Why operator of kinetic energy has a double derivative instead of square of single derivative?

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.

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:

$$\hat{p}^2=\left(\frac{\hbar}{i}\frac{d}{dx}\right)\left(\frac{\hbar}{i}\frac{d}{dx}\right)=\left(\frac{\hbar}{i}\right)^2\frac{d^2}{dx^2}=-\hbar^2\frac{d^2}{dx^2}$$

We have $$p^2\psi=\hat{p}^2\psi=\hat{p}\hat{p}\psi=\frac{\hbar}{i}\frac{d}{dx}\left(\frac{\hbar}{i}\frac{d\psi}{dx}\right)=-\hbar^2\frac{d^2\psi}{dx^2}.$$

• By using \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 Aug 5 '18 at 3:29

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.

• My answer is correct and equivalent to the other two answers. I just don't a) think I have to sp – my2cts Aug 6 '18 at 19:54