Commutator of $[\hat p, F(\hat x)]$ without using $\hat p=-i\hbar\frac\partial{\partial x}$? I have been able to prove this relation by using a certain method, but it uses the fact that $$\hat p=-i\hbar\frac\partial{\partial x},\tag{1}$$ which is a relation I have avoided so far, so I wish to prove it without using that ($\hat p$ has simply been defined as the generator of spatial translation). Is it possible to prove that $$[\hat p,F(\hat x)]=-i\hbar\frac{\partial F(\hat x)}{\partial x}.\tag{2}$$ without using this relation? I can't see how to get even the first step.
 A: Write $F(\hat{x})$ as a power series with arbitrary coefficients. Use linearity to express $[\hat{p}, F(\hat{x})]$ as a linear combination of commutators of the form $[\hat{p}, \hat{x}^n]$. You can calculate the latter commutator using induction and the identities $[A, BC] = B [A, C] + [A, B]C$ and $[\hat{x}, \hat{p}] = i \hbar$. Then recombine the power series into a single expression.
A: You can define the momentum operator as the generator of translations. Abstractly, the translation operator $T_a = \exp(-ia\hat{p})$ acts on eigenstates of position via $T_a|x\rangle = |x+a\rangle$. Hence, if $\langle x|\psi\rangle = \psi(x)$ then $\langle x| \exp(ia\hat{p})|\psi\rangle = \psi(x+a)$.
Expanding the exponential and the result in powers of $a$ tells us the action of the generator:
$$\langle x|(I+ia\hat{p})|\psi\rangle = \psi(x+a) = \psi(x)+a\frac{\partial\psi}{\partial x}$$
$$\Rightarrow \langle x| \hat{p}|\psi\rangle = -i\frac{\partial\psi}{\partial x}$$
In other words the $\hat{p}$ operator acting on a state with wavefunction $\psi$ yields a state with wavefunction $-i\frac{\partial\psi}{\partial x}$.
