Научная статья на тему 'The point spectrum of the three-particle Schrödinger operator for a system comprising two identical bosons and one fermion on Z'

The point spectrum of the three-particle Schrödinger operator for a system comprising two identical bosons and one fermion on Z Текст научной статьи по специальности «Математика»

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Schrödinger operator / dispersion functions / zero-range pair potentials / discrete spectrum / essential spectrum

Аннотация научной статьи по математике, автор научной работы — Zahriddin I. Muminov, Vasilau. Aktamova

We consider the Hamiltonian of a system of three quantum particles (two identical bosons and a fermion) on the one-dimensional lattice interacting by means of zero-range attractive or repulsive potentials. We investigate the point spectrum of the three-particle discrete Schrödinger operator H(K), K ∈ T which possesses infinitely many eigenvalues depending on repulsive or attractive interactions, under the assumption that the bosons in the system have infinite mass.

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Текст научной работы на тему «The point spectrum of the three-particle Schrödinger operator for a system comprising two identical bosons and one fermion on Z»

NANOSYSTEMS: Muminov Z.I., et al. Nanosystems:

PHYSICS, CHEMISTRY, MATHEMATICS Phys. Chem. Math., 2024, 15 (4), 438-447.

http://nanojournal.ifmo.ru

Original article DOI 10.17586/2220-8054-2024-15-4-438-447

The point spectrum of the three-particle Schrodinger operator for a system comprising two identical bosons and one fermion on Z

Zahriddin I. Muminov1,2'", VasilaU. Aktamova3'6

1Tashkent State University of Economics, 100066, Tashkent, Uzbekistan 2Institute of Mathematics named after V.I.Romanovsky, 100174, Tashkent, Uzbekistan 3Samarkand Institute of Veterinary Medicine, 140103, Samarkand, Uzbekistan

azimuminov@gmail.com, bvaktamova@mail.ru

Corresponding author: Z. I. Muminov, zimuminov@gmail.com

PACS 02.70.Hm, 02.30.-f

Abstract We consider the Hamiltonian of a system of three quantum particles (two identical bosons and a fermion) on the one-dimensional lattice interacting by means of zero-range attractive or repulsive potentials. We investigate the point spectrum of the three-particle discrete Schrodinger operator H(K), K e t which possesses infinitely many eigenvalues depending on repulsive or attractive interactions, under the assumption that the bosons in the system have infinite mass.

Keywords Schrodinger operator, dispersion functions, zero-range pair potentials, discrete spectrum, essential spectrum.

Acknowledgements The authors acknowledge support from the Innovative Development Agency of the Republic of Uzbekistan (Grant No. FZ-20200929224), and they thank the anonymous referee for reading the manuscript carefully and for making valuable suggestions.

For citation Muminov Z.I., Aktamova V.U. The point spectrum of the three-particle Schrodinger operator for a system comprising two identical bosons and one fermion on z. Nanosystems: Phys. Chem. Math., 2024, 15 (4), 438-447.

1. Introduction

In physics, stable composite objects often result from attractive forces, enabling the constituents to lower their energy by binding together. In contrast, repulsive forces tend to scatter particles when they are in free space. However, within structured environments such as a periodic potential, and in the absence of dissipation, stable composite objects can exist even with repulsive interactions arising from the lattice band structure [1].

The Bose-Hubbard model, which describes repulsive pairs, serves as the theoretical basis for various applications.

The work referenced by [1] exemplifies the crucial link between the Bose-Hubbard model [2,3] and atoms in optical lattices, and helps pave the way for many more interesting developments and applications, as discussed in [4]. In cold atom lattice physics, stable repulsively bound objects are expected to be common, which leading to potential composites with fermions [5] or Bose-Fermi mixtures [6], and can even be formed with more than two particles in a similar manner.

The Efimov effect, first discovered by V. Efimov in 1970 [7], is one of the most intriguing phenomena in physics. It occurs in three-body systems in three-dimensional space that interact through short-range pairwise potentials. It is always possible to ensure the couplings of the interactions in such a way that none of the particle pairs has a negative energy bound state, but at least two pairs exhibit a resonance at zero energy. The existence of this effect of a three-particle Schrodinger operator was discovered by Yafaev [8] and in the discrete case by others [9-14].

