was unintentionial.
It is a good reply, thank you, and even belatedly it is food for thought.

 

[DrChinese]

"Alice" and "Bob" aren't operators, so asking whether "Alice" and "Bob" commute makes no sense to me.

A and B, or Alice and Bob, or whatever we choose to call them, are intended to be entangled particles. Any measurement on canonical conjugate property pairs of those particles, assuming those specific properties are entangled, will not commute - no differently than if you were examining any single particle. That is the entire lesson of the EPR paradox: you cannot gain more information about one of those particles by examining properties of the other (than the HUP allows). Applying the HUP:
$$ \sigma A_p \space \sigma B_x \geq \frac \hbar 2 $$
$$ [A_p, B_x]| \psi AB_{entangled} \rangle \neq 0 $$
Whereas by definition, unentangled states are of the form:
$$ | \psi AB_{unentangled}\rangle = | \psi A_i \rangle \otimes | \psi B_j \rangle $$ I.e. They're tensor products of particular states describing the subsystems*.

Per Steven Weinberg (Lectures on Quantum Mechanics), on the quantum state of an entangled system:
"...no quantum state can have definite values for multiple non-commuting operators..."
"...according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem..."
"... [A] measurement, which gives particle 1 a definite position, does indeed prevent particle 2 from having a definite momentum, even though the two particles are far apart. The two particles are said to be entangled."

It would help to show me a citation that says that otherwise to the above. I suspect we are talking past each other, but I am not sure in what way. I am simply espousing the standard description of the EPR paradox, and its solution: quantum non-locality.*Format & description taken from a Lubos Motl blog post.

 

[vanhees71]

Of course, Weinberg is right, but I'd formulate a bit differently in the context of this discussion since it is obviously a big problem to understand the subtle balance how the causality structure of (special) relativity is completely fulfilled by local relativistic QFTs (local in the sense of interactions, i.e., fulfillment of the microcausality constraint for local observable operators) and at the same time implying the (sometimes) long-ranged correlations described by entanglement between far-distant parts of a quantum system.

The first sentence is unproblematic and follows directly from the formalism, though there are of course exception from this rule. E.g., the angular-momentum eigenstate for $j=0$ (implying $m=0$ too) is a common eigenstate for all angular-momentum components $J_k$, though the corresponding operators do not commute: $[\hat{J}_j,\hat{J}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{J}_l$.

The 2nd one is problematic, if interpreted in the sense of a "collapse" as a physical process. That's why I prefer the minimal statistical (ensemble) interpretation: The quantum state (which in general is a statistical operator rather than a state vector or rather a ray defined by it, which is a special case when the system or partial system is in a pure state) describes probabilities, and as in general probability theory the probability assignment to a given situation depends on the information one has about the system.

E.g., in the here discussed experiment, if A measures her two photons 1&2 she knows in which definite (pure) state Bob's pair 0&3 is. Bob cannot know this instantly but Alice has to provide her knowledge gained due to the measurement. For Bob's measurement outcome (in the sense of the statistics of these outcomes when repeated on an ensemble of equally prepared systems) nothing changes, and the description of both A and B at any stage provides the same statistics for the same ensemble. What's changed is the statistics for the partial ensemble chosen in dependence on A's measurement outcome. This sub-ensemble can also only selected (or even post-selected!) after A has provided her measurement result for each event to Bob (or another agent, called Victor in the paper, who then can choose the sub-ensemble).

The same holds for the 3rd sentence, which obviously refers to the original EPR example for momentum-entangled two-particle states (decay of a particle into two particles).

 

[Tendex]

The operators I wrote down commute regardless of whether events A and B are spacelike separated or not.

Now that I've seen the operators you wrote down for Bob and Alice's measurement I see they are of course commuting, aren't they written as commuting just because they refer to measurements of entangled particles that are not causally related(i.e. spacelike separated)? Otherwise if they referred to entangled particles causally related like DrChinese says they coudn't be written down as commuting.

 

[DrChinese]

Now that I've seen the operators you wrote down for Bob and Alice's measurement I see they are of course commuting, aren't they written as commuting just because they refer to measurements of entangled particles that are not causally related(i.e. spacelike separated)? Otherwise if they referred to entangled particles causally related like DrChinese says they coudn't be written down as commuting.

No entangled system can be written as the tensor product of 2 independent subsystems (assuming the basis you are describing is entangled). Obviously, there are entangled systems that are not entangled on every basis - those can be described as a product. There are 2 sets of statistics for a quantum system of 2 particles: Product State and Entangled State. And they apply just as you would expect.

