Factorized quantum Hamiltonian ($H=Q^{†}Q/(2m)$)

For $Q=\hat{p}-isA(x)$ with real $A$ and positive $s$, the factorization gives potential $(s^2{A}^2-s\hslash A^{\prime })/(2m)$ and a nonnegative energy form. Its zero mode is proportional to $\exp [-s\int Adx/\hslash ]$. Taking $A=x$ and optimizing positivity over $s$ gives the Heisenberg uncertainty principle for states with finite second moments.

For $Q=p-if(x)$, $H=Q^{†}Q/(2m)$ and $Q{\psi }_0=0$, the trial state ${\psi }_1=Q^{†}{\psi }_0$ obeys the displayed identity. It follows from $[Q,Q^{†}]=2\hslash f^{\prime }$ and $[p,f^{\prime }]=-i\hslash f^{\prime \prime }$. The coefficients are generally position-dependent multiplication functions, not constants. For linear positive $f$, the first term vanishes and the second becomes the harmonic-oscillator level spacing.

For real polynomial $f(x)=c{x}^n+\cdots$, $c\ne 0$, the zero mode of $H=(p+if)(p-if)/(2m)$ is normalizable on $\mathbb{R}$ exactly when $n$ is odd and $c>0$. Its squared modulus has leading exponential $\exp [-2c{x}^{n+1}/((n+1)\hslash )]$. Odd $n$ gives the same tail sign at both ends; positivity of $c$ makes both tails decay. This kernel is one-dimensional because $(p-if)\psi =0$ is a first-order differential equation. On the operator domain where integration by parts is valid, the energy form is $\Vert (p-if)\psi {\Vert }^2/(2m)\ge 0$.