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 $\frac{d}{dt}$?
Best Answer
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.