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This question already has an answer here:

If you can only measure either position and momentum in quantum mechanics how would one find the average value of $x^{2}p^{2}$ for an infinite square well?

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marked as duplicate by Rob Jeffries, John Rennie quantum-mechanics Nov 4 '18 at 11:16

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    $\begingroup$ It does not make sense: that operstor is not selfadjoint so it is not an observable. $\endgroup$ – Valter Moretti Nov 3 '18 at 17:41
  • $\begingroup$ Related, and also. $\endgroup$ – Cosmas Zachos Nov 3 '18 at 18:42
  • $\begingroup$ Before going on, what's the point of your question? It has a whiff of an intermediate step for something, but what? Have you brushed up on the pitfalls of the problem? $\endgroup$ – Cosmas Zachos Nov 3 '18 at 18:50
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What you are asking for can indeed be found. It is simply $\int\psi^*(x)\ \left(-x^2\hbar^2\frac{\partial^2}{\partial x^2}\right)\psi(x)dx$. Substitute whichever state of the infinite square well you like and work out the integral.

The problem is that this is a different result from say $p^2x^2$ or $xpxp$ because the order of $x$ and $p$ cannot be swapped.

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    $\begingroup$ You've made a typo. I don't think this is correct. $\endgroup$ – Rob Jeffries Nov 3 '18 at 16:08

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