need to require $g\in\Pi_r$. We denote the domain of the self-adjoint closure of $H(g)$ by $D(g)$, but keep the notation $H(g)$ for the closure. We now use the key identity (\[eiB\]) as in the proof of Lemma \[lemma2.3\] to deduce $$\begin{gathered}
\label{key2}
(\phi,H(g){{\cal I}}(g)\psi)=(H(g'){{\cal I}}(g')\phi,\psi),\qquad \forall \, \phi,\psi\in{{\cal H}}.\end{gathered}$$ Choosing $\lambda\in{{\mathbb C}}$ with $\Im \lambda >0$, this entails in particular $$\begin{gathered}
\label{ipH}
(\phi,(H(g)+\lambda){{\cal I}}(g)\psi)=((H(g')+\overline{\lambda}){{\cal I}}(g')\phi,\psi),\qquad \forall \, \phi\in D(g),\qquad \forall \, \psi\in D(g').\end{gathered}$$ Now for $\phi\in D(g)$ and $\psi\in D(g')$ there exist $e,f\in{{\cal H}}$ such that $$\begin{gathered}
\phi=(H(g)+\overline{\lambda})^{-1}e,\qquad \psi=(H(g')+\lambda)^{-1}f,\end{gathered}$$ so (\[ipH\]) implies $$\begin{gathered}
(e,{{\cal I}}(g)(H(g')+\lambda)^{-1}f)=({{\cal I}}(g')(H(g)+\overline{\lambda})^{-1}e,f),\qquad \forall \, e,f\in{{\cal H}}.\end{gathered}$$ Hence we obtain the relation $$\begin{gathered}
{{\cal I}}(g)(H(g')+\lambda)^{-1}=(H(g)+\lambda)^{-1}{{\cal I}}(g).\end{gathered}$$ From this we readily deduce $$\begin{gathered}
[(H(g)+\lambda)^{-1},T(g)]=0,\end{gathered}$$ where $$\begin{gathered}
\label{T}
T(g)\equiv {{\cal I}}(g){{\cal I}}(g)^{*}=\sum_{n=0}^{\infty}\nu_n^2(g)e_n(g)\otimes e_n(g).\end{gathered}$$

The upshot is that the self-adjoint trace class operator $T(g)$ commutes with the $H(g)$-resolvent. Since $T(g)$ has trivial kernel by (\[KerI\]), all of its eigenspaces are finite-dimensional, and the $H(g)$-resolvent leaves these finite-dimensional spaces invariant. Thus we can choose the eigenvectors of $T(g)$ to be eigenvectors of the $H(g)$-resolvent, hence of $H(g)$ as well. We now summarize and extend these findings.

\[theorem2.4\] Let $g\in\Pi_r$. Then the positive trace class operator $T(g)$  has an ONB $\{ e_n(g)\}_{n\in{{\mathbb 