Recently, the authors [15-17] considered perturbations of the system of three arbitrary quantum particles on lattices zd, d =1, 2, interacting through attractive zero-range pairwise potentials. They established that the three-particle Schrodinger operators posses infinitely many negative eigenvalues for all positive non-zero point interactions, under the assumption that two particles in the system have infinite mass. Also, note that the author of [17] obtained asymptotics for these eigenvalues.

The main goal of the paper is to investigate the existence of infinite number of bound states of the three-particle discrete Schrodinger operator associated with a system of two identical bosons and a fermion, where the bosons have infinite mass and the fermion has a finite mass. This investigation is performed on the one-dimensional lattice z and involves repulsive or attractive zero-range pairwise interactions. The problem is reduced to the study the Fredholm determinant of a diagonal operator.

It should be noted that, unlike the last three articles [15-17], we study eigenvalues below and above the essential spectrum of the unperturbed operator for all repulsive or attractive zero-range pairwise interactions.

© Muminov Z.I., Aktamova V.U., 2024

It is worth mentioning that in the continuous case, the authors of [18] studied a very similar system, specifically a system consisting of two infinitely heavy quantum particles and one light particle in three-dimensional space r3, interacting via long-range Coulomb pair potentials. Although the authors of this book are not interested in the number of eigenvalues, they briefly explained the scheme for solving problems of this type using the Coulomb spheroidal function decomposition. In the present paper, the problem is reduced to the study of eigenvalues of convolution type compact integral operators.

The paper is organized as follows. In Section 2, we introduce the three-particle discrete Schrodinger operator H(K) and the two-particle discrete Schrodinger operators associated with subsystems of the system of two identical bosons and a fermion. In Section 3, we study the essential spectrum of the three-particle discrete Schrodinger operator H(K). The eigenvalues of H(K) below and above the spectrum of the non-perturbed operator H0 (K) are investigated in Section 4. Section 5 is devoted to showing main result, Theorem 5.1.

2. Three-particle discrete Schrodinger operator on the lattice z

Let L2 (t2) be the linear subspace of the symmetric functions of the Hilbert space L2((t)2). Let us consider the discrete Schrodinger operator H(K), where K e t, associated with a system consisting of two identical bosons and a fermion moving on the one-dimensional lattice z (see [15,17]).

H(K) = Ho(K) - V

with zero-range attractive potentials

V = Vi + V2 + V3,

where

(Vif)(p,q) = 2^/ f (Vf )(p,q) = (i,9)dt

T T

(V3f)(p,q) = 2n/f(t,p + q - f e L2(t2),p,q e t,

T

and numbers A and ^ serve as the parameters of boson-fermion interaction and boson-boson interaction, respectively.

Here the numbers A and ^ indicate repulsive pair-wise interaction when A < 0 and ^ < 0, and attractive pair-wise interaction when A > 0 and ^ > 0. The operator H0 (K) is defined on the Hilbert space ¿S (t2) by

(Ho(K)f )(p, q) = E(K;p, q)f (p, q), f e ¿2(t2),

and

E(K;p, q) = £b(p) + £6(q) + £f(K - p - q), p, q e t. Here, the real-valued continuous function eb( ) and £f( ), referred to as the dispersion relation associated with the free boson and fermion, is defined as

£6 (p) = — £(p), £f(p) = — £(p), £(p) = 1 - cos(p), p e t, (1)

m m

respectively, and m and m represents the mass of the boson and fermion, respectively. Let k e t and h\ (t) be a linear subspace of the Hilbert space L2 (t) defined by

¿k(t) = {f e ¿2(t)|f(p) = f(k -p)}.

A two-particle discrete Schrodinger operator corresponding to the subsystem {bozon,fermion} and {bozon,bozon}, of the three-particle system acts on the Hilbert space L2 (t) and ¿k (t) as

h1(k) = h0(k) - v1; and h2(k) = h0(k) - v2, k e t, (2)

respectively.

