Bound States of a System of Two Fermions on Invariant Subspace ()

1. Introduction
The nature of bound states of two-particle cluster operators for small parameter values was first studied in detail by Minlos and Mamatov [1] and then in a more general setting by Minlos and Mogilner [2]. In [3], Howland showed that the Rellich theorem on perturbations of eigenvalues does not extend to the resonance theory. Studying bound states of a two-particle system Hamiltonian H on the d-dimensional lattice
reduces to studying [2] [4] [5] [6] [7] the eigenvalues of a family of Shrödinger operators
, where
is the total quasi-momentum of a system. Moreover, eigenfunctions of
are interpreted as bound states of the Hamiltonian H, and eigenvalues, as the bound state energies. The bound states of H of a system of two fermions on a one-dimensional lattice were studied in [4], a system of two bosons on a two-dimensional lattice was studied in [6], and perturbations of the eigenvalues of a two-particle Shrödinger operator on a one-dimensional lattice were studied in [8]. The finiteness of the number of eigenvalues of Shrödinger operator on a lattice was studied in the works [7] [9].
The discrete spectrum of the two-particle continuous Shrödinger operator
was studied by many authors, with the conditions for the potential V formulated in its coordinate representation. The condition for the finiteness of the set of negative elements of the spectrum and the absence of positive eigenvalues of
can be found in [10]. If
, then the number of negative eigenvalues
is a nondecreasing function of
, and each eigenvalue
decreases on the half-axis
. It is known that when the coupling constant
decreases, the bound state energies of
tend to the boundary of the continuous spectrum (see [10] ) and for some finite
are on the boundary. Two questions then arise: Does a bound or virtual state correspond to such a threshold state (i.e., is the corresponding wave function square-integrable)? And where do the bound states “disappear to” as
decreases further? The study of the first question was the subject in [11] [12]. Regarding the second question, it turns out that the bound state disappears by being absorbed into the continuous spectrum and becomes a resonance [5].
Here, we consider bound states of the Hamiltonian
(see (1)) of a system of two fermions on the three-dimensional lattice
with the special potential
(see (5)). In other words, we study the discrete spectrum of a family of the Shrödinger operators
,
, (see (3)) corresponding to
in the invariant subspace
.
Restriction of the operator
in the invariant subspace
is denoted by
.
In the case
, the operator
has an infinite number of eigenvalues of the form
and the essential spectrum consists of the single point 6. Here, the potential
is defined by (5) and
is a decreasing function on
and
. These eigenvalues
are arranged in ascending order,
, and the smallest eigenvalue
is threefold,
is sevenfold, and the other eigenvalues
are ninefold. All ninefold eigenvalues
of the operator
are simple eigenvalues for the operator
.
Further, we investigate eigenvalues and eigenfunctions of the restriction operator
.
In the case
the corresponding operator
has infinitely many invariant subspaces
. It is proved that the restriction
of the operator
in the invariant subspace
has no more than one eigenvalue. If exists, it can be calculated explicitly. For every
the operator
has only a finite number of eigenvalues.
For any perturbation
, the essential spectrum
of
becomes the essential spectrum
. If the potential
is of the form (5), the Shrödinger equation
can be exactly solved (see Theorem 1).
The Shrödinger equations
and
with small
are solved by using methods invariant subspaces and operator theory.
2. Description of the Hamiltonian and Expansion in a Direct Integral
The free Hamiltonian
of a system of two fermions on a three-dimensional lattice
usually corresponds to a bounded self-adjoint operator acting in the Hilbert space
by the formula
Here, m is the fermion mass, which we assume to be equal to unity in what follows,
and
, where I is the identity operator, and the lattice Laplacian
is a difference operator that describes a translation of a particle from a side to a neighboring side,
where
are unit vectors in
. The total Hamiltonian
acts in the Hilbert space
and is the difference of the free Hamiltonian
and the interaction potential
of the two fermions (see [8] [13] ):
(1)
where
Hereafter, we assume that
(2)
Under this condition, the Hamiltonian
is a bounded self-adjoint operator in
.
We pass to momentum representation using the Fourier transform [2] [4] [7]
The Hamiltonian
in the momentum representation commutes with the unitary operators
, given by
It follows that there exist decompositions of
and the operators
and H into direct integrals (see [7] [9] and [10] )
Here,
and
is an operator of multiplication by the function
in
. The fiber operator
of H also acts in
and is unitarly equivalent to
, which is called the Shrödinger operator. This operator acts in the Hilbert space
by the formula
(3)
The unperturbed operator
is an operator of multiplication by the function
(4)
From (3) and (4), it follows that
so we can assume
.
The perturbation operator V is an integral operator in
with the kernel
and belongs to the class of Hilbert-Schmidt operators
.
In this work, we consider the operator
with the potential
of the form
(5)
where
. Supporter is in the cylinder:
Since for every function
the equality
holds, then the value of the potential
at the origin can be set arbitrary, since it does not affect the result, for simplicity, we assume that
.
