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Let $H : \operatorname{dom}(H) \subset L^2(\Omega) \rightarrow L^2(\Omega)$, where $dom(H) \subset H^2(\Omega)$, $\Omega \subset \mathbb{R}$ should be a bounded open interval(so 1-d setting(!)) and $H$ be a self-adjoint Schrödinger operator $H = -\frac{d^2}{dx^2} + V$ and $V \in C^{\infty}$. Now assume that the Schrödinger equation has a ground-state solution $\psi$ (with $\psi > 0$- not sure if we actually need this, but it should definitely simplify quite something). By rescaling the potential, we can get that $H \psi = 0.$

Then we define $W(x) :=\frac{\psi'(x)}{\psi(x)}$ and closed(!) operators $A = \frac{d}{dx}+ W $ and $A^* = -\frac{d}{dx}+W$.

Now, my question is: Given this situation, are the domains of the operators $A,A^*$ fixed? Or is there at least a standard way to construct an appropriate domain of $A$ by using the domain of $H$ ?

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    $\begingroup$ The operator $H$ is elliptic (and with smooth coefficients), so for any bounded $\Omega$, and with Dirichlet boundary conditions, the eigenvectors form an orthonormal basis of $L^2(\Omega)$. Furthermore, these eigenvectors, and thus $\psi(x)$, are $C^\infty$. $\endgroup$ – yuggib Dec 11 '14 at 17:56
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In the applications of the method that I know of, one usually doesn't make explicit use of the operators $A, A^*$; rather, this is a method to go from the original potential $V=W^2-W'$ to a new potential $W^2+W'$ that has an extra eigenvalue.

However, if the original operator $H$ has Dirichlet boundary conditions $y(a)=y(b)=0$ (here I assume that $\Omega=(a,b)$), then you could take $$ D(A)= \{ y\in H^1: y(a)=y(b)=0 \} , $$ so $D(A^*)=H^1$, and $H=A^*A$ comes out right, with the correct domain. You now definitely want to assume that $0$ is below the spectrum of $H$; if $0$ is the bottom of the spectrum, then $\psi(a)=\psi(b)=0$ and $W$ is not locally integrable near $a,b$ and this doesn't work. The operator $H_2$ then has boundary conditions $y'=Wy$ at $a,b$.

For more general boundary conditions, this doesn't seem to work.

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