Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 1 / 15B / Solution

Using the momentum operator $\hat{p}=-i\hslash d/dx$ and its commutator with multiplication by $A$, expansion of the factorized quantum Hamiltonian givesThe first-order zero-mode equation integrates toFor $A=x^n$, its exponent is $-s{x}^{n+1}/[\hslash (n+1)]$. It decays at both ends exactly when $n+1$ is even. If $n$ is even it grows at negative infinity. Thus a nonzero square-integrable zero mode exists exactly for odd $n$.

For $n=1$ the Gaussian zero mode is $\chi =(s/(\pi \hslash ){)}^{1/4}{e}^{-s{x}^2/(2\hslash )}$. More generally, on normalized states with finite second moments and the usual integration-by-parts boundary conditions, $\hat{H}=Q^{†}Q/(2m)$ is nonnegative. With zero means this implies, for every $s>0$,Minimizing the quadratic in $s$ at $s=\hslash /[2(\Delta x{)}^2]$ yieldsThe Gaussian mode attains equality. The argument concerns the same state for the whole positive family of factorized operators, which permits the minimization.