= Solution
Using the <momentum operator> $\hat p=-i\hbar\,d/dx$ and its commutator with multiplication by $A$, expansion of the <factorized quantum Hamiltonian> gives
$$
\hat H=\frac1{2m}[\hat p^2+s^2A^2-s\hbar A'],\qquad
\boxed{V(x)=\frac{s^2A(x)^2-s\hbar A'(x)}{2m}.}
$$
The first-order zero-mode equation integrates to
$$
\boxed{\chi(x)=C\exp\left[-\frac s\hbar\int^x A(t)\,dt\right].}
$$
For $A=x^n$, its exponent is $-s x^{n+1}/[\hbar(n+1)]$. It decays at both ends exactly when $n+1$ is even. If $n$ is even it grows at negative infinity. Thus \b[a nonzero square-integrable zero mode exists exactly for odd $n$].
For $n=1$ the Gaussian zero mode is $\chi=(s/(\pi\hbar))^{1/4}e^{-s x^2/(2\hbar)}$. More generally, on normalized states with finite second moments and the usual integration-by-parts boundary conditions, $\hat H=Q^\dagger Q/(2m)$ is nonnegative. With zero means this implies, for every $s>0$,
$$
0\le (\Delta p)^2+s^2(\Delta x)^2-s\hbar.
$$
Minimizing the quadratic in $s$ at $s=\hbar/[2(\Delta x)^2]$ yields
$$
\boxed{\Delta x\Delta p\ge\hbar/2.}
$$
The Gaussian mode attains equality. The argument concerns the same state for the whole positive family of factorized operators, which permits the minimization.
Back to article page
