# I Confused by nonlocal models and relativity

#### PeterDonis

Mentor
the concept of entanglement admits a basis dependence
This is a non-standard usage of the term "basis" (see my response to @DrChinese just now).

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
Again, this is a non-standard usage of the term "basis", and this non-standard usage is confusing you (and it might be confusing @DrChinese as well). 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.

#### DrChinese

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

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.
Since you asked: The top state is good as being entangled in the spin degree of freedom (see I read your comment ).

I readily admit I don't follow what you are saying about $I_B$. That's an identity operator, correct? Can you help me to understand why you are using this operator instead of one that is conjugate (which would not be separable if A and B are entangled) ? In this context, I would have expected us to discuss something like $Z_A \otimes X_B$ .

#### PeterDonis

Mentor
That's an identity operator, correct?
Yes.

Can you help me to understand why you are using this operator instead of one that is conjugate (which would not be separable if A and B are entangled) ?
Because the operation "Alice measures her qubit" does nothing to Bob's qubit, i.e., it acts as the identity $I_B$ on that part of the state. And conversely, the operation "Bob measures his qubit" does nothing to Alice's qubit, so it acts as the identity $I_A$ on that part of the state.

In this context, I would have expected us to discuss something like $Z_A \otimes X_B$.
This would describe a single operation that measures both Alice's and Bob's qubit. But even if such an operation existed (I personally don't see how it could since the measurement events are spacelike separated), it is certainly not the operation that either Alice or Bob are performing. Alice and Bob each perform separate measurements on their own qubits, so we need two separate operators to describe those two separate measurements.

#### PeterDonis

Mentor
In this context, I would have expected us to discuss something like $Z_A \otimes X_B$
This is a single operator, which has to commute with itself (since any operator does), so if we were discussing this (which I don't think we are, see my previous post), I would still not understand what you were claiming didn't commute.

#### PeterDonis

Mentor
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.
I have no idea what you mean by this. The operators are what they are; there's only one way to write them down. Once you write them down, either they commute or they don't. There's no choice between "writing them down as commuting" and "writing them down as non-commuting".

Also, you are conflating "causally related" with "not spacelike separated", but that is precisely part of the point at issue. 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. 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.

#### PeterDonis

Mentor
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

Gold Member
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

Mentor
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

Mentor
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

Gold Member
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

Mentor
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

Mentor
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

Gold Member
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

Gold Member
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

Mentor
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.

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#### 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

Gold Member
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

Gold Member
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.

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#### 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 commute

isn't it ?

/Patrick

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#### vanhees71

Gold Member
Right, operator relations hold for all states.

"Confused by nonlocal models and relativity"

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