Talk:Greenberger–Horne–Zeilinger state

Edits

In Properties, it's mentioned that we get an unentangled mixed state for the partial density operator. Is this really the case? I don't really see that this is an unentangled state. Please cite — Preceding unsigned comment added by Landmark ni (talk • contribs) 02:11, 12 February 2019 (UTC)

The following are a few comments on this page: Content: In the introduction, a GHZ state is introduced as a certain type of entangled quantum state, but they do not define entanglement or the definition of a qubit. This may confuse a reader who is not previously familiar with quantum mechanics. In the second paragraph of the Properties section, I would like if the article went into more depth on how the GHZ states lead to a violation of Bell's Inequality. This displays the inconsistency between classical theory and quantum theory. The section Pairwise Entanglement is extremely clear and well-written: this was the most comprehensive section of the article in my opinion. The applications section should be added to with more specific applications. This section is very vague.

Tone: The tone is neutral throughout. The Applications section is extremely short and vague. This is by far the weakest section of the article. The previous three sections are clear.

Source: The second source does not seem to be a source. It is a sentence restating where it was quoted. The fourth article also does not appear to be a source. It is a sentence describing its own reference. They do not have titles, authors, or URLs. I am a bit confused as to why these are on the references list.Ssabir19 (talk) 03:41, 8 December 2018 (UTC)

Merge proposal

The GHZ experiment should be merged in to this article to create one strong article. These are just different aspects of the same topic. Johnjbarton (talk) 18:17, 16 November 2024 (UTC)

• Support I do not know how I never thought about this, I have been telling me that these two should complement each other better. A merge is clearly the good way to go.--ReyHahn (talk) 08:56, 17 November 2024 (UTC)

 Done Johnjbarton (talk) 00:31, 25 November 2024 (UTC)

dependence on measurement

I find this sentence very difficult to parse: "This example illustrates that, depending on which measurement is made of the GHZ state is more subtle than it first appears: a measurement along an orthogonal direction, followed by a quantum transform that depends on the measurement outcome, can leave behind a maximally entangled state." I suggest changing "depending" to "having decisions depend", i.e. changing the whole sentence to "This example illustrates that having decisions depend on which measurement is made of the GHZ state is more subtle than it first appears: a measurement along an orthogonal direction, followed by a quantum transform that depends on the measurement outcome, can leave behind a maximally entangled state." Would that convey a correct meaning? ☺Coppertwig (talk) 14:41, 18 December 2025 (UTC)

I think a better strategy is to delete that section altogether. We can't verify either sentence. Johnjbarton (talk) 16:57, 18 December 2025 (UTC)

If it helps, this sentence was mangled by this edit: https://en.wikipedia.org/w/index.php?title=Greenberger%E2%80%93Horne%E2%80%93Zeilinger_state&diff=prev&oldid=1056403295 . The other parts of this edit seem mostly good so I suspect human error. Bbbbbbbbba (talk) 01:11, 24 December 2025 (UTC)

Communication complexity references

One of the earliest edits (revision 88630891) added these sentences:

The correlations can be utilized in some quantum information tasks. These include multipartner quantum cryptography (1998) and communication complexity tasks (1997, 2004).

The "(1998)" seems to be referring to the paper by Hillery, Bužek & Berthiaume, which is currently in the reference section. However I cannot find any 1997 or 2004 references. I deleted the sentences while reorganizing the article, but could anyone find these references and see if they are worth adding again? Bbbbbbbbba (talk) 11:25, 27 December 2025 (UTC)

Never mind, I found them at https://arxiv.org/abs/quant-ph/9705033 and https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.92.127901. These seems to be quite theoretical results without much practical motivation, so maybe they do not belong in the "Application" section? Bbbbbbbbba (talk) 17:28, 27 December 2025 (UTC)

measuring the third qubit ?

The "Pairwise entanglement" section says

• "However, measuring the third qubit in the Pauli X or Y basis can leave the first two qubits in a maximally entangled Bell state"

I don't understand this sentence: the states do not come with little markers numbered 1,2,3. However we can number the observers. So in the Krenn Zeilinger paper they say

• By selecting the results of observers 1 and 2 with respect to the results of observer 3, one obtains two subensembles. Johnjbarton (talk) 23:38, 28 December 2025 (UTC)

I am not sure I understand your problem. The Krenn Zeilinger paper does talk about particles 1, 2, and 3, which are measured by the respective observers. The implementation of the measurements by different observers is a physically interesting detail, but is not essential when describing the properties of the abstract GHZ state.

