Basics on QM, uncertainity of a continous observable

Question

I know that this question may sound silly but i'm truly confused, but if i had a wave function just like one who describes a potential well, let's call it $\Psi(x)$ and I want to calculate the uncertainity of a momentum for example, we know that:

\begin{equation} \Delta P=\sqrt{\langle p^{2}\rangle-\langle p\rangle^2} \end{equation} Since $\Psi$ is a continous wave function I understand that $$ \langle P\rangle=\int_{\mathfrak{R}}\left( \Psi^{\ast}i\hbar\cdot - \frac{\partial}{\partial x}\Psi \right)dx$$ But, what about $\langle p^2 \rangle $? from the basics on QM I know that: $$ \langle\psi|p^2|\psi\rangle=\langle \psi|p(p|\psi\rangle) $$ the question is, for a continous variable the relation for $$\langle p^2 \rangle=\hbar \int _{\mathfrak{R}}\Psi^{\ast} \frac{\partial^2}{\partial x^2} \Psi dx$$ or it is?: $$\langle p^2 \rangle=-i \hbar \int _{\mathfrak{R}}\Psi^{\ast} \frac{\partial^2}{\partial x^2} \Psi dx$$

Answer 1

It is $$ p_x=-i\hbar\frac{\partial}{\partial x} $$ $$ p_y=-i\hbar\frac{\partial}{\partial y} $$ $$ p_z=-i\hbar\frac{\partial}{\partial z} $$ and $$ {\bf p}^2=-\hbar^2\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right). $$ The case you are considering is 1-dimensional and you can limit everything to $x$.

Answer 2

The momentum operator is a function: $\hat p (\psi) = -i\hbar \frac{\partial}{\partial x} \psi$. Applying this function twice:

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

Since $i\hbar$ is independent of $x$. Thus:

$$ \langle p^2 \rangle = -\hbar^2 \int \psi \frac{\partial^2 \psi}{\partial x^2} dx $$