andresB said:
...there is another theory in which the momentum operator is not identified with the translation operator and commute with the position operator, namely classical mechanics...
I'm not sure what you are trying to say here. I will sketch the standard argument (which can be found in, e.g., Varadarajan's "Geometry of Quantum Theory") so you can point out which are the points you find problematic.
In classical mechanics, linear momentum is indeed the generator of translations. Let ##M## be a Riemannian manifold representing space. Consider now the cotangent bundle of ##M##, ##T^{*}[M]##, as phase space. Let ##\varphi_{t}## be a one-parameter group of diffeomorphisms in ##M## and ##X^{\varphi}## its tangent field. We can associate an observable to it (called "
momentum associated to this particular one-parameter group"), ##h_{\varphi}:T^{*}[M]\longrightarrow\mathbb{R}##, via the following
definition: ##h_{\varphi}(P)\doteq v_{x}(X^{\varphi}\mid_{x})##, for all ##P\in T^{*}[M]## and remembering that points of phase space are actually cotangent vectors in ##M## and so we use that ##P\equiv v_{x}## with ##v_{x}\in T_{x}^{*}## for some ##x\in M##.
In this sense, given the action of a symmetry, we define an observable associated to it and call it momentum. In the particular case of the action in ##M## of the full translation group, we can form the corresponding one-parameter groups. The associated observables are called "linear momentum in the ##j##-direction". If we take the action in ##M## of the rotation group ##SO(3)##, we can form the corresponding one-parameter groups and the associated observables are called "angular momentum with respect to a rotation in the ##i-j## plane". And so on. The thing that determines the type of momentum (e.g., linear, angular) is the underlying group. Doing the calculation on some canonical chart, one obtains the familiar formulas for these observables.
If we take the Poisson bracket defined by the natural symplectic form in ##T^{*}[M]##, i.e., ##\Omega\doteq\mathrm{d}\tau## where ##\tau## is the tautological 1-form, we get: ##h_{aX+bY}=ah_{X}+bh_{Y}## and ##h_{\left[X,Y\right]}=\left\{ h_{X},h_{Y}\right\}##. Thus, we obtain a realization of the Lie algebra of the Lie group ##G## acting in ##M## (since the tangent field to the one-parameter groups is a representative of an element of this algebra via ##
(Xf)(x)\doteq\left[\frac{\mathrm{d}}{\mathbf{d}t}f\left(\left(exp\, tX\right)\cdot x\right)\right]\mid_{t=0}##) in terms of the previously defined momentum observables associated to the action.
This basic structure is completely analogous in QM. The action of the symmetry group is implemented in terms of, e.g., Wigner symmetries. If ##\pi## is a unitary representation of the Lie group ##G##, we take the corresponding one-parameter unitary groups and (using Stone's theorem) define the associated quantum momentum operators via ##\mathrm{d}\pi(X)(\varphi)=i\frac{\mathrm{d}}{\mathrm{d}t}\left[\pi\left(exp\left(tX\right)\right)\left(\varphi\right)\right]\mid_{t=0}##. Again, we get a Lie algebra representation: ##i\mathrm{d}\pi(\left[X,Y\right])=\left[\mathrm{d}\pi(X),\mathrm{d}\pi(Y)\right]##. So, again, the thing that determines the type of momentum is the underlying group
If the Galilei group acts as symmetries in the quantum system, we obtain unitary projective representations ##U_g## of this group. Using group extension techniques, one can get true unitary representations of the extension and therefore to use some theorems to classify all the possible representations. Done this, one finds that all representations are equivalent to certain special representations which have the familiar form ##
\left[U_{h,\overrightarrow{u}}f\right](\overrightarrow{x})=s(h)f\left(\delta^{-1}(h)\left(\overrightarrow{x}-\overrightarrow{u}\right)\right)## in ##
L^{2}(\mathbb{C}^{s};\mathbb{R}^{3})
##
for the part corresponding to the Euclidean subgroup. The linear momentum is as usual defined as the infinitesimal generator of the translation group part. So, we get the usual ##
P_{i}\propto\frac{\partial}{
\partial x_{i}
}
## in this realization.
