The Livšic function of a homogeneous symmetric operator and the multiplication theorem

Abstract

This paper presents a solution to the Jørgensen–Muhly problem for a homogeneous symmetric operator with deficiency indices (1, 1) that does not admit a homogeneous self-adjoint extension. Based on the Livšic function approach, we characterize the set of all the solutions of the Jørgensen–Muhly problem up to unitary equivalence and describe the complete set of the corresponding unitary invariants.

1 Introduction

In [17], Jørgensen and Muhly formulated the problem of complete (intrinsic) characterization (up to unitary equivalence) of all symmetric solutions $\dot{A}$, $\dot{A}\subset (\dot{A})^*$, of the infinitesimal Weyl commutation relations

$$\begin{aligned} U_t\dot{A}U_t^*=\dot{A}+t I \quad \text {on } \quad \textrm{Dom}(\dot{A}), \quad t\in {\mathbb {R}}, \end{aligned}$$

(1.1)

where $U_t$ is a strongly continuous group of unitary operators (see [24, 31, 36] for the background). In [29], we gave a complete solution to the Jørgensen–Muhly problem in the case where the symmetric operators $\dot{A}$ have deficiency indices (1, 1), and also classified the pairs $(\dot{A}, U_t)$ satisfying the relations (1.1) up to mutual unitary equivalence.

An analogous problem arises for the homogeneous commutation relations of the type

$$\begin{aligned} U_t\dot{A}U_t^*=e^t\dot{A} \quad \text {on } \quad \textrm{Dom}(\dot{A}), \quad t\in {\mathbb {R}}: \end{aligned}$$

(1.2)

Characterize all symmetric operators $\dot{A}$ satisfying (1.2) for some one-parameter strongly continuous unitary groups $U_t$ up to unitary equivalence.

In [30], we gave a solution to the Jørgensen–Muhly problem for the homogeneous commutation relations (1.2) under the hypothesis that the symmetric operator $\dot{A}$ has deficiency indices (1, 1), and admits a self-adjoint homogeneous extension with respect to the group $U_t$.

In the current paper, we classify symmetric solutions $\dot{A}$ of homogeneous commutation relations (1.2) under the additional hypothesis that $\dot{A}$ has no self-adjoint extensions with a $U_t$-invariant domain which, combined with the results of [30], completes the solution of the Jørgensen–Muhly Problem in the case of homogeneous commutation relations (1.2).

Notice that the homogeneous operators are also called scale-covariant and the unitary (scaling) group $U_t$ is often referred to as the group of symmetries. The obvious reason for using such terminology is that the domain of symmetric solutions to the commutation relations is required to be $U_t$-invariant. In this setting, the situation where a scale-covariant symmetric operator does not admit scale-covariant self-adjoint extensions is emphasized as the case of (spontaneously) broken symmetry.

It is worth mentioning that there is an extensive literature devoted to the study of concrete realizations of homogeneous operators, see, e.g., [6, 9, 11,12,13, 15, 19], and references therein, and also our recent work [25, 28, 30] where the study of homogeneous operators in the abstract setting has been undertaken.

The current paper approach is based on the following key observations.

If for a given homogeneous symmetric operator $\dot{A}$ with deficiency indices (1, 1) symmetry is broken, that is, $\dot{A}$ does not have homogeneous self-adjoint extensions with respect to the symmetry group, then one can always find (a unique) maximal dissipative extension ${{\widehat{A}}}$ of $\dot{A}$ which is homogeneous with respect to that group [28]. In this case, the characteristic function $S_{{{\widehat{A}}}}(z)$ of the dissipative operator ${{\widehat{A}}}$ can be shown to either coincide with:

• (i) the characteristic function $S_{{{\widehat{H}}}(\nu )}(z)$ of a unique dissipative realization ${{\widehat{H}}}(\nu )$ of the Schrödinger operator with a singular potential

  $$\begin{aligned} {{\widehat{H}}}(\nu )=-\frac{d^2}{dx^2}+\frac{\nu ^2-\frac{1}{4}}{x^2} \quad \text {for some}\quad \nu \in i{\mathbb {R}}_+, \end{aligned}$$

  that is homogeneous with respect to the unitary group $U_t$ of scaling transformations

  $$\begin{aligned} (U_tf)(x)=e^{-\frac{t}{4}}f(e^{-\frac{t}{2}}x), \quad f\in L^2({\mathbb {R}}_+); \end{aligned}$$

  or

• (ii) the characteristic function $S_{-({{\widehat{H}}}(\mu ))^*}(z)$ of the dissipative operator $-({{\widehat{H}}}(\mu ))^*$ for some $\mu \in i{\mathbb {R}}_+$;

  otherwise,

• (iii) $S_{{{\widehat{A}}}}(z)$ can be expressed as the product of the characteristic functions $S_{{{\widehat{H}}}(\nu )}(z)$ and $S_{-({{\widehat{H}}}(\mu ))^*}(z)$ from (i) and (ii), respectively.

Notice that whenever $\nu \in i {\mathbb {R}}_+$, we have

$$\begin{aligned} \nu ^2-\frac{1}{4}<-\frac{1}{4} \end{aligned}$$

and, therefore, the corresponding symmetric operator $\dot{H}(\nu )$ is not semi-bounded from below.

The core of our considerations is based on making use the multiplication theorem from [27]. In particular, we show that along with the symmetric operators $\dot{H}(\nu )$ and $-\dot{H}(\mu )$, the symmetric part ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ of an appropriate operator coupling of the dissipative operators $\widehat{H}(\nu )$ and $-({\widehat{H}}(\mu ))^*$ serves as a model solution of the homogeneous commutation relations in the case where scale symmetry is broken (see (6.17) and (6.18) for the definition of a symmetric part of a dissipative operator). (We refer to [27] where the concept of an operator coupling of two unbounded dissipative operators has been discussed in detail.)

The paper is organized as follows.

In Sect. 2, following our work [26], we recall the concept of the Weyl–Titchmarsh and Livšic functions associated with the pair ($\dot{A}, A)$, where $\dot{A}$ is a symmetric operator with deficiency indices (1, 1) and A is a reference self-adjoint extension of $\dot{A}$. Given a quasi-selfadjoint extension ${{\widehat{A}}}$ of $\dot{A}$, we also recall a relevant functional model for the triple $(\dot{A}, {{\widehat{A}}},A)$, see Theorem 2.1.

In Sect. 3, we find a necessary condition for a contractive function in the upper half-plane to be the characteristic function of the triple $(\dot{A}, {{\widehat{A}}},A)$ where both the symmetric operator $\dot{A}$ and its maximal dissipative non-self-adjoint extension ${\widehat{A}}$ are homogeneous operators with respect to some strongly continuous group of unitary operators.

In Sect. 4, we calculate the Livšic function associated with the symmetric operator $\dot{H}(\nu )$ on the positive semi-axis with a singular potential and its special self-adjoint (reference) extension $H(\nu )$, see Theorem 4.1. We also show that the corresponding Weyl–Titchmarsh function has a geometric progression of simple poles on the negative real axis which illustrates the well-known result that the self-adjoint operator $H(\nu )$, $\nu \in i{\mathbb {R}}_+$ has infinite negative discrete spectrum, and is not semi-bounded from below. We refer to [20, Sec. 35], [32], and [8] where physical aspects of the relevant “fall to the center–the Efimov effect" duality phenomenon are discussed in depth.

In Sect. 5, we provide a more detailed study of the uniquely determined quasi-selfadjoint dissipative ${{\widehat{H}}}(\nu )$ and cumulative $({{\widehat{H}}}(\nu ))^*$ homogeneous extensions of the symmetric operator $\dot{H}(\nu )$, and describe the domain of these extensions in terms of asymptotic boundary conditions.

In Sect. 6, in the framework of the operator coupling and triangular representations theory [22] (also see [2, 5, 23]), we introduce a two-parameter family of model symmetric operators ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ with deficiency indices (1, 1) (see eq. (6.9)). We also show that the operators ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ are homogeneous with respect to a modified group of scaling transformations ${\mathbb {U}}_t$ in $L^2({\mathbb {R}}_+)\oplus L^2({\mathbb {R}}_+)$, see Theorem 6.2.

In Sect. 7, we (i) evaluate the characteristic (Livšic) function of the symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$, (ii) prove that ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ is a prime operator, and (iii) establish that symmetry with respect to group ${\mathbb {U}}_t$ is broken (whenever $\mu \ne \nu$).

In Sect. 8, we show that along with the one-parameter families

$$\begin{aligned} i{\mathbb {R}}_+\ni \nu \mapsto \dot{H}(\nu )\quad \text {and}\quad i{\mathbb {R}}_+\ni \mu \mapsto -\dot{H}(\mu ), \end{aligned}$$

the two-parameter family of symmetric operators

$$\begin{aligned} i{\mathbb {R}}_+\times i{\mathbb {R}}_+\ni (\mu , \nu )\mapsto {{\dot{{\mathbb {H}}}}}(\mu , \nu ), \quad (\mu \ne \nu ) \end{aligned}$$

augmented by the differentiation operator on the real axis $\dot{{\mathbb {D}}}=i\frac{d}{dx}$ (see Eqs. (7.13) and (7.14)) exhaust the list of unitarily non-equivalent solutions to the homogeneous commutation relations (1.2) with broken scaling symmetry. The main result of the paper, Theorem 8.1, presents the solution of the homogeneous Jørgensen–Muhly problem in the case in question.

The obtained results combined with those in [30] (in the case where symmetry is not broken) complete the classification (up to unitary equivalence) of the set of prime symmetric solutions with deficiency indices (1, 1) to the homogenous commutation relations, regardless of whether scale symmetry is broken or not.

2 Preliminaries and basic definitions

Let $\dot{A}$ be a densely defined symmetric operator with deficiency indices (1, 1).

Given deficiency elements $g_\pm \in \textrm{Ker}((\dot{A})^*\mp iI)$ such that

$$\begin{aligned} \Vert g_\pm \Vert =1, \end{aligned}$$

suppose that A is the (reference) self-adjoint extension of $\dot{A}$ uniquely determined by the requirement

$$\begin{aligned} g_+-g_-\in \textrm{Dom}(A). \end{aligned}$$

(2.1)

Denote by $s(z)=s_{(\dot{A}, A)}(z)$ the Livšic function associated with the pair $(\dot{A}, A)$ [26],

$$\begin{aligned} s(z)=\frac{z-i}{z+i}\cdot \frac{(g_z, g_-)}{(g_z, g_+)}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(2.2)

where $g_z$ is a deficiency element such that

$$\begin{aligned} 0\ne g_z\in \textrm{Ker}( (\dot{A})^*- zI), \quad z\in {\mathbb {C}}_+. \end{aligned}$$

Recall that the Livšic function can alternatively be evaluated via one more important attribute of the extension theory for $\dot{A}$ which is the Weyl–Titchmarsh function M(z) associated with the pair $(\dot{A}, A)$ and defined as

$$\begin{aligned} M(z)= ((Az+I)(A-zI)^{-1}g_+,g_+), \quad z\in {\mathbb {C}}_+. \end{aligned}$$

(2.3)

Namely, we have the equality (see, e.g., [26])

$$\begin{aligned} s(z)=\frac{M(z)-i}{M(z)+i},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

(2.4)

which shows that both the Livšic and the Weyl–Titchmarsh functions are equally suitable for the classification of the pairs $(\dot{A}, A)$ (up to mutual unitary equivalence). Notice that when the reference self-adjoint operator A changes, the Weyl–Titchmarsh function undergoes a linear-fractional transformation determined by the von Neumann extension parameter, while the Livšic function acquires a z-independent phase factor only.

Next, suppose that ${{\widehat{A}}} \ne ({{\widehat{A}}} )^*$ is a maximal dissipative extension of $\dot{A}$,

$$\begin{aligned} {\textrm{Im}}({{\widehat{A}}} f,f)\ge 0, \quad f\in \textrm{Dom}({{\widehat{A}}} ). \end{aligned}$$

Recall that since $\dot{A}$ is symmetric, its dissipative extension ${{\widehat{A}}}$ is automatically quasi-selfadjoint [33], that is,

$$\begin{aligned} \dot{A} \subset {\widehat{A}} \subset (\dot{A})^*. \end{aligned}$$

In particular,

$$\begin{aligned} g_+-\varkappa g_-\in \textrm{Dom}({{\widehat{A}}} )\quad \text {for some } |\varkappa |<1. \end{aligned}$$

(2.5)

By definition, we call $\varkappa$ the von Neumann parameter of the triple $(\dot{A}, {{\widehat{A}}},A)$ and define the characteristic function $S(z)=S_{(\dot{A}, {{\widehat{A}}}, A)}(z)$ associated with the triple $(\dot{A}, {{\widehat{A}}},A)$ as follows (see [29], cf. [21])

$$\begin{aligned} S(z)=\frac{s(z)-\varkappa }{{\overline{\varkappa }}\,s(z)-1}, \quad z\in {\mathbb {C}}_+. \end{aligned}$$

(2.6)

In fact, the characteristic function S(z) of a triple of operators appears to be a convenient functional parameter of the triple. For instance, since $s(i)=0$, it is easily seen that the von Neumann parameter of the triple $\varkappa$ can be evaluated as

$$\begin{aligned} \varkappa =S(i) \end{aligned}$$

(2.7)

and then the Livšic function can be recovered from the characteristic function S(z) as

$$\begin{aligned} s(z)=\frac{S(z)-\varkappa }{\overline{ \varkappa }\,S(z)-1}=\frac{S(z)-S(i)}{\overline{ S(i)}\,S(z)-1}, \quad z\in {\mathbb {C}}_+. \end{aligned}$$

(2.8)

In this case, we also have the relation (see (2.4))

$$\begin{aligned} M(z)=\frac{1}{i}\cdot \frac{s(z)+1}{s(z)-1},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(2.9)

Next, recall that the function M(z) admits the representation

$$\begin{aligned} M(z)=\int _{\mathbb {R}}\left( \frac{1}{\lambda -z}-\frac{\lambda }{1+\lambda ^2}\right) d\mu (\lambda ), \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(2.10)

for some infinite Borel measure $\mu (d\lambda )$ such that

$$\begin{aligned} \mu ({\mathbb {R}})=\infty \end{aligned}$$

(2.11)

and (see, e.g., [26])

$$\begin{aligned} \int _{\mathbb {R}}\frac{d\mu (\lambda )}{1+\lambda ^2}=1. \end{aligned}$$

(2.12)

Notice that type $[\mu (d\lambda )]$ of the measure $d\mu (\lambda )$ coincides with the spectral measure type $[E_A(d\lambda )]$ from the Spectral Theorem

$$\begin{aligned} A=\int _{\mathbb {R}}\lambda \,dE_A(\lambda ) \end{aligned}$$

for the reference self-adjoint operator A, provided that the underling symmetric operator $\dot{A}$ is prime,¹ see, e.g., [34].

In particular, the Weyl–Titchmarsh function M(z) completely determines the character of the spectrum of the reference self-adjoint operator A, while the “singularities” of the characteristic function S(z) describe the spectrum of the dissipative operator ${{\widehat{A}}}$ (see [21] and [1, Theorem 1, Sec. 114]) from the triple $(\dot{A}, {{\widehat{A}}},A)$.

Moreover, the knowledge of the characteristic function S(z) as a complete unitary invariant of a triple of operators determines the triple up to mutual unitary equivalence.

To recall the corresponding uniqueness result, introduce the model triple $({{\dot{{{\mathcal {B}}}}}}, {{\widehat{{{\mathcal {B}}}}}},{{\mathcal {B}}})$ parameterized by the characteristic function $S_{\mathfrak {B}}(z)=S(z)$.

Let ${{\mathcal {B}}}$ be the (self-adjoint) multiplication operator by independent variable in the Hilbert space $L^2({\mathbb {R}};d\mu )$ defined on

$$\begin{aligned} \textrm{Dom}({{\mathcal {B}}})&=\left\{ f\in \,L^2({\mathbb {R}};d\mu ) \,\bigg | \, \int _{\mathbb {R}}\lambda ^2 | f(\lambda )|^2d \mu (\lambda )<\infty \right\} \nonumber \\&=\textrm{Dom}({{\dot{{{\mathcal {B}}}}}})\dot{+}\mathrm {lin\ span}\left\{ \,\frac{1}{\lambda -i}- \frac{1}{\lambda +i}\right\} , \end{aligned}$$

(2.13)

where $\mu (d\lambda )$ is the measure from the integral representation (2.10) for the corresponding Weyl–Titchmarsh function.

Also, denote by ${{\dot{{{\mathcal {B}}}}}}$ the symmetric restriction of ${{\mathcal {B}}}$ on

$$\begin{aligned} \textrm{Dom}({{\dot{{{\mathcal {B}}}}}})=\left\{ f\in \textrm{Dom}({{\mathcal {B}}})\, \bigg | \, \int _{\mathbb {R}}f(\lambda )d \mu (\lambda ) =0\right\} , \end{aligned}$$

(2.14)

and let ${{\widehat{{{\mathcal {B}}}}}}$ be the dissipative quasi-self-adjoint extension of ${{\dot{{{\mathcal {B}}}}}}$ on

$$\begin{aligned} \textrm{Dom}({{\widehat{{{\mathcal {B}}}}}})=\textrm{Dom}({{\dot{{{\mathcal {B}}}}}})\dot{+}\mathrm {lin\ span}\left\{ \,\frac{1}{\lambda -i}- \varkappa \frac{1}{\lambda +i}\right\} , \end{aligned}$$

(2.15)

where $\varkappa$ is the von Neumann parameter of the triple (2.7).