Here, the operators (k)

and where

The operators v1 and v2 are defined as

(h0(k)f (p) = E(1)(p)f(p), f e ¿2(t), (h2(k)f)(p) = Ek2)(p)f(p), f e Lk(t), E(1)(p) = £b(p) + £f(k - p), E(2)(p) = £b(p)+ £b(k -p), p e t. (3)

and

A

(vif)(p) = /f(q)dq, f e L2(t), p e t.

T

(V2f)(p) = /f (q)dq, f e ¿k(t), p e t,

respectively.

2.1. Spectral properties of the two-particle discrete Schrodinger operators when m = x and 0 < m < x

With m = to and 0 < m < to and the equality (1), the functions (3) can be written as

E(1)(p) = £f(h - p) = e(k - p)/m, E(2)(p) =0, p e t.

Consequently, since the potentials va, a =1, 2 have a convolution-type property, all two-particle Schrodinger operators do not depend on the quasi-momentum k e t,

h1 := h1(k) and h2 := h2(k).

Then, the operators h1(k) and h2(k) act as

h1(k)f (p) = £f(p)f (p) - (V1f)(p), f e L2(T) and h2(k)f (p) = -(V2f)(p), f e Li(t).

As v1 is a finite rank operator, according to the Weyl theorem, the essential spectrum aess(h1(k)) of the operator h1(k) in (2) coincides with the spectrum <7(h° (k)) of the non-perturbed operator h0(k). More specifically,

aess(h1(k)t = [E^k), Emax(k)],

where

E(„i)n(k) = min Ei1)(p), Emax(k) = max e[1) (p).

pet peT

Therefore, in our case, we have

(?ess{h1(k)t = [0, 2/m] and aess(h2(k)t = {0}. The Fredholm determinants associated with the operators h1(k) are defined as

[]

A(A; z) = 1 - Xdo(z), d0(z) = — —S-, z e c \ [0, 2/m].

2n J £f(s) - z

T

Lemma 2.1. (a) The number z e c \ [0, 2/m] is an eigenvalue of h1(k) if and only if A(A; z) = 0.

((b) If A < 0 and A> 0, then there exists a unique simple eigenvalue z = z0 of h1 (k) in the interval (- 2/m - to) and ( - to, 0), respectively. Moreover, z1 does not depend on k e t.

Proof. (a) The equation

h1(k)f = zf i.e., f = (h°(k) - z) — 1 V1f has a non-trivial solution if and only if

A(A; z)C = 0, C e C,

has a non-trivial solution.

Therein, the solutions C e c and f e L2 (t) are related as follows

C = V1 f and f = (h?(k) - z)—1C.

(b) Let A > 0. The function A(A; z) is monotonic decreasing in (-to, 0) and A(A; z) > 1 in ( - 2/m - ^, to). Since

lim A(A; z) = 1 and lim A(A; z) = -to,

the intermediate-value theorem implies the existence of a unique simple zero z = z0, z° e (-to, 0) U ( - 2/m - ^, to) of the function A(A; •), and furthermore z° e (-to, 0).

The lemma can be proven in a similar way when A < 0. □

Now, we can summarize the results of this section in the following lemma. Lemma 2.2. We have

*disc(h1(k)) = {z0}, f A = 0, a(h1(k)t = {z0} U [0, 2/m], if A = 0

and

^disc(h2(k)t = -}, a(h2(k)t = {-M} U {0}.

3. Essential spectrum of H(K)

One of the notable outcomes in the spectral theory of multi-particle continuous Schrodinger operators involves characterizing the essential spectrum of the Schrodinger operators in terms of cluster operators (the HVZ-theorem. See Refs. [19-23] for the discrete case and [24] for a pseudo-relativistic operator).

