Abstract

The problem of construction of self-adjoint Hamiltonian for quantum system consisting of three pointlike interacting particles (two fermions with mass 1 plus a particle of another nature with mass 𝑚>0) was studied in many works. In most of these works, a family of one-parametric symmetrical operators {𝐻𝜀,𝜀1} is considered as such Hamiltonians. In addition, the question about the self-adjointness of 𝐻𝜀 is equivalent to the one concerning the self-adjointness of some auxiliary operators {𝒯𝑙,𝑙=0,1,} acting in the space 𝐿2(1+,𝑟2𝑑𝑟). In this work, we establish a simple general criterion of self-adjointness for operators 𝒯𝑙 and apply it to the cases 𝑙=0 and 𝑙=1. It turns out that the operator 𝒯𝑙=0 is self-adjoint for any 𝑚, while the operator 𝒯𝑙=1 is self-adjoint for 𝑚>𝑚0, where the value of 𝑚0 is given explicitly in the paper.

1. Introduction and Statement of the Problem

This paper is continuation of works [14] studying the problem of construction of Hamiltonian for a quantum system which consists of two fermions with mass 1 interacting pointwise with a particle of another nature having mass 𝑚.

Originally, the construction of such Hamiltonian begins with introduction of the symmetric operator: 𝐻01=21𝑚Δ𝑦+Δ𝑥1+Δ𝑥2(1.1) acting in a Hilbert space =𝐿2(3)𝐿asym2(3×3). Here, 𝑥1, 𝑥23 are the positions of fermions, 𝑦 is the position of a separate particle, and Δ𝑦, Δ𝑥1, and Δ𝑥2 are Laplacians with respect to 𝑦, 𝑥1, and 𝑥2, respectively. The domain of definition of 𝐻0, 𝐷(𝐻0) consists of smooth rapidly decreasing functions 𝜓(𝑦,𝑥1,𝑥2) on infinity, antisymmetrical with respect to 𝑥1, 𝑥2 and satisfying the following conditions: 𝜓𝑦,𝑥1,𝑥2||𝑥𝑖=𝑦=0,𝑖=1,2.(1.2) Usually, some family {𝐻𝜀,𝜀1} of symmetric extensions of the operator 𝐻0 is proposed as a possible “true” Hamiltonian of the system (the so-called Ter-Martirosian-Skornyakov extensions, see [5]). These extensions were constructed in [14]. For some values of mass 𝑚, the extensions of Ter-Martirosian-Skornyakov are self-adjoint (for all values of the parameter 𝜀); however, for the other values of 𝑚 they are only symmetric with nonzero deficiency indexes (equal for all 𝜀). It turns out (see [3]) that the self-adjointness of all operators {𝐻𝜀} is equivalent to the one for some auxiliary symmetric operator 𝒯 acting in the space 𝐿2(3) (see below). This operator commutes with the operators {𝑈𝑔,𝑔𝑂3} of the representation of the rotation group 𝑂3 that acts in 𝐿2(3) by the usual formula: 𝑈𝑔𝑓𝑔(𝑘)=𝑓1𝑘,𝑔𝑂3,𝑓𝐿23.(1.3) Let us denote by 𝑙𝐿2(3) the maximal subspace, where the representation (1.3) is multiplied by the irreducible representation of 𝑂3 with weight 𝑙, 𝑙=0,1,2, (see [6]). Evidently, the space 𝑙 is invariant with respect to the operator 𝒯, and the restriction 𝒯𝑙=𝒯|𝑙 of this operator to the space 𝑙 is symmetric operator. The operator 𝒯 is self-adjoint if all the operators {𝒯𝑙,𝑙=0,1,} are self-adjoint. In this paper, we find general simple conditions of self-adjointness of 𝒯𝑙 and the form of the defect subspaces (with small exclusions) when these conditions are broken. Then, we apply these conditions to the cases 𝑙=0 and 𝑙=1 and get that the operator 𝒯𝑙=0 is self-adjoint for all values of 𝑚>0, while the operator 𝒯𝑙=1 is self-adjoint for 𝑚>𝑚0 and has nonzero deficiency indexes for 𝑚𝑚0, the constant 𝑚0>0 is indicated below (see (5.4)).

By the way, we note that the value of 𝑚0 obtained in this paper differs from that one given by mistake in [2].

2. A Short Explanation of the Constructions from Papers [13]

(1) After Fourier transformation: 𝜓𝑦,𝑥1,𝑥2𝜓𝑞,𝑘1,𝑘2=12𝜋9/233𝜓𝑦,𝑥1𝑥2𝑘exp𝑖(𝑞,𝑦)𝑖1,𝑥1𝑘𝑖2,𝑥2𝑑𝑦𝑑𝑥1𝑑𝑥2(𝜓)𝑞,𝑘1,𝑘2,(2.1) and change of variables: 𝑃=𝑞+𝑘1+𝑘2,𝑝𝑗=𝑃𝑚+2𝑘𝑗,𝑗=1,2,(2.2) the operator 𝐻0=𝐻01,(2.3) can be represented as a tensor sum: 𝐻0=𝐻0(1)+𝑚𝐻𝑚+10(2),(2.4) where 𝐻0(1) is a self-adjoint operator in 𝐿2(3): 𝐻0(1)𝑓𝑃(𝑃)=2𝑚+2𝑓(𝑃),𝑃3,𝑓𝐿23,(2.5) and 𝐻0(2) acts in 𝐿asym2(3×3) by formula 𝐻0(2)𝑔𝑝1,𝑝2𝑝=𝐺1,𝑝2𝑔𝑝1,𝑝2,𝑔𝐿asym23×3,(2.6) with 𝐺𝑝1,𝑝2=𝑝21+𝑝22+2𝑝𝑚+11,𝑝2>0.(2.7) The operator 𝐻0(2) is symmetric, and its domain is 𝐷𝐻0(2)=𝑔𝐿asym23×33𝑔𝑝1,𝑝2𝑑𝑝𝑗=0,𝑗=1,2,(2.8)