As a matter of fact, the states are marked as 1, 2, 3 — in an experimental setting, the trajectory of a particle (i.e. which observer it goes to) is part of the state of the particle, and we assume that the trajectories of the three particles are distinct enough that we can always distinguish them. Bbbbbbbbba (talk) 01:56, 29 December 2025 (UTC)

My understanding is that the "qubit" here is "polarization" or "spin", that is the property entangled. The state discussed in the article only takes two values, how can it represent the trajectory? The article says "In a GHZ experiment, first a GHZ state is prepared, and its three qubits are spatially separated." so I can't see out the trajectories can be distinct until they are separated.

My understanding of the source here is that state 3 corresponds to property measurement at observer 3, (ditto 1,2). Talking about particles "measured by the respective observers" is fine, because these are results of experiments. Krenn-Zeilinger discuss complete experiments and how the results from different observers are related.

I assume you would agree that the physics of the description does not change if we permute the numbers? The advantage of changing the numbering is that then the "measuring the third qubit" can be expressed as "once one of the qubits is measured in X or Y basis, the remaining two qubits are in a maximally entangled Bell state". This is close to the wording of the source given that the observers are spatially are presumably communicationally separated. This wording avoid implying any marking on the qubits. Johnjbarton (talk) 02:42, 29 December 2025 (UTC)

I would argue that the abstract concept of qubit already assumes that different qubits are distinguishable by some external property abstracted away; otherwise you are considering indistinguishable particles, not qubits. I am from a quantum computation background, so talking about distinct qubits is natural to me; no one would argue that Google's 105-qubit chip should be marketed as more than that because of the extra information needed to distinguish the qubits! Bbbbbbbbba (talk) 02:55, 29 December 2025 (UTC)

The qubits are distinguished by the location of the observers. It is the observers who are numbered. So 'third qubit' means a 'qubit measured by observer #3'. I understand this is a common shorthand, but the source didn't use it, and wisely IMO. Johnjbarton (talk) 03:32, 29 December 2025 (UTC)

The source did not use it because the source did not use the word "qubit". I am not sure if the authors even knew the word "qubit" (it existed at that time, but I would guess was only known in the quantum information theory circle). I would prefer Wikipedia to follow modern conventions that facilitate understanding. Bbbbbbbbba (talk) 04:20, 29 December 2025 (UTC)

"In a GHZ experiment, first a GHZ state is prepared, and its three qubits are spatially separated." Here "spatially separated" is supposed to mean "moved to observers with spacelike separations" (to rule out communication), not "given different trajectories". The trajectories of the three particles are already distinct when the GHZ state is prepared. For example, two photons would come out from the same beam splitter in different directions. Bbbbbbbbba (talk) 03:13, 29 December 2025 (UTC)

The GHZ state discussed here happens to be symmetric under qubit permutation, but this is because we are choosing a basis to make the results easier to understand. In fact, the Pan et al. 2000 paper says "for simplicity we assume the polarizations at D₃ to be defined at right angles relative to the others", which makes it clear that this symmetry is not inherent to the experimental setup. Bbbbbbbbba (talk) 03:26, 29 December 2025 (UTC)

I can see two ways of thinking that may lead to your objection:

1. You are worried about the particles carrying the qubits being indistinguishable. Technically this is valid, but then you must question the use of "particle 3" too.

2. You are insisting that qubits, being two-level properties, cannot be marked. To me this is absurd to saying that "bit 3" (in a bitstring) is problematic because bits cannot be marked. Bbbbbbbbba (talk) 04:44, 29 December 2025 (UTC)

I understand your point of view, but I think you have internalized an understanding of these systems that typical readers like myself may not have reached. A simple restatement will make the sentence clearer.

• However measuring the one of the three qubits in the Pauli X or Y basis leaves the other two qubits in a maximally entangled Bell state.

In this way we don't imply that qubit 3 is special. Better would be a human in the picture to anchor the qubit:

• However, once one observer measures a qubit in the Pauli X or Y basis, the other two qubits are in a maximally entangled Bell state.

