Spectra of the Energy Operator of Two-Electron System in the Impurity Hubbard Model ()
1. Introduction
In the early 1970s, three papers [1] [2] [3], where a simple model of a metal was proposed that has become a fundamental model in the theory of strongly correlated electron systems, appeared almost simultaneously and independently. In that model, a single nondegenerate electron band with a local Coulomb interaction is considered. The model Hamiltonian contains only two parameters: the parameter B of electron hopping from a lattice site to a neighboring site and the parameter U of the on-site Coulomb repulsion of two electrons. In the secondary quantization representation, the Hamiltonian can be written as
where
and
denote Fermi operators of creation and annihilation of an electron with spin
on a site m and the summation over
means summation over the nearest neighbors on the lattice.
Recall that the local form of Coulomb interaction was first introduced for an impurity model in a metal by Anderson [4].
The Hubbard model is currently one of the most extensively studied multielectron models of metals [5] [6]. But little is known about exact results for the spectrum and wave functions of the crystal described by the Hubbard model and impurity Hubbard model, and obtaining the corresponding statements is therefore of great interest.
In the work [7] is considered dominant correlation effects in two-electron atoms.
The spectrum and wave functions of the system of three electrons in a crystal described by the Hubbard Hamiltonian were studied in [8]. Correspondingly, the spectrum of the energy operator of system of four electrons for a crystal described by the Hubbard Hamiltonian in the triplet state was studied in [9]. For the four-electron systems are exists quintet state, and three type triplet states, and two type singlet states. The spectrum of the energy operator of four-electron systems in the Hubbard model in the quintet, and singlet states were studied in [10].
The use of films in various areas of physics and technology arouses great interest in studying a localized impurity state (LIS) of magnet. Therefore, it is important to study the spectral properties of electron systems in the impurity Hubbard model. The spectrum of the energy operator of three-electron systems in the impurity Hubbard model in the second doublet state was studied [11].
The spectrum and wave functions of the system of three electrons in a crystal described by the Impurity Hubbard Hamiltonian were studied in [12] in the quartet state of the system.
Naturally, we have analogous problem in the case of two-electron systems in the impurity Hubbard model. Here there are exist two states: triplet and singlet states. The investigation of the spectrum of Hamiltonian for this model in the case triplet state for three-dimensional lattice was given by Yu. Kh. Ishkobilov [13]. In the same time in this paper there are no exact values of Hamiltonian parameters for which exists the eigenvalues of corresponding operator.
In this paper we give a full description of the structure of the essential spectra and discrete spectrum of two-electron systems in the impurity Hubbard model for triplet and singlet states. The main result of this paper is Theorems 11 and 12, which describe the spectrum of considered model for singlet state. The results of sections 3 and 4 and 5 there are preliminary facts for the proof of Theorems 11 and 12, also Theorem 15 and 16.
2. Preliminaries
The Hamiltonian of two-electron systems in the impurity Hubbard model has the following form
(1)
Here A (A0) is the electron energy at a regular (impurity) lattice site; B (B0) is the transfer integral between electrons (between electron and impurity) in a neighboring sites (we assume that
,
),
, where
are unit mutually orthogonal vectors, which means that summation is taken over the nearest neighbors, U (U0) is the parameter of the on-site Coulomb interaction of two electrons, correspondingly in the regular (impurity) lattice site;
is the spin index,
or
,
and
denote the spin
values
and
, and
and
are the respective electron creation and annihilation operators at a site
, where
is a
-dimensional integer lattice.
It is known that the Hamiltonian H acts in the antisymmetric complex Foc’k space
. Suppose that
is the vacuum vector in the space
. The triplet state corresponds to the free motion of two electrons over the lattice with the basis functions
. The linear subspace
, corresponding to the triplet state is the set of all vectors of the form
,
, where
is the subspace of antisymmetric functions in the Hilbert space
.
Using the standard anticommutation relations between electron creation and annihilation operators at lattice sites,
,
, and also take into account that
, (
is the zero element) we get the following
Theorem 1. The subspace
is invariant with respect to the action of operator H, and the restriction
of operator H to the subspace
is a bounded self-adjoint operator. The operator
acts on a vector
as
(2)
where
is a bounded self-adjoint operator acting in the Hilbert space
as
(3)
(here
is the Kronecker symbol).
We need the following Lemma on the coincidence of the spectra of operators
and
.
Lemma 2. The spectra of the operators
and
coincide.
Proof. Let
. Since
is the bounded self-adjoint operators, it follows by the Weyl criterion (see, for example, Ch. VII, section 14 in [14]) that there exists a sequence of vectors
such that
,
, and
(4)
On the other hand, setting
, we have
and
Hence, by (4), we get that
as
. Using again the Weyl criterion we have
. Therefore,
.
