# A Special Class of Infinite Dimensional Dirac Operators on the Abstract Boson-Fermion Fock Space

Authors: Asao Arai
Journal: Journal of Mathematics
 (2014)

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## Abstract

Spectral properties of a special class of infinite dimensional Dirac operatorsQ(α) on the abstract boson-fermion Fock space ℱ(ℋ,𝒦) associated with the pair (ℋ,𝒦) of complex Hilbert spaces are investigated, where α∈C is a perturbation parameter (a coupling constant in the context of physics) and the unperturbed operator Q(0) is taken to be a free infinite dimensional Dirac operator. A variety of the kernel of Q(α) is shown. It is proved that there are cases where, for all sufficiently large |α| with α<0, Q(α) has infinitely many nonzero eigenvalues even if Q(0) has no nonzero eigenvalues. Also Fredholm property of Q(α) restricted to a subspace of ℱ(ℋ,𝒦) is discussed.


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## Body

## 1. Introduction

In a previous paper [1] (cf. [2]), the author introduced a general class of infinite dimensional Dirac type operators QA on the abstract boson-fermion Fock space F(H,K) associated with the pair (H,K) of complex Hilbert spaces (for the definition of F(H,K) and QA, see Section 2), where A is a densely defined closed linear operator from H to K. The operator QA gives an infinite dimensional and abstract version of finite dimensional Dirac type operators. In applications to physics, QA unifies self-adjoint supercharges (generators of supersymmetry) of some supersymmetric quantum field models (e.g., [3–6]).In the paper [1], basic properties of QA are discussed. As for the spectral properties of QA, only the zero-eigenvalue of QA is considered. Moreover, a class of perturbations V for QA is introduced and, under a suitable condition, a path (functional) integral representation for the index of the perturbed operator QA(V)∶=QA+V restricted to a subspace of F(H,K) is established. The (essential) self-adjointness of QA(V) is partially discussed in [7]. Analysis of other properties of QA(V) including spectral ones except for the zero-eigenvalue has been left open.In this paper, we undertake a comprehensive operator-theoretical analysis forQA(V). But, as a preliminary, we consider, in the present paper, only the case where V is a simple form and see what kind of phenomena occurs under the perturbation V.The outline of the present paper is as follows. In Section2 we review briefly some contents in [1]. In Section 3 we identify the spectra of QA. This is done via spectral analysis of a second quantization operator on the boson-fermion Fock space F(H,K). In Section 4, we introduce a perturbation V=Vg,v(α) with g∈H∖{0} and v∈K∖{0}, where α∈C is a perturbation parameter (a coupling constant in the context of physics). The perturbed operator Q(α)∶=QA+Vg,v(α) is the main object of our analysis in the present paper. We prove the self-adjointness of Q(α) and the essential self-adjointness of it on a suitable dense subspace of F(H,K) with some other properties (Theorem 14). Also the spectra of Q(α) are identified (Theorem 16). Moreover, it is shown that the domain of Q(α) is equal to that of Q(0)=QA for all sufficiently small |α| (Theorem 17). Section 5 is devoted to analysis of the kernel of Q(α). We see that the kernel property of Q(α) may be sensitive to conditions for (A,A*,α). In Section 6 we consider Fredholm property of Q(α)+, the restriction of Q(α) to a subspace of F(H,K), in comparison with that of (QA)+=Q(0)+. We obtain some classification on Fredholm property of Q(α)+ (Theorem 26). An interesting phenomenon occurs in the following sense: for a constant α0≠0 (resp., β0≠0), Q(α0)+ is not semi-Fredholm even if (QA)+ is Fredholm (resp., semi-Fredholm). In the last section, we consider nonzero eigenvalues of Q(α). We show that, under a suitable condition, for all sufficiently large |α| with α<0, Q(α) has infinitely many nonzero eigenvalues even if QA has no nonzero eigenvalues (Theorem 27). This structure also is interesting, which may be regarded as a strong coupling effect.


## 2. Brief Review on the Theory of Dirac Operators on the Abstract Boson-Fermion Fock Space

### 2.1. Definitions

For a linear operatorL on a Hilbert space, we use the following notation:(i)
D(L): the domain of L,(ii)
R(L): the range of L,(iii)
ker⁡⁡L∶={ψ∈D(L)∣Lψ=0}: the kernel of L,(iv)
ρ(L)⊂C: the resolvent set of L (λ∈ρ(L)⇔L is injective with R(L) being dense and the inverse operator (L-λ)-1 is bounded),(v)
σ(L)∶=C∖ρ(L): the spectrum of L,(vi)
σp(L): the point spectrum of L (the set of eigenvalues of L).For a complex Hilbert spaceH and a nonnegative integer n, we denote by ⊗snH (resp., ∧n(H)) the n-fold symmetric (resp., antisymmetric) tensor product Hilbert space of H with convention ⊗s0H:=C (the complex number field) and ∧0(H)∶=C. In terms of these Hilbert spaces, one can construct two types of larger Hilbert spaces. The one is theboson (symmetric) Fock space overH, denoted by Fb(H), which is defined to be the complete infinite direct sum of ⊗snH, n≥0:
(1)Fb(H)∶=⊕n=0∞⊗snH={ψ={ψ(n)}n=0∞∣ψ(n)∈⊗snH,n≥0,∑n=0∞‍∥ψ(n)∥2<∞},
where, for a complex Hilbert space X, we denote its inner product (resp., norm) by 〈·,·〉 (linear in the second variable) (resp., ∥·∥). The other is thefermion (antisymmetric) Fock space overH, denoted by Ff(H), which is defined to be the complete infinite direct sum of ∧n(H), n≥0:
(2)Ff(H)∶=⊕n=0∞∧n(H)={ψ={ψ(n)}n=0∞∣ψ(n)∈∧n(H),n≥0,∑n=0∞‍∥ψ(n)∥2<∞}.
For general theories of boson and fermion Fock spaces, see, for example, [8–10].LetK be a complex Hilbert space. Then theabstract boson-fermion Fock space associated with the pair (H,K) is defined as the tensor product Hilbert space
(3)F(H,K)∶=Fb(H)⊗Ff(K)
of Fb(H) and Ff(K).For eachf∈H, we denote by a(f) thebosonic annihilation operator on Fb(H) with test vector f; that is, a(f) is the densely defined closed linear operator acting in Fb(H) such that its adjoint a(f)* is given by the following form:
(4)(a(f)*ψ)(0)=0,(a(f)*ψ)(n)=nSn(f⊗ψ(n-1)),n≥1,ψ∈D(a(f)*),
where Sn is the symmetrization operator acting on the n-fold tensor product Hilbert space ⊗nH.Let(5)Ωb∶={1,0,0,…}∈Fb(H),
called thebosonic Fock vacuum. Then, for all f∈H,
(6)a(f)Ωb=0.For eachu∈K, there exists a unique linear operator b(u) on Ff(K), called thefermionic annihilation operator with test vector u, which has the following properties: (i) D(b(u))=Ff(K) and b(u) is bounded with operator norm ∥b(u)∥=∥u∥; (ii) its adjoint b(u)* is of the following form:
(7)(b(u)*ψ)(0)=0,(b(u)*ψ)(n)=nAn(u⊗ψ(n-1)),n≥1,ψ∈Ff(K),
where An is the antisymmetrization operator acting on the n-fold tensor product Hilbert space ⊗nH. The following canonical anticommutation relations (CAR) hold:
(8){b(u),b(v)*}=〈u,v〉,{b(u),b(v)}=0,u,v∈K,
where {X,Y}∶=XY+YX. In particular, one has
(9)b(u)2=0,u∈K.The vector(10)Ωf∶={1,0,0,…}∈Ff(K)
is called thefermionic Fock vacuum and obeys
(11)b(u)Ωf=0,u∈K.


