What does $\dot x$ mean as an operator in quantum mechanics? I've been looking at a paper titled "Feynman's proof of the Maxwell Equations" by Freeman Dyson (American Journal of Physics 58, 209 (1990); https://doi.org/10.1119/1.16188) and I'm confused by his notation, as in these equations (original equation numbering):
$$m\ddot x_j = F_j(x,\dot x,t) \tag{1}$$
$$[x_j, x_k] = 0 \tag{2} $$
$$m[x_j,\dot x_k] = i\hbar \delta_{jk} \tag{3} $$
$$[x_j, F_k] + m[\dot x_j, \dot x_k] = 0 \tag{9}$$
Everything inside a commutator has to be an operator to make sense.  I understand $x_j$ as the operator meaning "multiply by the scalar $x_j$", but what do $\dot x_j$ and $\ddot x_j$ mean as operators?  Is (2) trivially true, since multiplication by a scalar is commutative?
 A: $\dot{\hat{x}}$ means the same thing it does in Newtonian mechanics.
$$\dot{\hat{x}} = \frac{d\hat{x}}{dt}.$$
The trick is this is happening in the Heisenberg picture, not the Schrodinger picture that textbooks typically start with. There, you set an initial quantum state and then let the operators evolve with time. Hence that $\hat{x}$ acts on a positional wave function by multiplying by the position is only usually so at time zero.
The commutator relation,
$$m [\hat{x}, \dot{\hat{x}}] = i\hbar$$
establishes what $\dot{\hat{x}}$ is at time zero. Namely $-\frac{i\hbar}{m} \frac{\partial}{\partial x}$ when working on a positional wave function and where $\hat{x}$ is as you described (multiplying by position).
A: *

*In the Heisenberg picture operators evolve in time, and
dot denotes a time derivative
$$ \dot{\hat{x}}^j~:=~\frac{d\hat{x}^j}{dt}, \qquad 
\ddot{\hat{x}}^j~:=~\frac{d^2\hat{x}^j}{dt^2},\tag{A}$$
cf. the answer by The_Sympathizer.


*Normally in the Heisenberg picture we assume a Hamiltonian time evolution, so that we can effectively define the velocity operator via Heisenberg's EOM
$$ \dot{\hat{x}}^j~=~\frac{1}{i\hbar}[\hat{x}^j,\hat{H}],\tag{B}$$
cf. e.g. this Phys.SE post.


*However, it is an important point in Feynman's proof that he does not assume eq. (B) nor the standard CCR, cf. section II in Ref. 1. Instead he assumes eqs. (1)-(3).

*

*E.g. the force  $\hat{F}^j$ in eq. (1) could in principle contain dissipative forces incompatible with a Hamiltonian formulation.


*Also note that the commutator $[\dot{\hat{x}}^j,\dot{\hat{x}}^k]$ is not necessarily $0$, so we cannot directly apply the Stone-von Neumann theorem on eqs. (2)-(3).
References:

*

*F.J. Dyson, Feynman’s proof of the Maxwell equations, Am. J. Phys. 58 (1990) 209.

