Schrödinger operators with δ-interactions supported on conical surfaces

The purpose of this paper is to analyse the spectrum of the three−dimensional Schrödinger operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ with an attractive δ-interaction of constant strength $\alpha \gt 0$ supported on the conical surface

$$\begin{eqnarray*}&&{{\mathcal{C}}_{\theta }}:=\left\{ (x,y,z)\in {{\mathbb{R}}^{3}}:z:={\rm cot} (\theta )\sqrt{{{x}^{2}}+{{y}^{2}}} \right\},\quad \theta \in (0,\pi /2).\end{eqnarray*}$$

The Schrödinger operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ is defined via the first representation theorem [17, Theorem VI.2.1] as the unique self-adjoint operator in ${{L}^{2}}({{\mathbb{R}}^{3}})$ which is associated with the closed, densely defined, symmetric and semibounded quadratic form

$$\begin{eqnarray}&&{{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}[\psi ]=\left\|\nabla \psi \right\|_{{{L}^{2}}\left( {{\mathbb{R}}^{3}};{{\mathbb{C}}^{3}} \right)}^{2}-\alpha {{\int }_{{{\mathcal{C}}_{\theta }}}}{{\left| \psi \right|}^{2}}{\rm d}\sigma ,\quad {\rm dom}\;{{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}={{H}^{1}}\left( {{\mathbb{R}}^{3}} \right);\end{eqnarray} \tag{ 1 }$$

cf [1, 4]. In a short form the main result of this note is the following theorem.

Theorem.  For any $\theta \in (0,\pi /2)$ and $\alpha \gt 0$ the essential spectrum of the operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ is $[-{{\alpha }^{2}}/4,+\infty )$ , the discrete spectrum is infinite and accumulates to $-{{\alpha }^{2}}/4$.

In addition, we obtain an asymptotic estimate of the eigenvalues of $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ lying below $-{{\alpha }^{2}}/4$, see theorem 3.2. Roughly speaking, the infiniteness of the discrete spectrum is induced by global geometrical properties of the conical surface ${{\mathcal{C}}_{\theta }}$ and is not related to the singularity at the tip or other local geometrical properties. In fact, the same effect remains present after a local deformation of ${{\mathcal{C}}_{\theta }}$; cf theorem 3.3.

Various relations between the geometry and the bound states of quantum systems have been studied intensively in recent decades (see, e.g. [19]) after it had been realized in [14] that curvature can give rise to an effective attractive interaction. In addition to systems with a hard-wall confinement the so-called 'leaky' structures attracted attention, see the review paper [11]. Their advantage is that they make it possible to take quantum tunnelling into account. The model discussed in this paper can describe, for instance, a structure composed of two semiconductors: a conical substrate of one material on the top of which we have a thin layer of the second one, covered by a bulk mass of the former.

The proof of our main result is based on standard techniques in spectral theory of self-adjoint operators: we construct singular sequences and use Neumann bracketing in the spirit of [13] to show the assertion on the essential spectrum; for the infiniteness of the discrete spectrum we employ variational principles. The same approach was applied in [25] in the context of Schrödinger operators with slowly decaying negative regular potentials, see also [23, §XIII.3]. Similar arguments were also used in [10, 15] for the closely related question of infiniteness of the discrete spectrum for the Dirichlet Laplacian in a conical layer, see also [7, 18, 19, 21, 23] for further progress in this problem. We also point out [6, 9, 12] for related spectral problems for Schrödinger operators with δ-potentials.

In this section we show that the essential spectrum of the operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ is given by $[-{{\alpha }^{2}}/4,+\infty )$. The proof of the inclusion ${{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})\supseteq [-{{\alpha }^{2}}/4,+\infty )$ makes use of singular sequences and for the other inclusion a specially chosen Neumann bracketing is used. A similar type of argument was also employed in [1, 13] for δ and $\delta ^{\prime}$-interactions on broken lines in the two-dimensional setting.

Theorem 2.1. Let $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ be the self-adjoint operator in ${{L}^{2}}({{\mathbb{R}}^{3}})$ associated to the form (1) and let $\alpha \gt 0$ and $\theta \in (0,\pi /2)$. Then

$$\begin{eqnarray*}&&{{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})=[-{{\alpha }^{2}}/4,+\infty )\end{eqnarray*}$$

Remark 2.2. For completeness we mention that the above theorem is also valid in the case $\theta =\pi /2$, that is, the conical surface is a plane, and the statement can be shown directly via separation of variables.

Proof of theorem 2.1 Step 1. We verify the inclusion ${{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})\supseteq [-{{\alpha }^{2}}/4,+\infty )$ by constructing singular sequences for the operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ for every point of the interval $[-{{\alpha }^{2}}/4,+\infty )$. Let us start by fixing a function ${{\chi }_{1}}\in C_{0}^{\infty }(1,2)$ such that

$$\begin{eqnarray}&&\parallel {{\chi }_{1}}{{\parallel }_{{{L}^{2}}(1,2)}}=1,\end{eqnarray} \tag{ 2 }$$

and a function ${{\chi }_{2}}\in C_{0}^{\infty }(-\varepsilon ,\varepsilon )$ with some fixed $\varepsilon \in (0,{\rm tan} \theta )$, which satisfies

$$\begin{eqnarray}&&0\leqslant {{\chi }_{2}}\leqslant 1\quad {\rm and}\quad {{\chi }_{2}}(t)=1\;{\rm for}\;|t|\lt \varepsilon /2.\end{eqnarray} \tag{ 3 }$$

Define for all $p\in \mathbb{R}$ and $n\in \mathbb{N}$ the functions ${{\omega }_{n,p}}:\mathbb{R}_{+}^{2}\to \mathbb{C}$ as

$$\begin{eqnarray*}&&{{\omega }_{n,p}}(s,t):=\frac{1}{\sqrt{n}}\left( {{\chi }_{1}}(\frac{s}{n}){\rm exp} ({\rm i}ps) \right)\left( {{\chi }_{2}}(\frac{t}{n}){\rm exp} (-\frac{\alpha }{2}|t|) \right)\in C(\mathbb{R}_{+}^{2})\end{eqnarray*}$$

in the coordinate system (s, t) in figure 1. Here $\mathbb{R}_{+}^{2}$ denotes open right half-plane $\{(r,z)\in {{\mathbb{R}}^{2}}:r\gt 0\}$.

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Note that because of the choice $\varepsilon \in (0,{\rm tan} \theta )$ we have ${\rm supp}\;{{\omega }_{n,p}}\subset \mathbb{R}_{+}^{2}$ for all $n\in \mathbb{N}$ and, moreover, the distances between the z-axis and the supports of ${{\omega }_{n,p}}$ satisfy

$$\begin{eqnarray}&&{{\rho }_{n}}:={\rm inf} \{r:(r,z)\in {\rm supp}\;{{\omega }_{n,p}}\}\to +\infty ,\quad n\to \infty .\end{eqnarray} \tag{ 4 }$$

By dominated convergence, using (2) and (3), we get

$$\begin{eqnarray}\begin{array}{rcl} \parallel {{\omega }_{n,p}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2} & = & \left( \frac{1}{n}\int _{n}^{2n}{{\left| {{\chi }_{1}}(\frac{s}{n}){{e}^{{\rm i}ps}} \right|}^{2}}{\rm d}s \right)\left( \int _{-\varepsilon n}^{\varepsilon n}{{\left| {{\chi }_{2}}(\frac{t}{n}) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t \right) \\ {} & = & \int _{-\varepsilon n}^{\varepsilon n}{{\left| {{\chi }_{2}}(\frac{t}{n}) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t\to \int _{-\infty }^{\infty }{{e}^{-\alpha |t|}}{\rm d}t=\frac{2}{\alpha },\quad n\to \infty . \\ \end{array}\end{eqnarray} \tag{ 5 }$$

We denote by ${{\omega }_{n,p,\pm }}$ the restrictions of ${{\omega }_{n,p}}$ onto the open subsets

$$\begin{eqnarray*}&&{{S}_{+}}=\{(r,z)\in \mathbb{R}_{+}^{2}:z\gt r{\rm cot} \theta \}\quad {\rm and}\quad {{S}_{-}}=\{(r,z)\in \mathbb{R}_{+}^{2}:z\lt r{\rm cot} \theta \}\end{eqnarray*}$$

of $\mathbb{R}_{+}^{2}$. The partial derivatives of ${{\omega }_{n,p,\pm }}$ with respect to s and t are given by

$$\begin{eqnarray*}\begin{array}{rcl} {{\partial }_{s}}{{\omega }_{n,p,\pm }} & = & \frac{1}{\sqrt{n}}\left( \frac{1}{n}\chi _{1}^{\prime }(\frac{s}{n}){{e}^{{\rm i}ps}}+{\rm i}p{{\chi }_{1}}(\frac{s}{n}){{e}^{{\rm i}ps}} \right)\left( {{\chi }_{2}}(\frac{t}{n}){{e}^{\pm \frac{\alpha }{2}t}} \right), \\ {{\partial }_{t}}{{\omega }_{n,p,\pm }} & = & \frac{1}{\sqrt{n}}\left( {{\chi }_{1}}(\frac{s}{n}){{e}^{{\rm i}ps}} \right)\left( \frac{1}{n}\chi _{2}^{\prime }(\frac{t}{n}){{e}^{\pm \frac{\alpha }{2}t}}\pm \frac{\alpha }{2}{{\chi }_{2}}(\frac{t}{n}){{e}^{\pm \frac{\alpha }{2}t}} \right). \\ \end{array}\end{eqnarray*}$$

