# Quantum Mechanics – Generalization of Canonical Commutation Relation

commutatorquantum mechanics

The canonical commutation relation
$$[x,p] = i\hbar$$
can be generalized to
$$[p_i,F(\vec{x})] = -i\hbar\frac{\partial F(\vec{x})}{\partial x_i}, \ [x_i, F(\vec{p})] = i\hbar\frac{\partial F(\vec{p})}{\partial p_i}$$
according to Sakurai's Modern Quantum Mechanics.

The book requires readers to prove it in one problem, probably using power series method, since Sakurai claims it can be proved by repeatedly using $[A,BC]=[A,B]C+B[A,C]$.

I know the proof would be straightforward if
$F$ is an entire function, which can be expressed as a power series. But the situation seems to be not very clear if $F$ isn't an entire function?

Generally speaking functions of operators are defined as spectral functions, making use of the spectral theorem machinery.

The approaches based of power series usually have not a rigorous basis: Even identities like $$e^A = \sum_n \frac{1}{n!}A^n$$ are generally false if $A$ is an unbounded operator.

Nevertheless in most relevant physical cases some of the results obtained by formal (non-rigorous) manipulations can be obtained following alternative rigorous ways.

However the case you are considering is very easy. Suppose that $F$ is a smooth, generally non-analytic, complex-valued function bounded by some polynomial of $\vec{x}$. In this case both $F(\vec{x})$ and $p_i$ admit the space of Schwartz functions ${\cal S}(\mathbb R^3)$ as invariant space, so that $[p_i, F(\vec{x})]\psi$ is well defined for $\psi \in {\cal S}(\mathbb R^3)$. On that space $p_i$ coincides to $-i \hbar\frac{\partial}{\partial x_i}$. By direct computation: $$[p_i, F(\vec{x})]\psi = -i \hbar\frac{\partial}{\partial x_i} F(\vec{x}) \psi(x) + i \hbar F(\vec{x})\frac{\partial \psi}{\partial x_i}= -i\hbar \frac{\partial F}{\partial x_i} \psi\:.$$ We have obtained that $$\left([p_i, F(\vec{x})] + i\hbar \frac{\partial F}{\partial x_i} \right) \psi =0 \quad \forall \psi \in {\cal S}(\mathbb R^3)\:.$$ In other words, at least on the domain ${\cal S}(\mathbb R^3)$:
$$[p_i, F(\vec{x})] = -i\hbar \frac{\partial F}{\partial x_i}$$ In general, this identity holds on larger domains and, in fact, it can be extended exploiting some further know property of $F$ and some other properties like the fact that $\cal S(\mathbb R^3)$ is a core for $p_i$...

One could assume weaker hypotheses. The self-adjointness domain of $p_j$ is so made. $$D(p_j) := \{\psi \in L^2(\mathbb R^3)\:|\: \exists\: w\mbox{-}\partial_{x_j}\psi \in L^2(\mathbb R^3)\}$$ where $w\mbox{-}\partial_{x_j}\psi$ denotes the weak partial $j$-derivative of $\psi$. On that domain $p_j = w\mbox{-}\partial_{x_j}$ as expected.

If $F$ is, for instance, just $C^1$ and compactly supported, all above reasoning can be re-implemented with $\psi \in D(p_j)$, obtaining exactly the same result, using the fact that the weak derivative of a compactly supported $C^1$ function $F$ coincides with the standard one.