I Confused by nonlocal models and relativity

[PeterDonis]

I am simply espousing the standard description of the EPR paradox, and its solution: quantum non-locality.

This might be where there is a disconnect. As we've already established, "quantum non-locality" just means "violation of the Bell inequalities", and that in no way requires that the measurements on the two spacelike separated qubits cannot commute. Violation of the Bell inequalities just means the Bell inequalities are violated.

 

[microsansfil]

something like $Z_A \otimes X_B$ .

It should be noted that $Z_A \otimes X_B$ = ($Z_A \otimes I_B$)($I_A \otimes X_B$) corresponds to the ordinary product of the two operators of Hilbert space Ɛ = Ɛ _A ⊗ Ɛ _B. As say PeterDonis, is a single operator of Ɛ.

/Patrick

 

[DrChinese]

This might be where there is a disconnect. As we've already established, "quantum non-locality" just means "violation of the Bell inequalities", and that in no way requires that the measurements on the two spacelike separated qubits cannot commute. Violation of the Bell inequalities just means the Bell inequalities are violated.

We agree that measurements of X and Z spin on particle A don't commute. There is only one shared degree of freedom for the conjugate directions, correct? Same holds for $A_z$ and $B_x$ (A and B entangled as per your formula. That's the EPR paradox (actually the EPR-B paradox), solved in favor of QM and demonstrating that the nature of measurement on A alters the reality of distant B. In EPR's words:

"This makes the reality of P and Q [our $A_z$ and $B_x$] depend upon the process of measurement carried out on the first system, which does not disturb the second system in any way. No reasonable definition of reality could be expected to permit this."

The flaw in their thinking (sentence 1) being of course that they treat each entangled particle as an independent system, when they are in fact a single combined quantum system. And they double down in sentence 2 by excluding the possibility that they were wrong in sentence 1. They should have said something like: "If this were the case, then a distant reality is shaped by the observer - a seemingly unreasonable way for things to operate."

 

[PeterDonis]

We agree that measurements of X and Z spin on particle A don't commute

Yes. Mathematically, the operators $Z_A$ and $X_A$ don't commute. Nor do the operators $Z_B$ and $X_B$. But none of those by themselves can be the operators we are discussing, because none of them operate on the two-qubit Hilbert space.

There is only one shared degree of freedom for the conjugate directions, correct?

Meaning, "spin of Alice's qubit" is only one degree of freedom? Yes, that's right.

Same holds for $A_z$ and $B_x$

Not if those operators are the ones I wrote down; those operators manifestly commute. Nor do I see why that is a problem, since "spin of Alice's qubit" and "spin of Bob's qubit" are different degrees of freedom, and the operations "measure z-spin of Alice's qubit" and "measure x-spin of Bob's qubit" therefore operate on different degrees of freedom. The fact that those degrees of freedom are entangled in the particular state of the two-qubit system we are talking about does not change any of this.

If you think the operations of "measure z-spin on Alice's qubit" and "measure x-spin of Bob's qubit" are not described by the operators I wrote down, then what operators do you think should describe them? And note that, as I've already said, the operator $Z_A \otimes X_B$ is not two operators, it's one operator, and it has to commute with itself since any operator does, so this operator cannot be a valid answer to the question of what two operators correspond to the two operations "measure z-spin of Alice's qubit" and "measure x-spin of Bob's qubit".

That's the EPR paradox (actually the EPR-B paradox), solved in favor of QM and demonstrating that the nature of measurement on A alters the reality of distant B.

Please stop using vague ordinary language. "Alters the reality of" doesn't tell me what operator you are thinking of. We are not going to resolve this discussion unless we express what we are saying with precise math. I've already given the precise math I am using.

 

[PeterDonis]

That's the EPR paradox (actually the EPR-B paradox), solved in favor of QM and demonstrating that the nature of measurement on A alters the reality of distant B.

You keep saying this as if it were relevant to this discussion. It's not. The "solution" of the EPR paradox by Bell, as I've already said, nowhere assumes or requires that the operators describing the two spacelike-separated measurements on the two qubits of an entangled two-qubit system cannot commute. And in fact those operators do commute, for reasons I've already explained multiple times now. It is no answer to my repeated arguments to keep mentioning something that in no way refutes or even addresses them.

 

[DrChinese]

1. If "causality" means, as it does in QFT, that spacelike separated measurements commute, then it is perfectly possible for spacelike separated events to be "causally related", as long as they commute.

