If states are close together does there always exist a channel close to the identity mapping one to the other?

Question

Question: Given states $\rho,\omega\in\mathbb C^{n\times n}$ and $\varepsilon>0$ such that $\rho$ and $\omega$ are $\varepsilon$-close in trace norm does there exist a channel $\Phi$ with $\Phi(\rho)=\omega$ such that $\Phi$ is $\varepsilon$-close to the identity, say, in diamond norm? Or maybe if $\varepsilon$ is too small a bound does this maybe hold for a larger channel-distance bound (e.g., $2\varepsilon$ or $\sqrt\varepsilon$)?

One motivation behind this question could be that it is true for pure states:

Given any $\phi,\psi\in\mathbb C^n$, $\|\phi\|=\|\psi\|=1$ there exists $U\in\mathbb C^{n\times n}$ unitary such that $U|\phi\rangle\langle\phi|U^*=|\psi\rangle\langle\psi|$ and, more importantly, $$\|U(\cdot)U^*-{\rm id}\|_\diamond\leq\sqrt2\|\,|\phi\rangle\langle\phi|-|\psi\rangle\langle\psi|\,\|_1\tag{1}$$ so if the pure states are $\varepsilon$-close together, then the unitary channel can be chosen $\varepsilon\sqrt2$-close to the identity.

(This bound can probably be made even tighter but for our purpose that's good enough). While I don't have a reference at hand containing (1) this is not too difficult to show in a constructive manner by restricting the problem to the 2-dimensional subspace ${\rm span}\{\phi e^{i\alpha},\psi\}$—where $\alpha$ is chosen such that $\Re(e^{i\alpha}\langle\psi|\phi\rangle)=|\langle\psi|\phi\rangle|$—and then defining the unitary just on there. From this one readily computes the eigenvalues of $U$ to be $1$ and $|\langle\phi|\psi\rangle|\pm i\sqrt{1-|\langle\phi|\psi\rangle|^2}$ meaning $\|U-{\bf1}\|_\infty=\sqrt{2-2|\langle\phi|\psi\rangle|}$. Finally, for unitary channels the diamond norm distance is known to simplify considerably (cf. Proposition 18 in this paper / arXiv) to the point that $$\|U(\cdot)U^*-{\rm id}\|_\diamond\leq2\|U-{\bf1}\|_\infty\leq 2\sqrt{2(1-|\langle\phi|\psi\rangle|^2)}=\sqrt2\|\,|\phi\rangle\langle\phi|-|\psi\rangle\langle\psi|\,\|_1 $$ where the last step is due to the well-known trace distance identity for pure states, cf. also Eq.(1.1.86) in Watrous’ book (alt link).

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(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)

Answer 1

We shall see that for general mixed states no such upper bound can exist in the following precise sense: For no continuous function $c:[0,2]\to[0,2]$ with $c(0)=0$ does it hold that for all $\|\rho-\omega\|_1\leq \varepsilon$ there exists a channel $\Phi$ with $\Phi(\rho)=\omega$ such that $\|\Phi-{\rm id}\|_\diamond\leq c(\varepsilon)$. Equivalently, we will show:

Proposition. For every continuous function $c:[0,2]\to[0,2]$ do there exist $\varepsilon$ and states $\rho,\omega\in\mathbb C^{n\times n}$ with $\|\rho-\omega\|_1\leq \varepsilon$ such that every channel that maps $\rho$ to $\omega$ satisfies $\|\Phi-{\rm id}\|_\diamond> c(\varepsilon)$.

Indeed our proof will be constructive, i.e. one can explicitly construct states which are arbitrarily close together but the only channel that maps one to the other is the reset channel $X\mapsto{\rm tr}(X)\omega$ which is always "far away" from the identity (no matter how close $\rho,\omega$ were).

Proof. Because $c$ is continuous and $c(0)=0$ there exists $\delta\in(0,2)$ such that $c(\delta)<2$. Our task now is to construct states which are $\delta$-close together but all channels mapping between the states have the maximal distance $2$ from the identity; this would conclude the proof. The states we consider are $\omega=|0\rangle\langle 0|$ and $$ \rho_\delta:=\Big(1-\frac\delta2\Big)|0\rangle\langle 0|+\frac\delta{2(n-1)}({\bf1}-|0\rangle\langle 0|)\,. $$ It is easy to see that $\rho_\delta$ is indeed a state and, most importantly, $\rho_\delta$ is of full rank because $\delta\in(0,2)$. At this point we use the known result that the only channel which maps a given full-rank state to a pure state is the reset channel. Thus all that is left to do is to compute (or rather: bound) $\|\Phi-{\rm id}\|_\diamond$: \begin{align*} \|\Phi-{\rm id}\|_\diamond&\geq \|\Phi-{\rm id}\|_{1\to 1}\\ &\geq\| \Phi(|1\rangle\langle 1|)-|1\rangle\langle 1|\,\|_1\\ &\geq\| {\rm tr}(|1\rangle\langle 1|)|0\rangle\langle 0|-|1\rangle\langle 1|\,\|_1\\ &=\|\,|0\rangle\langle 0|-|1\rangle\langle 1|\,\|_1=2 \end{align*} so, altogether, $\|\Phi-{\rm id}\|_\diamond\geq 2>c(\delta)$, as desired. $\square$