Canonical commutation relation

Remarks

On $\mathcal{S}(\R)$ one has $(X\psi)(x) = x\psi(x)$ and $P\psi = -i\hbar\psi'$, and both maps preserve $\mathcal{S}(\R)$. By the product rule, $P(X\psi) = -i\hbar(\psi + x\psi')$, while $X(P\psi) = -i\hbar x\psi'$. The difference is $i\hbar\psi$.

In $\R^n$, with $X_j$ and $P_k = -i\hbar\, \partial/\partial x_k$, the same computation gives $[X_j, P_k] = i\hbar \delta_{jk} I$, $[X_j, X_k] = 0$ and $[P_j, P_k] = 0$ on $\mathcal{S}(\R^n)$.

No pair of bounded operators $A, B$ on a nonzero Hilbert space satisfies $AB - BA = cI$ with $c \ne 0$. In finite dimension $n$ this follows from the trace: $\operatorname{tr}(AB - BA) = 0 \ne cn$. So at least one of $X, P$ must be unbounded, and the relation can only hold on a subspace such as $\mathcal{S}(\R)$.

The relation mirrors the Poisson bracket $\set{x, p} = 1$ of classical mechanics, with $[\cdot, \cdot]/(i\hbar)$ in place of $\set{\cdot, \cdot}$. The Stone–von Neumann theorem says that, in the exponentiated Weyl form, the relation has essentially one irreducible representation, the Schrödinger representation on $L^2(\R^n)$.

Going deeper

For a student

The relation is the defining rule of canonical quantization: one replaces classical coordinates and momenta by operators whose commutators are $i\hbar$ times the classical Poisson brackets. With the Robertson inequality it gives $\sigma_X \sigma_P \ge \hbar/2$ for states in $\mathcal{S}(\R)$. Domain questions are real. Some pairs of self-adjoint operators satisfy the relation on a dense invariant domain but are not unitarily equivalent to the Schrödinger pair. This is why the Stone–von Neumann theorem uses the Weyl form.