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This question already has an answer here:

Can quantum entanglement make sending a message, whether audio, video, or even Morse code, instantaneous between two points (faster than it could travel normally at the speed of light)?

Let me first be clear, my question is not whether a message can travel faster than light. I simply mean whether a message can be generated here and then a copy of that message heard over there sooner than the message would normally be received.

Say a large number of entangled pairs are generated. One half the particles are given to Jack and the opposite ones to Sally. Jack then hops on a ship and travels 8 months to Mars. Sally then manipulates her particles in the form of a message. Do the corresponding opposite particles get manipulated simultaneously so that Jack receives the message instantly instead of the 16 light-minutes for a classically transmitted message?

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marked as duplicate by DanielSank, ACuriousMind, CR Drost, John Rennie, HDE 226868 Sep 2 '15 at 18:12

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You have a wrong understanding of quantum entanglement.

What is entanglement?

Quantum entanglement emerges naturally from the "only obvious way to do things" at the wavefunction level (distribute a wavefunction over all possibilities of two subsystems), and describes the fact that the general state of these systems cannot be "decoupled" into a pair of states for each of the subsystems.

Let me give you one example, qubits (quantum bits). These are variables which are allowed to be in the states $|0\rangle,\,|1\rangle,$ and any quantum superposition of them $\alpha|0\rangle + \beta|1\rangle$ where $\alpha,\beta\in\mathbb C$ and $|\alpha|^2 + |\beta|^2 = 1.$ All of those superpositions are "non-classical": there are experiments you can do on them which display strange outcomes.

The most famous strange outcome is to quantum-coherently map them to two different displays on a detector, the so-called "double slit experiment." If $|0\rangle$ becomes $|f_0(x)|^2$ and $|1\rangle$ becomes $|f_1(x)|^2$ then the state $\sqrt{\frac 12}|0\rangle + \sqrt{\frac 12}|1\rangle$ typically becomes $\frac 12 |f_0(x) + f_1(x)|^2,$ showing an "interference pattern". So if $|f_0(x)|^2$ is a bell curve and $|f_1(x)|^2$ is another bell curve, and $f_0$ and $f_1$ have complex phases which are not exactly the same, and the bell curves overlap, then in the overlap region you see some sort of "waviness" (intensity inhomogeneity) which is distinct from the "classical" overlapping-bell-curves result $\frac 12 |f_0(x)|^2 + \frac 12 |f_1(x)|^2.$

In the algebra of quantum mechanics, this comes from an "expectation value", which comes from flipping these symbols into their "duals"; I'm going to take that as a little too in-depth for this comment.

Now if you have two qubits, the general state becomes $\kappa |00\rangle + \lambda |01\rangle + \mu |10\rangle + \nu |11\rangle$ which is not necessarily representable in terms of a product of two individual states, $$(\alpha |0\rangle + \beta |1\rangle)\otimes(\gamma |0\rangle + \delta |1\rangle) = \alpha\gamma |00\rangle + \alpha\delta |01\rangle + \beta\gamma |10\rangle + \beta\delta |11\rangle.$$In particular, this is only possible when $\kappa\nu = \lambda\mu$ (they are both $\alpha\beta\gamma\delta$) but there is no reason to require that in the above "general" quantum state. So, we don't. That's where entanglement comes from; you have states which do not have this property.

Entanglement destroys coherence

Obviously, the product-states have a "quantum coherence" to both qubits: doing our double-slit experiment means that we see an interference pattern. Shockingly, entanglement weakens and sometimes eliminates this interference pattern. For example, the state $\sqrt{\frac 12} |00\rangle + \sqrt{\frac 12}|11\rangle$ describes an entangled state. If you pass the first qubit of this through the double-slit experiment, normal rules of quantum mechanics give the distribution $\frac 12 |f_0(x)|^2 + \frac 12 |f_1(x)|^2:$ classically overlapping bell curves!

This state also works a little like a classical hidden variable: two bags where I either put a toy in both of them or none of them. When you open up your bag, you do not know what you're going to get, but if there's a toy you know that the other person has a toy, or if there's no toy then the other person has no toy.

Entanglement can allow instantaneous quantum effects "over there".

Now suppose we split up the two qubits in this entangled $|00\rangle + |11\rangle$ state, where we've established that Alice is going to measure two overlapping bell curves with their double-slit experiment.

