On Pointlike Interaction between Three Particles : Two Fermions and Another Particle

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, ε ∈ R1} 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 R1 , 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 form > m0, where the value ofm0 is given explicitly in the paper.


Introduction and Statement of the Problem
This paper is continuation of works 1-4 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: ISRN Mathematical Physics of smooth rapidly decreasing functions ψ y, x 1 , x 2 ∈ H on infinity, antisymmetrical with respect to x 1 , x 2 and satisfying the following conditions: ψ y, x 1 , x 2 x i y 0, i 1, 2. 1.2 Usually, some family {H ε , ε ∈ R 1 } of symmetric extensions of the operator H 0 is proposed as a possible "true" Hamiltonian of the system the so-called Ter-Martirosian-Skornyakov extensions, see 5 .These extensions were constructed in 1-4 .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 {H ε } is equivalent to the one for some auxiliary symmetric operator T acting in the space L 2 R 3 see below .This operator commutes with the operators {U g , g ∈ O 3 } of the representation of the rotation group O 3 that acts in L 2 R 3 by the usual formula: Let us denote by H l ⊂ L 2 R 3 the maximal subspace, where the representation 1.3 is multiplied by the irreducible representation of O 3 with weight l, l 0, 1, 2, . . .see 6 .Evidently, the space H l is invariant with respect to the operator T, and the restriction T l T| H l of this operator to the space H l is symmetric operator.The operator T is selfadjoint if all the operators {T l , l 0, 1, . ..} are self-adjoint.In this paper, we find general simple conditions of self-adjointness of T l 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 T l 0 is self-adjoint for all values of m > 0, while the operator T l 1 is self-adjoint for m > m 0 and has nonzero deficiency indexes for m ≤ m 0 , the constant m 0 > 0 is indicated below see 5. 4 .
By the way, we note that the value of m 0 obtained in this paper differs from that one given by mistake in 2 .

A Short Explanation of the Constructions from Papers [1-3]
1 After Fourier transformation:

and change of variables:
can be represented as a tensor sum: where and The operator H 2 0 is symmetric, and its domain is consists of the functions of the form: where the function ℘ p belongs to Hilbert space with inner product

ISRN Mathematical Physics
Here W is some positive operator acting in L 2 R 3 see 3 .The domain of the operator
The following asymptotics holds for vectors g ∈ D H
Lemma 2.1.The operator T defined in the space L 2 R 3 by 2.16 is symmetric, and the selfadjointness of the operators H ε (for all ε) is equivalent to the self-adjointness of the operator T (see [2,3,5]).

ISRN Mathematical Physics 5
The operator T can be represented as a sum of two operators: where the symmetric operator T with the domain D T D T acts as follows: 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 H l ⊂ L 2 R 3 is invariant with respect to T; it has the form: where L l 2 S ⊂ L 2 S is the space of spherical functions of weight l see 6 on the unit sphere S ⊂ R 3 .In addition, the operator T l T| H l has the form where E l is the unit operator in L l 2 s , and M l acts in L 2 R 1 , r 2 dr by the formula: on the domain

2.25
The operators {M l , l 0, 1, . ..} are symmetric in L 2 R 1 , r 2 dr , and the self-adjointness of M l is equivalent to the self-adjointness of T l .Later on, we shall study the operators M l and derive a condition of self-adjointness.

Preparatory Constructions
For every function u ∈ V ⊂ L 2 R 1 , r 2 dr , we consider the family of functions which we call a chain with initial element u u 0 and the final one u 1 .All functions u α ∈ h u belong to L 2 R 1 , r 2 dr and have a uniformly bounded norm: Consider the unitary map Mellin's transformation 8 : and its inverse: For every set of functions B ⊂ L 2 R 1 , r 2 dr , we denote by B ⊂ L 2 R 1 , ds the set of their Mellin's transformations: B ωB.

3.5
For every chain h u , we denote by Γ u the family of functions: where γ α s ωu α s , u α ∈ h u .The family Γ u can be represented as a function Γ u z of a complex variable z s iα in the strip:

3.7
The function Γ u is said to be associated with the chain h u , and its values {γ α s } on the lines Proposition 3.1.For every chain h u , u ∈ V , the associated function {Γ u z , z ∈ I} is continuous in a closed strip I and analytic inside this strip.Moreover, its sections {γ α } satisfy the following inequality: Inversely, any function {Γ z , z ∈ I} which possesses these properties is associated with some (unique) chain Let call this chain generated by Γ.In addition, the functions {v α , α ∈ 0, 1 } of the chain h v are obtained by the inverse Mellin's transformation from the sections of Γ {γ α }: 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 {γ α } 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: Let N z , z ∈ I, be a bounded, continuous function in the strip I, which is analytic inside I.This function generates the family κ N α , α ∈ 0, 1 of bounded operators in L 2 R 1 which act as multiplication on the functions

3.11
Evidently, for any Γ ∈ G, the function N z Γ z belongs to G. If the chain h u is generated by Γ Γ u and the chain h v is generated by where