### 2.2. Infinite Dimensional Exterior Differential Operators

For a subsetD of a Hilbert space, we denote by L(D) the subspace algebraically spanned by all the vectors of D.LetA be a densely defined closed linear operator from H to K and define
(12)DA∞∶=L({a(f1)*⋯a(fn)*Ωb⊗b(u1)*⋯b(up)*Ωf∣n,p≥0,fj∈C∞(A*A),j=1,…,n,uk∈C∞(AA*),k=1,…,p{a(f1)*⋯a(fn)*Ωb⊗b(u1)*}),
where C∞(L)∶=∩n=1∞D(Ln) for a linear operator L on a Hilbert space (the symbol X1⋯XnΩ with n=0 should read Ω). It follows that DA∞ is dense in F(H,K).The following proposition is proved in [1].Proposition 1.
There exists a unique densely defined closed linear operatordA on F(H,K) such that the following (i) and (ii) hold: (i)
DA∞⊂D(dA) and DA∞ is a core of dA.(ii)
For each vectorΨ∈DA∞ of the form
(13)Ψ=a(f1)*⋯a(fn)*Ωb⊗b(u1)*⋯b(up)*Ωf,
d
A
acts as(14)dAΨ=0forn=0dAΨ=∑j=1n‍a(f1)*⋯a(fj)*^⋯a(fn)*Ωb⊗b(Afj)*b(u1)*⋯b(up)*Ωfforn≥1,
wherea(fj)*^ indicates the omission of a(fj)*. In particular, dA leaves DA∞ invariant.
Moreover, the following (iii)–(v) hold:(iii)
DA∞⊂D(dA*) and
(15)dA*Ψ=∑k=1p‍(-1)k-1a(A*uk)*a(f1)*⋯a(fn)*Ωb⊗b(u1)*⋯b(uk)*^⋯b(up)*Ωf
for all vectors Ψ of the form (13) with p≥1. In the case p=0, we have dA*Ψ=0. In particular, dA* leaves DA∞ invariant.(iv)
D(dA2)=D(dA) and, for all Ψ∈D(dA), dA2Ψ=0.(v)
LetB be a bounded linear operator from H to K with D(B)=H. Then, for all Ψ∈DA∞ and α,β∈C,
(16)αdAΨ+βdBΨ=dαA+βBΨ.The operatordA is an abstract version of finite dimensional exterior differential operators, including an infinite dimensional version of them if H and K are infinite dimensional.


### 2.3. Infinite Dimensional Dirac Operators

Using the operatordA, we define afree Dirac operator on F(H,K) by
(17)QA∶=dA+dA*
with D(QA)=D(dA)∩D(dA*). We set
(18)ΔA∶=dA*dA+dAdA*,
aLaplace-Beltrami operator on F(H,K).For a self-adjoint operatorS on H, we define an operator dΓb(n)(S) on ⊗snH by
(19)dΓb(0)(S)∶=0,dΓb(n)(S)∶=∑j=1n‍I⊗⋯⊗I⊗S⌣jth⊗I⊗⋯⊗I¯,n≥1,
where I denotes identity and, for a closable linear operator C, C¯ denotes the closure of C. For each n≥0, dΓb(n)(S) is self-adjoint. The second quantization dΓb(S) of S on Fb(H) is defined by
(20)dΓb(S)∶=⊕n=0∞dΓb(n)(S),
which is self-adjoint (see, e.g., [9, Section 5.2], [11, p.302, Example 2]). It follows that
(21)0∈σp(dΓb(S)),Ωb∈ker⁡(dΓb(S)).Similarly, for a self-adjoint operatorT on K, we can define the second quantization dΓf(T) of T on Ff(K) by
(22)dΓf(T)∶=⊕p=0∞dΓf(p)(T),
where
(23)dΓf(0)(T)∶=0,dΓf(p)(T)∶=∑j=1p‍I⊗⋯⊗I⊗T⌣jth⊗I⊗⋯⊗I¯,p≥1,
acting in ∧p(K). Similarly to the case of dΓb(S), one has
(24)0∈σp(dΓf(T)),Ωf∈ker⁡(dΓf(T)).The number operator onFf(K) is defined by
(25)Nf∶=dΓf(I).
Then the operator
(26)Γ:=I⊗(-1)Nf
on F(H,K) is self-adjoint with Γ2=I. The spectrum of Γ is {±1} with eigenspaces
(27)F+(H,K)∶=⊕p=0∞Fb(H)⊗∧2p(K),F-(H,K)∶=⊕p=0∞Fb(H)⊗∧2p+1(K),
giving the orthogonal decomposition
(28)F(H,K)=F+(H,K)⊕F-(H,K).LetP+ and P- be the orthogonal projections onto F+(H,K) and F-(H,K), respectively. Then
(29)Γ=P+-P-.For notational simplicity we set(30)Hb(S)∶=dΓb(S),Hf(T)∶=dΓf(T).
We introduce an operator
(31)HBF(A)∶=Hb(A*A)⊗I+I⊗Hf(AA*)
acting in F(H,K), which is nonnegative and self-adjoint. It follows from (21) and (24) that
(32)0∈σp(HBF),Ωb⊗Ωf∈ker⁡(HBF(A)).Theorem 2 (see [1]).
(i) The operatorQA is self-adjoint and essentially self-adjoint on every core of HBF(A). In particular, QA is essentially self-adjoint on DA∞.
(ii) The operatorΓ leaves D(QA) invariant and
(33)ΓQA+QAΓ=0
onD(QA).
(iii) The following operator equalities hold:(34)ΔA=HBF(A)=QA2.Remark 3.
The quadruple{F(H,K),QA,HBF(A),Γ} is an abstract form of models of free supersymmetric quantum field theory, where QA, HBF(A), and Γ are a self-adjoint supercharge, a supersymmetric Hamiltonian, and a sign operator of states (a “fermion number operator”), respectively [1].For a linear operatorL, we set
(35)nulL∶=dim⁡ker⁡L,
the nullity of L.Theorem 4 (see [1, Lemma 3.4]).
(36)
ker
⁡
Q
A
=
ker
⁡
H
B
F
(
A
)
=
⊕
n
,
p
=
0
∞
[
(
⊗
s
n
ker
⁡
A
)
⊗
∧
p
(
ker
⁡
A
*
)
]
.
Hence the following hold.(i)
LetnulA=0 and nulA*<∞. Then nulQA=2nulA*<∞.(ii)
LetnulA=0 and nulA*=∞. Then nulQA=∞.(iii)
LetnulA≥1. Then nulQA=∞.


### 2.4. Fredholm Property and Index

LetQ be a self-adjoint operator on F(H,K) such that Γ leaves D(Q) invariant satisfying ΓQ+QΓ=0 on D(Q). This class of operators Q may be an abstract category of Dirac operators on the abstract boson-fermion Fock space F(H,K).Then it is easy to show that the operatorQ+ defined by
(37)D(Q+)∶=D(Q)∩F+(H,K),Q+Ψ∶=QΨ,Ψ∈D(Q+),
is a densely defined closed linear operator from F+(H,K) to F-(H,K). We define an index indΓQ of Q by
(38)indΓQ∶=nulQ+-nulQ+*,
provided that at least one of nulQ+ and nulQ+* is finite. Note that
(39)indΓQ=indQ+,
the index of Q+ (e.g., [12, Chapter IV, Section 5]).As for the Fredholm property ofQA, one has the following theorem.Theorem 5 (see [1, Proposition 3.2]).
(i) LetA be injective and Fredholm. Then (QA)+ is Fredholm with
(40)indΓQA=δ0,nulA*.
(ii) LetA be semi-Fredholm with nulA≥1 and let A* be injective. Then (QA)+ is semi-Fredholm with
(41)indΓQA=∞.
(iii) LetA be semi-Fredholm with nulA≥1 and nulA*≥1. Then (QA)+ is not semi-Fredholm.


## 2.1. Definitions

For a linear operatorL on a Hilbert space, we use the following notation:(i)
D(L): the domain of L,(ii)
R(L): the range of L,(iii)
ker⁡⁡L∶={ψ∈D(L)∣Lψ=0}: the kernel of L,(iv)
ρ(L)⊂C: the resolvent set of L (λ∈ρ(L)⇔L is injective with R(L) being dense and the inverse operator (L-λ)-1 is bounded),(v)
σ(L)∶=C∖ρ(L): the spectrum of L,(vi)
σp(L): the point spectrum of L (the set of eigenvalues of L).For a complex Hilbert spaceH and a nonnegative integer n, we denote by ⊗snH (resp., ∧n(H)) the n-fold symmetric (resp., antisymmetric) tensor product Hilbert space of H with convention ⊗s0H:=C (the complex number field) and ∧0(H)∶=C. In terms of these Hilbert spaces, one can construct two types of larger Hilbert spaces. The one is theboson (symmetric) Fock space overH, denoted by Fb(H), which is defined to be the complete infinite direct sum of ⊗snH, n≥0:
(1)Fb(H)∶=⊕n=0∞⊗snH={ψ={ψ(n)}n=0∞∣ψ(n)∈⊗snH,n≥0,∑n=0∞‍∥ψ(n)∥2<∞},
where, for a complex Hilbert space X, we denote its inner product (resp., norm) by 〈·,·〉 (linear in the second variable) (resp., ∥·∥). The other is thefermion (antisymmetric) Fock space overH, denoted by Ff(H), which is defined to be the complete infinite direct sum of ∧n(H), n≥0:
(2)Ff(H)∶=⊕n=0∞∧n(H)={ψ={ψ(n)}n=0∞∣ψ(n)∈∧n(H),n≥0,∑n=0∞‍∥ψ(n)∥2<∞}.
For general theories of boson and fermion Fock spaces, see, for example, [8–10].LetK be a complex Hilbert space. Then theabstract boson-fermion Fock space associated with the pair (H,K) is defined as the tensor product Hilbert space
(3)F(H,K)∶=Fb(H)⊗Ff(K)
of Fb(H) and Ff(K).For eachf∈H, we denote by a(f) thebosonic annihilation operator on Fb(H) with test vector f; that is, a(f) is the densely defined closed linear operator acting in Fb(H) such that its adjoint a(f)* is given by the following form:
(4)(a(f)*ψ)(0)=0,(a(f)*ψ)(n)=nSn(f⊗ψ(n-1)),n≥1,ψ∈D(a(f)*),
where Sn is the symmetrization operator acting on the n-fold tensor product Hilbert space ⊗nH.Let(5)Ωb∶={1,0,0,…}∈Fb(H),
called thebosonic Fock vacuum. Then, for all f∈H,
(6)a(f)Ωb=0.For eachu∈K, there exists a unique linear operator b(u) on Ff(K), called thefermionic annihilation operator with test vector u, which has the following properties: (i) D(b(u))=Ff(K) and b(u) is bounded with operator norm ∥b(u)∥=∥u∥; (ii) its adjoint b(u)* is of the following form:
(7)(b(u)*ψ)(0)=0,(b(u)*ψ)(n)=nAn(u⊗ψ(n-1)),n≥1,ψ∈Ff(K),
where An is the antisymmetrization operator acting on the n-fold tensor product Hilbert space ⊗nH. The following canonical anticommutation relations (CAR) hold:
(8){b(u),b(v)*}=〈u,v〉,{b(u),b(v)}=0,u,v∈K,
where {X,Y}∶=XY+YX. In particular, one has
(9)b(u)2=0,u∈K.The vector(10)Ωf∶={1,0,0,…}∈Ff(K)
is called thefermionic Fock vacuum and obeys
(11)b(u)Ωf=0,u∈K.


