# Why this is classical correlation and not full (classical + quantum) correlation?

Let a quantum system be given which has two subsystems $$A$$ and $$B$$ so that the Hilbert space decomposes $$\mathscr{H}\simeq \mathscr{H}_A\otimes \mathscr{H}_B$$.

If the state of the system is $$\rho$$, one defines the classical correlation with respect to measurements on $$B$$ as

$$J_{\overleftarrow{AB}}(\rho)=\max_{\{1\otimes \Pi_i\}}\{S(\rho_A)-\sum_i p_i S(\rho_A^i)\}$$

where the maximum is taken over all measuremenst on $$B$$ with probabilities $$p_i$$ and with post-measurement states $$\rho^i$$, and where $$\rho_A = \operatorname{Tr}_A(\rho)$$ and $$\rho_A^i=\operatorname{Tr}(\rho^i)$$.

The idea here is simple:

1. Prior to measurement the state of $$A$$ is $$\rho_A$$ and the uncertainty on its specification is $$S(\rho_A)$$.

2. If upon a specific measurement $$\{1\otimes \Pi_i\}$$ is conducted on $$B$$ one has obtained result $$i$$ the post-measurement state of $$A$$ is $$\rho_A^i$$ and the uncertainty on its specification is $$S(\rho_A^i)$$. Therfore, the uncertainty on the state of $$A$$ has reduced by $$S(\rho_A)-S(\rho_A^i).$$ As such, the mean information about the state of $$A$$ gained in the process of measurement is $$\sum_i p_i (S(\rho_A)-S(\rho_A^i))=\sum p_i S(\rho_A)-\sum p_i S(\rho_A^i)=S(\rho_A)-\sum p_i S(\rho_A^i).$$

3. Finally, we may say that the maximum information we can obtain about $$A$$ by measuring $$B$$ is the maximum of such quantity over all measurements on $$B$$, which is $$J_{\overleftarrow{AB}}(\rho)$$.

That's fine, but why this is called "classical correlation" and not the total correlation?

I mean, for me correlation means the following

If whenever we observe $$B$$ we gain knowledge of $$A$$ then they are correlated.

The only way to observe $$B$$ is to measure something.

So why this quantity is classical correlation and not full correlation? In what sense quantum correlation differs from this and how then quantum correlations agree with the usual meaning of correlation?

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• can you add a reference for this usage of the term? – glS Feb 12 at 9:37
• Sure, one such reference is arxiv.org/abs/1006.2460. Many of the cited papers there also use it. – user1620696 Feb 12 at 10:22
• I'm a bit confused, shouldn't this definition give different results for example when considering $\lvert00\rangle\!\langle00\rvert+\lvert11\rangle\!\langle11\rvert$ vs $\lvert\Psi\rangle\!\langle\Psi\rvert$ with $\lvert\Psi\rangle=\lvert00\rangle+\lvert11\rangle$? According to this definition both have the same value of $J(\rho)$ ($\rho_A$ are the same and measuring in the computational basis we can get $\rho_A^i$ pure and thus $S(\rho_A^i)=0$), even though the way the states are correlated is clearly different – glS Feb 13 at 16:53
• @glS I think the idea behind this defintion is that both the maximally entangled and maximally mixed state have the same classical correlation (zero). The difference is only due to quantum correlations. Perhaps we need a different pair of states to try? – user1936752 Feb 15 at 10:44
• en.wikipedia.org/wiki/Quantum_discord - for an introduction to the distinction between classical and quantum correlations – nr2618 Feb 15 at 11:07