Theorem 2.1

([26, Theorems 1.4, 4.1], cf. [21]) Suppose that $\dot{A}$ is a prime, closed, densely defined symmetric operator with deficiency indices (1, 1). Assume that A and ${{\widehat{A}}}$ are self-adjoint and quasi-selfadjoint dissipative extensions of $\dot{A}$, respectively. Suppose that S(z) is the characteristic functions of the triple $(\dot{A},{{\widehat{A}}}, A)$ and let $({{\dot{{{\mathcal {B}}}}}}, {{\widehat{{{\mathcal {B}}}}}},{{\mathcal {B}}})$ be the model triple in the Hilbert space $L^2({\mathbb {R}};d\mu )$ parameterized by the characteristic function S(z).

Then, the triple $(\dot{A},{{\widehat{A}}}, A)$ is mutually unitarily equivalent to the model triple $({{\dot{{{\mathcal {B}}}}}}, {{\widehat{{{\mathcal {B}}}}}}, {{\mathcal {B}}})$ in the Hilbert space $L^2({\mathbb {R}};d\mu )$.

Moreover, arbitrary triples $(\dot{A},{{\widehat{A}}}, A)$ and $(\dot{B},{{\widehat{B}}}, B)$ in the Hilbert spaces ${{\mathcal {H}}}_A$ and ${{\mathcal {H}}}_B$, with $\dot{A}$ and $\dot{B}$ prime symmetric operators, are mutually unitarily equivalent² if, and only if, the corresponding characteristic functions of the triples coincide.

Throughout this paper, to facilitate some statement formulations, along with the Livšic function $s_{(\dot{A}, A)}(z)$ of the pair and the characteristic function $S_{(\dot{A},{{\widehat{A}}}, A)}(z)$ of the triple, we will also operate with the customary notion of characteristic function $s_{\dot{A}}(z)$ and $S_{\widehat{A}}(z)$ of the corresponding sole symmetric $\dot{A}$ and dissipative operator ${{\widehat{A}}}$, respectively, originally introduced modulo a phase factor, see [21]. For instance, we use the congruence notation

$$\begin{aligned} s_{\dot{A}}(z) \cong s(z)\quad \text {or}\quad S_{{\widehat{A}}}(z) \cong S(z), \quad z\in {\mathbb {C}}_+, \end{aligned}$$

whenever

$$\begin{aligned} s(z) =\Theta s_{(\dot{A}, A)}(z)\quad \text {or}\quad S(z) =\Theta S_{(\dot{A}, {{\widehat{A}}}, A)}(z), \quad z\in {\mathbb {C}}_+, \end{aligned}$$

for some z-independent unimodular factor $\Theta$. In this context, we also refer to [21] where it was shown that the knowledge (up to a unimodular constant factor) of the characteristic functions s(z) and S(z) of a (prime) symmetric operator $\dot{A}$ and its maximal dissipative extension ${{\widehat{A}}}$, respectively, characterizes the operators up to unitary equivalence.

3 The characteristic function of a homogeneous dissipative operator

Theorem 3.1

Assume that $\dot{A}$ is a closed, densely defined symmetric operator with deficiency indices (1, 1) and ${{\widehat{A}}}$ is a maximal dissipative extension of $\dot{A}$ such that ${{\widehat{A}}}\ne ({{\widehat{A}}})^*$. Suppose that the following commutation relations

$$\begin{aligned} U_t \dot{A}U_t^*=e^t\dot{A}\quad \text {on }\quad \textrm{Dom}( \dot{A}) \end{aligned}$$

(3.1)

and

$$\begin{aligned} U_t {{\widehat{A}}}U_t^*=e^t{{\widehat{A}}}\quad \text {on }\quad \textrm{Dom}( {{\widehat{A}}}) \end{aligned}$$

(3.2)

hold for some one-parameter strongly continuous group $U_t$ of unitary operators.

If A is a reference self-adjoint extension of $\dot{A}$, then the characteristic function $S_{\mathfrak {A}}(z)$ of the triple ${\mathfrak {A}}=(\dot{A},{{\widehat{A}}}, A)$ is either identically zero in the upper half-plane,

$$\begin{aligned} S_{\mathfrak {A}}(z)=0,\quad z\in {\mathbb {C}}_+, \end{aligned}$$

(3.3)

or it admits the representation

$$\begin{aligned} S_{\mathfrak {A}}(z)=e^{i\phi } z^\mu \left( -\frac{1}{z}\right) ^\nu ,\quad z\in {\mathbb {C}}_+, \end{aligned}$$

(3.4)

for some $\phi \in {\mathbb {R}}$ and some $\mu , \nu \in i{\mathbb {R}}_+\cup \{0\}$, not both zero.

Proof

From the invariance principle with respect to linear transformations for dissipative operators (see [29, Theorem F.1, Appendix F]), it follows that the characteristic functions $S_{{\mathfrak {A}}}(z)$ and $S_{e^t{\mathfrak {A}}}(z)$ of the triples ${\mathfrak {A}}=(\dot{A},{{\widehat{A}}}, A)$ and $e^t{\mathfrak {A}}=(e^t\dot{A},e^t{{\widehat{A}}}, e^t A),$ respectively, are related as

$$\begin{aligned} S_{e^t{\mathfrak {A}}}(e^tz)=\Theta _tS_{{\mathfrak {A}}}(z), \quad z\in {\mathbb {C}}_+,\quad t\in {\mathbb {R}}, \end{aligned}$$

(3.5)

where $\Theta _t$ is a unimodular continuous function in t.

From (3.1) and (3.2), it follows that

$$\begin{aligned} e^t{\mathfrak {A}}=(e^t\dot{A},e^t{{\widehat{A}}}, e^t A)=(U_t\dot{A}U_t^*,U_t\widehat{A}U_t^*, e^t A). \end{aligned}$$

Observing that the triples $e^t{\mathfrak {A}}$ and ${\mathfrak {A}}_t=(U_t\dot{A}U_t^*,U_t{{\widehat{A}}}U_t^*, U_t AU_t^*)$ differ only with their reference operators $e^t A$ and $U_t AU_t^*$, respectively, we conclude that their characteristic functions coincide up to a unimodular (z-independent) factor. That is,

$$\begin{aligned} S_{e^t{\mathfrak {A}}}(z)={{\widehat{\Theta }}}_tS_{{\mathfrak {A}}_t}(z),\quad z\in {\mathbb {C}}_+,\quad t\in {\mathbb {R}}, \end{aligned}$$

(3.6)

for some unimodular ${{\widehat{\Theta }}}_t$ function in t.

On the other hand, since the triples ${\mathfrak {A}}$ and ${\mathfrak {A}}_t$ are mutually unitarily equivalent, their characteristic functions coincide,

$$\begin{aligned} S_{{\mathfrak {A}}_t}(z)=S_{{\mathfrak {A}}}(z), \quad z\in {\mathbb {C}}_+. \end{aligned}$$

(3.7)

Combining (3.5), (3.6) and (3.7), we get

$$\begin{aligned} \Theta _tS_{{\mathfrak {A}}}(z)=S_{e^t{\mathfrak {A}}}(e^tz)=\widehat{\Theta }_tS_{{\mathfrak {A}}_t}(e^tz)={{\widehat{\Theta }}}_tS_{{\mathfrak {A}}}(e^tz),\quad z\in {\mathbb {C}}_+. \end{aligned}$$

Hence,

$$\begin{aligned} S_{{\mathfrak {A}}}(e^tz)=\Phi _tS_{{\mathfrak {A}}}(z), \quad z\in {\mathbb {C}}_+, \quad t\in {\mathbb {R}}, \end{aligned}$$

(3.8)

where

$$\begin{aligned} \Phi _t={{\widehat{\Theta }}}_t^{-1}\Theta _t, \quad t\in {\mathbb {R}}, \end{aligned}$$

is a one-dimensional character of the additive group ${\mathbb {R}}$.

Indeed,

$$\begin{aligned} S_{{\mathfrak {A}}}(e^{t+s}z)=\Phi _t\Phi _sS_{{\mathfrak {A}}}(z)=\Phi _{t+s}S_{{\mathfrak {A}}}(z), \quad z\in {\mathbb {C}}_+, \quad t\in {\mathbb {R}}, \end{aligned}$$

and, therefore,

$$\begin{aligned} \Phi _{t+s}=\Phi _t\Phi _s, \quad t,s\in {\mathbb {R}}. \end{aligned}$$

From (3.8), it also follows that the character $\Phi _t$ is continuous in t and, therefore, $\Phi _t=e^{i\alpha (t)}$, $t\in {\mathbb {R}}$, where $\alpha (t)$ is a solution to the functional equation

$$\begin{aligned} \alpha (t+s)=\alpha (t)+\alpha (s),\quad t,\,s\in {\mathbb {R}}. \end{aligned}$$

Due to continuity of the function $\alpha (t)$, we finally conclude to $\alpha (t)=\omega t$ for some $\omega \in {\mathbb {R}}$ (see, e.g., [10, Sec. 26]) so that

$$\begin{aligned} \Phi _t=e^{i\omega t},\quad t\in {\mathbb {R}}. \end{aligned}$$

(3.9)

Since $S_{{\mathfrak {A}}}(z)$ is a contractive analytic function on ${\mathbb {C}}_+$, then either $S_{{\mathfrak {A}}}(z)$ vanishes identically in the upper half-plane and, therefore, (3.3) holds, or $S_{{\mathfrak {A}}}(z)$ admits the representation

$$\begin{aligned} S_{{\mathfrak {A}}}(z)= B(z)e^{iN(z)},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

where B(z) is the Blaschke product associated with the zeroes (if any) of $S_{{\mathfrak {A}}}(z)$ in the upper half-plane and N(z) is an analytic map from ${\mathbb {C}}_+$ into itself (see, e.g., [18, Ch. VI]).

Assume, therefore, that $S_{{\mathfrak {A}}}(z)$ is not identically zero. Then, from the functional equation (3.8), it follows that the characteristic function has no zeroes in the upper half-plane and hence

$$\begin{aligned} S_{{\mathfrak {A}}}(z)= e^{iN(z)}, \quad z\in {\mathbb {C}}_+,\quad t\in {\mathbb {R}}. \end{aligned}$$

Combining (3.8) and (3.9), we obtain

$$\begin{aligned} S_{{\mathfrak {A}}}(e^{t}z)=e^{iN(e^tz)}= e^{i(\omega t +N(z))},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

Therefore, the Herglotz–Navanlinna function N(z) satisfies the functional equation

$$\begin{aligned} N(e^tz)=N(z)+\omega t, \quad z\in {\mathbb {C}}_+, \quad t\in {\mathbb {R}}. \end{aligned}$$

(3.10)

Taking the derivative of both sides with respect to t and then setting $t=0$ yields

$$\begin{aligned} N'(z)=\frac{\omega }{z}, \quad z\in {\mathbb {C}}_+. \end{aligned}$$

Therefore,

$$\begin{aligned} N(z)=\omega \log \left( \frac{z}{i}\right) +N(i)=\omega \log \left( \frac{z}{i}\right) +i\gamma +\phi , \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(3.11)

where

$$\begin{aligned} \gamma ={\textrm{Im}}(N(i))>0 \quad \text {and} \quad \phi ={\textrm{Re}}(N(i))\in {\mathbb {R}}. \end{aligned}$$

(Here $\log z$ is the principle branch of the logarithm with the cut on the negative semi-axis.)

Evaluating the non-tangential boundary values of N(z) on the real axis, and taking into account that ${\textrm{Im}}(N(z))>0$ for $z\in {\mathbb {C}}_+$, we get

$$\begin{aligned} 0\le {\textrm{Im}}(N(\lambda +i0))={\left\{ \begin{array}{ll}-\frac{\pi }{2}\omega +\gamma ,&{}\lambda >0 \\ \frac{\pi }{2}\omega +\gamma ,&{}\lambda <0 \end{array}\right. }, \end{aligned}$$

which shows that

$$\begin{aligned} |\omega |\le \frac{2}{\pi }\gamma . \end{aligned}$$

Therefore, we have the representation

$$\begin{aligned} \omega =\frac{2}{\pi }\gamma (p-q) \end{aligned}$$

for some $p, q\ge 0$, $p+q=1$. Using (3.11), we then get

$$\begin{aligned} N(z)&=\gamma \left[ (p-q)\frac{2}{\pi }\log \left( \frac{z}{i}\right) +i\right] +\phi \\&=\gamma \left[ p\frac{2}{\pi }\log z+q \frac{2}{\pi }\log \left( -\frac{1}{z}\right) \right] +\phi , \quad z\in {\mathbb {C}}_+. \end{aligned}$$

Therefore,

$$\begin{aligned} S_{\mathfrak {A}}(z)=e^{iN(z)}=e^{i\phi }[z^{i\frac{2}{\pi }\gamma }]^p\left[ \left( -\frac{1}{z}\right) ^{i\frac{2}{\pi }\gamma }\right] ^q,\quad z\in {\mathbb {C}}_+, \end{aligned}$$

which completes the proof of representation (3.4) with

$$\begin{aligned} \mu =i\frac{\pi }{2} p\gamma \quad \text {and}\quad \nu =i\frac{\pi }{2} q\gamma . \end{aligned}$$

$\square$

4 The characteristic function of the symmetric Schrödinger operator with a singular potential

In the Hilbert space $L^2({\mathbb {R}}_+)$, denote by $\dot{H}(\nu )$, $\nu \in {\mathbb {R}}\cup i{\mathbb {R}}$ the closure of the symmetric operator given by the differential expression

$$\begin{aligned} \tau _\nu =-\frac{d^2}{dx^2}+\frac{\nu ^2-\frac{1}{4}}{x^2} \end{aligned}$$

(4.1)

with the initial domain $C_0^\infty ({\mathbb {R}}_+)$.

It is well known that the symmetric operator $\dot{H}(\nu )$ is

• (i) self-adjoint for $\nu \in {\mathbb {R}}\setminus (-1,1)$,

• (ii) semi-bounded from below for $-1<\nu <1$,

• (iii) unbounded from below for $\nu \in i{\mathbb {R}}\setminus \{0\}$,

• (iv) the symmetric operator $\dot{H}(\nu )$ has deficiency indices (1, 1) whenever

  $$\begin{aligned} \nu \in (-1,1)\cup i{\mathbb {R}}. \end{aligned}$$

(For a textbook discussion of the properties (i)–(iv), we refer to [15], the proof of (i) and (iv) can be found in [35, Theorem X.10] see also [7, Ch. XIII, Sec. 6, Theorems 14, 23], (ii) follows from the Hardy inequality [16, Theorem 254], (iii) follows from [1, Theorem 5, Sec. 131].)

Notice that the symmetric operator $\dot{H}(\nu )$ is homogeneous with respect to the group of unitary scaling transformations $U_t$ given by

$$\begin{aligned} (U_tf)(x)=e^{-\frac{t}{4}}f(e^{-\frac{t}{2}}x), \quad f\in L^2({\mathbb {R}}_+). \end{aligned}$$

(4.2)

That is,

$$\begin{aligned} U_t \dot{H} (\nu )U_t^*=e^{t}\dot{H} (\nu )\quad \text {on}\quad \textrm{Dom}(\dot{H}). \end{aligned}$$

(4.3)

Starting from now on, we will be interested in case (iii) where $\dot{H}(\nu )$ is unbounded from below. We will assume that $\nu \in i(0,\infty )=i{\mathbb {R}}_+$, for definiteness.

Denote by

$$\begin{aligned} h_+^\nu (x)=x^{1/2}H_\nu ^{(1)}(\sqrt{i}x), \quad \quad x\in {\mathbb {R}}_+, \end{aligned}$$

(4.4)

the Weyl solution of the differential equation

$$\begin{aligned} -\frac{d^2}{dx^2}h_+^{\nu }{}(x)+\frac{\nu ^2-\frac{1}{4}}{x^2}h_+^\nu (x)=ih^\nu _+(x), \quad x\in {\mathbb {R}}_+. \end{aligned}$$

Here, $H_\nu ^{(1)}(x)$ is the Hankel function of the first order. Throughout this paper, we also set

$$\begin{aligned} h_-^\nu (x)=\overline{h_+^\nu (x)}. \end{aligned}$$

(4.5)

To simplify the notation, we omit the index $\nu$ and use shorthand notation $h_\pm$ instead of $h_\pm ^\nu$.

Theorem 4.1

Given $\nu \in i{\mathbb {R}}_+$, denote by $H(\nu )$ the self-adjoint extension of the symmetric operator $\dot{H}(\nu )$ such that

$$\begin{aligned} h_+-h_-\in \textrm{Dom}(H(\nu )), \end{aligned}$$

(4.6)

where $h_\pm$ are deficiency elements (4.4) and (4.5).

Then, the Livšic function s(z) associated with the pair $(\dot{H}(\nu ),H(\nu ))$ has the form

$$\begin{aligned} s(z)=\frac{S(z)-S(i) }{S(i)S(z)-1}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(4.7)

where

$$\begin{aligned} S(z)=-\left( -\frac{1}{z}\right) ^{\nu },\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(4.8)

Proof

In accordance with definition (2.2), from (4.6), it follows that the Livšic function s(z) associated with the pair $(\dot{H}(\nu ), H(\nu ))$ can be evaluated as

$$\begin{aligned} s(z)=\frac{z-i}{z+i}\cdot \frac{(h_z, h_-)}{(h_z, h_+)},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

where

$$\begin{aligned} h_z(x)=x^{1/2}H_\nu ^{(1)}(\sqrt{z}x),\quad z \in {\mathbb {C}}_+, \end{aligned}$$

is the Weyl solution (4.4), i.e., $h_z\in L^2({\mathbb {R}}_+)$.

