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Suppose I have a quantum operation $\mathcal{E}$ and a state $\rho$ such that:

$$ \operatorname{tr}(\Pi_z \rho \Pi_z) \le p $$

for some probability $p$ and some projection $\Pi_z$ onto some subspace of the Hilbert space. Let $\Pi_{-z} = \mathbb{1} - \Pi_z$.

I would like to prove (or disprove) that $\mathcal{E}(\rho)$ and $\mathcal{E}(\Pi_{-z} \rho \Pi_{-z})$ are close to each other, i.e. finding a bound for:

$$ || \mathcal{E}(\rho) - \mathcal{E}(\Pi_{-z} \rho \Pi_{-z}) ||_1 $$

The first thing it comes natural to do is to apply contractivity of quantum channels:

$$ || \mathcal{E}(\rho) - \mathcal{E}(\Pi_{-z} \rho \Pi_{-z}) ||_1 \le || \rho - \Pi_{-z} \rho \Pi_{-z} ||_1 $$

But now I can't go ahead. Can you help me?

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Using the triangle inequality, we have $$||\rho-\Pi_{-z}\rho\Pi_{-z}||_1\leq ||\rho||_1+||\Pi_{-z}\rho\Pi_{-z}||_1=1+\mathrm{Tr}(\Pi_{-z}\rho\Pi_{-z})$$ (the final equality holds because $\rho$ and $\Pi_{-z}\rho\Pi_{-z}$ are positive semidefinite). Then we can use $\Pi_{z}^2=\Pi_{z}$ and cyclicity of the trace to find $$\mathrm{Tr}(\Pi_{-z}\rho\Pi_{-z})=\mathrm{Tr}(\rho+\Pi_{z}\rho\Pi_{z}-\Pi_{z}\rho-\rho\Pi_{z})=\mathrm{Tr}(\rho-\Pi_{z}\rho\Pi_{z})=1-p.$$ Overall we thus have $$||\mathcal{E}(\rho)-\mathcal{E}(\Pi_{-z}\rho\Pi_{-z})||_1\leq 2-p.$$

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    $\begingroup$ isn't the first one $2 - p$? $\endgroup$ Oct 20 at 14:03
  • $\begingroup$ @NoImaginationGuy for sure, edited, thanks! $\endgroup$ Oct 20 at 14:04
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    $\begingroup$ The second equality is not correct. Consider $\rho = \begin{pmatrix} \alpha & \beta \\ \beta^* & 1-\alpha \end{pmatrix}$ and $\Pi_z = |0\rangle \langle 0|$. Then the LHS gives $\sqrt{\alpha^2 + |\beta|^2}$ but the RHS gives $\alpha$, These are clearly not the same when $beta \neq 0$. $\endgroup$
    – Rammus
    Oct 21 at 18:52
  • $\begingroup$ Great, thanks @Rammus. So we are stuck with the first answer $\endgroup$ Oct 21 at 19:22

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