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Oct 18, 2023 at 17:34 comment added terceira In questions of this type, it is often expedient to use the spectral theorem in the form that every (unbounded) self-adjoint operator can be regarded as multiplication by a messurable function on an $L^2$. They then sometimes become quite transparent, as here. In the specisl case you mention we have the bonus that this diagonalisation is effected by the Fourier transform..
Oct 17, 2023 at 20:18 answer added Christian Remling timeline score: 4
Oct 17, 2023 at 20:05 history became hot network question
Oct 17, 2023 at 15:00 answer added Michele Caselli timeline score: 3
Oct 17, 2023 at 13:13 history edited B.Hueber CC BY-SA 4.0
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Oct 17, 2023 at 12:02 history asked B.Hueber CC BY-SA 4.0