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Given two systems, correlations quantify how much information we can gain about system 1 by measuring system 2.

Entanglement is a type of correlations which only occurs in quantum mechanics.

There was discussion how this works, and Bell was able to show through his inequalities that it is not hidden variables responsible for it meaning that locality is violated.

I know when I have a entangled state like

$$\mid \psi \rangle = \frac{1}{\sqrt{2}}\mid 01 \rangle +\frac{1}{\sqrt{2}}\mid 10 \rangle$$

I can calculate the reduced density matrix

$$\rho_A = \mathrm{Tr}(\rho_{AB})$$

and then use the Von-Neumann entropy

$$S = -\mathrm{Tr}(\rho_A \mathrm{log}(\rho_A))$$

to quantify the entanglement present in $\mid \psi \rangle$. It's easy to see that entanglement is maximized if $\rho_A$ is maximally mixed.

What is here weird to me is the property that it seems that the less knowledge I have about the subsystems the more entangled they are.

What is even additionally confusing is the fact that this property seems also to be present for classical correlations.

Can someone give me maybe a nice analogy or something to help me to understand how less knowledge means more correlations/entanglement?

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What is here weird to me is the property that it seems that the less knowledge I have about the subsystems the more entangled they are.

I don't think this is very accurate. I guess you get to this conclusion because you interpret the entropy as lack of information (fair), and because the entanglement entropy $S$ is the entropy of the state you reach your conclusion.

However, note that $S$ is the entropy over the reduced state, not the total one. This means that it does not quantify the lack of knowledge about the state itself, but rather how much measuring one part of the state increases your uncertainty over the measurement results on the other part.

This makes sense because entanglement is all about information that is encoded and can be accessed only by operating on the full system, but is not accessible by only measuring its components. With the entanglement entropy you are therefore quantifying how much information is destroyed by operating locally, and therefore how "non-local" the system is.

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You cannot conclude that higher entropy of the reduced state implies correlations of the joint state. Consider the entangled state $$\vert\psi_{AB}\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)$$

and the maximally mixed state $$\rho_{AB} = I_A\otimes I_B$$

Both have the same reduced density matrices $\rho_A = \rho_B = I$, even though the maximally mixed state has no correlations between the subsystems while the maximally entangled state has perfect correlations between the subsystems.

If you restrict yourself to pure states, the only kind of correlation you can have is entanglement i.e. quantum correlations. Quantum correlations cannot be explained by a local hidden variable. So the partial state (which has all the local information you could ever extract) cannot be pure - it must be mixed. Indeed, the degree of mixedness is the amount of quantum correlations.

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  • $\begingroup$ Yes but obviously I was talking 1. Strictly about a pure global state $\mid \psi_{AB}\rangle$ and 2. one that is entangled in the sense that it cannot be rewritten in a separable way $\mid \psi_{AB}\rangle \ne \mid \psi_{A}\rangle \otimes \mid \psi_{B}\rangle$. $\endgroup$
    – CatoMaths
    Jan 18 '19 at 19:46
  • $\begingroup$ See edited answer $\endgroup$ Jan 18 '19 at 21:18
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The confusion here appears to be related to the meaning of statistical correlation in general rather than in quantum theory in particular. Consider two random variables $X$ and $Y$ that represent the outcomes of two measurements. The two variables are said to be correlated when learning the value of $X$ increases your knowledge about the value of $Y$. The amount of correlation is mathematically quantified by the mutual information. The point is that if you already know the value of $Y$ with certainty, then learning $X$ cannot increase your knowledge about $Y$ any further. Therefore, if $Y$ is not a random variable, then it is not correlated with anything else, by definition. The existence of correlations necessarily implies randomness.

In the context of quantum mechanics, this means that the marginal density matrix corresponding to a correlated state must be a mixed state. A pure reduced state (which has completely certain measurement outcomes in some basis) would imply zero correlations, as discussed above. On the other hand, a strongly correlated state has highly mixed marginals: in order for one's knowledge to increase, one must start out ignorant! However, as other answers have emphasised, the converse is not true: a random variable is not necessarily correlated with something else. Therefore, just because the reduced state is mixed, one cannot infer anything about the correlations: one needs to examine the full state for that.

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