(2) the deficiency subspace 1𝐿asym2(3×3) of the operator 𝐻0(2) consists of the functions of the form: 𝑈𝑝1,𝑝2=𝑝1𝑝2𝐺𝑝1,𝑝2+1,(2.9) where the function (𝑝) belongs to Hilbert space =3||||(𝑝)2𝑝2,+1𝑑𝑝<(2.10) with inner product 1,2𝑈=1,𝑈2𝐿2(3×3)𝑊1,2𝐿2(3).(2.11) Here 𝑊 is some positive operator acting in 𝐿2(3) (see [3]). The domain of the operator (𝐻0(2)), that is, a conjugate to 𝐻0(2), is 𝐷𝐻0(2)=𝑔𝐿asym23×3𝑝𝑔1,𝑝2𝑝=𝑓1,𝑝2+𝑈𝑝1,𝑝2+𝑈𝜓𝑝1,𝑝2𝐺𝑝1,𝑝2+1,(2.12) where 𝐻𝑓𝐷(0(2)), ,𝜓. In addition, the operator (𝐻0(2)) acts by the formula: 𝐻0(2)𝑔𝑝1,𝑝2𝑝=𝐺1,𝑝2𝑔𝑝1,𝑝2𝑝1𝑝2,(2.13) where is defined by (2.12).

The following asymptotics holds for vectors 𝐻𝑔𝐷((0(2)))𝑁: ||𝑝1||<𝑁𝑔𝑝1,𝑝2𝑑𝑝1=4𝜋𝑁𝑝2𝑝+𝑏2+𝑜(1).(2.14) Here 𝑏(𝑝)=(𝑇)(𝑝)+(𝑊𝜓)(𝑝),(2.15) where the operator 𝑊 is defined in (2.11), and (𝑇)(𝑝) is given by the following expression (𝜇=2/(𝑚+1)) (𝑇)(𝑝)=2𝜋2𝜇124𝑝2+1(𝑝)+3(𝑡)𝐺(𝑡,𝑝)+1𝑑𝑡,(2.16) defined on the set: 𝐷(𝑇)=𝐿23||𝑝||(𝑝)𝐿23.(2.17) The above-mentioned Ter-Martirosian-Skornyakov's extension 𝐻𝜀(2) of the operator 𝐻0(2) is obtained by requiring 𝑏(𝑝)=𝜀(𝑝),(2.18) where 𝜀1 is an arbitrary parameter.

Lemma 2.1. The operator 𝑇 defined in the space 𝐿2(3) by (2.16) is symmetric, and the self-adjointness of the operators 𝐻𝜀 (for all 𝜀) is equivalent to the self-adjointness of the operator 𝑇 (see [2, 3, 5]).

The operator 𝑇 can be represented as a sum of two operators: 𝑇=𝒯+𝑇,(2.19) where the symmetric operator 𝒯 (with the domain 𝐷(𝒯)=𝐷(𝑇)) acts as follows: (𝒯)(𝑝)=2𝜋2𝜇124||𝑝||(𝑝)+3(𝑡)𝑑𝑡𝐺(𝑡,𝑝)(2.20) and 𝑇 is a bounded self-adjoint operator. Since the deficiency indexes of 𝑇 coincide with the ones of 𝒯 (see [7]), we shall study the conditions of self-adjointness for the operator 𝒯;

(3) as we said, the space 𝑙𝐿2(3) is invariant with respect to 𝒯; it has the form: 𝑙=𝐿21+,𝑟2𝑑𝑟𝐿𝑙2(𝑆),(2.21) where 𝐿𝑙2(𝑆)𝐿2(𝑆) is the space of spherical functions of weight 𝑙 (see [6]) on the unit sphere 𝑆3. In addition, the operator 𝒯𝑙=𝒯|𝑙 has the form 𝒯𝑙=𝑀𝑙𝐸𝑙,(2.22) where 𝐸𝑙 is the unit operator in 𝐿𝑙2(𝑠), and 𝑀𝑙 acts in 𝐿2(1+,𝑟2𝑑𝑟) by the formula: 𝑀𝑙𝑓(𝑟)=2𝜋2𝜇124𝑟𝑓(𝑟)+2𝜋11𝑑𝑥𝑃𝑙(𝑥)0(𝑟)2𝑓(𝑟)𝑑𝑟𝑟2+(𝑟)2,+𝜇𝑟𝑟𝑥(2.23) on the domain 𝐷𝑀𝑙𝑉=𝑢𝐿21+,𝑟2𝑑𝑟𝑟𝑢(𝑟)𝐿21+,𝑟2.𝑑𝑟(2.24) Here 𝑃𝑙(𝑥), 𝑙=0,1,2,, 𝑥[1,1], are orthogonal polynomials (Legendre polynomials) satisfying 𝑃𝑙(1)=1: 𝑃𝑙1(𝑥)=2𝑙𝑑𝑙!𝑙𝑑𝑥𝑙𝑥21𝑙,𝑥(1,1).(2.25) The operators {𝑀𝑙,𝑙=0,1,} are symmetric in 𝐿2(1+,𝑟2𝑑𝑟), and the self-adjointness of 𝑀𝑙 is equivalent to the self-adjointness of 𝒯𝑙. Later on, we shall study the operators 𝑀𝑙 and derive a condition of self-adjointness.

