Timeline for Does the orthogonality of states apply when there is a potential?

12 events

when toggle format	what		by	license	comment
May 12, 2017 at 4:15	history	edited	loltospoon	CC BY-SA 3.0	added 32 characters in body
May 12, 2017 at 4:11	answer	added	Samapan Bhadury		timeline score: 0
May 12, 2017 at 4:10	answer	added	user12029		timeline score: 2
May 12, 2017 at 2:41	history	edited	Qmechanic♦		edited tags
May 12, 2017 at 2:40	comment	added	loltospoon		@knzhou updated the question to add the context. Is this method even valid?
May 12, 2017 at 2:39	history	edited	loltospoon	CC BY-SA 3.0	added 1010 characters in body
May 12, 2017 at 2:25	comment	added	knzhou		@loltospoon More generally, the first equation you have says that an integral is equal to zero (for $i \neq j$), while the second says that the same integral plus an arbitrary weighting factor $V(x)$ is also zero; that can't always hold.
May 12, 2017 at 2:23	comment	added	NickD		It depends on V, but in general, the answer is no.
May 12, 2017 at 2:21	comment	added	loltospoon		@knzhou ok well in that case, the end result is still somewhat what I was thinking - you get $0$ when $i\neq j$.
May 12, 2017 at 2:20	history	edited	loltospoon	CC BY-SA 3.0	added 117 characters in body
May 12, 2017 at 2:20	comment	added	knzhou		No, this isn't true. As the simplest possible example, if $V(x)$ is a constant $V_0$, then $\langle \psi_i \| V \| \psi_j \rangle = V_0 \delta_{ij}$ instead.
May 12, 2017 at 2:16	history	asked	loltospoon	CC BY-SA 3.0