We have

$$\begin{aligned} s(z)&=\frac{z-i}{z+i}\cdot \frac{(h_z,h_- )}{(h_z, h_+)}= \frac{(zh_z,h_-)- (h_z,(-i)h_-)}{(zh_z, h_+)-(h_z, ih_+)}\\&=\frac{(\tau _\nu [h_z],h_-)- (h_z,\tau _\nu [h_-])}{(\tau _\nu [h_z], h_+)-(h_z, \tau _\nu [h_+])} =\frac{(h_z'',h_-)- (h_z,h_-'')}{(h_z'', h_+)-(h_z, h_+'')} \\&=\frac{\int _0^\infty [h_z''(x)h_+ (x)-h_z( x)h_+''(x))]dx }{\int _0^\infty [h_z''( x)h_-( x)- h_z(x)h_-''(x)]dx },\quad z\in {\mathbb {C}}_+, \end{aligned}$$

with $\tau _\nu$ given by (4.1).

Integration by parts yields

$$\begin{aligned} s(z)=\lim _{x\downarrow 0}\frac{W[h_z, h_+](x)}{W[h_z, h_-](x)},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(4.9)

Here, W[f, g](x) stands for the Wronskian associated with the ordered pair of differentiable functions f and g,

$$\begin{aligned} W[f, g](x)=f(x)g'(x)-f'(x)g(x),\quad x\in (0,\infty ). \end{aligned}$$

Using the well-known relationship between the Hankel and Bessel functions,

$$\begin{aligned} H_\nu ^{(1)}(x)=\frac{J_{-\nu }(x)-e^{-i \pi \nu }J_\nu (x)}{i\sin \pi \nu }, \end{aligned}$$

and taking into account the series representation

$$\begin{aligned} J_\nu (z)=\left( \frac{1}{2}z\right) ^\nu \sum _{k=0}^\infty \frac{\left( -\frac{1}{4} z^2\right) ^k}{k!\Gamma (\nu +k+1)}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

for the Bessel function $J_\nu (z)$ (see [4, eq. 9.1.10]), we obtain the following asymptotic expansion (as $x\rightarrow 0$)

$$\begin{aligned} H_\nu ^{(1)}(x)= \frac{i}{\sin \pi \nu } \bigg \{ e^{-i\pi \nu } \frac{ (\frac{1}{2}x )^\nu }{\Gamma (1+\nu )} -\frac{ (\frac{1}{2}x )^{-\nu } }{\Gamma (1-\nu )} \bigg \} +O(x^{-\nu +2}),\quad \nu \ne 0.\nonumber \\ \end{aligned}$$

(4.10)

Now, it is easy to see that the Weyl solution $h_z(x)=x^{1/2}H_\nu ^{(1)}(\sqrt{z}x)$ has the asymptotic representation

$$\begin{aligned} h_z(x)=a_+(z)x^{\frac{1}{2}+\nu }+a_-(z)x^{\frac{1}{2}-\nu }+O(x^{\frac{3}{2}})\quad \text {as} \quad x\rightarrow 0, \end{aligned}$$

(4.11)

with the coefficients $a_\pm (z)$, $z\in {\mathbb {C}}_+$ given by

$$\begin{aligned} a_+(z)=-\frac{e^{-i\pi \nu }}{i\sin \pi \nu }\left[ \frac{1}{2^\nu \Gamma (1+\nu )}z^{\frac{\nu }{2}}\right] =\frac{e^{\pi |\nu |}}{\sinh \pi |\nu |}\left[ \frac{1}{2^\nu \Gamma (1+\nu )}\right] z^{\frac{\nu }{2}} \end{aligned}$$

(4.12)

and

$$\begin{aligned} a_-(z)=\frac{1}{i\sin \pi \nu }\left[ \frac{1}{2^{-\nu } \Gamma (1-\nu )}z^{-\frac{\nu }{2}}\right] =-\frac{1}{\sinh \pi |\nu |}\left[ \frac{1}{2^{-\nu } \Gamma (1-\nu )}\right] z^{-\frac{\nu }{2}}.\nonumber \\ \end{aligned}$$

(4.13)

By setting $z=i=e^{i\frac{\pi }{2}}$ in (4.12) and (4.13), for the deficiency element $h_+(z)$, we get the asymptotics

$$\begin{aligned} h_+(x)=A_+x^{\frac{1}{2}+\nu }+A_-x^{\frac{1}{2}-\nu }+O(x^{\frac{3}{2}})\quad \text {as} \quad x\rightarrow 0. \end{aligned}$$

(4.14)

Here,

$$\begin{aligned} A_+=\frac{e^{\frac{3}{4}\pi |\nu |}}{\sinh \pi |\nu |}\frac{2^{-\nu }}{\Gamma (1+\nu )} \end{aligned}$$

(4.15)

and

$$\begin{aligned} A_-=-\frac{e^{\frac{\pi }{4} |\nu |}}{\sinh \pi |\nu |}\frac{2^{\nu }}{\Gamma (1-\nu )}. \end{aligned}$$

(4.16)

Similarly, taking into account that ${\overline{\nu }}=-\nu$ for $\nu \in i{\mathbb {R}}_+$, for the deficiency element $h_-=\overline{h_+}$ we obtain the asymptotic expansion (with the help of (4.14)-(4.16))

$$\begin{aligned} h_-(x)=B_+x^{\frac{1}{2}+\nu }+B_-x^{\frac{1}{2}-\nu }+O(x^{\frac{3}{2}})\quad \text {as} \quad x\rightarrow 0, \end{aligned}$$

(4.17)

with

$$\begin{aligned} B_+=\overline{A_-}=-\frac{e^{\frac{\pi }{4} |\nu |}}{\sinh \pi |\nu |}\frac{2^{-\nu }}{\Gamma (1+\nu )} \end{aligned}$$

(4.18)

and

$$\begin{aligned} B_-=\overline{A_+}=\frac{e^{\frac{3}{4}\pi |\nu |}}{\sinh \pi |\nu |}\frac{2^{\nu }}{\Gamma (1-\nu )}. \end{aligned}$$

(4.19)

Using the shorthand notation

$$\begin{aligned} W[f,g](x)=W[f(x),g(x)], \end{aligned}$$

from (4.11), (4.14) and (4.17) we get ($z\in {\mathbb {C}}_+$)

$$\begin{aligned} \lim _{x\downarrow 0}\frac{W[h_z, h_+](x)}{W[h_z, h_-](x)}=\lim _{x\downarrow 0} \frac{W[a_+(z)x^{\frac{1}{2}+\nu }+a_-(z)x^{\frac{1}{2}-\nu },A_+x^{\frac{1}{2}+\nu }+A_-x^{\frac{1}{2}-\nu }]}{W[a_+(z)x^{\frac{1}{2}+\nu }+a_-(z)x^{\frac{1}{2}-\nu },B_+x^{\frac{1}{2}+\nu }+B_-x^{\frac{1}{2}-\nu }]}. \end{aligned}$$

Since $W[x^{\frac{1}{2}\pm \nu },x^{\frac{1}{2}\mp \nu }]=0$, representation (4.9) for the Livšic function s(z) simplifies to

$$\begin{aligned} s(z)&=\lim _{x\downarrow 0} \frac{W[a_+(z)x^{\frac{1}{2}+\nu },A_-x^{\frac{1}{2}-\nu }]+{W[a_-(z)x^{\frac{1}{2}-\nu },A_+x^{\frac{1}{2}+\nu }}]}{W[a_+(z)x^{\frac{1}{2}+\nu },B_-x^{\frac{1}{2}-\nu }]+{W[a_-(z)x^{\frac{1}{2}-\nu },B_+x^{\frac{1}{2}+\nu }}]}\\&=\lim _{x\downarrow 0} \frac{(a_+(z)A_--a_-(z)A_+)W[x^{\frac{1}{2}+\nu },x^{\frac{1}{2}-\nu }]}{(a_+(z)B_--a_-(z)B_+)W[x^{\frac{1}{2}+\nu },x^{\frac{1}{2}-\nu }]}\\&= \frac{a_+(z)A_--a_-(z)A_+}{a_+(z)B_--a_-(z)B_+},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

A closer look at (4.12), (4.13), (4.15), (4.16) and (4.18), (4.19) shows that the terms $a_\pm (z)A_\mp$ and $a_\pm (z)B_\mp$ share the same common factor

$$\begin{aligned} \frac{1}{\sinh \pi |\nu |}\frac{2^{-\nu }}{\Gamma (1+\nu )}\frac{1}{\sinh \pi |\nu |}\frac{2^{\nu }}{\Gamma (1-\nu )}=\frac{1}{\sinh ^2\pi |\nu |\Gamma (1+\nu )|^2}. \end{aligned}$$

Having this in mind, one obtains

$$\begin{aligned} s(z)&= \frac{a_+(z)A_--a_-(z)A_+}{a_+(z)B_--a_-(z)B_+} = \frac{-e^{\pi |\nu |} e^{\frac{\pi }{4}|\nu | }z^\frac{\nu }{2}+e^{\frac{3}{4}\pi |\nu |}z^{-\frac{\nu }{2}}}{e^{\pi |\nu |} e^{\frac{3}{4}\pi |\nu |}z^\frac{\nu }{2}-e^{\frac{\pi }{4} |\nu |}z^{-\frac{\nu }{2}}} \\ {}&= \frac{-e^{\pi |\nu |} e^{\frac{\pi }{4}|\nu | }+e^{\frac{3}{4}\pi |\nu |}z^{-\nu }}{e^{\pi |\nu |} e^{\frac{3}{4}\pi |\nu |}-e^{\frac{\pi }{4} |\nu |}z^{-\nu }} \\ {}&= \frac{- e^{\frac{\pi }{4}|\nu | }+e^{\frac{3}{4}\pi |\nu |}e^{-\pi |\nu |}z^{-\nu }}{ e^{\frac{3}{4}\pi |\nu |}-e^{\frac{\pi }{4} |\nu |} e^{-\pi |\nu |}z^{-\nu }} \\ {}&= \frac{- e^{-\frac{3}{4}\pi |\nu |}e^{\frac{\pi }{4}|\nu | }+e^{-\pi |\nu |}z^{-\nu }}{ 1-e^{-\frac{3}{4}\pi |\nu |}e^{\frac{\pi }{4} |\nu |} e^{-\pi |\nu |}z^{-\nu }} \\ {}&= -\frac{e^{-\pi |\nu |}z^{-\nu }- e^{-\frac{3}{4}\pi |\nu |}e^{\frac{\pi }{4}|\nu | }}{e^{-\frac{\pi }{2}|\nu |} e^{-\pi |\nu |}z^{-\nu }-1} \\ {}&= -\frac{\left( -\frac{1}{z}\right) ^\nu - e^{-\frac{\pi }{2}|\nu |}}{e^{-\frac{\pi }{2}|\nu |} \left( -\frac{1}{z}\right) ^\nu -1} , \quad z\in {\mathbb {C}}_+. \end{aligned}$$

In other words,

$$\begin{aligned} s(z)=\frac{S(z)-S(i) }{S(i) S(z)-1}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

where S(z) is given by (4.8), which completes the proof. $\square$

Remark 4.2

As it clearly seen from (4.7) and (4.8), the Livšic function s(z) associated with the pair $(\dot{H}(\nu ), H(\nu ))$ has a geometric progression of zeros located on the “positive” imaginary axis at the points

$$\begin{aligned} z_n=ie^{-\frac{2n}{|\nu |}\pi }, \quad n\in {\mathbb {Z}}. \end{aligned}$$

(4.20)

Notice that the zeros $z_n$ coincide with the complex eigenvalues of the maximal dissipative extension ${{\widetilde{H}}}(\nu )$ of the symmetric operator $\dot{H}(\nu )$ determined by the requirement

$$\begin{aligned} h_+\in \textrm{Dom}({{\widetilde{H}}}(\nu )). \end{aligned}$$

(4.21)

Indeed, membership (4.21) means that the von Neumann extension parameter $\varkappa$ (see Eq. (2.5)) of the triple ${\mathfrak {H}}=(\dot{H}(\nu ),{{\widetilde{H}}}(\nu ), H(\nu ))$ vanishes. In this case, in accordance with (2.6), the characteristic function $S_{\mathfrak {H}}(z)$ of the triple and Livšic function s(z) are related as

$$\begin{aligned} S_{\mathfrak {H}}(z)=-s(z), \quad z\in {\mathbb {C}}_+. \end{aligned}$$

Therefore, $S_{\mathfrak {H}}(z)$ has infinitely many zeros at the points (4.20) as well. It remains to recall that the zeroes of the characteristic function $S_{\mathfrak {H}}(z)$ of the triple determine the complex eigenvalues of the dissipative operator ${{\widetilde{H}}}(\nu )$ (see, e.g., [29, Appendix D]).

Corollary 4.3

The Weyl–Titchmarsh function M(z) associated with the pair of operators $(\dot{H}(\nu ),H(\nu ))$ referred to in Theorem 4.1 admits a meromorphic continuation to the complex plane ${\mathbb {C}}\setminus [0,\infty )$ with the cut on the positive semi-axis. The analytic continuation of M(z) has simple poles $\lambda _n\in (-\infty ,0)$ that form the two-sided geometric progression

$$\begin{aligned} \lambda _n=-e^{-\frac{2n}{|\nu |}\pi }, \quad n\in {\mathbb {Z}}. \end{aligned}$$

Moreover, in the relevant integral representation

$$\begin{aligned} M(z)=\int _{\mathbb {R}}\left( \frac{1}{\lambda -z}-\frac{\lambda }{1+\lambda ^2}\right) d\mu (\lambda ), \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(4.22)

the restriction of the measure $\mu (d\lambda )|_{{\mathbb {R}}_+}$ on the positive semi-axis is absolutely continuous.

Proof

Observe that the non-tangential boundary values of the function S(z) given by (4.8) on the negative semi-axis are unimodular,

$$\begin{aligned} S(\lambda +i0)=e^{-\nu \log |\lambda |}=e^{-i|\nu |\log |\lambda |},\quad \lambda \in (-\infty , 0). \end{aligned}$$

Therefore, the Livšic function $s(\lambda +i0)$ associated with the pair $(\dot{H}(\nu ),H(\nu ))$ and given by (4.7) is also unimodular on the negative real axis $(-\infty ,0)$, that is,

$$\begin{aligned} |s(\lambda +i0)|=1,\quad \lambda <0. \end{aligned}$$

(4.23)

As a consequence, from representation (2.9), it follows that the Weyl–Titchmarsh function M(z) associated with the pair $(\dot{H}(\nu ),H(\nu ))$ admits a meromorphic continuation to the complex plane ${\mathbb {C}}\setminus [0,\infty )$ with the cut on the positive semi-axis. On the negative semi-axis, thus obtained analytic continuation has simple poles $\lambda _n\in (-\infty ,0)$ determined by the roots of the equation

$$\begin{aligned} s(\lambda _n+i0)=1,\quad \lambda _n<0. \end{aligned}$$

Equivalently,

$$\begin{aligned} S(\lambda _n+i0)=-1,\quad \lambda _n<0, \end{aligned}$$

which, in view of (4.8), means that

$$\begin{aligned} |\nu | \log \left( -\frac{1}{\lambda _n}\right) =2n\pi , \quad n\in {\mathbb {Z}}, \end{aligned}$$

and therefore, the poles $\lambda _n$ form the two-sided geometric progression (cf. Remark 4.2)

$$\begin{aligned} \lambda _n=-e^{-\frac{2n}{|\nu |}\pi }, \quad n\in {\mathbb {Z}}. \end{aligned}$$

On the other hand, the boundary values $S(\lambda +i0)$ on the positive real axis are uniformly bounded,

$$\begin{aligned} |S(\lambda +i0)|\le e^{-\pi |\nu |}, \quad \lambda \in (0, \infty ). \end{aligned}$$

Therefore, for the boundary values of the Livšic function s(z) on the positive real axis we have the uniform estimate

$$\begin{aligned} |s(0+i\lambda )|\le r \quad \text {for some} \quad r<1, \quad \lambda \in (0, \infty ). \end{aligned}$$

To make the next step, we need a slightly stronger result: there exists $r<1$ such that for any $0<\alpha <\beta$ there exists $\varepsilon _0>0$ such that

$$\begin{aligned} |s(\lambda +i\varepsilon )|\le r, \quad \lambda \in (\alpha , \beta ), \quad \varepsilon \in [0,\varepsilon _0]. \end{aligned}$$

(4.24)

Recall that the Weyl–Titchmarsh function M(z) associated with the pair $(\dot{H}(\nu ),H(\nu ))$ and the corresponding Livšic function s(z) are related as (see (2.4))

$$\begin{aligned} M(z)=\frac{1}{i}\cdot \frac{s(z)+1}{s(z)-1},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(4.25)

By the Stieltjes inversion formula, we have

$$\begin{aligned} \frac{\mu \left( (\alpha ,\beta )\right) +\mu \left( [\alpha ,\beta ]\right) }{2}=\frac{1}{\pi }\overline{\lim _{\varepsilon \downarrow 0}}\int _\alpha ^\beta {\textrm{Im}}(M(\lambda +i\varepsilon ))d\lambda , \end{aligned}$$

(4.26)

where $\mu (d\lambda )$ is the measure from the integral representation (4.22). Combining (4.25) and (4.26), from (4.24) we get the bound

$$\begin{aligned} \frac{\mu \left( (\alpha ,\beta )\right) +\mu \left( [\alpha ,\beta ]\right) }{2}\le \frac{1}{\pi }\frac{2}{1-r}(\beta -\alpha ), \quad 0<\alpha <\beta , \end{aligned}$$

which shows that the restriction of the measure $\mu (d\lambda )|_{{\mathbb {R}}_+}$ on the positive semi-axis is absolutely continuous since $0<\alpha <\beta$ can be chosen arbitrarily. $\square$

Our next result shows that $\dot{H}(\nu )$, $\nu \in i {\mathbb {R}}_+$ is an example of a prime symmetric homogeneous operator while (scaling) symmetry is broken.

