# Does ${\rm tr}(\Pi_z\rho\Pi_z)\le p$ imply $\cal E(\rho)$ and $\cal E(\Pi_{-z}\rho\Pi_{-z})$ are close in trace distance?

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?

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 $$||\mathcal{E}(\rho)-\mathcal{E}(\Pi_{-z}\rho\Pi_{-z})||_1\leq 2-p.$$
• isn't the first one $2 - p$? Oct 20 at 14:03
• 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 $beta \neq 0$. Oct 21 at 18:52