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Commutation Watch

1. Im meant to show that if there is a complete set of mutual eigenkets of Hermitian operators A and B then [A,B] = 0

Not sure where i start..? Thanks for any help
2. Let be the eigenkets with and . Then every ket decomposes as a sum of (possibly infinitely many) eigenkets, and use that fact to compute . (To do it rigorously you will have to assume A and B are continuous operators.)
3. (Original post by Zhen Lin)
Let be the eigenkets with and . Then every ket decomposes as a sum of (possibly infinitely many) eigenkets, and use that fact to compute . (To do it rigorously you will have to assume A and B are continuous operators.)
Ahh great okay that makes sense. Thanks. How do i do it rigorously though..? would i replace eigenkets by eigenfunctions? How would it work out? Thanks!!
4. Rigorously means talking about norms and convergence and such. Given that you're talking about kets rather than vectors, I think this isn't something you need to be worried about.

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