# A Class ofd-Dimensional Dirac Operators with a Variable Mass

Authors: Asao Arai; Dayantsolmon Dagva
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

A class ofd-dimensional Dirac operators with a variable mass is introduced (d≥2), which includes, as a special case, the 3-dimensional Dirac operator describing the chiral quark soliton model in nuclear physics, and some aspects of it are investigated.


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## Body

## 1. Introduction

In the chiral quark soliton (CQS) model in nuclear physics (see, e.g., [1] and references therein), a Dirac operator of the following form appears (we use the physical unit system where the speed of lightcand ħ, the Planck constant divided by 2π, are equal to 1):
(1)HCQS≔-i∑j=13‍αjDj⊗12+mβ⊗12ei∑j=13Fγ5⊗τjnj,
acting in the tensor product Hilbert space L2(ℝ3;ℂ4)⊗ℂ2 of L2(ℝ3;ℂ4) (the Hilbert space of ℂ4-valued square integrable functions on ℝ3) and ℂ2. Hereiis the imaginary unit, α1, α2, and α3 are 4×4 Hermitian matrices obeying the anticommutation relations
(2){αj,αk}=2δjk14(j,k=1,2,3)
({A,B}≔AB+BA, δjk is the Kronecker delta and 1n denotes then×nunit matrix), Dj (j=1,2,3) is the generalized partial differential operator in the space variable xj (x=(x1,x2,x3)∈ℝ3), m>0 denotes the mass of a quark, β is a 4×4 Hermitian matrix satisfying
(3){αj,β}=0(j=1,2,3),β2=14,γ5≔-iα1α2α3,F:ℝ3→ℝ is a function called a profile function, τ1,τ2, and τ3 are the Pauli matrices, and nj:ℝ3→ℝ (j=1,2,3) is a Borel measurable function on ℝ3 such that ∑j=13‍nj(x)2=1 for a.e. (almost everywhere) x∈ℝ3.ComparingHCQSwith the usual free Dirac operator with mass, one notes that the term
(4)m(x)≔mei∑j=13F(x)γ5⊗τjnj(x)
corresponds to a mass, although it may depend on the space variable x in general. Hence, the CQS model may be regarded as a model of a Dirac particle with a variable mass. We also note that m(x) is not a scalar multiple of a constant matrix in general but may be a nontrivial matrix-valued function on ℝ3. This is one of the interesting features of the Dirac operator HCQS. From a general point of view, m(x) is a special case of the mass deformation of the form mL(x)≔mL(x) with L being a mapping from ℝ3 to the set of linear operators on ℂ4⊗ℂ2. To our best knowledge, mathematically rigorous analysis on Dirac operators with such a mass deformation seems to be few, although a Dirac operator with a mass given by a scalar function has been studied (e.g., [2]).In a paper [3], Arai et al. investigated spectral properties of the Dirac operator HCQS. These results have been extended to the case of a generalized CQS (GCQS) model in [4]. Miyao [5] proposed an abstract version of the CQS model and investigated a nonrelativistic limit of it; as an application of the abstract result to the CQS model, a Schrödinger operator with a binding potential was derived.As is pointed out in [3], under a condition for nj(x) (j=1,2,3), the CQS model has supersymmetry; that is, the Dirac operator HCQS may be a supercharge of a supersymmetric quantum mechanics (e.g., [6, Chapter5]). This structure is carried over to the GCQS model [4].In this paper, for each natural numberd≥2, we propose ad-dimensional version of the GCQS model and analyze some mathematical aspects of it including supersymmetric ones.The present paper is organized as follows. We first recall some basic facts in operator theory in Section2. In Section 3 we introduce a Dirac operator H which may be the Hamiltonian of ad-dimensional version of the GCQS model, as mentioned previously. A simple condition for H to be self-adjoint is given. In Section 4 we discuss supersymmetric aspects of H. We give a condition for H to be a supercharge of a supersymmetric quantum mechanical model. In that case, ker H, the kernel of H, describes the supersymmetric states. Hence, it is interesting and important to analyze ker H. In Section 5, we prove that, under a condition, ker His trivial: ker H={0}. In the case where H is a supercharge, this means that there is no supersymmetric states; namely, the supersymmetry is spontaneously broken. Section 6 is concerned with a unitary equivalence of H to a gauge theoretic Dirac operator. This may be physically interesting. Using this structure, we find another condition for the kernel of H to be trivial. In Section 7, we identify the essential spectrum ofHunder a suitable condition. In the last section, we discuss the number of eigenvalues ofHin the interval (-m,m) with m>0 being a constant.


