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Let $T: M_n \to M_n$ be a quantum channel. If I understand Definition 13.5.1 of the book "Quantum information theory" of Wilde, the coherent information $Q(T)=\max_{\phi_{AA'}} I(A \rangle B)_\rho$ of $T$ is given by \begin{equation} Q(T) =\sup_{\rho \textrm{ pure}} \Big\{H\big((\mathrm{tr} \otimes T)(\rho)\big)-H\big((\mathrm{Id} \otimes T)(\rho)\big)\Big\} \end{equation} where the supremum is taken over all bipartite pure states $\rho$ on $M_n \otimes M_n$.

It is true that we can replace the supremum by a supemum on all states (not necessarily pure) ?

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  • $\begingroup$ Yes, but its simply unnecessary, the state space is convex and so the sup value will be obtained on a pure state, since they are the extreme points of the space. $\endgroup$
    – Condo
    Oct 27 '20 at 14:46
  • $\begingroup$ @Condo You need to show the convexity of the objective function to use this argument. $\endgroup$
    – user13507
    Oct 27 '20 at 14:51
  • $\begingroup$ well Shannon entropy maximization is convex as per en.wikipedia.org/wiki/Entropy_maximization and I am pretty sure its the same for the von Neumann entropy but perhaps it an SDP or Conic program ill look for a reference.... $\endgroup$
    – Condo
    Oct 27 '20 at 14:59
  • $\begingroup$ It is well-known that Von Neumann entropy $H$ is concave. But it is not clear that $\rho \mapsto H\big((\mathrm{tr} \otimes T)(\rho)\big)-H\big((\mathrm{Id} \otimes T)(\rho)\big)$ is concave or convex. $\endgroup$
    – Aaron
    Oct 28 '20 at 6:13
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I follow Wilde's notation here. The coherent information of a channel $N:A' \rightarrow B$ is given by

$$Q(N) \equiv \max_{\phi_{A A^{\prime}}} I(A\rangle B)_{\rho},$$

where $\rho_{AB}=N_{A^{\prime} \rightarrow B}\left(\phi_{A A^{\prime}}\right)$. Notice that the channel only acts on the $A'$ register. The $A$ register is used to purify the input to the channel.

So yes, you can consider mixed states on $\phi_{AA'}$ but then you can just purify this to some $\phi_{RAA'}$ and relabel the $RA$ register as $A$.

Finally, a side point but note that the coherent information is obtained by taking the maximum over all input pure states, not the supremum.

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  • $\begingroup$ Thank you for the answer. Unfortunately, I am not able to understand the fifth sentence. $\endgroup$
    – Aaron
    Oct 28 '20 at 6:17
  • $\begingroup$ @Aaron, I meant that the input state achieving the coherent information always exists so you can take the maximum instead of the supremum (as you have written in the question) over all pure states. $\endgroup$
    – rnva
    Oct 28 '20 at 12:20

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