1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.