Suppose one is given a self-adjoint operator $A$ acting on an infinite dimensional separable Hilbert space $\mathcal{H}$. Under what conditions can one find an operator $B$ such that $[A,B] = i$? And if this is possible, how does one construct $B$?
For simplicity, assume that $H \simeq L_2(\mathbb{R})$ and that $A = A(X,P)$ is a function of the position operator $X$ and momentum operator $P$.
What I ultimately want to know is the following. Given a self-adjoint operator $A(X,P)$, by Stone’s theorem we know that $A(X,P)$ generates a one-parameter unitary group $U(t) = e^{-iA(X,P) t}$. What I want to know is if there is a self-adjoint operator (observable) $B(X,P)$, such that $U(t)$ generates translations of $B(X,P)$. And how to find $B(X,P)$ given $A(X,P)$. I know the the condition for $A(X,P)$ to generate translations of $B(X,P)$ is that these operators must satisfy the canonical commutation relation $[A,B] = i$, and hence the reason for phrasing the question as I did. This is analogous to how $P$ generates translations in $X$, that is, $e^{-i Px} \left|x_0\right\rangle = \left|x+x_0\right\rangle$.