Self-adjointness of the 2D Dirac Operator with Singular Interactions Supported on Star Graphs

Abstract

We consider the two-dimensional Dirac operator with Lorentz-scalar $\delta$-shell interactions on each edge of a star graph. An orthogonal decomposition is performed which shows such an operator is unitarily equivalent to an orthogonal sum of half-line Dirac operators with off-diagonal Coulomb potentials. This decomposition reduces the computation of the deficiency indices to determining the number of eigenvalues of a one-dimensional spin–orbit operator in the interval $(-1/2,1/2)$. If the number of edges of the star graph is two or three, these deficiency indices can then be analytically determined for a range of parameters. For higher numbers of edges, it is possible to numerically calculate the deficiency indices. Among others, examples are given where the strength of the Lorentz-scalar interactions directly change the deficiency indices, while other parameters are all fixed and where the deficiency indices are (2, 2), neither of which have been observed in the literature to the best knowledge of the authors. For those Dirac operators which are not already self-adjoint and do not have 0 in the spectrum of the associated spin–orbit operator, the distinguished self-adjoint extension is also characterized.

Appendix A. Closedness of the Operator \({\varvec{d}}\) \(_{\tilde{\lambda }}\)

Lemma A.1

The closure of the restriction ${\varvec{d}}_{\widetilde{\lambda }}\upharpoonright C_0^{\infty }({\mathbb {R}}_+,{\mathbb {C}}^2)$ is ${\varvec{d}}_{\widetilde{\lambda }}$ with domain given by (2.21). In particular, ${\varvec{d}}_{\widetilde{\lambda }}$ with domain given by (2.21) is closed.

Proof

We split the analysis into consideration of the two cases $\widetilde{\lambda }\ne 1/2$ and $\widetilde{\lambda }= 1/2$.

Let $\widetilde{\lambda }\ne 1/2$. For any $\psi \in C^\infty _0({\mathbb {R}}_+,{\mathbb {C}}^2)$, observe that

$$\begin{aligned} \begin{aligned}&\Vert {\varvec{d}}_{\widetilde{\lambda }}\psi \Vert ^2_{L^2({\mathbb {R}}_+,{\mathbb {C}}^2)} = \int _{{\mathbb {R}}_+}\left| \psi _2' + \frac{\widetilde{\lambda }\psi _2}{r}\right| ^2\hbox {d}r + \int _{{\mathbb {R}}_+}\left| \psi _1'- \frac{\widetilde{\lambda }\psi _1}{r}\right| ^2\hbox {d}r\\&\quad = \int _{{\mathbb {R}}_+}\left( |\psi _2'|^2 + \frac{2\widetilde{\lambda }\Re (\psi _2'\overline{\psi _2})}{r} +\frac{\widetilde{\lambda }^2|\psi _2|^2}{r^2}\right) \hbox {d}r\\&\qquad + \int _{{\mathbb {R}}_+}\left( |\psi _1'|^2 - \frac{2\widetilde{\lambda }\Re (\psi _1'\overline{\psi _1})}{r} +\frac{\widetilde{\lambda }^2|\psi _1|^2}{r^2}\right) \hbox {d}r\\&\quad = \int _{{\mathbb {R}}_+}\left( |\psi _2'|^2 + \frac{\widetilde{\lambda }(|\psi _2|^2)'}{r} +\frac{\widetilde{\lambda }^2|\psi _2|^2}{r^2}\right) \hbox {d}r\\ {}&\qquad + \int _{{\mathbb {R}}_+}\left( |\psi _1'|^2 - \frac{\widetilde{\lambda }(|\psi _1|^2)'}{r} +\frac{\widetilde{\lambda }^2|\psi _1|^2}{r^2}\right) \hbox {d}r\\&\quad = \int _{{\mathbb {R}}_+}\left( |\psi _2'|^2 +\frac{(\widetilde{\lambda }^2+\widetilde{\lambda })|\psi _2|^2}{r^2}\right) \hbox {d}r + \int _{{\mathbb {R}}_+}\left( |\psi _1'|^2 +\frac{(\widetilde{\lambda }^2-\widetilde{\lambda })|\psi _1|^2}{r^2}\right) \hbox {d}r,\\ \end{aligned} \end{aligned}$$

where in the last step we performed integration by parts. Combining the above formula with the one-dimensional Hardy inequality

$$\begin{aligned} \int _{{\mathbb {R}}_+}|\psi '|^2\hbox {d}r \ge \int _{{\mathbb {R}}_+}\frac{|\psi |^2}{4r^2}\hbox {d}r ,\qquad \forall \,\psi \in H^1_0({\mathbb {R}}_+), \end{aligned}$$

we conclude that the graph-norm induced by the operator ${\varvec{d}}_{\widetilde{\lambda }}$ on the space $C^\infty _0({\mathbb {R}}_+,{\mathbb {C}}^2)$ is equivalent to the $H^1$-norm. Indeed, we get the double sided estimate

$$\begin{aligned} \min \{1,(1-2\widetilde{\lambda })^2\}\Vert \psi \Vert ^2_{H^1({\mathbb {R}}_+,{\mathbb {C}}^2)}\le & {} \Vert {\varvec{d}}_{\widetilde{\lambda }}\psi \Vert ^2_{L^2({\mathbb {R}}_+,{\mathbb {C}}^2)} +\Vert \psi \Vert ^2_{L^2({\mathbb {R}}_+,{\mathbb {C}}^2)}\\\le & {} \left( 1+2\widetilde{\lambda }\right) ^2 \Vert \psi \Vert ^2_{H^1({\mathbb {R}}_+,{\mathbb {C}}^2)} \end{aligned}$$

for any $\psi \in C^\infty _0({\mathbb {R}}_+,{\mathbb {C}}^2)$. Hence, we have ${\varvec{d}}_{\widetilde{\lambda }} = \overline{{\varvec{d}}_{\widetilde{\lambda }}\upharpoonright C^\infty _0({\mathbb {R}}_+,{\mathbb {C}}^2)}$ and thus the operator ${\varvec{d}}_{\widetilde{\lambda }}$ is closed in this case.

Let $\widetilde{\lambda }= 1/2$. Using [16, Proposition 40] (see also [10, Proposition 3.1]), one can check that the domain of ${\varvec{d}}_{1/2}$ can be alternatively characterized as

$$\begin{aligned} {{\,\mathrm{dom}\,}}{\varvec{d}}_{1/2} = \left\{ \psi \in L^2({\mathbb {R}}_+,{\mathbb {C}}^2):\psi _1'-\frac{\psi _1}{2r}, \psi _2'+\frac{\psi _2}{2r}\in L^2({\mathbb {R}}_+)\right\} . \end{aligned}$$

Closedness of ${\varvec{d}}_{1/2}$ then follows from [13, Equation (1.10) and Theorem 1.1(ii)]. $\square$