What is quantum entanglement?

Please be pedagogical.

Edit: I have updated my background under my profile.

  • $\begingroup$ physics.stackexchange.com/q/54975 $\endgroup$
    – chharvey
    Commented Jul 5, 2015 at 3:40
  • $\begingroup$ As to Lubos good answer: Is this another way of saying that if a set of red dice has the same manufacturing defect (e.g. making them prone to show the 6-face) from company A but the white dice from company B are perfect then the red dice are all "entangled" as to this property but not the white? Assuming a casino in Monaco uses a red and a white at each throw and another casino in Mexico does the same then a record of the throw results at each casino would be cross-correlated and the individual records would be simply correlated. $\endgroup$
    – Jens
    Commented Feb 11, 2016 at 7:00
  • $\begingroup$ Possible duplicate of Quantum Entanglement - What's the big deal? $\endgroup$ Commented Jan 23, 2017 at 15:24

3 Answers 3


Entanglement is a quantum correlation between two (or many) objects - a correlation means that these two objects' properties are not independent of each other - which was created in the objects' common past when they were close to one another i.e. when they were two parts of the same physical system.

Quantum mechanics changes the character of possible "properties" that objects may have (the quantities describing objects's properties are usually called "observables" and they are represented by Hermitian operators on the Hilbert space), as well as the way how these properties are measured and predicted (just probabilistically), so it also changes the character and magnitude of correlations that the objects may exhibit.

In particular, quantum correlations may often be stronger - and affecting a large fraction of measurable properties of the objects - than what would be possible according to classical (i.e. non-quantum) physics. In classical physics, correlations have to satisfy e.g. the so-called Bell's inequalities in various situations but quantum mechanics - and the real world - can easily surpass these bounds.

Technically, objects and their properties in quantum mechanics are described by wave functions. To describe the state of two mostly independent objects, one has to take a wave function from the tensor product $H_1\otimes H_2$ of the Hilbert spaces describing the individual objects. The wave function in the tensor product implies probabilistic predictions for any pairs of properties of the first and second object; in general, they're not independent, and for each combination of the objects' properties, quantum mechanics (and the wave function) may remember an independent probability.

Any vector in the tensor product that can't be written as a tensor product of vectors from $H_1$ and $H_2$ (instead, it can only be written as a linear combination of such tensor products of vectors) is called entangled. In other words, it is non-entangled if it is a simple tensor product of two simpler vectors. If it is a simple tensor product, all probabilities of "coupled properties" of the pair of objects simply factorize to the probability of the first object, and probability of the second object, as you know from probabilities of independent phenomena.

The simplest pedagogical example of an entangled state (as well as the entangled state that is most often found in literature) is $$\frac{X_1\otimes Y_1 + X_2\otimes Y_2}{\sqrt{2}}$$ Because there are two terms with four different factors, you can't use the distributive law in any way that would allow you to rewrite it as a simple product. The letters $X,Y$ refer to the two objects and the labels $1,2$ refer to two different states of each of the two objects.

In this state, if the property "1 or 2" is measured on $X$, one obtains the answers 1 or 2 with 50% probability for each: the coefficient of the wave function is $1/\sqrt{2}$ because these complex coefficients have to be squared to obtain the probability. However, because $X_1$ is "coupled" to $Y_1$ and $X_2$ is coupled to $Y_2$, the state and the machinery of quantum mechanics predict that the object $Y$ will be measured to have the same property: if $X$ is in 1, $Y$ is in 1, and the same for the state 2.

Linear algebra - which is crucially important for quantum mechanics - allows one to reinterpret the state above as an "identity operator" so the correlation will exist regardless of the type of measurement that we perform both on $X$ and $Y$. For example, if the two states represent spins, the two particles will be correlated so that you will find out that they're polarized with respect to the same axis, if you measure both particles' polarizations with respect to the same, particular, but arbitrary axis.

This would be kind of impossible for two separated particles in classical physics that could only be perfectly correlated for one choice of the axis - but not another axis rotated by 45 degrees, for example - without a communication in between them. However, quantum mechanics predicts that such a 100% correlation "regardless of the axis" is not only possible but guaranteed by the state above and it requires no communication. Indeed, one can prove that relativistic theories in quantum mechanics - especially quantum field theory - don't allow one to transmit a single bit of information faster than light even though this would be needed in classical physics to guarantee the perfect correlation that quantum mechanics predicts for these experiments (and that the experiments confirm).

