0

Cohen-Tannoudji pp 223

The observable $\mathbf A$ which describes a classically defined physical quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively

From above we can say that there exists a velocity operator given as $\displaystyle\mathbf v=\mathbf {\frac{d\hat R}{dt}}$

I've never seen such an expression for velocity operator and I suspect it's wrong. If it's wrong why is it so?

Mauricio
  • 5,273
Kashmiri
  • 1,220
  • 13
  • 33
  • 4
    Does this cover your question: https://physics.stackexchange.com/q/430118/? (TL;DR: You can define such an operator, it will come out as $\mathbf{v} = \mathbf{p}/m$ in absence of a magnetic field, it is not really useful outside the semi-classical limit). – Sebastian Riese Jan 31 '23 at 16:49
  • 4
    Does this answer your question? Velocity definition in quantum mechanics? – Sebastian Riese Jan 31 '23 at 16:49
  • The formula in this case that is meant is $\vec{v} = \vec{p}/m$ where $\vec{p}$ is the momentum operator. I don't know of any variable for which the equation used to quantize includes derivatives of operator quantities. Maybe that's a rule of thumb you can go by. – doublefelix Jan 31 '23 at 16:51
  • @SebastianRiese, thanks for the links. My question is that why aren't the quantization rules followed to obtain velocity operator. Defining v=p/m isn't correct in a strict sense because the p used here isn't the mechanical momentum, it's the conjugate momentum – Kashmiri Jan 31 '23 at 17:00
  • 1
    More precisely: $v = d_t R = \frac{i}{\hbar} [H, P]$ (that's where my restriction the the "no magnetic field comes from" – which was an imprecise gloss for the case $H = P^2/2m + V(R)$). In my opinion thinking to much on "quantization rules" is the wrong direction – quantization rules are historical crutches to get the quantum equations of motion from the classical ones, while reality is the other way around (classical mechanics in an approximation of quantum systems). – Sebastian Riese Jan 31 '23 at 17:17
  • 2
    Part of the problem here is that the usual quantization scheme is following the Hamiltonian point-of-view where $P$ is the canonical momentum not just $p/m$. – Mauricio Jan 31 '23 at 17:33

0 Answers0