Suppose Bob likes wavy interference patterns. The rules of quantum mechanics allow Bob to do, on his qubit, any unitary transformation like $$\begin{align} |0\rangle\mapsto& \sqrt{\frac 12} |0\rangle + \sqrt{\frac 12} |1\rangle\\ |1\rangle\mapsto& \sqrt{\frac 12} |0\rangle - \sqrt{\frac 12} |1\rangle. \end{align}$$ This takes our state to:$$\sqrt{\frac 14} |00\rangle + \sqrt{\frac 14} |01\rangle + \sqrt{\frac 14} |10\rangle - \sqrt{\frac 14} |11\rangle $$ Now supposing that Bob measures his qubit as 0 or 1, then Alice must measure either the wavy interference patterns $\frac 12 |f_0(x) + f_1(x)|^2$ or $\frac 12 |f_0(x) - f_1(x)|^2.$ Bob can thereby instantaneously change, from a quantum perspective, what the outcomes of Alice's measurement are going to be.

Alice's wavefunction must change instantaneously and might even change retroactively: she may have already measured her qubit before Bob does this unitary transformation and measurement: nevertheless, to satisfy the predictions of quantum mechanics, her measurements must be consistent with Bob's manipulations.

But that can't send messages.

This thing that Bob has done is not directly visible to Alice, however. That's for a couple reasons, the first being that this only generates one photon of results on the double-slit screen, which isn't enough to see the pattern! But suppose we measure lots and lots of these qubits to try and see the pattern: then the problem is that Alice doesn't know which ones Bob measured as 0 or which ones Bob measured as 1. Since there was a 50/50 chance of Bob getting either, what Alice sees is therefore:$$\frac 14 |f_0(x) + f_1(x)|^2 + \frac 14 |f_0(x) - f_1(x)|^2 = \frac 12 |f_0(x)|^2 + \frac 12 |f_1(x)|^2.$$Alice therefore still measures two overlapping bell curves, overall!

Where are the interference patterns?! That is very simple: when Bob and Alice compare their measurements in the first case, Bob's 0-measurement can be used to "filter" Alice's patterns into $\frac 12 |f_0(x)|^2,$ the bell curve of photons which passed through only the first slit, and his 1-measurement filters the results to give $\frac 12 |f_1(x)|^2,$ the photons which passed through only the second slit.

Bob's transformation then changes how he can filter Alice's patterns: Alice's overlapping bell curves are now made up of the ones he measured $0$ for, which describe one wavy pattern, and the ones he measured $1$ for, which describe the other wavy pattern, and they add up into the non-wavy pattern.

And that's a general property of entanglement

What I've described is only one particular experiment, but its sneaky way of avoiding information-transfer is actually a very deep property of entanglement.

Entanglement manifests in "spooky" (non-classical) correlations of measurements, not in the measurements themselves.

You cannot observe the correlations without bringing the measurements back together, which is why you cannot transfer classical information via entanglement. You don't know anything about entanglement until you compare the two systems to see how they correlated.

Quantum teleportation, for example, uses an entangled state to "instantaneously" send an arbitrary quantum state from point A to point B. At least, that's the story. In fact, quantum teleportation sends it in a "garbled" way, and A must send a couple classical bits to B so that they can "ungarble" it and recover the full quantum state: these classical bits are absolutely necessary because the entangled state is fundamentally a correlation. It is still impressive that the arbitrary quantum bit, with its two-continuous-parameters of data, can be sent "mostly" through a quantum entanglement, reducing the actual data from "two infinities of bits" to just two bits, but it's not as magical as "teleportation" sounds to the layperson.

So that's the fundamental problem. Entanglement first destroys the quantum coherence (giving you something classical-looking) and then all of the things that you want to do with it happen within that classical veneer, and their quantum nature is only apparent when you compare two systems and say, "holy crap, you can't do that classically."

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  • $\begingroup$ I love you man! "Entanglement destroys coherence" this is the first time I hear this bit of information and it explains a lot of problems I have had. Would you say this is an oversimplified but good way to describe this: "Each particle is entangled with itself, until it either gets measured, or split into two. As long as its entangled with itself it will cause interferrence patterns and such, else it will not?" $\endgroup$ – Torge Sep 3 '15 at 21:41
  • $\begingroup$ @Torge I toyed around with that idea in college but I decided ultimately that I couldn't define easily a measurement of how "coherent" or "entangled" a subsystem was such that I could state some sort of "law of conservation of entanglement." $\endgroup$ – CR Drost Sep 3 '15 at 21:47
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No matter what spin measurement Jack performs on his particles, he's going to see half "up" and half "down" (or more precisely, each particle has a 50/50 chance of being up or down). This is so no matter what Sally chooses to do. So there's no way for her to send him a message.