3.13
Denote by Π the following self-adjoint operator in L 2 R 1 , r 2 dr : Πf r rf r , 3.14 with the domain D Π V .
It is clear that for any u ∈ V , the power Π β , 0 ≤ β ≤ 1 of the operator Π is applicable to an element u α ∈ h u if β α ≤ 1 and 3.15 For the function Γ u that is associated with h u ,the action of the operator Π β ωΠ β ω −1 on the sections {γ α } of Γ u has the form:

The Operator M l
The operator M l see 2.23 can be represented as where κ l κ l 1/2 is an operator in L 2 R 1 , r 2 dr acting by the formula: Proof.Pass to the operator: It follows from calculations in 2, 3 that κ l 1/2 is the operator of multiplication on the function: where for even l, for odd l, 4.5 and v x arcsinμx/2, 0 ≤ x ≤ 1.As we see the function n l 1/2 s , s ∈ R 1 , is bounded and real.The lemma is proved.
We see from 4.4 and 4.5 that the functions n l 1/2 s and λ l 1/2 are continued up to bounded, analytical functions N l z and Λ l z correspondingly, defined in the strip I {z ∈ Denote by κ l α the operators It is convenient to represent the operator M l in form of three sequential maps where v v 0 , v 1/2 , v 1 are elements of the chain h v generated by the function Note that the chain 4.9 can be rewritten in the following way: From 4.1 and self-adjointness of κ l 1/2 it follows that the operator M l with the domain D M l V is symmetric.For any α ∈ 0, 1 , a representation of M l similar to 4.1 is valid: as well as decomposition like 4.9 .
Let us now describe the domain

4.12
Here we use the representation 4.11 for α 1 and the equality 4.8 .Denote f r h r − Πκ l 0 g r and apply the following evident assertion.From this and 4.12 , it follows that Πκ l 0 g h.

4.14
Hence and h w 1 ∈ h w 0 is the final element of the chain h w 0 .Thus the domain D M * l of the operator

4.16
In the case when the operator κ l 0 has the inverse one, κ l 0 −1 , which is equivalent to the condition: n l 0 s / 0, for any s ∈ R 1 , 4.17 the following equality is true: Then for g ∈ D M * l , the following representation holds true: if condition 4.17 is fulfilled.Here w 0 s ωw 0 s where w 0 is defined in 4.15 .
Remarks. 1 Note that the function N l z is invariant with respect to reflection of the complex plane around the point z i/2: z −→ z * −z i.

4.20
Under this reflection, the strip I is mapped onto itself; hence, for every zero z ∈ I z / i/2 of the function N l , there exists another zero, z * ∈ I, of N l with the same multiplicity.The multiplicity of z i/2 z * is even; 2 Since N l z → 2π 2 1 − μ 2 /4 > 0 as z → ∞ inside I, the function N l z has finite number of zeros inside I.
We can now formulate the main criterion of self-adjointness of the operator M l .
Theorem 4.3.The operator M l is self-adjoint if and only if the function N l z has no zeros in the closed strip I.
Proof. 1 Assume N l z / 0 in the strip I. Then N l −1 z is bounded and continuous on I and analytical inside I. Let g ∈ D M * l .Since n l s / 0 for s ∈ R 1 , the representation 4.19 holds true.Since 2 assume now the function N l z has zeros z 1 , . . ., z k ∈ I. Consider first the case when all zeros are lying inside I and their multiplicities are equal to p 1 , . . ., p k , respectively.Again, let g ∈ D M * l .Since n l s / 0, the representation 4.19 holds true.The function N l z −1 Γ w 0 z is meromorphic in I with poles z 1 , . . ., z k having the order p 1 , . . ., p k respectively.For this function the usual canonical representation 10 is true: The proof of this lemma is given in The appendix.From 4.19 and 4.22 , for g ω −1 g ∈ D M * l , we have where the function v ∈ V is defined from relation

4.24
where A n m is an absolute constant, z n s n it n , 0 < t n < 1 and χ r 1, r > 1, 0, r ≤ 1.

4.25
Since linearly independent functions d m,n ∈ D M * l do not belong to V , due to 4.23 , they form the basis in the defect subspace V of the operator M l see 7 .Since the dimension of the subspace V is equal to k 1 p n and the operator M l is real, its deficiency indexes n ± are equal and have the form:

4.26
It follows from Remarks that the sum k 1 p n is even .Consider now the case when one of the zeros of N l z , say, z 0 s 0 ∈ R 1 , lies on the boundary of I and has multiplicity p in addition, there is a zero z * 0 s 0 i .In this case, in a neighborhood of z 0 , the function N l z has the form: where Q z is analytic in this neighborhood.Consider the function, 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 z 0 and does not belong to G.In addition, the function N l z G z now belongs to G as follows from 4.27 and 4.28 .Thus Consequently, g ω g∈V but κ l 0 g ∈ V , that is, g ∈ D M * l .Thus D M * l / V , and the operator M l has nonzero deficiency indexes.Theorem 4.3 is proved.

