Eigenvalue Curves for Generalized MIT Bag Models

Abstract

We study spectral properties of Dirac operators on bounded domains $\Omega \subset {\mathbb {R}}^3$ with boundary conditions of electrostatic and Lorentz scalar type and which depend on a parameter $\tau \in \mathbb {R}$; the case $\tau = 0$ corresponds to the MIT bag model. We show that the eigenvalues are parametrized as increasing functions of $\tau$, and we exploit this monotonicity to study the limits as $\tau \rightarrow \pm \infty$. We prove that if $\Omega$ is not a ball then the first positive eigenvalue is greater than the one of a ball with the same volume for all $\tau$ large enough. Moreover, we show that the first positive eigenvalue converges to the mass of the particle as $\tau \downarrow -\infty$, and we also analyze its first order asymptotics.

Appendices

Appendix A. Properties of the Spectrum

Here we prove the properties of the spectrum of ${\mathcal {H}}_\tau$ collected in Lemma 1.2. For a shorter notation, we will use

$$\begin{aligned} {\mathcal {B}} := -i \beta (\alpha \cdot \nu ), \end{aligned}$$

which defines a self-adjoint operator in $L^2(\partial \Omega )^4$ such that ${\mathcal {B}}^2 = I_4$. For every $\varphi \in \text {Dom}({\mathcal {H}}_\tau )$ it holds $\varphi = \sinh \tau {\mathcal {B}} \beta \varphi + \cosh \tau {\mathcal {B}} \varphi$ on $\partial \Omega$.

Proof of Lemma 1.2

From [21, Proposition 5.15] we know that ${\mathcal {H}}_\tau$ is self-adjoint in $L^2(\Omega )^4$. The proof of (i) follows from this and the compact embedding of $H^1(\Omega )^4$ into $L^2(\Omega )^4$.

Let us now show (ii). We first claim that for every $\varphi = (u,v)^\intercal \in \text {Dom}({\mathcal {H}}_\tau )$ it holds

$$\begin{aligned} \left\| { {\mathcal {H}}\varphi } \right\| ^2_{L^2(\Omega )^4} = \left\| { \alpha \cdot \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} + m^2 \left\| {\varphi } \right\| ^2_{L^2(\Omega )^4} + m e^\tau \left\| {u} \right\| ^2_{L^2(\partial \Omega )^2} + m e^{-\tau } \left\| {v} \right\| ^2_{L^2(\partial \Omega )^2}.\nonumber \\ \end{aligned}$$

(A.1)

To prove this formula, note first that expanding the square we have

$$\begin{aligned} \left\| { {\mathcal {H}}\varphi } \right\| ^2_{L^2(\Omega )^4}= & {} \langle -i \alpha \cdot \nabla \varphi , -i \alpha \cdot \nabla \varphi \rangle _{L^2(\Omega )^4} + m^2 \langle \beta \varphi , \beta \varphi \rangle _{L^2(\Omega )^4} \\{} & {} + 2 m \Re \langle \beta \varphi , -i \alpha \cdot \nabla \varphi \rangle _{L^2(\Omega )^4} \\= & {} \left\| { \alpha \cdot \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} + m^2 \left\| {\varphi } \right\| ^2_{L^2(\Omega )^4} + 2 m \Re \langle \beta \varphi , -i \alpha \cdot \nabla \varphi \rangle _{L^2(\Omega )^4} . \end{aligned}$$

Integrating by parts we get

$$\begin{aligned} \begin{aligned} \langle \beta \varphi , -i \alpha \cdot \nabla \varphi \rangle _{L^2(\Omega )^4}&= \langle -i \alpha \cdot \nabla (\beta \varphi ), \varphi \rangle _{L^2(\Omega )^4} - \langle -i (\alpha \cdot \nu ) \beta \varphi , \varphi \rangle _{L^2(\partial \Omega )^4} \\&= - \langle \beta (-i \alpha \cdot \nabla )\varphi , \varphi \rangle _{L^2(\Omega )^4} + \langle {\mathcal {B}} \varphi , \varphi \rangle _{L^2(\partial \Omega )^4} , \end{aligned} \end{aligned}$$

and using that $\beta$ is hermitian, we obtain $2 \Re \langle \beta \varphi , -i \alpha \cdot \nabla \varphi \rangle _{L^2(\Omega )^4} = \langle {\mathcal {B}} \varphi , \varphi \rangle _{L^2(\partial \Omega )^4}$. Thus,

$$\begin{aligned} \left\| { {\mathcal {H}}\varphi } \right\| ^2_{L^2(\Omega )^4} = \left\| { \alpha \cdot \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} + m^2 \left\| {\varphi } \right\| ^2_{L^2(\Omega )^4} + m \langle {\mathcal {B}} \varphi , \varphi \rangle _{L^2(\partial \Omega )^4}. \end{aligned}$$

(A.2)

Now, using that $\varphi \in \text {Dom}({\mathcal {H}}_{\tau })$, we see that

$$\begin{aligned} \langle {\mathcal {B}} \varphi , \varphi \rangle _{L^2(\partial \Omega )^4}= & {} \langle {\mathcal {B}} \varphi , \sinh \tau {\mathcal {B}} \beta \varphi + \cosh \tau {\mathcal {B}} \varphi \rangle _{L^2(\partial \Omega )^4} \\= & {} \langle \varphi , \sinh \tau {\mathcal {B}}^2 \beta \varphi + \cosh \tau {\mathcal {B}}^2 \varphi \rangle _{L^2(\partial \Omega )^4} \\= & {} \langle \varphi , \sinh \tau \beta \varphi + \cosh \tau \varphi \rangle _{L^2(\partial \Omega )^4}, \end{aligned}$$

where we have used that ${\mathcal {B}}$ is self-adjoint and that ${\mathcal {B}}^2 = I_4$. From here, and writing $\varphi = (u, v)^\intercal$, we easily see that

$$\begin{aligned} \langle {\mathcal {B}} \varphi , \varphi \rangle _{L^2(\partial \Omega )^4} = e^\tau \left\| {u} \right\| ^2_{L^2(\partial \Omega )^2} + e^{-\tau } \left\| {v} \right\| ^2_{L^2(\partial \Omega )^2}. \end{aligned}$$

Plugging this into (A.2) we obtain (A.1), proving the claim. Now, let $\varphi \in \text {Dom}({\mathcal {H}}_\tau )\setminus \{0\}$ be such that ${\mathcal {H}}\varphi = \uplambda \varphi$ in $\Omega$. Note that, by Lemma 2.4, $\varphi$ cannot vanish identically on $\partial \Omega$. Therefore, using (A.1) we obtain

$$\begin{aligned} \uplambda ^2 \left\| {\varphi } \right\| ^2_{L^2(\Omega )^4} = \left\| { {\mathcal {H}}\varphi } \right\| ^2_{L^2(\Omega )^4} > m^2 \left\| {\varphi } \right\| ^2_{L^2(\Omega )^4}, \end{aligned}$$

which yields $|\uplambda | > m$.

