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Assume $\Phi$ is a quantum channel, which is a CPTP map. Does the following equality holds and why? \begin{align} \Phi(0) = 0. \end{align}

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  • $\begingroup$ What do you mean by 0? A quantum channel describes the evolution of density matrices, which by the axioms of quantum mechanics should have unit trace. If you are just seeing it as a CPTP map for operators, I guess that 0 might be the null matrix, which would essentially fulfil what you state since $\sum_k E_k 0 E_k^\dagger = 0$ trivially. $\endgroup$ Mar 29, 2023 at 8:03

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Any linear function between any two vector spaces preserves the zero, so yes, any quantum map (and thus also any CPTP map) sends 0 to 0. Note that here "0" means the operator $\mathbb{C}^n\ni v\mapsto 0\in\mathbb{C}^n$.

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    $\begingroup$ Just to add, the proof is quite immediate. It's linear so $\Phi(\alpha A+\beta B) = \alpha \Phi(A) + \beta \Phi(B)$. So we have $\Phi(0) = \Phi(A + (-A)) = \Phi(A) - \Phi(A) = 0$. $\endgroup$
    – Rammus
    Mar 29, 2023 at 8:46

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