# Spectrum of a family of spin-orbit coupled Hamiltonians with singular
perturbation

R. Juršėnas  Vilnius University, Institute of Theoretical Physics and
Astronomy, A. Goštauto 12, Vilnius 01108, Lithuania<EMAIL_ADDRESS>
###### Abstract.

The present study is the first such attempt to examine rigorously and
comprehensively the spectral properties of a three-dimensional ultracold atom
when both the spin-orbit interaction and the Zeeman field are taken into
account. The model operator is the Rashba spin-orbit coupled operator in
dimension three. The self-adjoint extensions are constructed using the theory
of singular perturbations, where regularized rank two perturbations describe
spin-dependent contact interactions. The spectrum of self-adjoint extensions
is investigated in detail laying emphasis on the effects due to spin-orbit
coupling. When the spin-orbit-coupling strength is small enough, the
asymptotics of eigenvalues is obtained. The conditions for the existence of
eigenvalues above the threshold are discussed in particular.

###### Key words and phrases:

Rashba–Dresselhaus spin-orbit coupling, three-dimensional ultracold atom,
singular finite rank perturbation, self-adjoint extension, Krein Q-function

###### 2010 Mathematics Subject Classification:

47A55, 81Q10, 81Q15

##  1\. Introduction

The spin-orbit coupled quantum systems are described by the Hamiltonian which
is realized as the differential operator in the tensor product space
$\mathbb{R}^{3}\operatorname{\otimes}\mathbb{C}^{2}$:

(1.1)
$S:=-\Delta\operatorname{\otimes}I+i\alpha(\nabla_{1}\operatorname{\otimes}\sigma_{2}-\nabla_{2}\operatorname{\otimes}\sigma_{1})+\beta
I\operatorname{\otimes}\sigma_{3}\quad(\alpha,\beta\geq 0).$

In (1.1), $\alpha$ stands for the spin-orbit-coupling strength, $\beta$ is the
strength of the magnetic Zeeman field [23]. The symbol $\Delta$ denotes the
three-dimensional Laplace operator, $\nabla_{j}$ ($j=1,2$) is the gradient in
the $j$th component of a three-dimensional position-vector. The standard Pauli
matrices are represented by $(\sigma_{j})_{j=1,2,3}$. The imaginary unit
$i\equiv\sqrt{-1}$. Here and elsewhere we write $I$ for an identity operator.
The space in which $I$ acts is expected to be understood from the context.

###  1.1. Main goal

The main purpose of the current exposition is to study the spectrum of self-
adjoint extensions of the symmetric operator constructed from (1.1). Let $S$
be defined on a set of compactly supported smooth functions, with the support
outside the origin $0\in\mathbb{R}^{3}$:

(1.2)
$\mathop{\mathrm{dom}}S=C_{0}^{\infty}(\mathbb{R}^{3}\backslash\\{0\\})\operatorname{\otimes}\mathbb{C}^{2}.$

Then $S$ is symmetric with respect to the scalar product in the Hilbert tensor
product $L^{2}(\mathbb{R}^{3})\operatorname{\otimes}\mathbb{C}^{2}$. In
applications, the self-adjoint extensions of so defined $S$ are referred to as
the Rashba spin-orbit coupled Hamiltonians considered in the presence of the
out-of-plane magnetic field, with the impurity scattering treated via the
spin-dependent contact interaction. A good review on various types of disorder
in condensed-matter systems is given in [38]. For the analysis of long-range
interactions in electronic systems, as opposed to the present discussion where
the zero-range interactions are studied, the reader may refer to [21, 11].

###  1.2. Special cases

The special case $\alpha=0$ was discussed recently in [17, 16], where, among
other things, the authors examined the effects due to the in-plane magnetic
field. The corresponding term in (1.1) is then replaced by a unitarily
equivalent one $\beta I\operatorname{\otimes}\sigma_{1}$. In their Theorem 2
in [17] the authors showed under what conditions the point spectrum is empty.
Here we prove that under exactly the same conditions there is a subfamily of
self-adjoint extensions whose point spectrum is $\\{\pm\beta\\}$.

The case $\alpha=\beta=0$ is a classic one. The book [3] is much used as a
reference book for a standard (or von Neumann) operator extension theory; see
also an extensive list of references therein. The extensions constructed using
spaces of boundary values are discussed, for example, in [9, 36, 37, 24]. A
modern and comprehensive review of various operator techniques is given in
[15].

In one and two spatial dimensions, the spectral properties of self-adjoint
extensions of $S$, for $\alpha,\beta\geq 0$, were studied in [31, 20, 17, 16,
14, 27]. A mean field interpretation, which is commonly accepted in physics
literature, can be found in [32, 29, 34, 35].

###  1.3. Motivation and main tools

A recently proposed technique [12] (see also [46, 22, 19]) for producing the
Rashba-type spin-orbit coupling for a three-dimensional ultracold atom serves
as our main motive to examine a general case $\alpha,\beta\geq 0$. Many more
motivating aspects concerning the dynamics of quantum particles can be found
in [17, 16, 18], and we do not repeat them here. Contrary to [17], where the
modified Krein resolvent formula [10] has been used as a starting point for
calculating the resolvent of $S$ (with $\alpha=0$), here, we exploit the
theory of singular finite rank perturbations [42, 25, 33, 4, 7, 5]. On the one
hand, the two approaches lead to an identical description of the boundary
conditions that ensure self-adjointness [8]. On the other hand, the
singularities of Green function (see e.g. [1] and references therein) are
successfully eliminated using the so-called renormalization procedure [33].
Such a procedure is implemented in the theory of singular perturbations in its
most natural way [7].

As an illustration, let us consider a formal ”operator” $S+C\delta$, where $S$
in (1.1) is written in matrix representation, and where $\delta$ is the Dirac
distribution concentrated at the origin; for $\alpha=\beta=0$, this type of
operator, when restricted to one spatial dimension but extended to the many-
body case, is studied in [2]. The matrix $C$, which is Hermitian and
invertible, plays the role of the coupling parameter of contact interaction.
The Lipmann–Schwinger equation for the perturbed operator shows that the
solution of the integral equation is inconsistent, in that the Green function
for $S$ is singular at $x=0$ (see (3.5) for the details), meaning that the
equation does not have a unique solution. The phenomenon is usually called the
ultraviolet divergence. For, the Fourier integral of the Green function in the
momentum representation diverges when $x=0$. To avoid divergences, typically
one introduces an auxiliary UV momentum cut-off and regularizes the
interaction parameter. The details can be found in [13, Sec. 3.7], see also
[43, 45, 44, 39]. In the theory of singular perturbations, however, one deals
with a regularization and a renormalization procedure applied to $\delta$. In
this case the regularization is meant in the sense that the Dirac distribution
is extended to some subspace of a Hilbert space. Roughly speaking the action
of $\delta$ on the function which is undefined at the origin is replaced by
the action of some regularized $\delta$. The present discussion is justified
rigorously in Sec. 2. Next, the regularized and renormalized $\delta$ is
determined by the choice of an Hermitian matrix $R$, which is usually called
admissible; the matrix realization of $R$ is due to an extra $\mathbb{C}^{2}$
in (1.1). It turns out that the theory also deals with an additional parameter
(i.e. the matrix $R$). However, the precise operator realization of
$S+C\delta$ by extending $\delta$ gives one an advantage over the cut-off
practice in an obvious way. One shows that the self-adjoint realizations of
$S+C\delta$ are parametrized in terms of $\Gamma:=-C^{-1}-R$, with the
exception of the trivial extension corresponding to $C=0$ and the Friedrichs
extension corresponding to $C^{-1}=0$. For a general $\alpha,\beta\geq 0$, the
functional $\delta$ is of the class for which $R$ is not unique, but in some
cases $R$ can be found exactly; e.g. $R=-(4\sqrt{2}\pi)^{-1}\mathbb{I}$ for
$\alpha=\beta=0$. For some special classes of singular finite rank
perturbations, the uniqueness of $R$ is discussed in great detail in [28].

Another reason for choosing the approach of singular perturbations is that the
scattering theory for singular finite rank perturbations is fully established
[7, Chap. 4]: Using our results, the scattering matrix can be found directly
from [7, Eq. (4.34)]. For this particular reason we do not examine scattering
states but rather concentrate on analytic properties of the spectrum.

###  1.4. Solvability of the model

To the best of our knowledge, the model studied in the current paper is
investigated by means of the theory of singular perturbations for the first
time. The lack of rigorous results concerning the spectral analysis of the
perturbed $S$ may be explained by the complexity of the Green function for the
free Hamiltonian (or else the trivial extension of $S$). A typical strategy is
to write the Green function in the momentum representation. Then the
dispersion relation and the essential spectrum are easy to deduce. However,
the computation of the inverse Fourier transform of Green function is a rather
troublesome task. Recently [30], a hypergeometric series representation for
the Green function in the coordinate representation has been derived. In
contrast, the analogous result in two spatial dimensions possesses a closed
form [14, 20, 27].

(a) $\alpha=0$, $\Sigma=\beta\geq 0$

(b) $0<\alpha<\sqrt{2\beta}$, $\Sigma=\beta>0$

(c) $0<\sqrt{2\beta}\leq\alpha$, $\beta\leq\Sigma\leq 1$

Figure 1. Schematic of projected dispersion relations. Permitted eigenvalues
are situated along the vertical axis within the gray area. For example, (b)
shows that the point spectrum is empty above $-\beta$. The details are listed
in Theorems 6.1, 6.7, 6.8.

The absolute convergence of hypergeometric series causes the limitation on the
parameters $\alpha,\beta\geq 0$. The result is that the infimum of the
continuous spectrum $[-\Sigma,\infty)$ must satisfy the condition
$-\Sigma\geq-1$. Here $\Sigma$ equals $\beta$ if $\beta>\alpha^{2}/2$ and
$(\beta/\alpha)^{2}+(\alpha/2)^{2}$ otherwise.

For $\alpha$ arbitrarily small or zero, the restriction can be removed by
analytic continuation. Using the resolvent formula for a perturbed Hamiltonian
(or else nontrivial self-adjoint extension of $S$), we find that the
eigenvalues solve the transcendental equation. We then derive the solutions
analytically with the accuracy up to $O(\alpha^{4})$. In this particular case
we observe a tendency that the spin-orbit term moves the eigenvalues down from
the threshold $-\beta$. In general, the main conclusions following from our
results that concern eigenvalues can be schematically described as shown in
Fig. 1: For $\alpha>0$ small, there are no eigenvalues above the threshold
$-\beta$ no matter the form of a nonzero coupling parameter of contact
interaction ($C$). For $\alpha$ large, there exists a subfamily of
Hamiltonians having eigenvalues in $[\beta,E_{\nu,1}]$, for some lucid
$E_{\nu,1}\geq\beta$, but none of Hamiltonians has eigenvalues in
$[-\Sigma,\beta)$.

##  2\. Self-adjoint extensions

Let $H$ be a lower semi-bounded self-adjoint operator in a separable Hilbert
space $\mathcal{H}_{0}$. Let $(\mathcal{H}_{n})_{n\in\mathbb{Z}}$ be the scale
of Hilbert spaces associated to $H$. The scalar product and the induced norm
in $\mathcal{H}_{n}$ are denoted by $\braket{\cdot,\cdot}_{n}$ and
$\lVert\cdot\rVert_{n}$, respectively. Consider the orthonormal system
$(\psi_{s})_{s=\pm}\in\mathcal{H}_{-2}\backslash\mathcal{H}_{-1}$ and define
the duality pairing
$\braket{\cdot,\cdot}\colon\mathcal{H}_{-n}\times\mathcal{H}_{n}\to\mathbb{C}$,
for $n\in\mathbb{N}_{0}$, in a usual way [7, Sec. 1.2.2]. The orthogonality,
which is assumed with respect to the scalar product
$\braket{\cdot,\cdot}_{-2}^{*}:=\braket{(H-i)^{-1}\cdot,(H-i)^{-1}\cdot}_{0}$
in $\mathcal{H}_{-2}$, implies the one with respect to
$\braket{\cdot,\cdot}_{-2}$. We write $\psi$ for the column matrix
$(\psi_{s})$; $\braket{\psi,\cdot}$ is the column matrix
$(\braket{\psi_{s},\cdot})$.

Let $H^{0}$ be a symmetric and densely defined restriction of $H$ to the
domain

(2.1)
$\mathop{\mathrm{dom}}H^{0}=\\{f\in\mathcal{H}_{2}\colon\braket{\psi,f}=0\\}.$

For $z\in\mathop{\mathrm{res}}H$ (the resolvent set of $H$), we define the
functions

(2.2) $\Phi_{s}(z):=(H-z)^{-1}\psi_{s}$

and the column matrices $\Phi\equiv(\Phi_{s})$. The functions
$(\Phi_{s}(i))_{s=\pm}$ are orthonormal with respect to the scalar product
$\braket{\cdot,\cdot}_{0}$ in $\mathcal{H}_{0}$, for it holds
$\braket{\Phi_{s}(i),\Phi_{s^{\prime}}(i)}_{0}=\braket{\psi_{s},\psi_{s^{\prime}}}_{-2}^{*}$
($s,s^{\prime}=\pm$). Let $H^{0\,*}$ be the adjoint of $H^{0}$. Then
$\Phi_{s}(z)\in\mathop{\mathrm{ker}}(H^{0\,*}-z)$ and $H^{0\,*}$ has
deficiency indices (2,2). Considering $\mathop{\mathrm{dom}}H^{0\,*}$ as a
Banach space $\mathcal{H}_{2}\dotplus\mathbb{C}^{2}$ with the norm

$\lVert
f\rVert_{\mathop{\mathrm{dom}}H^{0\,*}}^{2}=\lVert\tilde{f}\rVert_{2}^{2}+\sum_{s}\lvert
b_{s}(f)\rvert^{2}$

($f\in\mathop{\mathrm{dom}}H^{0\,*}$; $\tilde{f}\in\mathcal{H}_{2}$;
$b_{s}(f)\in\mathbb{C}$), we write

(2.3) $f=\tilde{f}+\tfrac{1}{2}\sum_{s}b_{s}(f)(\Phi_{s}(i)+\Phi_{s}(-i)).$

The column matrix $b(f)\equiv(b_{s}(f))$, and we regard $b_{s}(\cdot)$ as a
functional $\mathop{\mathrm{dom}}H^{0\,*}\to\mathbb{C}$.

