Commutator of $x$ and $p^2$

I have a question: If I have to find the commutator $[x, p^2]$ (with $p= $) the right answer is: $[x,p^2]=x p^2 - p^2x = x p^2 -pxp + pxp - p^2x = [x,p]p + p[x,p] = 2hip$ But why can't I say: $[x,p^2]=x p^2 - p^2x = - x h^2 + h^2 x = 0$ ? Thank you for your reply.

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$\begingroup$ How do you derive '0' in the end? The point is that the 'multiply by $x$' operator and the 'differentiate with respect to $x$' operator don't commute, so you can't just blithely say that $xh^2\frac$ = $h^2\fracx$. $\endgroup$

Commented Sep 25, 2013 at 19:11 $\begingroup$ Of course, you can say that. But, it's is more involved. $\endgroup$ Commented Sep 25, 2013 at 20:25

5 Answers 5

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What you describe is a quite common situation which pops up when dealing with commutators of operators. On an appropriate space of functions $\mathcal D$ (like an $L^2$-space or the Schwartz space etc. ), the operators $x$ and $p$ are given by

for all $f\in \mathcal D$ and $x$ in the domain of $f$. In other words, $x(f)$ and $p(f)$ are elements in $\mathcal D$, i.e. functions. In particular $\frac$ is the derivative of $f$ w.r.t. $x$ at the point $x$, by convention.

is the operator that, evaluated at any $f$, gives the function $[x,p](f)$ s.t.

Similarly, $$[x,p^2]$$ is the operator that, once evaluated at any $f\in \mathcal D$, gives the function $[x,p^2](f)$, with