Expectation value of time derivative of operator vs. time derivative after operator

Problem 3.18 in Griffiths's Introduction to Quantum Mechanics (3rd ed.) asks to apply the generalised Ehrenfest theorem to operators like the Hamiltonian and momentum operator. The purpose of the exercise is to make classical formulas pop out of the equations. The general form is: $$\frac{d\langle Q\rangle}{dt} = \frac{i}{\hbar} [\hat H, \hat Q] + \left< \frac{\partial Q}{\partial t}\right>.$$ Now, when I applied this to the Hamiltonian in a stationary potential, my intuition told me it'd have to become:

$$\frac{d\langle H\rangle}{dt} = 0,$$

because this seems to reference conservation of energy. Similarly, for momentum, we should get:

$$m\langle a\rangle=\left<-\frac{\partial V}{\partial x}\right>,$$

which I know to resemble Newton's 2nd Law in a conservative force's potential. The problem I realised when solving these, was that it wasn't evident that $$\langle \partial \hat H/\partial t\rangle=0$$ or $$\langle \partial \hat p/\partial t\rangle=0$$: particularly, since linear operators (seem to) always act multiplicatively, I was interpreting $$\langle \partial \hat p/\partial t\rangle$$ as follows:

\begin{align} \left<\frac{\partial \hat p}{\partial t}\right>&=\left<\Psi(x,t)\mid\frac{\partial \hat p}{\partial t}\Psi(x,t)\right>\\&=\int^{+\infty}_{-\infty}\overline{\Psi(x,t)}\left(\frac{\partial \hat p}{\partial t}\right)\Psi(x,t)dx\\&=\int^{+\infty}_{-\infty}\overline{\Psi(x,t)}\frac{\partial}{\partial t}\Big(\hat p\:\Psi(x,t)\Big)dx \end{align}

I clearly am not the only one having trouble interpreting said derivative, and to that point, I think my worries have been answered in the linked threads (we should pretend like the derivative obliges us to look at $$\hat Q$$ as if it were a function that could explicitly depend on time, and derive the operator itself as such).

However, it got me wondering: what if I do want to express "the expected value of the operator that applies $$\partial/\partial t$$ after applying $$\hat Q$$"? The notation used in the generalised Ehrenfest theorem should not be interpreted as such, so the only other way I could see to express this, is writing $$\left<\frac{\partial}{\partial t}\hat Q\right>.$$ Is this correct? Why does the multiplicative notation of operators not apply in this theorem, but everywhere else (as far as I know from having read 130 pages), it does?

• Very short answer: $\partial/\partial t$ is not an operator. The space of states is the space of wavefunctions at a fixed time, not functions of $x$ and $t$. – Javier Apr 3 at 19:18
• @Javier: That makes sense for the most part. However, in the derivation of the theorem, we see terms of the form $\left<\frac{\partial\Psi}{\partial t}\mid\hat Q\Psi\right>$ and $\left<\Psi\mid\hat Q\frac{\partial\Psi}{\partial t}\right>$. I guess those don't count as expectation values of operators, then (just inner products between two functions)? Edit: you may add your comment as an answer below, and I will accept it. – Mew Apr 3 at 20:46

In the Schrödinger picture, the Hilbert space $$\mathcal{H}$$ is physically the set of states at a given time. A function like $$\psi(x,t)$$ is not a state, but a time evolution of a state. Operators are also a priori not time dependent: they take functions of $$x$$ and return functions of $$x$$. A time dependent operator is really an operator valued function; you have a time dependent operator if, to apply it to a wavefunction $$\psi(x)$$ you also need to know at what time you're taking the wavefunction. This is not the case for either $$X$$ or $$P$$.
This also shows that $$\partial/\partial t$$ is not an operator in the quantum sense of the word, because it acts on time evolutions of states, not on states. You can't apply $$\partial/\partial t$$ to $$\psi(x)$$. And like you say in your comment, things like
$$\langle \Psi | \hat{Q} \frac{\partial \Psi}{\partial t} \rangle$$
are not expectation values, just inner products; time dependent inner products, in fact. You need an evolving state $$|\Psi(t)\rangle$$ for it to make sense.