## 2. Preliminaries

Let𝒳 be a complex Hilbert space with inner product 〈·,·〉𝒳 (linear in the second variable) and norm ∥·∥𝒳 (we sometimes omit the subscript 𝒳 if there is no danger of confusion). For a linear operator A on 𝒳, we denote its domain by D(A). If A is densely defined, its adjoint is denoted by A*. For linear operators A and B on 𝒳, A⊂B means that B is an extension of A, that is, D(A)⊂D(B) and Aψ=Bψ, for all ψ∈D(A).We denote by𝔅(𝒳) the set of everywhere defined bounded linear operators on 𝒳. For T∈𝔅(𝒳), we denote the operator norm of T by ∥T∥.Definition 1.
LetA and B be self-adjoint operators on 𝒳. (i)
A and B are said to strongly commute if their spectral measures commute.(ii)
A and B are said to strongly anticommute [7, 8] if, for all t∈ℝ, eitBA⊂Ae-itB (it is shown that this definition is in fact symmetric in A and B).The next lemma summarizes some basic facts on strongly commuting (resp., anticommuting) self-adjoint operators.Lemma 2.
LetA and B be self-adjoint operators on 𝒳. (i)
A and B strongly commute if and only if, for all t,s∈ℝ, eitBeisA=eisAeitB.(ii)
A and B strongly commute if and only if, for all t∈ℝ, eitBA=AeitB.(iii)
LetA be bounded. Then A and B strongly commute if and only if AB⊂BA.(iv)
LetA be bounded. Then A and B strongly anticommute if and only if AB⊂-BA.Proof.
Part (i) is well known (e.g., [9, Theorem VIII.13]). Using (i), functional calculus, and strong differential calculus, one can easily prove (ii) and (iii). A proof of (iv) is similar to the proof of (iii).


## 3. Description of the Model

Letd≥2 be a natural number, and
(5)Nd≔{2d/2fordeven,2(d+1)/2fordodd.Let𝒦 be a separable complex Hilbert space, and
(6)ℋ≔L2(ℝd;ℂNd)⊗𝒦≅L2(ℝd;ℂNd⊗𝒦)≅∫ℝd⊕‍ℂNd⊗𝒦dx,
where each ≅ means the natural Hilbert space isomorphism and ∫ℝd⊕‍ℂNd⊗𝒦dx denotes the constant fibre direct integral with fiber ℂNd⊗𝒦 (e.g., [10, sectionXIII.16]).We denote byDjthe generalized partial differential operator in the variable xj (x=(x1,…,xd)∈ℝd), acting in L2(ℝd). Thed-dimensional generalized Laplacian
(7)Δ≔∑j=1d‍Dj2
on L2(ℝd) is a nonpositive self-adjoint operator. Each linear operator A on L2(ℝd) is extended as the direct sum ⊕NdA on L2(ℝd;ℂNd)=⊕NdL2(ℝd). For notational simplicity, we denote it by A again.Every densely defined closable linear operatorT on L2(ℝd,ℂNd) (resp., 𝒦) has a tensor product extension T⊗I (resp., I⊗T) to ℋ (I denotes identity). But we write itTsimply if there is no danger of confusion.We denote byℱs.a. the set of mappings Φ(·) from ℝd to the set of self-adjoint operators on ℂNd⊗𝒦 such that the mapping: ℝd∋x↦(Φ(x)+i)-1 is measurable. By a general theorem (e.g., [10, Theorem XIII.85(i)]), for each Φ(·)∈ℱs.a., the direct integral
(8)Φ≔∫ℝd⊕‍Φ(x)dx
is self-adjoint.Let{αj}j=1d+1 be Nd×Nd Hermitian matrices satisfying
(9){αj,αk}=2δjk1Nd,j,k=1,…,d+1.
