I assume that entanglement emerges from quantum mechanics because the idea was around before experimental verification (e.g the EPR paper). How then does entanglement emerge from the theory (please provide a less technical answer if possible)

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    $\begingroup$ Both. If that wasn't the case, quantum theory would be a failed theory. Entanglement emerges in the theory because the spatial distance makes no difference to the tensor product structure of the Hilbert space of quantum mechanical multi-particle systems. $\endgroup$ – CuriousOne Dec 19 '14 at 7:10
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    $\begingroup$ @CuriousOne: you're a devil for giving good answers as comments. Can't you expand this to an answer as a Christmas special? :-) $\endgroup$ – John Rennie Dec 19 '14 at 7:21
  • $\begingroup$ @JohnRennie: Thank you for the roses... if I had a better answer for the OP than others here, I would love to write it down... but how does one explain the crucial differences between the way degrees of freedom combine in classical mechanics vs. QM without mathematics? I thought about giving an example similar to Harry Levine below, but felt that the message might get lost among the forrest of symbols? $\endgroup$ – CuriousOne Dec 19 '14 at 8:01

Entanglement is simply a particular kind of quantum multiparticle state: it happens to be the "most common" kind of state in the sense that if you choose a random quantum superposition from a multiparticle state space, it will almost surely be (in the measure-theoretic sense) entangled, so it's a little curious why entanglement takes some effort to observe in the laboratory.

The technical details, through a simple example. We think of several quantum "particles", each with three-dimensional quantum state spaces: let's take two of them. Let's number each individual particle basis states $1,\,2,\,3$, so a general superposition for one particle is the vector $\alpha\left.\left|1\right.\right>+\beta\left.\left|2\right.\right>+\gamma\left.\left|3\right.\right>$.

The quantum state space of the combined system has nine, not six, basis states. Let $\left.\left|j,\,k\right.\right>$ stand for the basis state where the first particle is in basis state $j$, the second in basis state $k$. You should be able to see that there are nine such basis states: $\left.\left|1,\,1\right.\right>,\,\left.\left|1,\,2\right.\right>,\,\left.\left|1,\,3\right.\right>,\,\left.\left|2,\,1\right.\right>,\,\left.\left|2,\,2\right.\right>,\,\cdots,\,\left.\left|3,\,3\right.\right>$.

Some states are factorisable, that is they can be written in the form $\psi_1\otimes\psi_2$ where $\psi_1$ and $\psi_2$ are individual particle quantum states. So, let $\psi_1=\alpha\left.\left|1\right.\right>+\beta\left.\left|2\right.\right>+\gamma\left.\left|3\right.\right>$ and $\psi_2 = a\left.\left|1\right.\right>+b\left.\left|2\right.\right>+c\left.\left|3\right.\right>$. Then, on noting that in our notation above we have $\left.\left|j\right.\right>\otimes\left.\left|k\right.\right>\stackrel{def}{=}\left.\left|j,\,k\right.\right>$

$$\psi_1\otimes\psi_2 = \alpha\,a\,\left.\left|1,\,1\right.\right>+\alpha\,b\,\left.\left|1,\,2\right.\right>+\cdots+\gamma\,b\,\left.\left|2,\,3\right.\right>+\gamma\,c\,\left.\left|3,\,3\right.\right>\tag{1}$$

The point about this state is that if we measure particle 2 and force it into its base state say $\left.\left|2\right.\right>$, then we know that the particle 1 must be determined by the part of the superposition in (1) that contains only basis vectors of the form $\left.\left|j,2\right.\right>$, because we know particle 2 is in state 2. So, from (1), the system must be in state $\psi_1\otimes\left.\left|2\right.\right>$, i.e. our knowledge about particle one has not changed with our measurement. Our measurement tells us nothing about particle 1, so particle 1 is independent of particle 2.

Now let's choose any old superposition: let's choose:


and now let's measure particle 2. If our measurement forces particle 2 into state $\left.\left|2\right.\right>$, then from (2) we know particle 1 is in state $\left.\left|2\right.\right>$, because the only term in (2) with particle 2 in state 2 has particle 1 in state 2. Likewise, if our measurement forces particle 2 into state $\left.\left|1\right.\right>$, then we know for the same reasons that particle 1 must be in state $\left.\left|1\right.\right>$. Our measurement of particle 2 influences particle 1.

