quantum mechanics - Does the canonical commutation relation relate to the fact that momentum is the generator of spatial translations?

In classical mechanics momentum is the generator of spatial translations. This remains true in quantum mechanics. The way we define the momentum operator in one-dimension, for example, already shows that

$$\langle x |P |\psi\rangle =-i\hbar \dfrac{\partial}{\partial x}\langle x|\psi\rangle \, . $$

Now if we have a particle in one dimension and this particle is acted upon by a spatial translation, its position will change. In other words, its $x$ coordinate will change.

We have thus, two observables: the momentum $P$ and the position $X$. The momentum is the generator of translations. In that way, momentum generates transformations which directly affect the position $X$.

On the other hand we know that the canonical commutation relation (CCR) $$[X,P]=i\hbar$$ is enough to characterize $X$ and $P$.

Following my reasoning, is there a deeper connection between momentum being the generator of spatial translations and the CCR? How the two things relate? Can we interpret and make sense of the CCR by thinking about this point of view that momentum generates spatial translations?

Answer

Generically, given two self-adjoint operators $A,B$, the transformation of $A$ under the unitary transformation $U_B(t) := \mathrm{e}^{\mathrm{i}Bt}$ with parameter $t$ generated by $B$ through Stone's theorem is given for central $[X,Y]$ by a form of the BCH formula: $$ U_B(t)AU_B(t)^\dagger = \mathrm{e}^{\mathrm{i}Bt}A\mathrm{e}^{-\mathrm{i}Bt} = A + [B,A]t$$ that is, transforming $A$ by $B$ just shifts $A$ by the commutator of the two, and even if the commutator is not central, this still holds for infinitesimal $t$. This is the quantum version of the classical statement that the Poisson bracket of two functions on the phase space gives the infinitesimal shift of one by the transformation generated by the other, cf. this answer of mine.

In the case of $x$ and $p$, the commutator is unity, so transforming with parameter $t$ just shifts one observable by $t$. That is, the commutation relation is indeed the quantum mechanical version of the statement that the position operator generates translations in momentum, and the momentum generates translations in position.

Conversely, knowing that the translation operator is infinitesimally given by the momentum operator allows one to deduce the form of the momentum operator itself if the representation of the position operator is fixed, see this question. It is essentially the content of the Stone-von Neumann theorem that the commutation relations between position and momentum (or rather their exponentiated form, the Weyl relation between the translation in position and the translation in momentum) uniquely (up to commutation relation preserving unitary isomorphism) fixes the operators themselves.