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    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
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    Let \left| v_1 \right>, \left| v_2 \right>, \ldots be the eigenkets with A \left| v_k \right> = \lambda_k \left| v_k \right> and B \left| v_k \right> = \mu_k \left| v_k \right>. Then every ket decomposes as a sum of (possibly infinitely many) eigenkets, and use that fact to compute AB \left| v \right> - BA \left| v \right>. (To do it rigorously you will have to assume A and B are continuous operators.)
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    (Original post by Zhen Lin)
    Let \left| v_1 \right>, \left| v_2 \right>, \ldots be the eigenkets with A \left| v_k \right> = \lambda_k \left| v_k \right> and B \left| v_k \right> = \mu_k \left| v_k \right>. Then every ket decomposes as a sum of (possibly infinitely many) eigenkets, and use that fact to compute AB \left| v \right> - BA \left| v \right>. (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!!
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    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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