why there is no direct velocity operator on quantum mechanic while there is for mumentum ( $p_{x}=d/dx$ ) Also why use mumentum space not velocity?
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1$\begingroup$ So what is wrong with $\hat v = \frac{\hat p}{m}$? The kinetic energy operator is also defined via the momentum operator. $\endgroup$– FarcherCommented Feb 10, 2016 at 9:17
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$\begingroup$ No wrong, but just no use of classical form of $v$ for example. $\endgroup$– asaaCommented Feb 10, 2016 at 9:19
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$\begingroup$ @asaa can you shed more details on your question? $\endgroup$– Bruce LeeCommented Feb 10, 2016 at 9:33
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$\begingroup$ Edited, just as indirectly operator, $\endgroup$– asaaCommented Feb 10, 2016 at 9:43
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There is a speed operator in quantum mechanics, as there's a time derivative operator for all operators, using the Heisenberg equation :
\begin{equation} \frac{d}{dt}A(t)=\frac{i}{\hbar}[H,A(t)]+ \frac{\partial A}{\partial t} \end{equation}
For speed, this will be
\begin{equation} v = \frac{d}{dt}x=\frac{i}{\hbar}[H,x] \end{equation}
A simple Hamiltonian is $H = \frac{p^2}{2m} + V(x)$. $x$ will commute with the potential, leaving
\begin{equation} v = \frac{i}{\hbar2m}[p^2,x] = \frac{p}{m} \end{equation}
which is the same relation as in classical mechanics, except with operators. A similar relation exists for $F = ma$, which is $\frac{dp}{dt} = \frac{i}{\hbar}[H,p] = -\nabla V(x)$.
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$\begingroup$ except on heisenberg uncertainty principle and $Δx=<x^2>-<x>^2$ is there other place (in QM) that we see $Δx$? $\endgroup$– asaaCommented Feb 10, 2016 at 10:05