The expectation value of ${1\over 2}m\omega^2x^2$ in eigenstate

The expectation of $x^2$ will have some nonzero terms.

$$\begin{eqnarray*} \langle u_n\vert x^2\vert u_n\rangle &=&{\hbar\over 2m\omega}\... ...mega}\langle u_n\vert AA^\dagger+A^\dagger A\vert u_n\rangle \\ \end{eqnarray*}$$

We could drop the $AA$ term and the $A^\dagger A^\dagger$ term since they will produce 0 when the dot product is taken.

$$\begin{eqnarray*} \langle u_n\vert x^2\vert u_n\rangle &=&{\hbar\over 2m\omega}... ...mega}((n+1)+n)=\left(n+{1\over 2}\right){\hbar\over m\omega} \\ \end{eqnarray*}$$

With this we can compute the expected value of the potential energy.

$$\bgroup\color{black} \langle u_n\vert{1\over 2}m\omega^2x^2\v... ...er 2}\left(n+{1\over 2}\right)\hbar\omega={1\over 2}E_n \egroup$$