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Sep 12, 2018 at 3:01 history tweeted twitter.com/StackChemistry/status/1039710495496970243
Sep 11, 2018 at 11:09 answer added Feodoran timeline score: 3
Sep 11, 2018 at 5:00 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.
Aug 9, 2018 at 8:05 comment added Feodoran You can have a superposition of any number eigenstates, not just for same n and l. It just won't be an eigenstate anymore (except for a few special cases). What do you mean by "no m eigenvalue".?
Aug 9, 2018 at 5:26 comment added Ivan Neretin Well, if you have two or more degenerate eigenstates, then any their linear combination is also an eigenstate, and you have a continuum of those.
Aug 9, 2018 at 4:25 review First posts
Aug 9, 2018 at 6:45
Aug 9, 2018 at 4:22 history asked JOE CC BY-SA 4.0