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Sep 26, 2023 at 18:44 vote accept MBlrd
Sep 26, 2023 at 18:43 comment added ACuriousMind @MBlrd Yes (but I really wouldn't think about this in terms of "approximate" eigenstates, the claim that some version of those exists is true but not all that useful)
Sep 26, 2023 at 18:36 comment added MBlrd Thank you @ACuriousMind. So the logic is that, since $e^{i A t}$ is bounded it is in the C*-algebra, and moreover it also shares the same set of "approximated" eigenstates with $A$. Am I right?
Sep 26, 2023 at 18:26 history answered ACuriousMind CC BY-SA 4.0