Timeline for How to solve bound states of 2D finite rectangular square well?

10 events

when toggle format	what		by	license	comment
Mar 12, 2015 at 14:15	comment	added	mikuszefski		Hmm, I am not so sure about the 8 equations as for symmetry we know that in a square potential $f=g$. Even in a rectangular case I guess that re-scaling would work.
Mar 12, 2015 at 9:02	comment	added	user75205		The curvature can be considered to be $\frac{1}{f}\frac{d^2f}{dx^2}$ in one direction, of generally, as $\frac{1}{f}\nabla^2f$. (Belongs to the last post by A.Friend - but I don't have 50 reputation - whatever this means ;) )
Mar 10, 2015 at 9:47	review	Suggested edits
Mar 10, 2015 at 12:38
S Mar 6, 2015 at 12:08	history	suggested	CommunityBot	CC BY-SA 3.0	small correction for better clarity
Mar 6, 2015 at 11:23	review	Suggested edits
S Mar 6, 2015 at 12:08
Mar 5, 2015 at 13:05	comment	added	Kyle Kanos		What is curvature in this context? Also, if your solution gives a "clear contradiction," is it really a solution?
Mar 5, 2015 at 11:44	review	Late answers
Mar 5, 2015 at 13:05
Mar 5, 2015 at 11:34	review	First posts
Mar 5, 2015 at 13:09
Mar 5, 2015 at 11:32	history	edited	A.Friend	CC BY-SA 3.0	added 66 characters in body
Mar 5, 2015 at 11:27	history	answered	A.Friend	CC BY-SA 3.0