According to Arfken et al. Mathematical Methods p.277 $$[x,p^2]=xp^2 - pxp + pxp -p^2x =[x,p]p + p[x,p]= 2ip \, .$$ According to the text this follows solely from $[x,p]=i$.

I'm not understanding how the squared operator is being applied.

How is the commutator on squared operators supposed to work to get the middle terms $pxp$? Why would it not be $[x,p]p - p[x,p]$.


You can either think of the middle term as "adding zero" to an equality, or you can use the commutator relationship to replace $xp^2$ and $p^2 x$ with

$$xp^2 = (xp)p = (px + [x,p])p$$


$$p^2 x = p(px)= p(xp - [x,p]).$$

Using commutivity and substituting these into the original expression, you'll see the $pxp$ terms cancel:

$$xp^2 - p^2x = pxp + [x,p]p - pxp + p[x,p] = [x,p]p + p[x,p].$$

Since $[x,p]$ is a scalar, namely $i$, you can move $p$ to either side freely, and

$$[x,p]p + [x,p]p = 2[x,p]p = 2ip.$$

So you see, all we needed to prove this was the commutator $[x,p]$, and that $p$ commutes with it. You can follow a structurally identical line of reasoning to show by induction that

$$[x,p^n] = inp^{n-1}.$$


I think you're missing an $\hbar$, but it is due to the useful formula,

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

for which in your case, $n = 2$.

This formula I provided above is a special case of the more general formula, (in one dimension)

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

  • $\begingroup$ ah, I didn't notice that. They left off $\hbar$ and I think you left off a minus. (assuming the MIT youtube videos are the standard form). $\endgroup$ – user5389726598465 Oct 14 '18 at 19:43
  • $\begingroup$ They made the normalization that $\hbar=1$ $\endgroup$ – user5389726598465 Oct 14 '18 at 19:56

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