In quantum entanglement when something acts on one particle the other one reacts also, just in reverse (more or less). From what I've read though, anything acting on either particle will collapse the entanglement, right? So how do we know they were ever linked? Or is it just measurements that collapse it?

The reason I ask is because given the public impression of the topic it would suggest on of two things. That the information causing the reaction is superluminal or that the particles are occupying the same space since they are in different states. If the second were true then distance must be a human construct and somehow they must still be in the same place regardless of the virtual distance between them.

To clarify, I don't mean information in regards to communication such as QC. Just particle information.

  • $\begingroup$ Do you know much about quantum mechanics or linear algebra? $\endgroup$
    – ChrisM
    Mar 21, 2015 at 22:36
  • 2
    $\begingroup$ If you want a layman's answer without reference to the maths of quantum mechanics, perhaps look at Brian Greene's book, The Fabric of the Cosmos: Space, Time, and the Texture of Reality. This gives an excellent and detailed account of entanglement without recourse to the math. $\endgroup$
    – user70720
    Mar 21, 2015 at 22:49
  • 3
    $\begingroup$ "...given the public impression of the topic..." citation needed. What is the public impression? How are you formulating on opinion on "public impression"? $\endgroup$
    – DanielSank
    Sep 26, 2016 at 6:00
  • 5
    $\begingroup$ This post asks at least three different questions in the first paragraph, and none of them match the question posed in the title, which is itself a someone vacuous question since there's no distinction between "real" and "just math" in any way that I can think of. That being the case, I'm voting to close as unclear what you're asking. I hope the post will be edited and improved. $\endgroup$
    – DanielSank
    Sep 26, 2016 at 6:04
  • 1
    $\begingroup$ Possible duplicate of Quantum entanglement and spooky action at a distance $\endgroup$
    – stafusa
    Jan 21, 2018 at 21:57

6 Answers 6


Entanglement is a real property that can be shown by the violation of the Bell inequalities. How this is commonly done is that a pair of particles are created with entangled spin states in a configuration called Bell states. If entanglement is real, then measuring the state of one particle will give me definite knowledge of the state of the other particle. If there is no entanglement, then the measurement of one particle should not correlate so strongly with the measurement of the other.

What is found experimentally is that the Bell inequalities are violated and thus entanglement is real. The most popular view is that this means quantum mechanics is an accurate representation of reality and is not a statistical representation of an underlying deterministic mechanics. In this view, it does not permit superluminal transfer of information as you can't control which state you get and thus can't control what the other person sees.

An alternative view is that quantum mechanics is a description of a non-local hidden variable. I am not particularly well read on this particular viewpoint, but if you are interested in the implications, I would suggest looking into De Broglie-Bohm theory.

  • 3
    $\begingroup$ A loophole free violation of a Bell's inequality remains a daunting task as discussed here: en.wikipedia.org/wiki/Loopholes_in_Bell_test_experiments $\endgroup$
    – Mikhail
    Mar 21, 2015 at 22:57
  • 2
    $\begingroup$ +1 for the answer, I FERMLY don't believe QM is non-local. You cannot transmit any information non-locally, and non-local hidden variable theories are just monstrous imo. $\endgroup$
    – vsoftco
    Mar 21, 2015 at 23:23
  • 1
    $\begingroup$ If you have ever read the original paper of Bell deriving the inequalities you wouldn't be too sure about the fact that violation shows the impossibility of a classical local hidden-variable theory. (@letmethink, @ajctt). In short: There is premise that simply does not hold for classical, statistical systems. Therefore the inequalities are violated by classical systems as well. Hence, experimental violation shows nothing. I'll post an answer to this in the next days but this will take time. The first part is below. $\endgroup$
    – image357
    Mar 22, 2015 at 1:15
  • $\begingroup$ Also there is the possibility of non-local hidden variables. In other words think about parasites that swim up the pee stream of an animal and infect the urethra. What if the observation made by Alice disturbs the entire wave function across all its area? This is what we're arguing it does. But that doesn't have to mean it's not an effect that can be modeled physically with classical techniques like pilot-wave hydrodynamics $\endgroup$
    – CommaToast
    Apr 3, 2015 at 0:34
  • $\begingroup$ People like to spray vitriol on non-local hidden variables, but I don't think they truly understand what that implies. After all if one both presumes quantum theory is correct and forbids any underlying non-locality, then the only option is solipsism: indeed, you can't even argue the wavefunction is real, because that would also be non-local. One can't even argue experiments themselves are real, as they can also be described purely in terms of wavefunctions. Hence the only thing that can be 'real' is you observing it. (And being driven to that, does it even make sense to require 'locality'?) $\endgroup$ Oct 4, 2016 at 19:59

This problem has been pointed out historically in what is now universally abbreviated as the EPR paper, for which I'll simply refer you to an answer to a very similar question. This seemingly paradoxical effect has been observed experimentally.

