quantum mechanics - Deriving a QM expectation value for a square of momentum $langle p^2 rangle$

I alredy derived a QM expectation value for ordinary momentum which is:

$$\langle p \rangle= \int\limits_{-\infty}^{\infty} \overline{\Psi} \left(- i\hbar\frac{d}{dx}\right) \Psi \, d x$$

And i can read clearly that operator for momentum equals $\widehat{p}=- i\hbar\frac{d}{dx}$. Is there any easy way to derive an expectation value for $\langle p^2 \rangle$ and its QM operator $\widehat{p^2}$?

Answer

Well, $\widehat{p^2} = \hat{p}^2= \hat{p} \hat{p}$.

So, in the position basis it is $-\hbar^2 \frac{d^2}{dx^2}$, and $\langle p^2 \rangle = \int_{-\infty}^\infty \bar{\Psi}\left(-\hbar^2 \frac{d^2}{dx^2} \right)\Psi dx$.

Note: $\hat{p}$ is technically not equal to $-i\hbar d/dx$, but rather in the position basis $\langle x | \hat{p}| x' \rangle = -i\hbar d/dx \delta(x-x')$.