Then the free massless Dirac operator on L2(ℝd;ℂNd) is defined by
(10)H0≔-i∑j=1d‍αjDj.
The operator H0 is self-adjoint with D(H0)≔∩j=1dD(Dj) and
(11)H02=-Δ.To introduce a mass operator, letM(·)∈ℱs.a. such that, for a.e. x∈ℝd, M(x) is a bounded operator on ℂNd⊗𝒦, and set
(12)M≔∫ℝd⊕‍M(x)dx.
We use this self-adjoint operator as an extended mass (variable in the space ℝd) of the quantum particle of our model (a Dirac particle). Note that M is not necessarily bounded.The HamiltonianH of our model, a d-dimensional version of the GCQS model, is defined as follows:
(13)H≔H0+U,
with
(14)U≔αd+1eiΦM.
As remarked previously, the mass operator M in U can be variable spatially. This is a point different from the GCQS model.In this work, we do not intend to discuss essential self-adjointness ofHin full generality. In the present paper, we assume the following. (A.1)
αd+1Φ⊂-Φαd+1.(A.2)
αd+1M⊂Mαd+1.(A.3)
Φ and M strongly commute.(A.4)
The operatorM is (-Δ)1/2-bounded: D((-Δ)1/2)⊂D(M) and
(15)∥Mψ∥2≤a2∥(-Δ)1/2ψ∥2+b2∥ψ∥2,∀ψ∈D((-Δ)1/2)
with constants 0≤a<1 and b≥0.Remark 3.
In the abstract CQS model [5], the strong commutativity of M and H0 as well as the boundedness and the strict positivity of M is assumed. But, in our model, they are not assumed.Lemma 4.
(
i
)
Condition(A.1) holds if and only if αd+1 and Φ strongly anticommute.
(
ii
)
Condition(A.1) is equivalent to the operator equality αd+1Φ=-Φαd+1.
(
iii
)
Condition(A.2) holds if and only if αd+1 and M strongly commute.
(
iv
)
Condition(A.2) is equivalent to the operator equality αd+1M=Mαd+1.Proof.
(
i
)
This follows from Lemma2(iv).
(
ii
)
Assume(A.1). Let ψ∈D(Φαd+1). Then η≔αd+1ψ∈D(Φ). Hence, by (A.1), αd+1η∈D(Φ). But, since αd+12=I, we have αd+1η=ψ. Hence ψ∈D(Φ). Thereore D(Φαd+1)⊂D(αd+1Φ). Thus D(Φαd+1)=D(αd+1Φ). Hence the desired operator equality holds.
(
iii
)
This follows from Lemma2(iii).
(
iv
)
Simillar to the proof of part (ii).We define(16)HM≔H0+αd+1M.
If M is a constant operator m>0, then Hm represents the free Dirac operator with a constant massm. It is well known (e.g., [6, Theorem1.1]) that Hm is self-adjoint with D(Hm)=D(H0)=∩j=1dD(Dj) and bijective with ∥Hm-1∥=1/m.Lemma 5.
Assume(A.4). Let m>0 be a constant. Then MHm-1 is bounded with
(17)∥MHm-1∥≤max{a,bm}.Proof.
It is well known or easy to see that, for allψ∈D(Hm)=D(H0),
(18)∥Hmψ∥2=∥(-Δ)1/2ψ∥2+m2∥ψ∥2.
Hence, by(A.4), we have
(19)∥Mψ∥2≤a2∥Hmψ∥+(b2-a2m2)∥ψ∥.