2. What you can't do if the measurements commute is pick out one as the "cause" and the other as the "effect"; but that just means you have to accept that things can be "causally related" even if there is no invariant fact of the matter about which one came first (and is therefore the "cause" while the other is the "effect"). Any theoretical model that explains violations of the Bell inequalities is going to force you to accept something highly counterintuitive.

1. I don't follow this definition of causality, and have never seen it used this way in any paper I've read. Must be a QFT textbook thing.

2. I agree with this. And in fact I would say it's true even when there IS an invariant fact as to which came first. Selecting the first as the "cause" is [only] a convention.

 

[PeterDonis]

I don't follow this definition of causality

It's the same as the definition of "microcausality" that was repeatedly given in some thread or other; I'm unable to keep track at this point of all the arguments that have been going on recently about QM.

Must be a QFT textbook thing.

IIRC it does appear in multiple QFT textbooks, yes.

in fact I would say it's true even when there IS an invariant fact as to which came first. Selecting the first as the "cause" is [only] a convention.

Yes, I agree this is a necessary implication of the QFT definition of causality, although it's not one that is often pointed out or discussed.

 

[Tendex]

The operators I wrote down in #152 manifestly commute; you can compute it explicitly if you want, it's straightforward. But those operators acts on different degrees of freedom in the Hilbert space: the $A$ operators act on the "Alice's particle spin" degree of freedom (i.e., on Alice's qubit), while the $B$ operators act on the "Bob's particle spin" degree of freedom (i.e., on Bob's qubit).

In the language of the paper you linked to, these are measurements on different, disjoint, orthogonal subspaces of the Hilbert space, and with respect to each of these measurements, there is no "entanglement" in the sense the paper is using the term. Note that the definition of "entanglement" used in this paper is also non-standard, which is certainly going to be confusing when you try to relate what this paper is saying to the rest of the extensive literature on this topic, since the overall two-qubit state I described in post #152 is certainly entangled by the standard definition, yet by this paper's definition, the measurements on it that I described are not.

So if we agree that the product state you wrote in #152 has two different spin degrees of freedom that as subsystems are independent as written(since they commute), can you clarify why you insist the state is entangled?

 

[PeterDonis]

subsystems are independent as written(since they commute),

The fact that the operators commute does not mean the subsystems are independent.

can you clarify why you insist the state is entangled?

Because the state cannot be expressed as a single product of states of the subsystems. So the state is entangled.

In other words, you are confusing two distinct questions: whether the state is entangled, and whether the operators commute.

 

[Tendex]

The fact that the operators commute does not mean the subsystems are independent.
Because the state cannot be expressed as a single product of states of the subsystems. So the state is entangled.

In other words, you are confusing two distinct questions: whether the state is entangled, and whether the operators commute.

I see, so how exactly do the 2 spins depend on each other?

 

[Mentz114]

I see, so how exactly do the 2 spins depend on each other?

Written in the polarisation basis they share the same state. They will always show the correlation/anticorrelation whatever order they are measured.

 

[Tendex]

Written in the polarisation basis they share the same state. They will always show the correlation/anticorrelation whatever order they are measured.

Yes, but I mean in this particular case where there is no way to pick an order by construction.

 

[Mentz114]

Yes, but I mean in this particular case where there is no way to pick an order by construction.

Your question asks how the two photons depend on each other. But there is really only one thing described by a singlet state. So I don't know what you mean by dependence.

 

[PeterDonis]

how exactly do the 2 spins depend on each other?

This question is not well-defined. Are you talking about the entangled state? Then just look at the state vector. Are you talking about the correlations between measurement results? Then just compute them and see.

I'll make the same suggestion I made to @DrChinese: Please stop using vague ordinary language and express things in precise math. If you do I think you will find that the questions you are finding yourself wanting to ask will answer themselves--i.e., as soon as you've formulated them in precise math, the answer will be obvious.

 

[Mordred]

There is a bit to learn here, in some ways beyond the scope of a forum post.

Hopefully I can get this phone to properly link this MIT lesson plan

Article

Well that was painful, this forum used to be far easier to copy paste a link a well got it working.

 

[Mordred]

Anyways see the link under article last post. As mentioned by PeterDonis entangled states are basis independent. That detail is mentioned in the above article. As noted in previous posts one must be specific if the state is entangled or not. The reason will become clear in the article

 

[vanhees71]

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..."

