$x\in{\mathbb{R}}^2$. By $R^k_{\nu,\mu}$ we denote the integral operator from $L^2(\mu):=L^2({\mathbb{R}}^2,\mu)$ to $L^2(\nu)$ with the kernel $G_k$, i.e. $$R^k_{\nu,\mu} \phi = G_k \ast \phi\mu$$ holds $\nu$-a.e. for all $\phi\in D(R^k_{\nu,\mu}) \subset
L^2(\mu)$. In our case the two measures will be $m\equiv m_\Gamma$ introduced by (\[meas\]) and the Lebesgue measure ${\mathrm{d}}x$ on ${\mathbb{R}}^2$ in different combinations, which simply means that one or both variables in the kernel (\[freeG\]) are restricted to $\Gamma$. Using this notation we can state the following result:

\[BS\] (i) There is a $\kappa_0>0$ such that the operator $I-\alpha
R^{i\kappa}_{m,m}$ on $L^2(m)$ has a bounded inverse for any $\kappa \ge \kappa_0$.\
\[1mm\] (ii) Let ${\mathrm{Im\,}}k>0$. Suppose that $I-\alpha R^k_{m,m}$ is invertible and the operator $$R^k := R_0^k + \alpha R^k_{{\mathrm{d}}x,m} [I-\alpha R^k_{m,m}]^{-1}
R^k_{m,{\mathrm{d}}x}$$ from $L^2({\mathbb{R}}^2)$ to $L^2({\mathbb{R}}^2)$ is everywhere defined. Then $k^2$ belongs to $\rho(H_{\alpha,\Gamma})$ and $(H_{\alpha,\Gamma}-k^2)^{-1}= R^k$.\
\[1mm\] (iii) $\:\dim\ker(H_{\alpha,\Gamma}-k^2) = \dim\ker(I-\alpha
R^k_{m,m})$ for any $k$ with ${\mathrm{Im\,}}k>0$.\
\[1mm\] (iv) an eigenfunction of $H_{\alpha,\Gamma}$ associated with such an eigenvalue $k^2$ expresses as $$\psi(x)= \int_0^L R^k_{{\mathrm{d}}x,m}(x,s) \phi(s)\, {\mathrm{d}}s\,,$$ where $\phi$ is the corresponding eigenfunction of $\alpha
R^k_{m,m}$ with the eigenvalue one.

The result, which is in fact valid for any operator $\hat
H_{-\alpha m}$ of Sec. \[ss: qf&bc\], is 