## 2.2. Infinite Dimensional Exterior Differential Operators

For a subsetD of a Hilbert space, we denote by L(D) the subspace algebraically spanned by all the vectors of D.LetA be a densely defined closed linear operator from H to K and define
(12)DA∞∶=L({a(f1)*⋯a(fn)*Ωb⊗b(u1)*⋯b(up)*Ωf∣n,p≥0,fj∈C∞(A*A),j=1,…,n,uk∈C∞(AA*),k=1,…,p{a(f1)*⋯a(fn)*Ωb⊗b(u1)*}),
where C∞(L)∶=∩n=1∞D(Ln) for a linear operator L on a Hilbert space (the symbol X1⋯XnΩ with n=0 should read Ω). It follows that DA∞ is dense in F(H,K).The following proposition is proved in [1].Proposition 1.
There exists a unique densely defined closed linear operatordA on F(H,K) such that the following (i) and (ii) hold: (i)
DA∞⊂D(dA) and DA∞ is a core of dA.(ii)
For each vectorΨ∈DA∞ of the form
(13)Ψ=a(f1)*⋯a(fn)*Ωb⊗b(u1)*⋯b(up)*Ωf,
d
A
acts as(14)dAΨ=0forn=0dAΨ=∑j=1n‍a(f1)*⋯a(fj)*^⋯a(fn)*Ωb⊗b(Afj)*b(u1)*⋯b(up)*Ωfforn≥1,
wherea(fj)*^ indicates the omission of a(fj)*. In particular, dA leaves DA∞ invariant.
Moreover, the following (iii)–(v) hold:(iii)
DA∞⊂D(dA*) and
(15)dA*Ψ=∑k=1p‍(-1)k-1a(A*uk)*a(f1)*⋯a(fn)*Ωb⊗b(u1)*⋯b(uk)*^⋯b(up)*Ωf
for all vectors Ψ of the form (13) with p≥1. In the case p=0, we have dA*Ψ=0. In particular, dA* leaves DA∞ invariant.(iv)
D(dA2)=D(dA) and, for all Ψ∈D(dA), dA2Ψ=0.(v)
LetB be a bounded linear operator from H to K with D(B)=H. Then, for all Ψ∈DA∞ and α,β∈C,
(16)αdAΨ+βdBΨ=dαA+βBΨ.The operatordA is an abstract version of finite dimensional exterior differential operators, including an infinite dimensional version of them if H and K are infinite dimensional.


## 2.3. Infinite Dimensional Dirac Operators

Using the operatordA, we define afree Dirac operator on F(H,K) by
(17)QA∶=dA+dA*
with D(QA)=D(dA)∩D(dA*). We set
(18)ΔA∶=dA*dA+dAdA*,
aLaplace-Beltrami operator on F(H,K).For a self-adjoint operatorS on H, we define an operator dΓb(n)(S) on ⊗snH by
(19)dΓb(0)(S)∶=0,dΓb(n)(S)∶=∑j=1n‍I⊗⋯⊗I⊗S⌣jth⊗I⊗⋯⊗I¯,n≥1,
where I denotes identity and, for a closable linear operator C, C¯ denotes the closure of C. For each n≥0, dΓb(n)(S) is self-adjoint. The second quantization dΓb(S) of S on Fb(H) is defined by
(20)dΓb(S)∶=⊕n=0∞dΓb(n)(S),
which is self-adjoint (see, e.g., [9, Section 5.2], [11, p.302, Example 2]). It follows that
(21)0∈σp(dΓb(S)),Ωb∈ker⁡(dΓb(S)).Similarly, for a self-adjoint operatorT on K, we can define the second quantization dΓf(T) of T on Ff(K) by
(22)dΓf(T)∶=⊕p=0∞dΓf(p)(T),
where
(23)dΓf(0)(T)∶=0,dΓf(p)(T)∶=∑j=1p‍I⊗⋯⊗I⊗T⌣jth⊗I⊗⋯⊗I¯,p≥1,
acting in ∧p(K). Similarly to the case of dΓb(S), one has
(24)0∈σp(dΓf(T)),Ωf∈ker⁡(dΓf(T)).The number operator onFf(K) is defined by
(25)Nf∶=dΓf(I).
Then the operator
(26)Γ:=I⊗(-1)Nf
on F(H,K) is self-adjoint with Γ2=I. The spectrum of Γ is {±1} with eigenspaces
(27)F+(H,K)∶=⊕p=0∞Fb(H)⊗∧2p(K),F-(H,K)∶=⊕p=0∞Fb(H)⊗∧2p+1(K),
giving the orthogonal decomposition
(28)F(H,K)=F+(H,K)⊕F-(H,K).LetP+ and P- be the orthogonal projections onto F+(H,K) and F-(H,K), respectively. Then
(29)Γ=P+-P-.For notational simplicity we set(30)Hb(S)∶=dΓb(S),Hf(T)∶=dΓf(T).
We introduce an operator
(31)HBF(A)∶=Hb(A*A)⊗I+I⊗Hf(AA*)
acting in F(H,K), which is nonnegative and self-adjoint. It follows from (21) and (24) that
(32)0∈σp(HBF),Ωb⊗Ωf∈ker⁡(HBF(A)).Theorem 2 (see [1]).
(i) The operatorQA is self-adjoint and essentially self-adjoint on every core of HBF(A). In particular, QA is essentially self-adjoint on DA∞.
(ii) The operatorΓ leaves D(QA) invariant and
(33)ΓQA+QAΓ=0
onD(QA).
(iii) The following operator equalities hold:(34)ΔA=HBF(A)=QA2.Remark 3.
The quadruple{F(H,K),QA,HBF(A),Γ} is an abstract form of models of free supersymmetric quantum field theory, where QA, HBF(A), and Γ are a self-adjoint supercharge, a supersymmetric Hamiltonian, and a sign operator of states (a “fermion number operator”), respectively [1].For a linear operatorL, we set
(35)nulL∶=dim⁡ker⁡L,
the nullity of L.Theorem 4 (see [1, Lemma 3.4]).
(36)
ker
⁡
Q
A
=
ker
⁡
H
B
F
(
A
)
=
⊕
n
,
p
=
0
∞
[
(
⊗
s
n
ker
⁡
A
)
⊗
∧
p
(
ker
⁡
A
*
)
]
.
Hence the following hold.(i)
LetnulA=0 and nulA*<∞. Then nulQA=2nulA*<∞.(ii)
LetnulA=0 and nulA*=∞. Then nulQA=∞.(iii)
LetnulA≥1. Then nulQA=∞.


## 2.4. Fredholm Property and Index

LetQ be a self-adjoint operator on F(H,K) such that Γ leaves D(Q) invariant satisfying ΓQ+QΓ=0 on D(Q). This class of operators Q may be an abstract category of Dirac operators on the abstract boson-fermion Fock space F(H,K).Then it is easy to show that the operatorQ+ defined by
(37)D(Q+)∶=D(Q)∩F+(H,K),Q+Ψ∶=QΨ,Ψ∈D(Q+),
is a densely defined closed linear operator from F+(H,K) to F-(H,K). We define an index indΓQ of Q by
(38)indΓQ∶=nulQ+-nulQ+*,
provided that at least one of nulQ+ and nulQ+* is finite. Note that
(39)indΓQ=indQ+,
the index of Q+ (e.g., [12, Chapter IV, Section 5]).As for the Fredholm property ofQA, one has the following theorem.Theorem 5 (see [1, Proposition 3.2]).
(i) LetA be injective and Fredholm. Then (QA)+ is Fredholm with
(40)indΓQA=δ0,nulA*.
(ii) LetA be semi-Fredholm with nulA≥1 and let A* be injective. Then (QA)+ is semi-Fredholm with
(41)indΓQA=∞.
(iii) LetA be semi-Fredholm with nulA≥1 and nulA*≥1. Then (QA)+ is not semi-Fredholm.