Let $\psi_{s}^{R}$ be an extension of $\psi_{s}$ to
$\mathop{\mathrm{dom}}H^{0\,*}$ [28, 5]:

(2.4)
$\braket{\psi^{R},f}=\braket{\psi,\tilde{f}}+Rb(f),\quad\braket{\psi^{R},\tilde{f}}=\braket{\psi,\tilde{f}}$

where $f$ and $\tilde{f}$ are as above, and where
$\psi^{R}\equiv(\psi_{s}^{R})$. The matrix
$R\equiv(R_{ss^{\prime}})\in\mathbb{C}^{2\times 2}$ is the admissible matrix
for the functionals of class $\mathcal{H}_{-2}\backslash\mathcal{H}_{-1}$.
Then $\psi_{s}^{R}$ is an element of the topological dual
$(\mathop{\mathrm{dom}}H^{0\,*})^{\prime}=\mathcal{H}_{-2}\dotplus\mathbb{C}^{2}$
equipped with the supremum norm $\sup\lvert\braket{\psi_{s}^{R},f}\rvert$; the
supremum is taken over $s=\pm$ and $f\in\mathop{\mathrm{dom}}H^{0\,*}$ such
that $\lVert f\rVert_{\mathop{\mathrm{dom}}H^{0\,*}}\leq 1$.

Using another decomposition
$\mathop{\mathrm{dom}}H^{0\,*}=\mathop{\mathrm{dom}}H^{0}\dotplus\mathop{\mathrm{span}}\\{\Phi_{s}(\pm
i)\\}$, we have that, for $f\in\mathop{\mathrm{dom}}H^{0\,*}$,

(2.5) $f=f^{0}+\sum_{s}(a_{s}^{+}(f)\Phi_{s}(i)+a_{s}^{-}(f)\Phi_{s}(-i))$

($f^{0}\in\mathop{\mathrm{dom}}H^{0}$; $a_{s}^{\pm}(f)\in\mathbb{C}$). But
then, using (2.3) and (2.5),

(2.6)
$\tilde{f}=f^{0}+\tfrac{1}{2}\sum_{s}ib_{s}^{\prime}(f)\left(\Phi_{s}(i)-\Phi_{s}(-i)\right)$

($b_{s}^{\prime}(\cdot)$ is regarded as a functional
$\mathop{\mathrm{dom}}H^{0\,*}\to\mathbb{C}$; the column matrix
$b^{\prime}\equiv(b_{s}^{\prime})$) and

(2.7) $\begin{pmatrix}b_{s}(f)\\\
b_{s}^{\prime}(f)\end{pmatrix}=\begin{pmatrix}1&1\\\
-i&i\end{pmatrix}\begin{pmatrix}a_{s}^{+}(f)\\\ a_{s}^{-}(f)\end{pmatrix}.$

Using the normalization of $(\Phi_{s}(i))_{s=\pm}$, equations (2.5) and (2.7),
and

(2.8)
$H^{0\,*}f=H\tilde{f}+\tfrac{1}{2}\sum_{s}ib_{s}(f)\left(\Phi_{s}(i)-\Phi_{s}(-i)\right)$

one shows that $(\mathbb{C}^{2},b,b^{\prime})$ is a boundary triple for
$H^{0\,*}$.

The self-adjoint extensions $H(\Gamma)$ of $H^{0}$ in $\mathcal{H}_{0}$ are in
one-to-one correspondence with an Hermitian matrix
$\Gamma\equiv(\Gamma_{ss^{\prime}})\in\mathbb{C}^{2\times 2}$, and they are
defined as restrictions of $H^{0\,*}$ to the domain [7]

(2.9)
$\mathop{\mathrm{dom}}H(\Gamma)=\\{f\in\mathop{\mathrm{dom}}H^{0\,*}\colon\braket{\psi^{R},f}=-C^{-1}b(f)\\}$

($\Gamma:=-C^{-1}-R$; $C\in\mathbb{C}^{2\times 2}$; $\det C\neq 0$) provided
that $C^{-1}\neq 0$. When the coupling parameter $C\equiv 0$, we set
$H(\Gamma)\equiv H$. In view of (2.3), (2.8), (2.9), $H(\Gamma)$ is regarded
as

(2.10) $H(\Gamma)=H+V,\quad V\colon
f\mapsto\sum_{s}(C\braket{\psi^{R},f})_{s}\psi_{s}\quad(f\in\mathop{\mathrm{dom}}H(\Gamma))$

so that $V$ is a singular rank two perturbation. The resolvent operator of
$H(\Gamma)$ is given by

(2.11)
$(H(\Gamma)-z)^{-1}=(H-z)^{-1}+\sum_{s}((\Gamma-Q(z))^{-1}\braket{\Phi(\bar{z}),\cdot}_{0})_{s}\Phi_{s}(z)$

($z\in\mathop{\mathrm{res}}H(\Gamma)=\mathop{\mathrm{res}}H\cap\\{z\in\mathbb{C}\colon\det(\Gamma-Q(z))\neq
0\\}$). Here $Q\equiv(Q_{ss^{\prime}})\in\mathbb{C}^{2\times 2}$ is the Krein
$Q$-matrix function

(2.12)
$Q_{ss^{\prime}}(z):=\Braket{\psi_{s},\frac{I+zH}{H-z}\frac{1}{H^{2}+I}\psi_{s^{\prime}}}\quad(s,s^{\prime}=\pm).$

Let us show that (2.9) does not apply to the case $C^{-1}=0$. Let formally
$C^{-1}=0$. Then $\braket{\psi^{R},f}=0$ for all
$f\in\mathop{\mathrm{dom}}H(\Gamma=-R)$. But then, since there always exists
$f\in\mathop{\mathrm{dom}}H^{0\,*}$ such that $\braket{\psi^{R},f}\neq 0$, it
follows from the Hahn–Banach theorem that $\mathop{\mathrm{dom}}H(-R)$ is not
dense in $\mathop{\mathrm{dom}}H^{0\,*}$, and hence in $\mathcal{H}_{0}$. This
contradicts the fact that $H^{0}$ is densely defined and thus its all self-
adjoint extensions are operators.

Let $H_{F}:=H(-R)|\mathop{\mathrm{ker}}b$. Using (2.1), (2.6), and the
orthogonality of $(\Phi_{s}(i))_{s=\pm}$,
$\braket{\psi,\tilde{f}}=-b^{\prime}(f)$. Using (2.4), $H_{F}=H$,
$\mathop{\mathrm{dom}}H_{F}=\\{f\in\mathop{\mathrm{dom}}H^{0\,*}\colon
b(f)=b^{\prime}(f)=0\\}$.

######  Proposition 2.1.

$H_{F}$ is the Friedrichs extension of $H^{0}$.

###### Proof.

Let $H_{F}^{\prime}$ be the Friedrichs extension of $H^{0}$. We show that
$H_{F}=H_{F}^{\prime}$.

Let $\mathfrak{s}$, with
$\mathop{\mathrm{dom}}\mathfrak{s}=\mathop{\mathrm{dom}}H^{0}$, be the form
associated with $H^{0}$:

$\mathfrak{s}[f,g]:=\braket{H^{0}f,g}_{0}\quad(f,g\in\mathop{\mathrm{dom}}H^{0}).$

Then its closure $\bar{\mathfrak{s}}$ is defined on
$\mathop{\mathrm{dom}}\lvert H^{0}\rvert^{1/2}$, i.e.
$\mathop{\mathrm{dom}}\bar{\mathfrak{s}}=\\{f\in\mathcal{H}_{1}\colon\braket{\psi,f}=0\\}$,
and, by definition,
$H_{F}^{\prime}:=H^{0\,*}|\mathop{\mathrm{dom}}\bar{\mathfrak{s}}$. Put (2.3)
in $\braket{\psi,f}=0$, apply $\psi_{s}\notin\mathcal{H}_{-1}$, and deduce
that $b(f)=0$ and $\braket{\psi,\tilde{f}}=0$. But
$\braket{\psi,\tilde{f}}=-b^{\prime}(f)$ and hence
$H_{F}^{\prime}=H^{0\,*}|\mathop{\mathrm{ker}}b\cap\mathop{\mathrm{ker}}b^{\prime}$
as required. ∎

######  Remark 2.2.

According the theory of boundary triplets (see e.g. [41, Chap. 14]), the self-
adjoint extensions of $H^{0}$ are in one-to-one correspondence with the self-
adjoint relations on $\mathbb{C}^{2}$. One can show that, for $C^{-1}\neq 0$
and $C\neq 0$, the extensions correspond to the relations whose range is of
dimension $1$ and $2$. When $C=0$, the range of the corresponding relation on
$\mathbb{C}^{2}$ is $\\{0\\}$, i.e. $H=H^{0\,*}|\mathop{\mathrm{ker}}b$. When
$C^{-1}=0$, the relation is $\\{(0,0)\\}$, i.e.
$H_{F}=H^{0\,*}|\mathop{\mathrm{ker}}b\cap\mathop{\mathrm{ker}}b^{\prime}$.

##  3\. Free Hamiltonian

Throughout, $H^{m}$ ($m=1,2$) is the Sobolev space of $L^{2}$-functions whose
distributional derivatives of order $\leq m$ belong to $L^{2}$. The closure of
$C_{0}^{\infty}$ in $H^{m}$-norm is $H_{0}^{m}$.

The closure of the operator $S$ defined on the domain (1.2) is a densely
defined symmetric operator

(3.1) $H^{0}:=S\lvert
H_{0}^{2}(\mathbb{R}^{3}\backslash\\{0\\})\operatorname{\otimes}\mathbb{C}^{2}.$

Equivalently,

(3.2)
$H^{0}=S,\quad\mathop{\mathrm{dom}}H^{0}=\\{f\in\mathop{\mathrm{dom}}H\colon
f(0)=0\\}$

where

(3.3) $H:=S\lvert H^{2}(\mathbb{R}^{3})\operatorname{\otimes}\mathbb{C}^{2}.$

The operator $H$ in (3.3) is the maximal operator associated with $S$, and it
is the closure of the operator $S\lvert
C_{0}^{\infty}(\mathbb{R}^{3})\operatorname{\otimes}\mathbb{C}^{2}$. By Gauss
and Green formulas (see e.g. [41, Appendix D]) one deduces that the formal
adjoint of $S\lvert
C_{0}^{\infty}(\mathbb{R}^{3})\operatorname{\otimes}\mathbb{C}^{2}$ coincides
with itself. Hence the closure of $S\lvert
C_{0}^{\infty}(\mathbb{R}^{3})\operatorname{\otimes}\mathbb{C}^{2}$ is the
minimal operator associated with $S$. Now that the adjoint of the minimal
(resp. maximal) operator coincides with the maximal (resp. minimal) operator,
it follows that $H$ is self-adjoint. We call $H$ the free Hamiltonian. By
using exactly the same arguments one concludes that the adjoint operator
$H^{0\,*}$ of $H^{0}$, (3.1)–(3.2), is defined as follows:

(3.4) $H^{0\,*}=S\lvert
H^{2}(\mathbb{R}^{3}\backslash\\{0\\})\operatorname{\otimes}\mathbb{C}^{2}.$

The free Hamiltonian $H$ has lower bound $-\Sigma$, where $\Sigma$ equals
$\beta$ if $\beta>\alpha^{2}/2$ and $(\beta/\alpha)^{2}+(\alpha/2)^{2}$
otherwise. The Green function for $H$, or the free Green function, possesses a
representation

$\displaystyle(H-z)^{-1}(x)=$ $\displaystyle
G_{2}(x;z)\operatorname{\otimes}I-\beta
G_{1}(x;z)\operatorname{\otimes}\sigma_{3}+i\alpha\left(\nabla_{2}G_{1}(x;z)\operatorname{\otimes}\sigma_{1}\right.$
(3.5)
$\displaystyle-\left.\nabla_{1}G_{1}(x;z)\operatorname{\otimes}\sigma_{2}\right)\quad(x\in\mathbb{R}^{3}\backslash\\{0\\})$

provided that $z\in\mathop{\mathrm{res}}H$ meets at least one of the
following:

  1. (a)

$2\beta>\alpha^{2}$ and $\displaystyle\beta\leq\lvert
z\rvert<2(\beta/\alpha)^{2}$

  2. (b)

$\lvert z\rvert\geq\Sigma$; the equality is available only if $0\leq
2\beta<\alpha^{2}$

  3. (c)

$\displaystyle\lvert z\rvert>\max\\{\beta/(2\sqrt{R}),\alpha^{2}/(4S)\\}$ and
$\displaystyle R+\left(S-\tfrac{1}{2}\right)^{2}=\tfrac{1}{4}$ ($R,S>0$).

We always assume that $z\in\mathop{\mathrm{res}}H$ satisfies at least one of
(a)–(c) unless explicitly stated otherwise. In particular, $(H-i)^{-1}(\cdot)$
possesses a series representation (3.5) if $\Sigma\leq 1$ (put $\lvert
z\rvert=1$ in (a)–(c)). The functions $G_{j}(x;z)\equiv
G_{j}(x;\alpha,\beta,z)$ ($j=1,2$) obey a hypergeometric series
representation, and they are studied in [30]. Let

$G_{2}^{\mathrm{ren}}(x;z)\equiv
G_{2}^{\mathrm{ren}}(x;\alpha,\beta,z):=G_{2}(x;z)-\frac{e^{-\lvert
x\rvert\sqrt{-z}}}{4\pi\lvert x\rvert}.$

When $x=0$, one assumes the limit $\lvert x\rvert\downarrow 0$ on the right.
We have that

(3.6a) $\displaystyle G_{1}(0;\alpha,\beta,z)=$
$\displaystyle\frac{1}{4\pi\alpha}\mathrm{artanh}\left(\frac{\alpha}{\beta}\sqrt{\frac{-z}{2}\left(1-\sqrt{1-\left(\frac{\beta}{z}\right)^{2}}\right)}\right),$
$\displaystyle G_{2}^{\mathrm{ren}}(0;\alpha,\beta,z)=$
$\displaystyle\frac{\sqrt{-z}}{4\pi}-\frac{1}{4\pi}\sqrt{\frac{-z}{2}\left(1+\sqrt{1-\left(\frac{\beta}{z}\right)^{2}}\right)}$
(3.6b)
$\displaystyle+\frac{\alpha}{8\pi}\mathrm{artanh}\left(\frac{\alpha}{\beta}\sqrt{\frac{-z}{2}\left(1-\sqrt{1-\left(\frac{\beta}{z}\right)^{2}}\right)}\right).$

When $z\in\mathop{\mathrm{res}}H$, the inverse hyperbolic tangent
$\mathrm{artanh}$ is defined entirely on $\mathbb{C}\backslash\\{-1,1\\}$. Let
$U_{\epsilon}$ be an $\epsilon$-neighborhood of the origin
$0\in\mathbb{R}^{3}$. For $x\in U_{\epsilon}$ and for $\epsilon>0$ small,

(3.7)
$\nabla_{j}G_{1}(x;\alpha,\beta,z)=-\frac{\hat{x}_{j}}{8\pi},\quad\hat{x}_{j}:=\frac{x_{j}}{\lvert
x\rvert}\quad(j=1,2).$

##  4\. Orthonormal functionals

Let $\delta$ be the Dirac distribution concentrated at $0\in\mathbb{R}^{3}$.
Define the singular distribution

$\psi_{s}:=\mathcal{N}_{s}\delta\operatorname{\otimes}g_{s}\quad(s=\pm),\quad\text{where}\quad
g_{+}:=\begin{pmatrix}1\\\ 0\end{pmatrix},\quad g_{-}:=\begin{pmatrix}0\\\
1\end{pmatrix}.$

The normalization constant $\mathcal{N}_{s}>0$ is defined as

$\mathcal{N}_{s}:=\left(\mathop{\mathrm{Im}}G_{s}^{\mathrm{ren}}(0;i)+\frac{1}{4\sqrt{2}\pi}\right)^{-1/2}.$

Here we write, for simplicity,

$\displaystyle G_{s}^{\mathrm{ren}}(x;z)\equiv
G_{s}^{\mathrm{ren}}(x;\alpha,\beta,z):=$ $\displaystyle
G_{s}(x;z)-\frac{e^{-\lvert x\rvert\sqrt{-z}}}{4\pi\lvert
x\rvert}\quad(x\in\mathbb{R}^{3};\,z\in\mathop{\mathrm{res}}H),$
$\displaystyle G_{s}(x;z)\equiv G_{s}(x;\alpha,\beta,z):=$ $\displaystyle
G_{2}(x;z)-s\beta
G_{1}(x;z)\quad(x\in\mathbb{R}^{3}\backslash\\{0\\};\,z\in\mathop{\mathrm{res}}H).$

For $\alpha=\beta=0$, $\mathcal{N}_{s}=2\sqrt[4]{2}\sqrt{\pi}$ is as in [7,
Sec. 2.3], where the authors examine the Laplace operator in dimension three
(recall that $S=-\Delta\operatorname{\otimes}I$ for $\alpha=\beta=0$) with the
interaction determined using the model of generalized perturbations.