  • $\begingroup$ Thanks for good news that it was at an OK level. It was a pleasure. $\endgroup$ Commented Jan 21, 2011 at 13:40
  • $\begingroup$ @Lubos What is the other word for "object" in this context? As I have understand depending on framework you can talk about wave functions and particles. Dose quantum entanglement apply for both? $\endgroup$ Commented Jan 21, 2011 at 13:52
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    $\begingroup$ Just a minor comment: two particles can be entangled even if they never interacted in the past. The usual example is entanglement swapping. $\endgroup$ Commented Jan 21, 2011 at 13:58
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    $\begingroup$ Dear Amir, by an "object", I mean an electron, another particle, atom, molecule, photon, apparatus, human, or anything else. The state of objects (such as particles) is described by wave functions (or density matrices) - that's the relationship between them. So all these discussions are about both - objects and the wave functions. The wave functions are necessary to see what is exactly going on, at the mathematical level, but the purpose of mathematics is to say something about real objects we can observe in the reality. ... Thanks, space_cadet $\endgroup$ Commented Jan 21, 2011 at 18:38
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    $\begingroup$ Dear Anthony, OK, so the objects in entanglement swapping didn't have to interact with one another, but they had to interact with some other objects that interacted with each other :-). I just want to say that you can't create entanglement in the first place out of nothing. $\endgroup$ Commented Jan 21, 2011 at 18:39

A spin system has in the basis of the Pauli matrix $\sigma_z$ the states $|+\rangle$ and $|-\rangle$ for spin up and down. The Pauli matrix acts on these states as $$ \sigma_z|\pm\rangle~=~\pm|\pm\rangle. $$ Now these states are complex numbers, which means there are $2$ variables for each state and thus $4$ altogether. However, there are constraints, such as the probability Born rule $1~=~P_+~+~P_-$, $P_\pm~=~|a_\pm|^2$ for a state $|\psi\rangle~=~a_+|+\rangle~+~a_-|-\rangle$, and irrelevance of a phase in real valued measurements. So this reduces the number of variables from $4$ to $4~–~2~=~2$. That is just what we would expect.

Now let us consider two spin systems, say two electrons. The use of electron spin state is not concrete, for these arguments hold just as well for polarization direction of photons. So we have two sets of states and operators $\{\sigma_z, |\pm\rangle\}^1~ \{\sigma_z, |\pm\rangle\}^2$ denoted with an additional index $i~=~1,~2$ and we still have $$ \sigma ^i_z|\pm\rangle^i~=~\pm|\pm-\rangle^i. $$ We can form two independent states $|\psi\rangle^i~=~a^i_+|+\rangle^i~+~a^i_-|-\rangle^i$ for the two spin systems. For each there are $4$ variables and $2$ constraints. This gives $4$ degrees of freedom in total. Yet we can compose these spin states in various ways. One way of doing this is $$ |\psi\rangle~=~{1\over\sqrt{2}}(|+\rangle|-\rangle~+~e^{i\phi}|-\rangle |+\rangle), $$ where I have dropped the index $i$, and we just implicitly see the first and second $|\pm\rangle$ as $i~=~1$ and $2$. This makes reading things clearer. The $e^{i\phi}$ is a phase which for it equal $+$ and $–$ the state $|\psi\rangle$ is not an eigenstate of $\sigma^i$ and is an eigenstate of $\sigma ^i$ respectively. So these are singlet and triplet state configurations. This is an entangled state. If you have access to $|\pm\rangle^1$ then you also have access to $|\pm\rangle^2$, and this holds no matter how far apart these states end up as. You can entangle two electrons by overlapping their wave functions. One that is done you can separate them arbitrarily far and they are still entangled.

Now let us count the degrees of freedom for this state. We have again $4$ variables for each $|\pm\rangle^i$ but now we have one constraint from Born rule and another from the entanglement state. So you have $6$ independent variables with $4$ constraints giving $2$ in total. This is the basic bipartite entanglement. There are also n-partite entanglements, such as the W and GHZ states.

  • $\begingroup$ Not terribly "pedagogical" but very helpful nevertheless. +1 $\endgroup$
    – user346
    Commented Jan 21, 2011 at 16:02

Quantum entanglement is a quantum mechanical phenomenon in which the quantum states of two or more objects have to be described with reference to each other, even though the individual objects may be spatially separated.

  • $\begingroup$ That is wrong - classical systems can also show this phenomenon (imagine a system with correlations). The point of entanglement is that the correlations can be much stronger (in a sense). $\endgroup$
    – Martin
    Commented Apr 17, 2015 at 13:14
  • $\begingroup$ It would be nice if you could build on this answer, since this much is already subsumed in the existing answers (which are years old). Do you have anything to to add to them? $\endgroup$
    – 299792458
    Commented Apr 17, 2015 at 13:18

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