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

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?

• – CAF
Commented Sep 15, 2020 at 21:01
• @CAF so as far as I got what you replied to that old post , am I right? Commented Sep 16, 2020 at 8:58
• yes I had replied to that older post, hope it helps!
– CAF
Commented Sep 16, 2020 at 22:01

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