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Jul 5, 2021 at 20:38 comment added J.G. @yuggib Although a zero eigenvalue introduces subtleties.
Jul 5, 2021 at 20:30 history edited Qmechanic
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Jul 2, 2021 at 22:17 answer added Cosmas Zachos timeline score: 5
May 9, 2020 at 6:00 history tweeted twitter.com/StackPhysics/status/1259000212024111108
Sep 5, 2016 at 16:05 comment added yuggib The bottomline, as @ACuriousMind 's link suggests, is that it is possible to define the inverse of self-adjoint operators by means of the spectral calculus. So in particular it is possible to define the inverse of both position and momentum operators.
Sep 5, 2016 at 14:40 comment added ACuriousMind This is a subtle issue, for a closely related question (the inverse of the square of the position operator) see this question
Sep 5, 2016 at 14:33 history asked Sashwat Tanay CC BY-SA 3.0