# Closeness of purifications of states

Uhlmann's theorem states that if two states $$\rho_A, \sigma_A$$ satisfy $$F(\rho_A, \sigma_A)\geq 1 - \varepsilon$$, then there for any purification $$\Psi_{AR}$$ of $$\rho_A$$, one can find a purification $$\Phi_{AR}$$ of $$\sigma_A$$ such that

$$F(\Psi_{AR}, \Phi_{AR})\geq 1 - \varepsilon$$

The purification $$\Phi_{AR}$$ can be found by optimizing over unitaries on the purifying register alone i.e. the following holds for any choice of purification $$\Phi_{AR}$$

$$\sup_{U_R}F(\Psi_{AR}, (I_A\otimes U_R)\Phi_{AR})\geq 1- \varepsilon$$

Since the trace distance and fidelity are closely related, one can translate Uhlmann's theorem into the following. Given $$\|\rho_A - \sigma_A\|_1 \leq \varepsilon$$, for any purification $$\Psi_{AR}$$ of $$\rho_A$$ and $$\Phi_{AR}$$ of $$\sigma_A$$ , we have

$$\inf_{U_R}\|\Psi_{AR} - (I_A\otimes U_R)\Phi_{AR}\|_1\leq \delta(\varepsilon),$$

where $$\lim_{\varepsilon \rightarrow 0}\delta(\varepsilon) = 0$$. Crucially, $$\delta(\varepsilon)$$ has no dependence on the dimension of the state.

Question: Is the above statement true for any other Schatten p-norm. Given $$\rho_A, \sigma_A$$ such that $$\|\rho_A - \sigma_A\|_p\leq \varepsilon$$ and for any purifications $$\Psi_{AR}$$ of $$\rho_A$$ and $$\Phi_{AR}$$ of $$\sigma_A$$, is it true that

$$\inf_{U_R}\|\Psi_{AR} - (I_A\otimes U_R)\Phi_{AR} \|_p \leq \delta(\varepsilon)$$

I am particularly interested in the above statement for the operator norm i.e. $$p = \infty$$.

• When the trace distance between $\rho_A$ and $\sigma_A$ tends to zero, shouldn't the fidelity between then tend towards $1$? I.e. $\lim_{\epsilon \rightarrow 0} \delta(\epsilon) = 1$? – Rammus 2 days ago
• @Rammus sorry about that! Fixed now. – user1936752 2 days ago
• Remember that in finite dimensions all $p$-norms are equivalent, in the sense that they are always bound by eachother (times a constant). So your statement is trivially true, you just need the proof for $p =1$ and relate the other norms to it. More precisely, it holds for a vector $x$ of dimension $d$ that $\|x\|_{p}\leq \|x\|_{r}\le d^{(1/r-1/p)}\|x\|_{p}$, for $r < p$. – Mateus Araújo yesterday
• @MateusAraújo you are indeed correct but I was hoping for a tighter bound similar to the case of $p=1$, where $\delta(\varepsilon)$ is independent of the dimension of the states. – user1936752 yesterday
• Yes, I'm sure something better can be done. Perhaps even a dimension-independent bound is possible. – Mateus Araújo yesterday