Past exam of the quantum theory course of the University of Oxford / 2015 / 2 / b / iii / Solution

Write $E_0=\hslash \omega /2$. Repeated application of the annihilation operator lowers the energy eigenvalue in steps of $\hslash \omega$. The lower bound from part (ii) forces this process to stop. To see why it also quantizes the starting energy, writeThe first $n$ lowering steps have positive squared norms. If $\delta >0$, the next step is also nonzero by part (ii), and its energy would be $E_0+(\delta -1)\hslash \omega <E_0$, contradicting the bound. Thus $\delta =0$, andExistence of the ground state can also be checked directly in the position representation. The equation $a_{-}{\psi }_0=0$ iswhose solution is proportional to $e^{-m\omega x^2/(2\hslash )}$. This Gaussian function is square-integrable, so it gives an actual ground state. Its normalization can be chosen using the Gaussian integral.

All the displayed levels occur: for a nonzero state at $E_n$,Thus repeated raising creates a nonzero energy eigenstate at every $E_n$.