Similarly as in (5), using dominated convergence, we get

$$\begin{eqnarray}&&\parallel \nabla {{\omega }_{n,p}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2};{{\mathbb{C}}^{2}})}^{2}={{\int }_{\mathbb{R}_{+}^{2}}}\left( {{\left| {{\partial }_{s}}{{\omega }_{n,p}} \right|}^{2}}+{{\left| {{\partial }_{t}}{{\omega }_{n,p}} \right|}^{2}} \right){\rm d}s{\rm d}t\to \left( {{p}^{2}}+\frac{{{\alpha }^{2}}}{4} \right)\frac{2}{\alpha },\quad n\to \infty .\end{eqnarray} \tag{ 6 }$$

Let us define the sequence of functions ${{\psi }_{n,p}}:{{\mathbb{R}}^{3}}\to \mathbb{C}$ as

$$\begin{eqnarray}&&{{\psi }_{n,p}}(r,\varphi ,z):=\frac{{{\omega }_{n,p}}(r,z)}{\sqrt{2\pi r}},\quad n\in \mathbb{N},\end{eqnarray} \tag{ 7 }$$

where the functions ${{\omega }_{n,p}}:\mathbb{R}_{+}^{2}\to \mathbb{C}$ are interpreted as rotationally invariant functions on ${{\mathbb{R}}^{3}}$ in the cylindrical coordinate system $(r,\varphi ,z)$. The hypersurface ${{\mathcal{C}}_{\theta }}$ separates the Euclidean space ${{\mathbb{R}}^{3}}$ into two unbounded Lipschitz domains Ω₊ and Ω₋, where

$$\begin{eqnarray*}\begin{array}{rcl} {{\Omega }_{+}} & = & \left\{ (x,y,z)\in {{\mathbb{R}}^{3}}:z\gt {\rm cot} (\theta )\sqrt{{{x}^{2}}+{{y}^{2}}} \right\}, \\ {{\Omega }_{-}} & = & \left\{ (x,y,z)\in {{\mathbb{R}}^{3}}:z\lt {\rm cot} (\theta )\sqrt{{{x}^{2}}+{{y}^{2}}} \right\}. \\ \end{array}\end{eqnarray*}$$

We use the notation ${{\psi }_{n,p,\pm }}:={{\psi }_{n,p}}{{|}_{{{\Omega }_{\pm }}}}$. Then ${{\psi }_{n,p,\pm }}\in {{C}^{\infty }}({{\Omega }_{\pm }})$ and from (5) we obtain

$$\begin{eqnarray}&&\parallel {{\psi }_{n,p}}\parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}=\parallel {{\omega }_{n,p}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}\to \frac{2}{\alpha },\quad n\to \infty .\end{eqnarray} \tag{ 8 }$$

We claim that ${{\psi }_{n,p}}\in {\rm dom}\;(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})$. For this we still need to check that the boundary conditions

$$\begin{eqnarray}&&{{\psi }_{n,p,+}}{{\mid }_{C\theta }}={{\psi }_{n,p,-}}{{\mid }_{C\theta }}\quad {\rm and}\quad {{\partial }_{{{\nu }_{+}}}}{{\psi }_{n,p,+}}{{\mid }_{C\theta }}+{{\partial }_{{{\nu }_{-}}}}{{\psi }_{n,p,-}}{{\mid }_{C\theta }}=\alpha {{\psi }_{n,p}}{{\mid }_{C\theta }}\end{eqnarray} \tag{ 9 }$$

are satisfied; cf [1, Theorem 3.3(i)]. In fact, by the definition of ${{\omega }_{n,p}}$ we have ${{\omega }_{n,p,+}}{{\left| _{{{C}_{\theta }}}={{\omega }_{n,p,-}} \right|}_{{{C}_{\theta }}}}$, where ω_n,p,± are interpreted as rotationally invariant functions on Ω_±. This implies that the first condition (9) holds. Furthermore, one computes

$$\begin{eqnarray}&&{{\partial }_{{{\nu }_{+}}}}{{\omega }_{n,p,+}}{{\mid }_{{{C}_{\theta }}}}+{{\partial }_{{{\nu }_{-}}}}{{\omega }_{n,p,-}}{{\mid }_{{{C}_{\theta }}}}=\alpha \frac{1}{\sqrt{n}}\left( {{\chi }_{1}}(\frac{s}{n}){\rm exp} ({\rm i}ps) \right)=\alpha {{\omega }_{n,p}}{{\mid }_{{{C}_{\theta }}}}.\end{eqnarray} \tag{ 10 }$$

The gradient of ${{\psi }_{n,p,\pm }}$ can be expressed as

$$\begin{eqnarray*}&&\nabla {{\psi }_{n,p,\pm }}=\frac{1}{\sqrt{2\pi r}}\nabla {{\omega }_{n,p,\pm }}+{{\omega }_{n,p,\pm }}\nabla \left( \frac{1}{\sqrt{2\pi r}} \right),\end{eqnarray*}$$

where ∇ acts on the functions $(r,\varphi ,z)\mapsto {{\omega }_{n,p,\pm }}(r,z)$ and $(r,\varphi ,z)\mapsto \frac{1}{\sqrt{2\pi r}}$. Hence, we obtain

$$\begin{eqnarray*}\begin{array}{rcl} {{\partial }_{{{\nu }_{+}}}}{{\psi }_{n,p,+}}{{|}_{{{C}_{\theta }}}}+{{\partial }_{{{\nu }_{-}}}}{{\psi }_{n,p,-}}{{|}_{{{C}_{\theta }}}} & = & \left( \frac{1}{\sqrt{2\pi r}}{{|}_{{{C}_{\theta }}}} \right)\left( {{\partial }_{{{\nu }_{+}}}}{{\omega }_{n,p,+}}{{|}_{{{C}_{\theta }}}}+{{\partial }_{{{\nu }_{-}}}}{{\omega }_{n,p,-}}{{|}_{{{C}_{\theta }}}} \right) \\ {} & {} & +\left( {{\omega }_{n,p}}{{|}_{{{C}_{\theta }}}} \right)\left( {{\partial }_{{{\nu }_{+}}}}\left( \frac{1}{\sqrt{2\pi r}} \right){{|}_{{{C}_{\theta }}}}+{{\partial }_{{{\nu }_{-}}}}\left( \frac{1}{\sqrt{2\pi r}} \right){{|}_{{{C}_{\theta }}}} \right) \\ {} & = & \left( \frac{1}{\sqrt{2\pi r}}{{|}_{{{C}_{\theta }}}} \right)\alpha \left( {{\omega }_{n,p}}{{|}_{{{C}_{\theta }}}} \right)=\alpha {{\psi }_{n,p}}{{\mid }_{{{C}_{\theta }}}}, \\ \end{array}\end{eqnarray*}$$

where (10) was used in the second equality. Thus we have verified (9) and therefore ${{\psi }_{n,p}}\in {\rm dom}\;(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})$. Moreover, according to [1, Theorem 3.3(i)] we also have

$$\begin{eqnarray}&&-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}{{\psi }_{n,p}}=(-\Delta {{\psi }_{n,p,+}})\oplus (-\Delta {{\psi }_{n,p,-}}).\end{eqnarray} \tag{ 11 }$$

Using the expression for the three-dimensional Laplacian in cylindrical coordinates we find

$$\begin{eqnarray*}&&-\Delta {{\psi }_{n,p,\pm }}=-\frac{1}{r}{{\partial }_{r}}(r{{\partial }_{r}}{{\psi }_{n,p,\pm }})-\partial _{z}^{2}{{\psi }_{n,p,\pm }},\end{eqnarray*}$$

where the angular term is absent since the functions ${{\psi }_{n,p,\pm }}$ do not depend on φ. The above expression can be rewritten as

$$\begin{eqnarray}&&-\Delta {{\psi }_{n,p,\pm }}=-\partial _{r}^{2}{{\psi }_{n,p,\pm }}-\partial _{z}^{2}{{\psi }_{n,p,\pm }}-\frac{1}{r}({{\partial }_{r}}{{\psi }_{n,p,\pm }}).\end{eqnarray} \tag{ 12 }$$

Next we compute the first and second order partial derivatives of ${{\psi }_{n,p,\pm }}$ with respect to r:

$$\begin{eqnarray}\begin{array}{rcl} {{\partial }_{r}}{{\psi }_{n,p,\pm }} & = & \frac{{{\partial }_{r}}{{\omega }_{n,p,\pm }}}{\sqrt{2\pi r}}-\frac{{{\omega }_{n,p,\pm }}}{2\sqrt{2\pi }{{r}^{3/2}}}, \\ \partial _{r}^{2}{{\psi }_{n,p,\pm }} & = & \frac{\partial _{r}^{2}{{\omega }_{n,p,\pm }}}{\sqrt{2\pi r}}-\frac{{{\partial }_{r}}{{\omega }_{n,p,\pm }}}{\sqrt{2\pi }{{r}^{3/2}}}+\frac{3}{4}\frac{{{\omega }_{n,p,\pm }}}{\sqrt{2\pi }{{r}^{5/2}}}. \\ \end{array}\end{eqnarray} \tag{ 13 }$$