3. Preparatory Constructions

For every function 𝑢𝑉𝐿2(1+,𝑟2𝑑𝑟), we consider the family of functions 𝔥𝑢(𝑢)=𝛼=𝑟𝛼[]𝑢,𝛼0,1,𝑢0=𝑢,(3.1) which we call a chain (with initial element 𝑢=𝑢0 and the final one 𝑢1). All functions 𝑢𝛼𝔥(𝑢) belong to 𝐿2(1+,𝑟2𝑑𝑟) and have a uniformly bounded norm: 𝑢𝛼2𝑢02+𝑢12[].,𝛼0,1(3.2) Consider the unitary map (Mellin's transformation [8]): 𝜔𝐿21+,𝑟2𝑑𝑟𝐿211,𝑑𝑠𝑓(𝑟)𝑓(𝑠)=2𝜋0𝑟𝑖𝑠+1/2𝑓(𝑟)𝑑𝑟,𝑠1(3.3) and its inverse: 𝜔1𝑓1(𝑟)=2𝜋𝑟𝑖𝑠3/2𝑓(𝑠)𝑑𝑠.(3.4) For every set of functions 𝐵𝐿2(1+,𝑟2𝑑𝑟), we denote by 𝐵𝐿2(1,𝑑𝑠) the set of their Mellin's transformations: 𝐵=𝜔𝐵.(3.5) For every chain 𝔥(𝑢), we denote by Γ𝑢 the family of functions: Γ𝑢=𝛾𝔥(𝑢)=𝛼([],𝑠),𝛼0,1(3.6) where 𝛾𝛼(𝑠)=(𝜔𝑢𝛼)(𝑠), 𝑢𝛼𝔥(𝑢). The family Γ𝑢 can be represented as a function Γ𝑢(𝑧) of a complex variable 𝑧=𝑠+𝑖𝛼 in the strip: 𝐼=𝑧1[],Γ𝑧0,1𝑢1(𝑧)=2𝜋0𝑟𝑖𝑠1/2+𝛼1𝑢(𝑟)𝑑𝑟=2𝜋0𝑟𝑖𝑧𝑢(𝑟)𝑑𝑟.(3.7) The function Γ𝑢 is said to be associated with the chain 𝔥(𝑢), and its values {𝛾𝛼(𝑠)} on the lines 𝜉𝛼={𝑧=𝑠+𝑖𝛼,𝑠1,0𝛼1}𝐼 are called the sections of Γ𝑢.

Proposition 3.1. For every chain 𝔥(𝑢), 𝑢𝑉, the associated function {Γ𝑢(𝑧),𝑧𝐼} is continuous in a closed strip 𝐼 and analytic inside this strip. Moreover, its sections {𝛾𝛼} satisfy the following inequality: sup0𝛼1𝛾𝛼()𝐿2(1)<.(3.8) Inversely, any function {Γ(𝑧),𝑧𝐼} which possesses these properties is associated with some (unique) chain 𝔥(𝑣)Γ=Γ𝑣, 𝑣𝑉. Let call this chain generated by Γ. In addition, the functions {𝑣𝛼,𝛼[0,1]} of the chain 𝔥(𝑣) are obtained by the inverse Mellin's transformation from the sections of Γ={𝛾𝛼}: 𝑣𝛼=𝜔1𝛾𝛼.(3.9)

The proof of this proposition can be obtained by using the arguments given in the book by Paley and Wiener (see [9], Chapter I), which are related to the Fourier transformation of functions analytical in a strip in a complex plane. It is not difficult to reformulate these arguments in terms of Mellin's transformation.

Note that the estimate (3.8) for {𝛾𝛼} follows from the estimate (3.2) and the unitary Mellin's transformation. Denote by 𝒢 a linear space of functions Γ satisfying conditions of Proposition 3.1. Let us introduce two maps: Ω𝔥(𝑢)Γ𝑢𝒢,Ω1Γ𝑢𝔥(𝑢).(3.10) Let 𝑁(𝑧), 𝑧𝐼, be a bounded, continuous function in the strip 𝐼, which is analytic inside 𝐼. This function generates the family 𝜅𝑁𝛼,𝛼[0,1] of bounded operators in 𝐿2(1) which act as multiplication on the functions 𝑛𝑁𝛼(𝑠)=𝑁(𝑧)|𝑧=𝑠+𝑖𝛼, 𝑠1, 0𝛼1: 𝜅𝑁𝛼𝜓(𝑠)=𝑛𝑁𝛼(𝑠)𝜓(𝑠),𝜓𝐿21.(3.11) Evidently, for any Γ𝒢, the function 𝑁(𝑧)Γ(𝑧) belongs to 𝒢. If the chain 𝔥(𝑢) is generated by Γ=Γ𝑢 and the chain 𝔥(𝑣) is generated by 𝑁(𝑧)Γ(𝑧)=Γ𝑣(𝑧), then 𝑣𝛼=𝜅𝑁𝛼𝑢𝛼[],𝛼0,1,𝑢𝛼𝔥(𝑢),(3.12) where 𝜅𝑁𝛼=𝜔1𝜅𝑁𝛼𝜔.(3.13) Denote by Π the following self-adjoint operator in 𝐿2(1+,𝑟2𝑑𝑟): (Π𝑓)(𝑟)=𝑟𝑓(𝑟),(3.14) with the domain 𝐷(Π)=𝑉.

It is clear that for any 𝑢𝑉, the power Π𝛽, 0𝛽1 of the operator Π is applicable to an element 𝑢𝛼𝔥(𝑢) if 𝛽+𝛼1 and Π𝛽𝑢𝛼=𝑢𝛼+𝛽.(3.15) For the function Γ𝑢 that is associated with 𝔥(𝑢),the action of the operator Π𝛽=𝜔Π𝛽𝜔1 on the sections {𝛾𝛼} of Γ𝑢 has the form: Π𝛽𝛾𝛼=𝛾𝛼+𝛽.(3.16) (again if 𝛼+𝛽1).

