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Aug 18, 2020 at 7:01 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Apr 19, 2019 at 22:29 comment added Keith McClary @ValterMoretti is referring to Stone's theorem. Is that covered in your course? I don't see it in the Contents.
Apr 19, 2019 at 21:55 comment added Jiang-min Zhang @KeithMcClary page 118 and page 226.
Apr 19, 2019 at 14:18 comment added Keith McClary Is that part of the book in this "preview"?
Apr 19, 2019 at 12:51 comment added Valter Moretti However, all that has nothing to do with the time evolution of the state, which is always defined, even if the state does not beleng to the Hamiltonian domain, as it is implemented by a unitary operator. You should not rule out that state from QM! T
Apr 19, 2019 at 12:49 comment added Valter Moretti As the function does not belong to the domain of the Hamiltonian operator, the energy variance is not defined in the said state. The expectation value could be defined in any cases (it is enough to decompose the state along the eigenvectors of the Hamiltonian and sum the series of the energies...to check it).
Apr 19, 2019 at 11:27 answer added Halbeard timeline score: -1
Apr 19, 2019 at 10:55 comment added Norbert Schuch Can you quote precisely what they say in the book? Do they require this property to hold almost everywhere or sth like that? Otherwise, we are debating about your interpretation of what the book says, rather about what the book says.
Apr 19, 2019 at 10:49 history edited Qmechanic
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Apr 19, 2019 at 10:48 comment added Jiang-min Zhang it is finite. But the second moment of the hamiltonian diverges
Apr 19, 2019 at 10:03 comment added NLambert What's the energy of this state?
Apr 19, 2019 at 9:55 history asked Jiang-min Zhang CC BY-SA 4.0