# Time derivative of expectation value of observable is always zero (quantum mechanics)

I think this is a nice question. It ultimately boils down to the following:

If $i\hbar\frac{d}{dt}|\psi\rangle = H|\psi\rangle$, then why does $i \hbar\frac{d}{dt}\big(A|\psi\rangle\big) \neq H\big(A|\psi\rangle\big)$, since $A|\psi\rangle$ is also a valid state vector?

The answer is a bit subtle. The time evolution of a quantum mechanical state takes the form of a *path* through the underlying Hilbert space - that is, a function
$$\psi: \mathbb R\rightarrow \mathcal H$$
$$t \mapsto \psi(t)\in \mathcal H$$
The Schrodinger equation tells us that the physical paths through the Hilbert space are such that

$$i\hbar\psi'(t)= H\big(\psi(t)\big)$$
In particular, the time derivative acts on the **function** $\psi$, while the Hamiltonian operator acts on the **state vector** $\psi(t)$. The standard Dirac notation obscures this by writing
$$i\frac{d}{dt}|\psi\rangle = H|\psi\rangle$$
from which it is easy to get the mistaken impression that it makes sense to differentiate a state vector with respect to time.

Armed with this clarification, the answer is that $\psi(t)$ being a physical path does not guarantee that $A\big(\psi(t)\big)$ is a physical path. The latter is merely the *image* of a physical path under the action of the function (operator) $A$.

This concept is not reserved for quantum mechanics. Think about classical physics. Newton's law applied to a free particle yields $\frac{d^2}{dt^2} x = 0$. Does this imply that $\frac{d^2}{dt^2}f(x) = 0$ for some arbitrary function $f$? Certainly not - for example, consider $f(x)=x^2$.

If $\psi(t)$ is a physical path, then one has that $$\frac{d}{dt}(A\psi(t)) = \frac{\partial A}{\partial t} \psi(t) + A \psi'(t) = \frac{\partial A}{\partial t}\psi(t) + A\big(\frac{1}{i\hbar}H\psi(t)\big)$$

Inserting this into the expectation value then yields the correct result,

$$\begin{align}\frac{d}{dt}\langle \psi(t),A\psi(t)\rangle &= \langle \psi'(t),A\psi(t)\rangle + \langle \psi(t),\frac{\partial A}{\partial t}\psi(t)\rangle + \langle \psi(t),A\psi'(t)\rangle\\&=-\frac{1}{i\hbar}\langle H\psi,A\psi\rangle +\frac{1}{i\hbar}\langle \psi,AH\psi\rangle + \left\langle\frac{\partial A}{\partial t}\right\rangle\\&=-\frac{1}{i\hbar}\langle \psi,HA\psi\rangle +\frac{1}{i\hbar}\langle\psi,AH\psi\rangle + \left\langle\frac{\partial A}{\partial t}\right\rangle\\&=\frac{1}{i\hbar}\left\langle[A,H]\right\rangle + \left\langle\frac{\partial A}{\partial t}\right\rangle\end{align}$$