Even better:

• However, once one observer measures a qubit in the Pauli X or Y basis, observations of other two qubits will show the correlations of a maximally entangled Bell state.

This is close to the wording of the source.

I suppose these don't seem different to you, but thinking in terms as you mentioned earlier:

• "moved to observers with spacelike separations" (to rule out communication)

is not built in for readers. By writing the description in terms of the observers rather than a strange new thing called a qubit I think the concept is much clearer. When I started reading about qubits my brain associated them with particles and I suspect this is very common. Only once idea like communications sink in does one realize that qubits are associated with measurement points (observers). Johnjbarton (talk) 17:50, 29 December 2025 (UTC)

I do not see anything wrong with associating qubits with particles. On contrary, I think the idea that qubits are associated with measurements (observers) is a very advanced idea (in a large Hilbert space, two anti-commuting observable operators with eigenvalues ±1, representing Pauli Z and X operators, define a qubit), and comes with the caveat that it refers to potential measurements (the Z and X measurements are themselves exclusive, and one can also apply unitary gates to a qubit, etc.). For a beginner a much simpler approach is to first take it for granted that there are 3 qubits (carried by 3 particles), build a Hilbert space $(\mathbb {C} ^{2})^{\otimes 3}=\mathbb {C} ^{8}$ from them, and do all the math in that Hilbert space without worrying about alternative ways of decomposing the Hilbert space.

I feel that I see the direction you are going towards: The perspective of someone who do not yet firmly believe in quantum mechanics, which is also the perspective of many of those old papers (even if the authors are firm believers, they would be trying to convince other people). However, I am doing math in this section that requires taking the basic concepts of quantum mechanics for granted (even if only as a hypothesis to verify). Incidentally, the reason I prefer to single out the third qubit is also so that the text description agrees with the math formula; I think the symmetry of the state is evident enough that it is impossible to imply that qubit 3 is special, and the bigger danger is implying that this symmetry is inherent. Bbbbbbbbba (talk) 20:16, 29 December 2025 (UTC)

I agree that there is a generational difference, but not along the lines of belief in QM. Rather the split happens at "qubit". If you focus on qubits and think about them as parallel to classical bits, then all the QM is about predicting observations, as it should be. No time need be wasted on QM paradoxes and worrying about which-way particles go. I think that is what you are saying with "take it for granted".

The association of qubits with particles is fine but pointless. You won't need say, Schrödinger eqn, because all of the physics is now build into the qubit math. I would object to "carried by 3 particles" because this local realism is unnecessary: no aspect of the qubit results depends on this.

Sadly our quantum computing page are weak and I guess it may be trace to this generational split. Johnjbarton (talk) 23:57, 29 December 2025 (UTC)

I would also note that the sentence we have been discussing from the beginning is from the "Pairwise entanglement" section, which is not a part of the "GHZ experiment" section, and as such spatial separation is not necessarily assumed. It is true that the characteristics of entanglement are more evident when a spatial separation is involved, but it is also perfectly valid to just talk about entanglement between abstract qubits, as this article (GHZ state) is arguably ultimately about. This might be an unforeseen problem of merging the GHZ experiment page with this page: The GHZ experiment page was supposed to be the page to convince the non-firm believers and this page was supposed to be the page that takes quantum mechanics for granted.

To make myself clear: I do not strongly oppose the edits you are suggesting, but I feel that if you want to make these edits you might also want to make some more overarching edits instead of letting that one sentence stand out, and these overarching edits might ultimately be to the detriment of technical readers like me. Bbbbbbbbba (talk) 20:41, 29 December 2025 (UTC)

I will try some modest changes. Johnjbarton (talk) 23:57, 29 December 2025 (UTC)

I guess I will wait until I can get a copy of Uchida's paper. Johnjbarton (talk) 00:11, 30 December 2025 (UTC)

Fascinating discussion! It seems like the issue is happening because the term 'GHZ state' has become genericized beyond the original context of studying quantum weirdness and the GHZ experiment. And I see that bizarrely what is now currently called 'GHZ state' is actually a misnomer, in fact it is due to Mermin and should be called the Mermin state (or Mermin-GHZ state). Maybe two articles could be deserved: one about the Mermin state and one about the GHZ experiment.