Conversely, let
. By the Weyl criterion, there exists a sequence
such that
and
(5)
Setting
, we have
and by (5)
Hence by the Weyl criterion, we obtain that
. Therefore,
.
Bellow we will call the operator
as the operator two-electron triplet state operator.
Let
, be the Fourier transform, where
is the
-dimensional torus endowed with the normalized Lebesgue measure
, that is,
. Setting
we get that the operator
acts in the Hilbert space
, where
is the linear subspace of antisymmetric functions in
.
Using the equality (3) and properties of the Fourier transform we have the following
Theorem 3. The operator
acting in the space
as
(6)
where
,
, and
.
It is clear that spectral properties of energy operator of two-electron systems in the impurity Hubbard model in the triplet state are closely related to the spectral properties of its one-electron subsystems in the impurity Hubbard model. First we investigate the spectrum of one-electron subsystems.
3. Spectra of the Energy Operator of One-Electron System in the Impurity Hubbard Model
The Hamiltonian H of one-electron systems in the impurity Hubbard model also has the form (1). We let
denote the Hilbert space spanned by the vectors in the form
. It is called the space of one-electron states of the operator H. The space
is invariant with respect to action of operator H. Denote by
the restriction of H to the subspace
.
As in the proof of Theorem 1, using the standard anticommutation relations between electron creation and annihilation operators at lattice sites, we get the following
Theorem 4. The subspace
is invariant with respect to the action of the operator H, and the restriction H1 is a linear bounded self-adjoint operator, acting in
as
(7)
where
is a linear bounded self-adjoint operator acting in the space
as
(8)
Lemma 5. The spectra of the operators
and
coincide.
The proof of Lemma 5 is the same as the proof of the Lemma 2.
As in section 2 denote by
the Fourier transform. Setting
we get that the operator
acts in the Hilbert space
. Using the equality (8) and properties of the Fourier transform we have the following
Theorem 6 The operator
acting in the space
as
(9)
It is clear that the continuous spectrum of operator
is independent of the numbers
and
, and is equal to segment
, where
,
(here
).
To find the eigenvalues and eigenfunctions of operator
we rewrite (9) in following form:
(10)
where
.
Suppose first that
and denote
,
,
. From (10) it follows that
(11)
Now substitute (11) in expressing of a and b we get the following system of two linear homogeneous algebraic equations:
This system has a nontrivial solution if and only if the determinant
of this system is equal to zero, where
Therefore, it is true the following
Lemma 7. If a real number
then z is an eigenvalue of the operator
if and only if
.
The following Theorem describe of the exchange of the spectrum of operator
in the case
.
Theorem 8. Let
. Then
A) If
and
(respectively,
and
), then the operator
has a unique eigenvalue
, lying the below (respectively, above) of the continuous spectrum of operator
.
B) If
and
or
(respectively,
and
or
), then the operator
has a unique eigenvalue
(respectively,
), lying the below (respectively, above) of the continuous spectrum of operator
.
C) If
and
or
and
, then the operator
has a two eigenvalues
and
, where
, lying the below and above of the continuous spectrum of operator
.
D) If
(respectively,
), then the operator
has a unique eigenvalue
(respectively,
), where
, lying the above (respectively, below) of the continuous spectrum of operator
.
E) If
and
(respectively,
and
), then the operator
has a unique eigenvalue
, where
, and the real number
, lying the above of the continuous spectrum of operator
.
F) If
and
(respectively,
and
), then the operator
has a unique eigenvalue
, where
, and the real number
, lying the below of the continuous spectrum of operator
.
K) If
and
(respectively,
and
), then the operator
has a exactly two eigenvalues
and
, where
, and the real number
, lying correspondingly, the below and above of the continuous spectrum of operator
.
M) If
and
(respectively,
and
), then the operator
has a exactly two eigenvalues
and
, where
, and the real number
, lying correspondingly the below and above of the continuous spectrum of operator
.
N) If
, then the operator
has no eigenvalues lying the outside of the continuous spectrum of operator
.
Proof. In the case
, the continuous spectrum of the operator
coincide with segment
. Expressing all integrals in the equation
through the integral
, we find that the equation
is equivalent to the equation
(12)
Moreover, the function
is a differentiable function on the set
, in addition,
,
. Thus the function
is an monotone increasing function on
and on
. Furthermore,
as
,
as
,
as
, and
as
.
If
then from (12) follows that
The function
has a point of asymptotic discontinuity
. Since
for all
it follows that the function
is an monotone increasing (decreasing) function on
and on
in the case
(respectively,
), in addition, and if
, or
, then
as
,
as
,
as
,
as
(respectively, if
, then
as
,
as
,
as
,
as
).
