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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

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  • $\begingroup$ Here it is ;-) . $\endgroup$ – yuggib Mar 13 '15 at 12:55
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    $\begingroup$ $[x^2, p] = 2ix$ is a consequence of $p$ generating translations. $\endgroup$ – Robin Ekman Mar 13 '15 at 13:02
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    $\begingroup$ 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? $\endgroup$ – Chris Mueller Mar 13 '15 at 14:47
  • $\begingroup$ 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. $\endgroup$ – 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 $[AB,C]=A[B,C]+[A,C]B$ .

  • $\begingroup$ What did he do wrong in his derivation, his answer was reasonable yet by taking another path the answer seems completely different, are both answers valid? $\endgroup$ – Mark A. Ruiz Mar 30 '16 at 17:38
  • $\begingroup$ I don't know what he did wrong, but his result is inequivalent to mine, so at least one of the two is indeed incorrect $\endgroup$ – yuggib Mar 31 '16 at 9:54
  • $\begingroup$ Ok he is wrong, he wrongly identified p, he should've wrote at the last line 2.x(-1/ih)p which would give the same result as yours. $\endgroup$ – Mark A. Ruiz Apr 4 '16 at 11:01

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