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Aug 6, 2018 at 3:31 comment added user171780 I have found the answer to my previous question here.
Aug 6, 2018 at 3:13 comment added user171780 How did you found that $[p,x^j] =-i \hbar jx^{j-1}$?
Oct 19, 2017 at 18:10 vote accept John Doe
Oct 19, 2017 at 18:10 comment added John Doe I did it: $$F(x)=\sum_{j=0}^\infty f_j x^j;$$$$\begin{align} [p,F(x)]&=\sum_{j=0}^\infty f_j[p,x^j]\\ &=-i\hbar\sum_{j=0}^\infty f_j jx^{j-1}\\ &=-i\hbar\sum_{j=0}^\infty f_j \frac\partial{\partial x} x^j\\ &=-i\hbar \frac\partial{\partial x} \sum_{j=0}^\infty f_jx^j\\ &=-i\hbar \frac{\partial F(x)}{\partial x}\end{align}$$ Thanks!
Oct 19, 2017 at 18:07 history edited tparker CC BY-SA 3.0
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Oct 19, 2017 at 17:56 history answered tparker CC BY-SA 3.0