# I Confused by nonlocal models and relativity

#### DrChinese

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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.
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

Mentor
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.

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#### PeterDonis

Mentor
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

Mentor
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

Mentor
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.

#### Tendex

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.
I'm not sure what you mean by the operators in question. I'm no longer talking about the operators in #152.

#### PeterDonis

Mentor
I'm not sure what you mean by the operators in question. I'm no longer talking about the operators in #152.
I believe what I said is also true of the operators in the more complicated four-qubit experiment that is being discussed.

If what you're talking about is something other than that, then you'll need to clarify because I'm not aware of any other cases being discussed in this thread.

#### RUTA

Hello everyone!

Recently I saw this paper: https://arxiv.org/pdf/1304.4801.pdf ("Any nonlocal model assuming “local parts” conflicts with relativity " by Antoine Suarez).

He mentions standard experimental configuration with beam-splitters and detectors. Then he distinguishes possible models by assuming "decision" about the outcome at beam-splitters (like Bohmian mechanics); or by assuming "decision" at detection.

After that he shows that the first group of models conflict with relativity. Moreover, he uses the notion of "nonlocality at detection" as opposed to "Bell's nonlocality" etc. And this logic can be find in almost every paper by this author (especially, about the so called "before-before" experiments). For example,
-https://arxiv.org/abs/1204.1732
or
-https://www.unige.ch/gap/quantum/_media/publications:bib:suarez.pdf
or
-https://arxiv.org/pdf/0708.1997.pdf

One time the author claimed that "after the before-before experiment Bohm’s interpretation can hardly be considered a valid alternative to Copenhagen".

I have several questions and would be grateful if anybody made some clarification:
1. Somehow Suarez opposes predictions of standard QM and such models as Bohmian mechanics. I thought that at some foundamental level their predictions are the same. Did I make a mistake?

2. Suarez considers Bohmian mechanics to contain some "locality". Again, I was sure that nonlocality is an important feature of BM. What do I miss?

3. Finally, does the conclusions in those papers prevent any attempts to make BM relativistic?

Sorry if these questions are too amateurish.
Far from being incompatible, QM and SR share a deep underlying coherence. The mystery of the Bell basis states, e.g., the spin singlet state, violating Bell's inequality in QM arises from the same principle as the mystery of the relativity of simultaneity in SR. See this talk for an explanation.

#### vanhees71

Gold Member
Right, operator relations hold for all states.
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.
It's hopeless. You even always switch the notation, and obviously you are unwilling to understand the basic principles of QFT. I recommend to read a good textbook on quantum optics, like Garrison and Chiao, and also a good textbook on relativistic QFT like Weinberg and/or Duncan, where the fundamental principles are utmost clearly presented.

Just once more to make it clear for those willing to understand the scientific facts. In the discussed experiment (now switching to the new labelling of the photons again), what has been prepared is indeed the four-photon state (using the simplified notation ignoring the full symmetrization necessary to describe bosons, which is not important here!)
$$|\Psi \rangle=|\psi_{12}^- \rangle \otimes |\psi_{34}^{-} \rangle.$$
Then Alice does a measurement on the photons 2&3, using a beam splitter and coincidence photon detectors. If both detectors fire from the spatial part of the pair state, she can be sure to have found these photons to be in the state $|\psi_{23}^{-} \rangle$. The prepared initial state implies that for all these cases (which occur in 1/4 of all cases) Bob will find or has already found his photon pair (the temporal order of the two measurements is irrelevant here, which is a very important point to make the case that nothing is non-local in the interactions enabling the measurements, and it's based in the very foundations of QED, mathematically described as the microcausality constraint) in the state $|\psi_{14}^-\rangle$ either. This follows from the simple mathematical fact that the above initially prepared four-photon state can be decomposed as
$$|\Psi \rangle=\frac{1}{2} \left [|\psi_{14}^+ \rangle \otimes |\psi_{23}^+ \rangle -|\psi_{14}^- \rangle \otimes |\psi_{23}^- \rangle - |\phi_{14}^+ \rangle \otimes |\phi_{23}^+ \rangle + |\psi_{14}^- \rangle \otimes |\psi_{23}^- \rangle \right].$$
Here the two-photon Bell states are defined as
$$|\psi_{ij}^{\pm} \rangle = \frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_i,H) \hat{a}^{\dagger}(\vec{p}_j,V) \pm \hat{a}^{\dagger}(\vec{p}_i,V) \hat{a}^{\dagger}(\vec{p}_j,H) \right]|\Omega\rangle,\\ |\phi_{ij}^{\pm} \rangle = \frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_i,H)\hat{a}^{\dagger}(\vec{p}_j,H) \pm \hat{a}^{\dagger}(\vec{p}_i,V) \hat{a}^{\dagger}(\vec{p}_j,V)\right]|\Omega\rangle.$$
It is important to use the correct Bose-symmetrized states here to understand that the state $|\psi_{ij}^- \rangle$ is unsymmetric under either exchange of the photon momenta or the exchange of the polarizations (as a whole it's of course symmetric under exchange of the two single-photon states as it must be for photons). That's why it is particularly easy to identify this state, because it's the only one where the spatial part of the pair is antisymmetric and thus only in this state both of A's detectors register a photon. This happens on average for 1/4 of all such prepared states.