In general, the relation $\mathcal{N}_{s}>0$ can be shown as follows. Observe
that

(4.1a) $\displaystyle\mathop{\mathrm{Im}}G_{1}(0;i)=$
$\displaystyle\frac{1}{8\pi\alpha}\left(\mathop{\mathrm{Arg}}\left(1+c(1+i)\right)-\mathop{\mathrm{Arg}}\left(1-c(1+i)\right)\right),$
$\displaystyle\mathop{\mathrm{Im}}G_{2}^{\mathrm{ren}}(0;i)=$
$\displaystyle-\frac{1}{4\sqrt{2}\pi}+\frac{1}{8\pi}\sqrt{1+\sqrt{1+\beta^{2}}}$
(4.1b)
$\displaystyle+\frac{\alpha}{16\pi}\left(\mathop{\mathrm{Arg}}\left(1+c(1+i)\right)-\mathop{\mathrm{Arg}}\left(1-c(1+i)\right)\right).$

Here we define, for convenience,

$c:=\frac{\alpha}{2\beta}\sqrt[4]{2+\beta^{2}-2\sqrt{1+\beta^{2}}}\quad(c\geq
0),$

and $\mathop{\mathrm{Arg}}$ is the principal value of the argument; the range
of $\mathop{\mathrm{Arg}}$ is in $(-\pi,\pi]$. Then

(4.2)
$\mathcal{N}_{s}^{-2}=\frac{1}{8\pi}\left(\sqrt{1+\sqrt{1+\beta^{2}}}+\left(\frac{\alpha}{2}-\frac{s\beta}{\alpha}\right)\left(\mathop{\mathrm{Arg}}\left(1+c(1+i)\right)-\mathop{\mathrm{Arg}}\left(1-c(1+i)\right)\right)\right).$

Seeing that

$\displaystyle\mathop{\mathrm{Arg}}\left(1+c(1+i)\right)-\mathop{\mathrm{Arg}}\left(1-c(1+i)\right)\geq
0\,(=0\,\text{iff}\,\alpha=0),$
$\displaystyle\frac{\alpha}{2}-\frac{s\beta}{\alpha}\geq 0\quad\text{iff}\quad
s=-1\quad\text{or}\quad s=1\quad\text{and}\quad\beta\leq\frac{\alpha^{2}}{2},$
$\displaystyle\frac{\alpha}{2}-\frac{s\beta}{\alpha}<0\quad\text{iff}\quad
s=1\quad\text{and}\quad\beta>\frac{\alpha^{2}}{2},$

we conclude from (4.2) that the condition $\mathcal{N}_{s}^{-2}>0$ is
equivalent to the condition

$\displaystyle\sqrt{1+\sqrt{1+\beta^{2}}}>$ $\displaystyle
f(\alpha,\beta)\quad\text{for}\quad\beta>\frac{\alpha^{2}}{2},\quad\text{where}$
$\displaystyle f(\alpha,\beta):=$
$\displaystyle\left(\frac{\beta}{\alpha}-\frac{\alpha}{2}\right)\left(\mathop{\mathrm{Arg}}\left(1+c(1+i)\right)-\mathop{\mathrm{Arg}}\left(1-c(1+i)\right)\right).$

But, for a fixed $\beta\geq 0$,

$f(\alpha,\beta)\leq
f(0,\beta)=\sqrt[4]{2+\beta^{2}-2\sqrt{1+\beta^{2}}}<\sqrt{1+\sqrt{1+\beta^{2}}}.$

Therefore, $\mathcal{N}_{s}^{-2}>0$ and this last estimate implies that we can
always choose $\mathcal{N}_{s}>0$.

Next, using (2.2),

(4.3)
$\Phi_{s}(z)\equiv\Phi_{s}(x;z)=\mathcal{N}_{s}\left(G_{s}(x;z)\operatorname{\otimes}g_{s}+s\alpha
D_{s}G_{1}(x;z)\operatorname{\otimes}g_{-s}\right)\quad(x\in\mathbb{R}^{3}\backslash\\{0\\})$

with arguments and parameters as in (2.2). Here
$D_{s}:=\nabla_{1}+is\nabla_{2}$. Below we prove two equivalent relations:

  * •

$(\Phi_{s}(i))_{s=\pm}$ is an orthonormal system in
$\mathcal{H}_{0}=L^{2}(\mathbb{R}^{3})\operatorname{\otimes}\mathbb{C}^{2}$

  * •

$(\psi_{s})_{s=\pm}$ is an orthonormal system in
$\mathcal{H}_{-2}\backslash\mathcal{H}_{-1}$
($\mathcal{H}_{-n}=H^{-n}(\mathbb{R}^{3})\operatorname{\otimes}\mathbb{C}^{2}$;
$n=1,2$).

The important conclusion is that, if the above relations hold true, then the
operator $H^{0}$ in (3.1)–(3.2) may be treated similar to the operator $H^{0}$
in (2.1); subsequently, the functions $(\Phi_{s}(\pm i))_{s=\pm}$ in (4.3)
form deficiency subspaces of the adjoint operator $H^{0\,*}$ defined in (2.8),
(3.4).

Using (2.2) and

$\frac{H}{H^{2}+I}=\tfrac{1}{2}\left(\frac{1}{H-i}+\frac{1}{H+i}\right),$

we have that

$\Braket{\psi_{s},\frac{H}{H^{2}+I}\psi_{s}}=\tfrac{1}{2}\braket{\psi_{s},\Phi_{s}(i)+\Phi_{s}(-i)}.$

But

$\displaystyle\lvert\braket{\psi_{s},\Phi_{s}(i)+\Phi_{s}(-i)}\rvert=$
$\displaystyle 2\mathcal{N}_{s}^{2}\lvert\mathop{\mathrm{Re}}G_{s}(0;i)\rvert$
$\displaystyle=$ $\displaystyle
2\mathcal{N}_{s}^{2}\left\lvert\mathop{\mathrm{Re}}G_{s}^{\mathrm{ren}}(0;i)+\lim_{\lvert
x\rvert\downarrow 0}\mathop{\mathrm{Re}}\frac{e^{-\lvert
x\rvert\sqrt{-i}}}{4\pi\lvert x\rvert}\right\rvert=\infty.$

Then

$\left\lvert\Braket{\psi_{s},\frac{H}{H^{2}+I}\psi_{s}}\right\rvert\leq
2\lVert\psi_{s}\rVert_{-1}^{2}$

(cf. [6, Proof of Theorem 3.1]) implies that $\psi_{s}\notin\mathcal{H}_{-1}$.
Next, we show that

(4.4)
$\braket{\Phi_{s}(z),\Phi_{s^{\prime}}(z)}_{0}=\frac{\delta_{ss^{\prime}}\mathcal{N}_{s}^{2}}{\mathop{\mathrm{Im}}z}\mathop{\mathrm{Im}}\left(G_{s}^{\mathrm{ren}}(0;z)-\frac{\sqrt{-z}}{4\pi}\right)$

($s,s^{\prime}=\pm$; $z\in\mathop{\mathrm{res}}H$; $\mathop{\mathrm{Im}}z\neq
0$; $\delta_{ss^{\prime}}$ is the Kronecker symbol). When
$\mathop{\mathrm{Im}}z=0$, take the limit $\mathop{\mathrm{Im}}z\to 0$ in
(4.4):

$\displaystyle\braket{\Phi_{s}(z),\Phi_{s^{\prime}}(z)}_{0}=$
$\displaystyle\frac{\delta_{ss^{\prime}}\mathcal{N}_{s}^{2}}{8\sqrt{2}\pi\sqrt{z^{2}-\beta^{2}}}$
(4.5)
$\displaystyle\times\left(\sqrt{-z+\sqrt{z^{2}-\beta^{2}}}+\frac{\beta\left(\alpha^{2}-2s\beta\right)\sqrt{-z-\sqrt{z^{2}-\beta^{2}}}}{2\beta^{2}+\alpha^{2}\left(z+\sqrt{z^{2}-\beta^{2}}\right)}\right)$

($s,s^{\prime}=\pm$; $z<-\Sigma$).

Since

$\frac{1}{H-\bar{z}}\frac{1}{H-z}=\frac{1}{2i\mathop{\mathrm{Im}}z}\left(\frac{1}{H-z}-\frac{1}{H-\bar{z}}\right)$

it follows from (2.2) that

$\braket{\Phi_{s}(z),\Phi_{s^{\prime}}(z)}_{0}=\frac{1}{2i\mathop{\mathrm{Im}}z}\braket{\psi_{s},\Phi_{s^{\prime}}(z)-\Phi_{s^{\prime}}(\bar{z})}.$

Then, using (4.3),

(4.6)
$\braket{\Phi_{s}(z),\Phi_{s^{\prime}}(z)}_{0}=\frac{\mathcal{N}_{s}\mathcal{N}_{s^{\prime}}}{\mathop{\mathrm{Im}}z}\lim_{\lvert
x\rvert\downarrow
0}\left(\delta_{ss^{\prime}}\mathop{\mathrm{Im}}G_{s}(x;z)-s\alpha\delta_{s,-s^{\prime}}D_{-s}\mathop{\mathrm{Im}}G_{1}(x;z)\right).$

But

$\displaystyle\lim_{\lvert x\rvert\downarrow
0}\mathop{\mathrm{Im}}G_{s}(x;z)=$ $\displaystyle\frac{1}{2i}\lim_{\lvert
x\rvert\downarrow 0}\left(G_{s}(x;z)-G_{s}(x;\bar{z})\right)$ $\displaystyle=$
$\displaystyle\mathop{\mathrm{Im}}G_{s}^{\mathrm{ren}}(0;z)+\frac{1}{2i}\lim_{\lvert
x\rvert\downarrow 0}\left(\frac{e^{-\lvert x\rvert\sqrt{-z}}}{4\pi\lvert
x\rvert}-\frac{e^{-\lvert x\rvert\sqrt{-\bar{z}}}}{4\pi\lvert x\rvert}\right)$
(4.7) $\displaystyle=$
$\displaystyle\mathop{\mathrm{Im}}G_{s}^{\mathrm{ren}}(0;z)-\frac{\mathop{\mathrm{Im}}\sqrt{-z}}{4\pi}$

and, by (3.7),

$\displaystyle\lim_{\lvert x\rvert\downarrow
0}D_{s}\mathop{\mathrm{Im}}G_{1}(x;z)=$ $\displaystyle\frac{1}{2i}\lim_{\lvert
x\rvert\downarrow 0}D_{s}\left(G_{1}(x;z)-G_{1}(x;\bar{z})\right)$ (4.8)
$\displaystyle=$ $\displaystyle\frac{1}{2i}\lim_{\lvert x\rvert\downarrow
0}\left(-\frac{\hat{x}_{1}+is\hat{x}_{2}}{8\pi}+\frac{\hat{x}_{1}+is\hat{x}_{2}}{8\pi}\right)=0.$

Put (4.7) and (4.8) in (4.6) and deduce (4.4). Since
$\mathcal{H}_{-1}\subset\mathcal{H}_{-2}$ densely, we conclude that
$(\psi_{s})_{s=\pm}$ is an orthonormal system in
$\mathcal{H}_{-2}\backslash\mathcal{H}_{-1}$.

##  5\. Spectrum

Here we apply the resolvent formula (2.11) to the spectral analysis of the
operator $H(\Gamma)$ regarded as a perturbed Hamiltonian $H+V$ and where the
perturbation $V$ is referred to as the spin-dependent contact interaction
(recall (2.10)). We assume that $\Gamma\neq-R$. When $\Gamma=-R$, the
corresponding operator is $H_{F}$ (Proposition 2.1). Since the spectrum of
$H_{F}$ coincides with that of $H$, and the spectrum of $H$ is
$[-\Sigma,\infty)$ and it is absolutely continuous, we mainly concentrate on
the spectrum of $H(\Gamma)$. We adopt the classification of the spectrum in
[41, Chap. 9], [40, Sec. VII.3].

The continuous spectrum of $H(\Gamma)$ is given by $[-\Sigma,\infty)$. This
follows from an invariance of the continuous component of the spectrum under
singular finite rank perturbations [7, Theorem 4.1.4] and from the resolvent
formula.

