Violation of Bell Inequality with unentangled photons

[javisot]

TL;DR

Violation of Bell Inequality with unentangled photons

I would like to know the validity of the following criticism of one of Zeilinger's latest papers https://doi.org/10.48550/arXiv.2507.07756 "violation of bell inequality with unentangled photons"

The review is by Francis Villatoro, in Spanish, https://francis.naukas.com/2025/07/...e-pueden-incumplir-las-desigualdades-de-bell/

I will translate and summarize the criticism as follows:

-It is true that a Bell inequality is violated, but not a CHSH inequality. The inequality that is violated is:
S = |E(α0,β0)+E(α0,β1)+E(α1,β0)−E(α1,β1)| ≤ 2 √2. The title "Bell inequality violation with non-entangled photons" is loaded with ambiguity and sensationalism. The paper confirms that quantum correlations can exist between non-entangled photons (nothing new), but to violate a CHSH inequality, entanglement is required.

 

[Sambuco]

Perhaps this article by @Ken Wharton from last week will help contribute to the debate.

Lucas.

 

[Demystifier]

The effect is explained in the paper by quantum indistinguishability. But indistinguishability in QM is implemented by symmetrization of the wave function, and mathematically symmetrization looks just as a special kind of entanglement of particles that otherwise would be distinguishable. In other words, indistinguishability is a kind of entanglement, but for some technical reasons we usually don't call it "entanglement" in a narrow sense.

 

[javisot]

The effect is explained in the paper by quantum indistinguishability. But indistinguishability in QM is implemented by symmetrization of the wave function, and mathematically symmetrization looks just as a special kind of entanglement of particles that otherwise would be distinguishable. In other words, indistinguishability is a kind of entanglement, but for some technical reasons we usually don't call it "entanglement" in a narrow sense.

I understand that the technical reasons for not calling it entanglement are that, as shown in the paper, with indistinguishability we end up violating a certain inequality, but not the most restrictive inequality.

Sabine seems to also contribute to the debate

It would be a problem if Zeilinger's work were literally true and we could violate a CHSH-type inequality with non-entangled photons. In that case, we wouldn't be able to distinguish two entangled photons from two non-entangled photons.

 

[Demystifier]

Also note that even the QED vacuum, i.e., a state with zero number of photons, is highly entangled. Hence it is really impossible to have a truly non-entangled state when the notion of "entanglement" is understood in a more general sense.

 

[pines-demon]

If this experiment holds it looks like a competitor to DrChinese's entanglement swapping arguments

 

[PeterDonis]

indistinguishability is a kind of entanglement

It meets the standard definition of entanglement, correct? The joint state of the two particles can't be factorized into a product of single-particle states.

 

[martinbn]

The effect is explained in the paper by quantum indistinguishability.

What is indistiguishable here, the photons or something else?

 

[pines-demon]

It meets the standard definition of entanglement, correct?

I was under the impression that it did not but I could not figure it out from the original paper. Wharton pre-print above says it is separable yet it violates their Bell inequality (Wharton argues it violates statistical independence as an artifact of postselection).

 

[Demystifier]

What is indistiguishable here, the photons or something else?

The paper says that photon paths are indistinguishable. It's not the same as indistinguishability of photons themselves, but they define photon states with a second-quantized formalism, which implies that photons are indistinguishable too, even if they don't say it explicitly.

That being said, I have to admit that I'm not sure whether this more general notion of entanglement is essential for the effect. It could be that Wharton and Price are right that it's just postselection, so that an analogous effect can be obtained even by a classical system.

 

[Demystifier]

It meets the standard definition of entanglement, correct? The joint state of the two particles can't be factorized into a product of single-particle states.

Yes, but for indistinguishable particles one can define a more narrow notion of entanglement; the operator acting on the vacuum that creates the joint state can't be factorized into a product of single-particle creation operators.