4. The Operator 𝑀𝑙

The operator 𝑀𝑙 (see (2.23)) can be represented as 𝑀𝑙=Π1/2𝜅𝑙Π1/2,(4.1) where 𝜅𝑙=𝜅𝑙1/2 is an operator in 𝐿2(1+,𝑟2𝑑𝑟) acting by the formula: 𝜅𝑙1/2𝑓(𝑟)=2𝜋2𝜇124𝑓(𝑟)+2𝜋11𝑑𝑥𝑃0𝑙(𝑥)0𝑟2𝑓𝑟𝑑𝑟(𝑟𝑟)1/2𝑟2+(𝑟)2+𝜇𝑥𝑟𝑟.(4.2)

Lemma 4.1. Operator 𝜅𝑙1/2 is bounded and self-adjoint in 𝐿2(1+,𝑟2𝑑𝑟).

Proof. Pass to the operator: 𝜅𝑙1/2=𝜔𝜅𝑙1/2𝜔1,(4.3) acting in 𝐿2(1). It follows from calculations in [2, 3] that 𝜅𝑙1/2 is the operator of multiplication on the function: 𝑛𝑙1/2(𝑠)=2𝜋2𝜇124+𝜆𝑙1/2,(𝑠)(4.4) where 𝜆𝑙1/2(𝑠)=10𝑃𝑙(𝑥)ch(𝑠𝑣(𝑥))𝑑𝑥ch(𝑠𝜋/2)cos(𝑣(𝑥))foreven𝑙,10𝑃𝑙(𝑥)sh(𝑠𝑣(𝑥))𝑑𝑥sh(𝑠𝜋/2)cos(𝑣(𝑥))forodd𝑙,(4.5) and 𝑣(𝑥)=arcsin𝜇𝑥/2, 0𝑥1. As we see the function 𝑛𝑙1/2(𝑠), 𝑠1, is bounded and real. The lemma is proved.

We see from (4.4) and (4.5) that the functions 𝑛𝑙1/2(𝑠) and 𝜆𝑙1/2 are continued up to bounded, analytical functions 𝑁𝑙(𝑧) and Λ𝑙(𝑧) correspondingly, defined in the strip 𝐼={𝑧11/2𝑧1/2}. Let us define the functions 𝑁𝑙(𝑧)=𝑁𝑙(𝑧𝑖/2) which we shall consider in the strip 𝐼={𝑧0𝑧1}. The operator 𝜅𝑙1/2 coincides with the operator 𝑁𝜅𝑙1/2 from the family 𝑁{𝜅𝑙𝛼} generated by the function 𝑁𝑙 (see (3.11)). Any other operator of this family acts as multiplication on the function: ̂𝑛𝑙𝛼𝑁(𝑠)=𝑙|||(𝑧)𝑧=𝑠+𝑖𝛼.(4.6) Denote by 𝜅𝑙𝛼 the operators 𝜅𝑙𝛼=𝜔1𝜅𝑁𝑙𝛼𝜔,(4.7) acting in 𝐿2(1+,𝑟2𝑑𝑟).

Note that 𝜅𝑙𝛼=𝜅𝑙𝑖𝛼.(4.8) It is convenient to represent the operator 𝑀𝑙 in form of three sequential maps 𝑀𝑙𝑢0𝑢𝔥0Π1/2𝑢0=𝑢1/2𝜅𝑙1/2𝑢1/2=𝑣1/2Π1/2𝑣1/2=𝑣1𝔥(𝑣),(4.9) where 𝑣=𝑣0,𝑣1/2,𝑣1 are elements of the chain 𝔥(𝑣) generated by the function Γ𝑣=𝑁𝑙Γ𝑢𝒢. Note that the chain (4.9) can be rewritten in the following way: 𝑢0𝑢𝔥0ΩΓ𝑢0Γ𝑣=𝑁𝑙Γ𝑢0Ω1𝔥(𝑣)𝑣1𝔥(𝑣).(4.10) From (4.1) and self-adjointness of 𝜅𝑙1/2 it follows that the operator 𝑀𝑙 with the domain 𝐷(𝑀𝑙)=𝑉 is symmetric. For any 𝛼[0,1], a representation of 𝑀𝑙 similar to (4.1) is valid: 𝑀𝑙=Π1𝛼𝜅𝑙𝛼Π𝛼(4.11) as well as decomposition like (4.9).

Let us now describe the domain 𝐷(𝑀𝑙)𝑉 of the operator 𝑀𝑙 conjugated to 𝑀𝑙. Let 𝑔𝐷(𝑀𝑙) be a function from 𝐷(𝑀𝑙) and =𝑀𝑙𝑔𝐿2(1+,𝑟2𝑑𝑟). Then for every 𝑢𝑉=𝐷(𝑀𝑙), we can write 𝑀𝑙=𝜅𝑢,𝑔𝑙1=𝜅Π𝑢,𝑔Π𝑢,𝑙1𝑔=𝑢,Π𝜅𝑙0𝑔=(𝑢,).(4.12) Here we use the representation (4.11) for 𝛼=1 and the equality (4.8). Denote 𝑓(𝑟)=(𝑟)(Π𝜅𝑙0𝑔)(𝑟) and apply the following evident assertion.

Lemma 4.2. Let a measurable function 𝑓(𝑟) satisfies condition 0𝑓(𝑟)𝑢(𝑟)𝑟2𝑑𝑟=0,(4.13) for any 𝑢𝑉. Then 𝑓=0.

