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Let $\rho^{XA}$ be a classical-quantum state of the form $$ \rho^{XA} = \sum_{x\in X} p_x |x\rangle\langle x|\otimes \rho_x^A, $$ and let the accessible information be given by $$ I_{acc}(\rho^{XA}) = \sup_M I(X:Y), $$ where $M = \{M_y\}$ is a POVM and $Y$ is a random (classical) variable with joint probability distriubtion $\Pr(X = x, Y = y) = \operatorname{Tr} M_y p_x\rho_x$.

I have seen it referenced in papers (e.g. Uncertainty, Monogamy and Locking of Quantum Correlation, Proposition 6) that the accessible information is additive, that is $$ \frac{1}{n}I_{acc}(\rho^{\otimes n}) = I_{acc}(\rho), $$ however, they reference a russian paper by Holevo (A. S. Holevo. Information theoretical aspects of quantum measurements.), and from what I gather (I don't speak/read russian) he in fact shows $$ I_{acc}(\rho^{XA}) \leq I(X:A)_\rho. $$ Is additivity actually easy to see from this result, or am I missing something entirely?

I would appreciate any help.

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I'm assuming you are referring to this paper:

Uncertainty, Monogamy, and Locking of Quantum Correlations.

In proposition 6, it's not clear to me if they are referring to the same product state that you are considering. However, in that paper, where they mention the additivity of accessible information, along with the Holevo paper, they also cite the following paper:

Quantum Data Hiding.

In this paper, in section 7 part A, they explain the additivity of mutual information. My guess is, the additivity of accessible information would follow from that.

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  • $\begingroup$ Thanks, they do indeed explain additivity of mutual information, when you have all quantum-to-classical channels at hand! I am not quite able to follow the bound of the second term on the right hand side of the inequality at the very top of page 10 in "Quantum Data Hiding". Any chance you could explain? $\endgroup$
    – user16106
    Jun 11 '21 at 11:51
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$I_{acc}(\rho^{XA})=S(\rho_{A})-\sum_{x}p(x)S(\rho_{A}^x)$

$I(X:A)=S(\rho_{x})+S(\rho_{A})-S(\rho_{XA})=S(\rho_{x})+S(\rho_{A})-(S(\rho_{X})+\sum_{x}p(x)S(\rho_{A}^x))=S(\rho_{A})-\sum_{x}p(x)S(\rho_{A}^x)=I_{acc}(\rho^{XA})$

as the state is classical-quantum. The mutual information of product states is the sum of the mutual information of the individual states, it is additive. So since $I_{acc}(\rho^{XA})=I(X:A)$, it inherits this additive property.

I will point out, however, that accessible information is not additive. It can be super-additive in some cases.

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  • $\begingroup$ Interesting, could you please point me to a proof of this kind of super-additivity? $\endgroup$ Jun 8 '21 at 17:23
  • $\begingroup$ nature.com/articles/nphys1224 $\endgroup$ Jun 9 '21 at 15:17
  • $\begingroup$ I don't think your first statement is correct; or at least I don't think it is obvious. Could you explain $I_{acc}(\rho^{XA}) = I(X:A)_\rho$? I agree $\leq$ is true, though. $\endgroup$
    – user16106
    Jun 11 '21 at 11:45
  • $\begingroup$ @user16106 does my edit help? $\endgroup$ Jun 14 '21 at 15:25

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