Closeness of unitary dilations of CPTP maps

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.

• 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. Oct 29 at 17:39
• 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. Oct 30 at 23:09

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.

• +1 Very nice. I think $\le 2\|V_1-V_2\|_1$ can be written as an equality. Oct 27 at 17:09
• @Adam yes, thanks! Oct 27 at 17:21
• 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 $\|A\|_{op}\le\|A\|_1$. Oct 27 at 17:51
• Thank you! I do still wonder if one can obtain a dimension-independent bound, but this is very helpful. Oct 27 at 20:41
• @nickspoon yeah that's certainly a good question. Perhaps there is a result that bounds the Schatten $1$-norm of the unitaries by their diamond norm distance in a dimension-independent way... or someone knows a counterexample? Anyways, thanks for the question. Oct 27 at 20:53