From this and (4.12), it follows that Π𝜅𝑙0𝑔=.(4.14) Hence 𝑤0𝜅𝑙0𝑔𝑉,(4.15) and =𝑤1𝔥(𝑤0) is the final element of the chain 𝔥(𝑤0). Thus the domain 𝐷(𝑀𝑙) of the operator 𝑀𝑙 is 𝐷𝑀𝑙=𝑔𝐿21+,𝑟2𝑑𝑟𝜅𝑙0.𝑔𝑉(4.16) In the case when the operator 𝜅𝑙0 has the inverse one, (𝜅𝑙0)1, which is equivalent to the condition: ̂𝑛𝑙0(𝑠)0,forany𝑠1,(4.17) the following equality is true: 𝐷𝑀𝑙=𝜅𝑙01𝑉.(4.18) Let 𝑀𝑙=𝜔𝑀𝑙𝜔1 be an operator in 𝐿2(1) with domain 𝑀𝐷(𝑙)=𝜔𝐷(𝑀𝑙). Then for 𝑀̃𝑔𝐷(𝑙), the following representation holds true: ̃𝑔(𝑠)=̂𝑛𝑙0(𝑠)1𝑤0𝑁(𝑠)=𝑙(𝑧)1Γ𝑤0||||(𝑧)𝑧=𝑠,(4.19) if condition (4.17) is fulfilled. Here 𝑤0(𝑠)=(𝜔𝑤0)(𝑠) where 𝑤0 is defined in (4.15).

Remarks. (1) Note that the function 𝑁𝑙(𝑧) is invariant with respect to reflection of the complex plane around the point 𝑧=𝑖/2: 𝑧𝑧=𝑧+𝑖.(4.20) Under this reflection, the strip 𝐼 is mapped onto itself; hence, for every zero 𝑧𝐼(𝑧𝑖/2) of the function 𝑁𝑙, there exists another zero, 𝑧𝐼, of 𝑁𝑙 with the same multiplicity. The multiplicity of 𝑧=𝑖/2=𝑧 is even;

(2) Since 𝑁𝑙(𝑧)2𝜋21𝜇2/4>0 as 𝑧 inside 𝐼, the function 𝑁𝑙(𝑧) has finite number of zeros inside 𝐼.

We can now formulate the main criterion of self-adjointness of the operator 𝑀𝑙.

Theorem 4.3. The operator 𝑀𝑙 is self-adjoint if and only if the function 𝑁𝑙(𝑧) has no zeros in the closed strip 𝐼.

Proof. (1) Assume 𝑁𝑙(𝑧)0 in the strip 𝐼. Then (𝑁𝑙)1(𝑧) is bounded and continuous on 𝐼 and analytical inside 𝐼. Let 𝑀̃𝑔𝐷(𝑙). Since ̂𝑛𝑙(𝑠)0 for 𝑠1, the representation (4.19) holds true. Since 𝑁𝑙(𝑧)1Γ𝑤0(𝑧)=Γ𝑣𝒢,𝑣𝑉,(4.21) the element 𝑔=𝜔1̃𝑔𝐷(𝑀𝑙) coincides with 𝑣𝑉, that is, 𝐷(𝑀𝑙)=𝑉=𝐷(𝑀𝑙); it means the self-adjointness of 𝑀𝑙;
(2) assume now the function 𝑁𝑙(𝑧) has zeros 𝑧1,,𝑧𝑘𝐼. Consider first the case when all zeros are lying inside 𝐼 and their multiplicities are equal to 𝑝1,,𝑝𝑘, respectively. Again, let 𝑀̃𝑔𝐷(𝑙). Since ̂𝑛𝑙(𝑠)0, the representation (4.19) holds true. The function (𝑁𝑙(𝑧))1Γ𝑤0(𝑧) is meromorphic in 𝐼 with poles 𝑧1,,𝑧𝑘 having the order 𝑝1,,𝑝𝑘 respectively. For this function the usual canonical representation [10] is true: 𝑁𝑙(𝑧)1Γ𝑤0(𝑧)=𝐿𝑤0(𝑧)+𝑘𝑝𝑛=1𝑛𝑚=1𝑏𝑚(𝑛)𝑤0𝑧𝑧𝑛𝑚,(4.22) where 𝐿𝑤0(𝑧) is bounded, continuous function on 𝐼, and analytical inside 𝐼, and the coefficients 𝑏𝑚(𝑛)=𝑏𝑚(𝑛)(𝑤0) depend on 𝑤0.
Lemma  4.4.  The function 𝐿𝑤0(𝑧)in (4.22) belongs to the space 𝒢.
The proof of this lemma is given in The appendix.
From (4.19) and (4.22), for 𝑔=𝜔1̃𝑔𝐷(𝑀𝑙), we have 𝑔(𝑟)=𝑣(𝑟)+𝑚,𝑛𝑏𝑚(𝑛)𝜔11𝑧𝑛𝑚(𝑟),(4.23) where the function 𝑣𝑉 is defined from relation 𝐿𝑤0(𝑧)=Γ𝑣𝑑(𝑧)𝒢,𝑚,𝑛(𝑟)=𝜔11𝑧𝑛𝑚(𝑟)=𝐴𝑚(𝑛)𝑟3/2𝑡𝑛+𝑖𝑠𝑛(ln𝑟)𝑚1𝜒(𝑟),(4.24) where 𝐴𝑚(𝑛) is an absolute constant, 𝑧𝑛=𝑠𝑛+𝑖𝑡𝑛, 0<𝑡𝑛<1 and 𝜒(𝑟)=1,𝑟>1,0,𝑟1.(4.25) Since linearly independent functions 𝑑𝑚,𝑛𝐷(𝑀𝑙) do not belong to 𝑉, due to (4.23), they form the basis in the defect subspace 𝔙 of the operator 𝑀𝑙 (see [7]). Since the dimension of the subspace 𝔙 is equal to 𝑘1𝑝𝑛 and the operator 𝑀𝑙 is real, its deficiency indexes 𝑛± are equal and have the form: 𝑛+=𝑛=12𝑘1𝑝𝑛.(4.26) (It follows from Remarks that the sum 𝑘1𝑝𝑛 is even). Consider now the case when one of the zeros of 𝑁𝑙(𝑧), say, 𝑧0=𝑠01, lies on the boundary of 𝐼 and has multiplicity 𝑝 (in addition, there is a zero 𝑧0=𝑠0+𝑖). In this case, in a neighborhood of 𝑧0, the function 𝑁𝑙(𝑧) has the form: 𝑁𝑙(𝑧)=𝑧𝑧0𝑝𝑄(𝑧),(4.27) where 𝑄(𝑧) is analytic in this neighborhood. Consider the function, 1𝐺(𝑧)=𝑖𝑧𝑧01/31(𝑧+2𝑖)2,(4.28) whereby (𝑖𝑤)1/3 for 𝑤>0, we mean the branch of a many-valued function (𝑖𝑤)1/3 that takes positive values on the positive part of the imaginary axis. Evidently, the function 𝐺(𝑧) is analytic in the strip 𝐼 and satisfies condition (3.8). However, this function is discontinuous at 𝑧0 and does not belong to 𝒢. In addition, the function 𝑁𝑙(z)𝐺(𝑧) now belongs to 𝒢 as follows from (4.27) and (4.28). Thus ||̃𝑔(𝑠)=𝐺(𝑧)𝑧=𝑠𝑉=𝜔𝑉,(4.29) but ̂𝑛𝑙𝑁(𝑠)̃𝑔(𝑠)=𝑙|||(𝑧)𝐺(𝑧)𝑧=𝑠𝑉.(4.30) Consequently, 𝑔=𝜔̃𝑔𝑉 but 𝜅𝑙0𝑔𝑉, that is, 𝑔𝐷(𝑀𝑙). Thus 𝐷(𝑀𝑙)𝑉, and the operator 𝑀𝑙 has nonzero deficiency indexes. Theorem 4.3 is proved.

