What counts as information? In e.g. the EPR experiment why is one entangled particle knowing instantaneously the state of the other not counted as 'information'.


Following a discussion in the comments to giulio bullsaver's answer I have decided to summarise what is my main confusion:

Alice and Bob are scaled down to the quantum level* so they have analogous properties to an electron. They are also entangled so Bob will always go into the opposite state to Alice.

Alice flies of to a classical measuring device $M_A$ and Bob to another classical measuring device $M_B$.

Alice's state, $|+\rangle$ or $|-\rangle$, gets measured first with outcome $|+\rangle$ (say). Due to Bell's inequities there are no local hidden variables, so Alice must decide at the point of measurement (and not before) what state she is going to go into.

Bob get's measured at time** $t$ after Alice but he has to be measured in a state opposite to Alice (due to the entanglement) i.e. $|-\rangle$. Bob must therefore know (on the point of measurement) what state Alice decided to go into so he can go into the opposite state.

The 'information' of Alice's state must therefore be transferred to Bob faster than the speed of light. To deny this statement would be in contradiction to the Aspect experiment.

*Ignoring practicalities

**designed so the two measurement events are separated by a space-like interval.

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    $\begingroup$ I think the broadest answer I can give is "it is impossible to design an experiment such that the experimenter can tell that anything has changed" but I think someone will have a better explanation of quantum information so I leave this as a comment until I get around to expanding it. $\endgroup$
    – zeldredge
    Commented May 28, 2015 at 15:31
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    $\begingroup$ The "ignoring practicalities" is not valid - if you scale Alice and Bob down to the quantum level, there is no state of "Alice" or "Bob" in the system "Alice and Bob" - that is precisely what entanglement is about: The lack of well-defined states of the subsystems for an entangled state of the combined system. Your argument for information being transmitted crucially rests on the existence of such states, though. $\endgroup$
    – ACuriousMind
    Commented Jun 4, 2015 at 18:57
  • $\begingroup$ @ACuriousMind Let me put this another way. We have one entangled state, (so 'Alice and Bob' rather then the two states 'Alice' and 'Bob') then information of the collapse of the wavefunction (which happens due to an event at $M_A$) must be passed to the part of the wavefunction at $M_B$ instantaneously, then I think the rest of my argument should hold. $\endgroup$ Commented Jun 4, 2015 at 19:06
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    $\begingroup$ There are no "parts of the wavefunction". There's just one (in this case manifestly non-local) quantum state. It seems to me your question has actually nothing to do with entanglement - take a photon and fire it at a beam splitter. It has some chance to go through, and some chance to be deflected - the wavefunction is here naively also "split into two parts" that travel in different directions, and (non-)detection behind the splitter also collapses "the other part". You're troubled about how the "other part know when/where to collapse", right? $\endgroup$
    – ACuriousMind
    Commented Jun 4, 2015 at 19:20
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    $\begingroup$ @Joseph: you wrote "so Alice must decide at the point of measurement (and not before) what state she is going to go into" : it is false, Alice can't select a state. It's why we cannot speak of information. $\endgroup$
    – user46925
    Commented Jun 4, 2015 at 20:23

6 Answers 6


The particle does not know anything. The information is what we (physicists, experimentalists) can communicate to someone else. Ask yourself: what can $A$ do, sharing an EPR pair with $B$, to tell something (let's say $0$ or $1$) to his friend $B$ after they have left each other (but still sharing at a distance the EPR pair), in a superluminal way?

You will understand that the answer is "nothing", and that the EPR pair is for this purpose as helpful as sharing a pair of gloves ($A$ has either the left one or the right one, B has the other and none of them knows who has what when they leave each other.). From what concerns locality the EPR pair and the gloves are the same, and so you will agree with me (since none has ever published a superluminal information transfer protocol based on a pair of gloves) that entanglement is fine with relativity.

The difference between EPR and gloves is just a quantitative one, the EPR allows a bit more correlation (see Bell Inequalities) and since Nature allows that too (see Alain Aspect experiment) the EPR is much more realistic than a description based on hidden variable (i.e. gloves-like situation).

Just another comment, that I've found very useful, do not think of the wavefunction as "something that is really there" so that an istantaneous collapse of it may appear as violation of locality (when $B$ measure his qubit he make collapse the entire wavefunction...). The wavefunction is a mathematical tool to predict probabilities, the probabilities are the only thing that are "really there" since we actually see them in experiments. And you will notice, $B$ cannot change the probabilities for $A$'s measurement.

See also the reference frame blog for very nice posts about this stuff.


