# Commutator $[\hat{p},F(\hat{x})]$ of Momentum $\hat{p}$ with a Position dependent function $F(\hat{x})$?

I heard from my GSI that the commutator of momentum with a position dependent quantity is always $-i\hbar$ times the derivative of the position dependent quantity. Can someone point me towards a derivation, or provide one here?

• Write the position dependent quantity in terms of its power series $f(q) = \sum \frac{1}{n!} f^n(0) q^n$ and find its commutator with $p$. – nervxxx Nov 19 '13 at 4:40
• – Qmechanic Nov 19 '13 at 7:42

$[p,F(x)]\psi=(pF(x)-F(x)p)\psi$
knowing that $p=-i\hbar\frac{\partial}{\partial x}$ you'll get
$[p,F(x)]\psi=-i\hbar\frac{\partial}{\partial x}(F(x)\psi)+i\hbar F(x)\frac{\partial }{\partial x}\psi=-i\hbar\psi\frac{\partial}{\partial x}F(x)-i\hbar F(x)\frac{\partial}{\partial x}\psi+i\hbar F(x)\frac{\partial }{\partial x}\psi$
from where you find that $[p,F(x)]=-i\hbar\frac{\partial}{\partial x}F(x)$
• I'd like to point out that an implicit (quite reasonable) assumption in this answer is that $(F(\hat x)\psi)(x) = F(x)\psi(x)$, namely that $F(\hat x)$ can be treated as a multiplication operator in the position space representation. This assumption would hold if, for example, $F$ were an analytic function of its argument which is essentially the assumption suggested by nerxxx in his comment. – joshphysics Nov 19 '13 at 5:51
• Actually it is always true. The most general definition of $F(\hat{x})$ is obtained from the spectral theory. With this definition, a function of a multiplicative operator is always multiplicative as well. – Valter Moretti Oct 28 '14 at 17:23