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 $\widehat{p^2}$?

I already 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 an easy way to derive an expectation value for $\langle p^2 \rangle$ and its QM operator $\widehat{p^2}$?

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 $\widehat{p^2}$???

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 $\widehat{p^2}$?