 Solution

Use $[x,p]=i\hslash$. Acting on a smooth test wavefunction, $[f(x),p]\psi =f(-i\hslash {\psi }^{\prime })+i\hslash (f\psi {)}^{\prime }=i\hslash f^{\prime }\psi$. Take $f$ real and use the self-adjoint operator domain with vanishing integration-by-parts boundary terms. Writing $Q=p-if$ givesThe energy form is $⟨\psi ,H\psi ⟩=\Vert Q\psi {\Vert }^2/(2m)\ge 0$, so every normalizable energy eigenstate has $E\ge 0$. If a normalizable zero-energy state exists, its nonnegative norm must vanish, $Q{\psi }_0=0$. HenceZero energy is an additional existence assumption, not a consequence of factorization alone.

For ${\psi }_1=Q^{†}{\psi }_0$, the commutator $[Q,Q^{†}]=2\hslash f^{\prime }$ givesThus $\alpha (x)=-i{\hslash }^2{f}^{\prime \prime }(x)/m$ and $\beta (x)=\hslash f^{\prime }(x)/m$ in the requested relation. Generally these are multiplication functions, not constant coefficients, and ${\psi }_1$ is not automatically an energy eigenstate.

If $f(x)=c{x}^n+\cdots$, its integral has leading term $c{x}^{n+1}/(n+1)$. Decay at both infinities on the full real line requires odd $n$ and positive leading coefficient $c$. Odd degree alone is not sufficient. For $f=kx$ take $k=m\omega >0$. Then $H=H_{\mathrm{SHO}}-\hslash \omega /2$, $\alpha =0$ and $\beta =\hslash \omega$. After normalization and removal of an irrelevant phase,The corresponding factorized-Hamiltonian energies are zero and $\hslash \omega$; the constant shift must be restored for the stated oscillator Hamiltonian.