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Dec 5, 2021 at 6:01 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.
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Dec 4, 2019 at 20:00 review Close votes
Dec 7, 2019 at 16:27
Dec 4, 2019 at 19:55 answer added ZeroTheHero timeline score: 1
Dec 4, 2019 at 19:43 history edited DanielSank CC BY-SA 4.0
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Dec 4, 2019 at 19:40 answer added Kraigolas timeline score: 0
Dec 4, 2019 at 19:39 history edited Qmechanic CC BY-SA 4.0
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Dec 4, 2019 at 17:56 comment added march You find $\Psi_{nlms}$ by solving the time-independent Schrodinger equation, which is the same as finding the eigenstates of the Hamiltonian, which are interpreted as states of definite energy (since the Hamiltonian is the energy operator). Therefore, we call them \emph{energy eigenstates}.
Dec 4, 2019 at 17:13 comment added Danny Han Thank you for your answer. What I don't understand is, how that results in the 'energy' eigenstates having n,l,m,s as parameters
Dec 4, 2019 at 16:47 comment added John Rennie The eigenfunctions factor into radial, angular and spin parts: $\Psi_{nlms} = \psi_n(r) Y_{lm}(\theta, \phi) \eta(s)$.
Dec 4, 2019 at 16:19 history asked Danny Han CC BY-SA 4.0