The real and isolated singularities of the resolvent operator of $H(\Gamma)$
coincide with the points $E\in\mathbb{R}$ that solve $\det(\Gamma-Q(E))=0$.
Therefore, the singular spectrum of $H(\Gamma)$ consists of the points
$E\in\mathbb{R}$ such that

(5.1) $\det(\Gamma-Q(E))=0.$

The Krein $Q$-matrix function, which is defined in (2.12), is a diagonal
matrix $Q(z)\equiv(Q_{ss^{\prime}}(z))$ whose entries are given by

(5.2)
$Q_{ss^{\prime}}(z)=\delta_{ss^{\prime}}\mathcal{N}_{s}^{2}\left(\frac{1}{4\sqrt{2}\pi}-\frac{\sqrt{-z}}{4\pi}+G_{s}^{\mathrm{ren}}(0;z)-\mathop{\mathrm{Re}}G_{s}^{\mathrm{ren}}(0;i)\right)$

($s,s^{\prime}=\pm$). To prove (5.2), notice that

$\frac{I+zH}{H-z}\frac{1}{H^{2}+I}=\frac{1}{H-z}-\frac{1}{2}\frac{1}{H-i}-\frac{1}{2}\frac{1}{H+i}.$

Then, using (2.2) and (2.12),

$Q_{ss^{\prime}}(z)=\Braket{\psi_{s},\Phi_{s^{\prime}}(z)-\tfrac{1}{2}\Phi_{s^{\prime}}(i)-\tfrac{1}{2}\Phi_{s^{\prime}}(-i)}.$

By (4.3), it follows that

$\displaystyle Q_{ss^{\prime}}(z)=$ $\displaystyle\lim_{\lvert
x\rvert\downarrow
0}\left[\delta_{ss^{\prime}}\mathcal{N}_{s}^{2}\left(G_{s}(x;z)-\mathop{\mathrm{Re}}G_{s}(x;i)\right)\right.$
$\displaystyle\left.-s\alpha\delta_{s,-s^{\prime}}\mathcal{N}_{s}\mathcal{N}_{s^{\prime}}D_{-s}\left(G_{1}(x;z)-\mathop{\mathrm{Re}}G_{1}(x;i)\right)\right].$

But

$\displaystyle\lim_{\lvert x\rvert\downarrow
0}\left(G_{s}(x;z)-\mathop{\mathrm{Re}}G_{s}(x;i)\right)=$ $\displaystyle
G_{s}^{\mathrm{ren}}(0;z)-\mathop{\mathrm{Re}}G_{s}^{\mathrm{ren}}(0;i)$
$\displaystyle+\lim_{\lvert x\rvert\downarrow 0}\left(\frac{e^{-\lvert
x\rvert\sqrt{-z}}}{4\pi\lvert x\rvert}-\mathop{\mathrm{Re}}\frac{e^{-\lvert
x\rvert\sqrt{-i}}}{4\pi\lvert x\rvert}\right)$ $\displaystyle=$ $\displaystyle
G_{s}^{\mathrm{ren}}(0;z)-\mathop{\mathrm{Re}}G_{s}^{\mathrm{ren}}(0;i)+\frac{1}{4\sqrt{2}\pi}-\frac{\sqrt{-z}}{4\pi}$

and

$\lim_{\lvert x\rvert\downarrow
0}D_{-s}\left(G_{1}(x;z)-\mathop{\mathrm{Re}}G_{1}(x;i)\right)=0$

which proves (5.2). In particular,
$\mathop{\mathrm{Im}}Q_{ss}(z)=\lVert\Phi_{s}(z)\rVert_{0}^{2}\mathop{\mathrm{Im}}z$
and $Q_{ss}(\pm i)=\pm i$. For an arbitrary $z\in\mathbb{C}$, but for
$\alpha=\beta=0$, $Q_{ss}(z)=1-\sqrt{-2z}$ coincides with the $Q$-function
obtained in [7, Sec. 2.3.1].

######  Lemma 5.1.

The eigenspace $\mathop{\mathrm{ker}}(H(\Gamma)-E)$ consists of eigenfunctions

(5.3) $f(E+i\epsilon)=\sum_{s}b_{s}(f)\Phi_{s}(E+i\epsilon)$

where the coefficients $(b_{s}(f))_{s=\pm}$ are found from the boundary
condition defining the domain of $H(\Gamma)$. For $E<-\Sigma$, $\epsilon=0$.
For $E\geq-\Sigma$, $\epsilon>0$ and $f(E+i\epsilon)$ is understood in the
generalized sense, i.e. $f(E+i\epsilon)$ approaches $f(E)$ strongly in
$\mathcal{H}_{0}$ as $\epsilon\downarrow 0$.

###### Proof.

The proof of the lemma for $E<-\Sigma$ is a natural generalization of results
posted in [7, Sec. 2.3.1], so we concentrate on the case $E\geq-\Sigma$.

For convenience, define $E_{\epsilon}:=E+i\epsilon$. Let
$f(E)\in\mathop{\mathrm{ker}}(H(\Gamma)-E)$. Then $f(E)$ possesses
representations (2.3), (2.5) with an appropriate boundary condition. Plugging
(2.3) and (2.8) with $\tilde{f}\equiv\tilde{f}(E)$ into the eigenvalue
equation, we obtain the equation

(5.4)
$(H-E)\tilde{f}(E)=\tfrac{1}{2}\sum_{s}b_{s}(f)\left((E-i)\Phi_{s}(i)+(E+i)\Phi_{s}(-i)\right).$

By hypothesis, the operator $(H-E)^{-1}$ is unbounded in $\mathcal{H}_{0}$ and
it cannot be applied to (5.4). We apply $(H-E_{\epsilon})^{-1}$ instead.
Define

(5.5a) $\tilde{f}_{\epsilon}(E):=(H-E_{\epsilon})^{-1}(H-E)\tilde{f}(E).$ Put
(5.4) in (5.5a): (5.5b)
$\tilde{f}_{\epsilon}(E)=\tfrac{1}{2}\sum_{s}b_{s}(f)(H-E_{\epsilon})^{-1}\left((E-i)\Phi_{s}(i)+(E+i)\Phi_{s}(-i)\right).$

Let $f(E_{\epsilon})$ be as in (5.3). Similar to (5.5a), define

(5.6) $f_{\epsilon}(E):=(H-E_{\epsilon})^{-1}(H-E)f(E).$

Use $f(E)$ in (2.3) and plug the representation into (5.6). Then apply (2.2)
and (5.5):

$\displaystyle f_{\epsilon}(E)=$
$\displaystyle(H-E_{\epsilon})^{-1}(H-E)\left(\tilde{f}(E)+\tfrac{1}{2}\sum_{s}b_{s}(f)\left(\Phi_{s}(i)+\Phi_{s}(-i)\right)\right)$
$\displaystyle=$
$\displaystyle\tilde{f}_{\epsilon}(E)+\tfrac{1}{2}\sum_{s}b_{s}(f)(H-E_{\epsilon})^{-1}(H-E)\left(\Phi_{s}(i)+\Phi_{s}(-i)\right)$
$\displaystyle=$
$\displaystyle\tfrac{1}{2}\sum_{s}b_{s}(f)(H-E_{\epsilon})^{-1}\left((E-i)\Phi_{s}(i)+(E+i)\Phi_{s}(-i)\right)$
$\displaystyle+\tfrac{1}{2}\sum_{s}b_{s}(f)(H-E_{\epsilon})^{-1}(H-E)\left(\Phi_{s}(i)+\Phi_{s}(-i)\right)$
$\displaystyle=$
$\displaystyle\tfrac{1}{2}\sum_{s}b_{s}(f)(H-E_{\epsilon})^{-1}(\psi_{s}+\psi_{s})=f(E_{\epsilon}).$

Let $\mathcal{E}$ be the resolution of the identity for $H$, and let
$\mu(\cdot):=\braket{f(E),\mathcal{E}(\cdot)f(E)}_{0}$ be the spectral
measure. Using (5.6) and $f_{\epsilon}(E)=f(E_{\epsilon})$,

$\lVert
f(E_{\epsilon})-f(E)\rVert_{0}=\epsilon\lVert(H-E_{\epsilon})^{-1}f(E)\rVert_{0}$

or equivalently,

(5.7) $\lVert
f(E_{\epsilon})-f(E)\rVert_{0}^{2}=\epsilon\mathop{\mathrm{Im}}F(E_{\epsilon}),\quad
F(z):=\int_{\mathbb{R}}\frac{\mu(d\nu)}{\nu-z}\quad(z\in\mathop{\mathrm{res}}H).$

By [42, Theorem 11.6-(iii)], the Borel transform $F$ of $\mu$ fulfills

(5.8) $\mu(\\{E\\})=\lim_{\epsilon\downarrow
0}\epsilon\mathop{\mathrm{Im}}F(E+i\epsilon).$

Since $\mathcal{E}(\\{E\\})=0$ for all $E\geq-\Sigma$, we conclude that
$\mu(\\{E\\})=0$. Put this value in (5.8), and then deduce the result stated
in the lemma from (5.7). ∎

######  Theorem 5.2.

The singular continuous part is absent from the spectrum of $H(\Gamma)$.

###### Proof.

Consider the singular spectrum as a closed set of points $E_{n}\in\mathbb{R}$
that solve (5.1) for $n=1,2,\ldots,N$, for some $N\in\mathbb{N}$. Let
$\mathcal{H}_{\mathrm{s}}$, $\mathcal{H}_{\mathrm{p}}$, and
$\mathcal{H}_{\mathrm{sc}}$ be the reducing subspaces of $\mathcal{H}_{0}$
corresponding to the singular, discontinuous, and singular continuous parts of
$H(\Gamma)$. Then
$\mathcal{H}_{\mathrm{s}}=\mathcal{H}_{\mathrm{p}}\operatorname{\oplus}\mathcal{H}_{\mathrm{sc}}$.
For each function $f\in\mathcal{H}_{\mathrm{s}}$ , one finds functions
$g\in\mathcal{H}_{\mathrm{p}}$ and $h\in\mathcal{H}_{\mathrm{sc}}$ such that
$f=g+h$. Then the proof is accomplished if one shows that $h=0$ for each such
$f$.

Using the resolvent formula of $H(\Gamma)$, the integral representation of the
resolution of the identity $\mathcal{E}(\Gamma;\cdot)$ [26, Lemma XVIII.2.31],
and the residue theorem, we find that

(5.9)
$\mathcal{E}(\Gamma;\\{E\\})f=-\sum_{s}\sum_{n}\chi(\Gamma;E)\mathop{\mathrm{Res}}_{z=E_{n}}((\Gamma-Q(z))^{-1}\braket{\Phi(E_{n}),f}_{0})_{s}\Phi_{s}(E_{n}+i0)$

for arbitrary $f\in\mathcal{H}_{0}$. Here $\chi(\Gamma;\cdot)$ is the
characteristic function of the singular spectrum. When deriving (5.9) we used
$\mathcal{E}(\\{E\\})=0$ for the resolution of the identity $\mathcal{E}$ of
$H$. We write $\Phi_{s}(E_{n}+i0)$ to ensure that the right hand side of (5.9)
exists for $E_{n}\geq-\Sigma$. By Lemma 5.1, one safely replaces
$\Phi_{s}(E_{n}+i0)$ with $\Phi_{s}(E_{n})$ in the spectral measure
$\braket{f,\mathcal{E}(\Gamma;\\{E\\})f}_{0}$. For $E_{n}<-\Sigma$,
$\Phi_{s}(E_{n}+i0)=\Phi_{s}(E_{n})$.

The characteristic function in (5.9) indicates for which points
$E\in\mathbb{R}$ the function $\mathcal{E}(\Gamma;\\{E\\})f$ is nonzero. These
are the points that belong to the singular spectrum. Since out task is to
examine the functions in $\mathcal{H}_{\mathrm{s}}$, we identify $E$ with an
arbitrary $E_{n}$. Equation (5.9) is valid for an arbitrary
$f\in\mathcal{H}_{0}$, and now we take $f\in\mathcal{H}_{\mathrm{s}}$. For
each $E_{n}$, there exists $f\in\mathcal{H}_{\mathrm{s}}$ such that
$\mathcal{E}(\Gamma;\\{E_{n}\\})f=f$. If otherwise, since
$f\in\mathcal{H}_{\mathrm{s}}$, there must exist a Lebesgue null set
$e\subset\mathbb{R}$ such that $e$ is absent in the singular spectrum and
$\mathcal{E}(\Gamma;e)f=f$. But $\chi(\Gamma;E)=0$ for all $E\in e$, and hence
$\mathcal{E}(\Gamma;e)f=0$ by (5.9); this contradicts the definition of
$\mathcal{H}_{\mathrm{s}}$. Next, for each $f\in\mathcal{H}_{\mathrm{s}}$,
there exist $g\in\mathcal{H}_{\mathrm{p}}$ and $h\in\mathcal{H}_{\mathrm{sc}}$
such that $f=g+h$. By Lemma 5.1,
$g\in\mathop{\mathrm{span}}\\{\Phi_{s}(E_{n}+i0)\colon n=1,\ldots,N\\}$.
Therefore, since $\mathcal{H}_{\mathrm{sc}}$ is the orthogonal complement of
$\mathcal{H}_{\mathrm{p}}\subseteq\mathcal{H}_{\mathrm{s}}$, one has that
$\braket{\Phi_{s}(E_{n}),h}_{0}=0$ for all $n=1,\ldots,N$. But then, using
(5.9), $\mathcal{E}(\Gamma;\\{E_{n}\\})h=0$. Recalling that
$\mathcal{E}(\Gamma;\\{E_{n}\\})f=f$ and $f=g+h$, it follows from the latter
that $\braket{h,\mathcal{E}(\Gamma;\\{E_{n}\\})f}_{0}=\lVert
h\rVert_{0}^{2}=0$. Hence $f\in\mathcal{H}_{\mathrm{s}}\Rightarrow
f=g\in\mathcal{H}_{\mathrm{p}}$, and the proof of the theorem is accomplished.
∎

It follows from Theorem 5.2 that the point spectrum of $H(\Gamma)$ coincides
with the singular spectrum. Using the integral representation of the
resolution of the identity and the resolvent formula, one shows that, for $e$
in the set of Borel subsets of $\mathbb{R}$, and for $f\in\mathcal{H}_{0}$,
the spectral measures of $H(\Gamma)$ and $H$ agree up to sets of Lebesgue
measure zero. Since the absolutely continuous part of a measure is supported
by the set of $E\in\mathbb{R}$ such that
$\mathop{\mathrm{Im}}\braket{f,(H(\Gamma)-E-i0_{+})^{-1}f}_{0}\neq 0$ (see
e.g. [15]), we conclude that the absolutely continuous parts of spectral
measures of $H(\Gamma)$ and $H$ coincide. If we let $\mathcal{H}_{\mathrm{c}}$
and $\mathcal{H}_{\mathrm{ac}}$ be the reducing subspaces of $\mathcal{H}_{0}$
corresponding to the continuous and absolutely continuous parts of
$H(\Gamma)$, then we arrive at the same conclusion by noting that
$\mathcal{H}_{\mathrm{s}}=\mathcal{H}_{\mathrm{p}}$ and
$\mathcal{H}_{\mathrm{ac}}=\mathcal{H}_{\mathrm{s}}^{\bot}$, hence
$\mathcal{H}_{\mathrm{c}}=\mathcal{H}_{\mathrm{p}}^{\bot}=\mathcal{H}_{\mathrm{ac}}$.