A) If
and
(respectively,
and
), then the equation for eigenvalues and eigenfunctions (12) has the form
(13)
It is clear, that
for the values
. Therefore,
, i.e.,
. If
(
), then this eigenvalue lying the below (the above) of the continuous spectrum of operator
.
B) If
and
or
(respectively,
and
or
), then the equation for the eigenvalues and eigenfunctions has the form
. It is clear, what the integral
calculated in a quadrature, of the below (above) of continuous spectrum of operator
, the integral
(
), consequently,
(
). The calculated the integral
, the below of the continuous spectrum of operator
, we have the equation of the form
This equation has a solution
, lying the below of the continuous spectrum of operator
. In the above of continuous spectrum of operator
, the equation take the form
This equation has a solution of the form
, lying the above of the continuous spectrum of operator
.
C) If
and
or
and
, then the equation for the eigenvalues and eigenfunctions take in the form
or
Denote
. Then
, or
. In the below of the continuous spectrum of the operator
, we have the equation of the form
This equation has a solution
. It is obviously, that
. This eigenvalue lying the below of the continuous spectrum of operator
. In the above of the continuous spectrum of operator
, the equation for the eigenvalues and eigenfunctions has the form
From here, we find
. This eigenvalue lying the above of the continuous spectrum of operator
.
D) If
, then the equation for eigenvalues and eigenfunctions has the form
, from this
(14)
We denote
. In the first we consider Equation (14) in the below of continuous spectrum of
. In Equation (14) we find the equation of the form
From this, we find
and
. Now we verify the conditions
. The inequality
, is incorrectly, and inequality
, also is incorrectly. We now consider Equation (14) in the above of continuous spectrum of operator
. We have
In this equation we find the solutions above of continuous spectrum of operator
. Now we verify the conditions
. The inequality
, is correctly, and inequality
, is incorrectly. Consequently, in this case the operator
has a unique eigenvalue
, lying the above of continuous spectrum of operator
.
Let
, then the equation of eigenvalues and eigenfunctions take in the form
, where
.
In the below of continuous spectrum of
, we have equation of the form
From here we find
and
. The appear inequalities
, is correct, and
, is incorrect. In the above of continuous spectrum of operator
, we have equation of the form
It follows that, what
and
. The inequality
and
, are incorrectly. Therefore, in this case the operator
has a unique eigenvalue
, lying the below of continuous spectrum of operator
.
E) If
and
(respectively,
and
), then consider necessary, that
, where
real number. Then the equation for eigenvalues and eigenfunctions has the form
. From this
. We denote
, then
. In the first we consider this equation in the below of the continuous spectrum of operator
. Then
. This equation has the solutions
and
. Now, we verify the condition
. The solution
no satisfy the condition
, but
satisfy the condition
. We now verify the conditions
. The appear, this inequality is incorrectly. The appear inequalities
is correct, and
, is incorrect. We now verify the conditions
. So far as,
, the appear, this inequality is correctly. Consequently, in this case, the operator
has a unique eigenvalue
, above of continuous spectrum of
.
F) If
, and
(respectively,
, and
), then we assume that
, where
real number. The equation for eigenvalues and eigenfunctions take in the form
. From here
The introduce notation
. Then
(15)
In the below of the continuous spectrum of operator
, we have the equation
, from here
; this equation take the form
. We find
and
. We now verify the conditions
. The appear, that
, is correctly and
, is incorrectly. Now we consider Equation (15) in the above of the continuous spectrum of operator
. Then
. From this
We find
and
. We verify the conditions
. The appear
, it is not true, and the
, is true. We now verify the conditions
. The appear, this inequality is incorrectly. Consequently, in this case, the operator
have unique eigenvalue
, i.e., lying the below of the continuous spectrum of operator
.
K) If
and
(respectively,
and
), then we assume that
, where
real number. The equation for eigenvalues and eigenfunctions take in the form
(16)
We denote
. Then the Equation (16) receive the form
In the below of the continuous spectrum of
we have equation of the form
This equation has a solutions
and
. The inequalities
and
, is implements. The inequalities
, is correctly, and the inequality
, is incorrectly. We now verify the conditions
, since
, this inequality is true. We now consider Equation (16) in the above of the continuous spectrum of the operator
. We have the equation of the form
This equation has a solutions
and
. The inequalities
and
is true, as
, that the inequality
is correctly. The inequalities
and
is incorrectly. Consequently, in this case the operator
has a exactly two eigenvalues
and
, lying the below and above of the continuous spectrum of the operator
.
