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Let $\psi(x)=Ne^{iax -\frac{m^2x^2}{2} -ibt}$ and I want to compute the possibility of momentum $p$. By definition : $\langle p^2\rangle=\int_{-\infty}^{\infty}\psi^*p^2\psi dx$. Is that equivalent to $\langle p^2\rangle=\int_{-\infty}^{\infty}(p\psi^*)(p\psi) dx= \int_{-\infty}^{\infty}(p\psi)^*(p\psi) dx$? Is there any particular reason to choose one form from another ? Maybe easier computions? And is there any extra meaning/interpretation in physics for this equivalence?

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    $\begingroup$ See physics.stackexchange.com/q/57739 $\endgroup$
    – CAF
    Sep 15, 2020 at 21:01
  • $\begingroup$ @CAF so as far as I got what you replied to that old post , am I right? $\endgroup$ Sep 16, 2020 at 8:58
  • $\begingroup$ yes I had replied to that older post, hope it helps! $\endgroup$
    – CAF
    Sep 16, 2020 at 22:01

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No it's not equivalent. $p^2 \psi$ means that you apply the momentum operator twice to $\psi$. So it would be

$-i\hbar\frac{\partial}{\partial x}(-i\hbar\frac{\partial}{\partial x}\psi)=-\hbar^2\frac{\partial^2 \psi}{\partial x^2}$

Which is not the same as $(p\psi^*)(p\psi) = (-i\hbar\frac{\partial \psi^*}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})= -\hbar^2\frac{\partial \psi^*}{\partial x}\frac{\partial \psi}{\partial x}$

$\int_{-\infty}^{\infty}\psi^*p^2\psi dx$ means $-\hbar^2\int_{-\infty}^{\infty}\psi^*\frac{\partial^2 \psi}{\partial x^2} dx$

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    $\begingroup$ yes it is true , but some how it seems that the final results after intergration lead to the same values.. check the comment I got above $\endgroup$ Sep 16, 2020 at 8:59

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