We finally prove (iii) and (iv). First, by the compact embedding of $H^1(\Omega )^4$ into $L^2(\Omega )^4$ we have that the resolvent of ${\mathcal {H}}_\tau$ is a compact operator, which yields that every eigenvalue has finite multiplicity. Now, given $\psi \in \mathbb {C}^4$, consider the charge conjugation operator

$$\begin{aligned} {\mathcal {C}} \psi := i \beta \alpha _{2} {\overline{\psi }} \end{aligned}$$

and the time reversal-symmetry operator

$$\begin{aligned} T \psi : =-i \gamma _{5} \alpha _{2} {\overline{\psi }}, \quad \text {where}\quad \gamma _{5}:= \begin{pmatrix}0&{}I_2\\ I_2&{}0\end{pmatrix}. \end{aligned}$$

(A.3)

Then, simple computations show that ${\mathcal {H}}T=T {\mathcal {H}}$, ${\mathcal {H}}{\mathcal {C}}=-{\mathcal {C}} {\mathcal {H}}$, and $T {\mathcal {C}}= {\mathcal {C}} T$. In addition, setting

$$\begin{aligned} {\mathcal {B}}_\tau := \sinh \tau {\mathcal {B}} \beta + \cosh \tau {\mathcal {B}} = i (\sinh \tau - \cosh \tau \beta ) (\alpha \cdot \nu ), \end{aligned}$$

it is also easy to check that ${\mathcal {B}}_{\tau } T=T {\mathcal {B}}_{\tau }$ and ${\mathcal {B}}_{\tau } {\mathcal {C}}={\mathcal {C}} {\mathcal {B}}_{-\tau }$. Note that for every function $\varphi \in \text {Dom}({\mathcal {H}}_{\tau })$ it holds $\varphi = {\mathcal {B}}_\tau \varphi$ on $\partial \Omega$. As a consequence, given an eigenfunction $\varphi$ of ${\mathcal {H}}_\tau$ with eigenvalue $\uplambda$, $T\varphi$ is also an eigenfunction of ${\mathcal {H}}_\tau$ with eigenvalue $\uplambda$. Furthermore, ${\mathcal {C}}\varphi$ and $T{\mathcal {C}} \varphi$ are eigenfunctions of ${\mathcal {H}}_{-\tau }$ with eigenvalue $-\uplambda$. $\square$

To conclude this section, we establish a formula which relates the $L^2(\Omega )^4$-norms of $\nabla \varphi$ and $\alpha \cdot \nabla \varphi$ for functions $\varphi \in \text {Dom} ({\mathcal {H}}_\tau )$. Although we do not use this formula in this article, we think that it has its own interest, and it may be useful to present it here for future reference. The formula is a generalization of [7, formula (1.3)], in which the case $\tau = 0$ is considered, and the proof follows the same lines.

Lemma A.1

Let $\tau \in \mathbb {R}$ and $\varphi \in \text {Dom} ({\mathcal {H}}_\tau ) \cap H^1(\partial \Omega )^4$. Then,

$$\begin{aligned}{} & {} \left\| { \alpha \cdot \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} \\{} & {} = \left\| { \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} + \dfrac{1}{2} \int _{\partial \Omega } \kappa |\varphi |^2 d \upsigma + \sinh \tau \langle \gamma _5 (\alpha \cdot \nu ) \varphi , \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4}, \end{aligned}$$

where $\kappa$ denotes the mean curvature of $\partial \Omega$.

Proof

First, for every $\varphi \in H^2(\Omega )^4$, it holds

$$\begin{aligned} \left\| { \alpha \cdot \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} = \left\| { \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} - \langle \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4}, \end{aligned}$$

(A.4)

where $\gamma _5$ is defined in (A.3). This is proved in [7, Appendix A.2]. By density, it also holds for all $\varphi \in H^1(\Omega )^4 \cap H^1(\partial \Omega )^4$.

Let us now investigate the boundary term in the above expression. The crucial point is to use that the mean curvature of $\partial \Omega$ arises in our context through the formula

$$\begin{aligned}{}[-i \alpha \cdot (\nu \times \nabla ), {\mathcal {B}}]= - \kappa \gamma _5 {\mathcal {B}}, \end{aligned}$$

where $[\cdot , \cdot ]$ denotes the commutator of two operators, i.e., $[S,T] := ST - TS$; see [7, Lemma A.3]. Using this and the boundary condition for $\varphi \in \text {Dom}({\mathcal {H}}_\tau )\cap H^1(\partial \Omega )^4$ we get

$$\begin{aligned} \begin{aligned} \langle \gamma _5 \varphi&,-i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \\&= \sinh \tau \langle \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) {\mathcal {B}} \beta \varphi \rangle _{L^2(\partial \Omega )^4} + \cosh \tau \langle \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) {\mathcal {B}}\varphi \rangle _{L^2(\partial \Omega )^4} \\&= \sinh \tau \langle \gamma _5 \varphi , [-i \alpha \cdot (\nu \times \nabla ), {\mathcal {B}}] \beta \varphi \rangle _{L^2(\partial \Omega )^4} + \sinh \tau \langle \gamma _5 \varphi , {\mathcal {B}} ( -i \alpha \cdot (\nu \times \nabla ) ) \beta \varphi \rangle _{L^2(\partial \Omega )^4} \\&\quad \quad + \cosh \tau \langle \gamma _5 \varphi , [-i \alpha \cdot (\nu \times \nabla ), {\mathcal {B}}]\varphi \rangle _{L^2(\partial \Omega )^4} + \cosh \tau \langle \gamma _5 \varphi , {\mathcal {B}} (-i \alpha \cdot (\nu \times \nabla ))\varphi \rangle _{L^2(\partial \Omega )^4} \\&= -\sinh \tau \langle \gamma _5 \varphi , \kappa \gamma _5 {\mathcal {B}} \beta \varphi \rangle _{L^2(\partial \Omega )^4}+ \sinh \tau \langle {\mathcal {B}} \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) \beta \varphi \rangle _{L^2(\partial \Omega )^4} \\&\quad \quad - \cosh \tau \langle \gamma _5 \varphi , \kappa \gamma _5 {\mathcal {B}} \varphi \rangle _{L^2(\partial \Omega )^4} + \cosh \tau \langle {\mathcal {B}} \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \\&= - \langle \gamma _5 \varphi , \kappa \gamma _5 (\sinh \tau {\mathcal {B}} \beta + \cosh \tau {\mathcal {B}} )\varphi \rangle _{L^2(\partial \Omega )^4} + \sinh \tau \langle {\mathcal {B}} \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) \beta \varphi \rangle _{L^2(\partial \Omega )^4} \\&\quad \quad + \cosh \tau \langle {\mathcal {B}} \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \\&= - \langle \gamma _5 \varphi , \kappa \gamma _5 \varphi \rangle _{L^2(\partial \Omega )^4} - \sinh \tau \langle \gamma _5 {\mathcal {B}} \varphi , -i \alpha \cdot (\nu \times \nabla ) \beta \varphi \rangle _{L^2(\partial \Omega )^4} \\&\quad \quad - \cosh \tau \langle \gamma _5 {\mathcal {B}}\varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} .\\ \end{aligned} \end{aligned}$$

