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Apr 4, 2020 at 10:04 vote accept Mew
Apr 3, 2020 at 21:09 answer added Javier timeline score: 3
Apr 3, 2020 at 20:46 comment added Mew @Javier: That makes sense for the most part. However, in the derivation of the theorem, we see terms of the form $\left<\frac{\partial\Psi}{\partial t}\mid\hat Q\Psi\right>$ and $\left<\Psi\mid\hat Q\frac{\partial\Psi}{\partial t}\right>$. I guess those don't count as expectation values of operators, then (just inner products between two functions)? Edit: you may add your comment as an answer below, and I will accept it.
Apr 3, 2020 at 19:18 comment added Javier Very short answer: $\partial/\partial t$ is not an operator. The space of states is the space of wavefunctions at a fixed time, not functions of $x$ and $t$.
Apr 3, 2020 at 19:06 history edited Qmechanic
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Apr 3, 2020 at 17:59 history edited Mew
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Apr 3, 2020 at 17:54 history asked Mew CC BY-SA 4.0