@Johnjbarton regarding your original question, it is always implicitly the case that the three qubits are distinguishable. The formulas don't always include explicit 1,2,3 labels, but that is only notational shorthand. And sometimes they absolutely do include the explicit 1,2,3 markers. For example in Zeilinger's 1999 paper, they refer to their GHZ state as instead $|\psi \rangle ={\tfrac {1}{\sqrt {2}}}(|H\rangle _{1}|H\rangle _{2}|V\rangle _{3}+|V\rangle _{1}|V\rangle _{2}|H\rangle _{3})$. In that case the distinguished positions 1,2,3 refer to different arms of the optical path. I think this article could briefly include a formula with explicit subscripts like $|H\rangle _{1}|H\rangle _{2}|V\rangle _{3}$ or $|H_{1}H_{2}V_{3}\rangle$, before adopting shorthand like $|HHV\rangle$ or the somewhat specialized $\otimes$.

--Nanite (talk) 17:48, 30 December 2025 (UTC)

Thanks. Unfortunately your comment confuses me. I have no trouble at all with the implicit numbering, I think the subscripts are superfluous as long as the order implies the numbering. However, our article says

• The GHZ state can be written in bra–ket notation as $|\mathrm {GHZ} \rangle ={\frac {1}{\sqrt {2}}}(|\mathrm {HHH} \rangle +|\mathrm {VVV} \rangle ).$

which does not match the "their GHZ state". This seems like this difference should be explained in the article with an appropriate source.

My mistake is right in my opening post: "states do not come with little markers numbered 1,2,3". States do come with little markers, particles don't. The three observers (or arms of the optical path or detectors) are distinguishable and the states are representing possible observations. "measuring the third qubit" means that we detected one specific value in the third detector but we have not yet looked at the data for the first and second detectors. The state that represents our knowledge of that unexamined data is maximally entangled: we will find correlations between first and second when we look. Johnjbarton (talk) 23:28, 31 December 2025 (UTC)

What do you think of the second and third formulas in current "Definition" section? In my opinion, while they may feel more "concrete" to some people, ultimately it is not the best use of space to write the same formula three times with slightly different notations. Also while they are both directly from their respective sources, they both have subtle issues that could be awkward to clarify:

• For the photon implementation, as you have discovered, the actual experimental setup generates $(|\mathrm {HHV} \rangle +|\mathrm {VVH} \rangle )/{\sqrt {2}}$ (with a "natural" choice of coordinate system) instead of ⁠$(|\mathrm {HHH} \rangle +|\mathrm {VVV} \rangle )/{\sqrt {2}}$⁠, requiring a coordinate system change for the latter to make sense.
• For the spin implementation, Mermin's paper specifically mentions spin-1/2 particles (at the beginning of Section III), and yet in the appendix he uses 1 and −1 instead of ⁠1/2⁠ and −⁠1/2⁠. Alternatively we could use spin-1 particles, but then we have to explain why we do not use the s_z = 0 state. (Interestingly, if we use photons as the spin-1 particle, we get an alternative implementation with polarization qubits where the computational basis states are circular polarization states.)

Bbbbbbbbba (talk) 15:42, 1 January 2026 (UTC)

This article is short compared to the complexity of the topic. Thus I suppose most readers will seek either a quick idea of what the concept is or a gateway to more information. For the first group I agree the Definition is not helpful. For the second group questions of notation are vital because as we know the sources differ. Ideally we would rename "Definition" -> "Definitions" and explain your two points above, then move it down with page. Then in its current place we would have "Concept". However, Mermin and other have attempted compressed explanations with I guess modest success so such a section is a challenge. Johnjbarton (talk) 17:24, 1 January 2026 (UTC)

Unfortunately the sources differ even more than that and our examples here are not all that representative. For example, reference 1 in the current reference list denotes the 4-qubit state as:

$|1,0\rangle =(|{\uparrow }{\uparrow }{\downarrow }{\downarrow }\rangle -|{\downarrow }{\downarrow }{\uparrow }{\uparrow }\rangle )/{\sqrt {2}}.$

A lot is going on here: The basis states are denoted as $|{\uparrow }\rangle$ and $|{\downarrow }\rangle$ (which are also used in reference 6 even for the three-photon state), qubits 3 and 4 are flipped, the phase between the two states in superposition is also flipped (− instead of +), and the left-hand side $|1,0\rangle$ indicates that this state is decayed from an original particle with spin 1 and in the s_z = 0 state.