## 3. Spectra ofHBF(A) and QA

Now we go into analyzing new aspects in the operator theory ofQA and its perturbations. In the previous paper [1], the identification of the spectra of HBF(A) and QA was not discussed except for zero-eigenvalues. This section is devoted to identifying completely the spectra of HBF(A) and QA.In what follows, we assume thatH and K are separable.We say that a subsetJ of R issymmetric with respect to the origin if λ∈J implies that -λ∈J.The main results in this section are as follows.Theorem 6.
(42)
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The spectrumσ(QA) and the point spectrum σp(QA) of QA are symmetric with respect to the origin and(44)σ(QA)={0}∪(⋃n=1∞‍{±∑j=1n‍λj∣λj∈σ(A*A)∖{0},j=1,…,n}¯),σp(QA)={0}∪(⋃n=1∞‍{±∑j=1n‍λj∣λj∈σp(A*A)∖{0},j=1,…,n})with
(45)nul(QA-λ)=nul(QA+λ),λ∈σp(QA).To prove these theorems, we need a series of lemmas.Lemma 8.
LetS be a self-adjoint operator on a Hilbert space. (i)
Suppose thatσ(S) is symmetric with respect to the origin. Then
(46)σ(S)={±λ∣λ∈σ(S2)}.(ii)
Suppose thatσp(S) is symmetric with respect to the origin. Then
(47)σp(S)={±λ∣λ∈σp(S2)}.
Moreover, for each f∈ker⁡(S2-λ) with λ∈σp(S2), (S±λ)f∈ker⁡(S∓λ).Proof.
(i) LetΣS be the set on the right hand side of (46). Then, by the spectral mapping theorem, we have σ(S)⊂ΣS. Conversely, let μ∈ΣS, so that μ2∈σ(S2). Hence there exists a sequence {fn}n=1∞⊂D(S2) such that ∥fn∥=1 and (S2-μ2)fn→0(n→∞). Suppose that μ∉σ(S). Then S-μ is bijective with (S-μ)-1 bounded. Hence (S-μ)-1(S2-μ2)fn→0(n→∞), which means that (S+μ)fn→(n→∞). Hence -μ∈σ(S). Then, by the symmetry of σ(S), μ is in σ(S). But this is a contradiction. Hence μ∈σ(S). Thus ΣS⊂σ(S).
(ii) LetΣS,p be the set on the right hand side of (47). Then it is obvious that σp(S)⊂ΣS,p. Conversely, let μ∈ΣS,p, so that μ2∈σp(S2). Hence there exists a vector f∈D(S2) such that f≠0 and S2f=μ2f. The second equation is written as (S-μ)(S+μ)f=0. If ϕ∶=(S+μ)f=0, then -μ∈σp(S). Hence, by the symmetry of σp(S), μ is in σp(S). If ϕ≠0, then (S-μ)ϕ=0. Hence μ∈σp(S). Thus ΣS,p⊂σp(S). These arguments also imply the last statement.A sufficient condition for the spectra of a self-adjoint operator to be symmetric with respect to the origin is given in the following lemma.Lemma 9.
LetS be a self-adjoint operator on a Hilbert space X. Suppose that there exists a bounded self-adjoint operator τ on X such that τ2=I and τ leaves D(S) invariant satisfying
(48)τS+Sτ=0onD(S).
Then σ(S) and σp(S) are symmetric with respect to the origin and, for all λ∈σp(S),
(49)nul(S-λ)=nul(S+λ).Proof.
Note thatτ is unitary with τ-1=τ. Equation (48) implies the operator equality S=τ(-S)τ-1. Hence S is unitarily equivalent to -S. Thus the desired results follow.Lemma 10.
LetHn  (n=0,1,2,…) be a complex Hilbert space and An be a self-adjoint operator on Hn. Let ⊕n=0∞An be the direct sum of An’s acting in the direct sum Hilbert space ⊕n=0∞Hn. Then, ⊕n=0∞An is self-adjoint and
(50)σ(⊕n=0∞An)=∪n=0∞σ(An)¯,σp(⊕n=0∞An)=∪n=0∞σp(An).Proof.
See, for example, [8, Proposition 3.5-(ii), Theorem 3.7-(i)].Lemma 11.
LetS be a self-adjoint operator on H. Then
(51)σ(Hb(S))={0}⋃‍(⋃n=1∞‍{∑j=1n‍λj∣λj∈σ(S),j=1,…,n}¯),(52)σp(Hb(S))={0}⋃‍(⋃n=1∞‍{∑j=1n‍λj∣λj∈σp(S),j=1,…,n}).Proof.
The following are well known (e.g., [8, Proposition 4.12]): σ(dΓb(0)(S))={0}=σp(dΓb(0)(S)) and, for all n≥1,
(53)σ(dΓb(n)(S))={∑j=1n‍λj∣λj∈σ(S),j=1,…,n}¯,σp(dΓb(n)(S))={∑j=1n‍λj∣λj∈σp(S),j=1,…,n}.
By these facts and Lemma 10, we obtain (51) and (52).For a self-adjoint operatorT on a Hilbert space K and p=1,2,…, we define
(54)Σ(p)(T)∶={∑j=1p‍μj∣μj∈σ(T)(j=1,…,p),#{k∣μj=μk}≤nul(T-μj)ifμj∈σd(T)∑j=1p‍},
where, for a finite set F, #F denotes the number of the elements of F and σd(T) denotes the discrete spectrum of T (the set of isolated eigenvalues of T with finite multiplicity).Lemma 12 (see [13, Corollary 3.2]).
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.Proof of Theorem6.
By a theorem on the spectrum of the sum of tensor products of self-adjoint operators (e.g., [14, Corollary 7.25]),
(57)σ(HBF(A))={λ+μ∣λ∈σ(Hb(A*A)),μ∈σ(Hf(AA*))},
where we have used the fact that the set on the right hand side is closed (note that σ(Hb(A*A)), σ(Hf(AA*))⊂[0,∞)). By (51) and (55),
(58)σ(HBF(A))⊂{0}∪Σ¯1∪Σ¯2∪Σ¯3
with
(59)Σ1:=⋃n=1∞‍{∑j=1n‍λj∣λj∈σ(A*A),j=1,…,n},Σ2:=⋃p=1∞‍{∑k=1p‍μk∣μk∈σ(AA*)(k=1,…,p),#{l∣μk=μl}≤nul(AA*-μk)ifμk∈σd(AA*)∑k=1p‍},Σ3≔⋃n,p=1∞‍{∑j=1n‍λj+∑k=1p‍μk∣λj∈σ(A*A)(j=1,…,n),μk∈σ(AA*)(k=1,…,p),#{l∣μk=μl}≤nul(AA*-μk)ifμk∈σd(AA*)∑j=1n‍}.
By Deift’s theorem [15], we have
(60)σ(A*A)∖{0}=σ(AA*)∖{0},
Hence Σ2⊂{0}∪Σ1 and Σ3⊂Σ1. Therefore it follows that σ(HBF(A))⊂{0}∪Σ¯1. The converse inclusion relation {0}∪Σ¯1⊂σ(HBF(A)) is obvious. Thus σ(HBF(A))={0}∪Σ¯1. The set on the right hand side is equal to the set on the right hand side of (42). Thus (42) holds.
The proof of (43) is similar to that of (42), where we use another version of Deift's theorem [15]:
(61)σp(A*A)∖{0}=σp(AA*)∖{0}
with nul(A*A-λ)=nul(AA*-λ) for each positive eigenvalue λ>0 of A*A (AA*).Proof of Theorem7.
By Theorem2(ii) and Lemma 9, σ(QA) and σp(QA) are symmetric with respect to the origin and (45) holds. By Lemma 8 and Theorem 2(iii), we have
(62)σ(QA)={±λ∣λ∈σ(HBF(A))},σp(QA)={±λ∣λ∈σp(HBF(A))}.
By these facts and Theorem 6, we obtain (44).