Here we have used that $(\alpha \cdot x) \gamma _5 = \gamma _5 (\alpha \cdot x)$ for all $x\in \mathbb {R}^3$ (see [7, Lemma A.1]) and that $\beta$ anticommutes with $\gamma _5$, thus ${\mathcal {B}} \gamma _5 = - \gamma _5 {\mathcal {B}}$. Now, using that $(\alpha \cdot x) \beta = - \beta (\alpha \cdot x)$ for every $x\in \mathbb {R}^3$, and that $\beta$ anticommutes with ${\mathcal {B}}$ and $\gamma _5$, we have

$$\begin{aligned} \begin{aligned}&\langle \gamma _5 {\mathcal {B}} \varphi , -i \alpha \cdot (\nu \times \nabla ) \beta \varphi \rangle _{L^2(\partial \Omega )^4} \\&= - \langle \gamma _5 {\mathcal {B}} \varphi , \beta (-i \alpha \cdot (\nu \times \nabla ) ) \varphi \rangle _{L^2(\partial \Omega )^4} \\&= - \langle \beta \gamma _5{\mathcal {B}} \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} = -\langle \gamma _5 {\mathcal {B}} \beta \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4}. \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned}&\langle \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \\&= - \langle \gamma _5 \varphi , \kappa \gamma _5 \varphi \rangle _{L^2(\partial \Omega )^4} + \sinh \tau \langle \gamma _5 {\mathcal {B}} \beta \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \\&\quad \quad - \cosh \tau \langle \gamma _5 {\mathcal {B}}\varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \\&= - \langle \varphi , \kappa \varphi \rangle _{L^2(\partial \Omega )^4} + 2\sinh \tau \langle \gamma _5 {\mathcal {B}} \beta \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \\&\quad \quad - \langle \gamma _5 (\sinh \tau {\mathcal {B}} \beta + \cosh \tau {\mathcal {B}})\varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \end{aligned} \end{aligned}$$

and thus, using the boundary condition, we get

$$\begin{aligned}{} & {} \langle \gamma _5 \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4} \\{} & {} = - \dfrac{1}{2} \int _{\partial \Omega } \kappa |\varphi |^2 d \upsigma + \sinh \tau \langle \gamma _5 {\mathcal {B}} \beta \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4}, \end{aligned}$$

which combined with (A.4) gives

$$\begin{aligned}{} & {} \left\| { \alpha \cdot \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} \\{} & {} = \left\| { \nabla \varphi } \right\| ^2_{L^2(\Omega )^4} + \dfrac{1}{2} \int _{\partial \Omega } \kappa |\varphi |^2 d \upsigma - \sinh \tau \langle \gamma _5 {\mathcal {B}} \beta \varphi , -i \alpha \cdot (\nu \times \nabla ) \varphi \rangle _{L^2(\partial \Omega )^4}. \end{aligned}$$

Finally, using again that $(\alpha \cdot x) \beta = - \beta (\alpha \cdot x)$ for every $x\in \mathbb {R}^3$, we have ${\mathcal {B}} \beta = i \alpha \cdot \nu$, and inserting this into the above identity we conclude the proof. $\square$

Appendix B. The Ball

In this appendix we present a more explicit spectral analysis in the case that $\Omega \subset \mathbb {R}^3$ is a ball of radius $R>0$ centered at the origin, which will be denoted by $B_R$. To study this radially symmetric case we introduce spherical coordinates: if $x\in \mathbb {R}^3$ we write $x=r\theta$ with $r = |x| \in [0,+\infty )$ and $\theta = x/|x| \in \mathbb {S}^2$. Using separation of variables and the spherical harmonic spinors, we give the explicit equations for the eigenvalues and eigenfunctions of ${\mathcal {H}}_\tau$.

1.1 B.1 Decomposition using spherical harmonic spinors

Let $Y^\ell _n$ be the usual spherical harmonics on $\mathbb {S}^2$; here $n = 0, 1, 2, \ldots$ and $\ell = -n, -n+1, \ldots , n + 1, n$. They satisfy $\Delta _{\mathbb {S}^2} Y^\ell _n = -n(n + 1)Y^\ell _n$, where $\Delta _{\mathbb {S}^2}$ denotes the usual spherical Laplacian. Moreover, $Y^\ell _n$ form a complete orthonormal set in $L^2(\mathbb {S}^2)$.

Following [71, Section 4.6.4], the spherical harmonic spinors are defined as follows: for $j=1/2, 3/2, \ldots$ and $\mu _j = -j, -j+1, \ldots , j-1, j$, set

$$\begin{aligned} \psi _{j-1 / 2}^{\mu _j}= & {} \frac{1}{\sqrt{2 j}}\begin{pmatrix} \sqrt{j+\mu _j} Y_{j-1 / 2}^{\mu _j -1 / 2} \\ \sqrt{j-\mu _j} Y_{j-1 / 2}^{\mu _j +1 / 2} \end{pmatrix} \quad \text { and } \\ \quad \psi _{j+1 / 2}^{\mu _j}= & {} \frac{1}{\sqrt{2 j+2}}\begin{pmatrix} \sqrt{j+1-\mu _j} Y_{j+1 / 2}^{\mu _j -1 / 2} \\ -\sqrt{j+1+\mu _j} Y_{j+1 / 2}^{\mu _j +1 / 2} \end{pmatrix}. \end{aligned}$$

As shown in [71, Section 4.6.5], one can decompose the space $L^2(\mathbb {R}^3)^4$—and analogously $L^2(B_R)^4$—as

$$\begin{aligned} L^2(\mathbb {R}^3)^4 = \bigoplus _{j=1/2}^{+\infty } \, \bigoplus _{\mu _j=-j}^j L_{j, \mu _j}^+ \oplus L_{j, \mu _j}^-, \end{aligned}$$

where

$$\begin{aligned} L_{j, \mu _j}^\pm := \left\{ \varphi \in L^2(\mathbb {R}^3)^4 : \, \varphi (r\theta ) = \begin{pmatrix} i {\tilde{f}}(r) \psi _{j \pm 1/2}^{\mu _j} (\theta ) \\ {\tilde{g}}(r) \psi _{j {\mp } 1 / 2}^{\mu _j} (\theta ) \end{pmatrix} \text { with } {\tilde{f}}, {\tilde{g}} \in L^2(\mathbb {R}_+, r^2 d r) \right\} . \end{aligned}$$

