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It is well known that $\|\mathcal{E} \circ \mathcal{F} - \mathcal{E}\|_\lozenge \leq \|\mathcal{F} - \mathcal{I}\|_\lozenge$.

What if I have $\|\mathcal{A} \circ \mathcal{E} \circ \mathcal{F} - \mathcal{A} \circ \mathcal{E}\|_\lozenge$? Does this obey

$$\|\mathcal{A} \circ \mathcal{E} \circ \mathcal{F} - \mathcal{A} \circ \mathcal{E}\|_\lozenge \leq \|\mathcal{A} \circ \mathcal{F} - \mathcal{A} \circ \mathcal{I}\|_\lozenge$$

Here I am concerned only with CPTP maps.

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1 Answer 1

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TL;DR: No. The suggested bound fails to hold for any norm. Briefly, if we choose $\mathcal{A}=\mathcal{F}$ to be an idempotent channel then the right-hand side vanishes. However, if we choose $\mathcal{E}$ to be the Hadamard and $\mathcal{A}=\mathcal{F}$ to be the completely dephasing channel (which is idempotent) then the left-hand side remains positive.

Counterexample

Let's represent a quantum state in Bloch coordinates $(x, y, z)$, so that \begin{align} \rho=\frac12(I+xX+yY+zZ).\tag1 \end{align} Then a quantum channel is completely described by its action on a general real $3$-vector $(x, y, z)$.

Let's choose $\mathcal{A}$ and $\mathcal{F}$ to be the completely dephasing channel$^1$ \begin{align} \mathcal{A}(x, y, z)=\mathcal{F}(x, y, z)=(0, 0, z).\tag2 \end{align} We note that $\mathcal{A}\circ\mathcal{F}=\mathcal{A}=\mathcal{A}\circ\mathcal{I}$, so \begin{align} \|\mathcal{A}\circ\mathcal{F}-\mathcal{A}\circ\mathcal{I}\|_\diamond=0.\tag3 \end{align} Further, let's choose $\mathcal{E}$ to be the unitary Hadamard channel$^2$ \begin{align} \mathcal{E}(x, y, z)=(z, -y, x).\tag4 \end{align} Then \begin{align} (\mathcal{A}\circ\mathcal{E}\circ\mathcal{F})(x, y, z)&=\mathcal{A}(\mathcal{E}(\mathcal{F}(x, y, z)))\tag5\\ &=\mathcal{A}(\mathcal{E}(0, 0, z))\tag6\\ &=\mathcal{A}(z, 0, 0)\tag7\\ &=(0, 0, 0)\tag8 \end{align} so $\mathcal{A}\circ\mathcal{E}\circ\mathcal{F}$ is the completely depolarizing channel$^3$. Similarly, \begin{align} (\mathcal{A}\circ\mathcal{E})(x, y, z)&=\mathcal{A}(z, -y, x)\tag9\\ &=(0, 0, x)\tag{10} \end{align} so $\mathcal{A}\circ\mathcal{E}\circ\mathcal{F}\neq\mathcal{A}\circ\mathcal{E}$. But then the positive definiteness of the diamond norm implies that \begin{align} \|\mathcal{A}\circ\mathcal{E}\circ\mathcal{F}-\mathcal{A}\circ\mathcal{E}\|_\diamond > 0.\tag{11} \end{align} Finally, combining $(3)$ and $(11)$, we obtain \begin{align} \|\mathcal{A}\circ\mathcal{E}\circ\mathcal{F}-\mathcal{A}\circ\mathcal{E}\|_\diamond > \|\mathcal{A}\circ\mathcal{F}-\mathcal{A}\circ\mathcal{I}\|_\diamond\tag{12}. \end{align} Note that the only property of the norm used in the proof is that $\|a-b\|=0\iff a=b$. This is satisfied by every norm, so the counterexample proves that the bound suggested in the question fails to hold for any norm, not just the diamond norm.


$^1$ Kraus representation $\mathcal{A}(\rho)=\mathcal{F}(\rho)=\frac12\rho+\frac12 Z\rho Z$.
$^2$ Kraus representation $\mathcal{E}(\rho)=H\rho H$.
$^3$ Kraus representation $(\mathcal{A}\circ\mathcal{E}\circ\mathcal{F})(\rho)=\frac14\rho+\frac14 X\rho X+\frac14 Y\rho Y+\frac14 Z\rho Z$.

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