M) If
(respectively,
and
), the we take
, where
real number. Then the equation for eigenvalues and eigenfunctions has the form
We denote
. Then Equation (16) receive the form
. In the below of the continuous spectrum of the operator
we have equation of the form
This equation has a solutions
and
. The inequalities
and
, is implements. The inequalities
, is correctly, and the inequality
, is correctly. We now verify the conditions
, since
, this inequality it is not true. We now consider Equation (16) in the above of the continuous spectrum of the operator
. We have the equation of the form
This equation has a solutions
and
. The inequalities
and
it is not true, as
, that the inequality
is incorrectly. The inequalities
, and
is correctly. Consequently, in this case the operator
has a exactly two eigenvalues
, and
, lying the below and above of the continuous spectrum of operator
.
N). If
, then
, and the function
is a decreasing function in the intervals
and
; by,
the function
, and by
, the function
, and by
,
, and by
,
. The function
, by
, and by
, the function
, and by
, the function
, by
, the function
. Therefore, the equation
, that’s impossible the solutions in the outside the continuous spectrum of operator
. Therefore, in this case, the operator
has no eigenvalues lying the outside of the continuous spectrum of the operator
.
Now we consider the two-dimensional case. In two-dimensional case, we have, what the equation
, is equivalent to the equation of the form
, where
. In this case, also
, as
, and
, as
, and
, as
, and
, as
. In one- and two-dimensional case the behavior of function
be similarly. Therefore, we have the analogously results, what is find the one-dimensional case.
We consider the three-dimensional case. In the first We consider the Watson integral [15]
.
Theorem 9. Let
. Then
A). 1) If
and
(respectively,
and
), then the operator
has a unique eigenvalue
, lying the below (respectively, above) of the continuous spectrum of operator
.
2) If
and
(respectively,
and
), then the operator
has no eigenvalue, lying the below (respectively, above) of the continuous spectrum of operator
.
B) If
or
and
,
(respectively,
and
, and
), then the operator
has a unique eigenvalue
(respectively,
), lying the below (respectively, above) of the continuous spectrum of operator
. If
and
, and
(respectively,
and
, and
), then the operator
has no eigenvalue the outside of the continuous spectrum of operator
.
C) If
and
(respectively,
and
), then the operator
has a unique eigenvalue z (
), where
, lying the below (above) of the continuous spectrum of operator
. If
and
,
(respectively,
and
,
), then the operator
has no eigenvalues the outside of the continuous spectrum of operator
.
D) If
and
(respectively,
, and
), then the operator
has a unique eigenvalue z, (respectively,
), lying the above (respectively, below) of the continuous spectrum of operator
.
E) If
and
and
(respectively,
and
and
), then the operator
has a unique eigenvalue z, lying the above of the continuous spectrum of operator
.
F) If
and
and
(respectively,
and
and
), then the operator
has a unique eigenvalue
, lying the below of the continuous spectrum of operator
.
K) If
and
and
(respectively,
and
and
), then the operator
has a exactly two eigenvalues
and
, lying the above and below of the continuous spectrum of operator
.
M) If
and
and
(respectively,
and
and
), then the operator
has a exactly two eigenvalues
and
, lying the above and below of the continuous spectrum of operator
.
N) If
, then the operator
has no eigenvalues lying the outside of the continuous spectrum of operator
.
Proof. In the case
, the continuous spectrum of the operator
coincide with segment
. Expressing all integrals in the equation
through the integral
, we find that the equation
is equivalent to the equation
.
Moreover, the function
is a differentiable function on the set
, in addition,
.
In the three-dimensional case, the integral
have the finite value. Expressing these integral via Watson integral W, and taking into account, what the measure is normalized, we have, that
. Thus the function
is an monotone increasing function on
and on
. Furthermore, in the three-dimensional case
at
, and
as
, and
as
, and
as
.
If
then from (12) follows that
The function
has a point of asymptotic discontinuity
. Since
for all
it follows that the function
is an monotone increasing (decreasing) function on
and on
in the case
(respectively,
), in addition, and if
, or
, then
as
,
as
,
as
,
as
(respectively, if
, then
as
,
as
,
as
,
as
).
A) If
and
(respectively,
and
), then the equation for eigenvalues and eigenfunctions (12) has the form:
. It is clear, that
for the values
. Therefore,
, i.e.,
. If
, then this eigenvalue lying the below of the continuous spectrum of operator
, if
, then this eigenvalue lying the above of the continuous spectrum of operator
. If
(respectively,
,) then this eigenvalue not lying in the outside of the continuous spectrum of operator
.
B) If
or
and
(respectively,
or
and
), then the equation for the eigenvalues and eigenfunctions has the form
, that is,
. The equation
in the below (respectively, above) of continuous spectrum of operator
have the solution, one should implements the inequality
(respectively,
), i.e.,
,
(respectively,
,
). If
(respectively,
), then the operator
has no eigenvalues the outside the continuous spectrum of operator
.
