# Operator on Function of Momentum (QM)

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:

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

• Maybe you can do this in momentum space? What do the two operators look like there? – Prahar Mitra Jun 30 '13 at 2:51

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

• This is unnecessary given @Prahar's hint. (Also his hint yields a much quicker approach). – Will Jun 30 '13 at 11:54
• 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. – Guillermo Angeris Jul 29 '13 at 19:16

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.