Expectation energy for a quantum harmonic oscillator

Question

At 59:14 in this video, the expectation value of the energy of a harmonic oscillator is $$ \langle E \rangle = \int ||\tilde{\Psi}(p)||^2 \frac{p^2}{2m}\ \mathrm dp + \int ||\Psi(x)||^2\frac{m\omega^2}{2}x^2\ \mathrm dx\tag 1$$ My question is how was this equation reached? This was my attempt:$$\langle E \rangle = \int {\Psi}^*(x)~\hat{E}~\Psi(x)\ \mathrm dx=\int {\Psi}^*(x)\left(-\frac{\hbar ^2}{2m}\frac{\partial ^2 \Psi(x)}{\partial x ^2}\right)\ \mathrm dx + \int {\Psi}^*(x)\frac{m\omega^2}{2}\Psi(x)\ \mathrm dx $$

but I can't get any further. How can I reach equation $(1)$?

Answer 1

You missed off an $x^2$ term in the potential in the Schrödinger equation. $V(x)=\frac{1}{2}m\omega^2x^2$ for a harmonic oscillator. Then remember the definition of the momentum operator is $\hat{p}=-i\hbar\frac{\partial}{\partial x}$.

Answer 2

When you write down the expectation value of $H$, as its composed of two pieces:

$$H=\frac{P^2}{2m}+\frac12 m\omega^2 X^2$$

you have two different expectation values:

$$\langle H\rangle=\langle \frac{P^2}{2m}\rangle+\langle\frac12 m\omega^2 X^2\rangle$$

For the first one, you can proceed as you did but if you want to reproduce exactly the other answer you should introduce the momentum basis instead of the coordinate one:

$$\langle \frac{P^2}{2m}\rangle= \langle\psi| \frac{P^2}{2m}|\psi\rangle =\int_{-\infty}^\infty dp \langle \psi| \frac{P^2}{2m} |p\rangle\langle p|\psi\rangle= \text{have fun with the rest of calculation :)}$$

For the second term, the choice of coordinates is correct but as it has been already pointed out you’re missing a $x^2$