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Timeline for answer to Time evolution of the operators vs. the expectation values by ACuriousMind

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Jul 9, 2020 at 17:20 comment added HerpDerpington See also my comment to @CosmasZachos under the question.
Jul 9, 2020 at 17:06 comment added HerpDerpington "which you seem to think isn't true" that was a little bit misleading, what i actually wanted to say is that $\left\langle dA/df \right\rangle \neq d\left\langle A \right\rangle /df$. You cannot swap this in the general case. (Derivatives in $f$, not in $t$)
Jul 9, 2020 at 15:45 comment added ACuriousMind @HerpDerpington It follows directly from the second equation in my post, but note that it only holds in the context of the specific observable $A$. I'm not claiming that $f=\frac{\mathrm{i}}{\hbar}t$ holds in some general sense (which would be unclear to begin with), just that for this specific $A$ you get the expression for $A(f)$ by plugging $t = -\mathrm{i}\hbar f$ into $A(t)$.
Jul 9, 2020 at 15:40 comment added HerpDerpington Does that always work? What if, for example $f$ is the crystal momentum? (I have seen this in one lecture) Or what if $f$ is the spatial coordinate in a 1D system?
Jul 9, 2020 at 15:18 history answered ACuriousMind CC BY-SA 4.0