It's of course mathematically nonsense to say that a state contains a subspace. A vector doesn't contain any spaces, and that's not what I've ever said.

What I said is precisely what's encoded in the formalism above: One is able to define a specific subensemble of all so prepared four-photon states which is described by the state $|\psi_{14}^{-} \otimes |\psi_{23}^{-} \rangle$ by only taking under consideration those cases, where Alice's two detectors fire coincidently, ensuring that her pair must be in the state $|\psi_{23}^{-} \rangle$. That's the realization of a von Neumann filter measurement.

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#### Tendex

I believe what I said is also true of the operators in the more complicated four-qubit experiment that is being discussed.
It is indeed a bit more complicated and the 2-qubit tensor product Hilbert space leaves out some of the intricacies DrChinese is bringing up in the paper he linked. With four qubits there is room to make different subsystem arrangements in the product Hilbert space which is where swapping and postselection enter, and depending on which pairs of photons one picks for the different subsystem arrangements one is able or not of constructing them as either product states or entangled states. This can't be seen with only 2 spin photons.
But certainly your simplified 2-qubit model was right all along, it's just that it is too simple for conveying all the subtleties in the experiment linked by DrChinese.
In most of my posts from #166 I mixed considerations about the experiment and your simplified model and indeed were misleading in this sense and can be ignored.

#### DrChinese

Gold Member
But how does it change? ... 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. ... the operators in question can and do change which degrees of freedom are entangled.
OK, you say there is no change to (say) photon 4's probability (your Bob) alone, which is constant at 50-50. But clearly the relevant issue is that photons 1 & 4 go from being unentangled and uncorrelated to entangled and correlated. That is a physical change to that system, and depends on a decision to act (or not) on particles on a different location (2 &3). Now of course 1 & 4 cannot be considered as independent particles in the first place, because they are each part of entangled systems with spatio-temporal extent at all times in this experiment. But that is just saying that there is quantum non-locality, i.e. spooky action at a distance*.

*Some people may not like this term ("spooky action at a distance"), but I think it is fitting and it is in fact used frequently as being equivalent to the phrase "quantum non-locality". As the cited experiments show, norms of both time ordering and distance/separation are upended. I would say the adjective "spooky" is applicable.

#### vanhees71

Gold Member
Well, as I repeatedly emphasized there's no causal change of photons 1&4 by Alice's manipulations on photons 2&3. The only thing she does is to select a subensemble (which is about 1/4 the size of the original total ensemble) based on the measurement outcome "both detectors clicked coincidently". The "cause" for Bob's photons 1&4 being entangled for this subensemble lies in the properties of the initial state but not in a action-at-a-distance spooky influence of Alice's local manipulations on her photons 2&3 at Bob's far distant place.

#### PeterDonis

Mentor
the relevant issue is that photons 1 & 4 go from being unentangled and uncorrelated to entangled and correlated
Yes, I've already agreed that this happens.

That is a physical change to that system, and depends on a decision to act (or not) on particles on a different location (2 &3).
It's a physical change to the four-photon system, which, as you note, must be considered as a single system. The change is usually described as "entanglement swapping" since the switch is from "1&2 entangled, 3&4 entangled" to "1&4 entangled, 2&3 entangled".

The entanglement swapping is done by a single operator--the beam splitter that photons 2 and 3 pass through--so there is no question of whether anything commutes at that point. But the later measurements on photons 1 and 4, that show the correlations between them, are two separate measurements involving two operators, and those two operators do commute.