Proposition 4.4

(cf. [15]) The symmetric operator $\dot{H}(\nu )$, $\nu \in i{\mathbb {R}}_+$ with deficiency indices (1, 1) is prime.

Moreover, $\dot{H}(\nu )$ does not admit homogeneous self-adjoint extensions with respect to the group $U_t$ of scaling transformations (4.2).

Proof

To see that $\dot{H}(\nu )$ is a prime operator, we proceed as follows.

Observe (see, e.g., [29, Appendix B, Theorem B.2]) that the subspace

$$\begin{aligned} {{\mathcal {H}}}_1= \overline{\textrm{span}_{z\in {\mathbb {C}}\setminus {\mathbb {R}}}\textrm{Ker}((\dot{H}(\nu ))^*-zI)} \end{aligned}$$

reduces $\dot{H}(\nu )$ and that the part $A_0$ of $\dot{H}(\nu )$ associated with its orthogonal complement ${{\mathcal {H}}}_0={{\mathcal {H}}}_1^\perp$ is a self-adjoint operator.

From (4.3), it follows that

$$\begin{aligned} U_t\textrm{Ker}((\dot{H}(\nu ))^*-zI)=\textrm{Ker}((\dot{H}(\nu )))^*-e^{-t}zI), \quad t\in {\mathbb {R}}, \end{aligned}$$

where $U_t$ is the unitary group of scaling transformations (4.2). Hence, the subspace ${{\mathcal {H}}}_1$ is $U_t$-invariant for all $t\in {\mathbb {R}}$. Therefore, the subspace ${{\mathcal {H}}}_1$ reduces the representation ${\mathbb {R}}\ni t\mapsto U_t$, so does ${{\mathcal {H}}}_0$, and we have

$$\begin{aligned} V_t A_0V_t^*=e^{t} A_0\quad \text {on }\textrm{Dom}( A_0), \end{aligned}$$

(4.27)

where $V_t=U_t|_{{{\mathcal {H}}}_0}$ is the part of the representation ${\mathbb {R}}\ni t\mapsto U_t$ in its reducing subspace ${{\mathcal {H}}}_0$.

From commutation relations (4.27), it follows that

$$\begin{aligned} \textrm{spec}(A_0)=e^t\textrm{spec}(A_0) \quad \text {for all} \quad t\in {\mathbb {R}}. \end{aligned}$$

(4.28)

Therefore, there are the following four possibilities: either $\textrm{spec}(A_0)=[0, \infty )$ or $\textrm{spec}(A_0)=[-\infty , 0])$ or, $\textrm{spec}(A_0)={\mathbb {R}}$ or, finally,

$$\begin{aligned} \textrm{spec}(A_0)=\{0\}. \end{aligned}$$

The latter outcome does not occur since otherwise the point $\lambda =0$ would be an eigenvalue of $A_0$. In this case, $\dot{H}(\nu )$ has to possess a zero eigenvalue. However, any non-trivial solution to the differential equation

$$\begin{aligned} -f''(x)+\frac{\nu ^2-\frac{1}{4}}{x^2}f(x)=0, \quad x>0, \end{aligned}$$

does not belong to $L^2(0, \infty )$ and hence $\textrm{Ker}(\dot{H}(\nu ))=\{0\}.$ A contradiction.

Since the essential spectrum of the differential expression $\tau _\nu$ given by (4.1) coincides with $[0,\infty )$ [7, Ch. XIII, Sec. 7, Lemma 14], the essential spectrum on the negative semi-axis of any self-adjoint extension H of $\dot{H}(\nu )$ is empty. On the other hand, given the characterization of the spectrum of $A_0$ (see eq. (4.28)), we see that

$$\begin{aligned} \textrm{spec}(A_0)\subset \textrm{spec}_{\text {ess}}(H)= [0,\infty ). \end{aligned}$$

Hence, there is the only possibility left (provided that the reducing subspace ${{\mathcal {H}}}_0$ is non-trivial):

$$\begin{aligned} \textrm{spec}(A_0)=[0,\infty ). \end{aligned}$$

(4.29)

Since $\textrm{Ker}(A_0)=\{0\}$ and $A_0$ is non-negative, the self-adjoint operator $B=\log A_0$ is well defined by the functional calculus and, as it follows from the commutation relations (4.27), the self-adjoint operator B solves the canonical commutation relations

$$\begin{aligned} V_t BV_t^*=B+tI\quad \text {on }\textrm{Dom}( B). \end{aligned}$$

By the Stone–von Neumann uniqueness result,

$$\begin{aligned} \textrm{spec}(B)=\textrm{spec}_{ac}(B)={\mathbb {R}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \textrm{spec}(A_0)=\textrm{spec}_{\text {ac}}(A_0)=[0,\infty ). \end{aligned}$$

(4.30)

Denote by $\dot{A}_1$ the prime part of the symmetric operator $\dot{H}(\nu )$ in the Hilbert space ${{\mathcal {H}}}_1$. Observe that the part $A_1$ of the self-adjoint extension $H(\nu )$ of $\dot{H}(\nu )$ (determined by (4.6)) in the reducing subspace ${{\mathcal {H}}}_1$ is a self-adjoint extension of $\dot{A}_1$. It follows that the Weyl–Titchmarsh functions associated with the pairs $(\dot{H}(\nu ), H(\nu ))$ and $(\dot{A}_1, A_1)$ coincide. Taking into account that the restriction $\mu (d\lambda )|_{{\mathbb {R}}_+}$ of the measure $\mu (d\lambda )$ from the integral representation (4.22) for the Weyl–Titchmarsh function associated with the pairs $(\dot{H}(\nu ), H(\nu ))$) is absolutely continuous by Corollary 4.3, one concludes that the self-adjoint operator $A_1$ has simple absolutely continuous spectrum filling in $[0, \infty )$. Therefore, in view of (4.30), the self-adjoint operator $H(\nu )=A_1\oplus A_0$ has an absolutely continuous spectrum filling in $[0,\infty )$ of multiplicity at least two which is in contradiction with the well-known result that any self-adjoint extension of $\dot{H}(\nu )$ has a simple spectrum (see, e.g., [14, Corollary 5.6]). Therefore, ${{\mathcal {H}}}_0=\{0\}$, and hence $\dot{H}(\nu )$ is a prime symmetric operator.

To prove the last assertion of the theorem, observe that since $\dot{H}(\nu )$ is a prime symmetric operator, the poles of the Weyl–Titchmarsh function M(z) associated with the pair $(\dot{H}(\nu ), H(\nu ))$ determine the negative discrete spectrum of the self-adjoint extension $H(\nu )$. In particular, the extension $H(\nu )$ is not homogeneous for a homogeneous operator may not have eigenvalues different from zero: the spectrum of a homogeneous operator is a scale invariant set.

If H is an arbitrary self-adjoint extension of $\dot{H}(\nu )$ with $h_+-\Theta h_-\in \textrm{Dom}(H)$ (for some $|\Theta |=1$), then for the Livšic function $\check{s}(z)$ associated with the pair $(\dot{H}(\nu ), H)$ we have (see, e.g., [29, Appendix E, Lemma E1])

$$\begin{aligned} \check{s}(z)={\overline{\Theta }} s(z),\quad z\in {\mathbb {C}}_+. \end{aligned}$$

As above, one shows that the self-adjoint operator H has a geometric progression of negative eigenvalues $\check{\lambda }_n$, the roots of the equation

$$\begin{aligned} \check{s}(\check{\lambda }_n)={\overline{\Theta }} s(\check{\lambda }_n)=1, \quad n\in {\mathbb {Z}}, \end{aligned}$$

which ensures that H is not a homogeneous extension either. In other words, the symmetric operator $\dot{H}(\nu )$ does not have homogeneous self-adjoint extensions at all and, hence, scaling symmetry for $\dot{H}(\nu )$ with respect to the group (4.2) is broken for $\nu \in i{\mathbb {R}}_+$. $\square$

Remark 4.5

Notice that the symmetric operators $\dot{H}( \nu )$ and $\dot{H}( \nu ')$, $\nu , \nu '\in i{\mathbb {R}}_+$ are unitarily equivalent only if $\nu = \nu '$ which can easily be seen upon applying the uniqueness Theorem 2.1. Indeed, both $\dot{H}(\nu )$ and $\dot{H}(\nu ')$ are prime operators, and the characteristic (Livšic) functions of $\dot{H}(\nu )$ and $\dot{H}(\nu ')$ given by (4.7), respectively, are not congruent for $\nu \ne \nu '$,

5 Homogeneous quasi-selfadjoint extensions of $\dot{H}(\nu )$

Proposition  4.4 states that the homogeneous symmetric operator $\dot{H}(\nu )$ for $\nu \in i{\mathbb {R}}_+$ does not have homogeneous self-adjoint extensions (with respect to the group of scaling transformations $U_t$). In this situation, [28, Theorem 5.4] ensures the existence of a unique dissipative homogeneous extension of $\dot{H}(\nu )$ (with respect to $U_t$). The following result provides more detailed information about such an extension.

Theorem 5.1

The quasi-selfadjoint extension ${{\widehat{H}}}(\nu )$ of $\dot{H}(\nu )$ determined by the requirement

$$\begin{aligned} h_++e^{i\frac{\pi }{2} \nu }h_-\in \textrm{Dom}({{\widehat{H}}}(\nu )), \end{aligned}$$

(5.1)

with $h_\pm$ given by (4.4) and (4.5), is the unique dissipative homogeneous extension of $\dot{H}(\nu )$ with respect to the scaling group $U_t$.

Moreover, if $H(\nu )$ is the self-adjoint extension of $\dot{H}(\nu )$ such that

$$\begin{aligned} h_+-h_-\in \textrm{Dom}(H(\nu )), \end{aligned}$$

then the characteristic function $S_{\mathfrak {H}}(z)$ of the triple ${\mathfrak {H}}=(\dot{H}(\nu ), {\widehat{H}}(\nu ), H(\nu ))$ is given by

$$\begin{aligned} S_{\mathfrak {H}}(z)=-\left( -\frac{1}{z}\right) ^\nu ,\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(5.2)

In this case, the adjoint operator $({{\widehat{H}}}(\nu ))^*$ is characterized by the requirement

$$\begin{aligned} h_++e^{-i\frac{\pi }{2} \nu }h_-\in \textrm{Dom}(({{\widehat{H}}}(\nu ))^*). \end{aligned}$$

(5.3)

Moreover, the dissipative operator $-({\widehat{H}}(\nu ))^*$ is the unique dissipative homogeneous extension of the symmetric operator $-\dot{H}(\nu )$, with the characteristic function $S_{-{\mathfrak {H}}^*}(z)$ of the triple

$$\begin{aligned} -{\mathfrak {H}}^*=(-\dot{H}(\nu ), -({{\widehat{H}}}(\nu ))^*, -H(\nu )) \end{aligned}$$

given by

$$\begin{aligned} S_{-{\mathfrak {H}}^*}(z)=-z^\nu ,\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(5.4)

Proof

Membership (5.1) shows that $\varkappa =-e^{-\frac{\pi }{2}|\nu |}$ is the von Neumann parameter of the triple ${\mathfrak {H}}=(\dot{H}(\nu ), {\widehat{H}}(\nu ), H(\nu ))$. Therefore, for the characteristic function $S_{\mathfrak {H}}(z)$ of the triple ${\mathfrak {H}}$ we have the representation (see (2.6))

$$\begin{aligned} S_{\mathfrak {H}}(z)=\frac{s(z)-\varkappa }{{\overline{\varkappa }}\,s(z)-1}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

where s(z) the Livšic function s(z) associated with the pair $(\dot{H}(\nu ), H(\nu ))$. Solving for s(z) we get

$$\begin{aligned} s(z)=\frac{S_{\mathfrak {H}}(z)-\varkappa }{\overline{ \varkappa }\,S_{\mathfrak {H}}(z)-1}, \quad z\in {\mathbb {C}}_+. \end{aligned}$$

(5.5)

On the other hand, by Theorem 4.1,

$$\begin{aligned} s(z)=\frac{S(z)-S(i) }{\overline{S(i)}S(z)-1}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(5.6)

where

$$\begin{aligned} S(z)=-\left( -\frac{1}{z}\right) ^{\nu },\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(5.7)

Comparing (5.5) and (5.6) and taking into account that

$$\begin{aligned} \varkappa =S(i)=-e^{-\frac{\pi }{2}|\nu |}, \end{aligned}$$

one concludes that the characteristic function $S_{\mathfrak {H}}(z)$ coincides with the function S(z) given by (5.7) which proves (5.2).

To show that the dissipative operator ${{\widehat{H}}}(\nu )$ is homogeneous with respect to the scaling group $U_t$, we argue as follows.

Taking into account that $\dot{H}(\nu )$ does not have homogeneous self-adjoint extensions, denote by $\check{H}(\nu )$ the unique homogeneous dissipative extension of $\dot{H}(\nu )$ the existence of which is guaranteed by [28, Theorem 5.4].

Since $\check{H}(\nu )$ is a homogeneous extension, one can use Theorem 3.1 to see that the characteristic function associated with the triple $\check{{\mathfrak {H}}}=(\dot{H}(\nu ), \check{H}(\nu ), H(\nu ) )$ can be represented as

$$\begin{aligned} S_{\check{{\mathfrak {H}}}}(z)=e^{i\varphi }z^{\mu '}\left( -\frac{1}{z}\right) ^{\nu '},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

(5.8)

for some $\phi \in [0,2\pi )$ and some $\mu ', \nu '\in i{\mathbb {R}}_+\cup \{0\}$, not both zero.

We claim that $\mu '=0$. Indeed, otherwise, for the boundary values of the characteristic function $S_{\check{{\mathfrak {H}}}}(z)$ on the negative real axis we have the inequality

$$\begin{aligned} |S_{\check{{\mathfrak {H}}}}(\lambda +i0)|=e^{-\pi |\mu '|}<1, \quad \lambda \in {\mathbb {R}}_-. \end{aligned}$$

Therefore, the boundary values $s(\lambda +i0)$ of the Livšic function

$$\begin{aligned} s(z)=\frac{S_{\check{{\mathfrak {H}}}}(z)-S_{\check{{\mathfrak {H}}}}(i)}{\overline{S_{\check{{\mathfrak {H}}}}(i)}S_{\check{{\mathfrak {H}}}}(z)-1}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

associated with the pair $(\dot{H}(\nu ), H(\nu ))$ on the negative real axis are not unimodular. On the other hand, by Theorem 4.1 we have the representation

$$\begin{aligned} s(z)=-\frac{\left( -\frac{1}{z}\right) ^{\nu }-e^{-\frac{\pi }{2}|\nu |}}{e^{-\frac{\pi }{2}|\nu |}\left( -\frac{1}{z}\right) ^{\nu }-1}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(5.9)

which shows that $|s(\lambda +i0)|=1$ for $\lambda <0$ (cf. (4.23)). The obtained contradictions shows that $\mu '=0$, as claimed.

Now from (5.8), it follows that

$$\begin{aligned} S_{\check{{\mathfrak {H}}}}(z)=e^{i\varphi }\left( -\frac{1}{z}\right) ^{\nu '},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

and hence

$$\begin{aligned} s(z)=e^{i\varphi }\frac{\left( -\frac{1}{z}\right) ^{\nu '}-e^{-\frac{\pi }{2}|\nu '|}}{e^{-\frac{\pi }{2}|\nu '|}\left( -\frac{1}{z}\right) ^{\nu '}-1}, \quad z\in {\mathbb {C}}_+. \end{aligned}$$

(5.10)

Comparing representations (5.9) and (5.10), we conclude that $\nu =\nu '$ for $\nu , \nu '\in i{\mathbb {R}}_+$ and also that $e^{i\varphi }=-1$. Therefore,

$$\begin{aligned} S_{\check{{\mathfrak {H}}}}(z)=-\left( -\frac{1}{z}\right) ^{\nu }=S_{{\mathfrak {H}}}(z), \quad z\in {\mathbb {C}}_+. \end{aligned}$$

Since $\dot{H}(\nu )$ is a prime symmetric operator by Proposition 4.4, it follows from the uniqueness result, Theorem 2.1, that $\check{H}(\nu )$ and ${{\widehat{H}}}(\nu )$ coincide. Therefore, the operator ${{\widehat{H}}}(\nu )$ is homogeneous for $\check{H}(\nu )$ has been chosen to be the (unique) homogeneous dissipative extension of $\dot{H}(\nu )$.

