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I have exactly 0 clue on how to start this problem, but I would be forever grateful for a hint in the right direction.

Given the operators $\hat x=x$ and $\hat p=-i\hbar \frac{d}{dx}$, prove the following relation:

$$ [\hat x, g(\hat p)]=i\hbar\frac{dg}{d\hat p}. $$

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    $\begingroup$ Maybe you can do this in momentum space? What do the two operators look like there? $\endgroup$ – Prahar Mitra Jun 30 '13 at 2:51
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Deduce the general form of the commutator

$$[\hat{x},\hat{p}^n] $$

write your function as a power series of $\hat{p}$

$$g(\hat{p})=\sum_{n=0}^{\infty}g_{n}\hat{p}^n $$

apply linearity of the commutator and then you should get your result

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  • $\begingroup$ This is unnecessary given @Prahar's hint. (Also his hint yields a much quicker approach). $\endgroup$ – Will Jun 30 '13 at 11:54
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    $\begingroup$ Thank you; it's an interesting property which is still left as a later exercise for the problem set, but the momentum-space approach yields a simpler result, in this case. $\endgroup$ – Guillermo Angeris Jul 29 '13 at 19:16
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Like Prahar had said, the problem reduces fairly simply in momentum-space.

We note that, in such space: $\hat x = i\hbar\frac{\partial}{\partial p}$ and $\hat p=p$, thus, using some auxiliary function $f$: $$ [\hat x,\hat g(\hat p)]f=i\hbar\frac{\partial (\hat gf)}{\partial p}-i\hbar\, \hat g\frac{\partial f}{\partial p}=i\hbar\frac{\partial \hat g}{\partial p}f $$ By applying the product rule and reducing, this yields the correct result.

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