##  6\. Analytic properties of singular spectrum

According to the results of the previous paragraph, the notions of singular
spectrum and point spectrum (eigenvalues) are used interchangeably. As a rule,
we prefer the notion of singular spectrum to that of eigenvalues, because the
singular spectrum consists of the singular points $E\in\mathbb{R}$ of the
resolvent operator of $H(\Gamma)$ and because the singular points are
solutions of equation (5.1), which is our main target in the present section.

Define

$\displaystyle\omega_{s}:=$ $\displaystyle
4\pi(\tilde{\Gamma}_{ss}+\Lambda_{s}),\quad\gamma:=(4\pi\lvert\tilde{\Gamma}_{+-}\rvert)^{2},\quad\tilde{\Gamma}_{ss^{\prime}}:=\frac{\Gamma_{ss^{\prime}}}{\mathcal{N}_{s}\mathcal{N}_{s^{\prime}}},$
(6.1) $\displaystyle\Lambda_{s}:=$
$\displaystyle\mathop{\mathrm{Re}}G_{s}^{\mathrm{ren}}(0;i)-\frac{1}{4\sqrt{2}\pi}\quad(\omega_{s},\Lambda_{s}\in\mathbb{R};\,s,s^{\prime}=\pm;\,\gamma\geq
0).$

The matrix
$\tilde{\Gamma}\equiv(\tilde{\Gamma}_{ss^{\prime}})\equiv\Bigl{(}\begin{smallmatrix}\tilde{\Gamma}_{++}&\tilde{\Gamma}_{+-}\\\
\tilde{\Gamma}_{-+}&\tilde{\Gamma}_{--}\end{smallmatrix}\Bigr{)}$ is
Hermitian. Let also

(6.2)
$\xi(E):=\frac{1}{\beta}\sqrt{\frac{-E}{2}\left(1-\sqrt{1-\left(\frac{\beta}{E}\right)^{2}}\right)}\quad(E\in\mathbb{R};\,\beta\geq
0).$

When $\beta=0$, one takes the limit $\beta\downarrow 0$ on the right hand
side: $\xi(E)=1/(2\sqrt{-E})$. Using (6.1) and (6.2), (5.1) obeys the
following form:

(6.3)
$\gamma=\prod_{s}\left(\omega_{s}\pm\frac{1}{2\xi(E)}-\left(\frac{\alpha}{2}-\frac{s\beta}{\alpha}\right)\mathrm{artanh}(\alpha\xi(E))\right).$

The upper (resp. lower) sign is taken if $E<\beta$ (resp. $E\geq\beta$). Owing
to $\mathrm{artanh}$, equation (6.3) is transcendental and does not possess
analytic solutions. For the reason that we are unable to solve equation (6.3)
as it stands, we examine (6.3) in three specific cases:

  1. (A)

Case without spin-orbit coupling: $\alpha=0$, $\Sigma=\beta\geq 0$.

  2. (B)

Case with small spin-orbit coupling: $0<\alpha<\sqrt{2\beta}$ small,
$\Sigma=\beta>0$.

  3. (C)

Case with large spin-orbit coupling: $0<\sqrt{2\beta}\leq\alpha$,
$0<\beta\leq\Sigma\leq 1$, $E\geq-\Sigma$.

Although the $Q$-function itself is valid for all $\alpha,\beta\geq 0$, the
limitation on the parameters is caused by the series representation for the
free Green function (3.5). In fact, the validity of (4.4) is governed by
(a)–(c). In particular, putting $z=i$ in (4.4), the condition $\Sigma\leq 1$
follows from (a)–(c); this latter condition also ensures the existence of the
normalization constant $\mathcal{N}_{s}>0$.

For $\alpha$ zero or arbitrarily small (Cases (A) or (B), respectively),
conditions (a)–(c) can be safely removed without affecting the absolute
convergence of the series that defines the free Green function. This follows
from the fact that for $\alpha>0$ small, the confluent Horn function, which
defines the functions $G_{j}$ in (3.5), can be analytically continued to the
region $\beta\geq 1$; see [30] for more details. As a result one writes
$\Sigma\geq 0$ in (A) and $\Sigma>0$ in (B). For $E\geq-\Sigma$ in (C), the
eigenfunctions in Lemma 5.1 are restricted to (a) or (c).

###  6.1. Case without spin-orbit coupling

Using the asymptotic formula

(6.4)
$\left(\frac{\alpha}{2}-\frac{s\beta}{\alpha}\right)\mathrm{artanh}(\alpha\xi(E))=-s\beta\xi(E)+\alpha^{2}\xi(E)\left(\frac{1}{2}-\frac{s\beta\xi(E)^{2}}{3}\right)+O(\alpha^{4})$

as $\alpha\downarrow 0$, and picking the first term from the right hand side
of (6.4), we rewrite (6.3) as follows:

(6.5)
$\gamma=\prod_{s}\left(\omega_{s}\pm\frac{1}{2\xi(E)}+s\beta\xi(E)\right).$

It takes little effort to conclude from (6.2) and (6.5) the following result.

######  Theorem 6.1.

Let $\alpha=0$ and $0\leq\beta<\infty$. The part of the singular spectrum of
$H(\Gamma)$ below the threshold $-\beta$ is given by the set
$\\{E<-\beta\colon\gamma=\prod_{s}(\omega_{s}+\sqrt{s\beta-E})\\}$. For
$\beta>0$, the singular spectrum embedded in the continuous spectrum is
described by the following singletons: $\\{-\beta\\}$ iff
$\gamma=(\omega_{+}+\sqrt{2\beta})\omega_{-}$; $\\{\beta-\omega_{+}^{2}\\}$
iff $\gamma=0$ and $-\sqrt{2\beta}<\omega_{+}<0$; $\\{\beta\\}$ iff
$\gamma=\omega_{-}=0$. For $\beta=0$, there are no embedded eigenvalues.

It follows that $H(\Gamma)$ may have maximum three eigenvalues in the interval
$[-\beta,\beta]$. With regard to [17], assume that
$\omega_{+}=\omega_{-}\equiv\omega$. If $\gamma=0$, then Theorem 2 in [17]
says that the point spectrum is empty for $\omega\geq 0$, while it follows
from Theorem 6.1 that the point spectrum consists of the points $\pm\beta$
when $\omega=0$. If $-\sqrt{2\beta}<\omega<0$ and $\gamma>0$, then Theorem 3
in [17] says that there exists $\gamma_{0}>0$ such that, for
$0<\gamma<\gamma_{0}$, $H(\Gamma)$ does not have embedded eigenvalues. By
Theorem 6.1, however, there are no embedded eigenvalues for all $\gamma>0$;
recall that $\gamma=\infty$ and $\lvert\omega\rvert=\infty$ together suit $H$.

###  6.2. Case with small spin-orbit coupling

Here we study the behavior of eigenvalues of $H(\Gamma)\equiv
H(\Gamma,\alpha)$ in the limit $\alpha\downarrow 0$. The main conclusion
following from the results posted below is that, for arbitrarily small but
nonzero $\alpha$, the singular spectrum is empty above the threshold $-\beta$.
Our strategy is to expand $\gamma$ and $\omega_{s}$ in (6.1) as a series with
respect to $\alpha\downarrow 0$, and then, using the asymptotic formula (6.4),
to solve (6.3).

Using (4.2), we have that

(6.6)
$\mathcal{N}_{s}=\mathcal{N}_{s}^{(0)}-\alpha^{2}\mathcal{N}_{s}^{(1)}+O(\alpha^{4})$

as $\alpha\downarrow 0$, where

(6.7a) $\displaystyle\mathcal{N}_{s}^{(0)}:=$ $\displaystyle
2\sqrt[4]{2}\sqrt{\pi}\sqrt[4]{\sqrt{1+\beta^{2}}+s\beta},$ (6.7b)
$\displaystyle\mathcal{N}_{s}^{(1)}:=$
$\displaystyle\frac{\sqrt{\pi}}{6\sqrt[4]{2}\beta^{2}}\left(3\beta+s\left(1-\sqrt{1+\beta^{2}}\right)\right)\sqrt[4]{2+\beta^{2}-2\sqrt{1+\beta^{2}}}\left(\sqrt{1+\beta^{2}}+s\beta\right)^{\frac{3}{4}}.$

Using (3.6) and (6.1),

(6.8)
$\Lambda_{s}=\Lambda_{s}^{(0)}+\alpha^{2}\Lambda_{s}^{(1)}+O(\alpha^{4})$

as $\alpha\downarrow 0$, where

(6.9a) $\displaystyle\Lambda_{s}^{(0)}:=$
$\displaystyle-\frac{1}{8\pi}\left(\sqrt{\sqrt{1+\beta^{2}}+1}+s\sqrt{\sqrt{1+\beta^{2}}-1}\right),$
(6.9b) $\displaystyle\Lambda_{s}^{(1)}:=$
$\displaystyle\frac{1}{48\pi\beta^{2}}\left(3\beta-s\left(1-\sqrt{1+\beta^{2}}\right)\right)\sqrt[4]{2+\beta^{2}-2\sqrt{1+\beta^{2}}}.$

If we let

$\displaystyle\omega_{s}^{(0)}:=$ $\displaystyle
4\pi(\tilde{\Gamma}_{ss}^{(0)}+\Lambda_{s}^{(0)}),\quad\omega_{s}^{(1)}:=4\pi(\eta_{ss}\tilde{\Gamma}_{ss}^{(0)}+\Lambda_{s}^{(1)}),$
(6.10) $\displaystyle\gamma^{(0)}:=$
$\displaystyle(4\pi\lvert\tilde{\Gamma}_{+-}^{(0)}\rvert)^{2},\quad\eta_{ss^{\prime}}:=\frac{\mathcal{N}_{s}^{(1)}}{\mathcal{N}_{s}^{(0)}}+\frac{\mathcal{N}_{s^{\prime}}^{(1)}}{\mathcal{N}_{s^{\prime}}^{(0)}},\quad\tilde{\Gamma}_{ss^{\prime}}^{(0)}:=\frac{\Gamma_{ss^{\prime}}}{\mathcal{N}_{s}^{(0)}\mathcal{N}_{s^{\prime}}^{(0)}}$

($\omega_{s}^{(0)},\omega_{s}^{(1)},\Lambda_{s}^{(0)},\Lambda_{s}^{(1)}\in\mathbb{R}$;
$\eta_{ss^{\prime}}>0$; $s,s^{\prime}=\pm$; $\gamma^{(0)}\geq 0$), then

(6.11) $\displaystyle\omega_{s}=$
$\displaystyle\omega_{s}^{(0)}+\alpha^{2}\omega_{s}^{(1)}+O(\alpha^{4}),$
(6.12) $\displaystyle\gamma=$
$\displaystyle\gamma^{(0)}(1+2\alpha^{2}\eta_{+-})+O(\alpha^{4}),$ (6.13)
$\displaystyle\tilde{\Gamma}_{ss^{\prime}}=$
$\displaystyle\tilde{\Gamma}_{ss^{\prime}}^{(0)}(1+\alpha^{2}\eta_{ss^{\prime}})+O(\alpha^{4})$

as $\alpha\downarrow 0$ ($s,s^{\prime}=\pm$). It is clear that $\gamma^{(0)}$,
$\omega_{s}^{(0)}$ coincide with $\gamma$, $\omega_{s}$ in Theorem 6.1. Put
(6.11), (6.12), and (6.4) in equation (6.3) and get that

$\displaystyle\gamma^{(0)}=$
$\displaystyle\prod_{s}\left(\omega_{s}^{(0)}\pm\frac{1}{2\xi(E)}+s\beta\xi(E)\right)$
(6.14)
$\displaystyle+\alpha^{2}\left(\sum_{s}q_{-s}(E)\left(\omega_{s}^{(0)}\pm\frac{1}{2\xi(E)}+s\beta\xi(E)\right)-2\eta_{+-}\gamma^{(0)}\right)+O(\alpha^{4})$

as $\alpha\downarrow 0$. Here

(6.15)
$q_{s}(E):=\omega_{s}^{(1)}-\xi(E)(\tfrac{1}{2}-\tfrac{1}{3}s\beta\xi(E)^{2}).$

When $\alpha=0$, (6.14) reduces to (6.5). We look for the solutions of the
form $E=E^{(0)}+\epsilon$ for some small $\epsilon=\epsilon(\alpha)$;
$E^{(0)}\in\mathop{\mathrm{spec_{s}}}H(\Gamma,0)$ is in the singular spectrum
of $H(\Gamma,0)$, which is described in Theorem 6.1. It follows that
$\epsilon(0)=0$.

Let us pick $\Gamma\equiv\Gamma^{\circ}$ such that
$-\beta\in\mathop{\mathrm{spec_{s}}}H(\Gamma,0)$ and
$A\Gamma_{++}+B\Gamma_{--}+C=0$, where

$\displaystyle A:=$
$\displaystyle\mathcal{N}_{-}^{(0)\,2}(\Lambda_{-}^{(1)}-\eta_{--}\Lambda_{-}^{(0)}),$
$\displaystyle B:=$
$\displaystyle\mathcal{N}_{+}^{(0)\,2}\left(\Lambda_{+}^{(1)}-\eta_{++}\left(\Lambda_{+}^{(0)}+\frac{\sqrt{2\beta}}{4\pi}\right)\right),$
$\displaystyle C:=$
$\displaystyle(\mathcal{N}_{+}^{(0)}\mathcal{N}_{-}^{(0)})^{2}\left(\Lambda_{-}^{(0)}\Lambda_{+}^{(1)}+\left(\Lambda_{+}^{(0)}+\frac{\sqrt{2\beta}}{4\pi}\right)\left(\Lambda_{-}^{(1)}-(\eta_{++}+\eta_{--})\Lambda_{-}^{(0)}\right)\right).$

######  Lemma 6.2.

Let $0<\alpha<\sqrt{2\beta}$ be arbitrarily small and $\beta<\infty$. The
following holds: $-\beta\in\mathop{\mathrm{spec_{s}}}H(\Gamma)$ iff
$\Gamma=\Gamma^{\circ}$;
$-\beta+\epsilon\notin\mathop{\mathrm{spec_{s}}}H(\Gamma)$
$(\epsilon\in\mathbb{R}\backslash\\{0\\}\;\text{small})$;
$\beta+\epsilon\notin\mathop{\mathrm{spec_{s}}}H(\Gamma)$
$(\epsilon\in\mathbb{R}\;\text{small})$.