C) If
and
(respectively,
and
), then the equation for the eigenvalues and eigenfunctions take in the form
, or
. Denote
. Then
, or
. The equation
in the below (respectively, above) of continuous spectrum of operator
have the solution, one should implements the inequality
(respectively,
), i.e.,
. If
and
,
(respectively,
and
,
), then the operator
has no eigenvalues the outside the continuous spectrum of
.
D) If
, then the equation for eigenvalues and eigenfunctions has the form
, from this we have equation in the form (14):
. We denote
. In the first we consider Equation (14) in the below of continuous spectrum of operator
. In the below of continuous spectrum of operator
, the function
, as
,
, as
. Therefore, the below of continuous spectrum of operator
, the equation
has a unique solution, if
, i.e.,
. This inequality incorrectly. Therefore, the below of continuous spectrum of operator
, this equation has no solution.
We now consider the equation for eigenvalues and eigenfunctions
, in the above of continuous spectrum of operator
. In the above of continuous spectrum of operator
, the function
, as
,
, as
. Therefore, the above of continuous spectrum of operator
, the equation
has a unique solution, if
, i.e.,
. This inequality correctly. Therefore, the above of continuous spectrum of operator
, this equation has a unique solution z.
If
, then the equation for eigenvalues and eigenfunctions has the form
, from this we have the equation in the form (14):
. We denote
. In the first we consider Equation (14) in the below of continuous spectrum of operator
. In the below of continuous spectrum of operator
, the function
, as
,
, as
. Therefore, the below of continuous spectrum of operator
, the equation
has a unique solution, if
, i.e.,
. This inequality correctly. Therefore, the below of continuous spectrum of operator
, this equation has a unique solution.
We now consider the equation for eigenvalues and eigenfunctions
, in the above of continuous spectrum of operator
. In the above of continuous spectrum of operator
, the function
, as
,
, as
. Therefore, the above of continuous spectrum of operator
, the equation
has a unique solution, if
, i.e.,
. This inequality incorrectly. Therefore, the above of continuous spectrum of operator
, this equation has no solution.
E) If
and
(respectively,
and
), then consider necessary, that
, where
real number. Then the equation for eigenvalues and eigenfunctions has the form
. From this
. We denote
, then
. In the first we consider this equation in the below of the continuous spectrum of operator
. Then
, as
,
, as
,
, as
, and
, as
. The equation
have a unique solution, if
. From here
. This inequality is incorrect. Therefore, the below of continuous spectrum of operator
, the operator
has no eigenvalues.
The above of continuous spectrum of operator
, we have the
, if
,
, if
. Besides,
, as
,
, if
.
The equation
have a unique solution, if
. From here
. This inequality is correctly. Therefore, the above of continuous spectrum of operator
, the operator
has a unique eigenvalues
.
F) If
, and
(respectively,
, and
), then we assume that
, where
real number. The equation for eigenvalues and eigenfunctions take in the form
. From here
The introduce notation
. Then we have the equation in the form (15):
. In the below of the continuous spectrum of operator
, we have the equation
. In the below of continuous spectrum of operator
,
, as
,
, as
.
The equation
have a unique solution, if
. From here
. This inequality is correctly. Therefore, the below of continuous spectrum of operator
, the operator
has a unique eigenvalues.
In the above of continuous spectrum of operator
,
, as
,
, as
. Therefore, the above of continuous spectrum of operator
, the operator
has a unique eigenvalues, if
. From here
, what is incorrectly. Therefore, the above of continuous spectrum of operator
, the operator
has no eigenvalues.
K) If
and
(respectively,
and
), the we take
, where
positive real number. Then the equation for eigenvalues and eigenfunctions has the form (16):
,
.
We denote
. Then Equation (16) receive the form
In the below of continuous spectrum of operator
, we have
, as
, and
, as
. The equation
have a unique solution the below of continuous spectrum of operator
, if
. From here
. This inequality is correctly. Therefore, the below of continuous spectrum of operator
, the operator
has a unique eigenvalues
.
The above of continuous spectrum of operator
, we have
, as
, and
, as
. The equation
have a unique solution the above of operator
, if
, i.e.,
. This inequality is correctly.
Consequently, in this case the operator
have two eigenvalues
and
, lying the below and above of continuous spectrum of operator
.
M) If
and
(respectively,
and
), the we take
, where
positive real number. Then the equation for eigenvalues and eigenfunctions has the form (16):
,
.
We denote
. Then the Equation (16) receive the form
In the below of continuous spectrum of operator
, we have
, as
, and
, as
. The equation
have a unique solution the below of continuous spectrum of operator
, if
. From here
. This inequality is correctly. Therefore, the below of continuous spectrum of operator
, the operator
has a unique eigenvalues
.