This implies the following: (i) if am≤b, then ∥MHm-1∥2≤a2+(b2-a2m2)/m2=b2/m2; (ii) ifam≥b, then ∥MHm-1∥≤a. Thus (17) follows.Lemma 6.
Assume(A.1)–(A.4). Then (i)
U is self-adjoint with D(U)=D(M);(ii)
H is self-adjoint with D(H)=D(H0) and the subspace
(20)𝒟0≔C0∞(ℝd;ℂNd)⊗^𝒦,(⊗^means algebraic tensor product) is a core of H.Proof.
(i) SinceD(U)=D(M) and D(M) is dense, U is densely defined. Since αd+1eiΦ is bounded, it follows that U*=Me-iΦαd+1. By (A.1) and Lemma 2(iv), we have
(21)e-iΦαd+1=αd+1eiΦ.
By (A.2), (A.3), Lemma 2(ii), and Lemma 4(iv), we have
(22)U*=Mαd+1eiΦ=αd+1MeiΦ=αd+1eiΦM=U.
HenceUis self-adjoint.
(ii) By(A.4), we have for all ψ∈D((-Δ)1/2)(23)∥Uψ∥=∥Mψ∥≤a∥(-Δ)1/2ψ∥+b∥ψ∥.
Note that
(24)∥(-Δ)1/2ψ∥=∥H0ψ∥.
Hence ∥Uψ∥≤a∥H0ψ∥+b∥ψ∥. Here0≤a<1. Thus, by the Kato-Rellich theorem (e.g., [11, TheoremX.12]),His self-adjoint withD(H)=D(H0)and every core of H0 is a core of H. It is well known that the subspace C0∞(ℝd;ℂNd) is a core of H0 as a linear operator on L2(ℝd;ℂNd). Hence the subspace 𝒟0 defined by (20) is a core of H0 as a linear operator on ℋ. Thus it is a core of H too.Remark 7.
One of the other sufficient conditions forHto be essentially self-adjoint is as follows: assume(A.1)–(A.3)   andesssup|x|<R∥M(x)∥<∞  for allR>0. ThenHis essentially self-adjoint on𝒟0. The proof is similar to that of    [6, Theorem4.3].


## 4. Supersymmetric Aspects

As is well known, the standard free Dirac operator-i∑j=13‍αjDj+mβ on L2(ℝ3;ℂ4) with constant mass m≥0 and its suitably perturbed ones have supersymmetry; that is, they are, respectively, a supercharge with the grading operator iβγ5 [6, Section 5.5]. From this point of view, it would be interesting to investigate if the Hamiltonian H of the present model has supersymmetry. Indeed, it was shown that the Hamiltonian of the CQS model as well as that of the GCQS model has supersymmetry [3, 4]. In this section we see that a supersymmetric structure similar to that of the CQS (GCQS) model exists in our model.In this section, we consider only the case where
d
is odd. The matrix
(25)
γ
5
(
d
)
≔
i
d
(
d
-
1
)
/
2
α
1
⋯
α
d
is self-adjoint with
(26)(γ5(d))2=1Nd.
Sincedis odd, we have
(27)αjγ5(d)=γ5(d)αj(j=1,…,d),{αd+1,γ5(d)}=0.Letξ:ℝd→𝔅(𝒦) be Borel measurable such that, for a.e. x∈ℝd, ξ(x) is self-adjoint with
(28)ξ(x)2=I.
Then
(29)∥ξ(x)∥=1,a.e.x∈ℝd.We defineΓ(·):ℝd→𝔅(ℂNd⊗𝒦)by
(30)Γ(x)≔iγ5(d)αd+1⊗ξ(x),a.e.x∈ℝd.
Then
(31)Γ≔∫ℝd⊕‍Γ(x)dx
is self-adjoint with
(32)Γ2=I.
Hence Γ is a grading operator on ℋ. The following proposition shows that, under some additional condition for ξ(x), H has supersymmetry with respect to Γ.Proposition 8.
