# What exactly is the difference between entanglement and correlations?

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?

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 donâ€™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.

• 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 $\rho_{bw}$? – user2723984 Jun 24 at 11:28
• 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. – Alex Jun 24 at 13:10