Let me illustrate this by an example. Let $a^{\dagger}(k)$ be a bosonic creation operator that creates a particle of momentum $k$. Consider the 2-particle state
$$|\psi\rangle=a^{\dagger}(k_1)a^{\dagger}(k_2)|0\rangle \equiv |k_1,k_2\rangle
\neq |k_1\rangle \otimes |k_2\rangle$$
with $k_1\neq k_2$. The wave function associated with this state is a symmetrized wave function
$$\psi(x_1,x_2)=\frac{1}{\sqrt{2}} [e^{ik_1x_1}e^{ik_2x_2}+e^{ik_1x_2}e^{ik_2x_1}] \neq e^{ik_1x_1}e^{ik_2x_2}$$
Thus the same state $|\psi\rangle$ is "not entangled" in a narrow sense common in quantum optics, and "entangled" in a more general sense common in non-relativistic quantum mechanics.

Given that $|k_1,k_2\rangle$ is not entangled in this narrow sense, what does an entangled state in the narrow sense look like? An example of entangled state in the narrow sense is
$$ \frac{1}{\sqrt{2}} [ |k_1,k_2\rangle + |k_1-q,k_2+q\rangle ]$$
with $q\neq 0$ and $q\neq k_1-k_2$.

For a more detailed analysis see also https://arxiv.org/abs/2007.06253

 

[PeterDonis]

for indistinguishable particles one can define a more narrow notion of entanglement

But why would you want to? What's wrong with the standard definition? Why is it necessary to change it?

 

[Demystifier]

But why would you want to? What's wrong with the standard definition? Why is it necessary to change it?

With standard definition any two photons are entangled (because they are bosons so the wave function is always symmetrized), but it is not the case that you can use any two photons to violate Bell inequalities. In real experiments, the photons that violate Bell inequalities are prepared in a special way, e.g., by parametric down conversion. So in quantum optics it became common to think only about such special photon states as "truly" entangled.

 

[martinbn]

With standard definition any two photons are entangled (because they are bosons so the wave function is always symmetrized), but it it not the case that you can use any two photons to violate Bell inequalities. In real experiments, the photons that violate Bell inequalities are prepared in a special way, e.g., by parametric down conversion. So in quantum optics it became common to think only about such special photon states as "truly" entangled.

Doesn't this depend on which degrees of freedom you use? If you choose two random photons and ignore all but their polarizations, they will not be entangled. I don't think that the ability to violate Bell's inequalities is what defines "true" entanglement. In the EPR paper, the entangled particles cannot be used for Bell's inequality violations, and it was the fist paper about entanglement.

 

[pines-demon]

In the EPR paper, the entangled particles cannot be used for Bell's inequality violations,

why not?

 

[martinbn]

why not?

Can they? How?

 

[gentzen]

why not?

If you look at the actual combined state of the two subsystem that may be entangled, then you often end up with a density matrix instead of a pure state. Even so entanglement is well defined for such density matrices, it is often absent, because entanglement is easily to destroyed in that case.

I have not analyzed the EPR setup for how to preserve or destroy entanglement, but it is unclear to me which degrees of freedom you could measure that would keep their entanglement.

 

[pines-demon]

Can they? How?

What is the issue? There are Bell inequalities for continuous variables or you can discretize your space.

I have not analyzed the EPR setup for how to preserve or destroy entanglement, but it is unclear to me which degrees of freedom you could measure that would keep their entanglement.

I mean you measure the particle at three angles (mixtures of position and momentum). I am not saying it is easy but I see no argument against it.

 

[gentzen]

Can they? How?

What is the issue? There are Bell inequalities for continuous variables or you can discretize your space.

The issue is that your answer doesn't help me to see how the entangled particles can be used for Bell's inequality violations.

I have not analyzed the EPR setup for how to preserve or destroy entanglement, but it is unclear to me which degrees of freedom you could measure that would keep their entanglement.

I mean you measure the particle at three angles (mixtures of position and momentum). I am not saying it is easy but I see no argument against it.

What do you mean by position? The position where the particle was created? Or the position where the particle hits a spherical screen of given radius r around the expected position of particle creation? What do you mean by a mixture of position and momentum? Something like the x-component of the momentum and the y-component of the position? Or simply the position at some fixed time (or distance) after the creation of the particle? In that case, do you believe that two semispheres with separately adjustable radius would "in principle" allow to make measurements showing Bell's inequality violations?