The above of continuous spectrum of operator
, we have
, as
, and
, as
. The equation
have a unique solution the above of continuous spectrum of operator
, if
, i.e.,
. This inequality is correctly.
Consequently, in this case the operator
have two eigenvalues
and
, lying the below and above of continuous spectrum of operator
.
N) If
, then
, and the function
is a decreasing function in the intervals
and
; by,
the function
, and by
, the function
, and by
,
, and by
,
. The function
, by
, and by
, the function
, and by
, the function
, by
, the function
. Therefore, the equation
, that’s impossible the solutions in the outside the continuous spectrum of operator
. Therefore, in this case, the operator
has no eigenvalues lying the outside of the continuous spectrum of the operator
.
From obtaining results is obviously, that the spectrum of operator
is consists from continuous spectrum and no more than two eigenvalues. In turn the operator
is represented in the form
(17)
where I is the unit operator in the space
The spectrum of the operator
, where A and B are densely defined bounded linear operators, was studied in [16] [17] [18]. In this work explicit formulas were given there that express the essential spectrum
of
and the discrete spectrum
in terms of the spectrum
of A and the discrete spectrum
of A and in terms of the spectrum
of B and the discrete spectrum
of B.
4. Structure of the Essential Spectrum and Discrete Spectrum of Operator
Now, using the obtained results (Theorem 8) and representation (17), we describe the structure of the essential spectrum and the discrete spectrum of the operator
.
Theorem 10. Let
. Then
A) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where
, lying the below (above) of the essential spectrum of operator
.
B) If
or
and
(respectively,
or
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where
(respectively,
), lying the below (above) of the essential spectrum of operator
.
C) If
and
or
and
, then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of a two point:
, where
, and
, and
, in this case the eigenvalue
, lying the essential spectrum of operator
.
D) If
(respectively,
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where
(respectively,
), and
, lying the above (below) of the essential spectrum of operator
.
E) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where
and
, and real number
, lying the above of the essential spectrum of operator
.
F) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where
and
, and the real number
, lying the below of the essential spectrum of operator
.
K) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of a three points:
, where
and
, and
, and the real number
, lying the outside the essential spectrum of operator
.
M) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of a three points:
, where
and
, and
, and the real number
, lying the outside the essential spectrum of
.
N) If
, then the the essential spectrum of the operator
is consists is a single segment:
, and discrete spectrum of the operator
is empty set:
.
Proof. 1) It follows from representation (17), and from Theorem 8, that in one-dimensional case, the continuous spectrum of the operator
is consists from
, and discrete spectrum of the operator
is consists of unique eigenvalue
. Therefore, the essential spectrum of the operator
is consists of the union of two segments
and
, and discrete spectrum of the operator
consists is the unique point 2z. These is given to the proof of statement A) from Theorem 10.
The statements B), C), D), E), F), K) from Theorem 10 are proved similarly.
We now is proved the statement M) from Theorem 10. It can be seen from Theorem 8 (statement M) in one-dimensional case the operator
has exactly two eigenvalues
and
outside the domain of the continuous spectrum of
operator
. Therefore,
. The numbers
,
and
, are eigenvalues of operator
. These is given to the proof of statement M).
We now is proved the statement N) from Theorem 10. It can be seen from Theorem 8 (statement N) in one-dimensional case the operator
has no eigenvalues the outside the continuous spectrum of operator
. Therefore,
, and the operator also has no eigenvalues, i.e.,
.
The next theorems described the structure of essential spectrum of the operator
in a three-dimensional case.
Theorem 11. Let
. Then
A) 1) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where
.
2) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of a single segment:
, and discrete spectrum of the operator
is empty set.
B) If
or
and
,
(respectively,
or
and
,
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where z are the eigenvalue of operator
.
If
or
and
, and
(respectively,
or
and
, and
), then the essential spectrum of the operator
is consists of a single segment:
, and discrete spectrum of the operator
is empty set:
.
C) If
and
,
(respectively,
and
,
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where z is the eigenvalue of operator
, and
. If
and
,
, (respectively,
and
,
), then the essential spectrum of the operator
is consists of a single segment:
, and discrete spectrum of the operator
is empty set:
.
D) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where z is the eigenvalue of operator
.
E) If
and
and
(respectively,
and
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where z is the eigenvalue of
.
F) If
and
and
(respectively,
and
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of a single point:
, where
is the eigenvalue of
.
K) If
and
, and
(respectively,
and
, and
), then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of
is consists of a three point:
, where
and
, are the eigenvalues of
.
M) If
and
and
(respectively,
and
, and
), then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of a three point:
, where
and
, are the eigenvalues of operator
.