As another example, reference 3 uses $|0\rangle |0\rangle |0\rangle$ instead of ⁠$|000\rangle$⁠, and reference 8 even uses both these notations. Reference 9 uses an even more rigorous notation, $|R\rangle _{1}\otimes |R\rangle _{2}\otimes |R\rangle _{3}$.

I also suspect that Mermin's notation (the one with 1 and −1) was very ad hoc and would not be replicated in many other papers, if at all. It is almost like he did not want to commit to a notation system for the states at all, but changed his mind in the last second and picked something consistent with his previous notations (I did just realize that he mentioned "measure angular momentum in units of $\textstyle {\frac {1}{2}}\hbar$" previously so that everything being ±1 makes sense) to write the appendix.

Ultimately, switching the notations for basis states is just a complication of notation that is not even directly relevant to the GHZ state itself. Maybe this section could be better used to talk about the variations specific to the GHZ state, such as flipping some qubits or changing the phase factor, or even writing the entire state in another basis (as done this paper at Equation 5). Bbbbbbbbba (talk) 01:24, 2 January 2026 (UTC)

I believe the Uchida source calls the $|R\rangle _{1}\otimes |R\rangle _{2}\otimes |R\rangle _{3}$ version "right hand circular polarized" (an similar L for left) and then relates these to H,V states as linear polarized. This may be useful to add as it would explain the relationship between Mermin and Bouwmeester notations raised by Nanite above. If Uchida considered it notable we should. Unfortunately I only have access to snippets of that paper.

I think that $|0\rangle |0\rangle |0\rangle$ is equivalent to ⁠$|000\rangle$⁠ and to $|0\rangle _{1}\otimes |0\rangle _{2}\otimes |0\rangle _{3}$ See for example Tensor Product in bracket notation

I think we should avoid using the equal sign as GHZ did in their 4 particle paper. As I understand their description this is irreversible particle decay not equality. Johnjbarton (talk) 02:27, 2 January 2026 (UTC)

Just in case you did not realize, The GHZ state Uchida described is physically implemented in a fundamentally different basis than the one Zeilinger described. In Uchida's state, the circular polarization states of any two photons are perfectly correlated. while in Zeilinger's state, the correlation between circular polarization states only manifest when all three photons are measured. Bbbbbbbbba (talk) 03:12, 2 January 2026 (UTC)

Then how do we know these are both GHZ states? What makes them so? Johnjbarton (talk) 03:25, 2 January 2026 (UTC)

There always exists one basis or combination of bases under which the states of any two particles are perfectly correlated, i.e., the overall state could be written in the abstract form $(|000\rangle +|111\rangle )/{\sqrt {2}}$ if $\{|0\rangle ,|1\rangle \}$ is interpreted as that basis for each qubit. For Uchida's state this basis is the R/L basis, and for Zeilinger's state it is the H/V basis. Bbbbbbbbba (talk) 04:39, 2 January 2026 (UTC)

"$|0\rangle |0\rangle |0\rangle$ is equivalent to ⁠$|000\rangle$⁠ and to ⁠$|0\rangle _{1}\otimes |0\rangle _{2}\otimes |0\rangle _{3}$⁠" — this is certainly true, but since you mentioned how "questions of notation are vital" for readers looking for a gateway for more information, I am questioning whether it is more appropriate to demonstrate this fact in this article than to demonstrate that the symbols 0 and 1 could be substituted with many other symbols (admittedly with more physical meaning). Bbbbbbbbba (talk) 03:19, 2 January 2026 (UTC)

I agree that the symbology is not worth a lot of space, but the Definitions section actually covers GHZ state with different physical implementations as well. What I am against is presenting only one uniform notation in the article without comment when the sources have many. We can handle this in various ways. Johnjbarton (talk) 03:31, 2 January 2026 (UTC)

Unfortunately, from what I have seen, in many branches of mathematics sources having too many notations is the norm, not the exception. It is not uncommon that the best (original) source has the worst (ancient) notations. And I think a better solution is to explain the core logical concept so that readers can understand the source by reading the source, rather than addressing the notations of each source individually.