I would advise you to work through this example yourself in detail. You will then understand the following: measurement of particle 2 influences the state of particle 1 if and only if the initial state of the two particle system is not factorisable in the sense above.

So you can see that entanglement is a natural theoretical consequence of the tensor product, which in turn is really the only plausible way one would expect many particle systems to behave. Experiment has reproduced and confirmed this theoretical behaviour.

You are right insofar that that entanglement was theoretically foretold and discussed in the EPR paper and also by Schrödinger shortly afterwards. Our word "entanglement" was Schrödinger's own translation of his name for the phenomenon, "Verschränkung".

NOTE: educators who use two two-dimensional spaces to illustrate the tensor product deserve to be cut up into teeny-tiny little bits and be forced to do the total time that all their students they have wasted in dead-end understanding in purgatory, or at least some way bad place if you don't believe in purgatory

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    $\begingroup$ Brilliant! I love your note at the end... I never knew why I was always bothered by the 2x2 examples... when it's bloody obvious why one shouldn't do it that way. $\endgroup$ – CuriousOne Dec 19 '14 at 8:12
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    $\begingroup$ This answer could profit from trying to factorize the entangled state to show, why it is not possible. $\endgroup$ – M.Herzkamp Dec 19 '14 at 9:55
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    $\begingroup$ Rod, I'm curious why you think that 2D Hilbert spaces are bad for illustrating the tensor product, pedagogically speaking. Is it because $2^2 = 2\times 2$, in contrast to $3^2 \neq 3\times 3$, for example? $\endgroup$ – Mark Mitchison Dec 19 '14 at 10:53
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    $\begingroup$ @MarkMitchison That's the reason. I got muddled myself when I first came across the idea, and then many years afterwards I tried to explain the idea to a work colleague using the two D example - STUPIDLY forgetting my own difficulty, and it took me ages to understand she wasn't getting that it was something different from the Cartesian product. She was no slouch either: once I had lead her down the wrong path, it took quite a bit of explaining to undo my damage - I felt just like I do playing "Pictionary", having given my playing partner a clearly dud lead. $\endgroup$ – WetSavannaAnimal Dec 19 '14 at 10:59
  • $\begingroup$ Yeah I see your point and I quite agree that could cause some difficulty. I'll have to remember that! $\endgroup$ – Mark Mitchison Dec 19 '14 at 11:04

The underlying framework of physics is quantum mechanical, this is the current level of physics development. Classical theories emerge from this underlying framework in a consistent and understood manner.

Quantum mechanics has differential equations that have to be solved for specific situations . The square of the solutions of these equations give the probability of finding a particle in a particular state or specific location in space.

The function that will describe the probability of how two particles relate to each other is a solution of the quantum differential equation for this specific situation. This solution will have specific quantum numbers which need to be balanced one particle against the other, in all the possible situations. Quantum numbers are conserved and within the solution, if one for example has spin up, the other will be spin down ( if the total spin is zero) and vice versa. A measurement will give a point in the probability distribution, and if one is found be spin up, it is immediately from the mathematics clear that the other will have to be spin down , without need for measurement ( the conservation of quantum numbers is a distillation οf innumerable measurements). The distance where the measurement happened is irrelevant, if the measured particle has not interacted on the way.

This is called entanglement as a shorthand of the above description . It is inherent in quantum mechanics and the experimental rules of conservation of quantum numbers that have been established with numerous experiments.


Consider a two-state system. Particle A can either be in state $|\!\uparrow\rangle$ or state $|\!\downarrow\rangle$. If there are two independent particles in the system, then each particle can either be $|\!\uparrow\rangle$ or $|\!\downarrow\rangle$ so we label the overall system states $|\!\uparrow\uparrow\rangle$, $|\!\uparrow\downarrow\rangle$, $|\!\downarrow\uparrow\rangle$, $|\!\downarrow\downarrow\rangle$. Note this is just for independent particles. But we could also supposed that the combined system is somehow restricted to be in either $|\!\uparrow\uparrow\rangle$ or $|\!\downarrow\downarrow\rangle$. It is perfectly feasible for the particles to be created or captured in such a way that these are the only two possible states to be in. Then the particles are considered "entangled:" the state of the system must be considered as a whole; it cannot be thought of as the combination (technically, the "tensor product") of multiple independent particle states.

This idea is purely a thought experiment, and was considered long before experimental evidence confirmed that such states are possible.


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