Letd be odd. Assume (A.1)–(A.4). Suppose that ξ is strongly differentiable on ℝd with
(33)Cj≔supx∈ℝd∥Djξ(x)∥<∞,j=1,…,d.
Then
(34)ΓH⊂-HΓ
if and only if
(35)∑j=1d‍γ5(d)αd+1αjDjξ(x)=i(γ5(d)⊗ξ(x)eiΦ(x)M(x)-M(x)e-iΦ(x)γ5(d)⊗ξ(x)),a.e.x∈ℝd.
In that case, the spectrum σ(H) and the point spectrum σp(H) of H are, respectively, symmetric with respect to the origin 0∈ℝ.Proof.
Since the subspace𝒟0 given by (20) is a core of H by Lemma 6(ii), (34) is equivalent to that, for all ψ∈𝒟0, Γψ∈D(H) and
(36)ΓHψ=-HΓψ.
Letψ∈𝒟0. Then
(37)(Γψ)(x)=iγ5(d)αd+1⊗ξ(x)ψ(x),a.e.x∈ℝd.
It follows that the ℂNd⊗𝒦-valued function: x↦(Γψ)(x) is strongly differentiable on ℝd with
(38)Dj(Γψ)(x)=iγ5(d)αd+1⊗(Djξ(x))ψ(x)+iγ5(d)αd+1⊗ξ(x)Djψ(x),j=1,…,d.
Hence
(39)∥Dj(Γψ)(x)∥2≤2(Cj2∥ψ(x)∥2+∥Djψ(x)∥2),
which implies that DjΓψ∈ℋ and hence Γψ∈D(H0)=D(H). Moreover, we have
(40)(H0Γψ)(x)=-γ5(d)αd+1(∑j=1d‍αjDjξ(x))ψ(x)-(ΓH0ψ)(x),(UΓψ)(x)=-(ΓUψ)(x)+i(γ5(d)⊗ξ(x)eiΦ(x)M(x)-M(x)e-iΦ(x)γ5(d)⊗ξ(x))ψ(x).
Hence
(41)(HΓψ)(x)=-(ΓHψ)(x)-∑j=1d‍γ5(d)αd+1αj⊗(Djξ(x))ψ(x)+i(γ5(d)⊗ξ(x)eiΦ(x)M(x)-M(x)e-iΦ(x)γ5(d)⊗ξ(x))ψ(x).
Therefore, HΓψ=-ΓHψ for all ψ∈𝒟0 if and only if
(42)∑j=1d‍γ5(d)αd+1αj⊗(Djξ(x))ψ(x)=i(γ5(d)⊗ξ(x)eiΦ(x)M(x)-M(x)e-iΦ(x)γ5(d)⊗ξ(x))ψ(x),∀ψ∈𝒟0.
By the original assumption for M(·), M(x)∈𝔅(ℂNd⊗𝒦) for a.e. x∈ℝd. Therefore (42) is equivalent to (35).
By (32) and Γ*=Γ, one easily sees that (34) is in fact equivalent to operator equality Γ*HΓ=-H. Hence H is unitarily equivalent to -H. This implies the symmetry of σ(H) and σp(H) with respect to the origin.Remark 9.
Proposition8 gives a generalization of [4, Theorem1] and clarifies a condition for H to have supersymmetry.
It may be difficult in general to show the existence of self-adjoint, unitary solutionsξ(x) to operator equation (35). Here we only note the following fact.Lemma 10.
Letdbe odd. Assume (A.1)–(A.3). Suppose that
(43)γ5(d)Φ(x)⊂Φ(x)γ5(d),a.e.x∈ℝd,j=1,…,d,ξ=ξ(x) is independent of x∈ℝd and
(44)(I⊗ξ)Φ(x)⊂-Φ(x)(I⊗ξ),(45)(γ5(d)⊗ξ)M(x)⊂M(x)(γ5(d)⊗ξ),a.e.x∈ℝd.
Then ξ is a solution to (35).Proof.
