Do identical reduced system trajectories ρₛ(t) guarantee identical recoverability in small open quantum systems?

Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Stack Overflow for Teams is now called Stack Internal. Bring the best of human thought and AI automation together at your work. Bring the best of human thought and AI automation together at your work. Learn more TeamsQ&A for workConnect and share knowledge within a single location that is structured and easy to search.In a small open quantum system with a system qubit $S$ and an environment split into an accessible fragment $F$ and an inaccessible remainder $R$, consider the following situation.Two global evolutions $A$ and $B$ produce identical reduced system trajectories $$\rho_S^{A}(t) = \rho_S^{B}(t) \quad \forall t,$$ including after an entangling interaction between $S$ and $F$.However, their ability to recover coherence on $S$ (via any CPTP map acting on $S \otimes F$) differs maximally.Using the standard $\ell_1$-coherence measure: • World A: full echo recovery is possible ($C_{\mathrm{rec}} = 1$) • World B: no recovery is possible ($C_{\mathrm{rec}} = 0$) • Yet $\rho_S(t)$ is identical in both worlds at all times.A minimal example is the familiar 3-qubit construction: 1. I-phase Apply $\mathrm{CNOT}_{S\to F}$ to $\lvert + \rangle_S \lvert 0 \rangle_F \lvert 0 \rangle_R$. 2. World A Do nothing on $E = F \sqcup R$, then apply the reversal $\mathrm{CNOT}_{S\to F}$. Full recovery. 3. World B Apply a single $\mathrm{CNOT}{F\to R}$, then apply the same reversal $\mathrm{CNOT}{S\to F}$. No recovery; the reduced state of $S$ remains maximally mixed.In both worlds, the reduced state $\rho_S(t)$ is the same before the reversal.⸻My questionIs it correct that $\rho_S(t)$ alone is insufficient to determine recoverability, even when all CPTP maps on $S \otimes F$ are allowed?More precisely: • Does the quantity $$O = I(F : R \mid S)$$ (conditional mutual information) capture the structural difference between worlds A and B? • And is it true that any operation on $S \otimes F$ cannot change the Schmidt coefficients across the $(S \otimes F) : R$ bipartition? • Therefore making the recoverability gap $(\Delta C_{\mathrm{rec}} = 1)$ unavoidable?⸻What I’m looking for • A check of whether the argument is valid as stated • Any references discussing this specific limitation of subsystem-only descriptions • Any known theorems in quantum information theory that already imply thisI am assuming standard unitary quantum mechanics and CPTP maps only.Thanks for contributing an answer to Quantum Computing Stack Exchange!But avoid …Use MathJax to format equations. MathJax reference.To learn more, see our tips on writing great answers.Required, but never shown By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answersFind the answer to your question by asking.Explore related questionsSee similar questions with these tags.To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.11.20.37056