For a non-varying charge, your electric potential, $V$, is a function of the radial distance $r$. In Quantum Mechanics, the equivalence theorem permits us to derive a classical expression for a given physical observable, then convert it to an operator expression. We know a classical formula for the electric potential $V(r)$. To obtain a quantum mechanical operator for this potential, we simply convert $r$ to an operator, that is $r\to\hat{r}$. The operator for electric potential is then just $\hat{V} = V(\hat{r})$.
For the momentum operator, however, we can directly derive a quantum mechanical expression. Consider the free particle wave function:
$Ψ(x,t) = Ae^{i(kx - ωt)}$ where $A$ is a normalisation constant.
Using these facts: $E = hf = ħω$ ; $p = \frac{h}{λ} =\frac{2πħ}{λ}=ħk$,
we can rewrite $Ψ(x,t)$ as:
$Ψ(x,t) = Ae^{i\frac{(px - Et)}{ħ}}$
Now take the derivative of $Ψ$ with respect to $x$ (just to see if we can come up with something):
$\frac{∂Ψ}{∂x}=\frac{ip}{ħ}Ae^{i\frac{(px - Et)}{ħ}}=\frac{ip}{ħ}Ψ$
Now it's getting interesting. We rearrange the above expression into:
$\frac{ħ}{i}\frac{∂}{∂x}Ψ=pΨ$
Now this is an eigenvalue equation. We have an operator $\frac{ħ}{i}\frac{∂}{∂x}$ acting on $Ψ$ with eigenvalues $p$ which are momentum values. From this, we extract directly that the momentum operator must be:
$\hat{P}=\frac{ħ}{i}\frac{∂}{∂x}=-iħ\frac{∂}{∂x}$.
For simplicity,I have derived this expression in one dimension, but, of course, the partial derivative can be extended to 3D space as well. The general expression for the momentum operator is then:
$\hat{P}=-iħ(\frac{∂}{∂x}+\frac{∂}{∂y}+\frac{∂}{∂z}) = -iħ∇$.