#### DrChinese

Gold Member
Well, as I repeatedly emphasized there's no causal change of photons 1&4 by Alice's manipulations on photons 2&3. The only thing she does is to select a subensemble (which is about 1/4 the size of the original total ensemble) based on the measurement outcome "both detectors clicked coincidently".
And as I have already demonstrated, the original pre-projection state is something like
|Ψ1234original⟩=|ψ12⟩⊗|ψ34⟩ (1)|Ψ1234original⟩=|ψ12⟩⊗|ψ34⟩ (1)
|Ψ1234original⟩=12(|H1V2⟩±|V1H2⟩)⊗(|H3V4⟩±|V3H4⟩) (2)|Ψ1234original⟩=12(|H1V2⟩±|V1H2⟩)⊗(|H3V4⟩±|V3H4⟩) (2)​

And the post-projection state is
|ψ14⟩=1√2(|H1V4⟩±|V1H4⟩) (3)|ψ14⟩=12(|H1V4⟩±|V1H4⟩) (3)​
And the statistics are manifestly different, so the experiment shows. You cannot produce the post projection statistics from any "subensemble" of (1) without choosing to perform a projection, which in turn transforms (changes) the system.

Now I certainly respect that your knowledge (and PeterDonis') of QM/QED/QFT is vastly superior to mine. That includes your ability to solve working problems, something that I cannot do. However, you are missing something critical. Specifically, you are far missing the significance of the papers I am citing and the many experiments being conducted along similar lines - of which I endeavor to represent to you as best I am able in a fair manner.

None of the authors of these papers share your viewpoint on quantum non-locality, and I have NEVER seen one refer to entanglement swapping as merely "revealing" a pre-existing state as you describe. While my notation and terminology are (rightfully) a subject for you to criticize, I would reply that you do not see the forest for the trees. That QFT predicts (and experiments confirm) effects that cannot be described using local-only causal interaction reasoning (or however you describe your position) is generally accepted in physics today. But don't take my word for it.

Again, Weinberg (Lectures on Quantum Mechanics): "..according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem... " [describing EPR steering]

Or from here: "Each of the two photons belongs initially to one of two independent entangled photon pairs (e.g., photons 1 and 4 of the entangled pairs 1-2 and 3-4). The two other photons (2 and 3) are projected by a measurement onto a Bell state. As a result, the first two photons (1 and 4) become entangled even though they may be distant from each other. Entanglement swapping is the central principle used in quantum repeaters, whose purpose is to overcome the limiting effect of photon loss in long range quantum communication. " [describing entanglement swapping]

Or from here: Photon 2 and 3 undergo a joint Bell-measurement. In consequence, Photon 1 and 4 are projected into a Bell-state despite they do not locally interact. [describing traditional entanglement creation using 4 photons where 2 are used to project into a Bell state], and
... creation of entanglement in between two photons is possible, even if there is no direct local interaction between the involved photons. [describing a setup where entanglement is created from only 2 photons that don't interact locally]

Don't ignore the message because my math and terminology is weaker than yours. If you don't like my equations 1-3 above, then read the cited papers and see for yourself. And given that, it renders meaningless the idea that QFT is somehow formulated "local only". Quantum non-local experiments require a quantum non-local theory*, hopefully everyone can see why.

*Of course this has nothing to do with the various interpretations.

#### vanhees71

Gold Member
I read both papers you cited carefully. You ignore what the authors say, not I or @PeterDonis !

#### DrChinese

Gold Member
I read both papers you cited carefully. You ignore what the authors say, not I or @PeterDonis !
How about this quote: "Of course there's no change of photons 1&4 due to manipulations on photons 2&3." See anything like that?

I have represented my position (which is standard, generally accepted physics) as best I can at this point. In fact, I have probably repeated myself too much already. I will leave it to the readers to make their own judgments and assessments of these fabulous and groundbreaking experiments. Consequently, I will bow out of this thread and thank everyone for their time and comments - especially for helping me sharpen my LaTeX from terrible to poor.

#### Tendex

How about this quote: "Of course there's no change of photons 1&4 due to manipulations on photons 2&3." See anything like that?
Are you referring to the composite state that includes the 4 photons or singling out the just 1&4 system?
The former changes but not the latter, maybe that's where the disagreement lies.

#### PeterDonis

Mentor
Are you referring to the composite state that includes the 4 photons or singling out the just 1&4 system?
You can't single out the subsystem consisting of just photons 1 and 4 because that subsystem is entangled with the other subsystem, photons 2 and 3, for part of the experiment.

The states @DrChinese has been writing down include all four photons.

#### Tendex

You can't single out the subsystem consisting of just photons 1 and 4 because that subsystem is entangled with the other subsystem, photons 2 and 3, for part of the experiment.

The states @DrChinese has been writing down include all four photons.
Ok. And his quote of vanhees? Maybe vanhees can tell us.

#### PeterDonis

Mentor

"Confused by nonlocal models and relativity"

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