Independent of those concrete ambiguous and unclear points, your talk of "particle at three angles" when you don't mean actual angles is problematic. It often forces your discussion partners to lose time trying to clarify their meeting. And maybe even force them to conclude in the end: "Obviously, you didn‘t understand my remark, or what I would like to measure":

If you change frames in Minkowski spacetime, you change which set of transformations are "spatial rotations", because you change which spacelike hypersurfaces are surfaces of constant time. So I stand by my statement that in Minkowski spacetime you cannot invariantly separate spatial rotations and boosts.

I upvoted this, because I didn't realize it before.
This raised the question for me, what we actually measure in spin measurement, if rotations are not allowed.
In an actual measurement, we can "in principle" rotate the detector (or Stern Gerlach device) around the beam. This is only a 1-parameter group, so 2-parameters are still missing for general "spatial rotation". I think I have seen in the past how to get at least one more parameter, but maybe the details are not important at the moment.

But important for me is that it is not as trivial as simply rotating a detector in 3D space.

That's why you have three Pauli matrices. You can see how the Bell spin states are invariant wrt transformations generated by those matrices and what it means physically in the Methods section of this paper.

I read that method section now. Obviously, you didn‘t understand my remark, or what I would like to measure. This makes me wonder whether you actually understood what PeterDonis said.

Let me try to make the connection clear: if the measurement device can only be rotated around the beam, then the direction of the beam can also be used to restrict the allowed movements of the measurement device: It is only allowed to move parallel to the beam. This now allows us to define the 1-parameter group of rotations that has invariant meaning in this scenario, and provides a connection to the Lorentz group.

But the remaining 2-parameters for the full 3-parameter rotation group apparently don‘t have such a nice invariant connection to the Lorentz group.

 

[Demystifier]

Another paper appeared today arguing that it's just postselection.
https://arxiv.org/abs/2509.03127

 

[pines-demon]

What do you mean by position?

Take some state like $|\psi\rangle=\int \mathrm d x\, |x\rangle|-x\rangle$ (if you are worried that this is not normalized correctly, limit it to a given space), in momentum it looks something like $|\psi\rangle=\int \mathrm d p\, |p\rangle|p\rangle$

The position where the particle was created? Or the position where the particle hits a spherical screen of given radius r around the expected position of particle creation?

The position of detection, keep it 1D for now.

What do you mean by a mixture of position and momentum? Something like the x-component of the momentum and the y-component of the position?

No. It is only $x$ and $p=p_x$ the ones that do not commute.

Independent of those concrete ambiguous and unclear points, your talk of "particle at three angles" when you don't mean actual angles is problematic.

I mean in a simple "Mermin like" Bell test you need three angles to measure spin. Here I mean to measure linear combinations of $x$ and $p$. Note that you can render all this binary à la Bell by binning results (left vs right located), (positive vs negative $p$).

We can discuss how hard is to measure linear combinations of $x$ and $p$ but that is not my point to solve. You are discussing this as if it is somehow proven that the original EPR cannot be Bell tested, why?

 

[gentzen]

What do you mean by position?

Take some state like $|\psi\rangle=\int \mathrm d x\, |x\rangle|-x\rangle$ (if you are worried that this is not normalized correctly, limit it to a given space), in momentum it looks something like $|\psi\rangle=\int \mathrm d p\, |p\rangle|p\rangle$

OK, what has this to do with my question, or with anything you wrote in your answer?

The position where the particle was created? Or the position where the particle hits a spherical screen of given radius r around the expected position of particle creation?

The position of detection, keep it 1D for now.

Even in 1D, it makes an important difference whether position means to one at the moment of creation, or at some other time.
But I see your general issue with my question "What ...? ...? Or ...?": You interpret it as three separate questions, not as one question with some alternatives what you could have meant. I suggest you answer my questions again, and if staying in 1D makes this easier for you, no problem:

What do you mean by position? ...
What do you mean by a mixture of position and momentum? ...
In that case, do you believe ... would "in principle" allow to make measurements showing Bell's inequality violations?

Because I still have an urge to suggest concrete answers, here I go again, this time without question marks:

• Position could mean position at the moment of creation, it could mean the center of gravity of the entire system, it could mean position at a given time since the moment of creation, and it could also mean something like position at a "given distance" from the expected position of particle creation.
• Mixture of position and momentum could mean measurement of a complex superposition of position and momentum, i.e. $(\alpha|x\rangle+\beta|p\rangle)(\alpha\langle x|+\beta\langle p|)$ with complex numbers $\alpha, \beta$ with $|\alpha|^2+|\beta|^2=1$. It could mean some actual physical measurement procedure, like those I suggested in my initial post.
• For a concrete experimental arrangement, like the one described, the interesting question is whether somebody is willing to at least "guess" that it could work "in principle". Or any concrete scenario could always end in a retreat to abstract math.