In each subspace define the map** $U_{j,\mu _j}^\pm : L^\pm _{j, \mu _j} \rightarrow L^2(\mathbb {R}_+)^2$ by

$$\begin{aligned} (U_{j,\mu _j}^\pm \varphi ) (r) = \begin{pmatrix} r {\tilde{f}}(r)\\ r {\tilde{g}}(r) \end{pmatrix} =: \begin{pmatrix} f(r)\\ g(r) \end{pmatrix}, \end{aligned}$$

and also define the differential operator (see [71, equation (4.129)])

$$\begin{aligned} {\widehat{H}}_{j,\pm } := \begin{pmatrix} m &{} - \partial _r + \kappa _{j,\pm }/r\\ \partial _r + \kappa _{j,\pm }/r &{} -m \end{pmatrix}, \quad \text {where } \kappa _{j,\pm } := \pm (j + 1/2). \end{aligned}$$

Then, the differential operator ${\mathcal {H}}= -i \alpha \cdot \nabla + m \beta$ decomposes into the orthogonal sum of the operators $(U_{j,\mu _j}^\pm )^{-1} {\widehat{H}}_{j,\pm } U_{j,\mu _j}^\pm$. In particular, if $\phi = (f,g)^\intercal$ satisfies ${\widehat{H}}_{j,\pm } \phi = \uplambda \phi$ in (0, R), then

$$\begin{aligned} \varphi = \begin{pmatrix} u\\ v \end{pmatrix} = \begin{pmatrix} \dfrac{ i f(r)}{r} \psi ^{\mu _j}_{j \pm 1/2}\\ \dfrac{g(r)}{r} \psi ^{\mu _j}_{j {\mp } 1/2} \end{pmatrix} \end{aligned}$$

(B.1)

satisfies ${\mathcal {H}}\varphi = \uplambda \varphi$ in $B_R\setminus \{0\}$. As we will see, by further imposing that f(0) is finite, we can guarantee that ${\mathcal {H}}\varphi = \uplambda \varphi$ holds across the origin.

1.2 B.2. Eigenvalue equations

Our first goal is to find solutions to ${\widehat{H}}_{j,\pm } (f, g)^\intercal = \uplambda (f,g)^\intercal$. This equation rewrites as the system of ODE

$$\begin{aligned} \left\{ \begin{array}{ll} - g' + \frac{\kappa }{r} g &{} = (\uplambda - m) f, \\ f' + \frac{\kappa }{r} f &{} = (\uplambda + m) g, \end{array} \right. \end{aligned}$$

where $\kappa := \kappa _{j,\pm } := \pm (j + 1/2)$. For simplicity, let us assume first that $\kappa = j + 1/2$. To solve the system, note that from the second ODE we get

$$\begin{aligned} g = \dfrac{1}{\uplambda + m} \Big (f' + \frac{\kappa }{r} f \Big ) \end{aligned}$$

(B.2)

and, thus, inserting this into the first one we get the Bessel-type ODE

$$\begin{aligned} f'' + \Big ( \uplambda ^2 - m^2 - \frac{\kappa ^2 + \kappa }{r^2} \Big ) f = 0. \end{aligned}$$

Therefore, f is of the form

$$\begin{aligned} f(r) = b_1 \sqrt{r} J_{\kappa + 1/2}(\sqrt{\uplambda ^2 - m^2} r) + b_2 \sqrt{r} Y_{\kappa + 1/2}( \sqrt{\uplambda ^2 - m^2} r), \end{aligned}$$

where $b_1,b_2\in \mathbb {C}$, and $J_{\kappa + 1/2}$ and $Y_{\kappa + 1/2}$ denote the Bessel functions of the first and second kind of order ${\kappa + 1/2}$; see [1, Chapters 9 and 10]. Since the eigenfunctions are not allowed to be singular at $r=0$ (as the corresponding $\varphi$ given by (B.1) must solve an elliptic equation across the origin), we deduce that $b_2=0$, and thus f is of the form

$$\begin{aligned} f(r) = b_1 \sqrt{r} J_{\kappa + 1/2}(\sqrt{\uplambda ^2 - m^2} r). \end{aligned}$$

Now, note that for every real index p, one has the relation

$$\begin{aligned} \partial _r [J_{p}(\sqrt{\uplambda ^2 - m^2} r) ]= \sqrt{\uplambda ^2 - m^2} J_{p - 1}(\sqrt{\uplambda ^2 - m^2} r) - \dfrac{p}{r} J_{p}(\sqrt{\uplambda ^2 - m^2} r); \end{aligned}$$

see [1, formula (9.1.27)]. Using this and (B.2), we see that

$$\begin{aligned} g(r) = b_1 \dfrac{\sqrt{\uplambda ^2 - m^2}}{\uplambda + m}\sqrt{r} J_{\kappa - 1/2}(\sqrt{\uplambda ^2 - m^2} r). \end{aligned}$$

The case $\kappa = - j - 1/2$ follows by similar arguments. One isolates f instead of g and uses that, for a positive integer p, $J_{-(p + 1/2)} = (-1)^{p+1} Y_{p + 1/2}$ and $Y_{-(p+ 1/2)} = (-1)^{p} J_{p + 1/2}$.

As a conclusion, we obtain that every eigenfunction of ${\widehat{H}}_{j,\pm }$ with eigenvalue $\uplambda$ is, up to a multiplicative constant, of the form

$$\begin{aligned} \begin{pmatrix} f(r)\\ g(r) \end{pmatrix} = \sqrt{r} \begin{pmatrix} J_{\ell + 1/2}(\sqrt{\uplambda ^2 - m^2} r) \\ \pm \dfrac{\sqrt{\uplambda ^2 - m^2}}{\uplambda + m} J_{\ell '+1/2}(\sqrt{\uplambda ^2 - m^2} r) \end{pmatrix}, \end{aligned}$$

where $\ell = j \pm 1/2$ and $\ell ' = j {\mp } 1/2$.

To obtain the equation (1.4) that relates $\uplambda$ and $\tau$ by means of Bessel functions, it only remains to impose the boundary condition $v = i e^\tau (\sigma \cdot \nu ) u$ on $\partial B_R$ for $\varphi =(u,v)^\intercal$ as in (B.1) and satisfying ${\mathcal {H}}\varphi =\uplambda \varphi$ in $B_R$. Since

$$\begin{aligned} (\sigma \cdot \nu ) \psi _{j \pm 1/2}^{\mu _j} = \psi _{j {\mp } 1/2}^{\mu _j}, \end{aligned}$$

by [71, equation (4.121)], it follows from (B.1) that the boundary condition relating f and g is

$$\begin{aligned} g(R) = - e^\tau f(R). \end{aligned}$$

Therefore, for each $j = 1/2, 3/2, \ldots$, each $\mu _j = -j, -j + 1,\ldots , j$, and each subspace $L^\pm _{j, \mu _j}$, we obtain the eigenvalue equation

$$\begin{aligned} e^\tau J_{\ell + 1/2}(\sqrt{\uplambda ^2 - m^2} R) \pm \dfrac{\sqrt{\uplambda ^2 - m^2}}{\uplambda + m} J_{\ell '+1/2}(\sqrt{\uplambda ^2 - m^2} R) = 0, \end{aligned}$$

(B.3)

where $\ell = j \pm 1/2$ and $\ell ' = j {\mp } 1/2$. This corresponds to (1.4). Note that the equation is independent of the indexes $\mu _j$, accounting for the multiplicity of the eigenvalues.