I think the point of our disagreement is that you stick to a POV where the particle "chooses" a state after being measured and so it appears to you that the two particles have to exchange "information" in order to give coherent (correlated) answer when measured. Therefore you see a substantial difference from the classical case of gloves, where they already knew what they where (though the glove carriers didn't). This is what I think you think, but correct me if I am wrong.

Now, It seems to me that such a POV is a vestige of the so called “realism” of classical mechanics, i.e. That any physical observable has a definite value at any time. This pone a separation to what we, the experimenters, know and what the particle knows. So, for the case of the gloves, we do not know what glove it is, but the “glove” knows itself and so there is no FTL communication between them notwithstanding the same space-like perfect correlation (quoting your last comment).

In QM realism of course does not exists (as the Bell inequalities have shown) and so the above considerations have to be changed: No one, not even “the particle”, knows what will be the outcome of a measurement. Such a shift of paradigm requires to fix some terminology and boundaries on what is “physical” and what is not. The physical content of QM are the probabilities of measurement, that we , the experimenters, can predict and observe, nothing else. This is an idea deeply radicated in any QM theory we have about the world. Any different calculations or object that lead to the same probabilities may be thought as “redundant”, “not physical”. For example in QFT the fact that there is no privileged inertial reference frame is encoded in the fact that performing a Poincarè transformation correspond to act on states with an (anti-)unitary operator, that leaves unchanged the probabilities. To sum up: "we" are the experimenters, of a given QM system we can only predict and observe probabilities and nothing else, and causality must agree with this fact (and nothing else).

Having said this, I invite you to re-think about the EPR experience from this POV, where probabilities play the central role. The only thing that appear as “non-local” is that when A measures his particle he istantly changes the total wave-function of the two particles, from $|\mathrm{EPR}\rangle$ to $|0\rangle|1\rangle$ (if $A$ obtains $0$, $|\mathrm{EPR}\rangle$ is the usual entangled pair). It appears as non-local since the total wave-function involves both $A$'s and $B$'s particles and they are space-like separated. But the wave-function is not the true physical content of QM, only probabilities are. And $A$ cannot change any of $B$'s observed probabilities, so in no way $A$ can change the physics that $B$ will experience.

  • $\begingroup$ You say 'The particle does not know anything.' But it knows what state the other particle is in instantaneously. Granted it is not consciously there thinking this but it has to know otherwise we would not see the results that we get. $\endgroup$ Commented May 28, 2015 at 17:19
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    $\begingroup$ Ok but it is not what is meant by "information", we may say that the wavefunction contains information about the state of a sistem and so a measuring process, changing the wavefunction, changes the information about it. This is a completely reasonable statement (but it is not the focus in the EPR situation). What we mean by "information" in a EPR experience is something known by JUST ONE of the two experimenter that must be transferred to the other. And the point is that there is no paradox, entanglement does not aid you in anyway to do so in a superluminal way. $\endgroup$ Commented May 28, 2015 at 18:08
  • $\begingroup$ Maybe it could be useful for you to know that, if you compute the partial trace \rho_A = Tr_B[\rho], the result is the same no matter if B has already performed a measurement (provided that the outcome is not known to B, as it should be of course). $\endgroup$ Commented May 28, 2015 at 18:18
  • $\begingroup$ So even though the complete bipartite state changes after B measurement, for what concerns anything A can measure (with probabilities completely described by \rho_A) nothing change. So try not to think that "B has found 1 so now the particle of A is 0", think that "from the POV of A, B may have found 1 and A's particle is in 0 OR B may have found 0 and A's particle is in 1". But A has the same information about his particle than he had before an hypothetical measurement of B. $\endgroup$ Commented May 28, 2015 at 18:20
  • $\begingroup$ By contrast B, after his measurement, knows exactly what A will find if he measure too, but has no mean to use this fact to communicate anything to A. He is in the same situation he would have been into if they shared a pair of gloves: B looks at his glove and found it to be the left one, so instantly knows that A has the right one. But good luck in using this to communicate anything to him :) $\endgroup$ Commented May 28, 2015 at 18:20

Because they have less information that you imagine. They do not have separate bits of information describing them that gets somehow "transported" between them. From an information POV they are only one particle until measured and then they acquire separate but related information describing each separately.