5. The Operators 𝑀𝑙 in the Cases 𝑙=0 and 𝑙=1

Here, we apply Theorem 4.3 to the cases 𝑙=0 and 𝑙=1.

Theorem 5.1. (1) For 𝑙=0, the operator 𝑀𝑙=0  is self-adjoint for any 𝑚>0;
(2) the operator 𝑀𝑙=1 is self-adjoint for 𝑚>𝑚0 and has nonzero deficiency indexes for 𝑚𝑚0. In addition, for 𝑚<𝑚0 these indexes are equal to (1,1). The constant 𝑚0 is a unique zero of (5.4).

Proof. We need the following properties of the functions Λ𝑙=0(𝑧) and Λ𝑙=1(𝑧), 𝑧𝐼.
Lemma  5.2. (1) For any 𝑙=0,1,2,  the function Λ𝑙(𝑧)  is invariant with respect to reflection (4.20);
(2) The point 𝑧=𝑖/2𝐼  is a nondegenerate critical point for both functions Λ𝑙=0  and Λ𝑙=1;
(3) These functions take real values on the line: ̂𝜉1/2=𝑖𝑧=𝑠+2,𝑠1,(5.1)and on the segment: ̂𝜏={𝑧=𝑖𝑡,0𝑡1}.(5.2)Outside the set ̂𝜉𝐵=1/2̂𝜏,  both functions take nonreal values;
(4) the real values of Λ𝑙, 𝑙=0,1,  are between 0  and Λ𝑙Λ(0)=𝑙(𝑖). Every value of Λ𝑙|𝐵except Λ𝑙(𝑖/2) —is taken exactly at two points;
(5) the extreme values of Λ𝑙, 𝑙=0,1,   Λ𝑙Λ(0)=𝑙(𝑖)  are given byΛ𝑙=0(0)=82𝜋2𝜇11sin2𝜇arcsin2Λ>0,𝑙=1(0)=3232𝜋2𝜇2sin312𝜇arcsin2𝑞(𝜇)<0,(5.3)
(6) the function 𝑞(𝜇)  increases monotonically on the interval  0<𝜇<2.
The proof of this lemma is given in The appendix.
Corollary  5.3.  (1) The zeros of  𝑁𝑙(𝑧), 𝑙=0,1 can only lie in the set 𝐵;
(2) 𝑁𝑙=0(𝑧)|𝐵>0   for any value of 𝜇, and therefore the operator 𝑀𝑙=0  is self-adjoint for all 𝑚(0,2);
(3) The function 𝑁𝑙=1(𝑧)|𝐵  is positive if 2𝜋21𝜇2/4>𝑞(𝜇)  and vanishes at some point 𝑧𝐵 (and also at 𝑧𝐵) if 2𝜋21𝜇2/4𝑞(𝜇).
In Figure 1, the curves corresponding to the functions 2𝜋21𝜇2/4 and 𝑞(𝜇) are depicted. We see that they intersect at a unique point with abscissa 𝜇0(0,2) which satisfies the following equation: 2𝜋2𝜇1204𝜇=𝑞0.(5.4)
Thus, for 𝑚>𝑚0=2/𝜇01 the operator 𝑀𝑙=1 is self-adjoint, and for 𝑚<𝑚0 it has deficiency indexes (1,1). For 𝑚=𝑚0, the operator 𝑀𝑙=1 is not self-adjoint as well. Theorem 5.1 is proved.