The function
in (5) is decreasing in
i.e.,
(6)
and belongs to
. The kernel
, of the integral operator V, i.e., the Fourier transform
, of the potential
, has the form
(7)
Eigenvalues of the operator
. We note that the spectra of the operators
and V are known. The operator
does not have eigenvalues, its spectrum is continuous and coincides with the range of the function
:
The spectrum of V consists of the set
. Under condition (2), the operator V is a Hilbert-Schmidt operator and is hence compact. By the Weyl theorem [10], the essential spectrum of
coincides with the spectrum of
:
If
, then the spectrum of
consists of eigenvalues of the form
and the essential spectrum is
. If
(for some
), then there exists a potential
such that
has an infinite number of eigenvalues outside the continuous spectrum (see [4] [14] ).
We recall some notations and known facts. For any self-adjoint operator B acting in a Hilbert space
without an essential spectrum to the right of
, we let
denote the number of its eigenvalues to the right of
. We let
denote the number of eigenvalues of
to the left of
, i.e.,
. The number
in fact coincides with the number of eigenvalues outside the continuous spectrum of
. It follows from the self-adjointness of
and positivity of V that
and hence
. Therefore we seek only eigenvalues z less than
.
For any
and
, we define the integral operator
where
is the resolvent of the unperturbed operator
. Under condition (2), the operator V is positive, and we let
denote the positive square root of the positive operator V. A solution
of the Schrödinger equation
and the fixed points
of
are connected by the relations
The following proposition (the Birman-Schwinger principle) holds [9].
Lemma 1. The number of eigenvalues of
to the left of
coincides with the number of eigenvalues of
greater than unity, i.e., the equality
holds.
Lemma 2. If for some
the limit operator
exists and is compact, then the equality
(8)
holds.
Equality (8) states that the number of eigenvalues of
, to the left of
is equal to the number of eigenvalues of
greater than unity.
3. Invariant Subspaces of
In this section, we study the invariant subspaces with respect to the operator
.
Let
be a subspace of the space
, consisting of odd functions on
, and
be a subspace of
, consisting of even functions on
. In addition, we use the notation
Note that
is a subspace of the space
. It is natural to expect the invariance of the subspace
with respect to the operator
. It turns out that this subspace is invariant under the operator
, i.e. the following statement holds.
Lemma 3. Let the potential
have the form (5). Then the subspace
is invariant under the action of
.
Proof. We prove that this subspace is invariant first with respect to
, and then with respect to V. It follows from representation (4) that the function
belongs to the subspace
, and it follows from the inclusion
that
. This proves that
is invariant with respect to
.
Simple calculations show that the function (see (7))
belongs to the subspace
for
. Hence, we prove the invariance of
with respect to V, and it follows that
is invariant with respect to
.
denotes the restriction of
to the respective subspace
. The action of
is unchanged, the unperturbed operator
is an operator of multiplication by the function
. We present the formula for
operator V acts on the element
according to the formula
Note that for
, the spectrum of
consists only of the eigenvalues
and the essential spectrum
. Under condition (6) the number
is a threefold eigenvalue of
, with the corresponding eigenfunctions
the number
is a sevenfold eigenvalue with the corresponding eigenfunctions
for each
, the number
is a ninefold eigenvalue, and the corresponding eigenfunctions are
The number
is an eigenvalue of an infinite multiplicity, and the corresponding eigenfunctions are
All ninefold eigenvalues
of the operator
are simple eigenvalues for the operator
, and the number
is an eigenvalue of an infinite multiplicity.
If the third coordinate
of the total quasimomentum
is equal to
, then the operator
has infinitely many invariant subspaces
.
Next, we give a description of the invariant subspace
.
The system of functions
is an orthonormal basis in the space
. Let us denote by
the one-dimensional subspace spanned by the vector
. The space
can be decomposed into the direct sum
This decomposition produces another decomposition
where
Lemma 4. Let the potential
have the form (5). Then the subspace
is invariant under
for any
.
Proof. Let
, where
,
is an arbitrary element of
. We consider the action of
on
:
(9)
(10)
To obtain the last formula (10), we use the orthogonality of the system of functions
in
. Relations (9) and (10) imply the equality
(11)
which completes the proof of the lemma.
We denote by
restriction of the operator
in the invariant subspace
. Formula (11) shows that the restriction
to the subspace
has the form
(12)
where I is the identity operator and
,
, is a two-dimensional two-particle operator acting in
by the formula
where
, and
is a one-dimensional integral operator in
with the kernel
Studying the eigenvalues of
by representations (12) reduces to studying the eigenvalues of
i.e. the three-dimensional problem reduces to the two-dimensional problem.
4. Eigenvalues of the Operator
Our main goal in this section is to study the behavior of the nondegenerate eigenvalue
of
at small perturbations
(
or
), i.e. the eigenvalues of
(or
) at small perturbations
. The studying of the eigenvalues of
is reduced to study the eigenvalues of the operator
for each fixed
. In turn, the problem of studying the eigenvalues of the operator
by virtue of (12) is reduced to study of the discrete spectrum of the operator
Studying the eigenvalues of
and
reduces to studying the eigenvalues of
acting in
by the formula
(13)
It is known that the essential spectrum of
consists of a segment
, where
,
.