Lemma 3.1. The essential spectrum of H(K) satisfies the relation

aess(H(K)) = J |a(hi(K - k)) + eb(k)} U U {^(K - k)) + £f(k)}.

het het

Proof. The proof can be found in [17,21]. □

3.1. The essential spectrum of H(K) with m = ж, and m < ж

Due to Lemma 2.2 and the relations e b (p) = 0 and £f (p) = e (p) /m, we obtain

U {a(hi(K - k)) + eb(k)} = a(hi(k)) = jz?} U [0, 2/m],

he t

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U {^(h2(K - k)) + ef(k)} = U { j-м} U j0} + ef(k)} = [ - 2/m - M] U [0, 2/m].

keT heT

According to the last two relations and Lemma 3.1, we have Lemma 3.2.

aess (H(K)) = jz?} U ([-M, 2/m - U [0, 2/m]).

4. The point spectrum of H(K) for m = ж and m < ж

One can show easily that the subspace

A0 = jf G L2(t X t)|f (p,q) = g(p + q), g G L2(t)}

is invariant under the operator H(K), and so it is A^ = Lj; (t x t) 0 A0.

Therefore, we have ( ) ( ) ( )

app(H(K)) = app(A?(K)) U aw(Ai(K)),

where A0(K)and A1(K) are restrictions of H(K) on the linear subspaces A0 and A^, respectively.

Since A0 and L2(t) are isomorphic, the operator A0(K) is unitarily equivalent to the operator B0 on L2(t), where

B0 = E0(K) - - 2Av, (4)

E0(K) denotes the multiplication by the function ef(K - p), I is the identity operator, and v is an integral operator defined by

(vf)(p) = 2П / f(q)dq, f e l2(t), p e t.

T

The operator A1(K) takes the form

A1(K) = H0(K) - AVi - AV2. Let Uk : L2(t x t) ^ L2(t x t) be a unitary operator defined as

(Ukf )(p,q) = f ( - K/2+ p, -K/2 + q). (5)

It establishes a unitary equivalence between H(K) and H(0), and so we can prove the coming statements for H(0).

4.1. Spectrum of Ao(K)

The equivalence (4) and the Weyl theorem imply that

aess(A0(K)) = [ - 2/m - M].

The following lemma describes the behaviour of the eigenvector of H (K) in the linear space A0.

Lemma 4.1. (a) If A < 0, then A0(K) has an unique eigenfunction with the corresponding eigenvalue n, П G (2/m -to) for any ^ G r.

(b) If A = 0, then A0(K) has no eigenvalues for any ^ G r.

(c) If A > 0, then A0(K) has an unique eigenfunction with the corresponding eigenvalue n, n G (-to, -for any ^ G r.

Proof. The equation

H(0)f = zf, f e A0, has a solution if and only if the equation ( t

C(1 - 2Ad0(M + z)t =0, (6)

has one, and their solutions are related by

2A

f (p, q) := g(p + q) = -;-

£f(-p - q) - m - z

where

C = — [ g(t)dt. 2n J t

IT

Equation (6) has a nontrivial solution C if and only if the equation

A(2A; z) = 1 - 2Ad0(M + z) = 0

has a root. The function A(2A; z) is defined on (-to, -m) n (2/m - m, to) and according (o Lemma 2.1: (a) If A > 0, then A(2A; z) = 1 - 2Ad0(p + z) lias a unique zero in (-to, -m), but does not have one in (2/m - m, to) ; (c) If A < 0, then A(2A; z) has a unique zero in (2/m - m, to) , but does not have one in ( - to, -m) .

(b) If A = 0, then B0 = E0(K)-mI is a multiplication operator, and so A0(K) has no eigenvalues for any m e r. □

4.2. Spectrum of Ai(K)

As the operator A1 (K) does not contain the parameter m, Lemma 3.2 implies that

aess(A1(K)t = {z0} U [0, 2/m]. In order to study need, we define the following integral depending on n e z:

1 f eint

dn(z) = — -dt, z e c \ [0, 2/m].

