Time evolution of the operators vs. the expectation values

Question

The time evolution of a quantum mechanical operator $A$ (without explicit time dependence) is given by the Heisenberg equation

$$ \frac{d}{dt}A = \frac{i}{\hbar} \left[H,A\right] \tag{1}$$

where $H$ is the system's Hamiltonian. The time evolution of the corresponding expectation value is given by the Ehrenfest theorem

$$ \frac{d}{dt}\left\langle A\right\rangle = \frac{i}{\hbar} \left\langle \left[H,A\right]\right\rangle \tag{2} $$

However, as I have noticed, these can yield differential equations of different forms if $\left[H,A\right]$ contains expressions that do not "commute" with taking the expectation value. For example, let

$$\left[H,A\right]=\frac{dA}{df} \tag{3}$$

for some quantum number $f$. The states used for taking the expectation values in (2) clearly depend on $a$. From (1) and (2) the different time evolutions calculated using (3) are

$$ \frac{d}{dt}A = \frac{i}{\hbar}\frac{dA}{df} \tag{4}$$ and $$ \frac{d}{dt}\left\langle A\right\rangle = \frac{i}{\hbar}\left\langle \frac{dA}{df}\right\rangle \tag{5}$$

Note that clearly (5) is not the same as

$$ \frac{d}{dt}\left\langle A\right\rangle = \frac{i}{\hbar}\frac{d\left\langle A\right\rangle}{df}\tag{6}$$

which I would have expected, since then the equations looks equal to (4).

Did I make some mistake? Is this correct? What is the difference between (4)-(6)? Clearly, (4) and (6) give rise to the same solution (one for the operator and one for the expectation value) since the differential equations are the same, while this may not be the case for (5). In fact, from (5) alone, I don't see a clear way on how to construct a solution since $\left\langle A\right\rangle$ does not occur on the right hand side.

Answer 1

The "example" eq. (3) here is kind of a red herring. Let's just take the evolution equation eq. (1) and Ehrenfest's theorem eq. (2). What you're asking about is that taking the expectation value of eq. (1) and using eq. (2) results in $$ \frac{\mathrm{d} \langle A\rangle}{\mathrm{d}t} = \left \langle \frac{\mathrm{d}A}{\mathrm{d}t}\right\rangle,$$ which you seem to think isn't true, but the fact that taking the expectation value commutes with the time derivative is exactly the content of Ehrenfest's theorem.

Your example is just re-parametrizing the time evolution of the Heisenberg operator by $f = \frac{\mathrm{i}}{\hbar} t$, so this doesn't change anything. That you are really just reparametrizing you can see by plugging eq. (3) into eq. (1), yielding $$ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{i}}{\hbar}\frac{\mathrm{d}A}{\mathrm{d}f},$$ meaning $\frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\mathrm{i}}{\hbar}$ by the chain rule.

Answer 2

Note that $[H,A]$ is always the time derivative of $A$, modulo constants. If you want something physically different from a time derivative, then you should change the "generator" (of translations in time, i.e. time evolution) $H$. For example, the momentum operator $p$ generates translations, so $[p,A]$ gives you the spatial derivative of $A$ along the direction of $p$. That is, modulo constants, you always have $[p, A]=\nabla A$.