N) If
, then the essential spectrum of the operator
is consists of a single segment
, and discrete spectrum of the operator
is empty set:
.
Proof. The proof of Theorems 11 is similar to the proof of Theorems 10.
5. Structure of Essential Spectrum and Discrete Spectrum of the Operator
The singlet state corresponds two-electron bound states (or antibound states) to the basis functions:
. The subspace
, corresponding to the singlet state is the set of all vectors of the form
,
, where
is the subspace of symmetric functions in
.
In this case, the Hamiltonian H acts in the symmetric Fock space
. Let
be the vacuum vector in the symmetric Fock space
. The singlet state corresponds the free motions of two-electrons in the lattice and their interactions.
Theorem 12. The subspace
is invariant under the operator H, and the restriction
of operator H to the subspace
is a bounded self-adjoint operator. It generates a bounded self-adjoint operator
, acting in the space
as
(18)
where
Kronecker symbol. The operator
acts on a vector
as
(19)
Proof. The proof follows from describe of the acts HamiltonianH on vectors
, using the standard anticommutation relations between electron creation and annihilation operators at lattice sites:
,
, as well as the property
, where
is the zero element of
.
Lemma 13. The spectra of the operators
and
coincide.
Proof. The proof of Lemma 13 is similar to the proof of Lemma 2.
We call the operator
the two-electron singlet state operator.
We let
denote the Fourier transformation:
, where
is the
-dimensional torus endowed with the normalized Lebesgue measure
,
.
We set
. In the quasimomentum representation, the operator
acts in the Hilbert space
as
(20)
where
is the subspace of symmetric functions in
.
In turn, the operator
can be represented in the form
(21)
where
.
For the fixed value of total quasi-momentum of the two-electron system
the operator K is the finite-rank operator, i.e., the finite-dimensional operator. The rank of the operator K is equal to 2. Therefore, the essential spectrum of operators
and
coincide (chapter XIII, paragraph 4, in [19]).
We now, using the obtained results and representation (17) and (21), we describe the structure of essential spectrum and discrete spectrum of the operator
.
From the beginning, we consider the operator
.
Since, the family of the operators
are the family of bounded operators, that the
are the family of bounded operator valued analytical functions.
Therefore, in these family, one can the apply the Kato-Rellix theorem.
Theorem 14. (Kato-Rellix theorem) [19].
Let
is the analytical family in the terms of Kato. Let
is a nondegenerate eigenvalue of
. Then as
, near to
, the exist exactly one point
the near
and this point is isolated and nondegenerated.
is an analytical function of
as
, the near to
, and exist the analytical eigenvector
as
the near to
. If the as real
the operator
is a self-adjoint operator, then
can selected thus, that it will be normalized of real
.
Since, the operator
has a nondegenerate eigenvalue, such as, the near of eigenvalue
of the operator
, the operator
as U, near
, has a exactly one eigenvalue
the near
and this point is isolated and nondegenerated. The
is a analytical function of U as U, the near to
.
As the large values the existence no more one additional eigenvalue of the operator
is following from the same, what the perturbation
is the one-dimensional operator.
A new we consider the family of operators
.
As, the operator
has a nondegenerate eigenvalue, consequently, the near of eigenvalue
the operator
, operator
as
, the near of
, has a exactly one eigenvalue
the near
and this point is the isolated and nondegenerated. The
is a analytical function of
, as
, the near to
.
Later on via
, and
we denote the additional eigenvalues of operator
. Thus, we prove the next theorems, the described the spectra of operator
.
Theorem 15. Let
. Then
A) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and the discrete spectrum of the operator
is consists no more than three points
, where
,
and
are the additional eigenvalues of the operator
.
B) If
or
and
(respectively,
or
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists no more than three points:
, where
(respectively,
).
C) If
and
or
and
, then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of no more than four points:
, where
, and
, and
.
D) If
(respectively,
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of no more than three points:
, where
(respectively,
), and
.
E) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of no more than three points:
, where
and
, and the real number
.
F) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of no more than three points:
, where
and
, and the real number
.
K) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of no more than five points:
, where
and
, and
, and the real number
.
M) If
and
(respectively,
and
), then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of no more than five points:
, where
and
, and
, and the real number
.
N) If
, then the essential spectrum of the operator
is consists is a single segment
, and discrete spectrum of the operator
is consists of no more than two points:
.
Proof. A). From the representation (17), (23) and the formulas (18) and (19), and the Theorem 8, follow the in one-dimensional case, the continuous spectrum of the operator
is consists
, and the discrete spectrum of the operator
is consists of unique eigenvalue
. The operator K is a two-dimensional operator. Therefore, the essential spectrum of the operators
and
coincide (see. chapter XIII, paragraph 4, in [19]) and is consists from segments
, and
. Of extension the two-dimensional operator K to the operator
can appear no more then two additional eigenvalues
and
. These give the statement A) of the Theorem 15.
