Why $-ihbarvecnabla$ for momentum in quantum mechanics, while $mvec{v}$ in classical mechanics?

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.

Answer

$-i ħ \nabla$ 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. $E \Psi = - \frac{ħ^2}{2m}\nabla^2 \Psi = \frac{\left(-iħ\nabla\right)^2}{2m} \Psi = \frac{p^2}{2m} \Psi$