On page 560, it states that

$$C^{(1)} \geq S(\frac{\varepsilon(|{\psi}\rangle\langle{\psi}|) +\varepsilon(|{\varphi}\rangle\langle{\varphi}|)}{2} - \frac{1}{2}\varepsilon(|{\psi}\rangle\langle{\psi}|)-\frac{1}{2}\varepsilon(|{\varphi}\rangle\langle{\varphi}|)).$$

However, shouldn't this be

$$C^{(1)} \geq S(\frac{\varepsilon(|{\psi}\rangle\langle{\psi}|) +\varepsilon(|{\varphi}\rangle\langle{\varphi}|)}{2} - \frac{1}{2}S(\varepsilon(|{\psi}\rangle\langle{\psi}|))-\frac{1}{2}S(\varepsilon(|{\varphi}\rangle\langle{\varphi}|)).$$

as on the same page it states that if $\varepsilon(|{\psi_{j}}\rangle\langle{\psi_{j}}|) = p|{\psi_{j}}\rangle\langle{\psi_{j}}|+(1-p)\frac{I}{2}$ then $S(\varepsilon(|{\psi_{j}}\rangle\langle{\psi_{j}}|)) = H(\frac{1+p}{2})$, so $C(\varepsilon)=1-H(\frac{1-p}{2})$, which can't come about unless it's the entropy of the two states, not just the channel acting on them, or am I misreading this completely?

  • $\begingroup$ I think it has to be a typo, the term with $S(\cdot)$ is a scalar while the output of $\mathcal{E}(\cdot)$ is a state! So I think you're right, there is a $S(\cdot)$ missing. $\endgroup$ – keisuke.akira Jun 20 at 0:28

Adding my comment as an answer: I think this is a typo. The equation contains a scalar, namely $S(\cdot)$ and an operator, the output of the channel $\mathcal{E}(\cdot)$ in linear combination (their dimensions do not match). Therefore, this has to be a typo (there are no hidden Identity operators in the scalar terms as is easy to check).

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  • $\begingroup$ Alright, thanks, I wasn't sure if I was completely misinterpreting the meaning of the theorem. $\endgroup$ – GaussStrife Jun 20 at 14:39

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