I do agree that it is appropriate to talk about different physical implementations of the GHZ state; I'm just not too fond of the current presentation. Bbbbbbbbba (talk) 04:15, 2 January 2026 (UTC)

The GHZ experiment section

The argument as currently presented makes no sense. It assumes the same result for each spin component with each run of the experiment. Particularly,

$y_{1}y_{2}x_{3}\cdot y_{1}x_{2}y_{3}\cdot x_{1}y_{2}y_{3}\cdot x_{1}x_{2}x_{3}=x_{1}^{2}x_{2}^{2}x_{3}^{2}y_{1}^{2}y_{2}^{2}y_{3}^{2}$

makes this assumption implicitly, i.e. $y_{1}$ must have the same value when we measure $\langle \mathrm {GHZ} |Y_{1}Y_{2}X_{3}|\mathrm {GHZ} \rangle$ and $\langle \mathrm {GHZ} |Y_{1}X_{2}Y_{3}|\mathrm {GHZ} \rangle$. This is not required by the structure of the GHZ state itself, as each of these will be separate runs of the experiment with potentially different spin values on each axis. Geoff Beck (talk) 12:25, 6 January 2026 (UTC)

Do you understand what a local hidden-variable theory is? Bbbbbbbbba (talk) 16:39, 6 January 2026 (UTC)

I think the point is each of the contributions to the product are separate and a hidden variable explanation would predict an equal number of +1 and -1 values for eg $x_{1}$. So it could be +1 in one run and -1 in the next.

In any case Mermin does not say this. Johnjbarton (talk) 16:50, 6 January 2026 (UTC)

Right, Mermin has an argument that follows. Geoff Beck (talk) 16:51, 6 January 2026 (UTC)

Mermin's explanation of the QM version is much clearer and not longer. Mermin does say that the classical expectation value is opposite. He implies that the classical product for operators $y_{1}y_{2}x_{3}\cdot y_{1}x_{2}y_{3}\cdot x_{1}y_{2}y_{3}=x_{1}x_{2}x_{3}$. So I think we could rewrite the paragraph. Johnjbarton (talk) 17:10, 6 January 2026 (UTC)

Mermin's explanation of the local hidden-variable theory prediction is very much longer though. He basically dedicates the entire Section II to it. Bbbbbbbbba (talk) 17:19, 6 January 2026 (UTC)

Mermin's section II is long, but I think it can be shortened by instead just writing the state in the yyx and xxx bases and pointing out that xxx only has an even number of down spins whereas yyx (and other combinations) only has odd numbers of down spins. I think this functionally covers the whole red vs green spiel. Geoff Beck (talk) 17:26, 6 January 2026 (UTC)

Well, the problem is that if you reject any assumptions of the form "$y_{1}$ must have the same value when we measure $\langle \mathrm {GHZ} |Y_{1}Y_{2}X_{3}|\mathrm {GHZ} \rangle$ and $\langle \mathrm {GHZ} |Y_{1}X_{2}Y_{3}|\mathrm {GHZ} \rangle$", then you simply would not get a contradiction. And if you do, what was on this page before was a shortened form. Bbbbbbbbba (talk) 17:31, 6 January 2026 (UTC)

The contradiction is the claim that you cannot have one set of predefined values that contains both an odd and even number of down spins. Geoff Beck (talk) 17:33, 6 January 2026 (UTC)

Right, and those "predefined values" are the $x_{i}$ and $y_{i}$ variables in the previous revision of this page. So what was your objection again? Bbbbbbbbba (talk) 17:35, 6 January 2026 (UTC)

That $y_{i}$ could be +1 in one run and -1 in the next (from both QM and HV viewpoints). So the whole $y_{i}^{2}=1$ never follows. Geoff Beck (talk) 17:38, 6 January 2026 (UTC)

But the equation never talks about $y_{i}$ from different runs. It is all predefined values for the same run. Bbbbbbbbba (talk) 17:41, 6 January 2026 (UTC)

No, each composite operator is measured on a single run. Otherwise you would have to deal with successive non-commuting operations on each spin. Geoff Beck (talk) 17:44, 6 January 2026 (UTC)

Yes, a single run only factually measure one composite operator, but the point of hidden variable models is that the counterfactual predefined values also exist and satisfy the product relations. Otherwise there would be no contradiction. Bbbbbbbbba (talk) 17:49, 6 January 2026 (UTC)

Yeah, but that counterfactual would have to apply to the same preparation or it isn't actually a counterfactual. The individual runs do not constitute counterfactuals of each other. Each time you prepare the state the spins can take different values, even assuming predefinition is possible. Geoff Beck (talk) 17:52, 6 January 2026 (UTC)

