# Commutator identities and Fourier transform

Is it possible to derive one side of the arrow below from the other by using only the Fourier transform and its reciprocal?

$$[\hat{p},f(\hat{x})]=-i\hbar f'(\hat{x}) \leftrightarrow [\hat{x},f(\hat{p})]=i\hbar f'(\hat{p})$$

• I don't know why you'd bother. If you can prove one, proving the other is just a trivial modification. – Chris Jan 20 '18 at 5:30
• In particular, you don't need even a Fourier transform. Just the definition of the commutator is sufficient to prove one from the other. – Chris Jan 20 '18 at 5:35
• – Qmechanic Jan 20 '18 at 7:48

Under some hypotheses on $f$ the answer is positive. I consider the simplest case below.
If $U$ is a unitary operator on the Hilbert space $H$ and $A: D(A) \to H$ is a self-adjoint operator over the same Hilbert space, form spectral calculus it arises that $$Uf(A)U^{-1} = f(UAU^{-1})\tag{1}$$ for every measurable function $f : \mathbb R \to \mathbb R$.
Regarding momentum $P$ and position $X$ operators over $H= L^2(\mathbb R, dx)$, it holds $$U P U^{-1} =-X\:,\quad U X U^{-1} =P \tag{2}$$ where $U : L^2(\mathbb R, dx) \to L^2(\mathbb R, dx)$ is the unitary operator given by Fourier(-Plancherel) transform. If $f : \mathbb R \to \mathbb R$ is for instance in the space ${\cal S}(\mathbb R)$ of Schwartz' functions (also using the fact that (1) this space is invariant under $X,P$ and functions of each such operator $g(X)$, $g(P)$ for $g \in {\cal S}(\mathbb R)$, (2) the explicit form of $P$ over ${\cal S}(\mathbb R)$, and that (3) $f \in {\cal S}(\mathbb R)$ entails $f' \in {\cal S}(\mathbb R)$), $$[P, f(X)] =-i\hbar f'(X)\:.$$ As a consequence, since $U$ preserves ${\cal S}(\mathbb R)$, $$[UPU^{-1}, Uf(X)U^{-1}]= U[P, f(X)]U^{-1} =-i\hbar Uf'(X)U^{-1}\:.$$ From (1) and (2). the found result can be written $$[X, f(P)] =i\hbar f'(P)\:.$$
• The relation is this one: $X$ and $P$ have the same CCR as $-P$ and $X$ with the written order so (using an equivalent form of Stone von Neumann theorem), there must exists a unitary map transforming the first pair into the second one. This is Fourier transform... – Valter Moretti Jan 24 '18 at 21:33