Is expectation value of $p^2$ equivalent to this integral?

Question

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?

Answer 1

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