2n J £f(t) - z

£f(t) -•

T

Lemma 4.2. For any fixed n e z, the function dn(z) satisfies the equality

(1 - mz - \Jm2z2 - 2z/m)

|n|

dn(z)= m^-too*. — z e c \ [0, 2/m].

m2z2 - 2z/m

Proof. The theory of residues provides the proof (see [17, Lemma 6]). □

The operators V1 and V2 are represented as V1 = A<2<1 and V2 = A<2<2, respectively, where the operators

v 1, v2 : L2(t x t) ^ L2(t) are of the form

)(p) = tt^t I f (p,t)dt and (V2f)(q) = I f (t,q)dt.

(2n) 2 j (2n)2 J

TT

Lemma 4.3. The number z e c \ aess(A1(K)) is an eigenvalue of the operator A1(K) ifandonly if D(z) = 0, where

1

D(z)= a^IIa+(a; z), (7)

nez

with ( t

A+(A; z) = 1 - A(d0(z) + dn(z)).

If zn e c \ aes^A1 (K)) is an eigenvalue of A1(K), then Dn(zn) = 0, n e z and the corresponding eigenfunction is of the form

fn(p, q) = ^--(cos (n(-K/2 + p)t + cos (n(-K/2 + q)t) (8)

£f(K - p - q) - zn v '

Proof. Given the unitary equivalence of H(K) and A1(K) to H(0) and A1(0), respectively, we first establish the claim for the latter operators(. t

Let z e c \ aess(A^0)) be an eigenvalue of A1(0), and let f be the corresponding eigenfunction, then

f = R0(z)[AV1 + AV2] f, (9)

where R0(z) = (H0(0) - zl) —1 is a resolvent of H0(0).

This equation has a non-trivial solution if and only if the system of two linear equations

= <a(R0(z)[A« + A<2<2]), a = 1, 2, <1,<2 e L2(t) (10)

on the space L2 (t) © L2 (t) has a non-zero solution.

Solutions of equations (9) and (10) are linked by

f (p,q) = R0(z)[A^i + A^2], (11)

and

¿a = ¿af, a =1, 2.

Since f is a symmetric function, « and ¿2 are the same function, and only their arguments are different, that is, the argument of « is the variable p, while q is an independent variable of

¿2 (p) = ^(p)- (12)

We note that the functions

A„ := I - A^aR0(z)^a, z e c \ aess(H0(0)), a = 1, 2,

are nonzero for any z e c \ aess(A1 (0)), therefore, their inverses exist,

A-1 = (I - A^R0(zVar1. Then, the solutions ¿a, a =1, 2, of the equation (10) satisfy the following system of integral equations

¿i = AA-1 QC,52,

^ - (13)

¿2 = AA21 Q2«i, where

Q = ¿iR0(z)y>2

is the integral operator on L2(t) defined as

(q^)(p) = /

£f(-p -1) - z

T

and Q1 is the adjoint of Q.

Using the substitution method, system (13) can be reduced into the form

¿i = Q(z)«i, i.e., Q(z) = A2A-1A-1QQ2. (14)

Moreover, if $ = (<$1, ¿2) is a solution to (13), then « is an eigenfunction of Q(z) corresponding to the eigenvalue 1. Conversely, suppose that « is an eigenfunction corresponding to the eigenvalue 1 of the operator Q(z). Then $ =

(¿51, ¿2), with ¿2 = ———-Q« is a solution to (13) (i.e. (10)). Notice that the multiplicities of the linearly independent

A(A; z)

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eigenvectors « and $ coincide.

We also note that the function f defined in (11) is an eigenfunction of A1 (0) corresponding to an eigenvalue z e c \ <ress(A1(0^. Moreover, the multiplicity of the eigenvalues z of A1(0) is the same as the multiplicity of the eigenvalue

^ = 1 of Q(z).

The operator Q(z) is a convolution-type trace-class integral operator. The standard Fourier transform F1 : L2 (t) ^

l2(z),

(Fig)(n) = -V/einpg(p)dp, g e L2(t), n e z,

(2n) 2 Jt

establishes that Q(z) := F1Q(z)F"1* acts as a multiplication operator on the space I2 (z) by the function

A2

Kn(z) = ---2 d„(z)d-„(z), n e z. (15)

(A(A;z))

Thus, the spectrum of Q(z) consists of the following union

a(Q(z)) = {0}U J {«„(z)},

„ez

with the space of eigenfunctions

¿„(m) = i„,m, n, m e z, where is the Kronecker delta function on z.