The last two summands in the expression for $\partial _{r}^{2}{{\psi }_{n,p,\pm }}$ can be estimated in L²-norm as

$$\begin{eqnarray}\begin{array}{rcl} \left\|\frac{{{\partial }_{r}}{{\omega }_{n,p,\pm }}}{\sqrt{2\pi }{{r}^{3/2}}}\right\|_{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2} & \leqslant & \frac{1}{\rho _{n}^{2}}\parallel \nabla {{\omega }_{n,p}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2};{{\mathbb{C}}^{2}})}^{2}\to 0,\quad n\to \infty , \\ \frac{9}{16}\left\|\frac{{{\omega }_{n,p,\pm }}}{\sqrt{2\pi }{{r}^{5/2}}}\right\|_{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2} & \leqslant & \frac{9}{16\rho _{n}^{4}}\parallel {{\omega }_{n,p}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}\to 0,\quad n\to \infty , \\ \end{array}\end{eqnarray} \tag{ 14 }$$

where we have used (4), (5) and (6). The second order partial derivatives of ${{\psi }_{n,p,\pm }}$ with respect to z are

$$\begin{eqnarray}&&\partial _{z}^{2}{{\psi }_{n,p,\pm }}=\frac{\partial _{z}^{2}{{\omega }_{n,p,\pm }}}{\sqrt{2\pi r}}.\end{eqnarray} \tag{ 15 }$$

Using (13), (14), (15) and the invariance of the Laplacian under rotation of the coordinate system we obtain that

$$\begin{eqnarray}&&-\partial _{r}^{2}{{\psi }_{n,p,\pm }}-\partial _{z}^{2}{{\psi }_{n,p,\pm }}=-\frac{1}{\sqrt{2\pi r}}\left( \partial _{s}^{2}{{\omega }_{n,p,\pm }}+\partial _{t}^{2}{{\omega }_{n,p,\pm }} \right)+o(1),\;\;n\to \infty ;\end{eqnarray} \tag{ 16 }$$

here and in the following we understand $o(1)$ in the strong sense with respect to the corresponding L²-norm. With the help of (13) the norm of the last summand on the right hand side in (12) can be estimated as

$$\begin{eqnarray*}&&\left\|\frac{{{\partial }_{r}}{{\psi }_{n,p,\pm }}}{r}\right\|{{}_{{{L}^{2}}({{\mathbb{R}}^{3}})}}\leqslant \left\|\frac{{{\partial }_{r}}{{\omega }_{n,p,\pm }}}{\sqrt{2\pi }{{r}^{3/2}}}\right\|{{}_{{{L}^{2}}({{\mathbb{R}}^{3}})}}+\left\|\frac{{{\omega }_{n,p,\pm }}}{2\sqrt{2\pi }{{r}^{5/2}}}\right\|{{}_{{{L}^{2}}({{\mathbb{R}}^{3}})}},\end{eqnarray*}$$

and from (14) we conclude

$$\begin{eqnarray*}&&\left\|\frac{{{\partial }_{r}}{{\psi }_{n,p,\pm }}}{r}\right\|{{}_{{{L}^{2}}({{\mathbb{R}}^{3}})}}=o(1),\quad n\to \infty .\end{eqnarray*}$$

From (12), the latter result and (16) we obtain

$$\begin{eqnarray}&&-\Delta {{\psi }_{n,p,\pm }}=-\frac{1}{\sqrt{2\pi r}}\left( \partial _{s}^{2}{{\omega }_{n,p,\pm }}+\partial _{t}^{2}{{\omega }_{n,p,\pm }} \right)+o(1),\quad n\to \infty .\end{eqnarray} \tag{ 17 }$$

Again using dominated convergence we compute

$$\begin{eqnarray}\begin{array}{rcl} \partial _{s}^{2}{{\omega }_{n,p,\pm }} & = & \frac{1}{\sqrt{n}}\left( {{\chi }_{2}}(\frac{t}{n}){{e}^{\pm \frac{\alpha }{2}t}} \right)\left( \frac{1}{{{n}^{2}}}\chi _{1}^{\prime\prime }(\frac{s}{n}){{e}^{{\rm i}ps}}+\frac{2{\rm i}p}{n}\chi _{1}^{\prime }(\frac{s}{n}){{e}^{{\rm i}ps}}-{{p}^{2}}{{\chi }_{1}}(\frac{s}{n}){{e}^{{\rm i}ps}} \right) \\ {} & = & -{{p}^{2}}{{\omega }_{n,p,\pm }}+o(1),\quad n\to \infty , \\ \end{array}\end{eqnarray} \tag{ 18 }$$

and

$$\begin{eqnarray}\begin{array}{rcl} \partial _{t}^{2}{{\omega }_{n,p,\pm }} & = & \frac{1}{\sqrt{n}}\left( {{\chi }_{1}}(\frac{s}{n}){{e}^{{\rm i}ps}} \right)\left( \frac{1}{{{n}^{2}}}\chi _{2}^{\prime\prime }(\frac{t}{n}){{e}^{\pm \frac{\alpha }{2}t}}\pm \frac{\alpha }{n}\chi _{2}^{\prime }(\frac{t}{n}){{e}^{\pm \frac{\alpha }{2}t}}+\frac{{{\alpha }^{2}}}{4}{{\chi }_{2}}(\frac{t}{n}){{e}^{\pm \frac{\alpha }{2}t}} \right) \\ {} & = & \frac{{{\alpha }^{2}}}{4}{{\omega }_{n,p,\pm }}+o(1),\quad n\to \infty . \\ \end{array}\end{eqnarray} \tag{ 19 }$$

Finally, employing (11), (17), the definition of ${{\psi }_{n,p}}$ in (7) and (18), (19) we arrive at

$$\begin{eqnarray}&&-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}{{\psi }_{n,p}}=\left( -\frac{{{\alpha }^{2}}}{4}+{{p}^{2}} \right){{\psi }_{n,p}}+o(1),\quad n\to \infty .\end{eqnarray} \tag{ 20 }$$

Since the supports of ${{\psi }_{{{2}^{k}},p}}$ and ${{\psi }_{{{2}^{k^{\prime} }},p}}$, $k\ne k^{\prime}$, are disjoint, the sequence ${{\{{{\psi }_{{{2}^{k}},p}}\}}_{k}}$ converges weakly to zero. Moreover, by (8) we have ${\rm lim} {\rm inf} \parallel {{\psi }_{{{2}^{k}},p}}{{\parallel }_{{{L}^{2}}({{\mathbb{R}}^{3}})}}\gt 0$ and hence (20) implies that ${{\{{{\psi }_{{{2}^{k}},p}}\}}_{k}}$ is a singular sequence for the operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ corresponding to the point $-{{\alpha }^{2}}/4+{{p}^{2}}$. Therefore, $-{{\alpha }^{2}}/4+{{p}^{2}}\in {{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})$ for all $p\in \mathbb{R}$ (see, e.g. [3, Theorem 9.1.2] or [24, Proposition 8.11]) and it follows that $[-{{\alpha }^{2}}/4,+\infty )\subseteq {{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})$.

Step 2. In this step we show the inclusion ${{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})\subseteq [-{{\alpha }^{2}}/4,+\infty )$ using form decomposition methods. For sufficiently large $n\in \mathbb{N}$ we define three subsets of the closed half-plane $\overline{\mathbb{R}_{+}^{2}}:=\{(r,z)\in {{\mathbb{R}}^{2}}:r\geqslant 0,\;z\in \mathbb{R}\}$

$$\begin{eqnarray*}\begin{array}{rcl} \pi _{n}^{1}: & = & \left\{ (r(s,t),z(s,t))\in \overline{\mathbb{R}_{+}^{2}}:s\gt n,|t|\lt \sqrt{n} \right\}\subset \overline{\mathbb{R}_{+}^{2}}, \\ \pi _{n}^{2}: & = & \left\{ (r(s,t),z(s,t))\in \overline{\mathbb{R}_{+}^{2}}:s\lt n,|t|\lt \sqrt{n} \right\}\subset \overline{\mathbb{R}_{+}^{2}}, \\ \pi _{n}^{3}: & = & \left\{ (r(s,t),z(s,t))\in \overline{\mathbb{R}_{+}^{2}}:|t|\gt \sqrt{n} \right\}\subset \overline{\mathbb{R}_{+}^{2}}, \\ \end{array}\end{eqnarray*}$$

as shown in figure 2.