## 4. A Class of Perturbed Infinite Dimensional Dirac Operators

In this section we consider a class of perturbations ofQA. It would be natural to perturb QA through a perturbation of dA. Let g∈D(A)∖{0}, v∈D(A*)∖{0} and define
(63)d(α)∶=dA+αa(g)⊗b(v)*,
where α∈C is a perturbation parameter (a coupling constant in the context of physics). We have by direct computation
(64)d(α)2=0on⁡DA∞.
The adjoint d(α)* is densely defined with DA∞⊂D(d(α)*) and
(65)d(α)*=dA*+α*a(g)*⊗b(v)on⁡DA∞.
Hecne d(α) is closable. We denote the closure of d(α)↾DA∞ by d¯(α). Then, for all Ψ∈D(d¯(α)), d¯(α)Ψ is in D(d¯(α)) and
(66)d¯(α)2Ψ=0.The operatord¯(α) yields a perturbed Dirac operator:
(67)Q(α)∶=d¯(α)+d¯(α)*.
By an application of [2, Lemma 4.1], one sees that Q(α) is a closed symmetric operator. We have
(68)Q(α)=QA+Vg,v(α)on⁡DA∞
with
(69)Vg,v(α)∶=αa(g)⊗b(v)*+α*a(g)*⊗b(v).In what follows, we investigate properties ofQ(α).


### 4.1. Self-Adjointness ofQ(α)

We define a linear operatorTg,v from H to K by
(70)Tg,vf∶=〈g,f〉v,f∈H.
It is obvious that Tg,v is a bounded operator (a one-rank operator). Hence
(71)A(α)∶=A+αTg,v
is a densely defined closed linear operator with D(A(α))=D(A).The following lemma is a key fact.Lemma 13.
For allα∈C, the following operator equality holds:
(72)d¯(α)=dA(α).Proof.
LetΨ∈DA∞ be a vector of the form (13). Then, using the canonical commutation relations (CCR)
(73)[a(f),a(g)*]=〈f,g〉,f,g∈H,
on
(74)Fb,fin(H)∶=L{a(f1)*⋯a(fn)*Ωb∣n≥0,fj∈H,j=1,…,n}
and (6), we have
(75)a(g)⊗b(v)*Ψ=∑j=1n‍〈g,fj〉a(f1)*⋯a(fj)*^⋯a(fn)*Ωb⊗b(v)*b(u1)*⋯b(up)*Ωf=∑j=1n‍a(f1)*⋯a(fj)*^⋯a(fn)*Ωb⊗b(Tg,vfj)*b(u1)*⋯b(up)*Ωf=dTg,vΨ.
Hence, by Proposition 1-(v)
(76)d¯(α)Ψ=dAΨ+αdTg,vΨ=dA(α)Ψ.
By a limiting argument, we obtain the operator equality (72).Theorem 14.
(i) For allα∈C, Q(α) is self-adjoint and
(77)Q(α)=QA(α).
(ii) For allα∈C, Q(α) is essentially self-adjoint on DA∞.
(iii) For allα∈C,
(78)Q(α)=QA+Vg,v(α)¯.
(iv) The operatorΓ leaves D(Q(α)) invariant and
(79)ΓQ(α)+Q(α)Γ=0onD(Q(α)).
(v) For allΨ∈DA∞, the vector-valued function: α↦Q(α)Ψ is strongly continuous on C. Moreover, for all z∈C∖R, (Q(α)-z)-1 is strongly continuous in α∈C.Proof.
(i) Operator equality (77) follows from (72) and (67). Then, by Theorem 2-(i) with A replaced by A(α)), we conclude that Q(α) is self-adjoint.
(ii) We have the following operator equalities:(80)A(α)*A(α)=A*A+Xg,v(α),A(α)A(α)*=AA*+Yg,v(α),
where Xg,v(α) and Yg,v(α) are bounded linear operators on H and K, respectively, given as follows:
(81)Xg,v(α)f∶=α〈g,f〉A*v+α*〈A*v,f〉g+|α|2∥v∥2〈g,f〉g,1111111111111111111111111111111111111.f∈H,Yg,v(α)u∶=α〈Ag,u〉v+α*〈v,u〉Ag+|α|2∥g∥2〈v,u〉v,1111111111111111111111111111111111u∈K.
Hence, by the Kato-Rellich theorem ([12, Chapter V, Theorems 4.3, 4.4], [10, Theorem X.12]), A(α)*A(α) (resp., A(α)A(α)*) is essentially self-adjoint on each core of A*A (resp., AA*). In particular, C∞(A*A) (resp., C∞(AA*)) is a core of A(α)*A(α) (resp., A(α)A(α)*). Hence, by a core property of second quantization operators, DA∞ is a core of HBF(A(α)). By this fact and Theorem 2-(i), DA∞ is a core of QA(α)=Q(α).
(iii) This follows from (ii) and (68).
(iv) This follows from (77) and Theorem 2-(ii) with A replaced by A(α).
(v) LetΨ∈DA∞. Then, for all α,β∈C, we have
(82)∥Q(β)Ψ-Q(α)Ψ∥≤|β-α|(∥a(g)⊗b(v)*Ψ∥+∥a(g)*⊗b(v)Ψ∥)⟶0(β⟶α).
Since DA∞ is a common core for {Q(α)}α∈C as proved in part (ii), a general fact [11, Theorem VIII.25] implies the strong continuity of (Q(α)-z)-1 (z∈C∖R) in α∈C.Remark 15.
Since the operatorsXg,v(α) and Yg,v(α) are finite rank, they are compact. Hence it follows from (80) and a general theorem (e.g., [16, p.113, Corollary 2]) that, for all α∈C,
(83)σess(A(α)*A(α))=σess(A*A),σess(A(α)A(α)*)=σess(AA*),
where, for a self-adjoint operator T, σess(T) denotes the essential spectrum of T. In particular, if the spectrum σ(A*A) is purely discrete (in this case, σ(AA*) also is purely discrete by (60)), then so are σ(A(α)*A(α)) and σ(A(α)A(α)*).


### 4.2. Spectra ofQ(α)

We have (77). Hence, applying Theorem 7 with A replaced by A(α), we obtain the following result on the spectra of Q(α).Theorem 16.
For allα∈C, σ(Q(α)) and σp(Q(α)) are symmetric with respect to the origin and(84)σ(Q(α))={0}⋃‍(⋃n=1∞‍{±∑j=1n‍λj∣λj∈σ(A(α)*A(α))∖{0},j=1,…,n}¯),σp(Q(α))={0}⋃‍(⋃n=1∞‍{±∑j=1n‍λj∣λj∈σp(A(α)*A(α))∖{0},j=1,…,n})with
(85)nul(Q(α)-λ)=nul(Q(α)+λ),λ∈σp(Q(α)).This theorem shows that the spectrum and the point spectrum ofQ(α) are completely determined from those of A(α)*A(α)∖{0}.


### 4.3. Identification of the Domain ofQ(α)

The following theorem gives a sufficient condition for the domain ofQ(α) to be equal to that of the unperturbed operator QA.Theorem 17.
Suppose thatA is injective and g∈D(A-1). Then, for all |α|<1/∥v∥∥A-1g∥, Q(α) is self-adjoint with D(Q(α))=D(QA) and
(86)Q(α)=QA+Vg,v(α).
Moreover, Q(α) is essentially self-adjoint on any core for QA.To prove this theorem, we recall well known facts (e.g., [17, Lemma 2.1], [8]).Lemma 18.
LetS be a nonnegative self-adjoint operator on H with ker⁡S={0}. Then, for all f∈D(S-1/2), D(dΓb(S)1/2)⊂D(a(f))∩D(a(f)*) with
(87)∥a(f)ψ∥≤∥S-1/2f∥∥dΓb(S)1/2ψ∥,(88)∥a(f)*ψ∥2≤∥S-1/2f∥2∥dΓb(S)1/2ψ∥2+∥f∥2∥ψ∥2,ψ∈D(dΓb(S)1/2).Lemma 19.
The operatorVg,v(α) is a closed symmetric operator with D(Vg,v(α))=D(a(g)⊗b(v)*)∩D(a(g)*⊗b(v)).Proof.
We need only to prove the closedness ofVg,v(α). Using (9) and a limiting argument, we have, for all Ψ,Φ∈D(Vg,v(α))=D(a(f)⊗b(v)*)∩D(a(f)*⊗b(v)),
(89)(a(g)⊗b(v)*Ψ,a(g)*⊗b(v)Φ)=0.
Hence, by a general fact [2, Lemma 4.1], Vg,v(α) is closed.Lemma 20.
LetA be injective and g∈D(A-1). Then D(QA)⊂D(Vg,v(α)).Proof.
Under the present assumption,A*A is injective and (A*A)-1/2=|A|-1. Let Ψ∈DA∞. Then, using (87) and the well known fact that
(90)∥b(u)∥=∥b(u)*∥=∥u∥,u∈K,
we can show that
(91)∥a(g)⊗b(v)*Ψ∥≤∥v∥∥a(g)⊗IΨ∥≤∥v∥∥|A|-1g∥∥dΓb(A*A)1/2⊗IΨ∥≤∥v∥∥A-1g∥∥HBF(A)1/2Ψ∥.
By this estimate and the fact that
(92)∥HBF(A)1/2Φ∥=∥QAΦ∥,Φ∈D(QA),
we obtain
(93)∥a(g)⊗b(v)*Ψ∥≤∥v∥∥A-1g∥∥QAΨ∥.
Since DA∞ is a core of QA, this inequality extends to all Ψ∈D(QA) showing that D(QA)⊂D(a(g)⊗b(v)*). Similarly, using (88), we can show that D(QA)⊂D(a(g)*⊗b(v)) with
(94)∥a(g)*⊗b(v)Ψ∥2≤∥v∥2(∥A-1g∥2∥QAΨ∥2+∥g∥2∥Ψ∥2),Ψ∈D(QA).
Thus D(QA)⊂D(Vg,v(α)).Proof of Theorem17.
By Lemma20 and Theorem 14, the operator
(95)T∶=QA+Vg,v(α)
is essentially self-adjoint with D(T)=D(QA) and T¯=Q(α). By (89), the CCR (73), and the CAR (8), one can show that, for all Ψ∈D(QA),
(96)∥Vg,v(α)Ψ∥2=|α|2(∥a(g)⊗b(v)*Φ∥2+∥a(g)*⊗b(v)Φ∥2)=|α|2(∥v∥2∥a(g)⊗IΨ∥2+∥g∥2∥I⊗b(v)Ψ∥2)≤|α|2∥v∥2(∥a(g)⊗IΨ∥2+∥g∥2∥Ψ∥2).
Hence, using (87) (cf. the proof of Lemma 20), we obtain
(97)∥Vg,v(α)Ψ∥2≤|α|2∥v∥2(∥A-1g∥2∥QAΨ∥2+∥g∥2∥Φ∥2),Ψ∈D(QA).
Therefore, by the Kato-Rellich theorem, for all α∈C satisfying |α|∥v∥∥A-1g∥<1, T is self-adjoint on D(QA), implying that T=Q(α), and essentially self-adjoint on any core for QA.