Note that the compact operator Q(z) has eigenvalues k„(z), n e z with the corresponding eigenfunctions

^„(p) = einp, n e z, p e t. Therefore, the determinant of the operator I - Q(z) can be written as the following product

det(1 - Q(z))=H(1 - K„(z)), (16)

„ez

which takes the form (7), since d-n(z) = dn(z) and (15).

Let Kn(zn) = 1 be an eigenvalue of Q(zn), then ^n is the first component of the solution ipn = (^n,i}n) of equation (10), and the second component is defined as

V-n(q) := ^(q) = AA2-1Q>n(q) = V-n(q). (17)

A(A; zn)

Using the equality (12) in the last relation, we get

ei(n'q) = AA-1Q>n(q) = e-i(n'q),

A(A; zn)

which is a contradiction.

However, if ên(p) = (ei(n'p) + e-i(n'p))/2 = cos((n,p)), it satisfies (12) and (17) if and only if

Adn(zn\ = 1, i.e. A+(A; z) = 1 - A(do(zJ + dn(zj) = 0.

A(A; zn)

However, for the functions On(p) = (ei(n'p) — e-i(n'p))/2i = sin((n,p)) the relation

On = AA- 1Q*9n = — . n) On, A(A; zn)

holds, which contradicts with (12). It implies

Adn(zn)

A(A; zn)

= 1, i.e. A-(A; z) = 1 - A{d0(zn) - dn(zri))=0.

Since the system {$„, ви}, n e z is complete in l2(z), the last three relations and (15) allow us to use the Fredholm determinant (7) instead of (16).

Subsequently, the number zn is an eigenvalue of A1 (0) and the corresponding eigenfunction can be found by (11) as

fU = Ro(z„) [Х<р1Фи + Фи] ,

i.e.,

f°(p, q) = —-Г-(cos(np) + cos(nq)) .

£f( -p - q) - z V J

A

£f(-p - q) - z '

According to the relation H(0) = U*KH(K)UK, the number zn is also an eigenvalue of H(K) with the eigenfunction fn = UK fn in (8), where the unitary operator UK is defined in (5). □

Remark 4.4. Since A+(A; z) is even with respect to n e z, then zn = z—n, where zn is a zero of A+(A; z), and eigenfunctions (8) are same.

Lemma 4.5. For any fixed n e z, the function d0(z) + dn(z) is positive and monotonically increasing in the interval (-to, 0) as a function of z.

Additionally,

lim (do(z)+ dn(z)) = œ, (18)

z^Q —

andfor any z e (-to, 0) hold.

Proof. The equalities and

lim (do(z) + dn(z))=0. (19)

lim dn(z) = 0 (20)

dn(z) = 2П /

cos nt , dt

£f(t) - z

T

do(z) + dn(z) = 2- J

1 f 1 + cos nt , ' dt

£f(t) - z

T

imply the positivity and monotonicity of d0(z) + dn(z) in (-to, 0).

The limits (18), (19) and (20) follow from Lemma 4.2. □

4.3. Eigenvalues of H(K) below [0, 2/m]

Let z0 be a zero of Aa(-) = 0, i.e. an eigenvalues of hi(k) in the interval r \ [0, 6/m] (see Lemma 2.1).

Lemma 4.6. (a) Let A > 0. Then A+(A; z) has a unique zero zn in (-to, 0) such that zn < z0 < 0. Moreover, zn < zm if |m| > |n|.

(b) The following limit

lim zn = z0 (21)

holds.

Proof. (a) Since A+(A; z) is symmetric about the interval [0, 2/m], we prove part (a). According to Lemma 4.5, the function A+(A; z) is monotonically increasing in (-to, 0) and

lim A+(A; z) = 1,

A+(A; z0) < A(A; z0) = 0,

holds.