###### Proof.

By (6.2), $\xi(-\beta)=1/\sqrt{2\beta}$. Put $E=-\beta$ in (6.14) and get a
zero-valued quadratic polynomial in $\alpha>0$. The monomial of degree $0$ is
precisely the condition in Theorem 6.1 ensuring that
$-\beta\in\mathop{\mathrm{spec_{s}}}H(\Gamma,0)$. The monomial of degree $1$
is absent, and the monomial of degree $2$ reads

(6.16)
$2\eta_{+-}\gamma^{(0)}=q_{-}(-\beta)(\omega_{+}^{(0)}+\sqrt{2\beta})+q_{+}(-\beta)\omega_{-}^{(0)},\quad
q_{s}(-\beta)=\omega_{s}^{(1)}-\frac{1-s/3}{2\sqrt{2\beta}}$

($s=\pm$). Using (6.10),

(6.17)
$\omega_{s}^{(1)}=\eta_{ss}\omega_{s}^{(0)}+4\pi(\Lambda_{s}^{(1)}-\eta_{ss}\Lambda_{s}^{(0)}).$

Let $x:=\omega_{+}^{(0)}+\sqrt{2\beta}$, $y:=\omega_{-}^{(0)}$. It follows
from (6.16) and (6.17) that

$\displaystyle 2\eta_{+-}\gamma^{(0)}=$
$\displaystyle(\eta_{++}+\eta_{--})xy-4\pi(\eta_{--}\Lambda_{-}^{(0)}-\Lambda_{-}^{(1)})x$
(6.18)
$\displaystyle-4\pi\left(\eta_{++}\left(\Lambda_{+}^{(0)}+\frac{\sqrt{2\beta}}{4\pi}\right)-\Lambda_{+}^{(1)}\right)y.$

Since $\eta_{++}+\eta_{--}=2\eta_{+-}$ and
$-\beta\in\mathop{\mathrm{spec_{s}}}H(\Gamma,0)\Rightarrow\gamma^{(0)}=xy$, it
follows from (6.18) that
$\mathcal{N}_{+}^{(0)\,2}Ax+\mathcal{N}_{-}^{(0)\,2}By=0$, which is equivalent
to $A\Gamma_{++}+B\Gamma_{--}+C=0$. This proves the first statement of the
lemma.

Let $E=-\beta+\epsilon$, $\epsilon>0$ small. By (6.2) and (6.15),

$\xi(E)=\frac{1}{\sqrt{2\beta}}\left(1-i\sqrt{\frac{\epsilon}{2\beta}}\right)+O(\epsilon),\quad
q_{s}(E)=q_{s}(-\beta)+i\sqrt{\epsilon}\frac{1-s}{4\beta}+O(\epsilon)$

($s=\pm$). Put these expressions in (6.14) and get that

(6.19) $\gamma^{(0)}=0,\quad\omega_{+}^{(0)}=-\sqrt{2\beta},\quad
q_{+}(-\beta)=0.$

Then, by the second equality in (6.16),
$\omega_{+}^{(1)}=(3\sqrt{2\beta})^{-1}$. Using the latter and (6.17),

(6.20)
$4\pi(\Lambda_{+}^{(1)}-\eta_{++}\Lambda_{+}^{(0)})-(3\sqrt{2\beta})^{-1}-\eta_{++}\sqrt{2\beta}=0,$

which is false, since by (6.7), (6.9), (6.10), the function on the left has a
single maximum $\approx-0.14874$ at $\beta\approx 1.00553$. When $\epsilon<0$,
(6.14) also leads to (6.19), hence
$-\beta+\epsilon\notin\mathop{\mathrm{spec_{s}}}H(\Gamma)$ for small nonzero
$\epsilon$.

Next $E=\beta+\epsilon$ implies that $\gamma^{(0)}=\omega_{-}^{(0)}=0$
(Theorem 6.1). When $\epsilon\geq 0$, we have that

$\xi(E)=\frac{i}{\sqrt{2\beta}}\left(1-\sqrt{\frac{\epsilon}{2\beta}}\right)+O(\epsilon)$

and

$q_{s}(E)=q_{s}(\beta)+i\sqrt{\epsilon}\frac{1+s}{4\beta}+O(\epsilon),\quad
q_{s}(\beta)=\omega_{s}^{(1)}-i\frac{1+s/3}{2\sqrt{2\beta}}$

($s=\pm$). Put these expressions in (6.14) and deduce that (6.14) fails for
$\epsilon>0$. When $\epsilon=0$, we obtain the system
$3\omega_{-}^{(1)}\omega_{+}^{(0)}=-1$ and
$6\beta\omega_{-}^{(1)}=\omega_{+}^{(0)}$. The system does not have solutions
$\omega_{+}^{(0)}$, $\omega_{-}^{(1)}$ in $\mathbb{R}$, hence
$\beta+\epsilon\notin\mathop{\mathrm{spec_{s}}}H(\Gamma)$ for $\epsilon\geq 0$
small. Applying the above procedure for $\epsilon<0$ we arrive at the same
conclusion. ∎

As previously, let $E=E^{(0)}+\epsilon(\alpha)$. By Lemma 6.2,
$E^{(0)}\neq\pm\beta$ unless $\alpha=0$. We also conclude from Theorem 6.1 and
Lemma 6.2 that $E<\beta$. Next we look for $\epsilon(\alpha)$ of the form
$\alpha E^{(1)}+\alpha^{2}E^{(2)}+\alpha^{3}E^{(3)}$ for some
$E^{(j)}\in\mathbb{R}$ ($j=1,2,3$). Put this $E$ in (6.2):

(6.21) $\xi(E)=\sum_{j=0}^{3}\alpha^{j}\xi_{j}+O(\alpha^{4})$

as $\alpha\downarrow 0$, where $\xi_{0}:=\xi(E^{(0)})$ and

(6.22a) $\displaystyle\xi_{1}:=$
$\displaystyle-\frac{\xi_{0}E^{(1)}}{2(E^{(0)}+2\beta^{2}\xi_{0}^{2})},$
(6.22b) $\displaystyle\xi_{2}:=$
$\displaystyle\frac{\xi_{0}}{8}\frac{E^{(1)\,2}(2\xi_{0}^{2}E^{(0)}-1)+8\xi_{0}^{2}E^{(2)}(\beta^{2}-E^{(0)\,2})}{(2\xi_{0}^{2}E^{(0)}+1)(\beta^{2}-E^{(0)\,2})},$
$\displaystyle\xi_{3}:=$
$\displaystyle-\frac{1}{32\xi_{0}(\beta^{2}-E^{(0)\,2})^{3}}\bigl{\\{}8\beta^{4}[E^{(3)}+\xi_{0}^{2}(E^{(1)}E^{(2)}+2E^{(0)}E^{(3)})]$
$\displaystyle-E^{(0)\,2}[8E^{(0)}E^{(1)}E^{(2)}(1+\xi_{0}^{2}E^{(0)})-8E^{(0)\,2}E^{(3)}(1+2\xi_{0}^{2}E^{(0)})$
$\displaystyle-E^{(1)\,3}(3+2\xi_{0}^{2}E^{(0)})]+\beta^{2}[8E^{(0)}E^{(1)}E^{(2)}-16E^{(0)\,2}E^{(3)}(1+2\xi_{0}^{2}E^{(0)})$
(6.22c) $\displaystyle+E^{(1)\,3}(1+6\xi_{0}^{2}E^{(0)})]\bigr{\\}}.$

Notice that each denominator in (6.22) is nonzero since $E^{(0)}\neq\pm\beta$.
Put (6.21) in (6.15) and get that

(6.23)
$q_{s}(E)=\sum_{j=0}^{3}\alpha^{j}q_{s}^{(j)}+O(\alpha^{4})\quad(s=\pm)$

as $\alpha\downarrow 0$, where $q_{s}^{(0)}:=q_{s}(E^{(0)})$ and

(6.24a) $\displaystyle q_{s}^{(1)}:=$
$\displaystyle-\xi_{1}(\tfrac{1}{2}-s\beta\xi_{0}^{2}),$ (6.24b)
$\displaystyle q_{s}^{(2)}:=$
$\displaystyle-\xi_{2}(\tfrac{1}{2}-s\beta\xi_{0}^{2})+s\beta\xi_{0}\xi_{1}^{2},$
(6.24c) $\displaystyle q_{s}^{(3)}:=$
$\displaystyle-\xi_{3}(\tfrac{1}{2}-s\beta\xi_{0}^{2})+\tfrac{1}{3}s\beta\xi_{1}(\xi_{1}^{2}+6\xi_{0}\xi_{2}).$

Put (6.21) and (6.23) in (6.14), and then collect the terms with the same
powers $<4$ of $\alpha$. Since a cubic polynomial in $\alpha>0$ is zero, set
each monomial to zero: A zero-valued monomial of degree $0$ coincides with
(6.5), and the rest zero-valued monomials of degree $1,2,3$ are of the form

(6.25a) $\displaystyle 0=$
$\displaystyle\xi_{1}\sum_{s}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right)\left(\omega_{s}^{(0)}+\frac{1}{2\xi_{0}}+s\beta\xi_{0}\right),$
$\displaystyle 0=$
$\displaystyle-2\eta_{+-}\gamma^{(0)}+\xi_{1}^{2}\prod_{s}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right)$
(6.25b)
$\displaystyle+\sum_{s}\left(\omega_{s}^{(0)}+\frac{1}{2\xi_{0}}+s\beta\xi_{0}\right)\left(q_{-s}^{(0)}+\frac{\xi_{1}^{2}-\xi_{0}\xi_{2}}{2\xi_{0}^{3}}-s\beta\xi_{2}\right),$
$\displaystyle 0=$
$\displaystyle\sum_{s}\left(\xi_{1}\left(-\frac{1}{2\xi_{0}^{2}}+s\beta\right)\left(q_{-s}^{(0)}+\frac{\xi_{1}^{2}-\xi_{0}\xi_{2}}{2\xi_{0}^{3}}-s\beta\xi_{2}\right)\right.$
(6.25c)
$\displaystyle\left.+\left(\omega_{s}^{(0)}+\frac{1}{2\xi_{0}}+s\beta\xi_{0}\right)\left(q_{-s}^{(1)}-\frac{\xi_{1}^{3}-2\xi_{0}\xi_{1}\xi_{2}+\xi_{0}^{2}\xi_{3}}{2\xi_{0}^{4}}-s\beta\xi_{3}\right)\right).$

Equations (6.25a) and (6.25b) imply that

######  Proposition 6.3.

$E^{(1)}=0$.

###### Proof.

Let $E^{(0)}<-\beta$. Then (6.25a) simplifies thus:

$0=\xi_{1}\sum_{s}\sqrt{s\beta-E^{(0)}}\left(\sqrt{s\beta-E^{(0)}}+\sqrt{-s\beta-E^{(0)}}\right)\left(\omega_{s}^{(0)}+\sqrt{s\beta-E^{(0)}}\right).$

Since $s\beta-E^{(0)}>0$ ($s=\pm$), it follows that

$\sqrt{s\beta-E^{(0)}}\left(\sqrt{s\beta-E^{(0)}}+\sqrt{-s\beta-E^{(0)}}\right)>0$

for both $s=\pm$. But then, either $\xi_{1}=0$ or
$\omega_{s}^{(0)}+\sqrt{s\beta-E^{(0)}}=0$. If the former, then, by (6.22a),
$E^{(1)}=0$. If the latter, then

(6.26)
$E^{(0)}=\beta-\omega_{+}^{(0)\,2},\quad\gamma^{(0)}=0,\quad\omega_{+}^{(0)}<-\sqrt{2\beta},\quad\omega_{-}^{(0)}=-\sqrt{\omega_{+}^{(0)\,2}-2\beta}.$

Under these conditions, (6.25b) fails unless $\xi_{1}=0$. Therefore,
$E^{(1)}=0$ in either case.

If $-\beta<E^{(0)}\leq 0$, then
$-\sqrt{2\beta}<\omega_{+}^{(0)}\leq-\sqrt{\beta}$. But then (6.25a) implies
either $\xi_{1}=0$ or $\omega_{+}^{(0)}=\omega_{-}^{(0)}$ and $\beta=0$. Since
$\beta>0$, $E^{(1)}=0$ necessarily. Finally, when $0<E^{(0)}<\beta$, (6.25a)
also implies that $\xi_{1}=0$ necessarily. ∎

By (6.24a) and Proposition 6.3, $q_{s}^{(1)}=0$, and the system (6.25) reduces
to (6.25b)–(6.25c). It follows that $\xi_{2}$ and $\xi_{3}$ are uniquely
defined by the latter system unless (6.26) holds. Otherwise we need more terms
of expansion. Thus, $O(\alpha^{4})$ in (6.4) further reads

(6.27)
$O(\alpha^{4})=\alpha^{4}\xi(E)^{3}(\tfrac{1}{6}-\tfrac{1}{5}s\beta\xi(E)^{2})+\alpha^{6}\xi(E)^{5}(\tfrac{1}{10}-\tfrac{1}{7}s\beta\xi(E)^{2})+O(\alpha^{8}).$

By definition, the normalization constant $\mathcal{N}_{s}$ depends on even
powers of $\alpha$. Then $O(\alpha^{4})$ in (6.11) is given by

(6.28)
$O(\alpha^{4})=\alpha^{4}\omega_{s}^{(2)}+\alpha^{6}\omega_{s}^{(3)}+O(\alpha^{8}).$

Although $\omega_{s}^{(2)}$ and $\omega_{s}^{(3)}$ can be found similar to
$\omega_{s}^{(1)}$ in (6.10), their explicit representation is inessential
since our final goal is to express $E$ with the accuracy up to
$O(\alpha^{4})$. Next, if we supplement $\epsilon(\alpha)$ with terms indexed
by $j=4,5,6$, then the values of functions $\xi$ and $q_{s}$ in (6.21) and
(6.23), respectively, are also supplemented with terms indexed by $j=4,5,6$.
Put these values along with (6.27) and (6.28) in (6.4), (6.11), and then in
(6.3). Then collect the terms with the same powers $<7$ of $\alpha$, and
deduce a zero-valued polynomial in $\alpha>0$ of degree $6$. Set each monomial
to zero, apply Proposition 6.3, the relation $q_{s}^{(1)}=0$, and condition
(6.26), and get that the zero-valued monomials of degree $4,5,6$ are of the
form:

(6.29a) $\displaystyle 0=$
$\displaystyle\prod_{s}\left(q_{-s}^{(0)}-\xi_{2}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right)\right),$
(6.29b) $\displaystyle 0=$
$\displaystyle\xi_{3}\left(2\xi_{2}\prod_{s}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right)-\sum_{s}q_{s}^{(0)}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right)\right),$
$\displaystyle 0=$ $\displaystyle
30\xi_{3}^{2}\prod_{s}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right)-\sum_{s}\left(q_{-s}^{(0)}-\xi_{2}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right)\right)$
(6.29c)
$\displaystyle\times\left(\xi_{0}^{5}\left(\frac{5}{\xi_{0}^{2}}-6s\beta\right)+30\left(\xi_{0}^{2}\xi_{2}+\xi_{4}\right)\left(\frac{1}{2\xi_{0}^{2}}-s\beta\right)-\frac{15\xi_{2}^{2}}{\xi_{0}^{3}}-30\omega_{s}^{(2)}\right).$

######  Proposition 6.4.

$E^{(3)}=0$.