There is still a contradiction in that you can't define a single set of predefined values to account for it all Geoff Beck (talk) 17:53, 6 January 2026 (UTC)

OK, let me play devil's advocate. Why would a single set of predefined values need to account for all the product relations? Could not it be that, for example, in every run where we measure $Y_{1}Y_{2}X_{3}$, $y_{1}y_{2}x_{3}=-1$, but $y_{1}x_{2}y_{3}=+1$, and in every run where we measure $Y_{1}X_{2}Y_{3}$, $y_{1}x_{2}y_{3}=-1$, but $y_{1}y_{2}x_{3}=+1$? Bbbbbbbbba (talk) 17:57, 6 January 2026 (UTC)

Because I could change my mind and measure them in any order. So the state needs to carry all the instructions for how to respond to any of the operators in the same way. For hidden variables Mermin says this equates to a complete specification of the spins on each axis. Geoff Beck (talk) 18:03, 6 January 2026 (UTC)

So what was wrong with my claim that in the equation at the top of this discussion thread, all the $x_{i}$ and $y_{i}$ variables should refer to the same run, the same preparation? You said "No, each composite operator is measured on a single run." I agree with the second half of the sentence but I do not see why that is a "no". Bbbbbbbbba (talk) 18:07, 6 January 2026 (UTC)

You could never perform that experiment. How would you know the preparation was identical? It is "identical" in the sense we usually use in QM, but this just means it has the same statistics, not that all the values will match. Geoff Beck (talk) 18:11, 6 January 2026 (UTC)

Again, it is not "identical" preparation, it is literally the same preparation. You prepare one instance of the GHZ state, and all those six $x_{i}$ and $y_{i}$ variables exist simultaneously and we can talk about their values. In real life we do the repeated experiment to make sure that the system does not surprise us when we "change our mind" as you say, but the theoretical argument only works with a single instance of the GHZ state. Bbbbbbbbba (talk) 18:16, 6 January 2026 (UTC)

You could invoke superdeterminism here sure. Geoff Beck (talk) 18:04, 6 January 2026 (UTC)

On pg 152 of

• Peres, A. (2002). Quantum theory: concepts and methods. Dordrecht: Springer Netherlands.

Peres has a couple of pages which covers all of the issues discussed here including the same preparation. He says that $x_{1}$ deduced from $\sigma _{x}\sigma _{x}\sigma _{x}$ must be a different run from $\sigma _{x}\sigma _{y}\sigma _{y}$ because the measurements are mutually exclusive. He draws a parallel to the EPR setup in which Einstein wants to infer two properties on spatially separated particles but Bohr points out these measurements cannot be made.

In my own words I say that if the three of us measure a GHZ state and I have my arm set to measure $\sigma _{x}$, then your report of "I measured +1" does not allow me to predict with certainty my value unless you tell me how your arms are set. My prediction (state) is conditioned on your apparatus.

Now I wonder if the deleted equation was trying to say as much without using any words to say so. The deleted equation

$y_{1}y_{2}x_{3}\cdot y_{1}x_{2}y_{3}\cdot x_{1}y_{2}y_{3}\cdot x_{1}x_{2}x_{3}=x_{1}^{2}x_{2}^{2}x_{3}^{2}y_{1}^{2}y_{2}^{2}y_{3}^{2}$

makes no sense as QM since Mermin and Peres show that the LHS is

$(1)\cdot (1)\cdot (1)\cdot (-1)$

In QM the LHS is only meaningful as triplets and can't be rearranged.

If I ignore QM and say that all of the $x_{i}$ $y_{i}$ are "elements of reality" then we could claim that algebraically

$y_{1}y_{2}x_{3}\cdot y_{1}x_{2}y_{3}\cdot x_{1}y_{2}y_{3}\cdot x_{1}x_{2}x_{3}=y_{1}y_{2}x_{3}y_{1}x_{2}y_{3}x_{1}y_{2}y_{3}x_{1}x_{2}x_{3}=x_{1}^{2}x_{2}^{2}x_{3}^{2}y_{1}^{2}y_{2}^{2}y_{3}^{2}$

and assuming values $\pm 1$ it must be

$x_{1}^{2}x_{2}^{2}x_{3}^{2}y_{1}^{2}y_{2}^{2}y_{3}^{2}=1,$

a value opposite the experimentally verified QM value. Johnjbarton (talk) 18:13, 7 January 2026 (UTC)

This is basically accurate, except that the fact "your report of 'I measured +1' does not allow me to predict with certainty my value unless you tell me how your arms are set" is not so relevant here. Of course a report of "I measured +1" alone, without basis information, is worthless: You may have measured the opposite Pauli operator (e.g. −X instead of +X). What matters is that, in the same run that you measured X, you could not have also measured Y. That is what makes the argument counterfactual.

