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

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


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 $$[X, f(P)] =i\hbar f'(P)\:.$$

  • $\begingroup$ Valter - where can I read more about (1)? This seems like a very powerful result, I would like more details. Can you recommend a reference? $\endgroup$ – Frank Jan 20 '18 at 17:48
  • $\begingroup$ @Frank I think that every book on spectral theory (even mine) proves that identity. $\endgroup$ – Valter Moretti Jan 20 '18 at 18:21
  • $\begingroup$ How is this discussion related to this: en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem? $\endgroup$ – Frank Jan 24 '18 at 21:03
  • $\begingroup$ 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... $\endgroup$ – Valter Moretti Jan 24 '18 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.