Sinceξ is a constant operator, Djξ=0. By Lemma 2(iii), (43) implies the strong commutativity of γ5(d) and Φ. Hence γ5(d)eiΦ(x)=eiΦ(x)γ5(d) for a.e. x∈ℝd. By (44) and Lemma 2(iv), (I⊗ξ)eiΦ(x)=e-iΦ(x)(I⊗ξ). We also have (45) and the strong commutativity of Φ and M. Hence
(46)(γ5(d)⊗ξ)eiΦ(x)M(x)=M(x)(γ5(d)⊗ξ)eiΦ(x)=M(x)e-iΦ(γ5(d)⊗ξ).
Thus (35) holds with both sides being zero.Additionally we make a remark on the converse of Lemma10. For this purpose, we need a lemma.Lemma 11.
LetTj (j=1,…,d) be a densely defined closed linear operator on 𝒦. Suppose that
(47)∑j=1d‍αj⊗Tj=0on∩j=1dD(αj⊗Tj).
Then, for all j=1,…,d, Tj=0 on ∩j=1dD(Tj).Proof.
Equation (47) implies that, for all u∈∩j=1dD(Tj) and v∈𝒦, ∑j=1d‍〈v,Tju〉αj=0. Since {αj}j=1d is linearly independent, it follows that 〈v,Tju〉=0,j=1,…,d. Hence Tju=0,j=1,…,d.The following lemma gives a sufficient condition for a solution to (35) to be a constant operator.Lemma 12.
Letd be odd. Assume (A.1)–(A.3). Let ξ(x) be strongly differentiable on ℝd with (33) and be a solution to (35). Suppose that (43)–(45) hold. Then ξ is independent of x∈ℝd.Proof.
As in the proof of Lemma10, (43)–(45) imply (46). Hence the right-hand side of (35) vanishes, so that ∑j=1d‍γ5(d)αd+1αj⊗Djξ(x)=0, which implies that ∑j=1d‍αj⊗Djξ(x)=0. By Lemma 11, Djξ(x)=0,j=1,…,d, which implies that ξ is independent of x.We have from Proposition8 and Lemma 10 the following result.Corollary 13.
Letd be odd. Assume (A.1)–(A.4). Suppose that ξ=ξ(x) is independent of x∈ℝd and that (43)–(45) hold. Then H has supersymmetry with respect to Γ.


## 5. Vanishing Theorems of the Kernel ofH

In supersymmetric quantum mechanics with a superchargeQ, a nonzero vector in kerQ is called a supersymmetric states. If the kernel of Q vanishes, that is, kerQ={0}, then the supersymmetry is said to be spontaneously broken. It turns out that, in supersymmetric quantum mechanics, it is importanat to investigate kerQ. Thus we are led to consider kerH in view of Proposition 8. This would be interesting even if H does not have supersymmetry (note that H does not necessarily have supersymmetry).To investigate kerH, we also need an additional condition.(A.5)
(i) For eachf∈ℂNd⊗𝒦, the function: x↦M(x)f is strongly differentiable onℝdand, for all x∈ℝd, M(x) commutes with αj (j=1,…,d).   (ii) There exists a constant μ0>0 such that
(48)M(x)2-i∑j=1d‍αjαd+1DjM(x)≥μ02,∀x∈ℝd
as an operator inequality on ℂNd⊗𝒦 (note that, by the principle of uniform boundedness, the strong partial derivative DjM(x) is a bounded operator on ℂNd⊗𝒦 for each x∈ℝd and hence, under (A.2) and condition (A.5)(i), the operator M(x)2-i∑j=1d‍αjαd+1DjM(x) on ℂNd⊗𝒦 is a bounded self-adjoint operator).For a linear operatorL on a Hilbert space, we denote the resolvent set of L by ρ(L).Lemma 14.
Assume(A.2), (A.4), and (A.5). Then HM defined by (16) is self-adjoint with D(HM)=D(H0) and
(49)∥H0ψ∥2+μ02∥ψ∥2≤∥HMψ∥2,ψ∈D(H0).
