You have already got "practical" answers, so I intend to answer form another point of view.

There is a quite famous theorem due to Stone and von Neumann, later improved by Mackay, and finally by Dixmier and Nelson, roughly speaking establishing the following result within the most elementary version.
(Another version of the theorem focuses on the unitary groups generated by $X$ and $P$ avoiding problems with domains, however I stick here to the self-adjoint operator version.)

THEOREM. (rough statement "for physicists")
*If you have a couple of self-adjoint operators $X$ and $P$ defined on a Hilbert space $H$ such that are conjugated to each other:*

\begin{equation}
[X,P] = i \hbar I \quad\quad\quad (1)
\end{equation}
*and there is a cyclic vector for $X$ and $P$, then there exists a unitary operator $U : L^2(R, dx)\to H$ such that:*

$$(U^{-1} X U )\psi (x)= x\psi(x)\quad \mbox{and}\quad (U^{-1} P U )\psi (x)= -i\hbar \frac{d\psi(x)}{dx}\:.\quad (2)$$

(The rigorous statement, in this Nelson-like version is reads as follows

THEOREM. *Let $X$ and $P$ be a pair of self-adjoint operators on a complex Hilbert space $H$ such that (a) they veryfy (1) on a common invariant dense subspace $S\subset H$, (b) $X^2+P^2$ is essentially self-adjoint on $S$ and (c) there is a cyclic vector in $S$ for $X$ and $P$. Then there exists a unitary operator $U : L^2(R, dx)\to H$ such that (2) are valid for $\psi \in C_0^{\infty}(R)$.*

Notice that the operators defined in the right-hand sides of (2) admits unique self-adjoint extensions so they completely fix the operators representing respective observables. We can equally replace $C_0^\infty(R)$
for the Schwartz space ${\cal S}(R)$ in the last statement.)

Barring technicalities, all that means that *commutation relations actually fix position and momentum observables as well as the Hilbert space.* For instance, referring to Murod Abdukhakimov's answer, if the addition of $\partial f$ to the standard expressions of $X$ and $P$ gives rise to truly self-adjoint operators, then a unitary transformation (just that connecting $\psi$ to $\psi'$ in Murod Abdukhakimov's answer) gets rid of the deformation restoring the standard expression. Remember that unitary transformations do not alter all physical objects of QM.

The result extends to $R^n$, i.e., concerning particles in space for $n=3$.
Dropping the irreducibility requirement (there is a cyclic vector for $X$ and $P$) the thesis holds anyway but $H$ decompose into a direct sum (not direct integral!) of closed subspaces where
the strong statement is valid.

There are important consequences of this fundamental theorem.
First of all $H$ must be saparable as $L^2(R,dx)$ is. Moreover no time operator $T$ (conjugated with the Hamiltonian operator $H$) exists if the Hamiltonian operator id bounded below as physics requires. The latter statement is due to the fact that the theorem fixes the spectra of $X$ and $P$ as the whole real axes in both cases, so that the spectrum of $H$ would not be bounded below if $T,H$ were a conjugated pair of operators.
A similar no-go theorem arises concerning quantization of a particle on a circle when one tries to define position and impulse self-adjoint operators.
The attempt to solve these no-go results gave rise to more general formulation of quantum mechanics based on the notion of POVM and eventually turned out to be very useful in other contexts as quantum information theory.

An important observation is that Stone-von Neumann - MacKay - Dixmier -Nelson's result fails when dealing with infinite dimensional systems.
That is, roughly speaking, passing from the (symplectic space) of a finite number of particle to the (symplectic space) of a field. In that case the canonical commutation relations of $X_i$ and $P_j$ are replaced by those of the quantum fields. E.g:,

$$[\phi(t, x), \pi(t, y)] = i \hbar \delta(x,y) I$$

or more sophisticated versions of them. In this juncture, there exist infinitely many representations of the algebra of observables that cannot be connected by unitary operators. This is a well-known phenomenon in QFT or quantum statistical mechanics (in the thermodynamic limit).
For instance the free theory and the interacting theory of a given quantum field cannot be represented in the same Hilbert space once one assumes standard requirements on states and observables (the so called Haag's theorem and this is the deep reason why LSZ formalism uses the weak topology instead of the strong one as in standard quantum theory of the scattering).

If one includes superselections charges in the algebra of observables, non unitarily equivalent representations of the algebra arise automatically giving rise to sectors.

In QFT in curved spacetime the appearance of inequivalent representations of the algebra of observables is a quite common phenomenon due to the presence of curvature of the spacetime.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti