I Confused by nonlocal models and relativity

[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.

 

[PeterDonis]

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]

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

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]

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]

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]

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]

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.