###### Proof.

By Proposition 6.3 and (6.22c),

$\xi_{3}=\frac{E^{(3)}\left(2\xi_{0}^{2}E^{(0)}+1\right)}{4\xi_{0}\left(E^{(0)\,2}-\beta^{2}\right)}.$

Therefore, $\xi_{3}=0\Leftrightarrow E^{(3)}=0$. Since $q_{s}^{(1)}=0$ and
$\xi_{1}=0$, we see that (6.25c) coincides with (6.25a) but with $\xi_{1}$
replaced by $\xi_{3}$. If (6.26) does not hold, then we can redo all the steps
made up during the proof of Proposition 6.3 to conclude that $\xi_{3}=0$.
Hence $E^{(3)}=0$, as claimed. If (6.26) holds, then by (6.29b), either
$\xi_{3}=0$ or

(6.30)
$\xi_{2}=\tfrac{1}{2}\sum_{s}q_{s}^{(0)}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right)/\prod_{s}\left(\frac{1}{2\xi_{0}^{2}}+s\beta\right).$

Substitute (6.30) in (6.29a) and get that $\xi_{0}>0$ takes the value

(6.31)
$\xi_{0}=\sqrt{\frac{1}{2\beta}\frac{q_{-}^{(0)}-q_{+}^{(0)}}{q_{-}^{(0)}+q_{+}^{(0)}}}\quad\text{for}\quad\frac{q_{-}^{(0)}-q_{+}^{(0)}}{q_{-}^{(0)}+q_{+}^{(0)}}>0.$

Put this value in (6.30) and get that
$\xi_{2}=(q_{-}^{(0)}-q_{+}^{(0)})/(2\beta)$. Now put this $\xi_{2}$ and
(6.31) in (6.29c) and deduce that either $\xi_{3}=0$ or
$q_{+}^{(0)}q_{-}^{(0)}=0$. If the latter, then $q_{+}^{(0)}=0$, since the
inequality in (6.31) fails for $q_{-}^{(0)}=0$. If $q_{+}^{(0)}=0$, then
$\omega_{-}^{(0)}(\omega_{+}^{(0)}-\omega_{-}^{(0)})=0$. But
$\omega_{-}^{(0)}<0$ and therefore $\omega_{+}^{(0)}=\omega_{-}^{(0)}$, which
is true iff $\beta=0$, hence false. ∎

######  Proposition 6.5.

For $E^{(0)}<-\beta$,

$\displaystyle E^{(2)}=$
$\displaystyle\frac{2\sqrt{2}}{\beta}(\sqrt{E^{(0)\,2}-\beta^{2}}-E^{(0)})\sqrt{E^{(0)\,2}-\beta^{2}}\sqrt{-E^{(0)}-\sqrt{E^{(0)\,2}-\beta^{2}}}$
$\displaystyle\times\frac{\sum_{s}q_{-s}^{(0)}(\omega_{s}^{(0)}+\sqrt{s\beta-E^{(0)}})-2\eta_{+-}\gamma^{(0)}}{\sum_{s}\sqrt{s\beta-E^{(0)}}(\sqrt{s\beta-E^{(0)}}+\sqrt{-s\beta-E^{(0)}})(\omega_{s}^{(0)}+\sqrt{s\beta-E^{(0)}})}$
$\displaystyle\qquad\qquad\qquad(\gamma^{(0)}\neq 0);$ $\displaystyle=$
$\displaystyle-2\omega_{+}^{(0)}q_{+}^{(0)}\qquad(\gamma^{(0)}=0,\,\omega_{+}^{(0)}<-\sqrt{2\beta},\,\omega_{-}^{(0)}\neq-\sqrt{\omega_{+}^{(0)\,2}-2\beta});$
$\displaystyle=$
$\displaystyle-2\omega_{-}^{(0)}q_{-}^{(0)}\qquad(\gamma^{(0)}=0,\,\omega_{-}^{(0)}<0,\,\omega_{+}^{(0)}\neq-\sqrt{\omega_{-}^{(0)\,2}+2\beta}).$

Further, if $\gamma^{(0)}=0$, $\omega_{+}^{(0)}<-\sqrt{2\beta}$, and
$\omega_{-}^{(0)}=-\sqrt{\omega_{+}^{(0)\,2}-2\beta}$, then $E^{(2)}$ is
twofold, and it takes the values $-2\omega_{s}^{(0)}q_{s}^{(0)}$ $(s=\pm)$.

######  Remark 6.6.

It can be shown that, for $\Gamma$ diagonal and $E^{(0)}<-\beta$, the
following holds: $q_{-}^{(0)}<0$ for $E^{(0)}=-\beta-\omega_{-}^{(0)\,2}$,
$\omega_{-}^{(0)}<0$; $q_{+}^{(0)}<0$ for $E^{(0)}=\beta-\omega_{+}^{(0)\,2}$,
$\omega_{+}^{(0)}<-\sqrt{2\beta}$. By Proposition 6.5, this means the spin-
orbit term moves the eigenvalues down from the threshold $-\beta$.

###### Proof of Proposition 6.5.

Using Proposition 6.3, one finds from (6.22b) that

(6.32) $\xi_{2}=\frac{E^{(2)}\xi_{0}^{3}}{2\xi_{0}^{2}E^{(0)}+1}.$

When (6.26) does not hold, $E^{(2)}$ is found from (6.25b) and (6.32). If
(6.26) holds, then $E^{(2)}$ follows from (6.29a) and (6.32). ∎

######  Theorem 6.7.

Let $0<\alpha<\sqrt{2\beta}$ be arbitrarily small and $\beta<\infty$. Then the
part of the singular spectrum of $H(\Gamma,\alpha)$ that is below the
threshold $-\beta$ consists of the points
$E^{(0)}+\alpha^{2}E^{(2)}+O(\alpha^{4})$ where $E^{(0)}<-\beta$ belongs to
the singular spectrum of $H(\Gamma,0)$. Let $E\geq-\beta$ be in the singular
spectrum of $H(\Gamma,\alpha)$. Then $E=-\beta$ and $\Gamma=\Gamma^{\circ}$
necessarily; otherwise $E$ does not belong to the singular spectrum.

###### Proof.

By Propositions 6.3, 6.4, 6.5, the singular spectrum of $H(\Gamma,\alpha)$
consists of the points $E=E^{(0)}+\alpha^{2}E^{(2)}$ up to $O(\alpha^{4})$. To
prove the first part of the theorem, we need only show
$E^{(0)}<-\beta\Rightarrow E<-\beta$. Let $E^{(0)}=-\beta-\delta$, $\delta>0$
small. Then $E=-\beta+\epsilon$ where $\epsilon:=\alpha^{2}E^{(2)}-\delta$. By
Proposition 6.5, $E^{(2)}=c\sqrt{\delta}+O(\delta)$ for some real $c$ whose
explicit representation is inessential here. But then
$\epsilon<0\Leftrightarrow\delta>O(\alpha^{4})$, as claimed. To prove the
second part, we need to show $E^{(0)}\in(-\beta,\beta)\Rightarrow
E\notin\mathop{\mathrm{spec_{s}}}H(\Gamma,\alpha)$, as it follows from Lemma
6.2. By Theorem 6.1 and (6.25b),

(6.33) $q_{+}^{(0)}=\xi_{2}\left(\frac{1}{2\xi_{0}^{2}}-\beta\right)$

and we show that equation (6.33) does not have solutions with respect to
$\xi_{2}\in\mathbb{R}$. Let $-\beta<E^{(0)}<0$. The real and imaginary parts
of (6.33) imply that

(6.34a) $\displaystyle E^{(2)}=$
$\displaystyle-\frac{\omega_{+}^{(0)}}{3\beta^{2}}(6\beta^{2}\omega_{+}^{(1)}+\omega_{+}^{(0)}(5\beta-2\omega_{+}^{(0)\,2}))$
(6.34b) $\displaystyle=$
$\displaystyle\frac{1}{3\beta^{2}}(2\beta^{2}(2-3\omega_{+}^{(0)}\omega_{+}^{(1)})-\omega_{+}^{(0)}(7\beta-2\omega_{+}^{(0)\,2})).$

It follows that (6.34a) $=$ (6.34b) iff $2\beta=\omega_{+}^{(0)\,2}$, which
contradicts the initial hypothesis $-\sqrt{2\beta}<\omega_{+}^{(0)}<0$
(Theorem 6.1). If $E^{(0)}=0$, then $E^{(2)}=-\frac{1}{3}$ and
$\omega_{+}^{(1)}=\frac{1}{3}\beta^{-1/2}$. Put this $\omega_{+}^{(1)}$ along
with $\omega_{+}^{(0)}=-\sqrt{\beta}$ in (6.17) and get (6.20) with
$\sqrt{2\beta}$ replaced by $\sqrt{\beta}$. In this case the left-hand side
approaches $-\frac{5}{24}\beta^{-3/2}$ as $\beta\uparrow\infty$, hence false.
When $0<E^{(0)}<\beta$, $E^{(2)}$ satisfies (6.34), and we conclude that
$E^{(0)}+\alpha^{2}E^{(2)}\notin\mathop{\mathrm{spec_{s}}}H(\Gamma,\alpha)$
for all $E^{(0)}\in(-\beta,\beta)$. ∎

###  6.3. Case with large spin-orbit coupling

Here we wish to post some most important properties of solutions
$E\geq-\Sigma$ of (6.3) under hypothesis in (C). For this we find it
convenient to define the functions $(0,\nu]\to\mathbb{R}$ as

$\displaystyle U_{\nu}(x):=$
$\displaystyle\frac{1}{x}\left(\frac{\nu^{2}+1}{\nu^{2}}\arctan(x)-\frac{1}{x}\right),$
$\displaystyle V_{\nu}(x):=$
$\displaystyle\frac{\nu^{2}}{\nu^{2}+1}U_{\nu}(x)\left(2-\left(\nu^{2}-1\right)x^{2}U_{\nu}(x)\right).$

The parameter $\nu$ is defined as $\nu:=\alpha/\sqrt{2\beta}$, and it takes
the values

$1\leq\nu\leq\sqrt{\frac{1+\sqrt{1-\beta^{2}}}{\beta}}\quad(0<\beta\leq 1).$

Figure 2. The plots of functions $V_{\nu}$ (solid curves) and $U_{\nu}$
(dashed curves) at $0.6\leq x\leq 1.8$. The curve labeled by 1 shows the
values of $V_{1.0}=U_{1.0}$; the (solid and dashed) curves labeled by 2 show
the values of $V_{1.6}$ and $U_{1.6}$; the (solid and dashed) curves labeled
by 3 show the values of $V_{2.0}$ and $U_{2.0}$; the (solid and dashed) curves
labeled by 4 show the values of $V_{\infty}$ and $U_{\infty}$, i.e. $V_{\nu}$
and $U_{\nu}$ as $\nu\uparrow\infty$. The function $V_{\infty}$ takes the
value $-\infty$ everywhere except at $x_{\infty,1}\approx 1.16234$, which is
also a zero of $U_{\infty}$. For $\nu<\infty$, a zero of $U_{\nu}$ coincides
with the first zero, $x_{\nu,1}$, of $V_{\nu}$, and $V_{\nu}$ also has the
second zero, $x_{\nu,2}$, if $x_{\nu,2}\leq\nu$. In figure, $x_{1.6,2}\approx
1.61471>1.6$ and $x_{2.0,2}\approx 1.4018<2.0$.

The parameter $\nu$ is predetermined by the relations
$\alpha\geq\sqrt{2\beta}$ and $\Sigma\leq 1$, and by the definition of
$\Sigma$. Some typical plots of the functions $V_{\nu}$ and $U_{\nu}$ are
illustrated in Fig. 2. In particular, if $\beta>0$ is arbitrarily small, then
$\nu$ varies from $1$ to $\sqrt{2/\beta}$ depending on
$\sqrt{2\beta}\leq\alpha\leq 2$. Therefore, the curves in Fig. 2 labeled by 4
represent the case when the magnetic field $\beta$ approaches zero from above
and the spin-orbit-coupling strength $\alpha$ is large enough.

The function $V_{\nu}$ has at most two zeros, which we denote by $x_{\nu,1}$
and $x_{\nu,2}$. We assume that $x_{\nu,1}>0$ is the zero of $U_{\nu}$. Then
the second zero $x_{\nu,2}>0$ solves the equation
$U_{\nu}(x)=2/((\nu^{2}-1)x^{2})$ ($\nu>1$) provided that $x_{\nu,2}\leq\nu$.
The relation $x_{\nu,2}>\nu$ is meaningless because the domain of $V_{\nu}$ is
$(0,\nu]$, by definition. If $\nu=1$, then the function $V_{1}=U_{1}$ has only
one zero $x_{1,1}\approx 0.76538$. If $\nu\uparrow\infty$, we write
$V_{\infty}:=\lim V_{\nu}$, and $-V_{\infty}$ formally behaves like an ill-
defined Dirac delta function concentrated at a zero $x_{\infty,1}\approx
1.16234$ of $U_{\infty}$. In this latter case, the two zeros $x_{\infty,1}$
and $x_{\infty,2}$ coincide. In all other cases, i.e. for $1<\nu<\infty$, the
function $V_{\nu}$ has either two zeros, if $x_{\nu,2}\leq\nu$, or a single
zero, if otherwise. In general, $x_{\nu,1}\leq x_{\nu,2}$, and the equality
holds only if $\nu=\infty$.