In particular, 0∈ρ(HM) with operator-norm bound
(50)∥HM-1∥≤1μ0
and H0HM-1 is bounded with
(51)∥H0HM-1∥≤1.
Moreover, MHM-1 is bounded with
(52)∥MHM-1∥≤a+bμ0.Proof.
The self-adjointness ofHM follows from that of H with Φ=0. For all ψ∈𝒟0, using the anticommutativity of αj with αd+1 and the commutativity of M(x) with αd+1 and αj (j=1,…,d), we have
(53)∥HMψ∥2=∥H0ψ∥2+∥Mψ∥2+∑j=1d‍〈ψ,(-iDjM)αjαd+1ψ〉≥∥H0ψ|2+∫ℝd‍〈ψ(x),(-i∑j=1d‍αjαd+1DjM(x)M(x)2-i∑j=1d‍αjαd+1DjM(x))ψ(x)〉dx≥∥H0ψ∥2+μ02∥ψ∥2.
Hence (49) holds for all ψ∈𝒟0. Since 𝒟0 is a core of HM, this inequality extends to all ψ∈D(H0). In particular, we have
(54)μ0∥ψ∥≤∥HMψ∥,ψ∈D(H0).
This implies that the self-adjoint operator HM is bijective with (50).
Inequality (49) implies also that, for all ψ∈D(H0), ∥H0ψ∥≤∥HMψ∥. Hence H0HM-1 is bounded with (51).
By(A.4)and ∥(-Δ)1/2ψ∥=∥H0ψ∥ for all ψ∈D((-Δ)1/2)=D(H0), we have ∥Mψ∥≤a∥H0ψ∥+b∥ψ∥. Hence, for allϕ∈ℋ,
(55)∥MHM-1ϕ∥≤a∥H0HM-1ϕ∥+b∥HM-1ϕ∥≤(a∥H0HM-1∥+b∥HM-1∥)∥ϕ∥≤(a+bμ0)∥ϕ∥.
Thus (52) holds.Lemma 15.
LetA be a self-adjoint operator on a complex Hilbert space 𝒳. Then
(56)∥eiA-1∥=2∥sinA2∥.Proof.
By the functional calculus, one haseiA-1=2ieiA/2sin(A/2). Hence ∥eiA-1∥=2∥eiA/2sin(A/2)∥. Since eiA/2 is unitary, one has ∥eiA/2sin(A/2)∥=∥sin(A/2)∥. Thus (56) holds.Theorem 16.
Assume(A.1)–(A.5) and
(57)esssupx∈ℝd∥sinΦ(x)2∥<12(a+bμ0-1).
Then ker H={0} and 0∈ρ(H).
Moreover, the constant(58)γ(H)≔infψ∈D(H),∥ψ∥=1∥Hψ∥
is strictly positive, γ(H)∈σ(H) or -γ(H)∈σ(H), and
(59)σ(H)⊂(-∞,-γ(H)]∪[γ(H),∞).Proof.
The operatorH is written as
(60)H=HM+αd+1(eiΦ-1)M=KHM
with
(61)K≔I+αd+1(eiΦ-1)MHM-1.
By applying Lemma 15 with A=Φ(x), we have
(62)∥eiΦ(x)-1∥=2∥sinΦ(x)2∥.
Therefore, for all ψ∈ℋ,
(63)∥αd+1(eiΦ-1)MHM-1ψ∥≤2esssupx∈ℝd∥sin(Φ(x)2)∥∥MHM-1∥∥ψ∥.
By this estimate and (52), we obtain
(64)∥αd+1(eiΦ-1)MHM-1∥≤2(esssupx∈ℝd∥sin(Φ(x)2)∥)(a+bμ0-1).
Hence, by (57), we obtain ∥αd+1(eiΦ-1)MHM-1∥<1. This implies that K is bijective with bounded inverse K-1. Thus H is bijective with H-1=HM-1K-1 being bounded. Hence kerH={0} and 0∈ρ(H).