Now I can almost agree, but this needs a bit of qualification since otherwise we start again debating about Einstein causality:

It's not enough to state that the measurement changes the state description for the one, e.g., Alice who tests whether her photon pair 1&2 is found to be in the state $|\psi_{12}^{0} \rangle$, who did the measurement, but there's no way for Bob to know instantly what Alice knows. She has to provide the information. So Bob will still use the initial state $|\Psi \rangle$ to discribe the situation. This is no contradiction since he will just get his measurement outcome when measuring his pair 0&3 with the very probability given by $|\Psi \rangle$. He knows the probability as well as Alice. Only after Alice has provided the information that she found her photon pair 1&2 in the state $|\psi_{12}^{-} \rangle$, Bob knows that his photons 0&3 are also necessarily in the state $|\psi_{03}^- \rangle$ and he'll that he'll find these photons with certainty in this state when checking. All he knew before he got the information about Alices measurement is that with probability 1/4 he'll find his pair in this state. That's what also Alice concludes, given the initial state. Indeed she'll also find with probability 1/4 her pair to be in this very state before she does the measurement, i.e., there's no contradiction with this minimal (and epistemic!) interpretation of the quantum state.

If Bob does his Bell test prior to Alice (in her common rest frame) the same narrative of the minimal interpretation can be made with Alice and Bob interchanged. They can also do their measurements as space-like separated events. Nothing changes either, and everything is consistent and particularly consitent with Einstein causality.

 

[vanhees71]

Yes. Mathematically, the operators $Z_A$ and $X_A$ don't commute. Nor do the operators $Z_B$ and $X_B$. But none of those by themselves can be the operators we are discussing, because none of them operate on the two-qubit Hilbert space.
Meaning, "spin of Alice's qubit" is only one degree of freedom? Yes, that's right.
Not if those operators are the ones I wrote down; those operators manifestly commute. Nor do I see why that is a problem, since "spin of Alice's qubit" and "spin of Bob's qubit" are different degrees of freedom, and the operations "measure z-spin of Alice's qubit" and "measure x-spin of Bob's qubit" therefore operate on different degrees of freedom. The fact that those degrees of freedom are entangled in the particular state of the two-qubit system we are talking about does not change any of this.

If you think the operations of "measure z-spin on Alice's qubit" and "measure x-spin of Bob's qubit" are not described by the operators I wrote down, then what operators do you think should describe them? And note that, as I've already said, the operator $Z_A \otimes X_B$ is not two operators, it's one operator, and it has to commute with itself since any operator does, so this operator cannot be a valid answer to the question of what two operators correspond to the two operations "measure z-spin of Alice's qubit" and "measure x-spin of Bob's qubit".
Please stop using vague ordinary language. "Alters the reality of" doesn't tell me what operator you are thinking of. We are not going to resolve this discussion unless we express what we are saying with precise math. I've already given the precise math I am using.

This is the usual confusion, arising from not defining the operators properly. As you (@PeterDonis) correctly point out, if there are two dinstinguishable particles (which is a bit easier than discussing indistinguishable particles; so let's stick to this case for now) labelled $A$ and $B$ in the following are prepared in the state
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|A,H \rangle \otimes |B,V \rangle - |A,V \rangle \otimes |B,H \rangle,$$
A single-particle observable referring to particle $A$ only is given by
$$\hat{O}_A \otimes \hat{1}_B$$
and an observable referring only to particle $B$ by
$$\hat{1}_{A} \otimes \hat{O}_B.$$
Such operators always commute:
$$(\hat{O}_A \otimes \hat{1}_B) (\hat{1}_{A} \otimes \hat{O}_B) = \hat{O}_A \otimes \hat{O}_B = (\hat{1}_{A} \otimes \hat{O}_B) (\hat{O}_A \otimes \hat{1}_B).$$
This is indeed trivial, but the triviality gets unnoticed because often one tends to not write the trivial identity operators in the Kronecker products but abbreviates it somehow not writing the product.

 

[microsansfil]

There is no relationship between the following state :

$$|\psi \rangle=\frac{1}{\sqrt{2}} (|A,H \rangle \otimes |B,V \rangle - |A,H \rangle \otimes |B,V \rangle,$$

And the fact that such operators $\hat{O}_A \otimes \hat{1}_B$ and $\hat{1}_{A} \otimes \hat{O}_B$ always commuteisn't it ?

/Patrick

 

[vanhees71]

Right, operator relations hold for all states.

