I am a little bit confused when thinking of the momentum representation in QM and CM.
In QM, momentum is represented as $-i\hbar\vec\nabla$, while in classical, momentum is represented as $m\vec{v}$.
At least, where does the mass $m$ in CM gone when meets QM please?
Once I saw a sentence like "What quantum theory really unites is matter and information" from Prof. Xiao-Gang Wen's PPT file. Though I do not understand this sentence at all for the moment.
$-i ħ \nabla$ is the momentum operator. You have to apply it to a wave function to get the actual momentum.
Consider the plane wave solution to the Schrödinger equation: $\Psi = e^{i \mathbf{k} \cdot \mathbf{r} - \omega t}$. Applying the momentum operator gives $-i ħ \mathbf{k} \Psi$. You can see the eigenvalue has units of momentum. (If you can't see it, note that $\mathbf{k} \cdot \mathbf{r}$ in the exponent is dimensionless, so clearly $\mathbf{k}$ has units of inverse length. $ħ$ has units of angular momentum, so $ħ \mathbf{k}$ has units of momentum.)
As far as where the mass of the particle factors in, it's in the Schrödinger equation (and thus related to the wave function): $i ħ \frac{\partial}{\partial t} \Psi \left(\mathbf{r}, t \right) =\left[-\frac{ħ^2}{2m}\nabla^2 + V \left(\mathbf{r}, t \right)\right]\Psi \left(\mathbf{r}, t \right)$
In particular, the classical relationship between momentum and kinetic energy is $E = \frac{p^2}{2m}$. (That's the same as your $mv$, for $E = \frac{1}{2}mv^2$.) Note for the free particle in quantum mechanics, it's the same. $E \Psi = - \frac{ħ^2}{2m}\nabla^2 \Psi = \frac{\left(-iħ\nabla\right)^2}{2m} \Psi = \frac{p^2}{2m} \Psi $
I) On one hand $-i\hbar{\bf\nabla}$ is the Schrödinger (position) representation of the the canonical/conjugate momentum operator $\hat{\bf p}$ in order to satisfy the CCR
$$\tag{1} [x^i, p_j]~=~\hbar {\bf 1}~\delta^i_j. $$
II) On the other hand, $m\hat{\bf v}$ is the kinetic/mechanical momentum operator. (Let us for simplicity imagine a non-relativistic setting.)
III) These two momentum concepts are not the same, for instance in the presence of an electromagnetic field, see e.g. this Phys.SE answer.
IV) Let us now assume that we are in a situation where the canonical/conjugate momentum operator $\hat{\bf p}$ and the kinetic/mechanical momentum operator $m\hat{\bf v}$ coincide. Then we may represent the velocity operator as
$$\tag{2} \hat{\bf v}~=~\frac{\hbar}{im}{\bf\nabla}$$
in the Schrödinger (position) representation.