Some people insist the question of whether it is "real" is still unresolved. The main difficulty, however, is simply that the question what is "real" is a matter of philosophy, not of (natural) science. Stifflers for detail among physicists hence call the different ways of thinking that are not strictly speaking at odds with our mathematical description of quantum mechanics interpretations. In choosing an interpretation, you cannot simultaneously have locality (the idea that particles' states somehow reside in them, which implies superluminal effects, or a "spooky action at a distance" as Einstein allegedly called it), and realism (your question if it is "real"). CAVEAT: The wikipedia authors for the interpretation link appear to disagree, or at least the summary table suggests they do. I am basing my statement on Nielsen and Chuang's argument in their textbook Quantum Computation and Quantum Information.

If all you want is one interpretation to make sense of it all, I would recommend an informational approach: Do not think of states as being (local) properties of particles. Instead, they describe your information about the system: If you prepared an appropriate superposition between two particles, then you know that whatever state one particle is in, the other will be in the opposite state. That is global information created when the particles interacted, which does not require any superluminal effects. Instead of one particle suddenly changing its state, the effect lies in the (global) knowledge about the system.

The other aspect of your questions regards the issue of wave function collapse. Again, this is one interpretation, and as you apparently already realize, not a very consistent (or at least not comprehensive) one. The paradox disappears if you model your interaction not as some sort of measurement (whatever that is), but as a quantum-mechanical interaction with another particle. That creates additional entanglement (with that new particle, or else with the environment), which means that the wavefunctions describing only your particles without that one do not interfere the way they did before if you exclude the environment from your analysis. If this interference is completely destroyed, then you have, in the system excluding the environment, the same observable effects as if you were collapsing wavefunctions.


Entanglement isn't about interaction or information transfer betweeen entangled particles.

Consider spin-entaglement of two spin-$\frac{1}{2}$ particles: Let them be in singulet-state relative to an arbitrary axis (say z-axis):

$$ |\Psi \rangle = \frac{1}{\sqrt{2}} (\ |\uparrow_z, \downarrow_z \rangle - |\downarrow_z,\uparrow_z\rangle \ ) $$

The propability $P$ to measure both particles in state $|i,j \rangle$ with $i,j \in \{ \uparrow, \downarrow \}$ where the axis of both measurments enclose the angle $\theta$ is given by: $$ P_{i,j} = \| \langle i,j | \Psi \rangle \|^2 = \frac{1}{4} (1 - i \cdot j \cdot \cos \theta )$$ if we take $i,j$ to be 1 and -1 for $\uparrow$ and $\downarrow$, respectively.

The reduced propability $p_i$ of measuring only one particle (e.g. if we don't care about the other) is given by: $$ p_i = \sum_{j \in \{1,-1\}} P_{i,j} = \frac{1}{2} $$

The conditional propability of measuring the other particle (after we already know about the measurment of the first particle) is given by: $$ \tilde{p}_{j|i} = \frac{P_{i,j}}{p_i} = \frac{1}{2} (1 - i \cdot j \cdot \cos \theta ) $$

This does involve the angle $\theta$ and usually one starts here to argue about non-locality and instantanious actions changing the outcome of experiment when we change the angle $\theta$ at the first measurment apparatus.

This is however not true. If we are talking about conditional propabilities we have already performed a measurment and set the measurment axis of the first measurment. Changing this axis afterwards will not affect the propability as the angle $\theta$ is relative to the measured axis. Changing the axis of the second measurment only changes the propability predicting the outcome of the later measurment for the first observer because he has that extra knowledge.

