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It is well known that for if any two linear maps $V_1,V_2:\mathbb C^n\to\mathbb C^k\otimes\mathbb C^m$ (isometry or not) satisfy $$ {\rm tr}_{\mathbb C^m}(V_1(\cdot)V_1^\dagger)={\rm tr}_{\mathbb C^m}(V_2(\cdot)V_2^\dagger) \tag1 $$ then there exists $U\in\mathbb C^{m\times m}$ unitary such that $V_1=({\bf1}\otimes U)V_2$; cf. Corollary 2.24 in the book "The Theory of Quantum Information" by Watrous (alt link). In particular this shows that any two Stinespring isometries (cf. also here) which give rise to the same channel are locally unitarily equivalent in the above sense. Equivalently, any two sets of Kraus operators to the same channel can be transformed into each other using said unitary $U$, cf. here and here.

Equally as common as defining channels via Kraus operators is to use the environmental form (cf. also here and here), that is, to express $\Phi={\rm tr}_E(U((\cdot)\otimes\omega)U^\dagger)$ for some environment state $\omega$ and some system-environment unitary $U$. Here one can ask a similar

Question. Given two unitaries $U_1,U_2$ such that $$ {\rm tr}_E(U_1((\cdot)\otimes\omega)U_1^\dagger)={\rm tr}_E(U_2((\cdot)\otimes\omega)U_2^\dagger) \tag2 $$ in what sense are $U_1,U_2$ equivalent? (Actually, one can make things even more general by allowing for different environment states $\omega_1,\omega_2$ but let us keep things as are for now.)

A simple partial answer is the following: In the special case of a pure state $\omega=|\psi\rangle\langle\psi|$ the environmental form from (2) boils down to the isometry form from (1) because $\rho\otimes|\psi\rangle\langle\psi|=\iota_\psi\rho\iota_\psi^\dagger$ with $\iota_\psi|x\rangle:=|x\rangle\otimes|\psi\rangle$ meaning (1) holds with $V_j=U_j\iota_\psi$. Thus (2) implies that there exists a unitary $U$ on the environment such that $$ U_1({\bf1}\otimes|\psi\rangle)=U_1\iota_\psi=({\bf1}\otimes U)U_2\iota_\psi=({\bf1}\otimes U)U_2({\bf1}\otimes|\psi\rangle)\,.\tag3 $$ This is the best one can hope for because the other "columns" of $U_1,U_2$ do not contribute to (2) meaning they can be chosen arbitrarily, i.e. in general there exists no environment $U$ such that $U_1=({\bf1}\otimes U)U_2$.

There is, however, a special case where one may be able to say more: when $\omega$ is a full-rank state. Intuitively, in this case "all information from $U$ is used" to define the channel which is undermined by the fact that in the full-rank case the environmental form defines a unitary channel if and only if the system-environment unitary is of product form. Thus in this case $U_1=({\bf1}\otimes U)U_2$ does in fact hold. In general, however, it may happen that there is an additional local unitary "on the other side": trivially, if $[W,\omega]=0$, then $U_1$ and $U_1({\bf1}\otimes W)$ induce the same channel through different unitaries where $U_1({\bf1}\otimes W)\neq({\bf1}\otimes U)U_1$ for any $U$. This leads to the following

Question (refined). Given a full-rank state $\omega\in\mathbb C^{m\times m}$ as well as unitaries $U_1,U_2\in\mathbb C^{n\times n}\otimes\mathbb C^{m\times m}$ such that $$ {\rm tr}_E(U_1((\cdot)\otimes\omega)U_1^\dagger)={\rm tr}_E(U_2((\cdot)\otimes\omega)U_2^\dagger) $$ do there exist environment unitaries $U,W\in\mathbb C^{m\times m}$ with $[W,\omega]=0$ such that $U_1=({\bf1}\otimes U)U_2({\bf1}\otimes W)$?


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

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  • $\begingroup$ but if I understand you correctly, the titular question is still answered in the positive, yes? "Stinespring unitaries" implicitly assume a pure state in the environment, and from that perspective they're all related unitarily/isometrically in the usual way. This characterisation only fails for non-pure environments. In which case, you're saying the characterisation doesn't hold, even if you try adding a $W$ unitary in it to account for degeneracies in $\omega$? But I still don't get how $W$ is supposed to help: it would seem it just cancels out in the last equation? $\endgroup$
    – glS
    Commented May 29 at 9:11
  • $\begingroup$ To me personally Stinespring unitary does not presuppose any particular environment state, but seeing how the constructive dilation assumes a pure state I get where this is coming from. So I guess this point is up for interpretation, resp. up to definition of the term "Stinespring unitary" $\endgroup$ Commented May 29 at 10:14
  • $\begingroup$ (2/2) Regarding $W$: assume for example that the environment is maximally mixed and let a "generic" unitary $V$ be given. Then—trivially, as you pointed out—$V,V({\bf1}\otimes W)$ both give rise to the same channel but there will in general not exist an environment $U$ such that $({\bf1}\otimes U)V({\bf1}\otimes W)=V$. Really I introduced this $W$ to rule out such obvious counterexamples $\endgroup$ Commented May 29 at 10:14
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    $\begingroup$ the point about terminology is fair, I agree. Regarding $W$: ok, I get it now. The main point is that dilations wrt a given $\omega$ are not generally isometrically related, not the $W$. Though I feel like adding this $W$ then somewhat distracts from the most interesting bits of the problem; after all, it is a trivial calculation to check that it never helps $\endgroup$
    – glS
    Commented May 29 at 12:04

