@hokhani I'd say it is perfectly possible, as understood as the time-derivative of the velocity. You can work in the Heisenberg picture in which it is the operators (and not the Schrödinger wave functions) that now have explicit time-dependence. Then the Heisenberg equation for the dynamics of the momentum ##\mathbf{p}## is
$$
i\hbar\frac{\text{d} \mathbf{p}}{\text{d}t} = \left[\mathbf{p}, H\right] + i\hbar\frac{\partial \mathbf{p}}{\partial t} \rm{,}
$$
where ##\left[\cdot, H\right]## is the commutator with the Hamiltonian ##H##. Note that the operator ##\mathbf{p}## corresponds to the
canonical momentum, which can be different from the
kinetic momentum ##\mathbf{p}_\text{kin} = m\mathbf{v}## which is the "usual" product of the particle's mass and its velocity (which has a well-defined operational meaning). In particular, the canonical momentum of the particle with electric charge ##q## placed in the electromagnetic field described by a vector potential ##\mathbf{A}## is given by ##\mathbf{p} = \mathbf{p}_\text{kin} + q\mathbf{A} = m\mathbf{v} + q\mathbf{A} ##. Then the kinetic energy of the particle ##\frac{m\mathbf{v}^2}{2} = \frac{\mathbf{p}^2_\text{kin}}{2m}## (which has a well defined operational meaning as it is defined with the kinetic momentum) enters into the Hamiltonian as ##\frac{\mathbf{p}^2_\text{kin}}{2m} = \frac{\left(\mathbf{p}-q\mathbf{A}\right)^2}{2m}## and it is with respect to this Hamiltonian (including also the potential-energy terms) that the commutator in the Heisenberg equation is to be calculated. So you actually can make sense of the "accelerations" that you speak about, you just need (in the presence of the electromagnetic fields) to remember the difference between ##\mathbf{p}## and ##m\mathbf{v}## when calculating your time-derivatives.