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The ray ${{\Gamma }_{\theta }}$, which emerges from the origin and constitutes the angle θ with z-axis, is decomposed into

$$\begin{eqnarray*}\begin{array}{rcl} \Gamma _{\theta ,n}^{1}: & = & \{(r(s,t),z(s,t))\in {{\Gamma }_{\theta }}:s\gt n\}, \\ \Gamma _{\theta ,n}^{2}: & = & \{(r(s,t),z(s,t))\in {{\Gamma }_{\theta }}:s\lt n\}. \\ \end{array}\end{eqnarray*}$$

The splitting $\{\pi _{n}^{k}\}_{k=1}^{3}$ of $\overline{\mathbb{R}_{+}^{2}}$ induces the splitting of ${{\mathbb{R}}^{3}}$ into three domains

$$\begin{eqnarray*}&&\Omega _{n}^{k}:=\left\{ (r,\varphi ,z):(r,z)\in \pi _{n}^{k},\;\varphi \in \left[ 0,2\pi \right) \right\}\subset {{\mathbb{R}}^{3}},\quad k=1,2,3,\end{eqnarray*}$$

and the splitting of the conical surface ${{\mathcal{C}}_{\theta }}$ into two parts

$$\begin{eqnarray*}\begin{array}{rcl} \mathcal{C}_{\theta ,n}^{1}: & = & \{(r,\varphi ,z):(r,z)\in \Gamma _{\theta ,n}^{1},\;\varphi \in \left[ 0,2\pi \right)\}\subset {{\mathcal{C}}_{\theta }}, \\ \mathcal{C}_{\theta ,n}^{2}: & = & \{(r,\varphi ,z):(r,z)\in \Gamma _{\theta ,n}^{2},\;\varphi \in \left[ 0,2\pi \right)\}\subset {{\mathcal{C}}_{\theta }}. \\ \end{array}\end{eqnarray*}$$

We agree to denote the restriction of $\psi \in {{L}^{2}}({{\mathbb{R}}^{3}})$ onto Ω_n ^k with $k=1,2,3$ by ψ_k .

Consider the quadratic form

$$\begin{eqnarray*}\begin{array}{rcl} {{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }},n}}[\psi ]: & = & \mathop{\sum }\limits_{k=1}^{3}\parallel \nabla {{\psi }_{k}}\parallel _{{{L}^{2}}(\Omega _{n}^{k};{{\mathbb{C}}^{3}})}^{2}-\alpha \parallel {{\psi }_{1}}{{\mid }_{\mathcal{C}_{\theta ,n}^{1}}}\parallel _{{{L}^{2}}(\mathcal{C}_{\theta ,n}^{1})}^{2}-\alpha \parallel {{\psi }_{2}}{{\mid }_{\mathcal{C}_{\theta ,n}^{2}}}\parallel _{{{L}^{2}}(\mathcal{C}_{\theta ,n}^{2})}^{2}, \\ {\rm dom}\;{{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }},n}}: & = & \mathop{\oplus }\limits_{k=1}^{3}\;{{H}^{1}}(\Omega _{n}^{k}). \\ \end{array}\end{eqnarray*}$$

As in the proof of [1, Proposition 3.1] one verifies that the form ${{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }},n}}$ is closed, densely defined, symmetric and semibounded from below. Hence ${{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }},n}}$ induces a self-adjoint operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }},n}}$ in ${{L}^{2}}({{\mathbb{R}}^{3}})$ via the first representation theorem [17, Theorem VI.2.1]. The operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }},n}}$ can be decomposed into an orthogonal sum $\oplus _{k=1}^{3}{{H}_{n,k}}$ of self-adjoint operators ${{H}_{n,k}}$ in ${{L}^{2}}(\Omega _{n}^{k})$ with respect to the orthogonal decomposition ${{L}^{2}}({{\mathbb{R}}^{3}})=\oplus _{k=1}^{3}{{L}^{2}}(\Omega _{n}^{k})$, where ${{H}_{n,1}}$ and ${{H}_{n,2}}$ correspond to the quadratic forms

$$\begin{eqnarray*}\begin{array}{rcl} {{\mathfrak{a}}_{n,1}}[{{\psi }_{1}}] & = & \parallel \nabla {{\psi }_{1}}\parallel _{{{L}^{2}}(\Omega _{n}^{1};{{\mathbb{C}}^{3}})}^{2}-\alpha \parallel {{\psi }_{1}}{{\mid }_{\mathcal{C}_{\theta ,n}^{1}}}\parallel _{{{L}^{2}}(\mathcal{C}_{\theta ,n}^{1})}^{2},\quad {\rm dom}\;{{\mathfrak{a}}_{n,1}}={{H}^{1}}(\Omega _{n}^{1}), \\ {{\mathfrak{a}}_{n,2}}[{{\psi }_{2}}] & = & \parallel \nabla {{\psi }_{2}}\parallel _{{{L}^{2}}(\Omega _{n}^{2};{{\mathbb{C}}^{3}})}^{2}-\alpha \parallel {{\psi }_{2}}{{\mid }_{\mathcal{C}_{\theta ,n}^{2}}}\parallel _{{{L}^{2}}(\mathcal{C}_{\theta ,n}^{2})}^{2},\quad {\rm dom}\;{{\mathfrak{a}}_{n,2}}={{H}^{1}}(\Omega _{n}^{2}), \\ \end{array}\end{eqnarray*}$$

respectively, and ${{H}_{n,3}}$ corresponds to the quadratic form

$$\begin{eqnarray*}&&{{\mathfrak{a}}_{n,3}}[{{\psi }_{3}}]=\parallel \nabla {{\psi }_{3}}\parallel _{{{L}^{2}}(\Omega _{n}^{3};{{\mathbb{C}}^{3}})}^{2},\quad {dom}\;{{\mathfrak{a}}_{n,3}}={{H}^{1}}(\Omega _{n}^{3}).\end{eqnarray*}$$

Let us first estimate the spectrum of ${{H}_{n,1}}$. For this, note that ${{C}^{\infty }}(\Omega _{n}^{1})\cap {{H}^{1}}(\Omega _{n}^{1})$ is a core of ${{\mathfrak{a}}_{n,1}}$ and thus it suffices to use functions from this set in the estimates below (see, e.g. [8, Theorem 4.5.3]). For any ${{\psi }_{1}}\in {{C}^{\infty }}(\Omega _{n}^{1})\cap {{H}^{1}}(\Omega _{n}^{1})$ normalized as $\parallel {{\psi }_{1}}{{\parallel }_{{{L}^{2}}(\Omega _{n}^{1})}}=1$ we obtain

$$\begin{eqnarray*}\begin{array}{rcl} {{\mathfrak{a}}_{n,1}}[{{\psi }_{1}}] & \geqslant & \int _{0}^{2\pi }\left( \int _{n}^{+\infty }\int _{-\sqrt{n}}^{\sqrt{n}}r(s,t)\mid {{\partial }_{t}}{{\psi }_{1}}(s,t,\varphi ){{\mid }^{2}}{\rm d}t{\rm d}s \right. \\ {} & {} & \left. -\alpha \int _{n}^{+\infty }r(s,0)\mid {{\psi }_{1}}(s,0,\varphi ){{\mid }^{2}}{\rm d}s \right){\rm d}\varphi , \\ \end{array}\end{eqnarray*}$$

where we have used the form of the gradient in cylindrical coordinates and the invariance of the gradient with respect to rotations of the coordinate system, and the non-negative terms corresponding to the partial derivatives of ψ₁ with respect to φ and s were estimated from below by zero. Note that for simple geometric reasons we have $r(s,t)\geqslant r(s,-\sqrt{n})$ for all $(s,t)\in \pi _{n}^{1}$. Using this observation we get

$$\begin{eqnarray}\begin{array}{rcl} {{\mathfrak{a}}_{n,1}}[{{\psi }_{1}}] & \geqslant & \int _{0}^{2\pi }\int _{n}^{+\infty }r(s,-\sqrt{n})\left( \int _{-\sqrt{n}}^{\sqrt{n}}{{\left| {{\partial }_{t}}{{\psi }_{1}}(s,t,\varphi ) \right|}^{2}}{\rm d}t \right. \\ {} & {} & \left. -\frac{\alpha r(s,0)}{r(s,-\sqrt{n})}{{\left| {{\psi }_{1}}(s,0,\varphi ) \right|}^{2}} \right){\rm d}s{\rm d}\varphi . \\ \end{array}\end{eqnarray} \tag{ 21 }$$

Consider the closed, densely defined, symmetric and semibounded form

$$\begin{eqnarray*}&&\mathfrak{b}[h]=\int _{-\sqrt{n}}^{\sqrt{n}}{{\left| {{h}^{\prime }}(t) \right|}^{2}}{\rm d}t-\beta |h(0){{|}^{2}},\quad {\rm dom}\;\mathfrak{b}={{H}^{1}}((-\sqrt{n},\sqrt{n})),\end{eqnarray*}$$

and denote by $\mu (\beta ,2\sqrt{n})\lt 0$ the lower bound of the spectrum of the associated 1-D Schrödinger operator on the interval $(-\sqrt{n},\sqrt{n})$ with Neumann boundary conditions at the endpoints and attractive δ-interaction of strength $\beta \gt 0$ located at 0. Then

$$\begin{eqnarray*}&&\mathfrak{b}[h]\geqslant \mu (\beta ,2\sqrt{n})\int _{-\sqrt{n}}^{\sqrt{n}}|h(t){{|}^{2}}{\rm d}t\end{eqnarray*}$$

holds for all $h\in {{H}^{1}}((-\sqrt{n},\sqrt{n}))$ and hence (21) can be further estimated as

$$\begin{eqnarray}&&{{\mathfrak{a}}_{n,1}}[{{\psi }_{1}}]\geqslant \int _{0}^{2\pi }\int _{n}^{+\infty }\mu \left( \frac{\alpha r(s,0)}{r(s,-\sqrt{n})},2\sqrt{n} \right)\int _{-\sqrt{n}}^{\sqrt{n}}r(s,-\sqrt{n})|{{\psi }_{1}}(s,t,\varphi ){{\mid }^{2}}{\rm d}t{\rm d}s{\rm d}\varphi .\end{eqnarray} \tag{ 22 }$$