 

[DrChinese]

1. Yes. Mathematically, the operators $Z_A$ and $X_A$ don't commute. Nor do the operators $Z_B$ and $X_B$. But none of those by themselves can be the operators we are discussing, because none of them operate on the two-qubit Hilbert space.

Meaning, "spin of Alice's qubit" is only one degree of freedom? Yes, that's right.

2. Not if those operators are the ones I wrote down; those operators manifestly commute. Nor do I see why that is a problem, since "spin of Alice's qubit [in the ]" and "spin of Bob's qubit" are different degrees of freedom, and the operations "measure z-spin of Alice's qubit" and "measure x-spin of Bob's qubit" therefore operate on different degrees of freedom. The fact that those degrees of freedom are entangled in the particular state of the two-qubit system we are talking about does not change any of this.

If you think the operations of "measure z-spin on Alice's qubit" [$Z_A$] and "measure x-spin of Bob's qubit" [$X_B$]are not described by the operators I wrote down, then what operators do you think should describe them?

3. We are not going to resolve this discussion unless we express what we are saying with precise math.

1. I say there is 1 degree of freedom in the spin of an entangled singlet system as we have been discussing. Not 2, because for there to be 2, the particles would need to be separable. And they aren't if entangled.2. This is as specific as I can make it - and I am applying the same idea of the HUP as is formulated in EPR-B. The purpose of this representation is because we are analyzing the following statement by Weinberg, which summarizes my position in this thread as clearly and concisely as possible: ...according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem..." A sharp measurement on $Z_A$ here leads to $Z_B$ changing to match expectation, and a subsequent sharp measurement on $Z_B$ supports that.

Any particle called A:
$$ \sigma A_p \space \sigma A_x \geq \frac \hbar 2 $$
A pair of entangled particles (your example of A and B):
$$ \sigma A_p \space \sigma B_x \geq \frac \hbar 2 $$
Therefore, returning to the discussion on entanglement swapping/quantum teleportation (photons 1 to 4): When a measurement on photons 2 & 3 projects 1 & 4 into an entangled state, the 1 & 4 state vector changes quantum non-locally into one which could not possibly have existed prior to the swap. There is no revealing of a pre-existing entangled state for 1 & 4, because the swap was required to make that happen. 3. It seems awkward to me that I quote EPR, Weinberg et al verbatim to support the language I use. And yet you disagree, and show me representations of the math that appear to be diametrically the opposite of what I said. I am combing over what you are saying trying to pick out the source differences in our positions, but I admit it is as much a struggle for me as it is frustrating for you. Thanks for sticking with it.

 

[DrChinese]

There is no relationship between the following state :

$$|\psi \rangle=\frac{1}{\sqrt{2}} (|A,H \rangle \otimes |B,V \rangle - |A,H \rangle \otimes |B,V \rangle,$$

And the fact that such operators $\hat{O}_A \otimes \hat{1}_B$ and $\hat{1}_{A} \otimes \hat{O}_B$ always commute

That is how I see it too.

 

[DrChinese]

Now I can almost agree, but this needs a bit of qualification since otherwise we start again debating about Einstein causality:

It's not enough to state that the measurement changes the state description for the one, e.g., Alice who tests whether her photon pair 1&2 is found to be in the state $|\psi_{12}^{0} \rangle$, who did the measurement, but there's no way for Bob to know instantly what Alice knows. She has to provide the information.

LOL I said I wouldn't continue with you on this, and yet, I have succumbed to temptation.

I completely agree that there is no way for Bob to learn anything useful from Alice FTL. That doesn't change the assessment that the quantum state itself changes quantum non-locally. The original total state (which we agree upon)
$$|\Psi_{1234} \rangle = |\Psi_{12} \rangle \otimes |\Psi_{34} \rangle.$$
does not contain a subspace in which $|\Psi_{14} \rangle$ are entangled in any degree of freedom. That outcome requires the 2 separate subsystems $|\Psi_{12} \rangle$ and $|\Psi_{34} \rangle$to interact as part of the overall context.