"The phrase "entangled particle knowing instantaneously.." is a crude way to understand the phenomena. Consider a entangled state shared between Alice and Bob given by \begin{equation} |\psi\rangle_{AB}=\frac{1}{\sqrt{2}}(|z_+\rangle_{A}|z_-\rangle_{B}-|z_-\rangle_{A}|z_+\rangle_{B}) \end{equation} Alice can measure the spin of her particle in any direction. Then the state of Bob will reduce to a completely mixed state corresponding to the basis in which Alice measured her spin. For example, if Alice measured the spin along $x$ axis Bob's state will become \begin{equation} \rho_{B}=\frac{1}{2}\left(|x_+\rangle\langle x_+|+|x_-\rangle \langle x_-|\right) \end{equation} Similarly, if Alice measure her spin along $z$ axis Bob's state will reduce to a completely mixed state in $z$ basis. Hence, although we have mixtures of $2$ different kinds there are no objective difference between them. Which means one can not design any experiment to distinguish between completely mixed state in two different bases (consequence of GHJW theorem). Hence, Bob can not get any information on whether or not Alice actually measured her particle or not. Although the state shared between Alice and bob collapses after Alice's measurement Bob can not get any information about it. Hence, the information in this example or in EPR example it is not an usable information.


Let Alice and Bob shared $2n$ copies of the entangled state. Alice measured the spin in $\sigma_{x}$ direction for the n copies and get $|+\rangle$ or $|-\rangle$ as outcome. It is true that Bob's state will reduce to $|-\rangle$ or $|+\rangle$ respectively, which is really a mixed state of $|-\rangle$ and $|+\rangle$ . Now suppose Alice does measurement in $\sigma_{z}$ direction for the remaining $2n$ copies then Bob's state and get $|\uparrow\rangle$ or $|\downarrow\rangle$ then Bob's state will reduce to $|\downarrow\rangle$ or $|\uparrow\rangle$ respectively (again a mixed state of $|\downarrow\rangle$ and $|\uparrow\rangle$) . Now if Alice does not send the result of her experiment or the information about which basis she has done the measurement Bob cannot distinguish between these two mixed states statistically (consequence of GHJW theorem) i.e. Bob cannot do any type of measurement which distinguishes these mixed states in $x$ and $z$ direction. Because measurement outcomes are statistical property. Hence, although it seems like Bob's state has changed by faster than light signalling Bob cannot extract any information from his state.


In this answer we will stick totally to the Copenhagen interpretation of quantum mechanics.

  1. Before the system is measured, there is nothing physical (a consequence of realism not playing a part under this interpretation). There is only our mathematical, non-physical description of the possible measurement outcome called the wavefunction.
  2. On measurement the system is made to exist in a physical way. And is created in one of the eigenstates under which we are measuring.
  3. The system has never been in another state (since it did not exist before it was measured).
  4. Thus there is no need for one 'particle' to tell the other what state it is in because there is only one state which is physical.

One way to look at it is as follows:

Assume that in the EPR experiment, one entangled particle knowing instantaneously the state of the other not counts as 'information'. This assumption automatically entails that information has transferred from one point to another. Therefore, it has speed.

Now,considering EPR experiment and two parties involved. Alice makes a measurement A and Bob makes measurement B. Without a loss of generality, assume that Alice measures A first and then Bob measures. According to our assumption, Alice's outcome determines Bob's measurement outcome. Therefore, we get a cause-effect relation. However, this relation holds in the current reference frame. One could switch to some privileged reference frame where measuring B happens before A and then we get the effect determines the cause. Conducting the certain experiment, one can see in the second reference below, assuming the travel of information, one could achieve the lower bound for the "quantum information speed" to be $10^6C$, where $C$ is speed of light.

One could take a look at the following papers to learn more about the topic:

Testing spooky action at a distance D. Salart, A. Baas, C. Branciard, N. Gisin, H. Zbinden http://arxiv.org/abs/0808.3316


"Experimental Test of Relativistic Quantum State Collapse with Moving Reference Frames" H. Zbinden, J. Brendel, W. Tittel, N. Gisin http://arxiv.org/abs/quant-ph/0002031


Entanglement counts as information. In e.g. the EPR experiment one entangled particle "knows" the state of the other, because we humans tend to call entanglement "knowing" or "information".

After you have measured a particle, you "know something about the particle". In other words: After you have measured a particle, you and the particle are entangled.

When you "know" something about a particle, it means that if your idea about the state of the particle is examined, and the state of the particle is examined, it will be found that your idea about the state of the particle corresponds to the state of the particle, because there's that kind of entanglement between you and the particle.

Why do we say that one entangled particle knowing instantaneously the state of the other is not information? We say it for various reasons, quite often it's an attempt to set straight somebody dreaming about faster than light communication.

  • $\begingroup$ consider that you can replace any observer by a detector. The only remaining observer will record the answer to the question. In this case, will this observer be entangled whith an entanglement ( as observed statistical results ) ? $\endgroup$
    – user46925
    Commented Jun 8, 2015 at 12:58

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