No. It is only $x$ and $p=p_x$ the ones that do not commute.

I hope I already clarified above that my question should not have been interpreted as yes/no question, but as a request for general clarification. And to be perfectly clear: your answer above did not deliver that clarification for me. What it did deliver is "your mental model" of "how I think, and why that is wrong", but it did not deliver what you want to measure, neither in the abstract math, nor in concrete experimental arrangements.

I mean in a simple "Mermin like" Bell test you need three angles to measure spin. Here I mean to measure linear combinations of $x$ and $p$. Note that you can render all this binary à la Bell by binning results (left vs right located), (positive vs negative $p$).

So I get that you thought a bit more about what you are proposing, and noticed that position and momentum give continuous results, but "measurements showing Bell's inequality violations" work with binary results. So you suggest binning. My guess it that binning will probably destroy the delicate quantum entanglement. For a sufficiently well specified (thought) experiment, it should be possible to check whether this is the case or not.

We can discuss how hard is to measure linear combinations of $x$ and $p$ but that is not my point to solve. You are discussing this as if it is somehow proven that the original EPR cannot be Bell tested, why?

Let me check my words, whether they suggested anything like "it is somehow proven":

I have not analyzed the EPR setup for how to preserve or destroy entanglement, but it is unclear to me which degrees of freedom you could measure that would keep their entanglement.

 

[pines-demon]

OK, what has this to do with my question, or with anything you wrote in your answer?

Sorry if we are speaking past each other, I wanted an answer to a previous statement of another user, I am not even following the rest of the conversation if that has anything to do with your question.

Even in 1D, it makes an important difference whether position means to one at the moment of creation, or at some other time.

Because I still have an urge to suggest concrete answers, here I go again, this time without question marks:

• Position could mean position at the moment of creation, it could mean the center of gravity of the entire system, it could mean position at a given time since the moment of creation, and it could also mean something like position at a "given distance" from the expected position of particle creation.

Let's say a particle detector, the position is where we detect one of the entangled particles, where it is absorbed by the detector.

• Mixture of position and momentum could mean measurement of a complex superposition of position and momentum, i.e. $(\alpha|x\rangle+\beta|p\rangle)(\alpha\langle x|+\beta\langle p|)$ with complex numbers $\alpha, \beta$ with $|\alpha|^2+|\beta|^2=1$. It could mean some actual physical measurement procedure, like those I suggested in my initial post.

Let's say $S=X\cos(\theta)+P\sin(\theta)$ and choose three equally separated angles.

So you suggest binning. My guess it that binning will probably destroy the delicate quantum entanglement. For a sufficiently well specified (thought) experiment, it should be possible to check whether this is the case or not.

Let's say there are attempts at doing so. Note that if I write the state as $|\psi\rangle=(|\text{left}\rangle|\text{right}\rangle+|\text{right}\rangle|\text{left}\rangle)/\sqrt{2}$ (left and right are the sign of the position), things start looking exactly as a Bell test.

Let me check my words, whether they suggested anything like "it is somehow proven":

Sorry if this does not answer all of your questions. Again this tangent is not on you or related to the original topic of the thread. I may expand on my side with references later if we insist on focusing on that. However let me clarify I was just waiting for the rebuttal. It was stated that the original EPR scheme does not allow for a Bell test, I want to know why, I do not have to carry the burden of proof when I was first to ask for a clarification.

 

[gentzen]

OK, what has this to do with my question, or with anything you wrote in your answer?

Sorry if we are speaking past each other

Maybe read what you wrote exactly, see whether it makes any sense to you. And then ask yourself, how anybody who wants to have a serious discussion focused on physical details should respond to that.

Let's say a particle detector, the position is where we detect one of the entangled particles, where it is absorbed by the detector.

Wow, you still ignore the issue of when the detection happens, compared to the moment of creation.