Appendix

Proof of Lemma 4.4. The function (𝑁𝑙(𝑧))1, 𝑧𝐼 admits the canonical representation (see [10]) 𝑁𝑙(𝑧)1=𝑄𝑙(𝑧)+𝑘𝑝𝑛=1𝑛𝑚=1𝑎𝑚(𝑛)𝑧𝑧𝑛𝑚,(A.1) where 𝑧1,,𝑧𝑘𝐼 are zeros of 𝑁𝑙(𝑧) (with multiplicities 𝑝1,,𝑝𝑘), 𝑎𝑚(𝑛) are constants, 𝑎𝑝(𝑛)𝑛0, and 𝑄𝑙(𝑧) is a bounded, continuous analytic function in 𝐼. From this, it follows that for any 𝑣𝑉, 𝑄𝑙(𝑧)Γ𝑣(𝑧)𝒢. Consider some term of the sum (A.1) and write 𝑎𝑚(𝑛)𝑧𝑧𝑛𝑚Γ𝑣𝑃(𝑧)=(𝑛)𝑚,𝑣(𝑧)+𝑚𝑑=1𝑐(𝑛)𝑚𝑑𝑧𝑧𝑛𝑑𝑎𝑚(𝑛),(A.2) where 𝑃(𝑛)𝑚,𝑣1(𝑧)=𝑧𝑧𝑛𝑚Γ𝑣(𝑧)𝑚𝑑=1𝑐(𝑛)𝑚𝑑𝑧𝑧𝑛𝑚𝑑,𝑐𝑡(𝑛)=𝑐𝑡(𝑛)(1𝑣)=Γ𝑡!𝑣(𝑡)𝑧𝑛,𝑡=0,1,.(A.3) It is clear that 𝑃(𝑛)𝑚,𝑣(𝑧) is bounded, continuous analytic function in 𝐼. We are going to show that this function belongs to 𝒢. Let 𝑂𝐼 be a small neighborhood of 𝑧𝑛 and 𝜒𝑂(𝑧) the characteristic function of 𝑂. Obviously, the bounded function 𝜒𝑂𝑃(𝑛)𝑚,𝑣 satisfies condition (3.8). Every term of the sum 1𝜒𝑂𝑃(𝑛)𝑚,𝑣Γ(𝑧)=𝑣(𝑧)𝑧𝑧𝑛𝑚1𝜒𝑂𝑚𝑑=1𝑐(𝑛)𝑚𝑑(𝑣)𝑧𝑧𝑛𝑑1𝜒𝑂(A.4) satisfies this condition as well.
Thus for fixed 𝑧𝑛 and 𝑣𝑉, 𝑝𝑛𝑚=1𝑎𝑚(𝑛)Γ𝑣(𝑧)𝑧𝑧𝑛𝑚=𝐾𝑣(𝑛)(𝑧)+𝑝𝑛𝑑=1𝑏𝑑(𝑛)(𝑣)𝑧𝑧𝑛𝑑,(A.5) where 𝐾𝑣(𝑛)(𝑧)=𝑝𝑛𝑚=1𝑎𝑚(𝑛)𝑃(𝑛)𝑚,𝑣𝑏(𝑧),(A.6)𝑑(𝑛)(𝑣)=𝑝𝑛𝑚=1𝑎𝑚(𝑛)𝑐(𝑛)𝑚𝑑(𝑣),𝑑=1,,𝑝𝑛.(A.7) Thus, we get the representation (4.22) where 𝐿(𝑤0)(𝑧)=𝑄𝑙(𝑧)Γ𝑤0(𝑧)+𝑘𝑛=1𝐾𝑤(𝑛)0(𝑧)𝒢,(A.8) and the coefficients 𝑏𝑚(𝑛)(𝑤0) are given by formula (A.7). Lemma 4.4 is proved.