Consequently, the function A+(A; z) has a unique zero z„ in the interval (-to, z°) if A > 0. If |n| < |m|, then Lemma 4.2 implies that A+(A; z) < A+(z) and hence A+ (A; zm) < Am(zm) = 0. The last equality provides the inequality z„ < zm. (b) The equalities

A+(A; z0) = (A+ (A; £„))'(z0 - z„),

A+(A; z0) = -Ad„(z0), (A+ (A; z))' = -A(d0(z) + d„(z)),

where z„ < £„ < z0 is a number obtained due to the intermediate value theorem, imply

o _ d„(z0)

z„ - z — —

n 1 d0(e„) + <(£„)•

By Lemma 4.6, the inequality z0 < z„ < £„ < z0 and the monotonicity of the derivative of (A+ (A; z^',

(A+ (A; z0))' > (A+(A;C„))' > (A+ (A; z0))' Applying the limit (20) in the last inequality, we get the proof of the limit in (21). □

4.4. Eigenvalues of H(K) above [0, 2/m]

The unitary operator Un/2 in (5) is used to establish the equalities

2

Un/2H0(K)Un/2 = m - H0(K) and UV2VU^/2 = V

which implies the relation

2

Un/2 (H0 (K) - V) Un/2 = m - (H0 (K) + V). (22)

The final relationship enables us to shift the investigation of the eigenvalues of H(K) from above the interval [0,2/m] to below it.

Note that z„ is a zero of A+ (A; z), n e z, if it exists.

Lemma 4.7. Assume A < 0.

(a) The function A+(A; z) has a unique zero z„ in (2/m, to) such that 2/m < z0 < z„. Moreover, z„ > zm if

|m| > |n|.

(b) The following limit

zn = z0

holds.

Proof. The proof is a consequence of Lemma 4.6 and the identity (22). □

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5. Main theorem

Recall that z0 be a zero of A(A; •) = 0, i.e. an eigenvalues of h1(k), and z0 < 0 if A > 0 and 2/m < z0 if A < 0. Let n be an eigenvalue of H(K) mentioned in Lemma 4.1.

Now, we are ready to formulate the main result of the paper.

Theorem 5.1. Assume p e r and A e r. (a) Let A < 0. Then

aro(H(K)) = J {z„}U{n},

„ez

where 2/m < z0 < z„ and 2/m - p < n.

(b) Let A = 0. Then

*pp( H (K)) = 0.

(c) Let 0 < A. Then ( )

aro(H(K)) = (J {z„}U{n},

„e z

and z„ < z0 < 0 and n < - p.

Proof. Lemmas 4.1,4.6 and 4.7 provide the proof. For example, by combining the assertions (a) in Lemma 4.1 and (a) in Lemma 4.7, we get the proof of the part (a) of the theorem. □

Remark 5.2. According to Lemma 3.2

a ess (H (K)) = {z0} U ([ - p, 2/m - p] U [0, 2/m]) .

Theorem 5.1 demonstrates that z„ or n could be within aess (H(K)) or within a gap of the essential spectrum.

6. Conclusion

The discrete Schrodinger operator corresponding to the Hamiltonian of a system of three quantum particles (two identical bosons and a fermion) with masses m = to and m < to, respectively, is considered on the one-dimensional lattice for all non-zero point interactions. The point spectrum of the three-particle discrete Schrodinger operator, which possesses infinitely many eigenvalues, has been studied for all non-zero point interactions.

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Submitted 18 April 2024; revised 21 June 2024; accepted 26 June 2024

Information about the authors:

Zahriddinl. Muminov - Tashkent State University of Economics, 100066, Tashkent, Uzbekistan; Institute of Mathematics named after V.I.Romanovsky, 100174, Tashkent, Uzbekistan; ORCID 0000-0002-7201-6330; zimuminov@gmail.com

Vasila U. Aktamova - Samarkand Institute of Veterinary Medicine, 140103, Samarkand, Uzbekistan; ORCID 0009-00006587-0021; vaktamova@mail.ru

Conflict of interest: the authors declare no conflict of interest.

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