By the definition of $\pi _{n}^{1}$ one has

$$\begin{eqnarray}&&r(s,-\sqrt{n})=r(s,t)\left( 1+\mathcal{O}(\frac{1}{\sqrt{n}}) \right),\quad n\to \infty ,\end{eqnarray} \tag{ 23 }$$

for $(s,t)\in \pi _{n}^{1}$, where the remainder is uniform in s. Hence, we obtain from (22) and (23)

$$\begin{eqnarray}&&{{\mathfrak{a}}_{n,1}}[{{\psi }_{1}}]\geqslant \mu \left( \alpha \left( 1+\mathcal{O}\left( \frac{1}{\sqrt{n}} \right) \right),2\sqrt{n} \right)\left( 1+\mathcal{O}(\frac{1}{\sqrt{n}}) \right),\quad n\to \infty ,\end{eqnarray} \tag{ 24 }$$

where we used that

$$\begin{eqnarray*}&&\int _{0}^{2\pi }\int _{n}^{+\infty }\int _{-\sqrt{n}}^{\sqrt{n}}r(s,t)|{{\psi }_{1}}(s,t,\varphi ){{\mid }^{2}}{\rm d}t{\rm d}s{\rm d}\varphi =\parallel {{\psi }_{1}}\parallel _{{{L}^{2}}(\Omega _{n}^{1})}^{2}=1.\end{eqnarray*}$$

According to [16, Proposition 2.5] the following estimate

$$\begin{eqnarray*}&&\mu (\beta ,2\sqrt{n})\geqslant -\frac{{{\beta }^{2}}}{4}-C{{\beta }^{2}}{\rm exp} (-\frac{1}{2}\beta \sqrt{n})\end{eqnarray*}$$

holds with some constant $C\gt 0$ and n sufficiently large. Hence,

$$\begin{eqnarray*}&&\mu \left( \alpha \left( 1+\mathcal{O}\left( \frac{1}{\sqrt{n}} \right) \right),2\sqrt{n} \right)\geqslant -\frac{{{\alpha }^{2}}}{4}+\mathcal{O}\left( \frac{1}{\sqrt{n}} \right),\quad n\to \infty .\end{eqnarray*}$$

Plugging the above estimate into (24) we arrive at

$$\begin{eqnarray*}&&{{\mathfrak{a}}_{n,1}}[{{\psi }_{1}}]\geqslant -\frac{{{\alpha }^{2}}}{4}+\mathcal{O}\left( \frac{1}{\sqrt{n}} \right),\quad n\to \infty .\end{eqnarray*}$$

Hence, for any $\varepsilon \gt 0$ there exists a sufficiently large n for which

$$\begin{eqnarray}&&{\rm inf} \sigma ({{H}_{n,1}})\geqslant -\frac{{{\alpha }^{2}}}{4}-\varepsilon .\end{eqnarray} \tag{ 25 }$$

As ${{H}^{1}}(\Omega _{n}^{2})$ is compactly embedded into ${{L}^{2}}(\Omega _{n}^{2})$ the essential spectrum of ${{H}_{n,2}}$ is empty. The operator ${{H}_{n,3}}$ is non-negative and hence $\sigma ({{H}_{n,3}})\subseteq [0,+\infty )$. Due to the orthogonal decomposition $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }},n}}=\oplus _{k=1}^{3}{{H}_{n,k}}$ the property (25) implies that for any $\varepsilon \gt 0$ there exists a sufficiently large n for which

$$\begin{eqnarray}&&{\rm inf} {{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }},n}})\geqslant -\frac{{{\alpha }^{2}}}{4}-\varepsilon .\end{eqnarray} \tag{ 26 }$$

Finally, we apply a Neumann bracketing argument. Notice that the ordering ${{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }},n}}\leqslant {{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ holds in the sense of quadratic forms; cf [17, §VI.5]. Hence by [3, Theorem 10.2.4]

$$\begin{eqnarray}&&{\rm inf} {{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }},n}})\leqslant {\rm inf} {{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}).\end{eqnarray} \tag{ 27 }$$

In view of (27) the estimate (26) implies that for any $\varepsilon \gt 0$

$$\begin{eqnarray*}&&{\rm inf} {{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})\geqslant -\frac{{{\alpha }^{2}}}{4}-\varepsilon \end{eqnarray*}$$

and thus passing to the limit $\varepsilon \to 0+$ we arrive at

$$\begin{eqnarray*}&&{\rm inf} {{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})\geqslant -\frac{{{\alpha }^{2}}}{4},\end{eqnarray*}$$

which shows the inclusion ${{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}})\subseteq [-{{\alpha }^{2}}/4,+\infty )$ and finishes the proof of theorem 2.1. □

In this section we show that the discrete spectrum of the self-adjoint operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ below the bottom $-{{\alpha }^{2}}/4$ of the essential spectrum is infinite for all angles $\theta \in (0,\pi /2)$ and we estimate the rate of the convergence of these eigenvalues to $-{{\alpha }^{2}}/4$ with the help of variational principles. The following lemma will be useful.

Lemma 3.1. Let ${{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ be the form in (1). For $\omega \in {{H}^{1}}(\mathbb{R}_{+}^{2})$ with compact support ${\rm supp}\;\omega \subset \mathbb{R}_{+}^{2}$ define the function $\psi (r,\varphi ,z):=\frac{\omega (r,z)}{\sqrt{2\pi r}}$. Then $\psi \in {{H}^{1}}({{\mathbb{R}}^{3}})$ and

$$\begin{eqnarray}&&{{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}[\psi ]=\parallel \nabla \omega \parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2};{{\mathbb{C}}^{2}})}^{2}-{{\int }_{\mathbb{R}_{+}^{2}}}\frac{1}{4{{r}^{2}}}\mid \omega (r,z){{\mid }^{2}}{\rm d}r{\rm d}z-\alpha \parallel \omega {{\mid }_{{{\Gamma }_{\theta }}}}\parallel _{{{L}^{2}}({{\Gamma }_{\theta }})}^{2},\end{eqnarray} \tag{ 28 }$$

where ${{\Gamma }_{\theta }}$ is the ray in figure 1.

Proof. First of all observe that

$$\begin{eqnarray}&&\parallel \psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}={{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}\int _{0}^{2\pi }\frac{|\omega (r,z){{|}^{2}}}{2\pi r}\;r\;{\rm d}\varphi {\rm d}r{\rm d}z=\parallel \omega \parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}\lt \infty .\end{eqnarray} \tag{ 29 }$$

Moreover, we compute

$$\begin{eqnarray}&&{{\partial }_{r}}\psi =\frac{{{\partial }_{r}}\omega }{\sqrt{2\pi r}}-\frac{\omega }{2r\sqrt{2\pi r}}\quad {\rm and}\quad {{\partial }_{z}}\psi =\frac{{{\partial }_{z}}\omega }{\sqrt{2\pi r}},\end{eqnarray} \tag{ 30 }$$

and setting $\rho :={\rm inf} \{r:(r,z)\in {\rm supp}\;\omega \}\gt 0$ we obtain

$$\begin{eqnarray}\begin{array}{rcl} \parallel \nabla \psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}};{{\mathbb{C}}^{3}})}^{2} & = & \parallel \,{{\partial }_{r}}\psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}+\parallel {\mkern 1mu} {{\partial }_{z}}\psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2} \\ {} & \leqslant & 2\left\|\frac{{{\partial }_{r}}\omega }{\sqrt{2\pi r}}\right\|_{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}+2\left\|\frac{\omega }{2r\sqrt{2\pi r}}\right\|_{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}+\left\|\frac{{{\partial }_{z}}\omega }{\sqrt{2\pi r}}\right\|_{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2} \\ {} & \leqslant & 2\parallel {\mkern 1mu} {{\partial }_{r}}\omega \parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}+\frac{1}{2{{\rho }^{2}}}\parallel \omega \parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}+\parallel {\mkern 1mu} {{\partial }_{z}}\omega \parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}\lt \infty . \\ \end{array}\end{eqnarray} \tag{ 31 }$$

Hence (29) and (31) imply $\psi \in {{H}^{1}}({{\mathbb{R}}^{3}})$. Next we substitute ψ in the form ${{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ in (1). It follows from the form of ${{\partial }_{z}}\psi$ in (30) and $\parallel \psi {{\mid }_{{{\mathcal{C}}_{\theta }}}}\parallel _{{{L}^{2}}({{\mathcal{C}}_{\theta }})}^{2}=\parallel \omega {{\mid }_{{{\Gamma }_{\theta }}}}\parallel _{{{L}^{2}}({{\Gamma }_{\theta }})}^{2}$ that

$$\begin{eqnarray}\begin{array}{rcl} {{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}[\psi ] & = & {{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}|{{\partial }_{r}}\psi {{|}^{2}}2\pi r{\rm d}r{\rm d}z+{{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}|{{\partial }_{z}}\psi {{|}^{2}}2\pi r{\rm d}r{\rm d}z-\alpha \parallel \psi {{|}_{{{\mathcal{C}}_{\theta }}}}\parallel _{{{L}^{2}}({{\mathcal{C}}_{\theta }})}^{2} \\ {} & = & {{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}|{{\partial }_{r}}\psi {{|}^{2}}2\pi r{\rm d}r{\rm d}z+{{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}|{{\partial }_{z}}\omega {{|}^{2}}{\rm d}r{\rm d}z-\alpha \parallel \omega {{|}_{{{\Gamma }_{\theta }}}}\parallel _{{{L}^{2}}({{\Gamma }_{\theta }})}^{2}. \\ \end{array}\end{eqnarray} \tag{ 32 }$$

