What does it mean for an operator to depend on position or momentum?

Question

While trying to provide an answer to this question, I got confused with something which I think might be the root of the problem. In the paper the OP was reading, the author writes $$\frac{d\hat{A}}{dt}=\frac{\partial \hat{A}}{\partial t}+\sum_k \frac{\partial \hat{A}}{\partial x_k}\frac{dx_k}{dt}$$

for an observable $\hat{A}$. The author says that this chain rule expansion may be made, as well as the same thing but in momentum, but one may not have both at the same time, which was the source of OP's question. I argued that this was essentially a matter of function definition, and the fact that you can't have an observable depending on both position and momentum, because the operators of which these quantities are eigenvalues don't commute. But that raised the question which I came to pose here: what does it mean for the operator to depend on the $x_i$ or on the $p_i$? Is it something like $$\hat{A}: \Gamma \rightarrow Hom(H)$$ where the operator is mapping points in the phase space to observables in the Hilbert space? If so, is that meaningful? What is the interplay of the phase space here? My confusion lies in the fact that points in phase space as far as my understanding go would correspond to the eigenvalues of the position and momenta operators, and as such can I talk about them meaningfully in this way seeing as they do not commute? Furthermore, what is the meaning of $$\frac{\partial \hat{A}}{\partial x_i}$$

Essentially, what I'm asking about is: how does one define precisely the dependence of an operator on the "classical" degrees of freedom of a system and their conjugate momenta?

EDIT
Following d_b's answer, supposing that the derivative is taken in a classical setting and only then does quantization occur, take as an example the Hamiltonian for a 3D particle in some potential, $$H(x_i,p_i)=\frac{p^2}{2m}+V(x_i)$$ surely we'd all agree that $$\frac{dH}{dt}=\sum_k \frac{\partial H}{\partial x_k}\dot{x_k}+\frac{\partial H}{\partial p_k}\dot{p_k}$$ then "putting a hat on top of it" wouldn't alter the dependence on the $p$'s.
So either this isn't what's meant, or somehow the act of quantization removes one dependence or the other, which I can't see how.

Answer 1

The linked reference does not contain OP's first equation depending only on the coordinate operators. In fact, the equation in the reference is \begin{align} \frac{dA}{dt} = \frac{\partial A}{\partial t} + \sum_{k\neq0}\frac{\partial A}{\partial q_k}\frac{\partial q_k}{\partial t} \end{align} where the $q_k$ defined on page 5 of the reference:

Let $q = (q_0, q_1, q_2, q_3, \ldots, q_n)$ be the set of observables of a physical system such as a particle or group of particles and take them to be the coordinates of an $n$-dimensional vector $q$ in $q$-space. Spatial coordinates and time are included and placed on the same footing as the other observables.

Apparently this set includes all observables of the system, potentially including the momenta among the "other observables."

I am not sure of the meaning of $\partial A/\partial q_k$. If pressed to interpret it, I would treat $A(q)$ as a function of classical variables (or formal symbols), take the derivative as usual, then promote the $q_k$ to operators. See also this answer.

Actually, reading the reference a little more carefully, it is not clear that the $q_k$ are meant to be interpreted as operators at all. Later in the paper on page 24, an operator $Q$ is associated with the observable $q$. My conclusion is that we are meant to interpret the $q$ as classical observables until their promotion to operators $Q$, although I admit I find the formalism of the reference rather opaque.

Answer 2

1. When one says that an operator $\hat{A}$ depends on $\hat{q}$ and $\hat{p}$ it is meaningless/ambiguous to view $\hat{A}$ as an actual function of $\hat{q}$ and $\hat{p}$ because the arguments do not commute: $$[\hat{q},\hat{p}]=i\hbar\hat{\bf 1}.\tag{1}$$

2. Instead one often considers an algebra isomorphism between the algebra of operators $\hat{A},\hat{B},\ldots$ (using composition $\circ$) and an algebra of symbols $A,B,\ldots $ (which are functions on phase space equipped with a corresponding star product $\star$).

   The most common symbol is the Weyl/symmetric symbol, cf. e.g. this Phys.SE post. The corresponding star product $\star$ is the Groenewold-Moyal star product.

3. When a symbols $A$ is differential function, it makes mathematical sense to consider derivatives $$ \frac{\partial A}{\partial q}, \quad \frac{\partial A}{\partial p}, \quad\text{etc},\tag{2} $$ cf. OP's question.

4. The Heisenberg EOM for an operators$^1$ $$ i\hbar\frac{d\hat{A}}{dt}~=~[\hat{A},\hat{H}]~\equiv~\hat{A}\circ\hat{H}-\hat{H}\circ\hat{A} \tag{3}$$ can then be transformed into an EOM for a symbol $$ i\hbar\frac{dA}{dt}~=~[A\stackrel{\star}{,}H]~\equiv~A\star H-H\star A. \tag{4}$$

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$^1$ We assume for simplicity no explicit time dependence.