Next, it is straightforward to see that the membership

$$\begin{aligned} h_+-\varkappa h_-\in \textrm{Dom}({{\widehat{H}}}(\nu )) \end{aligned}$$

(with $\varkappa =-e^{-\frac{\pi }{2}|\nu |}$) implies

$$\begin{aligned} h_+-\frac{1}{{\overline{\varkappa }}} h_-\in \textrm{Dom}(({{\widehat{H}}}(\nu ))^*) \end{aligned}$$

and hence

$$\begin{aligned} h_++e^{-i\frac{\pi }{2} \nu }h_-\in \textrm{Dom}((\widehat{H}(\nu ))^*). \end{aligned}$$

In particular, the dissipative operator $-({{\widehat{H}}}(\nu ))^*$ is a unique dissipative quasi-selfadjoint homogeneous extension of the symmetric homogeneous operator $-\dot{H}(\nu )$ with respect to the scaling group $U_t$.

It follows from the transformation law [29, Lemma E.4, Appendix E] that the characteristic functions of the triples

$$\begin{aligned} {\mathfrak {H}}=(\dot{H}(\nu ), {\widehat{H}}(\nu ), H(\nu ))\quad \text {and}\quad -{\mathfrak {H}}^*=(-\dot{H}(\nu ), -({{\widehat{H}}}(\nu ))^*, -H(\nu )) \end{aligned}$$

are related as

$$\begin{aligned} S_{-{\mathfrak {H}}^*}(z)=\overline{S_{{\mathfrak {H}}} (-{\overline{z}})},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

Therefore,

$$\begin{aligned} S_{-{\mathfrak {H}}^*}(z)=- \overline{\left( -\frac{1}{{\overline{z}}}\right) ^\nu }=-z^\nu ,\quad z\in {\mathbb {C}}_+, \end{aligned}$$

(5.11)

and (5.4) follows. $\square$

Lemma 5.2

The domain of the homogeneous dissipative extension $\widehat{H}(\nu )$ can also be characterized in terms of the asymptotic boundary conditions as

$$\begin{aligned} \textrm{Dom}({{\widehat{H}}}(\nu ))=\left\{ g\in \textrm{Dom}((\dot{H}(\nu ))^*)\,|\,\lim _{x\downarrow 0}W[g(x),x^{\frac{1}{2}+\nu }]=0\right\} . \end{aligned}$$

(5.12)

Analogously,

$$\begin{aligned} \textrm{Dom}(({{\widehat{H}}}(\nu ))^*)=\left\{ g\in \textrm{Dom}(\dot{H}(\nu )^*)\,|\,\lim _{x\downarrow 0}W[g(x),x^{\frac{1}{2}-\nu }]=0\right\} . \end{aligned}$$

(5.13)

Proof

One can show (see, e.g., [15]) that the functions from the domain of the adjoint operator $(\dot{H}(\nu ))^*$ satisfy the asymptotic boundary conditions as $x\rightarrow 0$:

$$\begin{aligned}{} & {} g(x)=g_{\text {as}}(x)+O(x^{\frac{3}{2}}), \end{aligned}$$

(5.14)

$$\begin{aligned}{} & {} g'(x)=g_{\text {as}} '(x)+O(x^{\frac{1}{2}}), \end{aligned}$$

(5.15)

where the function $g_{\text {as}}(x)$ admits the representation

$$\begin{aligned} g_{\text {as}}(x)=a_{+}[g]x^{\frac{1}{2}+\nu }+a_-[g]x^{\frac{1}{2}-\nu } \end{aligned}$$

(5.16)

in a neighborhood of the origin and $a_{\pm }[g]\in {\mathbb {C}}$.

We claim that the unbounded functionals $a_\pm [g]$ on $\textrm{Dom}( (\dot{H}(\nu ))^*)$ can be evaluated as

$$\begin{aligned} a_\pm [g]=\mp \frac{1}{2\nu }\lim _{x\downarrow 0}W[g(x),x^{\frac{1}{2}\mp \nu }]. \end{aligned}$$

(5.17)

Indeed, we have,

$$\begin{aligned} \lim _{x\downarrow 0}W[g(x),x^{\frac{1}{2}+\nu }]&=W[g_{\text {as}}(x),x^{\frac{1}{2}+\nu }] =W\left[ a_{+}[g]x^{\frac{1}{2}+\nu }+a_-[g]x^{\frac{1}{2}-\nu },x^{\frac{1}{2}+\nu }\right] \\&=a_-[g]W [x^{\frac{1}{2}-\nu },x^{\frac{1}{2}+\nu }]=2\nu a_-[g]. \end{aligned}$$

Analogously,

$$\begin{aligned} \lim _{x\downarrow 0}W[g(x),x^{\frac{1}{2}-\nu }] =-2\nu a_+[g]. \end{aligned}$$

In the case in question, the requirement in (5.12) that

$$\begin{aligned} \lim _{x\downarrow 0}W[g(x),x^{\frac{1}{2}+\nu }]=0 \end{aligned}$$

simply means that the domain of the homogeneous dissipative extension $\widehat{H}(\nu )$ consists of all the functions from $\textrm{Dom}((\dot{H}(\nu ))^*)$ in the asymptotic expansion (5.16) of which the term $x^{\frac{1}{2}-\nu }$ is absent. That is,

$$\begin{aligned} g(x)=a_+x^{\frac{1}{2}+\nu }+O(x^\frac{3}{2}) \quad \text {and}\quad g'(x)=\left( \frac{1}{2}+\nu \right) a_+x^{-\frac{1}{2}+\nu }+O(x^\frac{1}{2})\quad \text {as }\quad x\rightarrow 0, \end{aligned}$$

with $a_+=a_+[g]$ the corresponding (unbounded) boundary functional.

Having in mind that

$$\begin{aligned} \textrm{Dom}({{\widehat{H}}}(\nu ))=\textrm{Dom}(\dot{H}(\nu ))\dot{+} \textrm{span}\{h_++e^{i\frac{\pi }{2}\nu }h_-\}, \end{aligned}$$

(5.18)

to prove (5.12) it suffices to show that

$$\begin{aligned} (h_++e^{i\frac{\pi }{2}\nu }h_-)(x)=ax^{\frac{1}{2}+\nu }+O(x^\frac{3}{2}), \quad \text {as} \quad x\rightarrow 0, \end{aligned}$$

(5.19)

for some $a\in {\mathbb {C}}$.

From (4.12) and (4.13), we get

$$\begin{aligned} (h_++e^{i\frac{\pi }{2}\nu }h_-)(x)=(A_++e^{i\frac{\pi }{2}\nu }B_+)x^{\frac{1}{2}+\nu }+ (A_-+e^{i\frac{\pi }{2}\nu }B_-)x^{\frac{1}{2}-\nu } +O(x^\frac{3}{2}), \end{aligned}$$

where $A_\pm$ and $B_\pm$ are given by (4.15), (4.16) and (4.18), (4.19), respectively.

Since

$$\begin{aligned} A_-+e^{i\frac{\pi }{2}\nu }B_-=-\frac{e^{\frac{\pi }{4} |\nu |}}{\sinh \pi |\nu |}\frac{2^{\nu }}{\Gamma (1-\nu )} +e^{-\frac{\pi }{2}|\nu |} \frac{e^{\frac{3}{4}\pi |\nu |}}{\sinh \pi |\nu |}\frac{2^{\nu }}{\Gamma (1-\nu )}=0, \end{aligned}$$

we obtain (5.19) with $a=(A_++e^{i\frac{\pi }{2}\nu }B_+)$, and (5.12) follows.

To prove (5.13), recall that from (5.3) it follows that

$$\begin{aligned} \textrm{Dom}(({{\widehat{H}}}(\nu ))^*)=\textrm{Dom}(\dot{H}(\nu ))\dot{+} \textrm{span}\{h_++e^{-i\frac{\pi }{2}\nu }h_-\}. \end{aligned}$$

(5.20)

Since

$$\begin{aligned} A_++e^{-i\frac{\pi }{2}\nu }B_+=\frac{e^{\frac{3}{4}\pi |\nu |}}{\sinh \pi |\nu |}\frac{2^{-\nu }}{\Gamma (1+\nu )} -e^{\frac{\pi }{2}|\nu |}\frac{e^{\frac{\pi }{4} |\nu |}}{\sinh \pi |\nu |}\frac{2^{-\nu }}{\Gamma (1+\nu )}=0, \end{aligned}$$

in the asymptotic representation

$$\begin{aligned} (h_++e^{-i\frac{\pi }{2}\nu }h_-)(x)= & {} (A_++e^{-i\frac{\pi }{2}\nu }B_+)x^{\frac{1}{2}+\nu }\\{} & {} +(A_-+e^{-i\frac{\pi }{2}\nu }B_-)x^{\frac{1}{2}-\nu }+O(x^\frac{3}{2})\quad \text {as } \rightarrow 0, \end{aligned}$$

the term proportional to $x^{\frac{1}{2}+\nu }$ is absent, which shows that

$$\begin{aligned} (h_++e^{-i\frac{\pi }{2}\nu }h_-)(x)=(A_-+e^{-i\frac{\pi }{2}\nu }B_-)x^{\frac{1}{2}-\nu }+O(x^\frac{3}{2})\quad \text {as } \rightarrow 0. \end{aligned}$$

Therefore, the domain of the accumulative operator $(\widehat{H}(\nu ))^*$ can be characterized by the “no $x^{\frac{1}{2}+\nu }$-terms" requirement:

$$\begin{aligned} g(x)=a_-x^{\frac{1}{2}-\nu }+O(x^\frac{3}{2}), \quad g\in \textrm{Dom}((\dot{H}(\nu ))^*, \end{aligned}$$

and

$$\begin{aligned} g'(x)=\left( \frac{1}{2}-\nu \right) a_-x^{-\frac{1}{2}-\nu }+O(x^\frac{1}{2})\quad \text {as }\quad x\rightarrow 0, \end{aligned}$$

with $a_-=a_-[g]$ the corresponding boundary functional. Hence

$$\begin{aligned} \lim _{x\downarrow 0}W[g(x),x^{\frac{1}{2}-\nu }]=0, \end{aligned}$$

which proves (5.13). $\square$

Next, we analyze the “time” evolution of the element $U_tf$ modulo $\textrm{Dom}(\dot{H}(\nu ))$ under various additional assumptions on f.

Corollary 5.3

Suppose that $f\in \textrm{Dom}({{\widehat{H}}}(\nu ))$ and $g\in \textrm{Dom}(({{\widehat{H}}}(\nu ))^*)$.

Then,

$$\begin{aligned} (U_tf)-e^{-\frac{t}{2}}e^{-\frac{t}{2}\nu }f\in \textrm{Dom}(\dot{H}(\nu )) \quad (f\in \textrm{Dom}({{\widehat{H}}}(\nu ))) \end{aligned}$$

(5.21)

and

$$\begin{aligned} (U_tg)-e^{-\frac{t}{2}}e^{\frac{t}{2}\nu }g\in \textrm{Dom}(\dot{H}(\nu )) \quad ( g\in \textrm{Dom}(({{\widehat{H}}}(\nu ))^*), \end{aligned}$$

(5.22)

where

$$\begin{aligned} (U_tf)(x)=e^{-\frac{t}{4}}f(e^{-\frac{t}{2}}x), \quad f\in L^2({\mathbb {R}}_+), \end{aligned}$$

is the unitary group $U_t$ of scaling transformations.

Proof

Based on the domain description (5.12) from Lemma 5.2, we see that if $f\in \textrm{Dom}({{\widehat{H}}}(\nu ))$, then in a neighborhood of the origin the following asymptotic representation

$$\begin{aligned} f(x)=a_{+}[f]x^{\frac{1}{2}+\nu }+O(x^{\frac{3}{2}}) \quad \text { as}\quad x\rightarrow 0 \end{aligned}$$

(5.23)

holds. Here, $a_{+}[f]$ is the boundary functional given by (5.17).

Moreover, since $\textrm{Dom}({{\widehat{H}}}(\nu ))\subset \textrm{Dom}((\dot{H}(\nu ))^*)$, one can differentiate the right-hand side of (5.23) term by term to get

$$\begin{aligned} f'(x)=\left( \frac{1}{2}+\nu \right) a_{+}[f]x^{-\frac{1}{2}+\nu }+O(x^{\frac{1}{2}}) \quad \quad \text { as}\quad x\rightarrow 0. \end{aligned}$$

(5.24)

Next, applying the scaling transformation $U_t$ to both sides of (5.23) we obtain

$$\begin{aligned} (U_tf)(x)= e^{-\frac{t}{4}}a_{+}[f](e^{-\frac{t}{2}}x)^{\frac{1}{2}+\nu }+O(x^{\frac{3}{2}}) \quad \text { as}\quad x\rightarrow 0, \end{aligned}$$

which after rearranging the terms yields

$$\begin{aligned} (U_tf)(x)= e^{-\frac{t}{2}}e^{-\frac{t}{2}\nu }a_{+}[f]x^{\frac{1}{2}+\nu }+O(x^{\frac{3}{2}}) \quad \text {as}\quad x\rightarrow 0. \end{aligned}$$

(5.25)

Combining (5.23) and (5.25), we get

$$\begin{aligned} (U_tf)(x)-e^{-\frac{t}{2}}e^{-\frac{t}{2}\nu }f(x)=O(x^{\frac{3}{2}}) \quad \text {as}\quad x\rightarrow 0. \end{aligned}$$

(5.26)

Since $\textrm{Dom}({{\widehat{H}}}(\nu ))$ is invariant under the action of the group $U_t$, and therefore $U_tf\in \textrm{Dom}(\widehat{H}(\nu ))\subset \textrm{Dom}((\dot{H}(\nu ))^*)$, in view of the characterization of the domain of the adjoint operator $(\dot{H}(\nu ))^*$ (see (5.14)–(5.16)), one can differentiate (5.25) term by term to get

$$\begin{aligned} (U_tf)'(x)= \left( \frac{1}{2}+\nu \right) e^{-\frac{t}{2}}e^{-\frac{t}{2}\nu }a_{+}[f]x^{-\frac{1}{2}+\nu }+O(x^{\frac{1}{2}}) \quad \text {as}\quad x\rightarrow 0, \end{aligned}$$

which together with (5.25) shows that

$$\begin{aligned} \frac{d}{dx}\left( (U_tf)(x)-e^{-\frac{t}{2}}e^{-\frac{t}{2}\nu }f(x)\right) =O(x^{\frac{1}{2}}) \quad \text {as}\quad x\rightarrow 0. \end{aligned}$$

(5.27)

From (5.26) and (5.27), we see that the boundary functionals $a_\pm [g_t]$ of the element

$$\begin{aligned} g_t=U_tf-e^{-\frac{t}{2}}e^{-\frac{t}{2}\nu }f,\quad t\in {\mathbb {R}}, \end{aligned}$$

vanish and, therefore, $g_t=U_tf-e^{-\frac{t}{2}}e^{-\frac{t}{2}\nu }f \in \textrm{Dom}(\dot{H}(\nu ))$ which proves (5.21).

The proof of (5.22) is analogous. $\square$

Remark 5.4

It is also worth noting that the symmetric operator $\dot{H}(\nu )$ is real in the sense that

$$\begin{aligned} {{\mathcal {J}}}\left( \textrm{Dom}(\dot{H}(\nu ))\right) = \textrm{Dom}(\dot{H}(\nu )) \end{aligned}$$

(5.28)

and

$$\begin{aligned} {{\mathcal {J}}}\dot{H}(\nu )=\dot{H}(\nu ){{\mathcal {J}}}, \end{aligned}$$

where ${{\mathcal {J}}}$ is the involution of complex conjugation,

$$\begin{aligned} ({{\mathcal {J}}}f)(x)=\overline{f(x)},\quad f\in L^2(0,\infty ). \end{aligned}$$

So is its adjoint,

$$\begin{aligned} {{\mathcal {J}}}(\dot{H}(\nu ))^*=(\dot{H}(\nu ))^*{{\mathcal {J}}}. \end{aligned}$$

Given (5.18) and (5.20) and taking into account that the deficiency elements $h_+$ and $h_-$ given by (4.4) and (4.5) are related as

$$\begin{aligned} h_-=\overline{h_+}={{\mathcal {J}}}h_+, \end{aligned}$$

we also see that

$$\begin{aligned} {{\mathcal {J}}}(\textrm{Dom}({\widehat{H}}(\nu )) )= \textrm{Dom}(({\widehat{H}}(\nu ))^* ). \end{aligned}$$

In particular, the maximal operators ${\widehat{H}}(\nu )$ and $({\widehat{H}}(\nu ))^*$ are anti-unitarily equivalent, that is,

$$\begin{aligned} ({\widehat{H}}(\nu ))^*={{\mathcal {J}}}{\widehat{H}}(\nu ) {{\mathcal {J}}}^{-1}. \end{aligned}$$

6 Homogeneous operators and an operator coupling

In this section, we enhance the list of model examples of homogeneous symmetric operators with broken scale symmetry, the family of Schrödinger symmetric operators $\dot{H}(\nu )$ and $-\dot{H}(\nu )$ for $\nu \in i{\mathbb {R}}_+$. Notice that in the case in question, the core of the spectrum of these operators fills in a semi-axis (the positive one for $\dot{H}(\nu )$ and negative one for $-\dot{H}(\nu )$).

