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Feb 19, 2019 at 3:03 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.
Nov 15, 2016 at 7:53 comment added Christian Remling The ev's are simple automatically, but there may not be any (for example, if $V\equiv 0$). Also, $0$ will never be an eigenvalue for rapidly decaying $V$, so it's unclear what you mean by Q2.
Nov 15, 2016 at 7:52 answer added Christian Remling timeline score: 3
Nov 15, 2016 at 5:35 review First posts
Nov 15, 2016 at 5:46
Nov 15, 2016 at 5:31 history asked user101142 CC BY-SA 3.0