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Jul 27, 2022 at 13:20 answer added ZeroTheHero timeline score: 2
Jul 27, 2022 at 12:58 history edited Qmechanic CC BY-SA 4.0
edited tags; edited title
Jul 27, 2022 at 12:47 comment added ZeroTheHero such systems certainly exist: the deuteron for instance has a single bound state.
Jul 27, 2022 at 12:22 comment added ZeroTheHero @hyportnex you should transform this into an answer.
Jul 27, 2022 at 12:17 review Close votes
Jul 27, 2022 at 12:20
S Jul 27, 2022 at 12:14 history edited ZeroTheHero CC BY-SA 4.0
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Jul 27, 2022 at 12:12 comment added hyportnex if the operator has only one eigenstate and eigenvalue then the operator is a scalar multiplier, there cannot be any evolution of it and the Mandelstam interpretation is not operative.
Jul 27, 2022 at 12:12 review Suggested edits
S Jul 27, 2022 at 12:14
Jul 27, 2022 at 11:59 comment added ZeroTheHero Does this answer your question? What is $\Delta t$ in the time-energy uncertainty principle?
Jul 27, 2022 at 11:59 history edited ZeroTheHero
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S Jul 27, 2022 at 11:52 review First questions
Jul 27, 2022 at 12:12
S Jul 27, 2022 at 11:52 history asked juggernugger CC BY-SA 4.0