Further we give some information about the eigenvalues and eigenfunctions of the operator
. Combining Theorem 6.3 in [6], Theorem 5.10 in [15] and Lemmas 1 and 2 we obtain the following statement about eigenvalues of the operator
.
Lemma 5. Let
.
a) If
, then the operator
has no eigenvalues lying outside of the essential spectrum.
b) If
, then the left edge
of essential spectrum of the operator
is a resonance.
c) If
, then the operator
has a unique nondegenerate eigenvalue
which lying in the left of the essential spectrum with corresponding normalized eigenfunction
(14)
Here
is the normalizing multiplicity.
d) The operator
has no embedded eigenvalues in the interval
.
Hilbert space
can be written as a direct sum:
The following lemma establishes a connection between the operators
and
.
Lemma 6. Let the potential
have the form (5). Then:
a) the subspace
and its orthogonal complement
are invariant under
.
b) restriction of the operator
to the invariant subspace
coinsides with the unperturbed operator
.
c) restriction of the operator
to the invariant subspace
can be represented as a tensor product:
(15)
Here, I is the identity operator, and
,
is a one-dimensional two-particle operator acting in
by the formula (13).
This lemma is proved in the same way as the Lemma 4. In particular, part b) of the lemma implies that the operator
has no eigenfunctions in
. Thus, studying the eigenvalues of the operator
is reduced to studying eigenvalues of the operator
.
From Lemmas 5 - 6 and tensor product (15) implies the following statement regarding operator
.
Theorem 1. Let
and
.
a) If
, then the operator
has no eigenvalues lying outside of the essential spectrum.
b) If
, then the left edge
of essential spectrum of the operator
is a resonance.
c) If
, then the operator
has a unique nondegenerate eigenvalue
(16)
which lies in the left of the essential spectrum and with the corresponding normalized eigenfunction
where
is the normalized eigenfunction of the operator
corresponding to the eigenvalue
, the operator
is defined by the formula (13).
d) The operator
has no embedded eigenvalues in the interval
.
Similar statement is true for the operator
. The eigenvalues of the operators
and
are same, but eigenfunctions differ with variable replacement
and
. In other words, the operators
and
are unitary equivalent. Therefore, the operators
and
are unitary equivalent too.
Similar statement can relatively be formulated for the operator
. For this purpose, we introduce the following notation. Through
we denote the Fredholm determinant of the operator
, where
is the resolvent of the operator
, and
is an integral operator with the kernel
Through
denote the value of the following integral:
Simple calculations reveal the following approximate value
.
Theorem 2. Let
,
.
a) If
, then the operator
has no eigenvalues lying outside of the essential spectrum.
b) If
, then the left edge
of the spectrum of the operator
is an eigenvalue.
c) If
, then the operator
has a unique nondegenerate eigenvalue
below the essential spectrum.
d) The operator
has no embedded eigenvalues in the interval
.
This theorem is proved in similar way as Lemma 5. There are some differences:
1) In the Theorem 2, the eigenvalue
was calculated with the accuracy of
:
and corresponding normalized eigenfunction has the form
(17)
where
is the normalizing multiplicity.
2) Left edge
of the essential spectrum is a resonance for the operator
, but for the operator
the left edge
of the essential spectrum is the eigenvalue, i.e. the equation
has a non-trivial solution
and it belongs to
.
5. Conclusions
1) We have shown that the operator
has infinitely many invariant subspaces
. It has been proved that if condition
holds then the operator
has a unique simple eigenvalue
of the form (16), otherwise, the operator has no eigenvalues outside of the essential spectrum. A similar statement holds for the operator
.
2) Without loss of generality it can be assumed that
. Since, if
then it follows from
that there exists a number
such that
and monotonicity of
implies that
for
, and in this case, the eigenvalues
of
exist for all
.
For a fixed
there exists
such that
and the operator
has m nondegenerate eigenvalues outside of the essential spectrum (see Theorem 1):
The corresponding normalized eigenfunctions are of the forms:
where,
is the normalized eigenfunction of the operator
corresponding to the eigenvalue
and the operator
is defined by the formula (13),
.
The eigenvalues of the operators
and
are same but eigenfunctions differ with variable replacement
and
. In other words, the operators
and
are unitary equivalent.
In the case
, the left edge
of the essential spectrum is a resonance of the operator
(see Theorem 1).
3) Let for some
the relation
hold then the operator
has m nondegenerate eigenvalues outside the essential spectrum (see Theorem 2) and for small
:
The corresponding normalized eigenfunctions are of the forms:
where,
is the normalized eigenfunction of the operator
corresponding to the eigenvalue
defined by the formula (17).
In the case
, the left edge
of the essential spectrum is the eigenvalue of
(see Theorem 2) with the corresponding eigenfunction
Remark 1. If the potential
is even in all arguments
and the condition
holds, then the statements of Lemmas 3 - 4 remain valid.
Remark 2. If
, then the subspaces
are not invariant under the operator
.
Acknowledgements
This work was supported by the Grant OT-F4-66 of Fundamental Science Foundation of Uzbekistan.