Let $\rho$ be a bipartite state. Its quantum mutual information is defined as $$\newcommand{\tr}{\operatorname{tr}}I(\rho) = S(\tr_B(\rho)) + S(\tr_A(\rho)) - S(\rho),$$ where $S(\sigma)$ is the von Neumann entropy of $\sigma$. If $\rho$ is pure, then it's easy to see that $I(\rho)=2S(\tr_A(\rho))$. If $\rho$ is also maximally entangled, then $I(\rho) = 2\log d$, with $d$ the dimension of the underlying space. This can also be written as $I(\rho)=2$, taking base-$d$ logarithms.

Now, let $\mu^A,\mu^B$ be some measurements for the first and second party. These will correspond to a joint probability distribution $p(a,b;\rho)\equiv \langle\mu^A_a\otimes\mu^B_b,\rho\rangle$. Let $p,p^A,p^B$ denote the corresponding probability distribution, and its two marginals, respectively. If $X,Y$ denote the random variables describing the corresponding outcomes, we can define a classical mutual information as $$I_{\mu}(\rho) \equiv I(X:Y)\equiv H(p^A) + H(p^B) - H(p),$$ with $H(p)$ the (classical) Shannon entropy of the probability distribution $p$. Note that the maximum value of the classical mutual information, obtained for fully correlated variables, is $\log d$ (without the factor of $2$).

When $I(\rho)=2\log d$, and thus $\rho$ is maximally entangled, the parties can observe correlations in infinitely many (projective) measurement bases: for any choice of measurement basis for $A$, there is a choice of measurement basis for $B$ which gives fully correlated outcomes. On the other hand, $I(\rho)=\log d$ can be obtained with the fully correlated mixed state: $\rho=\frac{1}{d}\sum_i |ii\rangle\!\langle ii|$, but said correlation can only be observed with a fixed choice of basis: unless the parties measure both in the computational basis, their outcomes won't be fully correlated.

Is the above intuition of any value in this context? In other words, can one think of the quantum mutual information being larger than its classical upper bound as being linked to the possibility of observing correlations in multiple measurement bases? Or is it just a red herring?

  • $\begingroup$ Shouldn't $I(\rho)=\log d$ instead of $1$ for the fully correlated mixed state? $\endgroup$ Oct 27, 2021 at 17:59
  • $\begingroup$ @AdamZalcman indeed, fixed, thanks $\endgroup$
    – glS
    Oct 27, 2021 at 19:22
  • 2
    $\begingroup$ I do think the intuition you describe is valuable in this context. My hunch is that the answer is some generalization of superdense coding where a shared quantum state allows one to send two bits of information instead of one. The excess bit is explained by entanglement and perhaps this is what's going to happen here, too. This is BTW a very interesting question that clarifies what it means that quantum correlations are stronger than classical ones. $\endgroup$ Oct 28, 2021 at 17:35
  • $\begingroup$ @AdamZalcman thanks. What I'm not sure about is how this works more generally for $1<I<2$. We know $I=2$ implies maximal entanglement, which is easy to characterise, But when $I<2$ things are trickier. In particular, I'm not sure when this freedom in measurement bases which give full correlations is or isn't there. Are there such examples for $I<2$? Can there be such examples for $I\le 1$? The discord is probably relevant to this discussion, but I don't enough about it to tell $\endgroup$
    – glS
    Oct 28, 2021 at 18:02
  • $\begingroup$ This paper connects quantum correlations with mutual information, and may be relevant: arxiv.org/pdf/quant-ph/0410091.pdf $\endgroup$ Nov 19, 2021 at 11:08


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