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 ) was studied in many works. In most of these works, a family of one-parametric symmetrical operators is considered as such Hamiltonians. In addition, the question about the self-adjointness of is equivalent to the one concerning the self-adjointness of some auxiliary operators acting in the space . In this work, we establish a simple general criterion of self-adjointness for operators and apply it to the cases and . It turns out that the operator is self-adjoint for any , while the operator is self-adjoint for , where the value of is given explicitly in the paper.
1. 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 interacting pointwise with a particle of another nature having mass .
Originally, the construction of such Hamiltonian begins with introduction of the symmetric operator: acting in a Hilbert space . Here, , are the positions of fermions, is the position of a separate particle, and , , and are Laplacians with respect to , , and , respectively. The domain of definition of , consists of smooth rapidly decreasing functions on infinity, antisymmetrical with respect to , and satisfying the following conditions: Usually, some family of symmetric extensions of the operator is proposed as a possible “true” Hamiltonian of the system (the so-called Ter-Martirosian-Skornyakov extensions, see ). These extensions were constructed in [1–4]. For some values of mass , the extensions of Ter-Martirosian-Skornyakov are self-adjoint (for all values of the parameter ); however, for the other values of they are only symmetric with nonzero deficiency indexes (equal for all ). It turns out (see ) that the self-adjointness of all operators is equivalent to the one for some auxiliary symmetric operator acting in the space (see below). This operator commutes with the operators of the representation of the rotation group that acts in by the usual formula: Let us denote by the maximal subspace, where the representation (1.3) is multiplied by the irreducible representation of with weight , (see ). Evidently, the space is invariant with respect to the operator , and the restriction of this operator to the space is symmetric operator. The operator is self-adjoint if all the operators are self-adjoint. In this paper, we find general simple conditions of self-adjointness of and the form of the defect subspaces (with small exclusions) when these conditions are broken. Then, we apply these conditions to the cases and and get that the operator is self-adjoint for all values of , while the operator is self-adjoint for and has nonzero deficiency indexes for , the constant is indicated below (see (5.4)).
By the way, we note that the value of obtained in this paper differs from that one given by mistake in .
(1) After Fourier transformation: and change of variables: the operator can be represented as a tensor sum: where is a self-adjoint operator in : and acts in by formula with The operator is symmetric, and its domain is
(2) the deficiency subspace of the operator consists of the functions of the form: where the function belongs to Hilbert space with inner product Here is some positive operator acting in (see ). The domain of the operator , that is, a conjugate to , is where , . In addition, the operator acts by the formula: where is defined by (2.12).
The following asymptotics holds for vectors : Here where the operator is defined in (2.11), and is given by the following expression () defined on the set: The above-mentioned Ter-Martirosian-Skornyakov's extension of the operator is obtained by requiring where is an arbitrary parameter.
The operator can be represented as a sum of two operators: where the symmetric operator (with the domain ) acts as follows: and is a bounded self-adjoint operator. Since the deficiency indexes of coincide with the ones of (see ), we shall study the conditions of self-adjointness for the operator ;
(3) as we said, the space is invariant with respect to ; it has the form: where is the space of spherical functions of weight (see ) on the unit sphere . In addition, the operator has the form where is the unit operator in , and acts in by the formula: on the domain Here , , , are orthogonal polynomials (Legendre polynomials) satisfying : The operators are symmetric in , and the self-adjointness of is equivalent to the self-adjointness of . Later on, we shall study the operators and derive a condition of self-adjointness.
3. Preparatory Constructions
For every function , we consider the family of functions which we call a chain (with initial element and the final one ). All functions belong to and have a uniformly bounded norm: Consider the unitary map (Mellin's transformation ): and its inverse: For every set of functions , we denote by the set of their Mellin's transformations: For every chain , we denote by the family of functions: where , . The family can be represented as a function of a complex variable in the strip: The function is said to be associated with the chain , and its values on the lines are called the sections of .
