In the present paper we consider a operator matrix acting in the direct sum of zero-particle and one-particle subspaces of Fock space. It is shown that this operator has no more than one positive and no more than two negative simple eigenvalues.
Keywords: Operator matrix, Fock space, eigenvalue, annihilation and creation operators, polynom, essential and discrete spectrum.
Block operator matrices are matrices the entries of which are linear operators between Banach or Hilbert spaces. Such operators often arise in mathematical physics, e.g. in fluid mechanics (see ), magnetohydrodynamics (see ) and quantum mechanics (see ). Spectral properties of the block operator matrices are studied in detail in . One of the special class of block operator matrices are Hamiltonians associated with a system describing -particles in interaction without conservation of the number of particles. Here off-diagonal entries of such block operator matrices are annihilation and creation operators. The study of systems describing -particles in interaction without conservation of the number of particles is reduced to the study of the spectral properties of self-adjoint operators acting in the zero-particle, one-particle,…, -particle subspaces of a Fock space.
In the present paper we consider a block operator matrix acting in the direct sum of zero-particle and one-particle subspaces of a Fock space. We prove that this operator has no more than one positive and no more than two negative simple eigenvalues.
Let be the field of complex numbers and be the Hilbert space of square integrable (complex) functions on . Denote by the direct sum of spaces and , that is, . The spaces and are zero- and one-particle subspaces of a Fock space over , respectively.
Let us consider the following operator matrix acting in the Hilbert space as
where the entries are defined by
Here ; is a fixed read number, the functions and are real-valued continuous functions on and denotes the adjoint operator to .
Under this assumptions the operator is bounded and self-adjoint in .
We remark that the operators and are called annihilation and creation operators, respectively.
We denote by , and the spectrum, essential spectrum and discrete spectrum of a bounded self-adjoint operator.
Lemma 1. The relation holds.
Proof. Since the operator is a bounded self-adjoint operator whose rank does not exceed three, we have . We show that . To this end, we consider the equation for , which is equivalent to the system of equations
where is the scalar product in . It is easy to see that the elements of the subspace
are solutions of system of equations (1). Then the fact implies that . The lemma is proved.
By Lemma 1 the operator may have only positive and negative discrete eigenvalues. The following theorem describes the number and location of these eigenvalues.
Theorem 1. The operator has no more than one positive and no more than two negative simple eigenvalues.
Proof. Let us consider the equation or the system of equations
Since from the second equation of (2) we find
Substituting the expression (3) for into the first equation of the system of equations (2) and the equality (4) we have that the system of equations (2) has a solution if and only if
where is the norm in .
We note that, if and are linear dependent, then . Therefore,
By the inequality we obtain that
There are three cases are possible: 1) and are orthogonal; 2) and are parallel; 3) and are neither orthogonal and nor parallel.
Let and be orthogonal. Then
In this case the numbers
are zeroes of , i.e., the eigenvalues of .
We remark that the numbers are also zeroes of in the case where and are not orthogonal.
Let and be parallel. Then
In this case the polynomial can be written in the form
For convenience we assume that . From here it follows that the numbers
are zeroes of , i.e., the eigenvalues of . In the case where we have and .
We remark that the numbers are also zeroes of in the case where and are not parallel.
Let and be neither orthogonal and nor parallel. Then
Set and . Without loss of generality (otherwise we would be prove the following facts in the same way) we assume that the inequalities hold. Then it follows that . Since the numbers and are zeroes of and , respectively, we have
i.e. on the boundary of the polynomial has a different sign. Hence, there exists a point , such that and . Analogously one can prove that there exist the numbers and , which are zeroes of the polynomial .
Since is a polynomial of degree 3 these zeroes are simple.
One can see that . Theorem 1 is completely proved.
Notice that Theorem 1 plays important role in the study the number of eigenvalues corresponding generalized Friedrichs model.
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