Proof of Lemma 5.2. (1) It is more convenient to consider the functions 𝑁𝑙(𝑧) and Λ𝑙(𝑧) in the strip 𝐼={𝑧|𝑧|<1/2} instead of the functions 𝑁𝑙(𝑧) and Λ𝑙(𝑧) in the strip 𝐼. Similarly, instead of the reflection 𝑧𝑧 we consider the reflection 𝑧𝑧 around the point 𝑧0=0. It is clear that the functions Λ𝑙(𝑧), 𝑙=0,1,2, are invariant with respect to the change 𝑧𝑧, and it means the invariance of Λ𝑙 with respect to reflection (4.20);
(2) it follows from (4.5) that 𝑧=0 is a nondegenerated critical point of Λ𝑙=0 and Λ𝑙=1, if we note that 0<𝑣(𝑥)𝜋/2. Correspondingly, 𝑧=𝑖/2 is a nondegenerated critical point for Λ𝑙(𝑧), 𝑙=0,1. The real axis 𝜉0={𝑧=𝑠;𝑠1} coincides with the saddle-point line at 𝑧=0 (see [10]) for Λ𝑙=0 and Λ𝑙=1. More precisely, these functions take real values on 𝜉0 and decrease monotonically to zero as |𝑠| increases from zero to infinity. On the contrary, Λ𝑙=0 and Λ𝑙=1 increase monotonically along imaginary axis as |𝑡| increases from zero to 1/2. The monotonicity of Λ𝑙=0 along real axis follows from (4.5), equality 𝑃0(𝑥)1, and inequality ch(𝑣(𝑥)𝑠)ch((𝜋/2)𝑠)𝑠𝜋<2sh(𝜋/2𝑣(𝑥))𝑠(ch((𝜋/2)𝑠))2<0,(A.9) for 𝑠>0 and a similar inequality for 𝑠<0. The proof of monotonicity of Λ𝑙=1 along real axis, and also monotonicity of both functions along imaginary axis is analogous if we note that 𝑃𝑙=1(𝑥)𝑥 on (0,1). Thus the functions Λ𝑙, 𝑙=0,1, take all values between 0 and Λ𝑙(𝑖/2)=Λ𝑙(𝑖/2) and every value except Λ𝑙(0) which is taken exactly twice;
(3) we will show now that the values of functions Λ𝑙(𝑧), 𝑙=0,1, on the set 𝐼𝐵 are nonreal. Let us represent this set as a union of four sets, 𝐼𝑖, 𝑖=1,2,3,4 as shown in Figure 2.
We consider the case 𝑙=0; the case 𝑙=1 is similar. Figure 3 shows the disposition of lines of levels for function 𝐾0(𝑧)=Λ𝑙=0(𝑧) which pass through the points 𝑖 and 𝑖 between lines 𝛽 and 𝛽, 𝛽={𝑧𝐾0(𝑧)=0,𝑧>0}, 𝛽={𝑧𝐾0(𝑧)=0,𝑧<0}.
All these lines have common tangents at points 𝑖 and 𝑖, and the line 𝛽 (resp. 𝛽) lies above (resp., below) the strip 𝐼. The picture represented in Figure 3 is obtained by detailed study of the explicit formula for Λ𝑙=0: Λ𝑙=0(𝑧)=4𝜋2𝜇sh(𝑧arcsin(𝜇/2)),𝑧ch(𝑧𝜋/2)(A.10) together with the proof that the lines 𝛽 and 𝛽 do not intersect the strip 𝐼. This proof is given below.
From Figure 3, we see that the set 𝐼1 lies inside the shaded domain 𝑈 that is bounded by the real semiaxis 𝜉+0={𝑧𝑧=𝑠,𝑠>0}, the segment (0,𝑖/2) on the imaginary axis and the part of line 𝛽 which lies in the right half-plane. From (A.10), it is easy to see that the function 𝑤=Λ𝑙=0(𝑧) maps the boundary 𝜕𝑈 of the domain 𝑈 into the boundary of the right lower quadrant 𝑀={𝑤𝑤>0,𝑤<0} of the plain 𝑤. Hence, the domain 𝑈 is mapped inside this quadrant, that is, all values of the function Λ𝑙=0 in 𝑈 are nonreal. It means the absence of real values of Λ𝑙=0 in 𝐼1. For the domains 𝐼2, 𝐼3, and 𝐼4, the proof is similar. Let us now prove that 𝛽 and 𝛽 do not intersect the line 𝜉1/2. It is sufficient to prove that Λ𝑙=0>0 on the line 𝜉1/2={𝑧𝑧=𝑠+𝑖/2,𝑠1} or, which is the same, that ch(𝑧𝑣(𝑥))||||ch(𝑧𝜋/2)𝑧=𝑠+𝑖/2>0,(A.11) for any 𝑠1 and 𝑥(0,1). Write []ch(𝑠+𝑖/2)𝑣(𝑥)[]=ch(𝑠+𝑖/2)𝜋/2ch(𝑠𝑣(𝑥))cos(𝑣(𝑥)/2)+𝑖sh(𝑠𝑣(𝑥))sin(𝑣(𝑥)/2)ch(𝑠𝜋/2)cos(𝜋/4)+𝑖sh(𝑠𝜋/2)sin(𝜋/4)=𝐷(𝑠,𝑥).(A.12)
Let 𝑠>0. Then the values of numerator and denominator of 𝐷(𝑠,𝑥) lie in the right upper quadrant of a complex plain, and hence 𝜋/2<arg𝐷(𝑠,𝑥)<𝜋/2, that is, 𝐷(𝑠,𝑥)>0. Similarly (A.11) can be proved in the case 𝑠<0 and for Λ𝑙=0|𝑧=𝑠𝑖/2;
(4) let us find the values Λ𝑙(𝑖/2), 𝑙=0,1:(I) the case 𝑙=0: Λ𝑙=0(𝑖/2)=2𝜋210cos(𝑣(𝑥)/2)cos𝑣(𝑥)cos(𝜋/4)𝑑𝑥.(A.13) After the change 𝑣(𝑥)=𝜉, the integral (A.13) becomes 42𝜋2𝜇0arcsin𝜇/2𝜉cos28𝑑𝜉=2𝜇𝜋21sin2𝜇arcsin2;(A.14)(II)The case 𝑙=1: Λ𝑙=1𝑖2=2𝜋210𝑥sin(𝑣(𝑥)/2)𝑑𝑥.cos𝑣(𝑥)sin(𝜋/4)(A.15) The same change 𝑣(𝑥)=𝜉 reduces to the integral 82𝜋2𝜇20arcsin𝜇/2𝜉sin𝜉sin2𝑑𝜉=3223𝜋2𝜇2sin312𝜇arcsin2;(A.16)
(5) let us show that the function: 𝑞(𝜇)=2𝜋210𝑥sin(𝑣(𝑥)/2)cos𝑣(𝑥)sin(𝜋/4)𝑑𝑥(A.17) decreases monotonically as 𝜇 changes from 0 to 2. We have sin(𝑣(𝑥)/2)cos𝑣(𝑥)𝜇0(A.18) because the numerator of (A.18) increases, while the denominator decreases with the growth of 𝜇. This implies that 𝑞(𝜇)0,(A.19) that is, 𝑞(𝜇) increases monotonically. Lemma 5.2 is proved.

Acknowledgment

This work is supported by RFBR Grant 11-01-00485a.