Our immediate goal to construct examples of homogeneous symmetric operators (with broken symmetry) that do not have the quasi-regular points on the real axis. To do so, we need some preliminaries.

In the Hilbert space $L^2({\mathbb {R}}_+)\oplus L^2({\mathbb {R}}_+)$ introduce the symmetric operator $\ddot{\mathbb {H}}(\mu ,\nu )$ with deficiency indices (2, 2) as the orthogonal sum of $-\dot{H}(\mu )$ and $\dot{H}(\nu )$:

$$\begin{aligned} \ddot{\mathbb {H}}(\mu ,\nu )=(-\dot{H}(\mu ))\oplus \dot{H}(\nu ),\quad \mu ,\, \nu \in i{\mathbb {R}}_+. \end{aligned}$$

(6.1)

Lemma 6.1

Given $\mu , \nu \in i{\mathbb {R}}_+$, introduce

$$\begin{aligned} G=f^\mu \oplus f^\nu \in L^2({\mathbb {R}}_+)\oplus L^2({\mathbb {R}}_+), \end{aligned}$$

(6.2)

where

$$\begin{aligned} f^\tau =\frac{1}{\sqrt{e^{\pi |\tau |}-1}}({{\hat{h}}}_+^\tau +e^{\frac{\pi }{2} |\tau |}{{\hat{h}}}_-^\tau ), \quad \tau =\mu , \nu , \end{aligned}$$

(6.3)

and

$$\begin{aligned} {{\hat{h}}}_\pm ^\tau =\frac{h_\pm ^\tau }{\Vert h_\pm ^\tau \Vert },\quad \tau =\mu ,\nu , \end{aligned}$$

(6.4)

are the normalized Weyl solutions $h_\pm ^\tau$, $\tau =\mu , \nu$, given by (4.4) and (4.5).

Then, $G\in \textrm{Dom}((\ddot{\mathbb {H}}(\mu ,\nu ))^*)$ and

$$\begin{aligned} {\textrm{Im}}\left( (\ddot{\mathbb {H}}(\mu ,\nu ))^* G, G)\right) =0. \end{aligned}$$

Proof

Recall that the Weyl solution $h_+^\nu$ given by (4.4) belongs to the deficiency subspace $\textrm{Ker}((\dot{H}(\nu ))^*-iI)$ of $\dot{H}(\nu )$ while $h_-^\nu \in \textrm{Ker}((\dot{H}(\nu ))^*+iI)$. Therefore, $G\in \textrm{Dom}((\ddot{\mathbb {H}}(\mu ,\nu ))^*)$.

We have

$$\begin{aligned} {\textrm{Im}}\left( ((\ddot{\mathbb {H}}(\mu ,\nu ))^* G, G )\right)&=\frac{1}{e^{\pi |\mu |}-1} {\textrm{Im}}\left( (-\dot{H}(\mu ))^* \left[ {{\hat{h}}}_+^\mu +e^{\frac{\pi }{2} |\mu |}{{\hat{h}}}_-^\mu \right] \right) \nonumber \\&\quad +\frac{1}{e^{\pi |\nu |}-1} {\textrm{Im}}\left( (\dot{H}(\nu ))^*\left[ {{\hat{h}}}_+^\nu + e^{\frac{\pi }{2} |\nu |}{{\hat{h}}}_-^\nu \right] \right) . \end{aligned}$$

(6.5)

Here, A[f] is a shorthand notation for the quadratic form (Af, f), $f\in \textrm{Dom}(A)$.

To evaluate the corresponding quadratic forms, observe that for any symmetric operator $\dot{A}$ with deficiency elements $g_\pm \in \textrm{Ker}((\dot{A})^*\mp iI)$ such that $\Vert g_\pm \Vert =1$ and any $\varkappa \in {\mathbb {C}}$ we have

$$\begin{aligned} {\textrm{Im}}\left( (\dot{A})^*[g_+- \varkappa g_-] \right)&={\textrm{Im}}\left( i (g_++ \varkappa g_-),g_+- \varkappa g_-\right) \nonumber \\&={\textrm{Re}}\left( g_++ \varkappa g_-,g_+- \varkappa g_-\right) \nonumber \\&=\Vert g_+\Vert ^2-|\varkappa |^2\Vert g_-\Vert ^2+{\textrm{Re}}\left( (\varkappa g_-,g_+)-(g_+,\varkappa g_-)\right) \nonumber \\&= 1-|\varkappa |^2. \end{aligned}$$

(6.6)

In particular, since ${{\hat{h}}}_\pm ^\nu \in \textrm{Ker}( (\dot{H}(\nu ))^*\mp iI)$ and the normalizing condition (6.4) holds, using (6.6), we have

$$\begin{aligned} {\textrm{Im}}\left( (\dot{H}(\nu ))^*\left[ {{\hat{h}}}_+^\nu +e^{\frac{\pi }{2} |\nu |}{{\hat{h}}}_-^\nu \right] \right) =1-e^{\pi |\nu |}, \end{aligned}$$

(6.7)

and also

$$\begin{aligned} {\textrm{Im}}\left( (-\dot{H}(\mu ))^*\left[ {{\hat{h}}}_+^\mu +e^{\frac{\pi }{2} |\mu |}{{\hat{h}}}_-^\mu \right] \right) =-(1-e^{\pi |\mu |}). \end{aligned}$$

(6.8)

Combining (6.7), (6.8), and (6.5), we finally get

$$\begin{aligned} {\textrm{Im}}\left( ((\ddot{\mathbb {H}}(\mu ,\nu ))^* G, G)\right) = -\frac{e^{\pi |\mu |}-1}{e^{\pi |\mu |}-1}+\frac{e^{\pi |\nu |}-1}{e^{\pi |\nu |}-1}=0, \end{aligned}$$

which completes the proof. $\square$

Introduce the restriction ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ of the adjoint operator $(\ddot{\mathbb {H}}(\mu ,\nu ))^*$ on

$$\begin{aligned} \textrm{Dom}({{\dot{{\mathbb {H}}}}}(\mu ,\nu ))=\textrm{Dom}( \ddot{\mathbb {H}}(\mu ,\nu ))\dot{+}\textrm{span}\left\{ G\right\} . \end{aligned}$$

(6.9)

Here, $\ddot{\mathbb {H}}(\mu , \nu )$ is the symmetric operator with deficiency indices (2, 2) in the direct sum of the Hilbert spaces $L^2({\mathbb {R}}_+)\oplus L^2({\mathbb {R}}_+)$ defined by (6.1) and G is given by (6.2).

Theorem 6.2

The operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ defined by (6.9) is a symmetric operator with deficiency indices (1, 1).

Moreover, ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ is a homogeneous symmetric operator with respect to the modified unitary group ${\mathbb {U}}_t$ given by

$$\begin{aligned} {\mathbb {U}}_t=e^{-i \frac{t}{2} |\mu |}U_t\oplus e^{-i \frac{t}{2} |\nu |}U_t,\quad t\in {\mathbb {R}}, \end{aligned}$$

(6.10)

with $U_t$ the scaling group (4.2).

Proof

By Lemma 6.1,

$$\begin{aligned} {\textrm{Im}}\left( ({{\dot{{\mathbb {H}}}}}(\mu , \nu ) G,G)\right) ={\textrm{Im}}\left( ((\ddot{\mathbb {H}}(\mu , \nu ))^* G,G)\right) =0, \end{aligned}$$

and, therefore, the linear space $\textrm{span}\{G\}$ is a neutral (Lagrangian) subspace for the adjoint operator $(\ddot{\mathbb {H}}(\mu , \nu ))^*$. Hence, the operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ is symmetric.

Applying [27, Theorem 3.1 (ii)], one can show that $\dot{{\mathbb {H}}}(\mu , \nu )$ has deficiency indices (1, 1). (We refer to [27] for the details.)

Since the adjoint operator $(\ddot{\mathbb {H}}(\mu ,\nu ))^*$ is obviously homogeneous with respect to the unitary group ${\mathbb {U}}_t$ and $\dot{{\mathbb {H}}}(\mu ,\nu )\subset (\ddot{\mathbb {H}}(\mu ,\nu ))^*$, to verify the homogeneity of ${{\dot{{\mathbb {H}}}}}(\mu ,\nu )$, it suffices to show that $\textrm{Dom}({{\dot{{\mathbb {H}}}}}(\mu ,\nu ))$ is ${\mathbb {U}}_t$-invariant.

To do so, recall that

$$\begin{aligned} G=f^\mu \oplus f^\nu , \end{aligned}$$

where

$$\begin{aligned} f^\tau =\frac{1}{\sqrt{e^{\pi |\tau |}-1}}({{\hat{h}}}_+^\tau +e^{\frac{\pi }{2} |\tau |}{{\hat{h}}}_-^\tau ) \in \textrm{Dom}(({{\widehat{H}}}(\tau ))^*), \quad \tau =\mu , \nu , \end{aligned}$$

(see Lemma 6.1, eq. (6.14)). Therefore, from Corollary 5.3 it follows that the “dynamics” of $f^\tau$ under the scaling transformations $U_t$ can be characterized as

$$\begin{aligned} e^{-i\frac{t}{2} |\tau |}U_tf^\tau =e^{-\frac{t}{2}}f^\tau +h^\tau , \quad \tau =\mu , \nu , \end{aligned}$$

(6.11)

where

$$\begin{aligned} h^\tau \in \textrm{Dom}(\dot{H}(\tau )),\quad \tau =\mu , \nu . \end{aligned}$$

Taking into account (6.10), from (6.11) we get

$$\begin{aligned} {\mathbb {U}}_tG-e^{-\frac{t}{2}} G\in \textrm{Dom}(\ddot{\mathbb {H}}(\mu , \nu ))\subset \textrm{Dom}({{\dot{{\mathbb {H}}}}}(\mu , \nu )),\quad t\in {\mathbb {R}}. \end{aligned}$$

(6.12)

In view of (6.9) we see that the membership ${\mathbb {U}}_tG\in \textrm{Dom}({{\dot{{\mathbb {H}}}}}(\mu ,\nu ))$ together with ${\mathbb {U}}_t$-invariance of $\textrm{Dom}( \ddot{\mathbb {H}}(\mu ,\nu ))$ proves that $\textrm{Dom}(\dot{{\mathbb {H}}}(\mu ,\nu ))$ is ${\mathbb {U}}_t$-invariant, which, as it has been explained above, implies that ${{\dot{{\mathbb {H}}}}}(\mu ,\nu )$ is homogeneous:

$$\begin{aligned} {\mathbb {U}}_t {{\dot{{\mathbb {H}}}}}(\mu ,\nu ){\mathbb {U}}_t^*=e^t{{\dot{{\mathbb {H}}}}}(\mu ,\nu ), \quad t\in {\mathbb {R}}. \end{aligned}$$

$\square$

Lemma 6.3

The following operator inclusion

$$\begin{aligned} {{\dot{{\mathbb {H}}}}}(\mu , \nu )\subset (-\widehat{H}(\mu ))^*\oplus ({{\widehat{H}}}(\nu ))^*,\quad \mu , \nu \in i{\mathbb {R}}_+, \end{aligned}$$

(6.13)

holds.

Proof

Recall that

$$\begin{aligned} \textrm{Dom}({{\dot{{\mathbb {H}}}}}(\mu ,\nu ))=\textrm{Dom}( \ddot{\mathbb {H}}(\mu ,\nu ))\dot{+}\textrm{span}\left\{ G\right\} , \end{aligned}$$

where $G=f^\mu \oplus f^\nu$ is given by (6.2) and $\ddot{\mathbb {H}}(\mu , \nu )$ is the symmetric operator with deficiency indices (2, 2) in the direct sum of the Hilbert spaces $L^2({\mathbb {R}}_+)\oplus L^2({\mathbb {R}}_+)$ (see eq. (6.1)).

By Theorem 5.1,

$$\begin{aligned} f^\tau \in \textrm{Dom}(({{\widehat{H}}}(\tau ))^*),\quad \tau =\mu , \nu . \end{aligned}$$

Therefore, since $G=f^\mu \oplus f^\nu$, we have

$$\begin{aligned} G\in \textrm{Dom}((-{{\widehat{H}}}(\mu ))^*\oplus ({{\widehat{H}}}(\nu ))^*), \end{aligned}$$

(6.14)

which shows that

$$\begin{aligned} \textrm{Dom}({{\dot{{\mathbb {H}}}}}(\mu , \nu ))\subset \textrm{Dom}\left( (-{{\widehat{H}}}(\mu ))^*\oplus ({{\widehat{H}}}(\nu ))^*\right) . \end{aligned}$$

(6.15)

Since both ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ and $(-{{\widehat{H}}}(\mu ))^*\oplus ({{\widehat{H}}}(\nu ))^*$ are restrictions of $( \ddot{\mathbb {H}}(\mu , \nu ))^*$, the domain inclusion (6.15) implies (6.13). $\square$

Remark 6.4

Notice that in the right-hand side of the operator inclusion (6.13) the operator $(-{{\widehat{H}}}(\mu ))^*$ is dissipative while $({{\widehat{H}}}(\nu ))^*$ is accumulative.

The operator inclusion in Lemma 6.3, allows one to apply [27, Theorem 5.4] to ensure the existence a unique maximal dissipative “triangular" extension ${{\widehat{{\mathbb {H}}}}} (\mu , \nu )$ of ${{\dot{{\mathbb {H}}}}} (\mu , \nu )$ with the property that

$$\begin{aligned} {\mathbb {P}}({{\widehat{{\mathbb {H}}}}} (\mu , \nu ))|_{L^2({\mathbb {R}}_+)}=(-{{\widehat{H}}}(\mu ))^*. \end{aligned}$$

(6.16)

Here, ${\mathbb {P}}$ is the orthogonal projection in the direct sum ${{\mathcal {H}}}=L^2({\mathbb {R}}_+)\oplus L^2({\mathbb {R}}_+)$ of the Hilbert spaces onto the subspace $L^2({\mathbb {R}}_+)\oplus \{0\}\subset {{\mathcal {H}}}$.

Following [27], we call ${{\widehat{{\mathbb {H}}}}} (\mu , \nu )$ the operator coupling of the dissipative operators $(-\widehat{H}(\mu ))^*$ and ${{\widehat{H}}}(\nu )$ extending ${{\dot{{\mathbb {H}}}}}(\mu ,\nu )$ and use the notation

$$\begin{aligned} {{\widehat{{\mathbb {H}}}}}(\mu , \nu )= (-{{\widehat{H}}}(\mu ))^*\uplus {{\widehat{H}}}(\nu ), \quad \mu , \nu \in i{\mathbb {R}}_+. \end{aligned}$$

Notice that the symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu ,\nu )$ can be recovered from the operator coupling ${{\widehat{{\mathbb {H}}}}}(\mu , \nu )$ as its symmetric part determined by the restriction of the operator coupling ${{\widehat{{\mathbb {H}}}}}(\mu , \nu )$ on

$$\begin{aligned} \textrm{Dom}({{\dot{{\mathbb {H}}}}}(\mu ,\nu ))=\textrm{Dom}({{\widehat{{\mathbb {H}}}}}(\mu ,\nu ))\cap \textrm{Dom}({{\widehat{{\mathbb {H}}}}}(\mu ,\nu ))^*. \end{aligned}$$

(6.17)

In writing,

$$\begin{aligned} \dot{{\mathbb {H}}}(\mu ,\nu )={{\widehat{{\mathbb {H}}}}}(\mu ,\nu )|_{\textrm{Dom}({{\dot{{\mathbb {H}}}}}(\mu ,\nu ))}, \quad \mu , \nu \in i{\mathbb {R}}_+. \end{aligned}$$

(6.18)

7 The characteristic function of the symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$

Our next goal is to evaluate the characteristic function of the (homogeneous) symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ introduced in Section 6.

Theorem 7.1

The characteristic function $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)$ of the sole symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$, $\mu , \nu \in i{\mathbb {R}}_+$ can be represented as

$$\begin{aligned} s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)\cong \frac{S(z)-S(i) }{\overline{S(i)}S(z)-1}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(7.1)

where the function S(z) is given by

$$\begin{aligned} S(z)=z^\mu \left( -\frac{1}{z}\right) ^{\nu },\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(7.2)

More explicitly,

$$\begin{aligned} s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)\cong \frac{z^\mu \left( -\frac{1}{z}\right) ^\nu -e^{-\frac{\pi }{2} (|\mu |+|\nu |)}}{e^{-\frac{\pi }{2} (|\mu |+|\nu |)}z^\mu \left( -\frac{1}{z}\right) ^\nu -1},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(7.3)

In particular, if $\mu =\nu$, we have

$$\begin{aligned} s_{\dot{{\mathbb {H}}}(\nu , \nu )}(z)\cong 0, \quad z\in {\mathbb {C}}_+. \end{aligned}$$

(7.4)

Proof

From the multiplication theorem (see [27, Theorem 6.1]), it follows that the characteristic function $S_{{{\widehat{{\mathbb {H}}}}}(\mu , \nu )}(z)$ of the operator coupling

$$\begin{aligned} {{\widehat{{\mathbb {H}}}}}(\mu , \nu )= (-{{\widehat{H}}}(\mu ))^*\uplus {{\widehat{H}}}(\nu ) \end{aligned}$$

extending ${{\dot{{\mathbb {H}}}}}(\mu ,\nu )$ and given by (6.16) and (6.18) splits into the product of the characteristic functions $S_{ (-{{\widehat{H}}}(\mu ))^*} (z)$ and $S_{ {{\widehat{H}}}(\nu )} (z)$ of the dissipative operators $(-{{\widehat{H}}}(\mu ))^*$ and $\widehat{H}(\nu )$, respectively.

