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Let $\Phi_1,\Phi_2 \colon S(\mathcal{H}) \to S(\mathcal{H})$ be CPTP maps on the same Hilbert space $\mathcal{H}$ which are $\varepsilon$-close in diamond norm, and let $U_1,U_2$ be respective unitary dilations on some larger space $\mathcal{H} \otimes \mathcal{K}$ (i.e. so that $\mathrm{Tr}_{\mathcal{K}}(U_i (\rho \otimes |0\rangle\langle0|) U_i^{\dagger}) = \Phi_i(\rho)$).

Now let $A$ be a unitary operating on $\mathcal{H}$, and let $$\rho_i := \mathrm{Tr}_{\mathcal{K}}(U_i^{\dagger} (A \otimes I_{\mathcal{K}}) U_i (\rho \otimes |0\rangle\langle0|) U_i^{\dagger} (A^{\dagger} \otimes I_{\mathcal{K}}) U_i).$$ That is, $\rho_i$ is obtained from $\rho$ by applying $U_i$, then $A$, then $U_i^{\dagger}$, then tracing out $\mathcal{K}$. Since all dilations are equivalent up to a local unitary, $\rho_i$ does not depend on the choice of $U_i$. Can we bound $\| \rho_1 - \rho_2 \|_1$?

I have tried to get a bound from results on the distance between Stinespring isometries, but this seems to not be enough here.

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  • $\begingroup$ Actually, it's not clear that your map giving the $\rho_i$'s is CPTP, the representation theorem for CPTP maps says there are unitaries such that $\Phi(\rho)=Tr_K(U^*(\rho\otimes |0\rangle\langle0|)U)$ is CPTP, it doesn't say you can pick these unitaries apriori. You can, however, pick an isometry, and then extend that to a unitary. $\endgroup$
    – Condo
    Oct 29, 2021 at 17:39
  • $\begingroup$ I am assuming here that we chose these unitaries so that the equality holds. However, I do believe that this map is CPTP for any choice of unitary. $\endgroup$
    – nickspoon
    Oct 30, 2021 at 23:09

1 Answer 1

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Suppose that $\Phi_i$ $i=1,2$ are CPTP maps with $\|\Phi_1-\Phi_2\|_\diamond\leq \epsilon$. Let $V_i=U_i^\dagger(A\otimes I_K)U_i$ and $\sigma=\rho\otimes |0\rangle\langle0|$, then we have that $V_i$ is unitary and

$$\|\rho_1-\rho_2\|_1=\|tr_K\left((V_1\sigma V_1^\dagger -V_2\sigma V_2^\dagger\right)\|_1\\ \leq \|V_1\sigma V_1^\dagger -V_2\sigma V_2^\dagger\|_1\\ =\|V_1\sigma V_1^\dagger -V_1\sigma V_2^\dagger+V_1\sigma V_2^\dagger-V_2\sigma V_2^\dagger\|_1\\ \leq \|V_1\sigma V_1^\dagger -V_1\sigma V_2^\dagger\|_1+\|V_1\sigma V_2^\dagger-V_2\sigma V_2^\dagger\|_1\\ \leq \|V_1\sigma\|_1\|V_1^\dagger-V_2^\dagger\|_1+\|V_1-V_2\|_1\|\sigma V_2^\dagger\|_1\\ =2\|V_1-V_2\|_1\\ \leq 2\dim(H\otimes K)\|V_1-V_2\|_\infty\\ \leq 2\dim(H\otimes K)\sqrt{\|\Phi_1-\Phi_2\|_\diamond}\\ \leq 2\dim(H\otimes K)\sqrt{\epsilon}$$

since, for $X\in L(H\otimes K)$ we have that $\|tr_K(X)\|_1\leq \|X\|_1$, $\|\sigma\|_1=1$ for any density matrix $\sigma$. We also use the fact that the $\|\cdot\|_1$ norm is unitarily invariant and relates to the operator norm via $\|\cdot\|_1\leq rank(X)\|\cdot\|_\infty$. At the end we use the result from this question about relating isometry distance to that of the diamond norm.

Note: In the case that $H$ or $K$ are infinite-dimensional Hilbert spaces this bound will of course be useless. There may be a subtler method that works in that case, but I don't see how it could work immediately.

Edit: Thank you to @FrederikvomEnde for pointing out that one can simply apply the $1$-norm bimodule property $\|ABC\|_1\leq \|A\|_\infty\|B\|_1\|C\|_\infty$ in the 4th line above to remove the dependence on the dimension of $H\otimes K$. It follows that $\|\rho_1-\rho_2\|_1$ is bounded by $2\sqrt{\epsilon}$ regardless of dimension.

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    $\begingroup$ +1 Very nice. I think $\le 2\|V_1-V_2\|_1$ can be written as an equality. $\endgroup$ Oct 27, 2021 at 17:09
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    $\begingroup$ Suggestion to better connect this with the linked answer: $$2\dim(H\otimes K)\|V_1-V_2\|_\infty\\\le2\dim(H\otimes K)\sqrt{\|\Phi_1-\Phi_2\|_{cb}}\\\leq 2\dim(H\otimes K)\sqrt{\|\Phi_1-\Phi_2\|_\diamond}$$ where the first inequality follows from that answer and the latter inequality is the "completely bounded" variant of $\|A\|_{op}\le\|A\|_1$. $\endgroup$ Oct 27, 2021 at 17:51
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    $\begingroup$ Thank you! I do still wonder if one can obtain a dimension-independent bound, but this is very helpful. $\endgroup$
    – nickspoon
    Oct 27, 2021 at 20:41
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    $\begingroup$ You can use the inequality $\|ABC\|_1\leq\|A\|_\infty\|B\|_1\|C\|_\infty$ to get rid of the dimension factor as then $\|V_1\sigma V_1^*-V_2\sigma V_2^*\|\leq 2\|\sigma\|_1\|V_1-V_2\|_\infty$ $\endgroup$ Jun 23, 2023 at 14:42
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    $\begingroup$ @FrederikvomEnde Yes, excellent! $\endgroup$
    – Condo
    Jun 23, 2023 at 20:24

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