1.3 B.3. Parametrization of the eigenvalues

Our goal now is to exploit the eigenvalue equations given by (B.3) to prove that the eigenvalues of ${\mathcal {H}}_{\tau }$ can be parametrized in terms of $\tau \in \mathbb {R}$, obtaining a family of increasing curves whose limits as $\tau \rightarrow \pm \infty$ are related with the zeroes of the Bessel functions (and thus with the eigenvalues of the Dirichlet Laplacian). In the following lemma we collect the results on the Bessel functions that we will use.

Lemma B.1

Let $J_p$ be the Bessel function of the first kind of order $p >0$, and denote the k-th positive zero of this function by $z_{p, k}$.

Then,

• (i) the positive zeroes of $J_p$ are simple and form an infinite increasing sequence,

• (ii) the zeroes of two consecutive Bessel functions are interlaced, meaning that

  $$\begin{aligned} 0< z_{p, 1}< z_{p + 1, 1}< z_{p, 2}< z_{p + 1, 2} < \ldots , \end{aligned}$$

• (iii) the quotient of two consecutive Bessel functions can be expressed as

  $$\begin{aligned} \dfrac{J_{p + 1}(x)}{J_{p}(x)} = \sum _{k \ge 1} \dfrac{2x}{z_{p, k}^2 - x^2} \quad \text { for } x \in \mathbb {R}\setminus \{z_{p, k}\}_{k\in \mathbb {N}}. \end{aligned}$$

  (B.4)

  As a consequence, $J_{p + 1}/J_{p}$ is odd, strictly increasing in each interval contained in $\mathbb {R}\setminus \{z_{p, k}\}_{k\in \mathbb {N}}$, it is positive in the intervals $(0,z_{p, 1})$ and $(z_{p + 1, k},z_{p, k + 1})$ for $k\ge 1$, and negative in the intervals $(z_{p, k},z_{p + 1, k})$ for $k\ge 1$.

Proof

The first two statements are shown in [72, Chapter XV]. Note that for $p \ge -1$ the zeroes of $J_p$ are real, and thus we can order them; see also [1, p. 372]. Last, (B.4) follows from formula (1) in [72, p. 498]. Note that this yields that $J_{p + 1}/J_{p}$ is an infinite sum of functions with singularities at $\pm z_{p, k}$ which have strictly positive derivative in their domain of definition. As a consequence, in each interval of the form $(z_{p, k},z_{p, k + 1})$ for $k\in \mathbb {N}$, the function $J_{p + 1}/J_{p}$ is well defined, smooth, and strictly increasing, and therefore has a unique zero which necessarily is $z_{p + 1, k}$. $\square$

With the help of the previous lemma we can now establish the following result on the parametrization of the eigenvalues.

Proposition B.2

For each index $j = 1/2, 3/2, \ldots$ there exists an infinite number of smooth and strictly increasing functions $\{\tau \mapsto \uplambda _{j, k}^\pm (\tau ) \}_{k \in \mathbb {Z}}$ such that, for each $\tau \in \mathbb {R}$, the real number $\uplambda _{j, k}^\pm (\tau )$ is an eigenvalue of ${\mathcal {H}}_{\tau }$. The functions $\uplambda _{j, k}^\pm$ are surjectively defined by

$$\begin{aligned} \begin{matrix} \uplambda _{j, k}^\pm : &{}\mathbb {R}&{}\rightarrow &{}I_{j, k}^\pm \\ &{}\tau &{}\mapsto &{} \uplambda _{j, k}^\pm (\tau ), \end{matrix} \end{aligned}$$

with

• $I_{j,0}^-= \big (m, \sqrt{(z_{j,1}/R)^2 + m^2}\,\big )$,

• $I_{j,k}^- = \big (\sqrt{(z_{j + 1,k}/R)^2 + m^2}, \sqrt{(z_{j,k + 1}/R)^2 + m^2}\,\big )$ for $k = 1, 2, \ldots$,

• $I_{j,k}^- = \big (\!-\!\sqrt{(z_{j + 1,|k|}/R)^2 + m^2}, - \sqrt{(z_{j,|k|}/R)^2 + m^2}\,\big )$ for $k=-1, -2, \ldots$,

and

• $I_{j,0}^+= \big (\!-\!\sqrt{(z_{j,1}/R)^2 + m^2}, -m\big )$,

• $I_{j,k}^+ = \big (\!-\!\sqrt{(z_{j,k + 1}/R)^2 + m^2}, -\sqrt{(z_{j + 1,k}/R)^2 + m^2}\,\big )$ for $k = 1, 2, \ldots$,

• $I_{j,k}^+ = \big (\sqrt{(z_{j,|k|}/R)^2 + m^2}, \sqrt{(z_{j + 1,|k|}/R)^2 + m^2}\,\big )$ for $k=-1, -2, \ldots$,

where $z_{p, k}$ denotes the k-th positive zero of $J_p$, the Bessel function of order p. As a consequence, for every $\tau \in \mathbb {R}$, the function

$$\begin{aligned} \varphi _\tau := \begin{pmatrix} u_\tau \\ v_\tau \end{pmatrix} = \dfrac{1}{\sqrt{r}} \begin{pmatrix} i J_{\ell + 1/2}\left( \sqrt{\uplambda _{j, k}^\pm (\tau )^2 - m^2} r \right) \, \psi ^{\mu _j}_{j \pm 1/2}(\theta ) \\ \pm \dfrac{\sqrt{\uplambda _{j, k}^\pm (\tau )^2 - m^2}}{\uplambda _{j, k}^\pm (\tau ) + m} J_{\ell '+1/2} \left( \sqrt{\uplambda _{j, k}^\pm (\tau )^2 - m^2} r\right) \, \psi ^{\mu _j}_{j {\mp } 1/2} (\theta ) \end{pmatrix} \end{aligned}$$

with $j = 1/2, 3/2, \ldots$, $\ell = j \pm 1/2$, $\ell ' = j {\mp } 1/2$, $\mu _j = -j, -j + 1,\ldots , j$, and $k\in \mathbb {Z}$, belongs to $L^\pm _{j, \mu _j}$ and is an eigenfunction of ${\mathcal {H}}_{\tau }$ with eigenvalue $\uplambda _{j, k}^\pm (\tau )$.

Note that the superindex in $\uplambda _{j, k}^\pm$ indicates to which invariant subspace belongs the associated eigenfunction. It should not be confused with the sign of the eigenvalue (as the superindex in $\uplambda ^\pm _1$ denotes).