The propability for the second observer stays the same, as this is the reduced propability (he doesn't know about the first measurment): $$ p_j = \sum_{i \in \{1,-1\}} P_{i,j} = \frac{1}{2} $$

In short: Without the extra knowledge of the first measurment, entanglement is not important for the second observer. To gain that extra knowledge there must be an additional information transfer to the second observer and this is restricted by means of relativity-causality ($v\le c$ etc.). So entanglement neither breaks causality nor can it transfer any information.


Sometimes one comes about the argument that the violation of Bell inequalities shows, that entanglement is still something more than classical perception would allow. So let's have a look at a certain expectation value. The axis for spin measurment shall be labeled by normalized vectors $\vec{a}$ and $\vec{b}$ such that $\vec{a}\cdot\vec{b} = \cos\theta$. Consider \begin{equation} \langle \Psi|\vec{a}\cdot\vec{S_1} \ \ \vec{b} \cdot \vec{S}_2 | \Psi \rangle = -\frac{\hbar^2}{4}\vec{a}\cdot\vec{b} = -\frac{\hbar^2}{4} \cos\theta \tag{1} \end{equation} , which is the expectation value of the product of both measurments results. Here we have $\vec{S} = \frac{\hbar}{2}(\sigma_x, \sigma_y, \sigma_z)^T$ with $\sigma_x, \sigma_y, \sigma_z$ the Pauli matrices. We now follow the reasoning of John Bell in his original work since other, similar inequalities are based on the same problem.

The argument goes like this: Assume a classical, statistical system with non-hidden and hidden variables all labeled by $\vec{\lambda} = (\lambda_1, \dots, \lambda_n)$ for some $n\in\mathbb N$. Furthermore there exists two functions $A(\vec{a},\vec{\lambda})$ and $B(\vec{b},\vec{\lambda})$ that give the results of spin measurment on particle 1 and 2, respectively. They can only yield $\pm\frac{\hbar}{2}$, since that is the only outcome of experiment. Those functions depend on one measurment axis only, because there shall be no action between measurment apparatus 1 and 2 (this is the assumed locality).


Because the system is studied on a statistical basis, there exists a propability density $ \varrho(\vec{\lambda}) $ that is a function of the system parameters $\vec{\lambda}$ and allows calculation of the expectation value $$ E(\vec{a},\vec{b}) = \int \varrho(\vec{\lambda}) \cdot A(\vec{a},\vec{\lambda}) B(\vec{b},\vec{\lambda}) \ d^n\lambda $$, which should equal the one from above (1) if it is to be interpreted on a classical, local basis (Note: one can incorporate discrete statistical variables by terms like $\sum_j \alpha_j \cdot \delta(c_j-\lambda_m)$). The malicious assumption here is that $\varrho$ is no function of the axis-vectors $\vec{a}$ and $\vec{b}$. This is, however, quite natural for classical systems with correlation. The point is: Allowing $\varrho(\vec{\lambda}, \vec{a}, \vec{b})$ or even just $\varrho(\vec{\lambda}, \vec{a} \cdot \vec{b})$, the Bell inequalities cannot be derived! Such propability densities can cause violation of the inequality. To understand that, I will now derive them and point out which step is not possible with the modified density:


