When quantum entanglement is explained in "layman's terms", it seems (to me) that the first premise, that we have to accept on faith, is that a particle doesn't have a certain property (the particle is not IN one state or another) until that property is measured.

For example, I have read that a particle may have "spin up" or "spin down" but -- here's where I get lost -- it's not that we don't KNOW which spin the particle has until we measure it, but actually the particle has NEITHER spin until we measure it.

Now, if we accept that premise, then we can see how spooky it is, when two particles are created with correlated spins -- if one has spin up, the other has spin down, but neither particle "has" either spin until one of them is measured. At that instant, the other particle "assumes" the other spin.

At least that's how I interpret things I have read.

Certainly, if we accept that a particle doesn't actually HAVE the property until it is measured, but rather the particle exists in a superposition of states until the measurement is made, then we "regular citizens" can understand what is meant by "spooky action at a distance". (As soon as one particle is measured, the other particle assumes the "other" spin, no matter how far away it is. How did the second particle know the first particle was measured? That, I think, is the spooky action at a distance (and the information transfer can be faster than light).)

However, from a layman's perspective, I want to cry out "but the particle does have a definite spin, we just don't KNOW what it is, until it is measured! Duh!"

Now, I can't believe that all of the physics community hasn't thought of that objection -- but -- here's my point -- although the spooky action at a distance can be explained to us, once we accept the premise of superposition -- Why isn't the fact of superposition explainable also, in terms mortals can understand? I have looked, but I haven't found a layman's explanation of why we should accept the fact that a particle doesn't have a particular spin, or other property, until it's measured. It sure seems that the particles "have" the properties, even though we haven't measured them yet.

I know that much of quantum mechanics is not "intuitive". If anyone can explain why particles don't have a definite property, even before we measure them, I would be grateful.

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    $\begingroup$ I think QM (and also relativity) can be also intuitive, but only for people knowing well its mathematics. $\endgroup$
    – peterh
    Commented Jul 15, 2014 at 14:55
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    $\begingroup$ Your question is well thought out. Nice job. The piece to your puzzle is Bell. $\endgroup$
    – BMS
    Commented Jul 15, 2014 at 15:34
  • $\begingroup$ You may like the Many Worlds Interpretation of quantum mechanics: en.wikipedia.org/wiki/Many-worlds_interpretation $\endgroup$ Commented Jul 15, 2014 at 16:08
  • $\begingroup$ What @BMS said. What you're looking for is the wonderful paper "Bringing home the quantum world: quantum mysteries for anyone" by Mermin which explains in wonderfully non-technical terms the essence of Bell's Theorem. Try this PDF link web.pdx.edu/~pmoeck/pdf/Mermin%20short.pdf $\endgroup$ Commented Jul 15, 2014 at 16:09
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    $\begingroup$ You may want to watch at this link. Recent experiments show that the usual interpretation of QM may be wrong. This link gives a possible solution to the failure of Bell-Inequalities, that allows a local realistic interpretation. There is also already a realistic interpretaion of QM called Bohmian Mechanics which in terms of classical mechanics seems to be more natural. $\endgroup$
    – image357
    Commented Jun 2, 2015 at 12:45

6 Answers 6


The assumption (if you cry it or not) "but the particle does have a definite spin, we just don't KNOW what it is, until it is measured! Duh!" is called realism, or in mathier speak, a theory of hidden variables.

Bell's inequalities now say that no theory that fulfills local realism (equivalently that has local hidden variables) can ever predict the correct results of a quantum mechanical experiment.

So we are faced with a problem: Do we give up locality or realism?

Most people choose realism, since giving up locality would totally destroy our conceptions of causality. It is possible that there is a non-local theory that assigns a definite value to every property at all times, but due to its non-locality, it would be even more unintuitive than "particles do not have definite properties".

There is no intuitive explanation for the non-realism of reality (there has to be a way to phrase that better...) because our intuitions have been forged in the macroscopic world which is, to good approximation, classical. But the non-realism is an effect that has no classical analogon, so we cannot understand it in pretty simple pictures or beautiful just-so stories.

