quantum mechanics - Does the canonical commutation relation fix the form of the momentum operator?

For one dimensional quantum mechanics $$[\hat{x},\hat{p}]=i\hbar. $$

Does this fix univocally the form of the $\hat{p}$ operator? My bet is no because $\hat{p}$ actually depends if we are on coordinate or momentum representation, but I don't know if that statement constitutes a proof. Moreover if we choose $\hat{x}\psi=x\psi$ is the answer of the following question yes?

For the second one

$$(\hat{x}\hat{p}-\hat{p}\hat{x})\psi=x\hat{p}\psi-\hat{p}x\psi=i\hbar\psi, $$

but I don't see how can I say that $\hat{p}$ must be proportional to $\frac{\partial}{\partial x}$. I don't know if trying to see that $\hat{p}$ must satisfy the Leibniz rule and thus it should be proportional to the $x$ derivative could help. Or using the fact that $\hat{x}$ and $\hat{p}$ must be hermitian

Any hint will be appreciated.

Answer

No. You can add an arbitrary constant shift (or an arbitrary operator commuting with $x$) without affecting the CCR.

For 1-dimensional QM, the general solution of the CCR with $\hat x$ represented as multiplication by $x$ on wave functions with argument $x$ is $\hat p=\hat p_0-A(\hat x)~~$, where $\hat p_0$ is the canonical momentum operator , and $A(x)$ is an arbitrary function of $x$.
Proof. The difference $\hat A:=\hat p_0-\hat p~$ commutes with $\hat x$, hence is a function of $\hat x$.