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Nov 24, 2022 at 19:27 comment added leftaroundabout FTR, this question doesn't really make sense if you write "for some function $f$". I could make it trivially true by defining $f(o) := B$, i.e. it ignores the input and maps any operator to $B$. What you actually mean is "for some scalar function $f$", and the (admittedly standard) notation $f(A)$ doesn't actually refer to function application, but rather to application of the operator-lifted version of the function (i.e., applied to the eigenvalues).
Nov 24, 2022 at 18:00 history tweeted twitter.com/StackPhysics/status/1595839697124839425
Nov 24, 2022 at 16:27 history became hot network question
Nov 24, 2022 at 10:28 vote accept peep
Nov 24, 2022 at 9:36 answer added Qmechanic timeline score: 5
Nov 24, 2022 at 9:35 answer added Valter Moretti timeline score: 9
Nov 24, 2022 at 9:26 history edited Qmechanic
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Nov 24, 2022 at 8:53 answer added peep timeline score: 3
Nov 24, 2022 at 8:26 history asked peep CC BY-SA 4.0