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  • $\begingroup$ 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? $\endgroup$ Commented Jul 9, 2020 at 15:40
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    $\begingroup$ @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 $A(t)$. $\endgroup$ Commented Jul 9, 2020 at 15:45
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    $\begingroup$ "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 $t$) $\endgroup$ Commented Jul 9, 2020 at 17:06
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    $\begingroup$ See also my comment to @CosmasZachos under the question. $\endgroup$ Commented Jul 9, 2020 at 17:20