## 4.1. Self-Adjointness ofQ(α)

We define a linear operatorTg,v from H to K by
(70)Tg,vf∶=〈g,f〉v,f∈H.
It is obvious that Tg,v is a bounded operator (a one-rank operator). Hence
(71)A(α)∶=A+αTg,v
is a densely defined closed linear operator with D(A(α))=D(A).The following lemma is a key fact.Lemma 13.
For allα∈C, the following operator equality holds:
(72)d¯(α)=dA(α).Proof.
LetΨ∈DA∞ be a vector of the form (13). Then, using the canonical commutation relations (CCR)
(73)[a(f),a(g)*]=〈f,g〉,f,g∈H,
on
(74)Fb,fin(H)∶=L{a(f1)*⋯a(fn)*Ωb∣n≥0,fj∈H,j=1,…,n}
and (6), we have
(75)a(g)⊗b(v)*Ψ=∑j=1n‍〈g,fj〉a(f1)*⋯a(fj)*^⋯a(fn)*Ωb⊗b(v)*b(u1)*⋯b(up)*Ωf=∑j=1n‍a(f1)*⋯a(fj)*^⋯a(fn)*Ωb⊗b(Tg,vfj)*b(u1)*⋯b(up)*Ωf=dTg,vΨ.
Hence, by Proposition 1-(v)
(76)d¯(α)Ψ=dAΨ+αdTg,vΨ=dA(α)Ψ.
By a limiting argument, we obtain the operator equality (72).Theorem 14.
(i) For allα∈C, Q(α) is self-adjoint and
(77)Q(α)=QA(α).
(ii) For allα∈C, Q(α) is essentially self-adjoint on DA∞.
(iii) For allα∈C,
(78)Q(α)=QA+Vg,v(α)¯.
(iv) The operatorΓ leaves D(Q(α)) invariant and
(79)ΓQ(α)+Q(α)Γ=0onD(Q(α)).
(v) For allΨ∈DA∞, the vector-valued function: α↦Q(α)Ψ is strongly continuous on C. Moreover, for all z∈C∖R, (Q(α)-z)-1 is strongly continuous in α∈C.Proof.
(i) Operator equality (77) follows from (72) and (67). Then, by Theorem 2-(i) with A replaced by A(α)), we conclude that Q(α) is self-adjoint.
(ii) We have the following operator equalities:(80)A(α)*A(α)=A*A+Xg,v(α),A(α)A(α)*=AA*+Yg,v(α),
where Xg,v(α) and Yg,v(α) are bounded linear operators on H and K, respectively, given as follows:
(81)Xg,v(α)f∶=α〈g,f〉A*v+α*〈A*v,f〉g+|α|2∥v∥2〈g,f〉g,1111111111111111111111111111111111111.f∈H,Yg,v(α)u∶=α〈Ag,u〉v+α*〈v,u〉Ag+|α|2∥g∥2〈v,u〉v,1111111111111111111111111111111111u∈K.
Hence, by the Kato-Rellich theorem ([12, Chapter V, Theorems 4.3, 4.4], [10, Theorem X.12]), A(α)*A(α) (resp., A(α)A(α)*) is essentially self-adjoint on each core of A*A (resp., AA*). In particular, C∞(A*A) (resp., C∞(AA*)) is a core of A(α)*A(α) (resp., A(α)A(α)*). Hence, by a core property of second quantization operators, DA∞ is a core of HBF(A(α)). By this fact and Theorem 2-(i), DA∞ is a core of QA(α)=Q(α).
(iii) This follows from (ii) and (68).
(iv) This follows from (77) and Theorem 2-(ii) with A replaced by A(α).
(v) LetΨ∈DA∞. Then, for all α,β∈C, we have
(82)∥Q(β)Ψ-Q(α)Ψ∥≤|β-α|(∥a(g)⊗b(v)*Ψ∥+∥a(g)*⊗b(v)Ψ∥)⟶0(β⟶α).
Since DA∞ is a common core for {Q(α)}α∈C as proved in part (ii), a general fact [11, Theorem VIII.25] implies the strong continuity of (Q(α)-z)-1 (z∈C∖R) in α∈C.Remark 15.
Since the operatorsXg,v(α) and Yg,v(α) are finite rank, they are compact. Hence it follows from (80) and a general theorem (e.g., [16, p.113, Corollary 2]) that, for all α∈C,
(83)σess(A(α)*A(α))=σess(A*A),σess(A(α)A(α)*)=σess(AA*),
where, for a self-adjoint operator T, σess(T) denotes the essential spectrum of T. In particular, if the spectrum σ(A*A) is purely discrete (in this case, σ(AA*) also is purely discrete by (60)), then so are σ(A(α)*A(α)) and σ(A(α)A(α)*).


## 4.2. Spectra ofQ(α)

We have (77). Hence, applying Theorem 7 with A replaced by A(α), we obtain the following result on the spectra of Q(α).Theorem 16.
For allα∈C, σ(Q(α)) and σp(Q(α)) are symmetric with respect to the origin and(84)σ(Q(α))={0}⋃‍(⋃n=1∞‍{±∑j=1n‍λj∣λj∈σ(A(α)*A(α))∖{0},j=1,…,n}¯),σp(Q(α))={0}⋃‍(⋃n=1∞‍{±∑j=1n‍λj∣λj∈σp(A(α)*A(α))∖{0},j=1,…,n})with
(85)nul(Q(α)-λ)=nul(Q(α)+λ),λ∈σp(Q(α)).This theorem shows that the spectrum and the point spectrum ofQ(α) are completely determined from those of A(α)*A(α)∖{0}.


