Past exam of the options II course of the University of Oxford / 2004 / J (Quantum Mechanics)

Set $a=\sqrt{m\omega /(2\hslash )}x+\sqrt{\hslash /(2m\omega )}d/dx$ and let $a^{†}$ have the opposite derivative sign. On the bound-state domain, $[a,a^{†}]=1$ and $H=\hslash \omega (a^{†}a+1/2)$. Thus every energy eigenvalue has $E/\hslash \omega -1/2=\nu \ge 0$.

The lowering operator maps an eigenstate of $a^{†}a$ of eigenvalue $\nu$ to one of eigenvalue $\nu -1$, with squared norm multiplied by $\nu$. Repeated lowering cannot pass below zero, so it must terminate at zero. Hence $\nu$ is a nonnegative integer. The equation $a{\psi }_0=0$ gives a Gaussian ground state, and repeated raising gives a nonzero square-integrable state for every integer $n\ge 0$. ThereforeThese are also the Gaussian-times-Hermite-polynomial solutions obtained by the suggested series method.

Put $\alpha =m\omega /\hslash$. Normalization gives ${\psi }_0=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, soThe expectation of $x$ itself is zero by parity; that cancellation does not apply to its absolute value.