 

[DrChinese]

To be more specific:

$$|\Psi_{1234_{original}} \rangle = |\Psi_{12} \rangle \otimes |\Psi_{34} \rangle ~~~~~~(1)$$
and therefore
$$|\psi_{1234_{original}} \rangle = \frac 1 2 (|H_1 V_2 \rangle \pm|V_1 H_2 \rangle) \otimes (|H_3 V_4 \rangle \pm |V_3 H_4 \rangle)~~~~~~(2)$$
Which cannot be re-arranged so that 1 & 4 are entangled. Before that entangled state can be achieved, 2 & 3 must be made indistinguishable exactly as your post says. When 2 & 3 are allowed to interact, the state vector of 1 & 4 changes to one of entanglement (such as below).
$$|\psi_{14} \rangle = \frac{1}{\sqrt 2} (|H_1 V_4 \rangle \pm |V_1 H_4 \rangle)~~~~~~(3)$$
Which would come from something like (I may have the terms a bit off for the entanglement of 2 & 3):
$$|\psi_{1234_{PostProjection}} \rangle = \frac{1}{2} (|H_1 V_4 \rangle \pm|V_1 H_4 \rangle) \otimes (|H_2 V_3 \rangle \pm|V_2 H_3 \rangle)~~~~~~(4)$$

"One, therefore, has to guarantee good spatial and temporal overlap at the beam splitter and, above all, one has to erase all kinds of path information for photon 2 and for photon 3."

Then the authors describe how they achieved this goal (here and in the following I do not explain this in detail; if necessary, I can try to do also this, but it's all standard textbook stuff concerning standard optical elements like half-wave plates, polarizers, (coincidence) photo detectors).

Concerning the measurement on the pair 1&4 for state verification after the projection:

"According to the entanglement swapping scheme, upon
projection of photons 2 and 3 into the $|\psi_{23}^- \rangle$ state, photons
1 and 4 should be projected into the $|\psi_{14}^- \rangle$ state. To
verify that this entangled state is obtained, we have to
analyze the polarization correlations between photons 1
and 4 conditioned on coincidences between the detectors
of the Bell-state analyzer."

and

View attachment 246043

Thus, indeed the authors state that by this procedure of coincidence measurments, i.e., the projection of the pair 2&3 to the said Bell state necessarily leads to entanglement of the pair 1&4 in the corresponding partial ensemble (which is the case in 1/4 of the cases, neglecting real-world inaccuracies of the equipment), and I agree with them.

Then this is your statement, which is contradicted by me above. Our agreed upon (2) cannot be re-factored to yield (3).

"Note that the projection acts only on the photons in the pair 2&3 and NOT on those in the pair 1&4, as shown in Eq. (Projector). On pair 1&4 nothing at all is done in the analysis, as indicated by the unit operator in the second factor of the Kronecker product in Eq. (Projector), and this is so provided the locality of interactions is as described by standard QED based on the microcausality constraint, and thus the experiment indeed precisely verifies the predictions of QED (note that in the description above the authors as well as I never have used anything contradicting standard QED). Nowhere is a causal* action at a distance which would be violating the very principles of relativity and also standard QED!*I disagree with this statement, if the word "causal" is removed and the word "spooky" inserted in its place. Signal locality is not in question, I agree QM/QED/QFT does not feature FTL signalling. Whatever can occur non-locally requires a classical signal to make sense of it, which obviously defeats the purpose.

 

[Tendex]

This question is not well-defined. Are you talking about the entangled state? Then just look at the state vector. Are you talking about the correlations between measurement results? Then just compute them and see.

I'll make the same suggestion I made to @DrChinese: Please stop using vague ordinary language and express things in precise math. If you do I think you will find that the questions you are finding yourself wanting to ask will answer themselves--i.e., as soon as you've formulated them in precise math, the answer will be obvious.

That's good advice.
You can obviously define an entangled state as you did in #152 and also the commuting operators you did. The important thing here in my opinion is to adapt the math to the experimental setup and commuting observables applies to a kind of trivial situation, when measurements are about to be made by Alice and Bob, while the usual setup in entanglement discussions and the one DrChinese seems to be referring to is the one where either Bob or Alice have already measured a spin and established an order for the observables of the singlet state.

 

[PeterDonis]

I say there is 1 degree of freedom in the spin of an entangled singlet system as we have been discussing.

The number of degrees of freedom is a property of the Hilbert space, not the state. The Hilbert space is that of a two-qubit system, i.e., it has two (spin) degrees of freedom. Entangling the two degrees of freedom doesn't make one of them disappear.

The entanglement of the two degrees of freedom does mean that the state of the two-qubit system is restricted to a particular subspace of the two-qubit Hilbert space. But I don't think "degrees of freedom" is the right term to use to describe that restriction. Nor does that restriction affect what I have been saying about the operators (see below).

