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Jan 22, 2023 at 22:28 comment added fedja I'm just an idiot: I'm looking at one operator and seeing another one. Don't know why. So, please, disregard everything I said and accept my apologies :-)
Jan 19, 2023 at 21:09 history edited Carlo Beenakker CC BY-SA 4.0
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Jan 19, 2023 at 15:10 comment added Carlo Beenakker I am not quite sure what you mean; the property of $a$ I need is that $a^\ast$ does not annihilate any state in the Hilbert space ($a^\ast|\psi\rangle=0$ has no nonzero solution).
Jan 19, 2023 at 14:56 comment added fedja That is more due to the lucky fact that the eigenvector of $a^*a$ corresponding to the eigenvalue $0$ is finitely supported than to the proposed idea itself. If it weren't, you would never see $0$ in any truncation despite all words about chiral symmetry and such would remain perfectly valid for a general compact $a$, woudn't they? I mean, the method definitely works in this case, but the explanation of why doesn't hold water IMHO. Am I saying nonsense? :-)
Jan 19, 2023 at 14:46 history edited Carlo Beenakker CC BY-SA 4.0
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Jan 19, 2023 at 14:44 vote accept António Borges Santos
Jan 19, 2023 at 14:38 history answered Carlo Beenakker CC BY-SA 4.0