Proof of Proposition B.2

From the arguments already presented in Appendix B.1 it only remains to show that from (B.3) we can obtain the aforementioned infinite number of parametrizations of $\uplambda$ in terms of $\tau \in \mathbb {R}$. To do it, the idea is to rewrite the eigenvalue equation (B.3) as

$$\begin{aligned} e^\tau = {\mp } \dfrac{\sqrt{\uplambda ^2-m^2}}{\uplambda +m} \dfrac{ J_{\ell ' + 1/2} ( \sqrt{\uplambda ^2-m^2} R ) }{J_{\ell + 1/2} ( \sqrt{\uplambda ^2-m^2} R ) } =: h(\uplambda ). \end{aligned}$$

Then, our goal will be to invert h in suitable intervals to get $\uplambda =\uplambda (\tau ):=h^{-1}(e^\tau )$.

First, note that we can restrict ourselves to the subspaces $L^-_{j, \mu _j}$, thanks to the odd symmetry mentioned in Remark 1.1. In this case the eigenvalue equation is written as

$$\begin{aligned} e^\tau = {{\,\textrm{sign}\,}}(\uplambda +m ) \sqrt{\dfrac{\uplambda -m}{\uplambda +m}} \dfrac{ J_{j + 1} ( \sqrt{\uplambda ^2-m^2} R ) }{J_{j} ( \sqrt{\uplambda ^2-m^2} R ) } = h(\uplambda ). \end{aligned}$$

Due to the fact that $e^\tau >0$ for all $\tau \in \mathbb {R}$, we are forced to work with h only on intervals $I\subset \mathbb {R}$ such that $h(I)\subset (0,+\infty )$ and where h is invertible. Since $\frac{\uplambda -m}{\uplambda +m}$ is positive and strictly increasing for $\uplambda \in (m, +\infty )$ and for $\uplambda \in (- \infty , -m)$, I must be such that

$$\begin{aligned} {{\,\textrm{sign}\,}}(\uplambda +m ) \dfrac{ J_{j + 1} ( \sqrt{\uplambda ^2-m^2} R ) }{J_{j} ( \sqrt{\uplambda ^2-m^2} R ) } \text { is positive and strictly increasing for } \uplambda \in I. \end{aligned}$$

Then, Lemma B.1 yields that all the possible intervals I are:

• $I= (m, \sqrt{(z_{j,1}/R)^2 + m^2})$,

• $I = (\sqrt{(z_{j + 1,k}/R)^2 + m^2}, \sqrt{(z_{j,k + 1}/R)^2 + m^2})$ for $k\ge 1$,

• $I = (-\sqrt{(z_{j + 1,k}/R)^2 + m^2}, - \sqrt{(z_{j,k}/R)^2 + m^2})$ for $k\ge 1$.

In each of these intervals I the function $h: I \rightarrow (0,+\infty )$ is of class $C^\infty$, strictly increasing, and surjective. Therefore, it can be inverted, obtaining a $C^\infty$ function $\tau \mapsto \uplambda (\tau ) := h^{-1}(e^\tau )$ which maps $\mathbb {R}$ into I surjectively and corresponds, for each $\tau \in \mathbb {R}$, to an eigenvalue of ${\mathcal {H}}_\tau$. In addition, the monotonicity of $\uplambda \mapsto \tau =\tau (\uplambda ):=\log (h(\uplambda ))$ yields that $\tau \mapsto \uplambda (\tau )$ is also strictly increasing. $\square$

The previous result yields that, for any given eigenvalue curve $\tau \mapsto \uplambda (\tau )$, it holds that $\lim _{\tau \rightarrow \pm \infty }|\uplambda (\tau )|$ is either m or a positive zero of the function $J_{k + 1/2} ( \sqrt{(\cdot )^2-m^2} R )$ for some $k=0,1,2,\ldots$; note that each of these zeroes corresponds to the square root of a Dirichlet eigenvalue of $-\Delta + m^2$ in $B_R$. The monotonicity and limiting values of $\tau \mapsto \uplambda (\tau )$ was observed in the curves plotted in Fig. 1. Furthermore, the alternation (with respect to the zeroes of the Bessel function) between positive and negative eigenvalue curves, which is given by the intervals $I_{j,k}^\pm$ defined in Proposition B.2, was already shown numerically in Figs. 2 and 3.

1.4 B.4. The first positive eigenvalue

In this section we focus on the first positive eigenvalue of ${\mathcal {H}}_{\tau }$ when $\Omega = B_R$. We provide a fine description of the associated eigenvalue curve, whose main properties are summarized in the following proposition. The reader may compare it with Theorem 1.5, which is the analogous result for general domains; see also Theorem 1.7 regarding (B.5).

Proposition B.3

The function $\tau \mapsto \uplambda _1^+(\tau )=\min (\sigma ({\mathcal {H}}_\tau )\cap (m,+\infty ))$ is of class $C^\infty$ in $\mathbb {R}$, and satisfies

$$\begin{aligned} \lim _{\tau \downarrow - \infty } \uplambda _1^+ (\tau ) = m \quad \text {and} \quad \lim _{\tau \uparrow + \infty } \uplambda _1^+ (\tau ) = \sqrt{\pi ^2/R^2 + m^2} = \sqrt{ \min \sigma (-\Delta _D) + m^2}, \end{aligned}$$

where $\min \sigma (-\Delta _D)$ denotes the first Dirichlet eigenvalue of $-\Delta$ in $B_R$. In addition, the corresponding eigenspace associated to $\uplambda _1^+(\tau )$ has always dimension 2.

Furthermore,

$$\begin{aligned} L^\star _{B_R} := \lim _{\tau \downarrow -\infty } (\uplambda _1^+(\tau )-m)e^{-\tau } = \frac{3}{R}=\frac{1}{{\mathscr {R}}_{B_R}}, \end{aligned}$$

(B.5)

where ${\mathscr {R}}_{B_R}$ is defined in (1.14).

Proof

Let $\uplambda : \mathbb {R}\rightarrow \big (m, \sqrt{\pi ^2/R^2 + m^2}\big )$ be the eigenvalue curve corresponding to $\uplambda _{1/2, 0}^-$ in the notation of Proposition B.2. This eigenfunction is associated to the two subspaces $L^-_{1/2, \mu _{1/2}}$, with either $\mu _{1/2} = 1/2$ or $\mu _{1/2} = -1/2$, and solves the implicit equation

$$\begin{aligned} e^\tau = \dfrac{\sqrt{\uplambda ^2-m^2}}{\uplambda +m} \dfrac{ J_{3/2} ( \sqrt{\uplambda ^2-m^2} R ) }{J_{1/2} ( \sqrt{\uplambda ^2-m^2} R ) }. \end{aligned}$$

(B.6)

We will show that $\uplambda = \uplambda _1^+$. The upper bound for $\uplambda (\tau )$ (and limit as $\tau \rightarrow +\infty$) is given by the fact that $z_{1/2, 1} = \pi$, which follows from the explicit expression of the Bessel functions involved in the above equation:

$$\begin{aligned} J_{1/2} (x) = \sqrt{\dfrac{2}{\pi x}} \sin (x) \quad \text { and } \quad J_{3/2} (x) = \sqrt{\dfrac{2}{\pi x}} \left( \dfrac{\sin (x)}{x} - \cos (x) \right) . \end{aligned}$$

As a matter of fact, these expressions can be used to show—after a tedious computation and using that $x>\sin (x)$ for all $x\in (0,\pi )$—that the right-hand side of (B.6) is a strictly increasing function of $\uplambda$ for $\uplambda \in \big (m, \sqrt{\pi ^2/R^2 + m^2}\big )$ without making use of Lemma B.1 (as done in the proof of Proposition B.2).