## 4.3. Identification of the Domain ofQ(α)

The following theorem gives a sufficient condition for the domain ofQ(α) to be equal to that of the unperturbed operator QA.Theorem 17.
Suppose thatA is injective and g∈D(A-1). Then, for all |α|<1/∥v∥∥A-1g∥, Q(α) is self-adjoint with D(Q(α))=D(QA) and
(86)Q(α)=QA+Vg,v(α).
Moreover, Q(α) is essentially self-adjoint on any core for QA.To prove this theorem, we recall well known facts (e.g., [17, Lemma 2.1], [8]).Lemma 18.
LetS be a nonnegative self-adjoint operator on H with ker⁡S={0}. Then, for all f∈D(S-1/2), D(dΓb(S)1/2)⊂D(a(f))∩D(a(f)*) with
(87)∥a(f)ψ∥≤∥S-1/2f∥∥dΓb(S)1/2ψ∥,(88)∥a(f)*ψ∥2≤∥S-1/2f∥2∥dΓb(S)1/2ψ∥2+∥f∥2∥ψ∥2,ψ∈D(dΓb(S)1/2).Lemma 19.
The operatorVg,v(α) is a closed symmetric operator with D(Vg,v(α))=D(a(g)⊗b(v)*)∩D(a(g)*⊗b(v)).Proof.
We need only to prove the closedness ofVg,v(α). Using (9) and a limiting argument, we have, for all Ψ,Φ∈D(Vg,v(α))=D(a(f)⊗b(v)*)∩D(a(f)*⊗b(v)),
(89)(a(g)⊗b(v)*Ψ,a(g)*⊗b(v)Φ)=0.
Hence, by a general fact [2, Lemma 4.1], Vg,v(α) is closed.Lemma 20.
LetA be injective and g∈D(A-1). Then D(QA)⊂D(Vg,v(α)).Proof.
Under the present assumption,A*A is injective and (A*A)-1/2=|A|-1. Let Ψ∈DA∞. Then, using (87) and the well known fact that
(90)∥b(u)∥=∥b(u)*∥=∥u∥,u∈K,
we can show that
(91)∥a(g)⊗b(v)*Ψ∥≤∥v∥∥a(g)⊗IΨ∥≤∥v∥∥|A|-1g∥∥dΓb(A*A)1/2⊗IΨ∥≤∥v∥∥A-1g∥∥HBF(A)1/2Ψ∥.
By this estimate and the fact that
(92)∥HBF(A)1/2Φ∥=∥QAΦ∥,Φ∈D(QA),
we obtain
(93)∥a(g)⊗b(v)*Ψ∥≤∥v∥∥A-1g∥∥QAΨ∥.
Since DA∞ is a core of QA, this inequality extends to all Ψ∈D(QA) showing that D(QA)⊂D(a(g)⊗b(v)*). Similarly, using (88), we can show that D(QA)⊂D(a(g)*⊗b(v)) with
(94)∥a(g)*⊗b(v)Ψ∥2≤∥v∥2(∥A-1g∥2∥QAΨ∥2+∥g∥2∥Ψ∥2),Ψ∈D(QA).
Thus D(QA)⊂D(Vg,v(α)).Proof of Theorem17.
By Lemma20 and Theorem 14, the operator
(95)T∶=QA+Vg,v(α)
is essentially self-adjoint with D(T)=D(QA) and T¯=Q(α). By (89), the CCR (73), and the CAR (8), one can show that, for all Ψ∈D(QA),
(96)∥Vg,v(α)Ψ∥2=|α|2(∥a(g)⊗b(v)*Φ∥2+∥a(g)*⊗b(v)Φ∥2)=|α|2(∥v∥2∥a(g)⊗IΨ∥2+∥g∥2∥I⊗b(v)Ψ∥2)≤|α|2∥v∥2(∥a(g)⊗IΨ∥2+∥g∥2∥Ψ∥2).
Hence, using (87) (cf. the proof of Lemma 20), we obtain
(97)∥Vg,v(α)Ψ∥2≤|α|2∥v∥2(∥A-1g∥2∥QAΨ∥2+∥g∥2∥Φ∥2),Ψ∈D(QA).
Therefore, by the Kato-Rellich theorem, for all α∈C satisfying |α|∥v∥∥A-1g∥<1, T is self-adjoint on D(QA), implying that T=Q(α), and essentially self-adjoint on any core for QA.


## 5. Kernel ofQ(α)

In this section we investigate the kernel ofQ(α). For this purpose, we need some conditions on {A,g,v}.(C.1)A is injective, v∈D(A-1), and 〈g,A-1v〉≠0. In this case we introduce a constant
(98)α0∶=-1〈g,A-1v〉.(C.2)A* is injective, g∈D(A*-1), and 〈v,A*-1g〉≠0. In this case we introduce a constant
(99)β0:=-1〈A*-1g,v〉.(C.3) (a) A is injective and v∉D(A-1) or (b) A is injective and v∈D(A-1) with 〈g,A-1v〉=0.(C.4) (a) A* is injective and g∉D(A*-1) or (b) A* is injective and g∈D(A*-1) with 〈v,A*-1g〉=0.Remark 21.
(i) If(C.1) and (C.2) hold, then it follows that α0=β0.
(ii) IfA is Fredholm, then A* is Fredholm [12, Chapter IV, Corollary 5.14]. Hence, if (C.1) and (C.2) hold and A is Fredholm, then H=R(A*)¯=R(A*), K=R(A)¯=R(A). Hence A and A* are bijective with D(A-1)=K and D(A*-1)=H. Thus, in this case, conditions v∈D(A-1) and g∈D(A*-1) are trivially satisfied.We first consider the kernel ofA(α) and A(α)*.Lemma 22.
(i) Suppose that(C.1) holds. Then
(100)ker⁡A(α)={0},α≠α0,(101)ker⁡A(α0)={cA-1v∣c∈C}.
(ii) Suppose that(C.2) holds. Then
(102)ker⁡A(α)*={0},α≠β0,ker⁡A(β0)*={cA*-1g∣c∈C}.
(iii) Suppose that(C.3) holds. Then, for all α∈C,
(103)ker⁡A(α)={0}.
(iv) Suppose that(C.4) holds. Then, for all α∈C,
(104)ker⁡A(α)*={0}.Proof.
(i) Letf∈ker⁡A(α). Then f∈D(A) and
(105)Af=-α〈g,f〉v.
Hence
(106)f=-α〈g,f〉f0
with f0∶=A-1v. This implies
(107)ker⁡A(α)⊂{cf0∣c∈C}.
Letα≠α0. Then, taking the inner product of (106) with g and using the fact that 〈g,f0〉=-1/α0, we have 〈g,f〉=0. Hence, by (106) again, f=0. Thus (100) follows.
Letα=α0. Then
(108)Af0=v=-α0〈g,f0〉v.
Hence f0∈ker⁡A(α0). Thus, by (107), we obtain (101).
(ii) We have(109)A(α)*=A*+α*〈v,·〉g.
Hence, by replacing {A,g,v,α} by {A*,v,g,α*} in the proof of (i), we can easily obtain (102).
(iii) For allf∈ker⁡A(α), we have (105). Let α≠0. Then (105) means that 〈g,f〉v∈D(A-1). Hence, if v∉(A-1), then 〈g,f〉=0, which implies via (105) again that Af=0; that is, f=0. Thus (103) follows.
Ifv∈D(A-1) with 〈g,A-1v〉=0, then f=-α〈g,f〉A-1v and hence 〈g,f〉=0. Hence f=0. Thus (103) follows.
(iv) It is similar to the proof of (iii).Theorem 23.
(i) Assume(C.1). Then
(110)ker⁡Q(α0)=⊕n,p=0∞[(⊗n{zA-1v∣z∈C})⊗∧p(ker⁡A(α0)*)],
and hence nulQ(α0)=∞.
Moreover, for allα≠α0,
(111)ker⁡Q(α)=⊕p=0∞C⊗∧p(ker⁡A(α)*).
(ii) Assume(C.2). Then
(112)ker⁡Q(β0)=⊕n=0∞{[⊗snker⁡(A(β0))]⊗[C⊕L({A*-1g})]},ker⁡Q(α)=⊕n=0∞[⊗snker⁡A(α)⊗C],α≠β0.
(iii) Assume(C.3). Then, for all α∈C,
(113)ker⁡Q(α)=⊕p=0∞[C⊗∧p(ker⁡A(α)*)].
(iv) Assume(C.4). Then, for all α∈C,
(114)ker⁡Q(α)=⊕n=0∞[⊗snker⁡A(α)⊗C].Proof.
Under condition(C.1), we have (100) and (101). Hence ⊗snker⁡A(α0)=⊗n{zA-1v∣z∈C}. Therefore, applying (77) and Theorem 4, we obtain (110).
In the caseα≠α0, (111) follows from (100), (77), and Theorem 4.
Using Lemma22, one can prove (ii)–(iv) in the same manner as in the proof of (i).Some compatible compositions of conditions(C.1)–(C.4) yield the following results.Corollary 24.
(i) Assume(C.1) and (C.2). Then
(115)ker⁡Q(α0)=L({a(A-1v)*nΩb⊗b(A*-1g)*jΩf∣n≥0,j=0,1})¯,ker⁡Q(α)={cΩb⊗Ωf∣c∈C},α≠α0.
(ii) Assume(C.1) and (C.4). Then
(116)ker⁡Q(α0)=L({a(A-1v)*nΩb⊗Ωf∣n≥0})¯,ker⁡Q(α)={cΩb⊗Ωf∣c∈C},α≠α0.
(iii) Assume(C.2) and (C.3). Then
(117)ker⁡Q(β0)=L({Ωb⊗b(A*-1g)*jΩf∣j=0,1}).ker⁡Q(α)={cΩb⊗Ωf∣c∈C},α≠β0.
(iv) Assume(C.3) and (C.4). Then, for all α∈C,
(118)ker⁡Q(α)={cΩb⊗Ωf∣c∈C}.


