following result:

\[dsexist\] Let $\alpha>0$ and suppose that $\gamma: {\mathbb{R}}\to{\mathbb{R}}^2$ satisfies the above assumptions. Then the essential spectrum is the same as for the straight line, $\sigma_\mathrm{ess}(H_{\alpha,\Gamma}) =
\left\lbrack -{1\over 4}\alpha^2, \infty \right)$, but $H_{\alpha,\Gamma}$ has at least one isolated eigenvalue below $-{1\over 4}\alpha^2$.

Observe first that in view of (a2) it is not difficult to construct a Weyl sequence to $H_{\alpha,\Gamma}$ showing that any non-negative number belongs to $\sigma_\mathrm{ess}$. To deal with the negative part, we use the generalized BS principle of Proposition \[BS\]; the idea is to treat the difference between the operator ${\mathcal{R}}^{\kappa}_{\alpha,\Gamma}:= \alpha
R^{i\kappa}_{m,m}$ on $L^2({\mathbb{R}})$ and its counterpart corresponding to $\Gamma_0$ as a perturbation. The integral kernel of the operator is $${\mathcal{R}}^{\kappa}_{\alpha,\Gamma}(s,s') = {\alpha\over 2\pi}\, K_0
\left( \kappa|\gamma(s)\!-\! \gamma(s')| \right) \,,$$ where $K_0$ is the Macdonald function; for $\Gamma=\Gamma_0$ one has to replace $|\gamma(s)\!-\! \gamma(s')|$ by $|s\!-\! s'|$. In the last named case the operator is of convolution type and using Fourier transformation it is easy to check that its spectrum is absolutely continuous covering the interval $[0,\alpha/2\kappa]$ in correspondence with (\[straightspec\]).

The key observation is that the kernel of ${\mathcal{D}}_{\kappa}:=
{\mathcal{R}}^{\kappa}_{\alpha,\Gamma} - {\mathcal{R}}^{\kappa}_{\alpha,\Gamma_0}$ satisfies $$\label{Dkern}
{\mathcal{D}}_{\kappa}(s,s') := {\alpha\over 2\pi}\, \bigg( K_0 \left(
\kappa|\gamma(s)\!-\! \gamma(s')| 