When a system is spacelike separated and entangled on some basis, it cannot be properly described as 2 independent systems on that basis. Ergo, quantum non-locality, regardless of any hand-waving. What happens to one leads to a decisive change in the reality of the other, regardless of distance. However, it is impossible to say which one "causes" the change to the other, except by assumption. Conventionally it is described that the earlier measurement "causes" the change to the state of the other (not yet measured particle). That convention is apparent in the Weinberg quote (hopefully he is regarded as a suitable authority):

"...according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem..."

 

[Tendex]

No entangled system can be written as the tensor product of 2 independent subsystems (assuming the basis you are describing is entangled). Obviously, there are entangled systems that are not entangled on every basis - those can be described as a product. There are 2 sets of statistics for a quantum system of 2 particles: Product State and Entangled State. And they apply just as you would expect.

When a system is spacelike separated and entangled on some basis, it cannot be properly described as 2 independent systems on that basis. Ergo, quantum non-locality, regardless of any hand-waving.

This seems right to me, and additionally to this quantum mechanical analysis that is conventionally done in a causal way with an earlier measurement "causing" the other by conventional assumption as you explain, there is a relativistic QFT analysis that uses products of field operators and where there is commuting or anticommuting(depending on the particles being bosons or fermions) for spacelike separated field operators and has been mixed with the former in this thread in an unnecesarily confusing way.

 

[PeterDonis]

Applying the HUP

I'm sorry, but this doesn't help. I need you to explain what's wrong with the math I wrote down in post #152. I wrote down an entangled state for a 2-qubit system and two sets of operators, the "Alice" set and the "Bob" set. Each operator in the "Alice" set manifestly commutes with each operator in the "Bob" set. (Note that these are just for a 2-qubit system, because I don't even want to try to tackle the 4-qubit system if I can't even understand what you're saying in the simpler case of the 2-qubit system.)

I want to know which of the following possibilities is correct:

(1) The math I wrote down in post #152 is correct. That means the "Alice" operators I wrote down do commute with the "Bob" operators I wrote down. If that is the case, then I need to understand what you are saying doesn't commute, since it must be something else.

(2) The math I wrote down in post #152 is not correct. If that is the case, I need you to tell me what the correct math is to describe the 2-qubit case, where Alice and Bob can each choose to measure either z spin or x spin on their respective qubits.

Neither of the above possibilities involves the HUP, because we can't even apply the HUP if we don't know which operators we are talking about or whether they commute. I need to know that first, and from what you've said so far, I don't.

 

[PeterDonis]

there are entangled systems that are not entangled on every basis - those can be described as a product.

This can't be right. Whether or not a state is entangled is basis independent.

 

[Mordred]

Not that I want to get seriously involved in this thread I find the discussion interesting. Particularly since I find myself having to testify several details on entanglement.

In the course of my self studies, I came across entanglement swapping via entanglement witness operators.

It occurred to me this may be involved in certain papers mentioned in this thread. Rather than make a statement that may cause more interference at this time.
Does anyone have any good papers covering both entanglement swapping and entanglement witness as these two details aren't typically found in standard textbooks lol.
Please do not hesitate due to math detail, I prefer the strong maths

In particular I wish to examine this possibility in the first paper posted by the OP on the tripartite system reference in paper 19

 

[Mordred]

Yeesh the Op's paper doesn't even address the possibility of entanglement swapping in the tripartite scenario. He extremely poorly discusses the communication from Alice to Bob or Charlie yet covers none of the relevant maths.

What a lousy examination on the authors part. He is clearly describing an entanglement swapping system with Alice as the entanglement witness without once mentioning those terms or the relevant math

 

[Tendex]

This can't be right. Whether or not a state is entangled is basis independent.

When including identical particles at some frame (in this case for instance spacelike separated prior to any measurement on either particle) the concept of entanglement admits a basis dependence (see for instance http://arxiv.org/abs/1302.3509) and I believe this is being used in DrChinese analysis. But then in the entangled basis by definition the operators won't commute (they describe causally distinguished particle events in an inseparable state and the HUP applies) and that's why the ones you wrote are not valid in the entangled basis.

The relativistic QFT analysis on the other hand is (as I've been insisting) more appropriate as the basis independence is always kept and prevents from speculations about any "spookiness" that sneaks in through basis-dependence.

 

[DrChinese]

This can't be right. Whether or not a state is entangled is basis independent.

I was using "basis" as meaning "property" or "observable". For example: a particle can be entangled with another on the spin or polarization basis without being entangled on position or momentum basis. Etc.

 

[PeterDonis]

I was using "basis" as meaning "property" or "observable".

Ok, that clarifies what you meant. I would use "degree of freedom" for that, since "basis" has a standard meaning in QM that is not what you meant.