Force quantization in physics? since we know from classical physics that
$$ \vec F= \frac{d\vec p}{dt} $$
and the quantization of momentum $ \hat p = -i\hbar\nabla $
can we expect that the force in quantum physics should appear as
$$ \hat F_{x_i}= -i\hbar \frac{\partial ^2}{\partial t \partial x_i}$$
for the compoments of the force . :D
 A: Stick to one space dimension, as you appear conflicted about the vector structure involved. In the Heisenberg picture,
$$
\hat F = \frac{d\hat p}{dt} =\frac{1}{i\hbar} [\hat p, \hat H]\\
=\frac{1}{i\hbar} [\hat p,V(\hat x)]= -V'(\hat x),
$$
which is what you apply in the Ehrenfest theorem, as stressed in the comments.
To generalize to more dimensions, you vectorize both the left hand side, and the right hand side, utilizing $\nabla$ instead of  $~~'\equiv\partial_x$.
Your gradient representation of $\hat p$ is strictly in the  coordinate representation, $\hat p=-i\hbar \int\! dx ~|x\rangle \partial_x \langle x|$, which you need not stick to. The above expression is valid in any and all representations!
In the coordinate representation, then, the above commutator presents trivially, $-[\partial_x, V(x)]= -V'(x)$. As almost always, time and space partial derivatives act dramatically differently in Hilbert space...
A: Well, let us see heuristically!
$$
\hat{F}=-i\hbar\frac{\partial}{\partial t}\nabla
$$
On the other hand, let us assume we are dealing with a conservative force field
$$
F=-\nabla V
$$
Now to equate these two, we need the second $F$ to be an operator. We are thus faced with the question of giving an operatorial definition for potentials. This is not a settled textbook dogma and people, following Vaidman's maxim, are trying to do this; see for example here; the issue is still open --Vaidman told me-- as of last year. Being short of any clue for operatorial definition of $V$ let us just proceed with the simplest possibility
$$
\hat{F}\psi \stackrel{?}{=}-(\nabla V)\psi
$$
yielding
$$
i\hbar\frac{\partial}{\partial t}\nabla\psi\stackrel{?}{=}(\nabla V)\psi;
$$
seems like a dead-end to me. But let us take
$$
\hat{F}\psi :=-\nabla (V\psi),
$$
then
$$
i\hbar\frac{\partial}{\partial t}\nabla\psi=\nabla (V\psi),
$$
as time and position(s) variables are independent, they commute on the left-hand-side --Uncertainty principle buzzers be calm, in your terms I am commuting energy and momentum, which is fine-- therefore
$$
i\hbar\nabla\frac{\partial}{\partial t}\psi=\nabla (V\psi),
$$
it seems we can 'cancel' nablas from both sides but this is a subtle issue outside the scope of this answer; I have explored it in detail here. Let us assume for the moment that we can do this, so
$$
i\hbar\frac{\partial}{\partial t}\psi=V\psi,
$$
which must immediately ring a bell: it is part of the Schrödinger equation. So what has gone wrong?
I found out it turns out that the `right way' (in the sense of leading to Schrödinger equation) to deal with the Newton Second Law in quantum mechanics is to see momentum as a function of trajectory and time, i.e.
$$
\textbf{p}=\textbf{p}\big(\textbf{x}(t),t\big)
$$
and proceed with the chain rule.
