Velocity operator, its expression and eigenvalues

Question

Cohen Tannoudji pp 215

Third postulate :The only possible result of the measurement of a physical quantity $\mathscr A$ is one of the eigenvalues of the corresponding observable $\mathbf A$.

pp 225

...there of course exists an operator associated with the velocity of the particle...

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 $\mathbf v=\frac{\mathbf p}{m}$ ,whose eigenvalues are the observed values of velocity.

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1. I've seen multiple times that we can't define velocity in quantum mechanics, but here I find that the eigenvalues of $\mathbf v$ can be the values of velocity

2. I've also seen that the velocity operator in Schrodinger picture is :

the $j$th velocity operator is defined as : $\frac{1}{i\hbar}[\hat{q}^j, \hat{H}]$ But we also have another expression for the velocity operator as $\mathbf v=\frac{\mathbf p}{m}$, so how do we reconcile the two.

Please help me, Thank you.

Answer 1

as the comments note, in non-relativistic quantum mechanics the velocity of a particle can be easily defined, just like in classical mechanics, as the time-derivative of the position. The corresponding operator is $\hat{v} = \frac{d}{dt} \hat{x}$ and using for example Heisenberg's equation one can get

$$ \hat{v} = -\frac{i}{\hbar}[\hat{x}, H] $$

which for a Hamiltonian of the form $H = \frac{\hat{p}^2}{2m} + V(\hat{x})$ will give $\hat{v} = -i [\hat{x}, \hat{p}^2] / (2m\hbar) = \hat{p}/m$ as we expect.

The eigenvalues of the velocity operator are the same ones as those of the momentum operator after dividing by the mass. And indeed it is tricky as the momentum operator doesn't have normalizable eigenmodes (in an infinite box), but one can define an appropriate limit and construct a wave packet with as small uncertainty in velocity as you like, such that for all intent and purposes a particle has a well-defined velocity.

You might be confused because it is often said that in quantum mechanics you can't know the velocity and the position of a particle, but this just means that as we reduce the uncertainty in the velocity, approaching a particle with a well-defined one, we are increasing the uncertainty in the position of the particle, and its wave-function in position space spreads wider and wider.