Proposition 3.1. For every chain , , the associated function is continuous in a closed strip and analytic inside this strip. Moreover, its sections satisfy the following inequality: Inversely, any function which possesses these properties is associated with some (unique) chain , . Let call this chain generated by . In addition, the functions of the chain are obtained by the inverse Mellin's transformation from the sections of :
The proof of this proposition can be obtained by using the arguments given in the book by Paley and Wiener (see , 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 follows from the estimate (3.2) and the unitary Mellin's transformation. Denote by a linear space of functions satisfying conditions of Proposition 3.1. Let us introduce two maps: Let , , be a bounded, continuous function in the strip , which is analytic inside . This function generates the family of bounded operators in which act as multiplication on the functions , , : Evidently, for any , the function belongs to . If the chain is generated by and the chain is generated by , then where Denote by the following self-adjoint operator in : with the domain .
It is clear that for any , the power , of the operator is applicable to an element if and For the function that is associated with ,the action of the operator on the sections of has the form: (again if ).
4. The Operator
The operator (see (2.23)) can be represented as where is an operator in acting by the formula:
Lemma 4.1. Operator is bounded and self-adjoint in .
Proof. Pass to the operator: acting in . It follows from calculations in [2, 3] that is the operator of multiplication on the function: where and , . As we see the function , , is bounded and real. The lemma is proved.
We see from (4.4) and (4.5) that the functions and are continued up to bounded, analytical functions and correspondingly, defined in the strip . Let us define the functions which we shall consider in the strip . The operator coincides with the operator from the family generated by the function (see (3.11)). Any other operator of this family acts as multiplication on the function: Denote by the operators acting in .
Note that It is convenient to represent the operator in form of three sequential maps where are elements of the chain generated by the function . Note that the chain (4.9) can be rewritten in the following way: From (4.1) and self-adjointness of it follows that the operator with the domain is symmetric. For any , a representation of similar to (4.1) is valid: as well as decomposition like (4.9).
Let us now describe the domain of the operator conjugated to . Let be a function from and . Then for every , we can write Here we use the representation (4.11) for and the equality (4.8). Denote and apply the following evident assertion.
Lemma 4.2. Let a measurable function satisfies condition for any . Then .
From this and (4.12), it follows that Hence and is the final element of the chain . Thus the domain of the operator is In the case when the operator has the inverse one, , which is equivalent to the condition: the following equality is true: Let be an operator in with domain . Then for , the following representation holds true: if condition (4.17) is fulfilled. Here where is defined in (4.15).
Remarks. (1) Note that the function is invariant with respect to reflection of the complex plane around the point : Under this reflection, the strip is mapped onto itself; hence, for every zero of the function , there exists another zero, , of with the same multiplicity. The multiplicity of is even;
(2) Since as inside , the function has finite number of zeros inside .
We can now formulate the main criterion of self-adjointness of the operator .
Theorem 4.3. The operator is self-adjoint if and only if the function has no zeros in the closed strip .
Proof. (1) Assume in the strip . Then is bounded and continuous on and analytical inside . Let . Since for , the representation (4.19) holds true. Since
the element coincides with , that is, ; it means the self-adjointness of ;
(2) assume now the function has zeros . Consider first the case when all zeros are lying inside and their multiplicities are equal to , respectively. Again, let . Since , the representation (4.19) holds true. The function is meromorphic in with poles having the order respectively. For this function the usual canonical representation  is true: where is bounded, continuous function on , and analytical inside , and the coefficients depend on .
Lemma 4.4. The function in (4.22) belongs to the space .
The proof of this lemma is given in The appendix.
From (4.19) and (4.22), for , we have where the function is defined from relation where is an absolute constant, , and Since linearly independent functions do not belong to , due to (4.23), they form the basis in the defect subspace of the operator (see ). Since the dimension of the subspace is equal to and the operator is real, its deficiency indexes are equal and have the form: (It follows from Remarks that the sum is even). Consider now the case when one of the zeros of , say, , lies on the boundary of and has multiplicity (in addition, there is a zero ). In this case, in a neighborhood of , the function has the form: where is analytic in this neighborhood. Consider the function, whereby for , we mean the branch of a many-valued function that takes positive values on the positive part of the imaginary axis. Evidently, the function is analytic in the strip and satisfies condition (3.8). However, this function is discontinuous at and does not belong to . In addition, the function now belongs to as follows from (4.27) and (4.28). Thus but Consequently, but , that is, . Thus , and the operator has nonzero deficiency indexes. Theorem 4.3 is proved.