B) In this case the operator
has a one eigenvalue
, lying the outside of the continuous spectrum of operator
. Therefore, the essential spectrum of the operators
is consists of the union of two segments and discrete spectrum of the operator
is consists of single point. These give the statement B) of the Theorem 15. The other statements of the Theorem 15 the analogously is proved.
The next theorems is described the structure of essential spectrum of the operator
in the three-dimensional case.
Theorem 16. Let
. Then
A) 1) If
and
(respectively,
and
,) then the essential spectrum of the operator
is consists of the union of two segments
, and discrete spectrum of the operator
is consists of no more then three points:
, where
,
and
are the additional eigenvalues of the operator
.
2) If
and
(respectively,
and
,) then the essential spectrum of the operator
is consists of a single segment
, and discrete spectrum of the operator
is consists of no more then two points:
.
B) If
or
, and
,
, (respectively,
or
, and
,
), then the essential spectrum of the operator
is consists of the union of two segments
(respectively,
), and discrete spectrum of the operator
is consists of no more then three points:
(respectively,
), where
(respectively,
) are the eigenvalue of operator
.
If
(respectively,
), then the essential spectrum of the operator
is consists of a single segment
, and discrete spectrum of the operator
is consists of no more then two points:
.
C) If
and
,
(respectively,
and
,
), then the essential spectrum of the operator
is consists of the union of two segments:
(respectively,
), and discrete spectrum of the operator
is consists of no more then three points:
(respectively,
), where z (respectively,
), is the eigenvalue of operator
, and
. If
and
,
(respectively,
and
,
), then the essential spectrum of the operator
is consists of a single segment:
, and discrete spectrum of the operator
is consists no more then two points:
.
D) If
(respectively,
), then the essential spectrum of the operator
is consists of the union of two segments:
(respectively,
), and discrete spectrum of the operator
is consists of no more then three points:
(respectively,
), where z (respectively,
) is the eigenvalue of operator
.
E) If
and
and
(respectively,
and
and
), then the essential spectrum of the operator
is consists of the union of two segments
, and discrete spectrum of the operator
is consists of no more then three points:
, where
is the eigenvalue of operator
.
F) If
and
and
(respectively,
and
and
), then the essential spectrum of the operator
is consists of the union of two segments:
, and discrete spectrum of the operator
is consists of no more then three points:
, where
is the eigenvalue of operator
.
K) If
and
and
(respectively,
and
and
), then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of no more then five points:
, where
and
are the eigenvalues of operator
.
M) If
and
and
(respectively,
and
and
), then the essential spectrum of the operator
is consists of the union of three segments:
, and discrete spectrum of the operator
is consists of no more then five points:
, where
and
are the eigenvalues of operator
.
N) If
, then the essential spectrum of the operator
is consists of a single segment:
, and discrete spectrum of the operator
is consists of no more then two points:
.
Proof. A) 1) From the Theorem 9 is follows, that, if
and
and
(respectively,
and
), the operator
has a unique eigenvalue
, the outside the continuous spectrum of the operator
. Furthermore, the continuous spectrum of the operator
is consists of the segment
, therefore, the essential spectrum of the operator
is consists of a union of two segments:
. The number 2z is the eigenvalue for the operator
. In the representation (17) the operator K is a two-dimensional operator. Therefore, the operator
can have no more then two additional eigenvalues
and
. Consequently, the operator
can have no more then three eigenvalues 2z,
and
.
2) From the Theorem 9 is follows, that, if
and
and
(respectively,
and
), then the operator
has no eigenvalues, the outside the continuous spectrum of the operator
. Furthermore, the continuous spectrum of the operator
is consists of the segment
, therefore, the essential spectrum of the operator
is consists of a single segment:
. In the representation (17) the operator K is a two-dimensional operator. Therefore, the operator
can have no more then two additional eigenvalues
and
. Consequently, the operator
can have no more then two eigenvalues
and
.
M) From the Theorem 9 is follows, that, if
and
and
and
(respectively,
and
and
), the operator
has a exactly two eigenvalues
and
, lying the below and above of the continuous spectrum of the operator
. Furthermore, the continuous spectrum of the operator
is consists of the segment
, therefore, then the essential spectrum of the operator
is consists of the union of three segments:
, and point
,
and
, are the eigenvalues of the operator
, and in the representation (17) the operator K is a two-dimensional operator. Therefore, the operator
can have no more then two additional eigenvalues
and
. Consequently, the operator
can have no more then five eigenvalues
,
,
,
and
.
The other statements of the Theorem 16 the analogously is proved.