# Momentum operator ambiguous?

In nonrelativistic quantum mechanics, are different operators possible as a candidate for the momentum operator, given that one has fixed one position operator and a hilbert space that this position operator acts uppon?

If I choose the Hilbert space to be the space of square integrable functions, and the position operator to be $x$, then the usual choice for the momentum operator is $- i \frac{\partial}{\partial x}$. However, there are other choices, that would also satisfy the commutation relations:

\begin{align} -i \frac{\partial}{\partial x} + f(x) \end{align}

With an abitrary function $f$ would satisfy the communtation relations just as well. Is there some kind of "gauge freedom" that is fixed by choosing $\hat{p} = - \frac{\partial}{\partial x}$, or are there additional assumptions that $\hat{p}$ has to satisfy?

EDIT Iread in other answers that the momentum operator has to generate translations, and this answer mentions that the property of being the generator of translations is linked to the property of the commutation relation. It however also states that the choice of the momentum operator is a unique choice. Am I missing something? If the ccr's are satisfied, then the given formula of a unitary shift of the position operator can be used, resulting in the said translation.

The canonical commutation relations in the Lie algebra form $[x,p]=i$ do not generate necessarily the Heisenberg group, and therefore in that form there is no uniqueness result.