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I want to show that $$[\hat{x}^n,f(\hat{p})] = i\hbar\left(n\hat{x}^{n-1}\frac{\partial f(\hat{p})}{\partial \hat{p}}\right) + \mathcal{0}(\hbar^2)$$ where the last term means of order $\hbar^2$.

I have already shown that $[\hat{x},f(\hat{p})] = i\hbar \frac{\partial f(\hat{p})}{\partial \hat{p}}$

I think what I can do first is to expand the commutator like so: $[\hat{x}^n,f(\hat{p})] = \hat{x}^{n-1}[\hat{x},f(\hat{p})] + [\hat{x}^{n-1},f(\hat{p})]\hat{x} $

Then I expand the 2nd term. However, if do this some more times, I don't seem to get a closed form expression so I am stuck here.

I would be grateful for any advice

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We will show the sought for expression by induction. Assume that $$[\hat{x}^n,f(\hat{p})] = i\hbar\left(n\hat{x}^{n-1}\frac{\partial f(\hat{p})}{\partial \hat{p}}\right) + \mathcal{0}(\hbar^2)$$ which is true for $n=1$.

The $(n+1)$-th case is $$[\hat{x}^{n+1},f(\hat{p})] = \hat{x}^{n}[\hat{x},f(\hat{p})] + [\hat{x}^{n},f(\hat{p})]\hat{x}\\ =i\hbar x^n \frac{\partial f(\hat{p})}{\partial \hat{p}}+i\hbar\left(n\hat{x}^{n-1}\frac{\partial f(\hat{p})}{\partial \hat{p}}\right)\hat x + \mathcal{0}(\hbar^2) .$$ But using the same relation, $\frac{\partial f(\hat{p})}{\partial \hat{p}}\hat x=\hat x\frac{\partial f(\hat{p})}{\partial \hat{p}}-\left[\hat x,\frac{\partial f(\hat{p})}{\partial \hat{p}}\right] =\hat x\frac{\partial f(\hat{p})}{\partial \hat{p}}+ \mathcal{0}(\hbar)$, and thus $$[\hat{x}^{n+1},f(\hat{p})] = i\hbar\left((n+1)\hat{x}^{n}\frac{\partial f(\hat{p})}{\partial \hat{p}}\right) + \mathcal{0}(\hbar^2),$$ which completes the proof.

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  • $\begingroup$ Didn't even think about induction, yet it is so elegant. Thanks alot! But just for case n=1: I know it is true if the $\mathcal{0}(\hbar^2)$ is missing, but why is it still true with that term? $\endgroup$ – Eren May 17 '17 at 16:04

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