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 $\Delta = 0$?