Yet another quantity which we use to describe our main results is a function
$E_{\nu}\colon(0,\nu]\to[\beta,\infty)$ defined as

$E_{\nu}(x):=\frac{\nu^{4}+x^{4}}{2\left(\nu
x\right)^{2}}\beta\quad\text{and}\quad
E_{\nu,n}:=E_{\nu}(x_{\nu,n})\quad(n=1,2).$

The relations $x_{\nu,1}\leq x_{\nu,2}\leq\nu$ imply that $\beta\leq
E_{\nu,2}\leq E_{\nu,1}$. In particular, we have that $E_{1,1}\approx
1.14643\beta$. In the limit $\beta\downarrow 0$, the function $E_{\nu}(x)$
ranges from $0$ to $x^{-2}$ when $\nu$ ranges from $1$ to $\sqrt{2/\beta}$.
Choosing $\nu=\sqrt{2/\beta}\uparrow\infty$ (i.e. $\alpha=2$, $\beta\downarrow
0$) one deduces $E_{\infty,1}=E_{\infty,2}\approx 0.74018$.

######  Theorem 6.8.

Let $\sqrt{2\beta}\leq\alpha\leq\sqrt{2(1+\sqrt{1-\beta^{2}})}$, $0<\beta\leq
1$. The singular spectrum of $H(\Gamma)$, which is embedded in the continuous
spectrum $[-\Sigma,\infty)$, is given by the set

$\displaystyle\bigl{\\{}E_{\nu}(x)\in[\beta,E_{\nu,1}]\colon$ $\displaystyle
x\in[x_{\nu,1},\nu],\,\gamma=\omega_{+}\omega_{-}+(\beta/2)V_{\nu}(x),$
$\displaystyle
2\omega_{-}=x^{2}U_{\nu}(x)\sum_{s}\omega_{s}\left(\nu^{2}+s\right)\bigr{\\}}.$

For example, if $\beta$ is arbitrarily small and $\alpha$ large, say
$\alpha=2$, then $\nu$ approaches $\infty$ from below, and Theorem 6.8 tells
us that there is an eigenvalue, $E_{\infty,1}$($\approx 0.74018$), above the
threshold $-1$ iff $\gamma=\omega_{+}\omega_{-}$. This particular eigenvalue
fails to satisfy (b) and (c), so the corresponding eigenfunction in Lemma 5.1
has to be analytically continued applying the methods similar to those used in
[30].

###### Proof of Theorem 6.8.

Since $\mathrm{artanh}(\alpha\xi(-\Sigma))=\infty$, $E=-\Sigma$ does not solve
(6.3). If $-\Sigma<E\leq-\beta$, then $\alpha\xi(E)>1$,
$\alpha>\sqrt{2\beta}$, and $\mathrm{artanh}(\alpha\xi(E))$ is of the form
$r-i\pi/2$, where $r>0$ is the real part. Put this form in (6.3) and deduce
the system of equations

(6.35) $\gamma=\prod_{s}a_{s}-\prod_{s}b_{s},\quad 0=\sum_{s}a_{s}b_{s}$

where

$a_{s}:=\omega_{s}+\frac{1}{2\xi(E)}-r\left(\frac{\alpha}{2}-\frac{s\beta}{\alpha}\right),\quad
b_{s}:=\frac{\pi}{2}\left(\frac{\alpha}{2}+\frac{s\beta}{\alpha}\right)\quad(s=\pm).$

Now that $a_{s}\in\mathbb{R}$ and $b_{s}>0$, it follows that
$\gamma=-(b_{+}/b_{-})(a_{+}^{2}+b_{+}^{2})<0$, in contradiction to
$\gamma\geq 0$. Next, suppose that $-\beta<E<\beta$. In this case
$\xi(E)=R+iT$ where the real part $R>0$ while the imaginary part $T<0$ for
$-\beta<E<0$, and $T>0$ for $0\leq E<\beta$. If we let

$\displaystyle r:=$
$\displaystyle\mathop{\mathrm{Re}}\mathrm{artanh}(\alpha\xi(E))=\tfrac{1}{4}\ln\left(\frac{(1+\alpha
R)^{2}+(\alpha T)^{2}}{(1-\alpha R)^{2}+(\alpha T)^{2}}\right)\quad(r>0),$
$\displaystyle t:=$
$\displaystyle\mathop{\mathrm{Im}}\mathrm{artanh}(\alpha\xi(E))=\tfrac{1}{2}(\mathop{\mathrm{Arg}}(1+\alpha
R+i\alpha T)-\mathop{\mathrm{Arg}}(1+\alpha R-i\alpha T))$

then $t<0$ for $-\beta<E<0$, and $t>0$ for $0\leq E<\beta$. Put $\xi(E)=R+iT$
and $\mathrm{artanh}(\alpha\xi(E))=r+it$ in (6.3) and deduce the system (6.35)
where now

$a_{s}:=\omega_{s}+\frac{R}{2\lvert\xi(E)\rvert^{2}}-r\left(\frac{\alpha}{2}-\frac{s\beta}{\alpha}\right),\quad
b_{s}:=\frac{T}{2\lvert\xi(E)\rvert^{2}}+t\left(\frac{\alpha}{2}+\frac{s\beta}{\alpha}\right).$

We see that $a_{s}\in\mathbb{R}$ and $b_{s}<0$ for $-\beta<E<0$, and $b_{s}>0$
for $0\leq E<\beta$. But then (6.35) leads to the previous
$\gamma=-(b_{+}/b_{-})(a_{+}^{2}+b_{+}^{2})<0$. Finally, if $E\geq\beta$, then
$\xi=iT$, where $T(E)>0$ reads

$T(E):=\frac{1}{\sqrt{2}\beta}\sqrt{E-\sqrt{E^{2}-\beta^{2}}},\quad 0<T(E)\leq
T(\beta)=1/\sqrt{2\beta}.$

Then $\mathrm{artanh}(\alpha\xi)=i\arctan(\alpha T)$, and (6.3) reduces to
(6.35) where

$a_{s}:=\omega_{s},\quad
b_{s}:=\frac{1}{2T(E)}-\left(\frac{\alpha}{2}+\frac{s\beta}{\alpha}\right)\arctan(\alpha
T(E))$

and $a_{s},b_{s}\in\mathbb{R}$. Using the definition of $T$, we write
$E=E_{\nu}(x)$ where $x:=\nu T(E)\sqrt{2\beta}$ ($0<x\leq\nu$). When written
in terms of $\nu$, $x$, $U_{\nu}$, $V_{\nu}$, the system (6.35) coincides with
that defining the singular spectrum in the theorem. It remains to show that
$x\geq x_{\nu,1}$, i.e. $E_{\nu}(x)\leq E_{\nu,1}$. Let $\nu=1$. Since
$V_{1}=U_{1}$, it follows from (6.35) that
$\gamma=U_{1}(x)((\omega_{+}x)^{2}+(\beta/2))\geq 0$, hence $U_{1}(x)\geq 0$.
But then $x\in[x_{1,1},1]$ and $E_{1}(x)\in[\beta,E_{1,1}]$. Let $\nu>1$ and
$x\in(0,x_{\nu,1})$. In this case $U_{\nu}(x),V_{\nu}(x)<0$ and
$(\nu^{2}-1)x^{2}U_{\nu}(x)<2$. On the other hand, the system (6.35) implies
that $\omega_{-}=-c\omega_{+}$, $c>0$, and thus
$\gamma=-(\omega_{+}^{2}c+(\beta/2)\lvert V_{\nu}(x)\rvert)<0$. We conclude
that $x\in[x_{\nu,1},\nu]$. ∎

##  7\. Discussion

In the current paper, the main results concerning the eigenvalues (or else the
singular points) are summarized in Theorems 6.1, 6.7, 6.8. Other parts of the
spectrum, including the eigenfunctions, are discussed in Sec. 5 and
particularly in Lemma 5.1 and Theorem 5.2. Although we said nothing about the
eigenvalues below the threshold $-\Sigma$ when $\alpha\geq\sqrt{2\beta}$,
their existence can be seen from a general equation (6.3). For example, taking
$\Gamma:=-C^{-1}-R$ diagonal with the entries such that
$\omega_{+}=\omega_{-}\equiv\omega$, one finds that

(7.1)
$\omega+\sqrt{-E}=\frac{\alpha}{2}\mathrm{artanh}\left(\frac{\alpha}{2\sqrt{-E}}\right)$

(a) $\beta\sim 0$

(b) $\beta=1/2$

Figure 3. The eigenvalue $E$ of the perturbed Hamiltonian (2.10) as a function
of impurity scattering strength $C=c\mathbb{I}$ when the admissible matrix
$R=r\mathbb{I}$ with $r\approx-0.17850$. The spin-orbit-coupling strength
$\alpha=2$. When $c=0$, the perturbed Hamiltonian coincides with the free
Hamiltonian whose spectrum $[-1-\beta^{2}/4,\infty)$ is absolutely continuous.
When $c^{-1}=0$, the perturbed Hamiltonian is described by the Friedrichs
extension whose spectrum is that of the free Hamiltonian. In (a) the threshold
approaches $-1$ and the dashed line shows the eigenvalue $E\approx-1.43923$
corresponding to the coupling parameter $c$ whose absolute value is
arbitrarily large but finite. In (b) the threshold is $-17/16=-1.06250$, the
black (upper) solid curves represent the eigenvalue for $s=+$ in (6.3), the
blue (lower) solid curves represent the eigenvalue for $s=-$, and the black
dashed (upper) and blue dashed (lower) lines show the eigenvalues
$E\approx-1.37956$ (for $s=+$) and $E\approx-1.60313$ (for $s=-$)
corresponding to $c$ whose absolute value is arbitrarily large but finite.

provided that the magnetic field $\beta$ is arbitrarily close to $0$. In this
regime, the eigenvalue $E<-\Sigma$ exists if $\Gamma$ is of the form
$v\mathbb{I}$ for some real $v$, where $\mathbb{I}$ is the identity matrix;
this follows from the conditions $\omega_{+}=\omega_{-}$ and $\beta\sim 0$ and
the definition in (6.1), which in turn gives a one-to-one correspondence
between $\omega$ and $v$. Note that, when $\alpha=0$, (7.1) gives the solution
$E=-\omega^{2}$, $\omega<0$, which is (by no accident) in exact agreement with
that described in Theorem 6.1 for $\beta=0$. Also, $v\mathbb{I}\neq-R$,
because $v\mathbb{I}=-R$ may correspond only to the Friedrichs extension
(Proposition 2.1) whose spectrum is only absolutely continuous. However, if
one takes the coupling parameter of contact interaction of the form
$C=c\mathbb{I}$ with $c$ nonzero and real, then necessarily $R=r\mathbb{I}$
with $r$ real. It then follows that $v\sim-r$ for $\lvert c\rvert$ arbitrarily
large but finite. For a given $r$, there exists $\omega$ corresponding to
$v\sim-r$, so (7.1) shows that the perturbed Hamiltonian with arbitrarily
large but finite impurity scattering strength $\lvert c\rvert$ has an
eigenvalue below the threshold. For example, when $\alpha=2$ and $\omega=0$,
the parameter $-v=(\mathrm{arcosh}(3)-2\sqrt{2})/(2\sqrt{2}+\pi)$ approximates
$r$, and the eigenvalue $E\approx-1.43923$ (Fig. 3(a)). When $\omega=0$ and
$\beta\sim 0$, there also exists an eigenvalue $E_{\infty,1}\approx 0.74018$
above the threshold (Theorem 6.8), but the same does not apply to the case
when $\beta=0$. This explains why we chose $\beta$ arbitrarily small but
nonzero. However, the above considerations remain valid for $\beta=0$ as well,
with the exception that now the hypothesis of Theorem 6.8 fails. For
comparison reasons, in Fig. 3(b) we plot the eigenvalues corresponding to
$s=\pm$ in (6.3) as functions of $c$ when the magnetic field is nonzero. The
present discussion is easy to extend to a general case when
$\omega_{+}\neq\omega_{-}$ and $\gamma\neq 0$. Moreover, the considerations
apply to the case when $\alpha<\sqrt{2\beta}$. One can show similar plots
either solving equation (6.3) numerically or applying analytic expressions in
Theorem 6.7 for $\alpha$ sufficiently small. For the mean field analysis of
eigenvalues at large $\lvert c\rvert$, but in dimensions one and two, the
reader may also refer to [32, 29, 35], where the case when
$\alpha<\sqrt{2\beta}$ is studied.

We recall that in our model, when $\alpha\geq\sqrt{2\beta}$, $2$ is the
maximum possible value of $\alpha$, as well as $1$ is the maximum possible
value of $\beta$. This is due to the restriction $\Sigma\leq 1$ caused by the
absolute convergence of the hypergeometric series which defines the Green
function for the free Hamiltonian. However, an additional procedure of
analytic continuation, similar to that when $\alpha<\sqrt{2\beta}$ is small,
may probably lead to larger $\alpha$ as well. When applied to concrete schemes
that propose the creation of spin-orbit coupling in cold atomic gases, the
parameters $\alpha$ and $\beta$ are expressed in terms of the quantities which
are measurable experimentally either directly or indirectly. For example,
using $\hbar=1$ and $m=1/2$ ($m$ is the mass of the particle), the spin-orbit
coupling for the Rashba Hamiltonian constructed in [12] is characterized by
$\alpha=k_{\mathrm{eff}}/2$, where $k_{\mathrm{eff}}$ ($\approx 1\mu$m-1 in SI
units) points to the strength of the magnetic field gradient. Without the
additional procedure of continuation our model can therefore be applied
straightforwardly up to $k_{\mathrm{eff}}=4$. Likewise, in the experimental
setup proposed in [22], the spin-orbit coupling is given by
$\alpha=\sqrt{E_{R}}/2$, where $E_{R}$ is the two-photon recoil energy due to
Raman process. Hence, in the regime $\alpha\geq\sqrt{2\beta}$, one would
expect $E_{R}\leq 16$. In [35] one can find some realistic parameters
suggested for observing bound states when using the scheme in [46] with
$\alpha\approx 2\sqrt{E_{R}}$ and $\beta\gtrsim 4.2E_{R}$.

While the point spectrum of the Rashba Hamiltonian with spin-dependent delta-
like impurity scattering depends on the admissible matrix $R$, the main
conclusion, which is independent of $R$, is that: 1) The Hamiltonian does not
have eigenvalues above the threshold provided that the spin-orbit-coupling
strength is small enough ($\alpha<\sqrt{2\beta}$) but nonzero. 2) The
Hamiltonian does not have eigenvalues located in-between the threshold and the
minimum of the upper branch of dispersion provided that the spin-orbit-
coupling strength is large enough ($\alpha\geq\sqrt{2\beta}$). 3) The maximum
possible eigenvalue that the Hamiltonian can reach is $E_{\nu,1}$, and
$E_{\nu,1}\geq\beta$ depends only on $\alpha\geq\sqrt{2\beta}$ and $\beta>0$.

## Acknowledgments

The author acknowledges the anonymous referees whose remarks have helped in
improving the presentation. The author also thanks the referee of the previous
versions of the paper.

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