Since $\pi = z_{1/2, 1}$ is the smallest positive zero among all the positive zeroes of the Bessel functions of half-integer index—as shown by Lemma B.1 (ii)—, it follows that $\uplambda (\tau )$ coincides with $\uplambda ^+_1(\tau )$ at least for big enough values of $\tau$. To show that indeed $\uplambda (\tau )$ is the first positive eigenvalue $\uplambda _1^+ (\tau )$ for all $\tau \in \mathbb {R}$, it suffices to show that $\uplambda$ cannot cross any other eigenvalue curve. On the one hand, taking into account the possible intervals $I_{j,k}^+$ given in Proposition B.2, it follows that any positive eigenvalue curve associated to the spaces $L^+_{j, \mu _{j}}$ must lie above $\sqrt{\pi ^2/R^2 + m^2}$, and thus it cannot cross $\uplambda (\tau )$. On the other hand, if there was a crossing between $\uplambda (\tau )$ and another positive eigenvalue curve associated to $L^-_{j_\circ , \mu _{j_\circ }}$ for some half-integer $j_\circ > 1/2$, then by (B.6) there would exist a point $x_\circ \in (0, z_{1/2, 1})$ such that

$$\begin{aligned} \dfrac{ J_{3/2} (x_\circ ) }{J_{1/2} (x_\circ ) } = \dfrac{ J_{j_\circ + 1} (x_\circ ) }{J_{j_\circ } (x_\circ ) }. \end{aligned}$$

(B.7)

However, since the zeroes of the Bessel functions are ordered (see Lemma B.1), for every half-integer $j\ge 1/2$ it follows that $z_{j, k} < z_{j + 1,k}$ for all $k \ge 1$, and therefore

$$\begin{aligned} \dfrac{2x}{z_{j, k}^2 - x^2} > \dfrac{2x}{z_{j + 1, k}^2 - x^2} \quad \text { for every } x \in (0, z_{1/2, 1}). \end{aligned}$$

Hence, by (B.4) it follows that

$$\begin{aligned} \dfrac{ J_{j + 1} (x) }{J_{j} (x ) } > \dfrac{ J_{j+2} (x) }{J_{j+1} (x ) } \quad \text { for every } x \in (0, z_{1/2, 1}) \text { and for every half-integer } j \ge 1/2, \end{aligned}$$

thus there cannot exist such a $x_\circ \in (0, z_{1/2, 1})$ satisfying (B.7). In conclusion, $\uplambda$ does not cross any other eigenvalue curve. Therefore, $\uplambda (\tau )$ is the first positive eigenvalue of ${\mathcal {H}}_\tau$ for all $\tau \in \mathbb {R}$. As a byproduct, since $\uplambda (\tau )$ is associated to $L^-_{1/2, \mu _{1/2}}$ with either $\mu _{1/2} = 1/2$ or $\mu _{1/2} = -1/2$, it follows that the first positive eigenvalue of ${\mathcal {H}}_\tau$ has multiplicity 2 for all $\tau \in \mathbb {R}$.

To conclude the proof, we are only left to show that $L_{B_R}^* = 3/R=1/{\mathscr {R}}_{B_R}$. First, note that from the rescaling properties of the operators $K_m$ and $W_m$ defined in (2.12), it follows readily that ${\mathscr {R}}_{B_R} = R{\mathscr {R}}_{B_1}$. Moreover, $L_{B_R}^\star = L_{B_1}^\star /R$. To show this second equality, one notices that if $u_1$ and $u_R$ denote the boundary values of the upper component of the first eigenfunction in $B_1$ and $B_R$ respectively, both associated to the same subspace $L^-_{1/2, \mu _{1/2}}$, then after a normalization one can choose them in such a way that $u_R(\cdot ) = u_1(\cdot /R)$. Hence, from (2.19) and taking the limit $\tau \downarrow -\infty$, using again the scaling of $K_m$, and that $\{W_m, \sigma \cdot \nu \} = 0$ by Lemma 4.7, it follows that $L_{B_R}^\star = L_{B_1}^\star /R$; see the argument to get to (B.9) below for more details. As a consequence of all this, it is enough to prove the result for $R=1$.

Recall that $\uplambda (\tau )$ is associated, for all $\tau \in \mathbb {R}$, to the two subspaces $L^-_{1/2, \mu _{1/2}}$ with either $\mu _{1/2}= 1/2$ or $\mu _{1/2}= -1/2$. We will work in one of these two subspaces (the precise choice will be completely irrelevant for the rest of the argument), and therefore, after a normalization, we can take an eigenfunction $\varphi _\tau = (u_\tau , v_\tau )^\intercal$ associated to $\uplambda (\tau )$ such that $u_\tau$ at the boundary of $B_1$ is given by $\psi ^{\mu _{1/2}}_0$ for all $\tau \in \mathbb {R}$. By Lemma 2.9, it holds

$$\begin{aligned} \begin{aligned} \Big ( \frac{1}{2} -i W_\uplambda ({\sigma }\cdot \nu ) \Big )\psi ^{\mu _{1/2}}_0&= - (\uplambda +m) e^\tau K_\uplambda \psi ^{\mu _{1/2}}_0\quad \text {and}\\ \Big ( \frac{1}{2} -i ({\sigma }\cdot \nu ) W_\uplambda \Big )\psi ^{\mu _{1/2}}_0&= (\uplambda -m)e^{-\tau } ({\sigma }\cdot \nu ) \, K_\uplambda ({\sigma }\cdot \nu ) \psi ^{\mu _{1/2}}_0 \end{aligned} \end{aligned}$$

(B.8)

in $L^2(\mathbb {S}^2)^2$. We will take the limit $\tau \downarrow -\infty$ in (B.8), taking into account that $\uplambda \downarrow m$ as $\tau \downarrow -\infty$. Letting $\tau \downarrow -\infty$ in (B.8), and using that then $K_\uplambda \rightarrow K_m$ and $W_\uplambda \rightarrow W_m$ as bounded operators in $L^2(\mathbb {S}^2)^2$ (which follows from the explicit expressions of the operators), we obtain

$$\begin{aligned} \begin{aligned} P_- \, \psi ^{\mu _{1/2}}_0 = 0 \quad \text {and} \quad (P_+)^*\, \psi ^{\mu _{1/2}}_0 = L_{B_1}^\star ({\sigma }\cdot \nu ) \, K_m ({\sigma }\cdot \nu ) \psi ^{\mu _{1/2}}_0 \quad \text { in } L^2(\mathbb {S}^2)^2, \end{aligned}\nonumber \\ \end{aligned}$$

(B.9)

where $L^\star _{B_1} =\lim _{\tau \downarrow -\infty } (\uplambda (\tau )-m)e^{-\tau }$, and $P_- = \frac{1}{2} -i W_m ({\sigma }\cdot \nu )$ and $(P_+)^*= \frac{1}{2} -i ({\sigma }\cdot \nu ) W_m$, as defined in Sect. 4.2. In particular, (B.9) shows that $L^\star _{B_1}$ is finite, since all the involved operators are bounded and $K_m$ is injective. It is worth pointing out that this last argument leading to $L_{B_1}^\star < +\infty$ works thanks to the fact that, on $\partial \Omega$, the eigenfunction $u_\tau = \psi ^{\mu _{1/2}}_0$ is indeed independent of $\tau$, something that may not be guaranteed on a general domain $\Omega$.

