Quantum mechanics, Fourier transformation

Question

Why do we use $p=-i\hbar\frac{\partial}{\partial x}$ in quantum physics? (I know the reason for $i\hbar$, quantization). Is this right to say we can't measure velocity and position of electrons at the same time, so we use mathematical method, Fourier transformation?

Answer 1

We define the quantum Poisson bracket for the operators corresponding to dynamical variables $u$ and $v$ as $${\{u,v\}}=uv-vu=i\hbar[u,v]$$ where $[u,v]$ is the classical Poisson bracket. For the simple case of momentum and position (considering one dimension for simplicity), we get $$xp-px=i\hbar.\tag{*}$$ We now define the differentiation operator $\frac{d}{dx}$ acting on a ket $|\psi\rangle$ as $$\frac{d}{dx}|\psi\rangle=|\frac{d\psi}{dx}\rangle.$$ Using this we get $$\frac{d}{dx}x-x\frac{d}{dx}=1.$$ From here we can see that $$p=-i\hbar\frac{d}{dx}$$ satisfies the commutation relation $(*)$. Now it is not necessary for us to take $p$ as defined above but with a bit more work it can be shown that choosing a suitable representation (basis) allows (forces) us to take $p$ as above.

Refer to section $22$ of Principle of Quantum Mechanics by Dirac ($4^{th}$ edition) for more details.