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The derivation for the Ehrenfest Theorem that I've seen uses the chain rule across $\frac{d}{dt}\langle\psi|\hat{A}|\psi\rangle$ giving three terms: $(\frac{d}{dt}\langle\psi|)\hat{A}|\psi\rangle + \langle\psi|\frac{\partial\hat{A}}{\partial t}|\psi\rangle + \langle\psi|\hat{A}(\frac{d}{dt}|\psi\rangle)$ Among these, the one in the middle includes a derivative of the operator $\hat{A}$.

After using the TDSE, we eventually get:

$\frac{d}{dt}\langle\psi|\hat{A}|\psi\rangle=\frac{1}{i\hbar}\langle[\hat{A},\hat{H}]\rangle+\langle\frac{\partial \hat{A}}{\partial t}\rangle \tag{0}$

Now, due to my lack of intuition as to what the derivative of an operator signifies, I tried to apply the chain rule instead like this:

$\frac{d}{dt}\langle\psi|\hat{A}\psi\rangle = (\frac{d}{dt}\langle\psi|)\hat{A}\psi\rangle + \langle\psi|\frac{d}{dt}(\hat{A}|\psi\rangle) \tag{1}$

Now, if $\frac{d}{dt}$ is just another operator, the following should hold:

$\frac{d}{dt}\hat{A} = [\frac{d}{dt},\hat{A}]+\hat{A}\frac{d}{dt} \tag{2}$

Therefore:

$\frac{d}{dt}\langle\psi|\hat{A}\psi\rangle = (\frac{d}{dt}\langle\psi|)\hat{A}\psi\rangle + \langle\psi|\hat{A}\frac{d}{dt}|\psi\rangle + \langle\psi|[\frac{d}{dt},\hat{A}]|\psi\rangle \tag{3}$

Now, using the TDSE, the first and second terms can be written as such:

$(\frac{d}{dt}\langle\psi|)\hat{A}\psi\rangle + \langle\psi|\hat{A}\frac{d}{dt}|\psi\rangle = \langle\psi|\frac{-\hat{H}}{i\hbar}\hat{A}|\psi\rangle + \langle\psi|\hat{A}\frac{\hat{H}}{i\hbar}|\psi\rangle = \frac{1}{i\hbar}\langle\psi|[\hat{A},\hat{H}]|\psi\rangle \tag{4}$

Thus:

$\frac{d}{dt}\langle\psi|\hat{A}\psi\rangle = \frac{1}{i\hbar}\langle\psi|[\hat{A},\hat{H}]|\psi\rangle + \langle\psi|[\frac{d}{dt},\hat{A}]|\psi\rangle = \frac{1}{i\hbar}\langle[\hat{A},\hat{H}]\rangle + \langle[\frac{d}{dt},\hat{A}]\rangle \tag{5}$

Now, if The Ehrenfest Theorem holds and my derivation is correct, this would seem to suggest that $\langle\frac{\partial \hat{A}}{\partial t}\rangle = \langle[\frac{d}{dt},\hat{A}]\rangle$

Is this relation correct? Can I, in general understand the time derivative of an operator in terms of its commutator with $\frac{d}{dt}$?

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    $\begingroup$ The time-derivative is not an operator on the Hilbert space. Your commutator is therefore of no meaning. See for example this PSE post and the links therein. $\endgroup$ Aug 10, 2021 at 20:30
  • $\begingroup$ I've descended into the rabbit hole and have come out much richer in understanding. Thanks for pointing me in the right direction @Jakob . Since your response is a comment and not an answer, I cannot mark the question as answered. Should I delete this question, then, or just leave it as is? $\endgroup$
    – Fawxl
    Aug 12, 2021 at 8:40
  • $\begingroup$ I've reposted the comment as an answer. Do not delete the question, because it could be helpful for future readers. You can also answer your own question, i.e. write an answer by yourself and include more details etc. $\endgroup$ Aug 12, 2021 at 9:08
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    $\begingroup$ Gotcha'. Not super familiar with SE, so thanks for helping on that side too! Marked your answer as correct. $\endgroup$
    – Fawxl
    Aug 12, 2021 at 9:37

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The time derivative is not an operator on the Hilbert space. Your commutator is therefore of no meaning. See for example this PSE post and the links therein.

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