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As it turns out the statement in question is wrong. For a counterexample consider the full-rank environment state $$ \omega=\begin{pmatrix} \frac12&0&0\\0&\frac14&0\\0&0&\frac14 \end{pmatrix} $$ as well as the unitary matrices $$ U_1= \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix}\,, \quad U_2= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ \end{pmatrix}\,. $$ A straightforward computation shows that $U_1,U_2$ together with $\omega$ give rise to the same mixed unitary channel: $$ {\rm tr}_{E}(U_1((\cdot)\otimes\omega)U_1^*)= \frac12\big( (\cdot)+\sigma_x(\cdot)\sigma_x \big) ={\rm tr}_{E}(U_2((\cdot)\otimes\omega)U_2^*) $$ Now assume there existed $U,W\in\mathbb C^{m\times m}$ unitary such that $U_1=({\bf1}\otimes U)U_2({\bf1}\otimes W)$. Then \begin{align*} &\begin{pmatrix} |1\rangle\langle1|+|2\rangle\langle2|&|0\rangle\langle0|\\ |0\rangle\langle0|&|1\rangle\langle1|+|2\rangle\langle2| \end{pmatrix}\\ &\qquad\qquad=U_1=({\bf1}\otimes U)U_2({\bf1}\otimes W)\\ &\qquad\qquad=\begin{pmatrix}U&0\\0&U\end{pmatrix} \begin{pmatrix} |0\rangle\langle0|&|1\rangle\langle1|+|2\rangle\langle2|\\ |1\rangle\langle1|+|2\rangle\langle2|&|0\rangle\langle0| \end{pmatrix} \begin{pmatrix}W&0\\0&W\end{pmatrix}\\ &\qquad\qquad=\begin{pmatrix} U|0\rangle\langle0|W&U(|1\rangle\langle1|+|2\rangle\langle2|)W\\ U(|1\rangle\langle1|+|2\rangle\langle2|)W&U|0\rangle\langle0|W \end{pmatrix}. \end{align*} But this would mean that the rank-1 matrix $U|0\rangle\langle0|W$ is equal to the rank-2 matrix $|1\rangle\langle1|+|2\rangle\langle2|$ which is obviously impossible, regardless even of whether or not $W$ commutes with $\omega$.


What one usually does in these situations is to purify the environment state $\omega={\rm tr}_R(|\Omega\rangle\langle\Omega|)$ using some additional environment $R\simeq\mathbb C^r$. Then Eq. (3) applies and yields $U\in\mathbb C^{m\times m}\otimes\mathbb C^{r\times r}$ such that $$ (U_1\otimes{\bf1}_R)({\bf1}\otimes|\Omega\rangle)=({\bf1}\otimes U)(U_2\otimes{\bf1}_R)({\bf1}\otimes|\Omega\rangle)\,. $$ Note that these expressions do not simplify because they feature three-party tensor products (system $\mathbb C^n$, original environment $\mathbb C^m$, purification environment $\mathbb C^r$) with incompatible product forms.

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  • $\begingroup$ I'm a bit confused by this. How does $W$ not cancel out in $U_2\underbrace{(I\otimes W)(\rho\otimes \omega)(I\otimes W^\dagger)}_{\rho\otimes W\omega W^\dagger} U_2^\dagger$? $\endgroup$
    – glS
    Commented May 29 at 8:24
  • $\begingroup$ Not quite sure I understand but if you're asking about the discrepancy between the refined question (where we required $[W,\omega]=0$ so $W$ does cancel for the channel) and the counterexample (where $W$ was arbitrary): I waived the additional condition because it doesn't change anything about the example. If you were asking about something else feel free to elaborate further $\endgroup$ Commented May 29 at 10:21
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    $\begingroup$ no it's precisely that. As per the other comments, allowing for $W$ commuting with $\omega$ trivially doesn't help. But then, here you're showing an even stronger statement: that even allowing for arbitrary $W$ doesn't help; which is certainly not as a priori obvious. $\endgroup$
    – glS
    Commented May 29 at 12:07

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