Let's say there are attempts at doing so. Note that if I write the state as $|\psi\rangle=(|\text{left}\rangle|\text{right}\rangle+|\text{right}\rangle|\text{left}\rangle)/\sqrt{2}$ (left and right are the sign of the position), things start looking exactly as a Bell test.

It doesn't look like a Bell test to me. And I have no idea what you mean by "if I write the state as".

Sorry if this does not answer all of your questions.

Why do you write mathematical expression "with unclear meaning" instead of directly saying what you expect from your discussion partner? I learned now that you don't expect him to point out why your words and mathematical expressions remain unclear to him.

However let me clarify I was just waiting for the rebuttal.

My intial answer why it may not work was

If you look at the actual combined state of the two subsystem that may be entangled, then you often end up with a density matrix instead of a pure state. Even so entanglement is well defined for such density matrices, it is often absent, because entanglement is easily to destroyed in that case.

together with the hint that it depends on "which degrees of freedom you could measure".

It was stated that the original EPR scheme does not allow for a Bell test, I want to know why, I do not have to carry the burden of proof when I was first to ask for a clarification.

I didn't ask you for a proof. I did ask you for clarification what you meant by your words and mathematical expressions.

But let us stop here, until you get some answer from martinbb. Otherwise, your answers to me will just steal my time, because you are annoyed that somebody else tried to answer, and thereby deprive you of the answer from martinbb.

 

[pines-demon]

Wow, you still ignore the issue of when the detection happens, compared to the moment of creation.

Does it matter? just prepare the state.

But let us stop here, until you get some answer from martinbb.

Agreed

 

[RUTA]

Independent of those concrete ambiguous and unclear points, your talk of "particle at three angles" when you don't mean actual angles is problematic. It often forces your discussion partners to lose time trying to clarify their meeting. And maybe even force them to conclude in the end: "Obviously, you didn‘t understand my remark, or what I would like to measure":

I guess I don't understand what you're talking about then. The rotation is always in the plane perpendicular to the beam but the beam direction is arbitrary. The Pauli matrices allow you to specify that rotation plane in any direction relative to your Cartesian coordinates. This is all in accord with homogeneous spacetime per the Lorentz group.

 

[gentzen]

Take some state like $|\psi\rangle=\int \mathrm d x\, |x\rangle|-x\rangle$ (if you are worried that this is not normalized correctly, limit it to a given space), in momentum it looks something like $|\psi\rangle=\int \mathrm d p\, |p\rangle|p\rangle$

Now I understand what you wanted to say: The initial state $|\psi\rangle$ for the EPR experiment can be assumed to be ψ... I guess |x>|-x> as starting point is "unusual" for the EPR experiment, I would rather have started with the integral over |p>|-p> (possibly restricted by some max energy). What this looks like in position space (for infinite max energy) is a nice question (probably with a known simple answer). I guess the integral over |x>|x> looks reasonable, because the two particles were probably created at the same position.

Does it matter? just prepare the state.

The state (at least as I interpreted it above, with reversed signs compared to yours) is not time invariant. At the time of creation, both particles have the same position. As time goes by, the distance between the particles increases.

Agreed

Sorry that I answered again, but I thought it was important that I figured out what you meant with "Take some state like ..."

 

[martinbn]

But let us stop here, until you get some answer from martinbb.

Agreed

I don't have a prove that something is impossible. My statement is based only on the fact that I have not seen it done. @pines-demon said that it is possible, and i asked how, as in how is it done. I wanted to see the details. I am not sure what answer is expected from me!

 

[pines-demon]

I don't have a prove that something is impossible. My statement is based only on the fact that I have not seen it done. @pines-demon said that it is possible, and i asked how, as in how is it done. I wanted to see the details. I am not sure what answer is expected from me!

Thanks for stepping in. I hoped that there was something stronger that I was missing. Trying to perform Bell tests on the original EPR set up is a hard task, but there is no argument to discard it right out of the bat.

 

[pines-demon]

The state (at least as I interpreted it above, with reversed signs compared to yours) is not time invariant. At the time of creation, both particles have the same position. As time goes by, the distance between the particles increases.

What is the issue with that? just take a finite time after that? Why are you focusing on that point? You could instead start with some weird potential where that is its prepared ground-state, at least to first approximation.

Also why take a different sign convention to mine?