Common observables and associated operators: operator momentum [duplicate]

Question

Starting from my previous question Commutators in quantum mechanics and considering that the commutator

$$\left[i\hbar\frac{\partial}{\partial x},x\right]=i\hbar, \tag{1}$$ the associated linear operator momentum (for example the momentum $p_x$ of the $x$-axis) is:

$$p_x\longrightarrow -i\hbar\frac{\partial}{\partial x}=\frac{\hbar}{i}\frac{\partial}{\partial x} \tag{2}$$ The association of $p_x$ with $-i\hbar\partial/\partial x$ is it a postulate or exist a proof that $$\left[i\hbar\frac{\partial}{\partial x},x\right]=i\hbar\color{red}{\boldsymbol{\equiv}[p_x,x]\,\,?} \tag{3}$$

Answer 1

The canonical commutation relation

$$[\hat x, \hat{p}_x]=i\hbar$$

can be considered a postulate of quantum mechanics. In the position representation where wave functions are functions of position, and the position operator is just multiplication by $x$, the momentum operator then can be chosen as

$$\hat{p}_x=-i\hbar\frac{\partial}{\partial x}$$

in order to satisfy the commutation relation.

A more physical way to think about this choice for the momentum operator is to consider a plane wave, $e^{i(k_xx-\omega t)}$. Operating on this with a momentum operator should give the momentum egenvalue $\hbar k_x$, and it does with that choice for $\hat{p}_x$.