Assume $$ E(\vec{a},\vec{b}) = -\frac{\hbar^2}{4} \vec{a} \cdot \vec{b} \tag2 $$, so that quantum mechanical description is in agreement with the classical one. For $\vec{a} = \vec{b}$: \begin{equation} \begin{aligned} -\frac{\hbar^2}{4} & = \int \underbrace{\varrho(\vec{\lambda})}_{\ge 0} \cdot \underbrace{A(\vec{a},\vec{\lambda}) B(\vec{a},\vec{\lambda})}_{\ge -\frac{\hbar^2}{4}} \, d^n\lambda \\ & \Leftrightarrow \\ 0 & = \int \underbrace{\varrho(\vec{\lambda})}_{\ge 0} \cdot \left( \underbrace{A(\vec{a},\vec{\lambda}) B(\vec{a},\vec{\lambda}) + \frac{\hbar^2}{4}}_{\ge 0} \right) \, d^n\lambda \end{aligned} \end{equation} because $\varrho$ is a normalized propability density. It follows that \begin{equation} \begin{aligned} A(\vec{a},\vec{\lambda}) B(\vec{a},\vec{\lambda}) = -\frac{\hbar^2}{4} \end{aligned} \end{equation} is a valid equation under the integral with $\varrho$. This can only hold if \begin{equation} \begin{aligned} B(\vec{a},\vec{\lambda}) = - A(\vec{a},\vec{\lambda}) \end{aligned} \tag3 \end{equation}. Note that this holds for any vector $\vec{a}$. Now take another normalized vector $\vec{c}$ and do the following calculations: \begin{align} \frac{\hbar^2}{4}|(-\vec{a}\cdot\vec{b}) - (-\vec{a}\cdot\vec{c})| & = |E(\vec{a},\vec{b}) - E(\vec{a},\vec{c}) | \\ & = \left| - \int \varrho(\vec{\lambda}) \cdot (A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) - A(\vec{a},\vec{\lambda}) A(\vec{c},\vec{\lambda})) \, d^n\lambda \right| \\ & = \left| \int \varrho(\vec{\lambda}) \cdot A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) \cdot (1 - \frac{4}{\hbar^2}A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda})) \, d^n\lambda \right| \\ & \le \int | \varrho(\vec{\lambda}) | \cdot | A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) | \cdot |1 - \frac{4}{\hbar^2}A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda})| \, d^n\lambda \\ & = \int \varrho(\vec{\lambda}) \cdot (\frac{\hbar^2}{4} - A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda})) \, d^n\lambda \\ & = \frac{\hbar^2}{4} + E(\vec{b},\vec{c}) = \frac{\hbar^2}{4} - \frac{\hbar^2}{4}\vec{b}\cdot\vec{c} \tag4 \end{align}

In the first equality we used (2). In the second we used (3). In the third we used $A(\vec{b},\vec{\lambda})^2 = \frac{\hbar^2}{4}$. The fourth step is the triangle inequality for integrals. In the fifth step we used $A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) = \pm \frac{\hbar^2}{4}$ and $\varrho(\vec{\lambda}) \ge 0$. In the last step we used (2) and the fact that $\varrho$ is normalized.

So we finaly have Bell's inequality \begin{equation} \begin{aligned} |\vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c}| + \vec{b}\cdot \vec{c} \le 1 \, , \end{aligned} \tag5 \end{equation}, which can be violated for some choise of $\vec{a},\vec{b},\vec{c}$. This usually shows that our first assumption (2) is false. Therefore, no classical, local system should be able to describe the expectation value (1).


With the modified probability density the steps in (4) look like this: \begin{align} \frac{\hbar^2}{4}|(-\vec{a}\cdot\vec{b}) - (-\vec{a}\cdot\vec{c})| & = |E(\vec{a},\vec{b}) - E(\vec{a},\vec{c}) | \notag \\ & = \left| - \int \varrho(\vec{\lambda}, \vec{a}, \vec{b}) \cdot A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) - \varrho(\vec{\lambda}, \vec{a}, \vec{c}) \cdot A(\vec{a},\vec{\lambda}) A(\vec{c},\vec{\lambda}) \, d^n\lambda \right| \notag \\ & = \left| \int A(\vec{a},\vec{\lambda}) A(\vec{b},\vec{\lambda}) (\varrho(\vec{\lambda}, \vec{a}, \vec{b}) - \varrho(\vec{\lambda}, \vec{a}, \vec{c}) \frac{4}{\hbar^2}A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda})) \, d^n\lambda \right| \notag \\ & \le \int \frac{\hbar^2}{4} \cdot \left| \varrho(\vec{\lambda}, \vec{a}, \vec{b}) - \varrho(\vec{\lambda}, \vec{a}, \vec{c}) \frac{4}{\hbar^2}A(\vec{b},\vec{\lambda}) A(\vec{c},\vec{\lambda}) \right| \, d^n\lambda \end{align} Note that one cannot proceed from here since in general $\varrho(\vec{\lambda}, \vec{a},\vec{b}) \ne \varrho(\vec{\lambda}, \vec{a},\vec{c})$. Also the second equality shouldn't work here anyway since (3) is only vaild when multiplied by $\varrho(\vec{\lambda},\vec{a},\vec{a})$. For instance, when $\varrho(\vec{\lambda},\vec{a},\vec{a}) = 0$ equation (3) can be violated in general. Nevertheless, one could only try to use another triangle equation on the term $|\dots|$, leaving us finally with the inequality \begin{equation} \begin{aligned} |\vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c}| \le 2 \, , \end{aligned} \end{equation}, which is not to be violated by any choise of $\vec{a},\vec{b},\vec{c}$.