The Operators M l 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 M l 0 is self-adjoint for any m > 0; (2) the operator M l 1 is self-adjoint for m > m 0 and has nonzero deficiency indexes for m ≤ m 0 .In addition, for m < m 0 these indexes are equal to 1, 1 .The constant m 0 is a unique zero of 5.4 .
Proof.We need the following properties of the functions Λ l 0 z and Λ l 1 z , z ∈ I. Lemma 5.2. 1 For any l 0, 1, 2, . . . the function Λ l z is invariant with respect to reflection 4.20 ; 2 The point z i/2 ∈ I is a nondegenerate critical point for both functions Λ l 0 and Λ l 1 ; 3 These functions take real values on the line: and on the segment: Outside the set B ξ 1/2 ∪ τ, both functions take nonreal values; 4 the real values of Λ l , l 0, 1, are between 0 and Λ l 0 5 the extreme values of Λ l , l 0, 1, Λ l 0 Λ l i are given by 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 N l z , l 0, 1 can only lie in the set B; 2 N l 0 z | B > 0 for any value of μ, and therefore the operator M l 0 is self-adjoint for all m ∈ 0, 2 ; In Figure 1, the curves corresponding to the functions 2π 2 1 − μ 2 /4 and q μ are depicted.We see that they intersect at a unique point with abscissa μ 0 ∈ 0, 2 which satisfies the following equation: Thus, for m > m 0 2/μ 0 − 1 the operator M l 1 is self-adjoint, and for m < m 0 it has deficiency indexes 1, 1 .For m m 0 , the operator M l 1 is not self-adjoint as well.Theorem 5.1 is proved.
satisfies this condition as well.
Thus for fixed z n and v ∈ V , and the coefficients b n m w 0 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 N l z and Λ l z in the strip I {z : | z| < 1/2} instead of the functions N l z and Λ l z in the strip I. Similarly, instead of the reflection z → z * we consider the reflection z → −z around the point z 0 0. It is clear that the functions Λ l z , l 0, 1, 2, . . .are invariant with respect to the change z → −z, and it means the invariance of Λ l with respect to reflection 4.20 ; 2 it follows from 4.5 that z 0 is a nondegenerated critical point of Λ l 0 and Λ l 1 , if we note that 0 < v x ≤ π/2.Correspondingly, z i/2 is a nondegenerated critical point for Λ l z , l 0, 1.The real axis ξ 0 {z s; s ∈ R 1 } coincides with the saddle-point line at z 0 see 10 for Λ l 0 and −Λ l 1 .More precisely, these functions take real values on ξ 0 and decrease monotonically to zero as |s| increases from zero to infinity.On the contrary, Λ l 0 and −Λ l 1 increase monotonically along imaginary axis as |t| increases from zero to 1/2.The monotonicity of Λ l 0 along real axis follows from 4.5 , equality P 0 x ≡ 1, and inequality for s > 0 and a similar inequality for s < 0. The proof of monotonicity of Λ l 1 along real axis, and also monotonicity of both functions along imaginary axis is analogous if we note that P l 1 x ≡ x on 0, 1 .Thus the functions Λ l , l 0, 1, take all values between 0 and Λ l i/2 Λ l −i/2 and every value except Λ l 0 which is taken exactly twice; 3 we will show now that the values of functions Λ l z , l 0, 1, on the set I \ B are nonreal.Let us represent this set as a union of four sets, I i , 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 K 0 z Λ l 0 z which pass through the points i and −i between lines β and β * , β {z : 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 Λ l 0 : 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 I 1 lies inside the shaded domain U that is bounded by the real semiaxis ξ 0 {z : z 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 Λ l 0 z maps the boundary ∂U of the domain U into the boundary of the right lower quadrant M {w : w > 0, w < 0} of the plain w.Hence, the domain U is mapped inside this quadrant, that is, all values of the function Λ l 0 in U are nonreal.It means the absence of real values of Λ l 0 in I 1 .For the domains I 2 , I 3 , and I 4 , the proof is similar.Let us now prove that β and β * do not intersect the line ξ 1/2 .It is sufficient to prove that Λ l 0 > 0 on the line ξ 1/2 {z : z s i/2, s ∈ R 1 } or, which is the same, that ch zv x ch zπ/2 z s i/2 > 0, A.11 for any s ∈ R 1 and x ∈ 0, 1 .Write ch s i/2 v x ch s i/2 π/2 ch 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 < arg D s, x < π/2, that is, D s, x > 0. Similarly A.11 can be proved in the case s < 0 and for Λ l 0 | z s−i/2 ; 4 let us find the values Λ l i/2 , l 0, 1: I the case l 0: 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.

C 1 :
−1/2 ≤ z ≤ 1/2}.Let us define the functions N l z N l z − i/2 which we shall consider in the strip I {z ∈ C : 0 ≤ z ≤ 1}.The operator κ l 1/2 coincides with the operator κ N l 1/2 from the family { κ N l α } generated by the function N l see 3.11 .Any other operator of this family acts as multiplication on the function:

Lemma 4 . 2 . 0 f r u r r 2
Let a measurable function f r satisfies condition ∞ dr 0, 4.13for any u ∈ V .Then f 0.

4 .
where L w 0 z is bounded, continuous function on I, and analytical inside I, and the coefficients b The function L w 0 z in 4.22 belongs to the space G.