Denote the first integral by ${{I}_{\psi }}$. Making use of ${{\partial }_{r}}\psi$ in (30) we rewrite ${{I}_{\psi }}$ as

$$\begin{eqnarray}&&{{I}_{\psi }}={{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}\mid {{\partial }_{r}}\omega {{\mid }^{2}}{\rm d}r{\rm d}z+{{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}\frac{1}{4{{r}^{2}}}|\omega {{|}^{2}}{\rm d}r{\rm d}z-{{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}\frac{1}{r}\ \operatorname{Re}({{\partial }_{r}}\omega \bar{\omega }){\rm d}r{\rm d}z\end{eqnarray} \tag{ 33 }$$

and the last term can be further rewritten as

$$\begin{eqnarray}&&{{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}\frac{1}{r}\ \operatorname{Re}({{\partial }_{r}}\omega \bar{\omega }){\rm d}r{\rm d}z={{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}\frac{1}{2r}{{\partial }_{r}}\left( |\omega {{|}^{2}} \right){\rm d}r{\rm d}z={{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}\frac{1}{2{{r}^{2}}}|\omega {{|}^{2}}{\rm d}r{\rm d}z,\end{eqnarray} \tag{ 34 }$$

where we integrated by parts and used the fact that ${\rm supp}\;\omega$ is contained in the open half-plane $\mathbb{R}_{+}^{2}$. Hence, (33) and (34) imply

$$\begin{eqnarray*}&&{{I}_{\psi }}={{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}|{{\partial }_{r}}\omega {{|}^{2}}{\rm d}r{\rm d}z-{{\int }_{\mathbb{R}}}{{\int }_{{{\mathbb{R}}_{+}}}}\frac{1}{4{{r}^{2}}}|\omega {{|}^{2}}{\rm d}r{\rm d}z.\end{eqnarray*}$$

Substituting this expression for the first integral in (32) we obtain (28). □

Now we are ready to formulate and prove our main result on the infiniteness of the discrete spectrum of $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ below the bottom of the essential spectrum for all $\alpha \gt 0$ and $\theta \in (0,\pi /2)$. Recall that $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ is bounded from below, and hence it also follows that the discrete spectrum has a single accumulation point, namely $-{{\alpha }^{2}}/4$. This result illustrates the typical phenomenon that curvature induces bound states. The peculiarity in this three-dimensional system is that the global geometry of the interaction support plays an important role. We point out that in the case $\theta =\pi /2$ the conical surface ${{\mathcal{C}}_{\theta }}$ coincides with a plane, in which case it follows by separation of variables that the discrete spectrum is empty.

Theorem 3.2. Let $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ be the self-adjoint operator in ${{L}^{2}}({{\mathbb{R}}^{3}})$ associated to the form (1) and let $\alpha \gt 0$ and $\theta \in (0,\pi /2)$. Then the discrete spectrum of $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ below $-{{\alpha }^{2}}/4$ is infinite, accumulates at $-{{\alpha }^{2}}/4$, and the eigenvalues ${{\lambda }_{k}}\lt -{{\alpha }^{2}}/4$ (enumerated in non-decreasing order with multiplicities taken into account) satisfy the estimate

$$\begin{eqnarray}&&{{\lambda }_{k}}\leqslant -\frac{{{\alpha }^{2}}}{4}-\frac{\gamma (\theta )}{n_{k}^{4}},\quad k\in \mathbb{N},\end{eqnarray} \tag{ 35 }$$

where $\gamma (\theta )\gt 0$, ${{n}_{k+1}}:=n_{k}^{2}+{{n}_{k}}$ for $k\in \mathbb{N}$, and ${{n}_{1}}=N$ with $N\in \mathbb{N}$ sufficiently large.

Proof. Let us pick a function ${{\chi }_{1}}\in H_{0}^{1}(0,1)$ with $\parallel {{\chi }_{1}}{{\parallel }_{{{L}^{2}}(0,1)}}=1$ such that

$$\begin{eqnarray}&&\parallel {\mkern 1mu} \chi _{1}^{\prime }\parallel _{{{L}^{2}}(0,1)}^{2}\lt \frac{1}{4{{{\rm sin} }^{2}}\theta }\int _{0}^{1}\frac{\mid {{\chi }_{1}}(t){{\mid }^{2}}}{{{t}^{2}}}{\rm d}t\end{eqnarray} \tag{ 36 }$$

holds; [5, Lemma in §1]. Let us fix $\varepsilon \gt 0$ and choose ${{\chi }_{2}}\in C_{0}^{\infty }(-\varepsilon ,\varepsilon )$ such that $0\leqslant {{\chi }_{2}}\leqslant 1$ and ${{\chi }_{2}}(t)=1$ for $|t|\leqslant \varepsilon /2$. In the coordinate system (s, t) in figure 1 we define the sequence of functions

$$\begin{eqnarray}&&{{\omega }_{n}}(s,t):=\frac{1}{n}{{\chi }_{1}}(\frac{s-n}{{{n}^{2}}}){{\chi }_{2}}(\frac{t}{\sqrt{n}}){\rm exp} (-\frac{\alpha }{2}|t|)\in H_{0}^{1}(\mathbb{R}_{+}^{2}),\end{eqnarray} \tag{ 37 }$$

where the support of ω_n satisfies

$$\begin{eqnarray}&&{\rm supp}\;{{\omega }_{n}}\subset [n,n+{{n}^{2}}]\times [-\varepsilon \sqrt{n},\varepsilon \sqrt{n}],\quad n\in \mathbb{N},\end{eqnarray} \tag{ 38 }$$

in the coordinate system (s, t).

For sufficiently large $n\in \mathbb{N}$ the functions ω_n satisfy the conditions of lemma 3.1. The function ω_n can also be viewed as a function in r and z; cf figure 1. Then we define

$$\begin{eqnarray}&&{{\psi }_{n}}(r,\varphi ,z):=\frac{{{\omega }_{n}}(r,z)}{\sqrt{2\pi r}},\quad n\in \mathbb{N}.\end{eqnarray} \tag{ 39 }$$

Using lemma 3.1 we compute the values

$$\begin{eqnarray}\begin{array}{rcl} {{S}_{n}}: & = & {{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}[{{\psi }_{n}}]+\frac{{{\alpha }^{2}}}{4}\parallel {{\psi }_{n}}\parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2} \\ {} & = & \parallel \nabla {{\omega }_{n}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2};{{\mathbb{C}}^{2}})}^{2}-{{\int }_{\mathbb{R}_{+}^{2}}}\frac{1}{4{{r}^{2}}}\mid {{\omega }_{n}}{{\mid }^{2}}{\rm d}r{\rm d}z-\alpha \parallel {{\omega }_{n}}{{\mid }_{{{\Gamma }_{\theta }}}}\parallel _{{{L}^{2}}({{\Gamma }_{\theta }})}^{2}+\frac{{{\alpha }^{2}}}{4}\parallel {{\omega }_{n}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}. \\ \end{array}\end{eqnarray} \tag{ 40 }$$

The choice of χ₁ in (36) together with a subtle treatment of the second term in (40) will finally lead to ${{S}_{n}}\lt 0$ for sufficiently large $n\in \mathbb{N}$. First of all it is not difficult to check the asymptotics

$$\begin{eqnarray}&&\int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}{{\left| {{\chi }_{2}}\left( \frac{t}{\sqrt{n}} \right) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t=\frac{2}{\alpha }+\mathcal{O}({{e}^{-c\sqrt{n}}}),\quad n\to \infty ,\end{eqnarray} \tag{ 41 }$$

$$\begin{eqnarray}&&\int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}{{\left| \chi _{2}^{\prime }\left( \frac{t}{\sqrt{n}} \right) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t=\mathcal{O}({{e}^{-c\sqrt{n}}}),\quad n\to \infty ,\end{eqnarray} \tag{ 42 }$$

$$\begin{eqnarray}&&\int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}{{\chi }_{2}}\left( \frac{t}{\sqrt{n}} \right)\chi _{2}^{\prime }\left( \frac{t}{\sqrt{n}} \right){{e}^{-\alpha |t|}}{\rm d}t=\mathcal{O}({{e}^{-c\sqrt{n}}}),\quad n\to \infty ,\end{eqnarray} \tag{ 43 }$$

with some constant $c\gt 0$. Using (41) we get

$$\begin{eqnarray}\begin{array}{rcl} \frac{{{\alpha }^{2}}}{4}\parallel {{\omega }_{n}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2} & = & \frac{{{\alpha }^{2}}}{4}\left( \frac{1}{{{n}^{2}}}\int _{n}^{n+{{n}^{2}}}|{{\chi }_{1}}(\frac{s-n}{{{n}^{2}}}){{|}^{2}}{\rm d}s \right)\left( \int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}{{\left| {{\chi }_{2}}(\frac{t}{\sqrt{n}}) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t \right) \\ {} & = & \frac{\alpha }{2}+\mathcal{O}({{e}^{-c\sqrt{n}}}),\quad n\to \infty , \\ \end{array}\end{eqnarray} \tag{ 44 }$$