In summary: If one allows propability densities $\varrho(\vec{\lambda}, \vec{a}, \vec{b})$, that depend on some parameters of the measurment, the derivation of an inequality which is violated by quantum mechanical expectation values is not possible in the usual way. Above, I already argued that the dependence on $\vec{a}, \vec{b}$ is in general no cause for non-local behaviour as long as the reduced propability of a subsystem is only depended on its own parameters. This problem is inherent to inequalities that are derived on the same arguments like Bell's inequality: see for example the CHSH-inequality on page 527 equation 2, which is frequently used in experiments!


So if we would find some functions $A$ and $B$ that satisfy our locality-conditions from above there is no reason to think of the expectation value (1) as a non-local one. Take \begin{align} p_{i,j}(\vec{a},\vec{b}) & = \frac{1}{4} (1 - i j \ \vec{a}\cdot\vec{b}) \\ A(i,\vec{a}) & = \frac{\hbar}{2} \ i \\ B(j,\vec{b}) & = \frac{\hbar}{2} \ j \end{align} Then we have $$ E(\vec{a}, \vec{b}) = \sum_{i,j \in \{1,-1 \}} p_{i,j}(\vec{a},\vec{b}) \cdot A(i,\vec{a}) B(j,\vec{b}) = - \frac{\hbar^2}{4} \ \vec{a}\cdot\vec{b} = - \frac{\hbar^2}{4} \ \cos\theta$$, which equals (1) on a pure, local and classical basis.

  • 4
    $\begingroup$ From your comment on ajctt's answer above I understand that one of your main points is that Bell's inequalities could be violated even by classical (local, real) systems. While you are extending your answer it would be great if you could illustrate this point further, ideally with a concrete example. That would be very interesting. $\endgroup$
    – Emil
    Mar 22, 2015 at 9:51
  • 2
    $\begingroup$ FYI only - I imagine that this answer is useful but it is too "dense" to be readily useful to someone not used to the field, terminology and symbology. (eg I have a Masters degree in an unrelated technical area and a 'moderately high' IQ but I'd be included in those categories). That is fine enough if you are aiming at a very limited audience but if you want it to be accessible to intelligent competent scientists (and engineers (to include me :-))) who do not 'tick all the above boxes' then it needs additional explanatory material.Whether this is considered worth your doing is up to you. $\endgroup$ Mar 23, 2015 at 1:49
  • 5
    $\begingroup$ Nice answer, I am looking forward to the thought experiment, because, right now, $\rho(\lambda, \vec a , \vec b)$ looks very unclassical to me - a probability distribution that depends on what we want to measure?! $\endgroup$
    – ACuriousMind
    Mar 24, 2015 at 22:54
  • 2
    $\begingroup$ @ACuriousMind: The propability distribution does not depend on what we want to measure, since $\vec{a}$ is not part of the measurment result. It only accords to the fact that the setting of the measurment apparatus can (not must!) influence the propability. take for instance a particle that is always directed to the x axis. If we measure along the y-axis we wont find it, thus $\varrho(\lambda,x-axis) \ne \varrho(\lambda,y-axis) = 0$ $\endgroup$
    – image357
    Mar 28, 2015 at 13:31
  • 1
    $\begingroup$ @Marcel If you look into the last paragraph in section III, "Illustration", of Bell's original work (philoscience.unibe.ch/documents/TexteHS10/bell1964epr.pdf), you will find that he excluded the case that you proposed, as the non-local case. In his words, "there is no difficulty in reproducing the quantum mechanical correlation (3) if the results A and B in (2) are allowed to depend on $\vec{b}$ and $\vec{a}$ respectively as well as $\vec{a}$ and $\vec{b}$." $\endgroup$
    – user36125
    Apr 10, 2015 at 20:42

It is real, and experimentally verified. If you measure the spin of entangled particle A, particle B when measured will always have the opposite spin. Some physicists believe it has to do with superluminal communication, but there are many other theories.


So how do we know they were ever linked?

Because the results violate Bell's inequalities.

Or is it just measurements that collapse it?

Measuring an entangled system destroys the entanglement.

