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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?

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    $\begingroup$ 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 $p$. $\endgroup$
    – nervxxx
    Nov 19, 2013 at 4:40
  • $\begingroup$ Related: physics.stackexchange.com/q/78222/2451 $\endgroup$
    – Qmechanic
    Nov 19, 2013 at 7:42

1 Answer 1

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You start from this

$[p,F(x)]\psi=(pF(x)-F(x)p)\psi$

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 $[p,F(x)]=-i\hbar\frac{\partial}{\partial x}F(x)$

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    $\begingroup$ 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. $\endgroup$ Nov 19, 2013 at 5:51
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    $\begingroup$ 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. $\endgroup$ Oct 28, 2014 at 17:23

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