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Jan 28, 2013 at 14:25 comment added Ondřej Černotík @Manishearth Glad it helped :)
Jan 28, 2013 at 14:24 comment added Manishearth Great edit, the whole concept makes a lot more sense to me now. I like the "weighted expectation value" interpretation :)
Jan 28, 2013 at 14:21 vote accept Manishearth
Jan 28, 2013 at 14:13 history edited Ondřej Černotík CC BY-SA 3.0
added 1659 characters in body
Jan 28, 2013 at 9:53 comment added Ondřej Černotík @daaxix Thanks for the observation, I'll edit it to the answer :)
Jan 28, 2013 at 0:23 comment added daaxix The interpretation of the first integral is exactly the sum $\sum C_k\langle \hat{A}\rangle$, or the sum of expectation values over a canonical orthogonal wavefunction basis, where both $\phi$ and $\psi$ can be represented in that orthogonal basis. So it is sort of like a weighted expectation value, weighted by the respective wavefunctions $\phi$ and $\psi$.
Jan 27, 2013 at 18:31 comment added Ondřej Černotík It's not a big deal, there are just several possibilities and I think it's better to treat them separately..
Jan 27, 2013 at 17:29 comment added Manishearth @Ondejernotík: "Not a completely general one" -- that's intriguing, my interest is piqued :) No problem, add it whenever you have the time..
Jan 27, 2013 at 17:28 comment added Ondřej Černotík Not a completely general one, as far as I know. I'll try and add something more on the topic tomorrow, don't have much time now...
Jan 27, 2013 at 17:24 comment added Manishearth Thanks for your answer (I'll accept tomorrow after giving others a chance to answer). Is there any way to interpret $\hat A|\psi\rangle$? Even a slightly mathematical/complicated explanation would do (editing it into your answer would be nice)
Jan 27, 2013 at 13:52 history answered Ondřej Černotík CC BY-SA 3.0