Sometimes, we just have to take the world the way it is. (I have assumed that you do not want the whole QM story of non-commuting observables and eigenbases and so on to explain why we, formally from QM principles, expect realism to be false. If I have erred in that respect, just tell me)

  • $\begingroup$ On a personal note, how do you reconcile Bell's inequalities with this claim concerning the proof of impossiblity: "While an impossibility assertion in science can never be absolutely proven, it could be refuted by the observation of a single counterexample."? $\endgroup$ Commented Nov 15, 2014 at 13:52
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    $\begingroup$ @brightmagus: Strictly speaking, Bell's theorem is not a physical statement, but a mathematical one. Locality and realism are properties of theories, not of the real world, and the theorem says that no theory constrained to give the measured output of the world (i.e. the outputs of QM) can be local and real. It's not a statement about reality, but about theories, and, as such, it is indeed a proof of impossibility. $\endgroup$
    – ACuriousMind
    Commented Nov 15, 2014 at 14:26
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    $\begingroup$ You are correct in your description of Bell's inequalities (that no theory can fulfill local realism), but incorrect in your implication that we have a choice about whether to give up locality or realism. The Einstein, Podolsky, Rosen (EPR) argument says local QM theory necessarily implies realism. Taken together with Bell's inequalities, this means you can't have any local theory at all. You're still able to choose between realism (e.g. Bohmian mechanics) and non-realism (e.g. QM/Copenhagen interpretation), but you're stuck with nonlocality. @Anixx Bohmian mechanics is extremely nonlocal. $\endgroup$
    – pwf
    Commented Aug 3, 2015 at 19:44
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    $\begingroup$ @pwf: Hm, you're right in the sense in which you use locality, I guess. There's a lot of confusion about this because one shows in QFT that the theory is local in a certain technical sense of the propagators, but this is indeed not the locality meant in this context. $\endgroup$
    – ACuriousMind
    Commented Aug 3, 2015 at 21:47
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    $\begingroup$ I think you should edit your answer to address the fact that Bell's inequalities violations indicate that no local theory can exist, whether you want to keep realism or not. I find it confusing as it's currently standing. $\endgroup$
    – elelias
    Commented Apr 17, 2018 at 17:24

At that instant, the other particle "assumes" the other spin.

This is where you go wrong, because things are actually weirder than that. I'll try to keep this simple, especially because if I don't I'd have to make extremely rigorous arguments given that this is a matter where different physicists may see things a bit differently.

A historical note : the view of QM that you now have was solidified with experiments that tested Bell inequalities. To put it shorty, these inequalities had to hold if QM was true, but if they did hold they definitively proved that "local realism" couldn't hold. "Local" (again to keep things simple) you can think of as meaning that the theory will never predict event B happening as a result of event A unless a signal moving at the speed of light has had the time to travel from A to B ("spooky action at a distance" violates this because it has to happen instantaneously). Realism you can think of as meaning your very objection to QM :

but the particle does have a definite spin, we just don't KNOW what it is, until it is measured!

Now here's the thing, laymen such as yourself might be tempted to give up on locality rather than realism. It is only in that case that you can even speak of "spooky action at a distance", because the fact is that if the particle doesn't have a definitive spin then no "spooky action" is required. If we accept that the probabilities predicted by QM are an intrinsic part of the very nature of the particle, then all we're seeing are the predicted correlations, not the result of one particle changing state because another was measured.

Giving up realism seems to bother you, so instead you may be tempted to say "Let's give up locality!". Unfortunately for you my friend, this does not work. QM you can think of as a sort of simplistic version of QFT, which is explicitly local in its first principles (at least for the Standard Model Lagrangian it is). Furthermore, it actually makes more sense to give up realism, if only because that's what the theory is actually telling you. The only reason we consider giving up locality is because a layman's intuition (or physicists before we finally just accepted that this was the nature of things) might be tempted to invent some convoluted non-local mechanism like "spooky action at a distance", but there's no reason to do this, and it just adds a needless and untestable layer of complications to your theory. In other words, it's actually far more natural to give up realism, because, again, that's what the theory is actually telling you.

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    $\begingroup$ "because the fact is that if the particle doesn't have a definitive spin then no "spooky action" is required" This is not true at all. What Alice reads depends on what choice of measurement is performed by Bob. Meaning, Bob's choice has to necessarily affect Alice's photon measurement, or else you cannot recover the observed correlations. You always need spooky action at a distance. $\endgroup$
    – elelias
    Commented Apr 17, 2018 at 17:28
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    $\begingroup$ "Bob's choice has to necessarily affect Alice's photon measurement, or else you cannot recover the observed correlations." This is completely false. It's backwards in fact. The statistical distribution for each outcome Alice can measure is completely independent of anything Bob does on his side. After the fact, correlations between the two observations can be established, and the nature of those correlations will depend on what Bob did. But Alice's measurements are completely independent of Bob's actions. $\endgroup$
    – ticster
    Commented Apr 17, 2018 at 20:01
  • $\begingroup$ @elelias, ticster is right. Theoretically, within QM, the independence of Alice's measurement on Bob's actions (and vice versa) is guaranteed by the fact that the observables available to Alice and Bob commute with those of the other person. In quantum field theory, the locality is expressed by the fact that field operators at space-like-separated points commute with each other. This fact holds perfectly accurately in relativistic QFTs which is why there is no action at a distance. $\endgroup$ Commented Apr 18, 2018 at 5:08
  • $\begingroup$ Entanglement experiments only prove correlation which was there from the beginning when the two entangled particles were in touch - when they were simultaneously born, or communicated with each other due to their proximity. These very early events are the actual reason for the correlation. The later measurements are not the reason of the correlation - and they are not the reason of anything that happens or is seen in the distant lab of the other person. $\endgroup$ Commented Apr 18, 2018 at 5:10

I would just mention here that every probabilistic system, even classical, may exhibit a kind of "entanglement" or "spooky action at a distance" features.

