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

Question

I already derived a QM expectation value for ordinary momentum which is:
$$
langle p rangle=  intlimits_{-infty}^{infty} overline{Psi}  left(- ihbarfrac{d}{dx}right) Psi , d x
$$
And I can read clearly that operator for momentum equals $widehat{p}=- ihbarfrac{d}{dx}$. Is there an easy way to derive an expectation value for $langle p^2 rangle$ and its QM operator $widehat{p^2}$?

nervxxx · Accepted 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 $-ihbar d/dx$, but rather in the position basis $langle x | hat{p}| x' rangle = -ihbar d/dx delta(x-x')$.

dannygoldstein · Answer

The expectation value for some operator $A$ is given by

$$langle Arangle = int Psi^*APsi.$$

If we set $A=p$ then we get the expression you've written above. Now just set $A = p^2$ to get what you want.

CAF · Answer

Between any two vectors $| psi_1 rangle$, $| psi_2 rangle in mathcal H$ $$langle psi_1 | hat p^2 | psi_2 rangle = langle psi_1 | hat p | psi_2' rangle = langle hat p psi_1 | psi_2' rangle = langle hat p  psi_1 | hat p psi_2 rangle.$$
In the first equality, $|psi_2'rangle = hat p | psi_2 rangle in mathcal H$ by definition of a linear operator acting on a vector space. In the second, we used that $hat p$ is hermitian. In particular, for an expectation, $$langle psi | hat p^2 | psi rangle = langle hat p  psi | hat p psi rangle = int_{mathcal D} ( hat p psi)^* (hat p psi),mathrm{d}x,$$ so one need only really compute $hat p psi(x)$.

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

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