Timeline for Why is $\frac{d}{dt} \langle \psi | = (\frac{d}{dt} | \psi \rangle )^{\dagger} $?

11 events

when toggle format	what		by	license	comment
Feb 21, 2018 at 17:55	history	edited	J. Murray	CC BY-SA 3.0	added 71 characters in body
S Feb 20, 2018 at 20:58	history	suggested	secavara	CC BY-SA 3.0	corrected a typo
Feb 20, 2018 at 20:42	review	Suggested edits
S Feb 20, 2018 at 20:58
Feb 20, 2018 at 20:35	comment	added	J. Murray		@StarBucK I've added to my answer to justify the exchange of the limit and the adjoint operation.
Feb 20, 2018 at 20:34	history	edited	J. Murray	CC BY-SA 3.0	Added section RE OP's comments
Feb 20, 2018 at 20:14	comment	added	J. Murray		That's a reasonable question, and the answer depends on how deeply you want to go. The operator norm of some operator $A$ is equal to the operator norm of $A^\dagger$ - you can use this fact to demonstrate that if you have a sequence of operators $\{a_n\}$ which converges to $A$, then the sequence $\{a_n^\dagger\}$ converges to $A^\dagger$, and take the limit that way.
Feb 20, 2018 at 20:12	history	edited	J. Murray	CC BY-SA 3.0	deleted 15 characters in body
Feb 20, 2018 at 19:55	comment	added	anon		I tried to do this but here you use the fact you can intervet the limit and the dagger. How do we know we can do this ?
Feb 20, 2018 at 19:31	history	undeleted	J. Murray
Feb 20, 2018 at 19:31	history	deleted	J. Murray		via Vote
Feb 20, 2018 at 19:30	history	answered	J. Murray	CC BY-SA 3.0