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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 $\Delta = 0$?

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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 $k\in \mathbb N$.

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  • $\begingroup$ it might perhaps be instructive to consult this article (paywall) where a class of potentials is constructed for which the Witten index can be calculated exactly: dx.doi.org/10.1016/0550-3213(84)90296-7 $\endgroup$ – Carlo Beenakker Jan 15 '15 at 21:11
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In cases where you consider differential operators on compact manifolds instead of $\mathbb{R}^n$, this is related to the Atiyah-Singer Theorem. It turns out that in many cases, the index of your operator $H = D^*D$ is equal to a topological invariant of your manifold.

For examples, if $D$ is the Dirac operator acting on spinors, one has $$\mathrm{ind}(H) = \hat{A}(M),$$ where the right-hand side is the so-called A-roof-genus of $M$, which can be computed in purely topological terms or alternatively can be represented as an integral over purely local quantities (curvature terms).

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