By Theorem 5.1, we know that the characteristic function $S_{ {{\widehat{H}}}(\nu )} (z)$ of the (sole) dissipative operator ${{\widehat{H}}}(\mu )$ can be represented as

$$\begin{aligned} S_{ {{\widehat{H}}}(\nu )} (z)\cong \left( -\frac{1}{z}\right) ^\nu ,\quad z\in {\mathbb {C}}_+, \end{aligned}$$

while the characteristic function $S_{ (-{{\widehat{H}}}(\mu ))^*} (z)$ of the dissipative operator $- {{\widehat{H}}}(\mu )^*$ is given by

$$\begin{aligned} S_{ (-{{\widehat{H}}}(\mu ))^*} (z)\cong \overline{S_{ {{\widehat{H}}}(\mu )} (-{\overline{z}})}\cong z^\mu ,\quad z\in {\mathbb {C}}_+. \end{aligned}$$

Therefore,

$$\begin{aligned} S_{{{\widehat{{\mathbb {H}}}}}(\mu , \nu )}(z)\cong z^\mu \left( -\frac{1}{z}\right) ^\nu ,\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(7.5)

In particular, the characteristic function $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)$ of the symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ admits the representation

$$\begin{aligned} s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)\cong \frac{S_{{{\widehat{{\mathbb {H}}}}}(\mu , \nu )}(z)-S_{{{\widehat{{\mathbb {H}}}}}(\mu , \nu )}(i)}{\overline{S_{{{\widehat{{\mathbb {H}}}}}(\mu , \nu )}(i)}S_{{{\widehat{{\mathbb {H}}}}}(\mu , \nu )}(z)-1}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

which together with (7.5) proves (7.1). $\square$

Remark 7.2

As it follows from explicit representation (7.3), for $\mu \ne \nu$ the characteristic function $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)$ has a geometric progression of zeros $z_n$, $n\in {\mathbb {Z}}$ on the imaginary semi-axis $i{\mathbb {R}}_+$ (cf. Remark 4.2, eq. (4.20))

$$\begin{aligned} z_n=i\exp \left[ -\frac{2\pi }{{|\mu |-|\nu |}}n\right] ,\quad n\in {\mathbb {Z}}, \end{aligned}$$

which, as we have already seen, is a harbinger of the loss of initial (scaling) symmetry in the problem in question (cf. Proposition 4.4).

The next goal is to show rigorously that the symmetric operators ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ form a two-parameter family of homogeneous symmetric operators with broken scaling symmetry. More precisely, we will show that the symmetric operator $\dot{{\mathbb {H}}}(\mu , \nu )$ does not admit a homogeneous self-adjoint extension with respect to the modified scaling group ${\mathbb {U}}_t$.

Our reasoning relies on the results of [30] where a complete list of characteristic functions for homogeneous operators that do admit a homogeneous self-adjoint extension are presented. We will simply show that the characteristic function (7.1) is not in that list.

Theorem 7.3

Assume that $\mu , \nu \in i{\mathbb {R}}_+$ and $\mu \ne \nu$. Then, the homogeneous symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ does not have self-adjoint extensions that are homogeneous with respect to the modified scaling group ${\mathbb {U}}_t$.

Proof

Suppose to the contrary that ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ $(\mu \ne \nu )$ has a self-adjoint homogeneous extension with respect to the group ${\mathbb {U}}_t$. In this case, from [30, Theorem 4.1], it follows that the characteristic function $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)$ of the symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ has to be either of the form

• $$\begin{aligned} s_\tau (z)=\frac{\left( \frac{z}{i}\right) ^\tau -1}{\left( \frac{z}{i}\right) ^\tau -\Theta _\tau (p,q)}, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

  (7.6)

  where

  $$\begin{aligned} \Theta _\tau (p,q)= \frac{(p-q) \cos \frac{\pi }{2} \tau -i\sin \frac{\pi }{2} \tau }{(p-q) \cos \frac{\pi }{2} \tau +i\sin \frac{\pi }{2} \tau } \end{aligned}$$

  (7.7)

  for some $\tau \in (0,1)$ and some weights $p,q\ge 0$, $p+q=1$,

  or $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)$ coincides with

  $$\begin{aligned} s_0(z)=\frac{(q-p)\log \left( \frac{z}{i}\right) }{(q-p)\log \left( \frac{z}{i}\right) -i\pi },\quad z\in {\mathbb {C}}_+. \end{aligned}$$

  (7.8)

Here, (for $\tau \ne 0)$ the power function $\left( \frac{z}{i}\right) ^\tau$ is defined as

$$\begin{aligned} \left( \frac{z}{i}\right) ^\tau =e^{\tau \log \left( \frac{z}{i}\right) }, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

via the standard branch of the logarithmic function $\log z$ with the cut on the negative real axis, $z\in {\mathbb {C}}{\setminus }{\mathbb {R}}_-$ and $\log \lambda \in {\mathbb {R}}$ for $\lambda >0$.

One observes that under the hypothesis $\mu \ne \nu$, the characteristic function $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)$ does not vanish identically. Moreover, the function $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(iy)$ (see (7.1)) is log-periodic function in y on the semi-axis $(0, \infty )$ with the (log)-period T given by

$$\begin{aligned} T= \frac{2\pi }{|\mu -\nu |}. \end{aligned}$$

Therefore, the congruence

$$\begin{aligned} s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)\cong s_\tau (z), \quad z\in {\mathbb {C}}_+, \end{aligned}$$

does not take place for any $\tau \in [0,1)$ (the corresponding functions are not log-periodic). In other words, the assumption that the homogeneous symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu ,\nu )$ admits a self-adjoint homogeneous extension fails to hold. $\square$

We also need the following result.

Lemma 7.4

The homogeneous symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$, $\mu , \nu \in i{\mathbb {R}}_+$ is prime.

Proof

Consider the self-adjoint (reference) extension ${\mathbb {H}}$ of the symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ such that the Livšic function s(z) associated with the pair $({{\dot{{\mathbb {H}}}}}(\mu , \nu ),{\mathbb {H}})$ coincides with $\frac{S(z)-S(i) }{\overline{S(i)}S(z)-1}$, where S(z) is given by (7.2). From Theorem 7.1, it follows that such an extension does exist.

Observe that

$$\begin{aligned} |S(z)|\le e^{-\pi \min \{|\mu |, |\nu |\}}<1,\quad z\in {\mathbb {C}}_+. \end{aligned}$$

Therefore, the range of the the Livšic function s(z) associated with the pair $({{\dot{{\mathbb {H}}}}}(\mu , \nu ),{\mathbb {H}})$ lies within a closed circle of the radius $r<1$.

As in the proof Corollary 4.3, one shows that in the integral representation

$$\begin{aligned} M(z)=\int _{\mathbb {R}}\left( \frac{1}{\lambda -z}-\frac{\lambda }{1+\lambda ^2}\right) d\mu (\lambda ), \quad z\in {\mathbb {C}}_+, \end{aligned}$$

for the corresponding Weyl–Titchmarsh function M(z) associated with the pair $({{\dot{{\mathbb {H}}}}}(\mu , \nu ),{\mathbb {H}})$ and given by (see (2.9))

$$\begin{aligned} M(z)=\frac{1}{i}\cdot \frac{s(z)+1}{s(z)-1},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

the measure $\mu (d\lambda )$ is absolutely continuous.

Arguing exactly in the same way as in the proof of Proposition 4.4, one observes that the subspace

$$\begin{aligned} {{\mathcal {H}}}_1= \overline{\textrm{span}_{z\in {\mathbb {C}}\setminus {\mathbb {R}}}\textrm{Ker}\left( ({{\dot{{\mathbb {H}}}}}(\mu , \nu ))^*-zI\right) } \end{aligned}$$

reduces ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ and that the part $A_0$ of associated with its orthogonal complement ${{\mathcal {H}}}_0={{\mathcal {H}}}_1^\perp$ is a homogeneous self-adjoint operator with respect to the unitary group of modified scaling transformations (6.10).

Suppose that ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ is not a prime operator and therefore the reducing subspace ${{\mathcal {H}}}_0$ is non-trivial.

It follows that either $\textrm{spec}(A_0)=\{0\}$, or the self-adjoint operator $A_0$ has absolutely continuous spectrum filling in at least one of the semi-axes ${\mathbb {R}}_\pm$. Since $\textrm{Ker}(\dot{{\mathbb {H}}}(\mu ,\nu ))$ is obviously trivial, the first outcome does not occur. The hypothesis that the subspace ${{\mathcal {H}}}_0$ is non-trivial necessarily implies that the self-adjoint reference operator ${\mathbb {H}}$ has absolutely continuous spectrum of multiplicity at least two on one of the semi-axes.

However, the direct sum $(-H(\mu ))\oplus H(\nu )$ of the self-adjoint operators $-H(\mu )$ and $H(\nu )$ (referred to in Theorem 5.1) has simple spectrum of Lebesgue type (filling in the whole real axis). (As it follows from the proof of Proposition  4.4, the self-adjoint operators $-H(\mu )$ and $H(\nu )$ have simple absolute continuous spectrum filling in the negative and positive semi-axes, respectively.) Both the reference operator ${\mathbb {H}}$ and the direct sum of the self-adjoint operators $(-H(\mu ))\oplus H(\nu )$ are self-adjoint extensions of the symmetric operator $\ddot{\mathbb {H}}(\mu , \nu )=(-\dot{H}(\mu ))\oplus \dot{H}(\nu )$ with deficiency indices (2, 2). Therefore, the multiplicities of the absolute continuous spectra of these operators must coincide due to stability of absolutely continuous spectrum under finite rank perturbations. So, the self-adjoint operator ${\mathbb {H}}$ has simple absolutely continuous spectrum on the real axis.

The obtained contradiction shows that the hypothesis that ${{\mathcal {H}}}_0$ is non-trivial fails to hold which proves that ${{\dot{{\mathbb {H}}}}}(\mu ,\nu )$ is a prime symmetric operator $\square$

Lemma 7.5

Suppose that $\mu ,\nu \in i{\mathbb {R}}_+$ and $\mu \ne \nu$. Then the symmetric operators ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ and ${{\dot{{\mathbb {H}}}}}(\mu ', \nu ')$ are unitarily equivalent only if $\mu =\mu '$ and $\nu = \nu '$.

Proof

Suppose that ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ and ${{\dot{{\mathbb {H}}}}}(\mu ', \nu ')$ are unitarily equivalent. Then for the characteristic functions $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)$ and $s_{{{\dot{{\mathbb {H}}}}}(\mu ', \nu ')}(z)$ of the symmetric operators ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ and $\dot{{\mathbb {H}}}(\mu ', \nu ')$, the congruence

$$\begin{aligned} s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)\cong s_{{{\dot{{\mathbb {H}}}}}(\mu ', \nu ')}(z),\quad z\in {\mathbb {C}}_+, \end{aligned}$$

(7.9)

holds. Under the hypothesis that $\mu \ne \nu$, the function $s_{\dot{{\mathbb {H}}}(\mu , \nu )}(z)$ does not vanish identically, so does $s_{\dot{{\mathbb {H}}}(\mu ', \nu ')}(z)$ and, therefore, $\mu '\ne \nu '$ as it clearly seen from the explicit representation (7.3).

For the boundary values of the characteristic function $s_{\dot{{\mathbb {H}}}(\mu , \nu )}(z)$ on the positive real axis, we have the congruence

$$\begin{aligned} s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(\lambda +i0)\cong \frac{\lambda ^\mu \left( -\frac{1}{\lambda }\right) ^\nu -\beta }{\beta \lambda ^\mu \left( -\frac{1}{\lambda }\right) ^\nu -1}, \quad \lambda \in {\mathbb {R}}_+, \end{aligned}$$

where

$$\begin{aligned} \beta =e^{-\frac{\pi }{2}(|\mu |+|\nu |)}. \end{aligned}$$

Observing that

$$\begin{aligned} \lambda ^\mu \left( -\frac{1}{\lambda }\right) ^\nu =e^{-\pi |\nu |}e^{i(|\mu |-|\nu |)\log \lambda }, \quad \lambda \in {\mathbb {R}}_+, \end{aligned}$$

we get

$$\begin{aligned} s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(e^x+i0)\cong \frac{e^{-\pi |\nu |}e^{i\omega x}-\beta }{ \beta e^{-\pi |\nu |} e^{i\omega x}-1},\quad x\in {\mathbb {R}}, \end{aligned}$$

with

$$\begin{aligned} \omega =|\mu |-|\nu |\ne 0. \end{aligned}$$

Similarly,

$$\begin{aligned} s_{{{\dot{{\mathbb {H}}}}}(\mu ', \nu ')}(e^x+i0)\cong \frac{e^{-\pi |\nu '|}e^{i\omega ' x}-\beta ' }{ \beta e^{-\pi |\nu '|} e^{i\omega ' x}-1},\quad x\in {\mathbb {R}}, \end{aligned}$$

with

$$\begin{aligned} \beta ' =e^{-\frac{\pi }{2}(|\mu '|+|\nu '|)}\quad \text {and}\quad \omega '=|\mu '|-|\nu '|\ne 0. \end{aligned}$$

From congruence (7.9), it follows that there exists a $\Theta$, $|\Theta |=1$ such that

$$\begin{aligned} \frac{e^{-\pi |\nu |}e^{i\omega x}-\beta }{ \beta e^{-\pi |\nu |} e^{i\omega x}-1}=\Theta \frac{e^{-\pi |\nu '|}e^{i\omega ' x}-\beta ' }{ \beta 'e^{-\pi |\nu '|} e^{i\omega ' x}-1},\quad x\in {\mathbb {R}}. \end{aligned}$$

(7.10)

Both the right- and left-hand sides of (7.10) (as functions in x) admit meromorphic continuation to the whole complex plane ${\mathbb {C}}$. In particular, the equations

$$\begin{aligned} \beta e^{-\pi |\nu |} e^{i\omega z}=1\quad \text {and}\quad \beta 'e^{-\pi |\nu '|} e^{i\omega 'z}=1, \quad z\in {\mathbb {C}}\end{aligned}$$

have the same set of roots. It follows then that

$$\begin{aligned} \omega =\omega '\quad \text { and}\quad \beta e^{-\pi |\nu |}=\beta 'e^{-\pi |\nu '|}. \end{aligned}$$

Since $\omega =\mu -\nu$ and $\omega '=\mu '-\nu '$, we have

$$\begin{aligned} |\mu |-|\nu |=|\mu '|-|\nu '|. \end{aligned}$$

(7.11)

Taking into account that

$$\begin{aligned} \beta e^{-\pi |\nu |}=e^{-\frac{\pi }{2}(|\mu |+|\nu |)}e^{-\pi |\nu |}=e^{-\frac{\pi }{2}(|\mu |+3|\nu |)} \end{aligned}$$

and also that

$$\begin{aligned} \beta e^{-\pi |\nu |}=e^{-\frac{\pi }{2}(|\mu '|+3|\nu '|)}, \end{aligned}$$

we obtain

$$\begin{aligned} |\mu |+3|\nu |=|\mu '|+3|\nu '|. \end{aligned}$$

(7.12)

Comparing (7.11) and (7.12) shows that

$$\begin{aligned} |\mu |=|\mu '|\quad \text {and}\quad |\nu |=|\nu '|, \end{aligned}$$

and, hence, $\mu =\mu '$ and $\nu =\nu '$ for $\mu , \nu \in i{\mathbb {R}}_+$, which completes the proof. $\square$

Remark 7.6

Notice that for $\mu =\nu$, from Theorem 7.1 (see eq. (4.20)) it follows that the characteristic function $s_{{{\dot{{\mathbb {H}}}}}(\mu , \nu )}(z)=s_{{{\dot{{\mathbb {H}}}}}(\nu , \nu )}(z)$ vanishes identically in the upper half-plane. Therefore, the symmetric operators $\dot{{\mathbb {H}}}(\nu , \nu )$ are unitarily equivalent to each other for all $\nu \in i{\mathbb {R}}_+$ (recall that ${{\dot{{\mathbb {H}}}}}(\nu , \nu )$ are prime operators by Lemma 7.4). In fact, the operators $\dot{{\mathbb {H}}}(\nu , \nu )$, $\nu \in i{\mathbb {R}}_+$ are unitarily equivalent to the symmetric differentiation operator

$$\begin{aligned} \dot{{\mathbb {D}}}=i\frac{d}{dx} \end{aligned}$$

(7.13)

on the real axis defined on

$$\begin{aligned} \textrm{Dom}({{\dot{{\mathbb {D}}}}})=\left\{ f\in W_2^1(-\infty ,0)\oplus W_2^1(0,\infty )\,|\,f(0+)=f(0-)=0\right\} . \end{aligned}$$

(7.14)

Here, we have used the well known fact that ${{\dot{{\mathbb {D}}}}}$ is a prime symmetric operator with deficiency indices (1, 1) and $s_{ {{\dot{{\mathbb {D}}}}}}(z)=0$, $z\in {\mathbb {C}}_+$, see, e.g., [1].