I found the book, but I am not sure how many page you read starting from page 152 and I am not sure what gave you the impression that needing the basis information was an important part. Bbbbbbbbba (talk) 11:10, 9 January 2026 (UTC)

Page 152 has a section entitled "Three particle model" and the discussion extends as far as pg 154. Johnjbarton (talk) 18:05, 9 January 2026 (UTC)

I still have not seen what you might have seen, but I did notice that "If the first observer got a bad education in quantum theory and believes that the pair of particles has, at each instant, a definite wave function..." I wonder if the author was subtly bringing up the point that in fact the observer is entangled with the system after observation. Then again, maybe he was just saying that quantum field theory is necessary in this case, even though in that case I think it is more of a "bad education in relativity" thing than a quantum theory thing. Bbbbbbbbba (talk) 15:48, 13 January 2026 (UTC)

This comment appears in the discussion on relativity, so I think your last suggestion is what is meant. I interpret the "bad education" comment to mean: the QM model for the 3 particle system has no time axis, it only predicts the outcomes of 3-particle experiments. It cannot make any predictions related to simultaneity and any attempt to do so will generate confusion characteristic of a "bad education". Johnjbarton (talk) 18:09, 13 January 2026 (UTC)

So I have been hard and at work and I think Mermin's argument is actually equivalent to form with the squared spin values you argued for @Bbbbbbbbba. It does still suffer from being untestable, as Peres argues too. This is simply because you can't guarantee the counterfactual relationships required between runs of the experiment.

However, provides a way to make the empirical connection: relating the operator eigenvalues (3-spin correlation function) to functions that assign definite outcomes to all axes as a function of angle. Unfortunately, this argument is not as neat and simple. Geoff Beck (talk) 09:28, 15 January 2026 (UTC)

The untestability problem is fundamental: There is no way to guarantee the counterfactual relationships between the function values ⁠$A_{\lambda }(\mathbf {{\hat {n}}_{1}} )$⁠ and ⁠$B_{\lambda }(\mathbf {{\hat {n}}_{2}} )$⁠ either. In fact, the squared spin values formulation is exactly the same as this formulation, just with ⁠$A_{\lambda }(0)$⁠ written as ⁠$x_{1}$⁠ and ⁠$B_{\lambda }\left({\frac {\pi }{2}}\right)$⁠ written as ⁠$y_{2}$⁠, etc. The main difference is that the squared spin values equation omitted the hidden variable ⁠$\lambda$⁠. Bbbbbbbbba (talk) 13:59, 15 January 2026 (UTC)

Yes, the deleted paragraph, the squared spin values equation as you call it, omitted the hidden variable which ultimately causes the confusion. The paragraph was entirely about the incorrect model. The "Bell’s theorem without inequalities" paper says explicitly

• Of course, any specific triple of particles can be subjected to only one of the four choices of phase angles, and therefore the entire ensemble of triples emitted from the aperture is subdivided into four mutually exclusive and exhaustive subensembles. But since the triples are emitted before encountering the phase plates, where the subdivi-sion into subensembles takes place, it is reasonable to as-sume that the same probability measure p governs all four of the subensembles.

I think this addresses the original concern directly and explains why the subsequent multiplication remains valid: the three sets of values may be from different runs but specifically in the hidden variable case the results should not be affected. I know that I was incorrectly applying QM reasoning to the analysis when its not a QM problem, but the fact that the paper calls this out means this is not a trivial point. Johnjbarton (talk) 19:47, 15 January 2026 (UTC)

Hmmm, I suspect you are right about this. I was under the apprehension that the idea must be testable and that the Wiki article wasn't presenting the testable version. Geoff Beck (talk) 07:00, 16 January 2026 (UTC)