I know that operator for $p = {h\over i} {d\over dx}$. so $p = {h\over i} {d\psi\over dx}$ where $\psi$ is the wave function.
So, $T$ (kinetic energy) $ = {p^2 \over 2m} = {-h^2\over 2m} {d\psi \over dx}^2$, so the operator for $T$ should be $-{h^2\over 2m} \left({d\over dx}\right)^2$ and not $d^2\over dx^2$.
Please explain where I am wrong.
We have $$p^2\psi=\hat{p}^2\psi=\hat{p}\hat{p}\psi=\frac{\hbar}{i}\frac{d}{dx}\left(\frac{\hbar}{i}\frac{d\psi}{dx}\right)=-\hbar^2\frac{d^2\psi}{dx^2}.$$
\left( and \right) you can get re-sizing parenthesis. It also work with other brackets: say \left[ x^{x^x}\right\} to get $\left[ x^{x^x}\right}$ (note that you have to escape the {).
– dmckee --- ex-moderator kitten
Aug 05 '18 at 03:29
Momentum operator is defined as:
$$\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$$
This means that if you want to get the momentum from the wavefunction $\psi$, you must apply the momentum operator to the wavefunction. Let's define the wavefunction as:
$$\psi=A\exp\left[\frac{i}{\hbar}\left(px-Et\right)\right]$$
It is easy to see that:
$$\frac{\hbar}{i}\frac{d}{dx}\psi=p\psi$$
If you want to get $p^2$ you have to apply the momentum operator to the wavefunction twice.
$$\hat{p}^2\psi=\hat{p}\hat{p}\psi=\frac{\hbar}{i}\frac{d}{dx}\left(\frac{\hbar}{i}\frac{d}{dx}\psi\right)=p\left(\frac{\hbar}{i}\frac{d}{dx}\psi\right)=p^2\psi$$
So the $\hat{p}^2$ operator is:
$$\hat{p}^2=\left(\frac{\hbar}{i}\frac{d}{dx}\right)\left(\frac{\hbar}{i}\frac{d}{dx}\right)=\left(\frac{\hbar}{i}\right)^2\frac{d^2}{dx^2}=-\hbar^2\frac{d^2}{dx^2}$$
You are correct. However the difference is zero for the expectation value as the space integral over both expressions differs only by a partial integration.