The results from experiments on entangled particles are only possible with:

A) information travelling FTL (i put "space isn't what we think it is" theories in the same category because there's a lot of grey area between the two). I'll also mention that i don't feel like theories that try to get around this (ie. pilot wave theory) actually do. As far as i'm concerned non-locality is encompassed within the broader category of FTL.

or B) superdeterminism, "particle A knows how we will decide to measure particle B before we measure it, and yes i know it's far fetched but it has to be mentioned as the only other alternative.

Entanglement is definitely real, and not just math. Reading through that wiki link might provide some more insight.


OP wrote "In quantum entanglement when something acts on one particle the other one reacts also, just in reverse (more or less)."

This observation may be hinting at "perfect anti-correlation". i.e. when spin of two "entangled" particles are measured in same angle, they are guaranteed to be opposite. There are some specific state(s), when entangled pairs created/prepared in that state, the particles of the pair exhibit perfect anti correlation. (If there is no such state(s), knowledgeable QM folks, please do comment). The state may be one of the Bell states - https://en.wikipedia.org/wiki/Bell_state#The_Bell_states

Anti correlation is very different from statistical correlation and mixing of the two causes all the confusion.

QM assigns averages/probabilities. There is nothing like 0 or 1 probability, meaning there is no guaranteed outcome in probability. That means perfect anti correlation (of spin) is enforced by a law rather than by probability.

That law is conservation of angular momentum. When the particles are prepared, at that time, they are prepared in such a way, that to conserve angular momentum in each and every direction, they are coded (hidden variables) with opposite spin in each and every direction. Whenever and wherever they are measured. So, distance is not even a factor here.

Therefore, they do not need to occupy same space, and they do not need to communication at all, let alone at FTL speeds. So anti correlation is a direct consequence of conservation laws. Anti correlation is a relation between outcomes of two particles of same pair.

The problem arises when people use anti correlation phenomena to explain statistical correlations. They enumerate hidden variables based upon anti correlation and compute Bell's inequality based upon these enumerations and apply the inequality to statistical correlations. At this point, they forget to consider that statistical correlation is between outcomes of VARIOUS PAIRS, while anti correlation is between outcomes of the SAME PAIR. Therefore applying Bell's inequality to statistical correlations is a gross misuse of a mathematical theorem.

Anti correlation is explained by conservation laws.

Statistical correlation should be scrutinized and explained independent of anti correlations. Mixing of anti correlation with statistical correlation, is the root of whole mystery that surrounds entanglement.

Statistical correlation is a game of averages and it must be scrutinized, and explained in terms of factors local to the experiment. Factors which are termed loopholes These factors have to shape up statistical correlations over duration of the experiment. If that is found to be the case, then there is no such thing like entanglement.

As a first step, I analyzed data from a recent entanglement experiment and that analysis gives an indication that statistical correlations are not necessarily guided by probabilities. It can be read at http://vixra.org/abs/1609.0237.

  • 2
    $\begingroup$ "when spin of two "entangled" particles are measured in same angle, they are guaranteed to be opposite." That is certainly false. $\endgroup$
    – DanielSank
    Sep 26, 2016 at 6:05
  • 3
    $\begingroup$ What about $|\uparrow \uparrow \rangle + |\downarrow \downarrow \rangle $. $\endgroup$
    – DanielSank
    Sep 26, 2016 at 15:43
  • 2
    $\begingroup$ @kpv If the point that Daniel just raised isn't clear to you, then you should seriously consider the possibility that you just don't understand enough about the topic to answer this question. (In case you missed it, Daniel just gave you a clear example of a maximally entangled state that is not perfectly anticorrelated.) As this stands, this is an appallingly poor answer to the question and little more than an advertising (spam?) for the paper you keep pushing on this site. $\endgroup$ Oct 1, 2016 at 19:18
  • 1
    $\begingroup$ @EmilioPisanty: page 100 of books.google.com/… talks about it too. $\endgroup$
    – kpv
    Oct 1, 2016 at 23:01
  • 3
    $\begingroup$ @kpv What are you on about? Are you incapable of seeing the difference between the claims "all entangled states represent anti-correlated spins" and "some entangled states represent anti-correlated spins"? Why do you feel entitled to me devoting my time to explaining entanglement to you when you refuse to pick up a textbook? $\endgroup$ Oct 2, 2016 at 9:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.