= Solution
Take the <momentum operator> $p=-i\hbar\,d/dx$ and sufficiently regular real $f$. Acting on a test <wave function>, the <commutator> is $[p,f]=-i\hbar f'$. Expanding the <factorized quantum Hamiltonian> gives
$$
(p+if)(p-if)=p^2+f^2+i(fp-pf)=p^2+f^2-\hbar f',
$$
so \b[$V(x)=(f(x)^2-\hbar f'(x))/(2m)$], which is real.
Set $Q=p-if$. Its zero-mode equation is
$$
Q\psi_0=0\quad\Longleftrightarrow\quad\hbar\psi_0'+f\psi_0=0,
$$
with solution
$$
\boxed{\psi_0(x)=C\exp\!\left[-\frac1\hbar\int_0^x f(t)\,dt\right].}
$$
It satisfies $H\psi_0=Q^\dagger Q\psi_0/(2m)=0$ as a differential equation. It is an actual quantum state only when it is square-integrable. If $f(x)=cx^n+\cdots$, $c\ne0$, its antiderivative has leading term $cx^{n+1}/(n+1)$. For even $n$, this leading term has opposite signs at the two ends of the line, so the <wave function> grows at one end and cannot be normalized. For odd $n$, it has the same sign at both ends. Hence \b[normalizability requires odd $n$, and in fact holds exactly when $n$ is odd and $c>0$]. A negative leading coefficient makes it grow at both ends; $f=0$ gives a nonnormalizable constant. These are the <normalizable zero mode of a polynomial factorized Hamiltonian> conditions.
For a normalized state in the <operator domain>, integration by parts, with the appropriate vanishing boundary term, gives
$$
\langle\psi,H\psi\rangle=\frac1{2m}\int_{\mathbb R}|Q\psi|^2\,dx\geq0.
$$
Thus every <energy eigenvalue> is nonnegative. Equality forces $Q\psi=0$ almost everywhere, whose first-order equation has the one-dimensional solution space above. Whenever its zero mode is normalizable, \b[every <energy eigenstate> not proportional to $\psi_0$ has strictly positive energy]. If the formal zero mode is not normalizable, there is no zero-energy eigenstate; the positivity statement still applies to every normalizable eigenstate. This argument concerns eigenvalues and does not assert a strictly positive bound on an arbitrary continuous spectrum.
For the <quantum harmonic oscillator>, take $f(x)=m\omega x$ with $m,\omega>0$. Then
$$
H=\frac{p^2}{2m}+\frac12m\omega^2x^2-\frac12\hbar\omega
=H_{\mathrm{osc}}-\frac12\hbar\omega.
$$
Its zero mode is normalizable and has no competitor of lower energy. Using the <Gaussian integral> for its normalization gives the concise answer
$$
\boxed{E_{\min}=\frac12\hbar\omega,\qquad
\psi_{\min}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/(2\hbar)},}
$$
up to an arbitrary constant phase.
Back to article page
