Let $H$ be a self-adjoint Hamiltonian and $H$ admits a decomposition into closed operators $D,D^\*$, such that we have $H = D^\*D$.


I will now consider the one-dimensional case on a compact set:


So assume that $H$ has a ground-state wavefunction $\psi\_0$ that does not have any nodes (this is possible for most boundary conditions by Sturm-Liouville theory).


Now there is a superpotential $\Phi = -\frac{\psi\_0'}{\psi\_0}.$


It is now easy to see that $D = \frac{d}{dx} + \Phi$ and $D^\* = - \frac{d}{dx}+\Phi$ will do it. 


The so-called Witten index $\Delta$ for our operator $H$ is now defined as $\Delta = \operatorname{dim Ker(D)}-\operatorname{dim Ker(D^\*)}.$ 


Now, I was wondering: Isn't this index most of the times zero? At-least if $\phi$ is well-behaved, then both differential equations $Df=0$ and $D^\*f=0$ will give one-dimensional spaces by Picard-Lindelöf.


Therefore, I don't really understand in which cases we are interested in calculating this Witten index, such that we do not end up with zero. Is it in higher-dimensional spaces(for which the Hamiltonian acts on two or three-dimensional spaces) or is it for superpotentials that are somehow pathological? 


Are there supotentials for this 1d-case for example that would not give me $\Delta = 0$?



Witten index non-trivial in the context of Quantum Mechanics?
Consider the most classical example, $D=\frac{d}{dx}-x$, which is the creation operator. Note that $D$ is injective since, with $L^2$ norms and dot-products and say $u$ in the Schwartz class,
$$
\Vert{Du}\Vert^2\ge \langle[-x, \frac{d}{dx}] u, u\rangle=\Vert{u}\Vert^2.
$$
On the other hand, the range of $D$ has codimension 1 since
$$
D^\*u=0 \Longleftrightarrow u(x)=\lambda e^{-x^2/2}.
$$
As a result, the index $\Delta=-1$. Note that we have here
$$
H=(-\frac{d}{dx}-x)(\frac{d}{dx}-x)=-\frac{d^2}{dx^2}+x^2+1=1+\text{harmonic oscillator.}
$$
Some variations could be made on this prototypic situation with 
$D\_k=\frac{d}{dx}-x^{2k+1}$ where $k\in \mathbb N$.