We will use (B.9) to establish (B.5) for $R=1$. First, let $L\in {\mathcal {L}}_{B_1}$—recall that ${\mathcal {L}}_{B_1}$ is defined in (1.11), see also (4.20). We claim that 1/L is an eigenvalue of $K_m$. This claim can be proven by adding the two equations from which ${\mathcal {L}}_{B_1}$ is defined and using that $\{W_m, \sigma \cdot \nu \} = 0$ as operators in $L^2(\mathbb {S}^2)^2$; see Lemma 4.7. As a consequence, and since $1/{\mathscr {R}}_\Omega = \min {\mathcal {L}}_{\Omega }$ for every $\Omega \subset \mathbb {R}^3$ by Theorem 1.7, ${\mathscr {R}}_{B_1}$ is the maximum of the eigenvalues of $K_m$ among eigenfunctions of the form $(\sigma \cdot \nu )u$ with $u\in P_+(L^2(\mathbb {S}^2)^2)$, where $P_+=\frac{1}{2}+ iW_m(\sigma \cdot \nu )$.

Let us now compute explicitly the eigenvalues of $K_m$ as an operator in $L^2(\mathbb {S}^2)^2$. Using [10, Lemma 4.3] and [58, Theorem 3.6] on $K_a$ for $a>0$ and letting²$a \uparrow m$, it follows that the spectrum of $K_m$ is given by a sequence $\{ d_{j\pm 1/2} \}_{j = 1/2, \ 3/2, \ldots }$ whose associated eigenfunctions are $\psi _{j \pm 1 / 2}^{\mu _j}$, i.e.,

$$\begin{aligned} K_m \psi _{j \pm 1 / 2}^{\mu _j} = d_{j\pm 1/2}\, \psi _{j \pm 1 / 2}^{\mu _j}, \end{aligned}$$

(B.10)

and the eigenvalues $d_{j\pm 1/2}$ are given by the expression

$$\begin{aligned} d_{j\pm 1/2} = \lim _{t\downarrow 0} {\mathcal {I}}_{j \pm 1/2 +1/2}(t) {\mathcal {K}}_{j \pm 1/2 +1/2}(t). \end{aligned}$$

Here ${\mathcal {I}}_\kappa$ and ${\mathcal {K}}_\kappa$ are the modified Bessel functions of the first and second kind of order $\kappa$, respectively. By [1, formulas 9.6.7 and 9.6.9] we have

$$\begin{aligned} \lim _{t\downarrow 0} {\mathcal {I}}_{\kappa }(t) {\mathcal {K}}_{\kappa }(t) = \dfrac{\Gamma (\kappa )}{2 \Gamma (\kappa + 1)} = \dfrac{1}{2 \kappa } \quad \text {for } \kappa > 0. \end{aligned}$$

Hence, for $k= 0, 1, \ldots ,$ we have

$$\begin{aligned} d_{k} = \dfrac{1}{2k +1}. \end{aligned}$$

(B.11)

We claim that $(\sigma \cdot \nu ) \psi _0^{\mu _j} = \psi _1^{\mu _j}$ is orthogonal to $P_+(L^2(\mathbb {S}^2)^2)$. Once this is proved, it follows that $1/d_0 \notin {\mathcal {L}}_{B_1}$ and, as a consequence, ${\mathscr {R}}_{B_1} \le d_1 = 1/3$. To prove the claim, notice that for all $\tau \in \mathbb {R}$, after a normalization, $\psi _1^{\mu _j}$ is the upper component of the eigenfunction of ${\mathcal {H}}_{\tau }$ associated to $-\uplambda (-\tau )$, i.e., the first (larger) negative eigenvalue of ${\mathcal {H}}_{\tau }$. Indeed, this can be shown with exactly the same arguments as we did at the beginning of the proof of the proposition for the first positive eigenvalue, considering in this case the subspace $L^+_{1/2, \mu _{1/2}}$; see also Remark 1.1. Therefore, since such eigenvalue converges to $-m$ as $\tau \uparrow +\infty$, using the second equation in (2.18) with $u= \psi _1^{\mu _j}$ and taking the limit $\tau \uparrow +\infty$—analogously as how we proceeded before to show (B.9)—, it follows that $(P_+)^* \psi _1^{\mu _j} = 0$, establishing our claim and, as a byproduct, the inequality ${\mathscr {R}}_{B_1} \le d_1 = 1/3$.

We shall now prove that $L_{B_1}^\star = 1/ d_1 = 3$, which will finish the proof of (B.5) since $1/{\mathscr {R}}_{B_1} \le L_{B_1}^*$ by Lemmas 4.10 and 4.11. By adding the two equations in (B.9) and using again that $\{W_m,\sigma \cdot \nu \} = 0$ (or equivalently, that $(P_+)^* = P_+$ by Lemma 4.7 since the underlying domain is a ball), we obtain

$$\begin{aligned} \psi ^{\mu _{1/2}}_0 = L_{B_1}^\star ({\sigma }\cdot \nu ) \, K_m ({\sigma }\cdot \nu ) \psi ^{\mu _{1/2}}_0 \quad \text {in } L^2(\mathbb {S}^2)^2. \end{aligned}$$

Finally, using that $({\sigma }\cdot \nu ) \psi ^{\mu _{1/2}}_{j\pm 1/2} = \psi ^{\mu _{1/2}}_{j {\mp } 1/2}$ for all half-integers j (see [71, equation (4.121)]), and that $\psi ^{\mu _{1/2}}_1$ is an eigenfunction of $K_m$ with eigenvalue $d_1$ by (B.10), we get

$$\begin{aligned} \psi ^{\mu _{1/2}}_0 = L_{B_1}^\star d_1 \psi ^{\mu _{1/2}}_0 \quad \text { in } L^2(\mathbb {S}^2)^2. \end{aligned}$$

Therefore, we have $L_{B_1}^\star = 1/d_1 = 3$ by (B.11). This concludes the proof. $\square$