What is velocity in QM?

Question

I've always thought that velocity is the quantity $\vec v=\frac {d \vec x} {dt}$ by definition. That is, velocity is a quantity whose measurement is the above operation of the quantities $\vec x$ and $t$.

Then I studied quantum mechanics and I've seen the indetermination principle:it is impossible to know position and momentum $P$ simultaneously.

Taking $P=m \vec v$ as the definition of momentum involves that velocity can't be the quantity $\vec v=\frac {d \vec x} {dt}$ by definition, because it would violate the indetermination principle, indeed:

1. To know velocity I need to know $x(t)$ and $x(t+dt)$ that means I know the position.
2. Moreover, imagine to have a system (whose position is undetermined), if the above definition of velocity is correct than when we measure velocity the result would be infinite, because $x(t)$ can be completely different respect to $x(t+dt)$

So, what is the definition of velocity?

Answer 1

The easiest way to compute time derivatives in QM is in the Heisenberg picture. We know that for any operator $\hat{A}$ the Heisenberg equation of motion holds: $$\frac{d}{dt} \hat{A} = \frac{1}{i\hbar} [\hat{A},\hat{H}] + \frac{\partial}{\partial t} \hat{A}$$ More specifically for the position operator, the partial derivative will be zero, giving us $$\frac{d}{dt} \hat{x} = \frac{1}{i\hbar} [\hat{x},\hat{H}]$$ and, considering an Hamiltonian of the type $\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat x)$, this further simplifies to $$\frac{d}{dt} \hat{x} = \frac{1}{i\hbar} \left[\hat{x},\frac{\hat{p}^2}{2m}\right]$$ because $V(\hat x)$ commutes with $\hat{x}$. Using the commutation relation $[\hat{x},\hat{p}^2] = 2 i \hbar \hat{p}$ we obtain $$\frac{d}{dt} \hat{x} = \frac{\hat{p}}{m}$$ Which is consistent with the classical view that $p = m \dot{x}$.