# How to justify the conclusion $|E_{sq}(\rho)-E_{sq}(\sigma)|\le f(\epsilon)$, when proving the continuity of the squashed entanglement?

I am following the paper by Christandl and Winter introducing squashed entanglement. My question is particularly on the continuity proof of squshed entanglement mentioned after conjecture 14 and before remark 15 (page 6 of the arXiv version).

To make the question more or less self-contained, squashed entanglement of a density matrix $$\rho^{AB}$$ is defined by $$E_{sq}(\rho){=}\inf_E I(A:B|E)_{\rho}/2$$, where the infimum of quantum conditional mutual information (QCMI) is obtained for all possible extensions of $$\rho^{AB}$$. More precisely, if a purification of $$\rho^{AB}$$ is given by $$|\rho\rangle\rangle^{ABC_1}$$, we have to take infimum over all possible CPTP maps $$\Lambda^{C_1{\rightarrow}E_1}$$ acting on the purified extension, $$\rho^{ABE_1}{=}\Lambda^{C_1{\rightarrow}E_1}(|\rho\rangle\rangle\langle\langle\rho|^{ABC_1})$$.

Now Christandl and Winter attempt to show the continuity of squashed entanglement, which asks if the trace distance between two density matrices is small, $$\parallel\rho^{AB}-\sigma^{AB}\parallel_1{\leq}\epsilon$$, does it imply $$|E_{sq}(\rho)-E_{sq}(\sigma)|{\leq}f(\epsilon)$$ with $$\lim_{\epsilon {\rightarrow} 0}f(\epsilon){=}0$$?

Brief proof-sketch: $$\parallel\rho^{AB}-\sigma^{AB}\parallel_1{\leq}\epsilon$$ implies the purifications are also close, $$\parallel|\rho\rangle\rangle^{ABC_1}-|\sigma\rangle\rangle^{ABC_2}\parallel_1{\leq}2\sqrt{\epsilon}$$. Now trace-distance is contractive under a CPTP map, i.e. for a CPTP map $$\Phi(.)$$, $$\parallel \Phi(\rho)-\Phi(\sigma)\parallel_1{\leq}\parallel \rho-\sigma\parallel_1$$. Hence if on the purified systems, same CPTP maps, $$\Lambda^{C_1{\rightarrow}E_1}$$ and $$\Lambda^{C_2{\rightarrow}E_2}$$ are applied, we have $$\parallel \rho^{ABE_1}-\sigma^{ABE_2}\parallel_1{\leq}2\sqrt{\epsilon}$$. From that continuity of QCMI follows (see corollary 1 of the Shirokov paper for a better bound).

However, I have trouble understanding the next part of the proof. The part where after getting $$|I(A:B|E_1)_\rho-I(A:B|E_2)_\sigma|{\leq}f(\epsilon)$$, they argued since this applies to any quantum operation $$\Lambda$$ and thus to every state extension of $$\rho_{AB}$$ and $$\sigma_{AB}$$ respectively, we obtain" $$|E_{sq}(\rho)-E_{sq}(\sigma)|{\leq}f(\epsilon)$$.

Question: Note the optimal extensions may have different dimensions, $$\dim(E_1){\neq}\dim(E_2)$$. Even if this is not the case, the extensions $$\rho^{ABE_1}$$ and $$\sigma^{ABE_2}$$ may be obtained by different CPTP maps (e.g., $$\Lambda_1$$ for $$\rho$$ and $$\Lambda_2$$ for $$\sigma$$), in which case we cannot apply the contractivity of trace distance. Given these issues, how can we justify the conclusion $$|E_{sq}(\rho)-E_{sq}(\sigma)|{\leq}f(\epsilon)$$?

## 1 Answer

Regarding the dimensions, this is not a problem because we can purify $$\rho$$ and $$\sigma$$ using the same system and then we anyway consider applying the same map to both systems and hence the dimensions also remain the same. (Actually it's a bit strange that you write different system labels for the same map). Regardless, if you encounter a situation where $$\mathrm{dim}(E_2) > \mathrm{dim}(E_1)$$ then you can just view $$E_1$$ as a subspace of $$E_2$$ and extend everything.

Now I'm assuming you're happy up to the statement that $$\|\rho_{AB} - \sigma_{AB}\|_1 \leq \epsilon$$ implies that $$|I(A:B|E)_{\rho} - I(A:B|E)_{\sigma}|\leq \epsilon'$$ when we apply the same extension map $$\Lambda$$ to the purifications of both $$\rho$$ and $$\sigma$$. Taken this as a given let's assume without loss of generality that $$E_{\mathrm{sq}}(\rho) \leq E_{\mathrm{sq}}(\sigma)$$. I'm also going to make the simplifying assumption that the infimum for $$E_{\mathrm{sq}}(\sigma)$$ is achieved and $$\Lambda$$ is the achieving map (a more rigorous analysis would need a limiting argument). More precisely, I'll assume there exists $$\Lambda$$ such that $$E_{\mathrm{sq}}(\sigma) = I(A:B|E)_{\Lambda(|\sigma\rangle \langle \sigma|)}.$$

So then \begin{aligned} |E_{\mathrm{sq}}(\rho) - E_{\mathrm{sq}}(\sigma)| &= E_{\mathrm{sq}}(\rho) - E_{\mathrm{sq}}(\sigma) \\ &= \inf I(A:B|E)_{\rho} - \inf I(A:B|E)_{\sigma} \\ &= \inf I(A:B|E)_{\rho} - I(A:B|E)_{\Lambda(|\sigma\rangle \langle \sigma|)} \\ &\leq I(A:B|E)_{\Lambda(|\rho\rangle \langle \rho|)} - I(A:B|E)_{\Lambda(|\sigma\rangle \langle \sigma|)} \\ &\leq \epsilon' \end{aligned}

• Great, this helps. Could you please explain why it is a simplifying assumption to consider the achievability of infimum? Regarding the system labels, I wanted to label the systems according to the density matrices which, as you rightly mentioned, is redundant after appending the lower dimensional system with the necessary ancilla to get equal system dimensions for both states.
– Abir
Jun 27, 2023 at 12:00
• It simplifies the answer I wrote but it is not without loss of generality. In general there does not have to exist a point achieving the infimum. E.g., take $\inf x$ for $0 < x < 1$ the infimum is 0 but it is never achieved by any point in the considered domain. You'd thus need to adapt the agument to consider some limit of a sequence of extensions if you want a full proof. Jun 27, 2023 at 14:03