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. $p = \infty$.

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    $\begingroup$ When the trace distance between $\rho_A$ and $\sigma_A$ tends to zero, shouldn't the fidelity between then tend towards $1$? I.e. $\lim_{\epsilon \rightarrow 0} \delta(\epsilon) = 1$? $\endgroup$
    – Rammus
    Commented Nov 23, 2020 at 16:48
  • $\begingroup$ @Rammus sorry about that! Fixed now. $\endgroup$ Commented Nov 23, 2020 at 16:56
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    $\begingroup$ 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 $r < p$. $\endgroup$ Commented Nov 24, 2020 at 12:55
  • $\begingroup$ @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. $\endgroup$ Commented Nov 24, 2020 at 13:49
  • $\begingroup$ Yes, I'm sure something better can be done. Perhaps even a dimension-independent bound is possible. $\endgroup$ Commented Nov 24, 2020 at 22:18

1 Answer 1


No dimension-independent bound is possible.

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 $\varepsilon$.


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