Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 17B / f / Solution

Start with a nonzero eigenfunction of energy $E$ and apply the lowering ladder operator repeatedly. Whenever the resulting wavefunction is nonzero, part (b) makes it an eigenfunction of energy $E-k\hslash \omega$. This cannot remain nonzero for every $k$, since part (e) rules out negative energies. Let $\psi ={\hat{A}}^n\varphi$ be the last nonzero member. Then $\hat{A}\psi =0$, and part (d) forces its energy to be $\hslash \omega /2$. Therefore every energy eigenvalue has the formThese levels actually occur: solving $\hat{A}{\psi }_0=0$ with $\hat{p}=-i\hslash d/dx$ gives the normalizable Gaussian ${\psi }_0\propto \exp (-m\omega x^2/(2\hslash ))$. The raising operator $\hat{B}$ increases the energy by $\hslash \omega$, and $\Vert \hat{B}\psi {\Vert }^2=2m(E+\hslash \omega /2)\Vert \psi {\Vert }^2>0$. Repeated raising therefore produces a nonzero normalizable state at every listed level.