A pair of entangled particles (your example of A and B):
$$
\sigma A_p \space \sigma B_x \geq \frac \hbar 2
$$

The subscripts $p$ and $x$ here make it seem like you're thinking of position and momentum rather than spin. I think the case of spin measurements on qubits is much easier to handle mathematically so I would prefer to discuss that case. However, even for the position and momentum case, the operators $A_p$ and $B_x$ commute, as can be shown easily. I'll do it for the case of each particle moving only in one dimension to simplify the math.

We have a general two-particle configuration space wave function $\psi(x_A, x_B)$ in the position representation. The operators in question are $A_p = - i \hbar \partial / \partial x_A$ and $B_x = x_B$ (i.e., multiply by $x_B$). So we have:

$$
A_p B_x \psi = - i \hbar \frac{\partial}{\partial x_A} \left( x_B \psi \right) = - i \hbar x_B \frac{\partial \psi}{\partial x_A}
$$

$$
B_x A_p \psi = x_B \left( - i \hbar \frac{\partial}{\partial x_A} \psi \right) = - i \hbar x_B \frac{\partial \psi}{\partial x_A}
$$

So I think you have a basic misconception about the behavior of operators that operate on different degrees of freedom of Hilbert spaces; you think they don't commute if the corresponding operators on the same degree of freedom would not commute, but in fact operators on different degrees of freedom always commute (see below). In the example above, the different degrees of freedom are $x_A$ and $x_B$, i.e., the configurations of the two particles.

How do we know that operators on different degrees of freedom always commute? Because when we write the operators correctly, we have to include the fact that they act as the identity on the degrees of freedom they don't operate on. So the operator $A_p$ that you wrote should actually be written $P_A \otimes I_B$, and the operator $B_x$ as $I_A \otimes X_B$ (where I have put the degree of freedom in the subscript since that is a more commonly used notation), where $I_A$ and $I_B$ are the identity operators on the noted degrees of freedom. And those operators manifestly commute, as they would regardless of which operators we put in place of $P$ and $X$, provided that the $A$ and $B$ degrees of freedom (i.e., subspaces of the Hilbert space) are disjoint. And this does not mean the state must be separable; operator equations are valid for all states, so the above operators commute even on a state which entangles the $A$ and $B$ degrees of freedom.

 

[PeterDonis]

That doesn't change the assessment that the quantum state itself changes quantum non-locally.

But how does it change? Let's look at how it changes for the two-qubit case.

We have the state $| \psi \rangle = |z+ \rangle_A |z- \rangle_B - |z- \rangle_A |z+ \rangle_B$, and the operator $Z_A \otimes I_B$. Applying the operator to the state gives $Z_A \otimes I_B | \psi \rangle = |z+ \rangle_A |z- \rangle_B + |z- \rangle_A |z+ \rangle_B$, i.e., it flips the sign of the second term. This is a "non-local" change because the state is entangled: flipping the sign of the second term affects the relative phase on both qubits. But it doesn't change the expected measurement probabilities for Bob. Nor does the operator $Z_A \otimes I_B$ fail to commute with the operator $I_A \otimes X_B$, even though both operators induce "non-local" changes because they change the relative phase on both qubits. The phase changes are commutative (as can easily be shown for this particular case).

In this simple case, the operator doesn't change which degrees of freedom are entangled. In the four-qubit case of the experiment you describe, the operators in question can and do change which degrees of freedom are entangled. But operators on different degrees of freedom will still commute.

 

[PeterDonis]

The important thing here in my opinion is to adapt the math to the experimental setup and commuting observables applies to a kind of trivial situation, when measurements are about to be made by Alice and Bob, while the usual setup in entanglement discussions and the one DrChinese seems to be referring to is the one where either Bob or Alice have already measured a spin and established an order for the observables of the singlet state.

These aren't different situations; they're the same situation. We are talking about spacelike separated measurements, so there is no invariant order in which they are made. And this is fine since the operators commute, so the results don't depend on the order in which the measurements are made.

 

[Tendex]

These aren't different situations; they're the same situation. We are talking about spacelike separated measurements, so there is no invariant order in which they are made. And this is fine since the operators commute, so the results don't depend on the order in which the measurements are made.

In the second case I was referring to timelike separated measurements referred to just one(either Alice's or Bob's) spin, it is a different state.

 

[PeterDonis]

In the second case I was referring to timelike separated measurements referred to just one(either Alice's or Bob's) spin, it is a different state.

Actually it turns out not to make any difference in this particular case; the operators in question commute regardless of the spacetime interval between the measurement events.