If a quantum channel is an isometry, is the channel also isometric?

Question

This question may appear trivial at first, but comes down to an interesting naming convention in quantum information. Recall that

• a quantum channel $\Phi:\mathbb C^{m\times m}\to\mathbb C^{n\times n}$ is said to be isometric if there exists an isometry $V\in\mathbb C^{n\times m}$ such that $\Phi(X)=VXV^\dagger$ for all $X\in\mathbb C^{m\times m}$, cf. Chapter 2.2.3 in the book "The Theory of Quantum Information" by Watrous (alt link) or, alternatively, this or this stackexchange answer.
• a channel—or really any linear map—$\Phi:\mathbb C^{m\times m}\to\mathbb C^{n\times n}$ is called an isometry (with respect to, say, the trace norm) if $\|\Phi(X)\|_1=\|X\|_1$ for all${}^1$ $X\in\mathbb C^{m\times m}$. As a consequence, such channels preserve arbitrary state distances, that is, $\|\Phi(\rho)-\Phi(\sigma)\|_1=\|\rho-\sigma\|_1$ for all states $\rho,\sigma$.

As an example, every isometric channel is certainly an isometry because $VXV^\dagger$ has the same singular values as $X$ (as $(VXV^\dagger)^\dagger(VXV^\dagger)=VX^\dagger XV^\dagger$ and $X^\dagger X$ have the same eigenvalues) so the trace norm is always preserved. This begs the question whether the converse also holds and, as a consequence, whether these two notions are equivalent? In other words:

Question. If a channel $\Phi:\mathbb C^{m\times m}\to\mathbb C^{n\times n}$ satisfies $\|\Phi(X)\|_1=\|X\|_1$ for all $X\in\mathbb C^{m\times m}$ (i.e., $\Phi$ is an isometry), does there always exist an isometry $V\in\mathbb C^{n\times m}$ such that $\Phi(X)=VXV^\dagger$ for all $X\in\mathbb C^{m\times m}$ (i.e., $\Phi$ is isometric)?

There is one special case where is known to be true, and that is $m=n$. In that case one can take the high road and use an old functional analysis result of Russo (Proc. Amer. Math. Soc. 23 (1969), 213)—or, in a slightly more general and more readable form, by Arazy (Isr. J. Math., 22 (1975), 247)—which states that every trace-norm isometry between spaces of the same dimension is either of the form $\Phi=U(\cdot)W$ or $\Phi=U(\cdot)^TW$ for some unitaries $U,W$. Combining this with the fact that $\Phi$ is completely positive yields $\Phi(X)=UXU^\dagger$ for all $X$ which is precisely what we had to show (recall that for $m=n$, every isometry $V\in\mathbb C^{n\times n}$ is unitary by the rank-nullity theorem). But what about the general case where $m\neq n$?

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_{${}^1$: To be clear, every channel restricted to states is an isometry (because $\|\Phi(\rho)\|_1={\rm tr}(\Phi(\rho))={\rm tr}(\rho)=\|\rho\|_1$)—the non-trivial property here is that $\Phi$ has to preserve the trace norm regardless of the input $X\in\mathbb C^{m\times m}$, which is something not all channels satisfy. The simplest example which highlights this crucial distinction is the reset channel $\Phi(X)={\rm tr}(X)|0\rangle\langle 0|$ because $$\|\Phi(\sigma_z)\|_1=\|{\rm tr}(\sigma_z) |0\rangle\langle 0|\,\|_1=0\neq 2=\|\sigma_z\|_1$$}

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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

While the two concepts coincide for $m=n$, it turns out that in general not every channel which is an isometry is also isometric in the above sense. For a simple counterexample consider the channel $\Phi$ which extends any qubit by a second qubit in the maximally mixed state: \begin{align*} \Phi:\mathbb C^{2\times 2}&\to\mathbb C^{4\times 4}\\ X&\mapsto X\otimes\frac12{\bf1} \end{align*} Because the trace norm is a cross norm—i.e., $\|A\otimes B\|_1=\|A\|_1\|B\|_1$ for all $A,B$—$\Phi$ is a trace-norm isometry. However, its Choi matrix of $\Phi$ reads $$ \mathsf C(\Phi)=\frac12\begin{pmatrix}1&0&0&0&0&0&1&0\\0&1&0&0&0&0&0&1\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\1&0&0&0&0&0&1&0\\0&1&0&0&0&0&0&1\end{pmatrix}\simeq \frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\oplus\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\oplus{\bf 0}_4 $$ so ${\rm rank}(\mathsf C(\Phi))=2$. This means that $\Phi$ has at least two Kraus operators, hence it cannot be of the form $V(\cdot)V^\dagger$ for any $V$. Indeed, a set of Kraus operators of $\Phi$ is given by $\{\frac1{\sqrt2}\iota_0,\frac1{\sqrt2}\iota_1\}$ where $\iota_j|x\rangle:=|x\rangle\otimes|j\rangle$ for $j=1,2$.