5. The Operators in the Cases and
Here, we apply Theorem 4.3 to the cases and .
Theorem 5.1. (1) For , the operator is self-adjoint for any ;
(2) the operator is self-adjoint for and has nonzero deficiency indexes for . In addition, for these indexes are equal to . The constant is a unique zero of (5.4).
Proof. We need the following properties of the functions and , .
Lemma 5.2. (1) For any the function is invariant with respect to reflection (4.20);
(2) The point is a nondegenerate critical point for both functions and ;
(3) These functions take real values on the line: and on the segment: Outside the set , both functions take nonreal values;
(4) the real values of , , are between and . Every value of —except —is taken exactly at two points;
(5) the extreme values of , , are given by
(6) the function increases monotonically on the interval .
The proof of this lemma is given in The appendix.
Corollary 5.3. (1) The zeros of , can only lie in the set ;
(2) for any value of , and therefore the operator is self-adjoint for all ;
(3) The function is positive if and vanishes at some point (and also at ) if .
In Figure 1, the curves corresponding to the functions and are depicted. We see that they intersect at a unique point with abscissa which satisfies the following equation:
Thus, for the operator is self-adjoint, and for it has deficiency indexes . For , the operator is not self-adjoint as well. Theorem 5.1 is proved.
Proof of Lemma 4.4. The function , admits the canonical representation (see )
where are zeros of (with multiplicities ), are constants, , and is a bounded, continuous analytic function in . From this, it follows that for any , . Consider some term of the sum (A.1) and write
It is clear that is bounded, continuous analytic function in . We are going to show that this function belongs to . Let be a small neighborhood of and the characteristic function of . Obviously, the bounded function satisfies condition (3.8). Every term of the sum
satisfies this condition as well.
Thus for fixed and , where Thus, we get the representation (4.22) where and the coefficients 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 and in the strip instead of the functions and in the strip . Similarly, instead of the reflection we consider the reflection around the point . It is clear that the functions , are invariant with respect to the change , and it means the invariance of with respect to reflection (4.20);
(2) it follows from (4.5) that is a nondegenerated critical point of and , if we note that . Correspondingly, is a nondegenerated critical point for , . The real axis coincides with the saddle-point line at (see ) for and . More precisely, these functions take real values on and decrease monotonically to zero as increases from zero to infinity. On the contrary, and increase monotonically along imaginary axis as increases from zero to . The monotonicity of along real axis follows from (4.5), equality , and inequality for and a similar inequality for . The proof of monotonicity of along real axis, and also monotonicity of both functions along imaginary axis is analogous if we note that on . Thus the functions , , take all values between and and every value except which is taken exactly twice;
(3) we will show now that the values of functions , , on the set are nonreal. Let us represent this set as a union of four sets, , as shown in Figure 2.
We consider the case ; the case is similar. Figure 3 shows the disposition of lines of levels for function which pass through the points and between lines and , , .
All these lines have common tangents at points and , and the line (resp. ) lies above (resp., below) the strip . The picture represented in Figure 3 is obtained by detailed study of the explicit formula for : together with the proof that the lines and do not intersect the strip . This proof is given below.
From Figure 3, we see that the set lies inside the shaded domain that is bounded by the real semiaxis , the segment on the imaginary axis and the part of line which lies in the right half-plane. From (A.10), it is easy to see that the function maps the boundary of the domain into the boundary of the right lower quadrant of the plain . Hence, the domain is mapped inside this quadrant, that is, all values of the function in are nonreal. It means the absence of real values of in . For the domains , , and , the proof is similar. Let us now prove that and do not intersect the line . It is sufficient to prove that on the line or, which is the same, that for any and . Write
Let . Then the values of numerator and denominator of lie in the right upper quadrant of a complex plain, and hence , that is, . Similarly (A.11) can be proved in the case and for ;
(4) let us find the values , :(I) the case : After the change , the integral (A.13) becomes (II)The case : The same change reduces to the integral
(5) let us show that the function: decreases monotonically as changes from to . We have because the numerator of (A.18) increases, while the denominator decreases with the growth of . This implies that that is, increases monotonically. Lemma 5.2 is proved.
This work is supported by RFBR Grant 11-01-00485a.
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