## 6. Fredholm Property and the IndexindΓQ(α)

Theorem23 has also implications for Fredholm property of Q(α). Since we want to see how Fredholm property of QA may change under the perturbation Vg,v(α), we take our consideration under the same conditions as in Theorem 5.We first note the following fact.Lemma 25.
LetA be semi-Fredholm. Then, for all α∈C, A(α) is semi-Fredholm and
(119)indA=indA(α).Proof.
We have (71) and Tg,v is a one-rank operator, which is compact. Thus, by a general theorem [12, Chapter IV, Theorem 5.26], the lemma holds.Theorem 26.
(i) Suppose that(C.1) holds and A is Fredholm. Then Q(α0)+ is not semi-Fredholm and, for all α∈C∖{α0}, Q(α)+ is Fredholm with
(120)indΓQ(α)=δ0,indA*=indΓQA.
(ii) Suppose that(C.3) holds and A is Fredholm. Then, for all α∈C, Q(α) is Fredholm and (120) holds.
(iii) Suppose that(C.2) holds and A is semi-Fredholm. Then Q(β0)+ is not semi-Fredholm and, for all α∈C∖{β0}, the following hold. (a) IfA is injective, then Q(α)+ is Fredholm with indΓQ(α)=1.(b) IfA is not injective, then Q(α)+ is semi-Fredholm with indΓQ(α)=∞.
(iv) Suppose that(C.4) holds and that A is semi-Fredholm. (a) IfA is injective, then, for all α∈C, Q(α)+ is Fredholm and indΓQ(α)=1.(b) IfA is not injective, then, for all α∈C, Q(α)+ is semi-Fredholm and indΓQ(α)=∞.
(v) LetA be semi-Fredholm with nulA≥1 and nulA*≥1. Suppose that g∈(ker⁡A)⊥ and v∈(ker⁡A*)⊥. Then Q(α)+ is not semi-Fredholm.Proof.
In the proofs below, (77) is used without mentioning explicitly.
(i) By Lemma22-(i), nulA(α0)=1. By this fact and Lemma 25, we have
(121)nulA(α0)*=1+nulA*≥1.
Hence, applying Theorem 5-(iii) with A replaced by A(α0), one concludes that Q(α0)+ is not semi-Fredholm.
Letα≠α0. Then, by Lemma 25, we have nulA*=nulA(α)*. Hence, by Theorem 5-(i), Q(α)+ is Fredholm and indΓQ(α)=δ0,nulA(α)*=δ0,nulA*. Thus (120) holds.
(ii) It is similar to (i).
(iii) In the present case, we havenulA*=0 and nulA(β0)*=1. Hence, by Lemma 25, nulA(β0)=nulA+1≥1. Hence, by applying Theorem 5-(iii) with A replaced by A(β0), we conclude that Q(β0)+ is not semi-Fredholm.
Letα≠β0. Then, by Lemma 22-(ii), A(α)* is injective. By Lemma 25, nulA=nulA(α).(a) LetA be injective. Then A(α) is injective. Hence, we can apply Theorem 5-(i) with A replaced by A(α) to conclude that Q(α)+ is Fredholm with indΓQ(α)=1.(b) LetA be not injective. Then nulA(α)≥1. Hence, we can apply Theorem 5-(ii) with A replaced by A(α) to conclude that Q(α)+ is semi-Fredholm with indΓQ(α)=∞.
(iv) In the present case, we havenulA(α)*=0 for all α∈C. Hence, by Lemma 25, nulA=nulA(α) for all α∈C.(a) In this case,nulA(α)=0. Hence, by Theorem 5-(i) with A replaced by A(α), Q(α) is Fredholm and indΓQ(α)+=1.(b) In this case, we can apply Theorem5-(ii) with A replaced by A(α) to conclude that Q(α)+ is semi-Fredholm and indΓQ(α)+=∞.
(v) In the present case, we haveker⁡A⊂ker⁡A(α) and ker⁡A*⊂ker⁡A(α)*. Hence nulA(α)≥1 and nulA(α)*≥1. Therefore, we can apply Theorem 5-(iii) with A replaced by A(α) to conclude that Q(α)+ is not semi-Fredholm.


## 7. Nonzero Eigenvalues ofQ(α)

Finally we consider existence of nonzero eigenvalues ofQ(α). In this section we need the following condition.Hypothesis (I). (i) H=K;(ii)A is an injective and nonnegative self-adjoint operator;(iii)g=v∈D(A-1).Under Hypothesis (I), the constantα0 defined by (98) takes the form
(122)α0=-1〈v,A-1v〉<0.Theorem 27.
Let Hypothesis (I) be satisfied andα<α0(<0). Then, there exists a unique constant x0(α)<0 such that α〈v,(x0(α)-A)-1v〉=1 and, for all n∈{0}∪N,
(123)±nx0(α)∈σp(Q(α)).
with eigenvectors
(124)[Q(α)±nx0(α)]{a(ϕα)*n-pΩb⊗b(ϕα)*pΩf}∈ker⁡(Q(α)∓nx0(α))(n≥p≥0),
where
(125)ϕα∶=(x0(α)-A)-1v.
Moreover,x0(α), as a function of α<α0, is strictly monotone increasing on (-∞,α0) with lim⁡α→-∞x0(α)=-∞ and lim⁡α→α0x0(α)=0.Note that Theorem27 holds even if QA has no nonzero eigenvalues. This is an interesting phenomenon. Since the condition α<α0<0 implies that |α|>|α0|, the phenomenon may be regarded as astrong coupling effect.To prove Theorem27, we prepare two lemmas.We consider a condition slightly more general than Hypothesis (I).Hypothesis (II). In addition to (i) and (ii) in Hypothesis (I), the following holds: v∈D(Aa)∩D(A-1) for some a∈R and
(126)g=Aav.Under Hypothesis (II), the constantα0 defined by (98) takes the form
(127)α0=-1〈Aav,A-1v〉<0.Lemma 28.
Let Hypothesis (II) be satisfied.(i)
Letα0<α. Then there exist no solutions x≤0 to the equation
(128)α〈Aav,(x-A)-1v〉=1.(ii)
Letα≤α0(<0). Then there exists a unique constant x0(α)≤0 such that
(129)α〈Aav,(x0(α)-A)-1v〉=1.
The constant x0(α) is, as a function of α≤α0(<0), strictly monotone increasing and differentiable on (-∞,α0) with x0(α0)=0 and lim⁡α→-∞x0(α)=-∞.Proof.
(i) SinceA is injective, nonnegative, and self-adjoint, we can define a function F on (-∞,0] by
(130)F(x)∶=-〈Aav,(x-A)-1v〉>0,x∈(-∞,0].
This function is differentiable on (-∞,0) with
(131)dF(x)dx=〈Aav,(x-A)-2v〉>0,x∈(-∞,0).
Hence F is strictly monotone increasing. It is easy to see that lim⁡x→-∞F(x)=0 and lim⁡x→0F(x)=-1/α0>0 (note that α0<0). Hence, if α>α0, then -αF(x)=1 has no solutions x in (-∞,0].
(ii) Letα≤α0. Then, by the proof of (i), there exists a unique constant x0(α)≤0 such that -αF(x0(α))=1. The inverse function F-1 of F exists on (0,-1/α0] and is differentiable on (0,-1/α0). Hence x0(α)=F-1(-1/α), α≤α0. This implies the assertion on x0(α).Lemma 29.
Let Hypothesis (II) be satisfied.(i)
Letα>α0. Then
(132)σp(A(α))∩(-∞,0]=∅.(ii)
Letα≤α0 and ϕα be defined by (125). Then
(133)σp(A(α))∩(-∞,0]={x0(α)},(134)ker⁡(A(α)-x0(α))={cϕα∣c∈C}.Proof.
Letx∈σp(A(α))∩(-∞,0]. Then there is a nonzero vector ϕ∈D(A) such that Aϕ+α〈Aav,ϕ〉v=xϕ. It follows that
(135)ϕ=α〈Aav,ϕ〉(x-A)-1v.
If 〈Aav,ϕ〉=0, then ϕ=0, which is a contradiction. Hence 〈Aav,ϕ〉≠0. Taking the inner product of (135) with Aav, we obtain
(136)α〈Aav,(x-A)-1v〉=1.
(i) Ifα>α0, then, by Lemma 28-(i), (136) has no solutions x in (-∞,0]. Hence (132) holds.
(ii) Letα≤α0. Then, by Lemma 28-(ii), we have x=x0(α). Thus σp(A(α))∩(-∞,0]⊂{x0(α)}. It is easy to check that A(α)ϕα=x0(α)ϕα. Thus (133) follows. By (135), every vector ϕ∈ker⁡(A(α)-x0(α)) satisfies ϕ=α〈Aav,ϕ〉ϕα. Hence ϕ∈{cϕα∣c∈C}. Thus (134) follows.Proof of Theorem27.
Under the present assumption,A(α) is self-adjoint. Hence, by Lemma 29, x0(α)2∈σp(A(α)2). Hence Theorem 16 implies (123). We have A(α)2ϕα=x0(α)2ϕα. Hence
(137)Φ(n,p)(α)∶=a(ϕα)*n-pΩb⊗b(ϕα)*pΩf(n≥p≥0)
are eigenvectgors of QA(α)2=Q(α)2 with eigenvalue nx0(α)2. Hence, applying Lemma 8 with S=Q(α), we see that (Q(α)±nx0(α))Φ(n,p)(α)∈ker⁡(Q(α)∓nx0(α)).



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