For instance, imagine that you have $2$ boxes, and one bowl in each box. The bowl could have only a white or black color, and the two bowls have the same color. The boxes are closed, then one box stay on Earth, and the other box is sent to the planet March.

Imagine you are an external observer which do not put the bowls in the boxes, your analysis of the system is the following : one has the probability $\frac{1}{2}$ to find a white bowl in the two boxes, and a probability $\frac{1}{2}$ to find a black bowl in the two boxes. This is a probabilistic classical system.

Note this means, for you, that the bowls in the boxes have not a definite color.

Now you may open the box which stays on Earth. If you see a white bowl, you know immediately that the bowl in the box which is on March is white too, and you may imagine a "spooky action at a distance".

Of course, there is no "spooky action at a distance", and correlations are not causal relations. Here the correlations are about degrees of freedom (the color) which are completely independent of position degrees of freedom. So the correlations are exactly the same is the boxes are near each other, or very far from each other.

Quantum mechanics is a probabilistic theory, so much of the arguments above are correct for QM too. However, they are some specific features, because in QM, we are working with probablity amplitudes instead of probabilites, with vectorial states instead of "point" states, so the value of the correlations are very specific too, and you cannot always obtain these correlations even by a probabilistic classical system (Bell's theorem). So, quantum entanglement is special, sure, but in some sense, it is an extension of a probabilistic classical "entanglement".

Finally, I would say that it is far more interesting to compare probabilistic classical systems, and (probabilistic) quantum systems. It is not very interesting to compare determistic classical systems and QM.

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    $\begingroup$ Yes, but the layman's explanations of QM that I read, says that it's not true that the bowl in the box has either a black or a white color until someone looks at it. So this analogy doesn't quite seem to match what I have read. $\endgroup$
    – DWalker
    Commented Dec 29, 2014 at 20:21
  • $\begingroup$ The bowls in your example are not entangled. $\endgroup$
    – WillO
    Commented Jul 14 at 20:46

You have not described entanglement properly. What sort of thing do you have to explain to understand what's going on in an entanglement experiment? The problems often look somewhat like this.

(1) There are observables on A and B, call them Acorr, Bcorr such that if you compare the results of the measurements after they have been completed and the information about the measurement results has been transmitted to the same location, you will find that they are correlated. So it might be that if you're measuring electron spin the spins will be found to be opposite with probability 1, or with some probability that differs from 1/2 when they are compared.

(2) There are observables on A and B, call them Anoncorr, Bnoncorr such that if you compare the results of the measurements after they have been completed and the information about the measurement results has been transmitted to the same location, you will find that they are not correlated. So it might be that if you're measuring electron spin the spins will be found to be opposite with probability 1/2 and the same with probability 1/2.

(3) There are intermediate cases. And overall the correlations are such that they do not match what you would get if you had two systems represented by local classical stochastic variables.

There might be complications due to measurement error or whatever in assessing the results of a particular experiment. You can also come up with formulae for precisely how the probabilities of matching differ depending on what observables you pick. The basic problem is that the probability of seeing correlations when you compare the results differs from what you would expect from a local theory using classical stochastic variables. Specifically the probability of a correlation between the compared results of measurements on entangled systems depends on what measurement was conducted on each system.

What people commonly do next is leap to the conclusion that quantum mechanics is non-local (spooky), but this conclusion is false. The reason that you can't account for the results of experiments on entanglement using local classical stochastic variables is that quantum systems cannot be described by classical stochastic variables. They are described by Heisenberg picture observables. These observables represent many versions of a given system, one for each possible measurement result that can interfere with one another. The system does not have a single value until you measure it, which is why you can't say what the value is. We can rule out the idea that a measurable quantity has a single value by conducting experiments whose results cannot be explained without invoking multiple versions of the systems involved, these include entanglement experiments and single particle interference experiments. The reason you don't see multiple versions of, say, your chair, is that any interaction that makes another system depend on where you chair is prevents interference: this is called dechorerence. Your chair affects the light around it, exerts pressure on the floor and so on and this prevents interference.

