Nondegeneracy propagated by an oscillator ladder ($\dim \ker (H-E_n)=1\text{if}\dim \ker (H-E_0)=1$)

For the quantum harmonic oscillator with $a_{-}=P-im\omega X$, $a_{+}=a_{-}^{\ast }$ and $\gamma =2m\hslash \omega$, factorization gives $a_{+}{a}_{-}=2m(H-E_0)$, where $E_0=\hslash \omega /2$. For an energy eigenstate $\psi$ at $E_n=E_0+n\hslash \omega$,The first identity shows that lowering to the ground state is injective on the $E_n$ eigenspace; the second shows how to invert it by raising. If the ground eigenspace is one-dimensional, $(a_{-}{)}^n\psi =c{\psi }_0$ and therefore $\psi =c(a_{+}{)}^n{\psi }_0/({\gamma }^{n}n!)$. Thus every level is also nondegenerate. This proof explicitly rules out a second independent energy eigenstate at an excited level rather than merely constructing one ladder of states.