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    $\begingroup$ Even if each one is isolated from the others? Say you have pair A which includes A1 and A2 particles and pair B with B1 and B2 etc. Jack has all the "2" particles and Sally has all the "1" particles. Say these are all kept isolated from each other. So then Sally manipulates the H1, E1, and Y1 particles. Does Jack see the manipulation in his H2, E2, and Y2 particles? $\endgroup$ – SkyPharaoh Sep 2 '15 at 11:39
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    $\begingroup$ Didn't I already answer this? $\endgroup$ – WillO Sep 2 '15 at 12:50
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Quantum entanglement can't be used to send any information faster than light. Suppose you have two electrons with entangled spins, and that you can measure the spin along the x,y or z directions. Regardless of which direction you measure half of the results will be spin up, the other half spin down.

If you measure the spin for both electrons in the same direction, then you will find that the results are opposite when you compare them. For example, if you measure the z direction for both electrons, when you compare them you will see if one is spin up, the other is spin down, and vice versa.

If you measure in two different directions, the probability of finding a match when you compare them will be 1/2. For example, if you get spin up for one electron, you will find the other can spin up or spin down, each with probability 1/2.

So the correlations depend on whether the measured quantities match. And if you do the maths you find that there is no local explanation in which the descriptors of the system are represented by stochastic variables, i.e. - a single number picked out of a hat with probability 1/2.

However, quantum mechanical systems aren't described by stochastic variables. They are described by Hermitian operators (Heisenberg picture observables), each of whose eigenvalues represents a possible measurement result. For each electron, both of the possible results of a measurement on continue to exist after the measurement. The reason why you don't see both results is that decoherence prevents interference between them. The correlation is established when the results are compared, not at the time of the measurement:

arxiv.org/abs/quant-ph/9906007

arxiv.org/abs/1109.6223.

The observables change entirely locally and so there is no prospect of non-locally transferring information from one system to another by entanglement.

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FTL Communication with Quantum Entanglement?

Sorry, no. I know people talk about this sort of thing and others find it very exciting. But there's an element of deception I'm afraid. There are professional scientists who "bait" their papers and articles with fantastic claims in order to attract media and public attention, and increase their profile and further their careers. You can't trust everything you read. Some of it is "woo". You might think that woo comes only from religious types and obvious nutters, but it isn't true.

Can quantum entanglement make sending a message, whether audio, video, or even Morse code, instantaneous between two points (faster than it could travel normally at the speed of light)?

No. Note how WillO said the answer was no, and your response indicated that you wanted the answer to be yes? The people who peddle woo rely on this. Like medieval bishops, they know that you want to believe.

Let me first be clear, my question is not whether a message can travel faster than light. I simply mean whether a message can be generated here and then a copy of that message heard over there sooner than the message would normally be received.

The answer is still no I'm afraid. There is no magic.

Say a large number of entangled pairs are generated. One half the particles are given to Jack and the opposite ones to Sally. Jack then hops on a ship and travels 8 months to Mars. Sally then manipulates her particles in the form of a message. Do the corresponding opposite particles get manipulated simultaneously so that Jack receives the message instantly instead of the 16 light-minutes for a classically transmitted message?

No. Now you can read articles that suggest otherwise, such as this one in Nature. But when you dig into it, you always find that the claims melt away, and there's no actual evidence of any FTL messages being sent. IMHO it's always like that for entanglement and spooky action at a distance. There's just no substance to it. Then when you dig deeper what comes as a surprise is the way historic stuff like the Einstein-Rosen paper is nothing like what you were led to believe. See J.S. Bell's Concept of Local Causality by Travis Norsen who makes it clear that Bell didn't think what people say: "Many textbooks and commentators report that Bell's theorem refutes the possibility of supplementing ordinary quantum theory with additional ("hidden") variables... On this view, Bell's theorem supports the orthodox Copenhagen interpretation. Bell's own view of his theorem, however, was quite different".

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