# Commutator between square position and square momentum [duplicate]

This question already has an answer here:

I need (as a part of one exercise) to find commutator between $\hat{x}^2$ and $\hat{p}^2$ and my derivation goes as follows:

$$[\hat{x}^2,\hat{p}^2]\psi = [\hat{x}^2\hat{p}^2 - \hat{p}^2\hat{x}^2]\psi = - \hbar^2 x^2 \cdot \psi'' + \hbar^2 \frac{\partial^2}{\partial x^2}(x^2 \cdot \psi)$$

Now: $$\frac{\partial}{\partial x}(x^2 \cdot \psi) = 2x\cdot \psi + x^2 \cdot \psi'$$

$$\frac{\partial}{\partial x}(2x\cdot \psi + x^2 \cdot \psi') = 2 \cdot \psi + 2x \cdot \psi' + 2x \cdot \psi' + x^2 \cdot \psi''$$

And then

$$[\hat{x}^2,\hat{p}^2]\psi = (2 \hbar^2 + 4 \hbar^2 x \cdot \frac{\partial}{\partial x})\psi$$

So I can derive, that

$$[\hat{x}^2,\hat{p}^2] = 2 \hbar^2 \cdot (1 + 2\hat{x}\hat{p})$$

I can not found this derivation anywhere and wonder: am I correct? Can there be other way to derive this?

I can not deduce any physical meaning from it, so any subtle mathematical error may go unnoticed.

## marked as duplicate by Kyle Kanos, ACuriousMind♦, Chris Mueller, Qmechanic♦Mar 13 '15 at 15:21

• Here it is ;-) . – yuggib Mar 13 '15 at 12:55
• $[x^2, p] = 2ix$ is a consequence of $p$ generating translations. – Robin Ekman Mar 13 '15 at 13:02
• Hi Konstantin, welcome to phys.SE. 'Check my work' type questions are off-topic on our site. Can you rephrase the question to have a more concrete answer? – Chris Mueller Mar 13 '15 at 14:47
• Chris, actually my question is "what is commutator between square position and square momentum". I just tried to show my attempt to got correct answer. – Konstantin Vladimirov Mar 13 '15 at 15:00

$$[x^2,p^2]=x[x,p^2]+[x,p^2]x=x[x,p]p+xp[x,p]+[x,p]px+p[x,p]x=i\hbar(2xp+2px)=2i‌​\hbar[x,p]_+=4i\hbar xp +2\hbar^2\; ;$$
using the fact that $[x,p]=i\hbar$ and $[AB,C]=A[B,C]+[A,C]B$ .