and

$$\begin{eqnarray*}\begin{array}{rcl} \parallel {\mkern 1mu} {{\partial }_{s}}{{\omega }_{n}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2} & = & \left( \int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}{{\left| {{\chi }_{2}}(\frac{t}{\sqrt{n}}) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t \right)\left( \frac{1}{{{n}^{4}}}\frac{1}{{{n}^{2}}}\int _{n}^{n+{{n}^{2}}}{{\left| \chi _{1}^{\prime }(\frac{s-n}{{{n}^{2}}}) \right|}^{2}}{\rm d}s \right) \\ {} & = & \frac{2}{\alpha }\frac{1}{{{n}^{4}}}\parallel \chi _{1}^{\prime }\parallel _{{{L}^{2}}(0,1)}^{2}+\mathcal{O}({{e}^{-c\sqrt{n}}}),\quad n\to \infty , \\ \end{array}\end{eqnarray*}$$

and from (42) and (43) we obtain

$$\begin{eqnarray}\begin{array}{rcl} \parallel {\mkern 1mu} {{\partial }_{t}}{{\omega }_{n}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2} & = & \left( \frac{1}{{{n}^{2}}}\int _{n}^{n+{{n}^{2}}}{{\left| {{\chi }_{1}}\left( \frac{s-n}{{{n}^{2}}} \right) \right|}^{2}}{\rm d}s \right) \\ {} & {} & \times \left( \int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}{{\left| \frac{1}{\sqrt{n}}\chi _{2}^{\prime }\left( \frac{t}{\sqrt{n}} \right)-\frac{\alpha {\rm sign}\;(t)}{2}{{\chi }_{2}}\left( \frac{t}{\sqrt{n}} \right) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t \right) \\ {} & = & \frac{\alpha }{2}+\mathcal{O}({{e}^{-c\sqrt{n}}}),\quad n\to \infty , \\ \end{array}\end{eqnarray} \tag{ 45 }$$

that is,

$$\begin{eqnarray}&&\parallel \nabla {{\omega }_{n}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2};{{\mathbb{C}}^{2}})}^{2}=\frac{2}{\alpha }\frac{1}{{{n}^{4}}}\parallel \chi _{1}^{\prime }\parallel _{{{L}^{2}}(0,1)}^{2}+\frac{\alpha }{2}+\mathcal{O}({{e}^{-c\sqrt{n}}}),\quad n\to \infty .\end{eqnarray} \tag{ 46 }$$

It is straightforward to see that

$$\begin{eqnarray}&&\alpha \parallel {{\omega }_{n}}{{\mid }_{{{\Gamma }_{\theta }}}}\parallel _{{{L}^{2}}({{\Gamma }_{\theta }})}^{2}=\frac{\alpha }{{{n}^{2}}}\int _{n}^{n+{{n}^{2}}}|{{\chi }_{1}}(\frac{s-n}{{{n}^{2}}}){{|}^{2}}{\rm d}s=\alpha \parallel {{\chi }_{1}}\parallel _{{{L}^{2}}(0,1)}^{2}=\alpha ,\end{eqnarray} \tag{ 47 }$$

and hence it remains to estimate the term ${{\int }_{\mathbb{R}_{+}^{2}}}\frac{1}{4{{r}^{2}}}|{{\omega }_{n}}{{|}^{2}}$ in (40). For that we make the following splitting

$$\begin{eqnarray}&&{{\int }_{\mathbb{R}_{+}^{2}}}\frac{1}{4{{r}^{2}}}{{\left| {{\omega }_{n}}(r,z) \right|}^{2}}{\rm d}r{\rm d}z=\int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}\int _{n}^{n+{{n}^{2}}}\frac{1}{4r{{\left( s,t \right)}^{2}}}{{\left| {{\omega }_{n}}\left( s,t \right) \right|}^{2}}{\rm d}s{\rm d}t={{I}_{n}}+{{J}_{n}},\end{eqnarray} \tag{ 48 }$$

where

$$\begin{eqnarray}&&{{I}_{n}}:=\int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}\int _{n}^{n+{{n}^{2}}}\frac{1}{4r{{\left( s,0 \right)}^{2}}}{{\left| {{\omega }_{n}}(s,t) \right|}^{2}}{\rm d}s{\rm d}t\end{eqnarray} \tag{ 49 }$$

and

$$\begin{eqnarray*}&&{{J}_{n}}:=\int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}\int _{n}^{n+{{n}^{2}}}\left( \frac{1}{4r{{(s,t)}^{2}}}-\frac{1}{4r{{(s,0)}^{2}}} \right)\mid {{\omega }_{n}}(s,t){{\mid }^{2}}{\rm d}s{\rm d}t.\end{eqnarray*}$$

The term J_n can be further rewritten as

$$\begin{eqnarray}&&{{J}_{n}}=\int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}\int _{n}^{n+{{n}^{2}}}\frac{(r(s,0)-r(s,t))(r(s,0)+r(s,t))}{4r{{(s,t)}^{2}}r{{(s,0)}^{2}}}\mid {{\omega }_{n}}(s,t){{\mid }^{2}}{\rm d}s{\rm d}t.\end{eqnarray} \tag{ 50 }$$

For geometric reasons we have $|r(s,0)-r(s,t)|\leqslant a\sqrt{n}$ with some $0\lt a\leqslant \varepsilon$ and $r(s,t)\gt bn$ with some $b\gt 0$ for all $(s,t)\in {\rm supp}\;{{\omega }_{n}}$. We first conclude from (50) that

$$\begin{eqnarray*}&&|{{J}_{n}}|\leqslant a\sqrt{n}\int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}\int _{n}^{n+{{n}^{2}}}\left| \frac{2}{4r(s,t)r{{\left( s,0 \right)}^{2}}}+\frac{r(s,0)-r(s,t)}{4r{{\left( s,t \right)}^{2}}r{{\left( s,0 \right)}^{2}}} \right||{{\omega }_{n}}\left( s,t \right){{|}^{2}}{\rm d}s{\rm d}t\end{eqnarray*}$$

and hence

$$\begin{eqnarray}&&|{{J}_{n}}|\leqslant \left( \frac{2a}{b\sqrt{n}}+\frac{{{a}^{2}}}{{{b}^{2}}n} \right){{I}_{n}}\end{eqnarray} \tag{ 51 }$$

follows together with (49). For I_n we have

$$\begin{eqnarray}\begin{array}{rcl} {{I}_{n}} & = & \left( \frac{1}{{{n}^{2}}}\int _{n}^{n+{{n}^{2}}}\frac{1}{4\;{{s}^{2}}{\rm si}{{{\rm n}}^{2}}\theta }|{{\chi }_{1}}(\frac{s-n}{{{n}^{2}}}){{|}^{2}}{\rm d}s \right)\left( \int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}{{\left| {{\chi }_{2}}\left( \frac{t}{\sqrt{n}} \right) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t \right) \\ {} & = & \left( \frac{1}{{{n}^{4}}}\int _{0}^{1}\frac{|{{\chi }_{1}}(u){{|}^{2}}}{4{\rm si}{{{\rm n}}^{2}}(\theta ){{\left( u+1/n \right)}^{2}}}{\rm d}u \right)\left( \int _{-\varepsilon \sqrt{n}}^{\varepsilon \sqrt{n}}{{\left| {{\chi }_{2}}\left( \frac{t}{\sqrt{n}} \right) \right|}^{2}}{{e}^{-\alpha |t|}}{\rm d}t \right), \\ \end{array}\end{eqnarray} \tag{ 52 }$$

and the choice of χ₁ (see (36)) together with monotone convergence yields

$$\begin{eqnarray*}&&\int _{0}^{1}\frac{|{{\chi }_{1}}(u){{|}^{2}}}{{{(u+1/n)}^{2}}}{\rm d}u=\int _{0}^{1}\frac{|{{\chi }_{1}}(u){{|}^{2}}}{{{u}^{2}}}{\rm d}u+o(1),\quad n\to \infty .\end{eqnarray*}$$

Hence we conclude from (41) and (52) that

$$\begin{eqnarray*}&&{{I}_{n}}=\frac{2}{\alpha }\frac{1}{{{n}^{4}}}\frac{1}{4{{{\rm sin} }^{2}}(\theta )}\int _{0}^{1}\frac{|{{\chi }_{1}}(u){{|}^{2}}}{{{u}^{2}}}{\rm d}u+o\left( \frac{1}{{{n}^{4}}} \right),\quad n\to \infty ,\end{eqnarray*}$$

and from (51) we find

$$\begin{eqnarray*}&&{{J}_{n}}=o\left( \frac{1}{{{n}^{4}}} \right),\quad n\to \infty .\end{eqnarray*}$$