We conclude this section by the observation that the unique maximal dissipative homogeneous extension of ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$, $\mu \ne \nu$ with respect to the modified group of scaling transformations guaranteed by [28, Theorem 5.4] coincides with the operator coupling

$$\begin{aligned} {{\widehat{{\mathbb {H}}}}}(\mu , \nu )= (-{{\widehat{H}}}(\mu ))^*\uplus {{\widehat{H}}}(\nu ) \end{aligned}$$

(7.15)

extending $\dot{{\mathbb {H}}}(\mu , \nu )$.

Indeed, by Theorem 7.3 scaling symmetry for the operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ is broken, and one can use [28, Theorem 5.4] to assert that there exists a a unique homogeneous maximal dissipative extension $\check{{\mathbb {H}}}$ of $\dot{{\mathbb {H}}}(\mu ,\nu )$. By Theorem 3.1, the characteristic function of the dissipative operator $\check{{\mathbb {H}}}$ admits the representation

$$\begin{aligned} S_{\check{{\mathbb {H}}}}(z)\cong z^{\mu '}\left( -\frac{1}{z}\right) ^{\nu '},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(7.16)

for some $\mu '$ and $\nu '$, $\mu ',\nu '\in i{\mathbb {R}}_+$.

On the other hand, the characteristic function of the operator coupling ${{\widehat{{\mathbb {H}}}}}(\mu , \nu )$ can be represented as (see eq. (7.5))

$$\begin{aligned} S_{{{\widehat{{\mathbb {H}}}}}(\mu , \nu )}(z)\cong z^\mu \left( -\frac{1}{z}\right) ^\nu ,\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(7.17)

Since $\mu \ne \nu$, by Lemma 7.5, we conclude that $\mu =\mu '$ and $\nu =\nu '$. Hence,

$$\begin{aligned} S_{{{\widehat{{\mathbb {H}}}}}(\mu , \nu )}(z)\cong S_{\check{{\mathbb {H}}}}(z), \quad z\in {\mathbb {C}}_+, \end{aligned}$$

which shows that the dissipative extensions $\check{{\mathbb {H}}}$ and ${{\widehat{{\mathbb {H}}}}}(\mu , \nu )$ are unitarily equivalent by the uniqueness Theorem (2.1). Here, we have used that the symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ is a prime operator by Lemma 7.4.

However, a symmetric operator with deficiency indices (1, 1) admits distinct unitarily equivalent quasi-self-adjoint extensions if and only if its characteristic function vanishes identically in the upper half-plane, see, e.g., [3]. By Theorem 7.1, the characteristic function of the (sole) symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$, $\mu \ne \nu$ does not vanish on ${\mathbb {C}}_+$. Therefore, unitary equivalence of $\check{{\mathbb {H}}}(\nu )$ and ${{\widehat{{\mathbb {H}}}}}(\mu ,\nu )$ implies $\check{{\mathbb {H}}} =\widehat{\mathbb {H}}(\mu , \nu )$. Since $\check{{\mathbb {H}}}$ is homogeneous, so is ${{\widehat{{\mathbb {H}}}}}(\mu , \nu )$, which proves the claim.

8 The homogeneous Jørgensen–Muhly problem in the case of broken symmetry

Extending the family of symmetric operators ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$, $\mu , \nu \in i{\mathbb {R}}_+,$ with $\mu \ne \nu$ to the borderline case where exactly one of the exponents $\mu$ or $\nu$ vanishes by setting

$$\begin{aligned} {{\dot{{\mathbb {H}}}}}(0, \nu )=\dot{H}(\nu ),\quad \nu \in i{\mathbb {R}}_+, \end{aligned}$$

(8.1)

and

$$\begin{aligned} {{\dot{{\mathbb {H}}}}}(\mu ,0)=-(\dot{H}(\mu ))^*,\quad \mu \in i{\mathbb {R}}_+. \end{aligned}$$

(8.2)

we come to the two-parameter family

$$\begin{aligned} i{\mathbb {R}}_+\cup \{0\}\ni (\mu , \nu )\mapsto \dot{{\mathbb {H}}}(\mu , \nu ) \quad (\mu \ne \nu ) \end{aligned}$$

(8.3)

of unitarily non-equivalent homogeneous symmetric operators that do not allow self-adjoint homogeneous extensions with respect to the given symmetry group, see Proposition 4.4, Theorem 7.3 and Lemma 7.6.

The obtained list of model operators should be supplemented with the symmetric differentiation operator given by (7.13)

$$\begin{aligned} {{\dot{{\mathbb {D}}}}}=i\frac{d}{dx} \end{aligned}$$

with

$$\begin{aligned} \textrm{Dom}({{\dot{{\mathbb {D}}}}})=\left\{ f\in W_2^1(-\infty ,0)\oplus W_2^1(0,\infty )\,|\,f(0+)=f(0-)=0\right\} . \end{aligned}$$

Clearly, the symmetric operator ${{\dot{{\mathbb {D}}}}}$ is homogeneous with respect to the unitary (modified) group of scaling transformations

$$\begin{aligned} (V_tf)(x)=e^{i\, \textrm{sgn}(x) t}e^{-\frac{t}{2}}f(e^{-t}x), \quad f\in L^2({\mathbb {R}}),\quad t\in {\mathbb {R}}, \end{aligned}$$

and has no homogeneous self-adjoint extensions with respect to the group $V_t$.

Indeed, the domain of the self-adjoint extension ${\mathbb {D}}_\Theta$ of ${{\dot{{\mathbb {D}}}}}$ on

$$\begin{aligned} \textrm{Dom}({\mathbb {D}}_\Theta )=\left\{ f\in W_2^1(-\infty ,0)\oplus W_2^1(0,\infty )\,|\,f(0+)=\Theta f(0-)\right\} \end{aligned}$$

(8.4)

is not $V_t$-invariant for any $|\Theta |=1$ but the one-parameter family ${\mathbb {D}}_\Theta$, $|\Theta |=1$ describes all self-adjoint extensions of ${{\dot{{\mathbb {D}}}}}$. So, symmetry is broken for this particular choice of the unitary group.

The main result of this paper below states that the two-parameter family (8.3) together with the differentiation symmetric operator ${{\dot{{\mathbb {D}}}}}$ on the real axis presents the complete list of non-unitarily equivalent solutions to the homogeneous Jørgensen–Muhly problem in the case where symmetry is broken.

Theorem 8.1

Suppose that $\dot{A}$ is a closed, densely defined, prime symmetric operator with deficiency indices (1, 1) in the Hilbert space ${{\mathcal {H}}}$.

Assume that $\dot{A}$ is homogeneous (scale-coinvariant) with respect to some group of unitary operators. Assume, in addition, that $\dot{A}$ does not admit homogeneous self-adjoint extensions with respect to the given group.

Then $\dot{A}$ is either unitarily equivalent to

• (i) the differentiation operator ${{\dot{{\mathbb {D}}}}}$ defined by Eqs. (7.13) and (7.14)

  or to

• (ii) the symmetric operator ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ for some uniquely determined

  $$\begin{aligned} \mu , \nu \in i{\mathbb {R}}_+\cup \{0\} \quad (\mu \ne \nu ). \end{aligned}$$

In the first case, the characteristic function $s_{\dot{A}}(z)$ of the symmetric operator $\dot{A}$ vanishes identically in the upper half-place,

$$\begin{aligned} s_{\dot{A}}(z)=0,\quad z\in {\mathbb {C}}_+. \end{aligned}$$

In the second case, $s_{\dot{A}}(z)$ admits the representation

$$\begin{aligned} s_{\dot{A}}(z)\cong \frac{z^\mu \left( -\frac{1}{z}\right) ^{\nu }-e^{-\frac{\pi }{2}(|\mu |+|\nu |)}}{e^{-\frac{\pi }{2}(|\mu |+|\nu |)}z^\mu \left( -\frac{1}{z}\right) ^{\nu }-1},\quad z\in {\mathbb {C}}_+. \end{aligned}$$

(8.5)

Proof

Suppose that $\dot{A}$ is a homogeneous operator such that $\dot{A}$ does not admit homogeneous self-adjoint extensions with respect to the symmetry group. Theorem 5.4 in [28] ensures the existence of a (unique) homogeneous quasi-self-adjoint dissipative extension ${{\widehat{A}}}$ of $\dot{A}$. By Theorem 3.1, the characteristic function $S_{{{\widehat{A}}}}(z)$ of of the sole dissipative operator $\widehat{A}$ either vanishes, that is,

$$\begin{aligned} S_{\widehat{A}}(z)\cong 0, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(8.6)

or the congruence

$$\begin{aligned} S_{{{\widehat{A}}}}(z)\cong z^\mu \left( -\frac{1}{z}\right) ^{\nu }, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(8.7)

takes place, with some $\mu , \nu \in i{\mathbb {R}}_+\cup \{0\}$, not both zero.

If (8.6) takes place, we also have that $s_{\dot{A}}(z)\equiv 0$ and, therefore, $\dot{A}$ is unitarily equivalent to the symmetric differentiation operator ${{\dot{{\mathbb {D}}}}}$, since both $\dot{A}$ and $\dot{{\mathbb {D}}}$ are prime symmetric operators and their characteristics functions coincide.

Now, suppose that (8.7) holds. If $\mu =\nu$, then the characteristic function $S_{{{\widehat{A}}}}(z)$ is a constant function,

$$\begin{aligned} S_{{{\widehat{A}}}}(z)\cong S_{{{\widehat{A}}}}(i) \cong e^{-\pi |\nu |},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

and then again $s_{\dot{A}}(z)\equiv 0$ and, therefore, $\dot{A}$ is unitarily equivalent to ${{\dot{{\mathbb {D}}}}}$ in this case as well.

Assume, therefore, that (8.7) holds with $\mu \ne \nu$. If neither $\mu$ nor $\nu$ is equal to zero, comparing (7.17) and (8.7), we see that the characteristic function $S_{{{\widehat{A}}}}(z)$ is congruent to the characteristic function of the operator coupling ${{\widehat{{\mathbb {H}}}}}(\mu , \nu )$ given by (7.15). Since $\dot{A}$ is a prime symmetric operator by the hypothesis and the symmetric part ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ of ${{\widehat{{\mathbb {H}}}}}(\mu , \nu )$ (see (6.17) and (6.18)) is also a prime operator by Lemma 7.4, one can apply the uniqueness Theorem 2.1 to conclude that $\dot{A}$ and ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ are unitary equivalent.

In the borderline case where $\mu =0$, we have

$$\begin{aligned} S_{{{\widehat{A}}}}(z)\cong \left( -\frac{1}{z}\right) ^{\nu }, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(8.8)

which by Theorem 5.1 coincides with the characteristic function of the dissipative operator $\widehat{H}(\nu )={{\widehat{{\mathbb {H}}}}}(0, \nu )$. Therefore, the dissipative operators ${{\widehat{H}}}(\nu )={{\widehat{{\mathbb {H}}}}}(0, \nu )$ and ${\widehat{A}}$ are unitarily equivalent. Consequently, the symmetric operator $\dot{A}$ and the symmetric part ${{\dot{{\mathbb {H}}}}}(0, \nu )$ of the dissipative operator ${{\widehat{{\mathbb {H}}}}}(0, \nu )$ are unitarily equivalent. (Here we use that $\dot{H}(\nu )= {{\dot{{\mathbb {H}}}}}(0, \nu )$ is a prime symmetric operator by Proposition 4.4.)

In a similar way, one shows that in the remaining borderline case with $\nu =0$, we have

$$\begin{aligned} S_{{{\widehat{A}}}}(z)\cong z^{\mu }, \quad z\in {\mathbb {C}}_+, \end{aligned}$$

(8.9)

and the symmetric operator $\dot{A}$ is unitarily equivalent to the symmetric part $\dot{{\mathbb {H}}}(\mu ,0)=-\dot{H}(\mu )$ of the dissipative operator $\widehat{\mathbb {H}}(\mu ,0)=-({{\widehat{H}}}(\mu ))^*$.

Finally, combining (8.7), (8.8), and (8.9), we obtain

$$\begin{aligned} s_{\dot{A}}(z)\cong \frac{S_{{{\widehat{A}}}}(z)-S_{\widehat{A}}(i)}{\overline{S_{{{\widehat{A}}}}(i)}S_{{{\widehat{A}}}}(z)-1}\cong \frac{z^\mu \left( -\frac{1}{z}\right) ^{\nu }-e^{-\frac{\pi }{2}(|\mu |+|\nu |)}}{e^{-\frac{\pi }{2}(|\mu |+|\nu |)}z^\mu \left( -\frac{1}{z}\right) ^{\nu }-1},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

which proves (8.5).

To ensure the uniqueness of the choice of the exponents $\mu$ and $\nu$ in the statement of the theorem, we observe that for $\mu \ne \nu$ the operators ${{\dot{{\mathbb {H}}}}}(\mu , \nu )$ and ${{\dot{{\mathbb {H}}}}}(\mu ', \nu ')$ are unitarily equivalent only if $\mu =\mu '$ and $\nu = \nu '$ (see Lemma 7.5 if $\mu \nu \ne 0$ and Remark 4.5 in the borderline case when exactly one of the exponents $\mu$ and $\nu$ vanishes: the symmetric operators $\dot{H}(\tau )$ and $\dot{H}(\tau ')$ are not unitary equivalent for $\tau \ne \tau '$, $\tau , \tau '\in i {\mathbb {R}}_+$). $\square$

In conclusion, we would like to emphasize the (dual) role that the symmetric differentiation operator ${{\dot{{\mathbb {D}}}}}=i\frac{d}{dx}$ plays in representation theory of the homogeneous as well as canonical commutation relations.

• (i). As we have already seen, the differentiation operator $\dot{{\mathbb {D}}}$ is homogeneous with respect to the unitary group of scaling transformations

  $$\begin{aligned} (V_tf)(x)=e^{i\, \textrm{sgn}(x) t}e^{-\frac{t}{2}}f(e^{-t}x), \quad f\in L^2({\mathbb {R}}),\quad t\in {\mathbb {R}}. \end{aligned}$$

  Moreover, there are no homogeneous self-adjoint extensions of $\dot{{\mathbb {D}}}$ with respect to that group. That is, in accordance with our terminology symmetry is broken in this case.

At the same time, the operator ${{\dot{{\mathbb {D}}}}}$ is also homogeneous with respect to the standard group of scaling transformations

$$\begin{aligned} (U_tf)(x)=e^{-\frac{t}{2}}f(e^{-t}x), \quad f\in L^2({\mathbb {R}}), \quad t\in {\mathbb {R}}: \end{aligned}$$

the domain (8.4) of the self-adjoint extension ${\mathbb {D}}_\Theta$ of ${{\dot{{\mathbb {D}}}}}$ is obviously $U_t$-invariant, any self-adjoint (even any quasi-selfadjoint) extension of ${{\dot{{\mathbb {D}}}}}$ is homogeneous with respect to the scaling group of unitary transformations $U_t$. Therefore, for the symmetry group of unitary transformations $U_t$ symmetry is not broken.

• (ii). We also notice that if $\dot{A}$ is unitarily equivalent to ${{\dot{{\mathbb {D}}}}}$, then $\dot{A}$ solves both the classical commutation relations

  $$\begin{aligned} V_t\dot{A}V_t^*=\dot{A}+t I \quad \text {on } \quad \textrm{Dom}(\dot{A}), \quad t\in {\mathbb {R}}, \end{aligned}$$

  (8.10)

  and the homogeneous commutation relations

  $$\begin{aligned} U_t\dot{A}U_t^*=e^t\dot{A} \quad \text {on } \quad \textrm{Dom}(\dot{A}), \quad t\in {\mathbb {R}}, \end{aligned}$$

  (8.11)

  for some unitary groups $V_t$ and $U_t$.

Indeed, for the differentiation operator ${{\dot{{\mathbb {D}}}}}$, the commutation relations (8.10) and (8.11) hold with the group of shifts $V_t$,

$$\begin{aligned} (V_tf)(x)=f(x-t), \quad f\in L^2({\mathbb {R}}), \end{aligned}$$

and with the scaling group $U_t$,

$$\begin{aligned} (U_tg)(x)=e^{-t/2}g(e^{-t}x), \quad g\in L^2({\mathbb {R}}), \end{aligned}$$

respectively.

• (iii). Conversely, if a prime symmetric operator $\dot{A}$ with deficiency indices (1, 1) satisfies both the classical and homogeneous commutation relations (8.10) and (8.11), then $\dot{A}$ is unitarily equivalent to the differentiation operator ${{\dot{{\mathbb {D}}}}}$ (see [30] where this result was stated without proof).

Indeed, otherwise, the characteristic functions of $\dot{A}$ does not vanish identically. then, in view of (8.11), one can apply Theorem 8.1 and see that the characteristic function $s_{\dot{A}}(z)$ is log-periodic. On the other hand, if commutation relations (8.10) hold, then from [29, Corollary 3.7] it follows that the characteristic function $s_{\dot{A}}(z)$ can be represented as

$$\begin{aligned} s_{\dot{A}}(z)\cong \kappa \frac{e^{iz\ell }-e^{-\ell }}{\kappa ^2e^{-\ell }e^{iz\ell }-1},\quad z\in {\mathbb {C}}_+, \end{aligned}$$

for some $0\le \kappa \le 1$ and $\ell >0$. The case $\kappa =0$ does not occur, since $s_{\dot{A}}(z)$ does not vanish identically in the upper half-plane. But then $s_{\dot{A}}(z)$ is periodic with the period $T=\frac{2\pi }{\ell }$. Thus, $s_{\dot{A}}(z)$ is periodic and log-periodic simultaneously. A contradiction.