# 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. $\lim_{\epsilon \rightarrow 0} \delta(\epsilon) = 1$? Nov 23, 2020 at 16:48
• @Rammus sorry about that! Fixed now. Nov 23, 2020 at 16:56
• 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 $r < p$. Nov 24, 2020 at 12:55
• @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. Nov 24, 2020 at 13:49
• Yes, I'm sure something better can be done. Perhaps even a dimension-independent bound is possible. Nov 24, 2020 at 22:18

Consider states $$\rho_A$$ and $$\sigma_A$$ that are close in $$p$$-norm (for $$p>1$$) but have relatively low fidelity. Specifically, assume $$\|\rho_A - \sigma_A\|_p = \varepsilon$$ and $$\operatorname{F}(\rho_A,\sigma_A) = \bigl\|\sqrt{\rho_A}\sqrt{\sigma_A}\bigr\|_1 = \delta,$$ where $$\varepsilon$$ is small and $$\delta$$ is bounded away from 1. I'll give a specific example below.
The maximal fidelity between purifications $$\Phi_{AB}$$ and $$\Psi_{AB}$$ is also equal to $$\delta$$, so the minimal trace norm of the difference between purifications is bounded as follows: $$\bigl\|\Phi_{AB} - \Psi_{AB}\bigr\|_1 \geq 2 \sqrt{1 - \delta^2},$$ with equality when the purifications are chosen optimally.
Now the key is that the operator $$\Phi_{AB} - \Psi_{AB}$$ has rank equal to 2 (assuming the two states are not equal, which we get from $$\delta <1$$). Thus, for any choice of $$p\in[1,\infty]$$, we obtain $$\bigl\|\Phi_{AB} - \Psi_{AB}\bigr\|_p \geq \frac{1}{2} \bigl\|\Phi_{AB} - \Psi_{AB}\bigr\|_1 \geq \sqrt{1 - \delta^2},$$ which is not small when $$\delta$$ is bounded away from 1.
As an extreme example, choose $$n$$ to be a large positive integer and define states in $$2n$$ dimensions like this: $$\rho_A = \frac{1}{n}\sum_{k=1}^n |k\rangle\langle k|$$ and $$\sigma_A = \frac{1}{n}\sum_{k=n+1}^{2n} |k\rangle\langle k|.$$ These states are close in $$\infty$$-norm when $$n$$ is large, $$\varepsilon = \bigl\|\rho_A - \sigma_A\|_{\infty} = \frac{1}{n},$$ and because they are orthogonal their fidelity is zero: $$\delta = 0$$. The minimal $$\infty$$-norm between two purifications is therefore at least $$1$$, which obviously exceeds any constant factor times $$\varepsilon$$.