It follows that (48) becomes

$$\begin{eqnarray}&&{{\int }_{\mathbb{R}_{+}^{2}}}\frac{1}{4{{r}^{2}}}|{{\omega }_{n}}(r,z){{|}^{2}}{\rm d}r{\rm d}z=\frac{2}{\alpha }\frac{1}{{{n}^{4}}}\frac{1}{4{{{\rm sin} }^{2}}(\theta )}\int _{0}^{1}\frac{|{{\chi }_{1}}(u){{|}^{2}}}{{{u}^{2}}}{\rm d}u+o\left( \frac{1}{{{n}^{4}}} \right),\end{eqnarray} \tag{ 53 }$$

as $n\to \infty$. Finally, (44), (46), (47) and (53) yield

$$\begin{eqnarray}&&{{S}_{n}}=\frac{2}{\alpha }\frac{1}{{{n}^{4}}}\left( \parallel \chi _{1}^{\prime }\parallel _{{{L}^{2}}(0,1)}^{2}-\int _{0}^{1}\frac{|{{\chi }_{1}}(u){{|}^{2}}}{4{{{\rm sin} }^{2}}(\theta ){{u}^{2}}}{\rm d}u \right)+o\left( \frac{1}{{{n}^{4}}} \right),\quad n\to \infty ,\end{eqnarray} \tag{ 54 }$$

for S_n in (40). In view of the above asymptotics and according to (36) there exists $N\in \mathbb{N}$ such that for all $n\geqslant N$ we have

$$\begin{eqnarray}&&{{S}_{n}}\leqslant -\frac{2\gamma (\theta )}{\alpha {{n}^{4}}}\end{eqnarray} \tag{ 55 }$$

for some constant $\gamma (\theta )\gt 0$. Let us consider a sequence ${{\{{{n}_{k}}\}}_{k}}$, where ${{n}_{1}}:=N$ and ${{n}_{k+1}}:=n_{k}^{2}+{{n}_{k}}$ for $k\in \mathbb{N}$. Then by (38) the measure of ${\rm supp}\;{{\omega }_{{{n}_{k}}}}\cap {\rm supp}\;{{\omega }_{{{n}_{l}}}}$ is zero for all $k,l\in \mathbb{N}$, $k\ne l$, and hence it follows from the definition (39) that the measure of ${\rm supp}\;{{\psi }_{{{n}_{k}}}}\cap {\rm supp}\;{{\psi }_{{{n}_{l}}}}$ is zero for all $k,l\in \mathbb{N}$, $k\ne l$, and, in particular, the functions ${{\psi }_{{{n}_{k}}}}$ are orthogonal in ${{L}^{2}}({{\mathbb{R}}^{3}})$. The space

$$\begin{eqnarray*}&&{{F}_{k}}:={\rm span}\left\{ {{\psi }_{{{n}_{1}}}},{{\psi }_{{{n}_{2}}}},...,{{\psi }_{{{n}_{k}}}} \right\}\subset {{H}^{1}}({{\mathbb{R}}^{3}}),\end{eqnarray*}$$

has dimension k and for an arbitrary $\psi =\sum _{l=1}^{k}{{a}_{l}}{{\psi }_{{{n}_{l}}}}\in {{F}_{k}}$, ${{a}_{l}}\in \mathbb{C}$, we get

$$\begin{eqnarray}&&\parallel \psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}=\mathop{\sum }\limits_{l=1}^{k}|{{a}_{l}}{{|}^{2}}\parallel {{\psi }_{{{n}_{l}}}}\parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}=\mathop{\sum }\limits_{l=1}^{k}|{{a}_{l}}{{|}^{2}}\parallel {{\omega }_{{{n}_{l}}}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}\leqslant \frac{2}{\alpha }\mathop{\sum }\limits_{l=1}^{k}|{{a}_{l}}{{|}^{2}},\end{eqnarray} \tag{ 56 }$$

where we have also used the estimate $\parallel {{\omega }_{{{n}_{l}}}}\parallel _{{{L}^{2}}(\mathbb{R}_{+}^{2})}^{2}\leqslant \frac{2}{\alpha }$. Employing (55) we obtain

$$\begin{eqnarray*}&&{{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}[\psi ]+\frac{{{\alpha }^{2}}}{4}\parallel \psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}=\mathop{\sum }\limits_{l=1}^{k}|{{a}_{l}}{{|}^{2}}{{S}_{{{n}_{l}}}}\leqslant -\frac{2\gamma (\theta )}{\alpha n_{k}^{4}}\mathop{\sum }\limits_{l=1}^{k}|{{a}_{l}}{{|}^{2}},\end{eqnarray*}$$

where we have again used that the mutual intersections of the supports of $\{{{\psi }_{{{n}_{l}}}}\}_{l=1}^{k}$ are of measure zero.

Combining the above estimate with (56) we get

$$\begin{eqnarray}\begin{array}{rcl} \frac{{{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}[\psi ]}{\parallel \psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}} & = & -\frac{{{\alpha }^{2}}}{4}+\frac{{{\mathfrak{a}}_{\alpha ,{{\mathcal{C}}_{\theta }}}}[\psi ]+({{\alpha }^{2}}/4)\parallel \psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}}{\parallel \psi \parallel _{{{L}^{2}}({{\mathbb{R}}^{3}})}^{2}} \\ {} & \leqslant & -\frac{{{\alpha }^{2}}}{4}-\frac{\gamma (\theta )}{n_{k}^{4}}\lt -\frac{{{\alpha }^{2}}}{4}. \\ \end{array}\end{eqnarray} \tag{ 57 }$$

Hence, according to [3, Theorem 10.2.3] the operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ has at least k eigenvalues below the bottom of the essential spectrum $-{{\alpha }^{2}}/4$. The above construction works for any $k\in \mathbb{N}$, so that the operator $-{{\Delta }_{\alpha ,{{\mathcal{C}}_{\theta }}}}$ has infinitely many eigenvalues below $-{{\alpha }^{2}}/4$. The eigenvalue estimate (35) follows from [3, Theorem 10.2.3] and (57). □

Let $\theta \in (0,\pi /2)$ and ${{\mathcal{C}}_{\theta }}$ be the conical surface as above. A hypersurface $\Sigma \subset {{\mathbb{R}}^{3}}$, which for some compact set $K\subset {{\mathbb{R}}^{3}}$ satisfies the condition $\Sigma \backslash K={{\mathcal{C}}_{\theta }}\backslash K$ and which splits the space ${{\mathbb{R}}^{3}}$ into two unbounded Lipschitz domains, is called a local deformation of ${{\mathcal{C}}_{\theta }}$; cf [1, Section 4.2]. Below we consider the self-adjoint Schrödinger operator $-{{\Delta }_{\alpha ,\Sigma }}$ with an attractive δ-interaction of constant strength $\alpha \gt 0$ supported on the Lipschitz hypersurface Σ. This Schrödinger operator is defined via the closed, densely defined, symmetric and semibounded quadratic form

$$\begin{eqnarray}&&{{\mathfrak{a}}_{\alpha ,\Sigma }}[\psi ]=\left\|\nabla \psi \right\|_{{{L}^{2}}\left( {{\mathbb{R}}^{3}};{{\mathbb{C}}^{3}} \right)}^{2}-\alpha {{\int }_{\Sigma }}{{\left| \psi \right|}^{2}}{\rm d}\sigma ,\quad {\rm dom}\;{{\mathfrak{a}}_{\alpha ,\Sigma }}={{H}^{1}}\left( {{\mathbb{R}}^{3}} \right).\end{eqnarray} \tag{ 58 }$$

The assertion on the essential spectrum in the next theorem is a consequence of [1, Theorem 4.7]; the infiniteness of the discrete spectrum can be shown as in the proof of theorem 3.2 using the same functions ψ_n in (39) and $n\in \mathbb{N}$ sufficiently large.

Theorem 3.3. Let $\theta \in (0,\pi /2)$ and $\alpha \gt 0$. Let Σ be a local deformation of the cone ${{\mathcal{C}}_{\theta }}$ and let $-{{\Delta }_{\alpha ,\Sigma }}$ be the self-adjoint operator in ${{L}^{2}}({{\mathbb{R}}^{3}})$ associated to (58). Then

$$\begin{eqnarray*}&&{{\sigma }_{{\rm ess}}}(-{{\Delta }_{\alpha ,\Sigma }})=[-{{\alpha }^{2}}/4,+\infty ),\end{eqnarray*}$$

the discrete spectrum below $-{{\alpha }^{2}}/4$ is infinite, accumulates at $-{{\alpha }^{2}}/4$, and the eigenvalues ${{\lambda }_{k}}\lt -{{\alpha }^{2}}/4$ (enumerated in non-decreasing order with multiplicities taken into account) satisfy the estimate

$$\begin{eqnarray*}&&{{\lambda }_{k}}\leqslant -\frac{{{\alpha }^{2}}}{4}-\frac{\gamma (\theta )}{n_{k}^{4}},\quad k\in \mathbb{N},\end{eqnarray*}$$

where $\gamma (\theta )\gt 0$, ${{n}_{k+1}}:=n_{k}^{2}+{{n}_{k}}$ for $k\in \mathbb{N}$, and ${{n}_{1}}=N$ with $N\in \mathbb{N}$ sufficiently large.

The authors gratefully acknowledge financial support by the Austrian Science Fund (FWF), project P 25162-N26, Czech Science Foundation (GAČR), project 14–06818S, and the Austrian Agency for International Cooperation in Education and Research (OeAD), Austria-Czech Republic cooperation grant CZ01/2013.