How do we explain the results of an entanglement experiment? Suppose there are two experimentalists Alice and Bob who have quantum systems that are entangled with one another. What happens when Bob measures his system? The measurement apparatus differentiates into multiple versions, each of which has recorded one of the possible outcomes. The records of measurements on a become correlated with records of measurement on Alice's system when the measurement results are compared because the decoherent systems that carry the measurement outcomes also carry locally inaccessible information, which helps bring about the correlation. The observables of the systems carrying the measurement results from Bob depend on what Bob has measured but the expectation values of those observables do not depend on the measurement, see



The expectation value of an observable is just the sum of (probability of each outcome)x(value of the observable for each outcome) for all the outcomes and this tells you all information that quantum mechanics allows you to get about that observable. Because it is impossible to get the locally inaccessible information about the correlations by measuring just Bob's system or just Alice's system, this information can produce quantum mechanical effects that involve multiple versions of each system when the measurements results compared despite decoherence. All of the relevant comparisons happen locally because the equations that describe how the observables evolve are local. That is a system A only changes in a way that is dependent on another system B if they interact with one another either directly or by interacting with some third system C that interacts with both.


"I know that much of quantum mechanics is not "intuitive". If anyone can explain why particles don't have a definite property, even before we measure them, I would be grateful. Thanks."

I think the short answer is, "Because, that's the way it is." We do the experiment and are then stuck with the result.

I always go back to the two slit experiment. How does an electron/particle go through both slits, and then interfere with itself?

  • $\begingroup$ "A particle interfering with itself" is not quite how I would describe the electron/double-slit interactions. If you run one electron at a time through the apparatus, it definitely lands on a screen as a single electron (a particle). Each electron does NOT provide a faint interference pattern. However, when a lot of electrons land on the screen, they distribute in a way that produces the expected interference pattern. Somehow, each electron has some type of interaction with both slits as it is going through one or the other slit, leading to an interference pattern. $\endgroup$ Commented Aug 3, 2015 at 20:06
  • $\begingroup$ There is a whole bunch of electrons going through the slits and producing the interference pattern. Not just one. If just one went through, there wouldn't be an interference pattern, but just a single spot on the screen. $\endgroup$
    – jjack
    Commented Aug 3, 2015 at 20:57

I'm really late to this party but I don't think the answers above answer what you are asking.

There are two concepts you are working with here. The first one is the fundamental indeterminism of quantum measurements and the other one has to do with non-locality.

  1. How do we know photons don't have a hidden variable that unambiguously determines the outcome of a measurement? I don't think people have addressed this question. Here's a great video explaining it. The gist of it is that if you assume they do, you cannot possibly get the results one gets in experiments measuring the polarization in a sequential manner.

  2. Non-locality. A lot of people seem confused about what the result of Bell's theorem means. They seem to imply that one cannot have a "local realist theory" as per Bell's inequalities being violated. So one needs to decide which one to throw away, realism or locality.

    This is bollocks. One cannot have a local theory period. A local theory in what sense? in the sense that choices on what to measure on photon A should not affect the result of photon B's measurement. However, this is exactly what is observed. Somehow, photon B knows what question Alice decided to ask her photon, even when that decision is done outside of B's cone of light.

Interestingly, when the measurement of B and Alice's choice are out of each other's cone of light, there is a reference frame in which first the measurement on B occurs and only then Alice chooses what to measure. However, in this reference frame we would interpret things as the choice performed by Bob having an effect on A's measurement. So which is the cause and which is the consequence?

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    $\begingroup$ Your conclusions are the exact opposite of the overwhelming consensus on the matter. You can have a local theory if you abandon realism, and the two photons don't need to "know" anything about each other once you've abandoned realism. $\endgroup$
    – ticster
    Commented Apr 17, 2018 at 20:05
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    $\begingroup$ I'm not familiar with that particular work, but what I can tell you is that while some non local interpretations of QM are out there, it is simply false to imply that quantum entanglement demands non localism. $\endgroup$
    – ticster
    Commented Apr 17, 2018 at 20:31
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    $\begingroup$ You seem to fundamentally misunderstand what correlation is. I'd encourage you to read my answer here $\endgroup$
    – ticster
    Commented Apr 17, 2018 at 21:39
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    $\begingroup$ Correlation is not causation. Nothing you do measuring one particle causes any change to the other measurement. The two measurements are simply correlated with one another. Your statement that one photon "knows" the state the other polarizer is in seems to imply that there is some measurement one could make on the photon to determine the state of the polarizer, which is decidely not the case. $\endgroup$
    – Chris
    Commented Apr 17, 2018 at 21:48
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    $\begingroup$ Again, if you give up on realism then it becomes entirely un-necessary to rely on the photons "knowing" about each other. If you don't see that, then all I can do is again point you to this explanation. There's no need to get defensive, I'm not trying to shut you up, I'm genuinely just trying to point out your mistake here. $\endgroup$
    – ticster
    Commented Apr 17, 2018 at 22:17

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