= Solution
Use $[x,p]=i\hbar$. Acting on a smooth test <wavefunction>, $[f(x),p]\psi=f(-i\hbar\psi')+i\hbar(f\psi)'=i\hbar f'\psi$. Take $f$ real and use the <self-adjoint operator> domain with vanishing integration-by-parts boundary terms. Writing $Q=p-if$ gives
$$
H=\frac{Q^\dagger Q}{2m}=\frac{p^2+f^2-\hbar f'}{2m},
\qquad\boxed{V(x)=\frac{f(x)^2-\hbar f'(x)}{2m}.}
$$
The <energy> form is $\langle\psi,H\psi\rangle=\|Q\psi\|^2/(2m)\ge0$, so every normalizable <energy eigenstate> has $E\ge0$. If a normalizable zero-energy state exists, its nonnegative norm must vanish, $Q\psi_0=0$. Hence
$$
\hbar\psi_0'+f\psi_0=0,\qquad
\boxed{\psi_0(x)=N\exp\left[-\frac1\hbar\int_0^xf(t)dt\right].}
$$
Zero <energy> is an additional existence assumption, not a consequence of factorization alone.
For $\psi_1=Q^\dagger\psi_0$, the <commutator> $[Q,Q^\dagger]=2\hbar f'$ gives
$$
H\psi_1=\frac\hbar m Q^\dagger(f'\psi_0)
=-\frac{i\hbar^2}{m}f''\psi_0+\frac\hbar m f'\psi_1.
$$
Thus \b[$\alpha(x)=-i\hbar^2f''(x)/m$ and $\beta(x)=\hbar f'(x)/m$] in the requested relation. Generally these are multiplication functions, not constant coefficients, and $\psi_1$ is not automatically an <energy eigenstate>.
If $f(x)=cx^n+\cdots$, its integral has leading term $cx^{n+1}/(n+1)$. Decay at both infinities on the full real line requires \b[odd $n$ and positive <leading coefficient> $c$]. Odd degree alone is not sufficient. For $f=kx$ take $k=m\omega>0$. Then $H=H_{\rm SHO}-\hbar\omega/2$, $\alpha=0$ and $\beta=\hbar\omega$. After normalization and removal of an irrelevant phase,
$$
\boxed{\psi_0=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/(2\hbar)},\quad
\psi_1=\sqrt{\frac{2m\omega}{\hbar}}\,x\psi_0,\quad
E_0^{\rm SHO}=\frac12\hbar\omega,\quad E_1^{\rm SHO}=\frac32\hbar\omega.}
$$
The corresponding factorized-Hamiltonian energies are zero and $\hbar\omega$; the constant shift must be restored for the stated oscillator <Hamiltonian>.
Back to article page
