Often when describing entanglement in an informal way, we talk about perfect correlation or anticorrelation of measurements of faraway particles, but I don't see how such correlation is particular to entanglement. Imagine I had a black ball and a white ball, I put them into two separate opaque boxes and I mix them up. One goes to Alice, who stays on Earth, the other goes to Bob, who leaves for the other side of the galaxy.

At some point Alice opens the box and finds the white ball (and she had a 1/2 probability of finding it), so she instantly knows that Bob will find the black ball. Obviously there is no entanglement here, this is just a correlated classical distribution

$$ \frac{1}{2}|bb\rangle\langle bb|+\frac{1}{2}|ww\rangle\langle ww|$$

where $b$ stands for black and $w$ for white. If instead the balls were particles in a maximally entangled state $\frac{1}{\sqrt{2}}(|bb\rangle+|ww\rangle)$, we can think of $b$ as spin up and $w$ as spin down. Then the state would be

$$ \frac{1}{2}|bb\rangle\langle bb|+\frac{1}{2}|ww\rangle\langle ww|+\frac{1}{2}|bb\rangle\langle ww|+\frac{1}{2}|ww\rangle\langle bb| $$

If Alice and Bob repeat the same experiment with these entangled balls, they will observe the same correlation, each time Alice measures $b$, Bob will measure $w$, and vice versa.

What is then the observable difference between these two distributions?


2 Answers 2


This is an excellent question with a subtle answer. To answer it, let us consider the following two states:

\begin{align} |\psi \rangle_{\textrm{bw}} &= \sqrt{p}\,|bb \rangle + \sqrt{1-p}\,|ww \rangle \, , \\ \rho_{\textrm{bw}} &= p\,|bb\rangle\langle bb|+ (1-p)\,|ww\rangle\langle ww|\,. \end{align}

Both states are correlated, but they are fundamentally different. The first is a pure state. There is no uncertainty associated with the global state. And it is also entangled. The second, on the other hand, is just a statistical mixture of two possibilities. It has no entanglement nor any other type of quantum correlations. However, in both cases, the probability that Alice finds the black ball is $p$. And, in both cases, if she does find it, then the state of Bob will be updated to $|b\rangle$. So in this sense, it seems these two states behave quite similarly.

The state $\rho_{\textrm{bw}}$ represents our degree of ignorance about the configurations of Alice and Bob. We don’t know in which configuration they are, $|bb\rangle$ or $|ww\rangle$. If Alice measures and happens to find out, then we update our information. The state $|\psi\rangle_{\textrm{bw}}$, on the other hand, contains no ignorance at all. We know exactly which state the two "ball-bits" are and the randomness associated with $|\psi\rangle_{\textrm{bw}}$, has nothing to do with ignorance. It is intrinsic.

Let me know if I answered your question.


  • $\begingroup$ Hi Alex, I understand the mathematical difference between the two states, and also the interpretation of mixed states as classical mixtures of pure states. What I was asking about it, what does this difference entail operationally? It seems that these two situations give the same results experimentally, what experiments could Alice and Bob do to distinguish between $|\psi\rangle_{bw}$ and $\rho_{bw}$? $\endgroup$ Jun 24, 2020 at 11:28
  • $\begingroup$ Well, the easiest experiment without carrying about entanglement in what you can think is in the Stern-Gerlach experiment. If initially, you let only spin up particles to be selected and measure the observable $Oz$, the only outcome is particles prepared with spin up. But now, if you measure the observable $Ox$ and then $Oz$ again the outcome won't be only spin up, but also spin down. $\endgroup$
    – Alex
    Jun 24, 2020 at 13:10

Bell's theorem is a lot more subtle. First replace the coloured balls with gyroscopes, and assume that when you measure the spin direction of a gyroscope you also give it a kick, such that the gyroscope automatically aligns with the apparatus used to measure it, with a probability for aligning either way which depends on the angle between prior gyroscope spin and the orientation of the apparatus. The two observers can measure spin on any axes. Bell's inequality then shows that the probabilities found for the correlations of spin in quantum mechanics are not explainable by any classical mechanism such as gyroscopes.

  • $\begingroup$ What you are saying is that there is no operational difference unless we consider multiple measurement bases, and that the correlations between measurements in different bases would violate some Bell inequality. So Bell violations are what's special about entanglement, not correlation between measurement results, am I correct? $\endgroup$ Jun 24, 2020 at 11:31
  • 1
    $\begingroup$ Yes, it should not be surprising that there is a correlation between events which have common cause, (typically particles created with an equal and opposite property, e.g. opposite momentum or opposite spin). The issue is that the exact correlation for many measurements cannot be explained classically, but this only applies to measurements for which there is no classical explanation anyway. For the case you gave there is indeed no operational difference. $\endgroup$ Jun 24, 2020 at 18:21

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