We setbH≔∥H-1∥. If |λ|<1/bH, then λ is in ρ(H). Therefore
(65)σ(H)⊂(-∞,-bH-1]∪[bH-1,∞).
It is obvious that, for all ψ∈D(H) with ∥ψ∥=1, 1≤bH∥Hψ∥. This implies that bHγ(H)≥1. On the other hand, we have from (58) ∥ψ∥≥γ(H)∥H-1ψ∥, for all ψ∈ℋ. Hence bHγ(H)≤1. Therefore bH-1=γ(H). Thus (59) holds and γ(H)>0. Since bH∈σ(H-1) or -bH∈σ(H-1), it follows that γ(H)=bH-1∈σ(H)or-γ(H)∈σ(H).Remark 17.
Under the same assumption as in Theorem16,His Fredholm (the proof is easy).We next consider a perturbation ofΦ(·). Let η(·)∈ℱs.a.such that, for a.e. x∈ℝd, η(x) is bounded and strongly commutes with Φ(x) and M(x). Then, for a.e. x∈ℝd,
(66)Φη(x)≔Φ(x)+η(x)
is self-adjoint on ℂNd⊗𝒦 and
(67)Φη≔∫ℝd⊕‍Φη(x)dx
is a self-adjoint operator on ℋ.The quantity(68)κ(H)≔supψ∈D(H),∥ψ∥=1∥Mψ∥∥Hψ∥,
may be infinite. But we have the following.Lemma 18.
Under the assumption of Theorem16, 0<κ(H)<∞.Proof.
SinceHis closed with D(H)=D(H0)∩D(U)(=D(H0)) and ∥Uψ∥=∥Mψ∥,ψ∈D(H), it follows from the closed graph theorem that there exists a constant c>0 such that
(69)∥Mψ∥≤c(∥Hψ∥+∥ψ∥),ψ∈D(H).
Let ψ∈D(H) with ∥ψ∥=1. Then, by Theorem 16, we have ∥Hψ∥≥γ(H)>0. Hence
(70)∥Mψ∥∥Hψ∥≤c+cγ(H).
Therefore κ(H)≤c+c/γ(H)<∞. If κ(H)=0, then ∥Mψ∥=0 for all ψ∈D(M)=D(H0). Hence M=0. But this contradicts condition (A.5).Theorem 19.
Assume(A.1)–(A.5) and (57). Suppose that
(71)esssupx∈ℝd∥sinη(x)2∥<12κ(H).
Let
(72)Hη≔H0+αd+1eiΦηM.
Then ker  Hη={0} and 0∈ρ(Hη). Moreover, the last statement on γ(H) and σ(H) in Theorem 16 holds with H being replaced by Hη.Proof.
We write(73)Hη=H+W,W≔αd+1(eiΦη-eiΦ)M.
By the strong commutativity of Φ(x) and η(x), we have for a.e. x∈ℝd(74)eiΦη(x)-eiΦ(x)=eiΦ(x)(eiη(x)-1)=2ieiΦ(x)eiη(x)/2sin(η(x)2).
Hence, for all ψ∈D(H0)(75)∥Wψ∥≤2esssupx∈ℝd∥sinη(x)2∥∥Mψ∥.
We have ∥Mψ∥≤κ(H)∥Hψ∥. Hence
(76)∥Wψ∥≤Cη∥Hψ∥
with
(77)Cη≔2κ(H)esssupx∈ℝd∥sinη(x)2∥.
Hence W is H-bounded. By Remark 17, H is Fredholm. Condition (71) is equivalent to Cη<1. Hence, by a stability theorem (e.g., [12, ChapterIV, Theorem5.22]), Hη is Fredholm and dim ker Hη≤ dim ker H=0. Therefore ker Hη={0}. It follows from this fact and the self-adjointness of Hη that Ran (Hη)=ℋ. Hence 0∈ρ(Hη). Then the last statement of the present theorem can be proved in the same way as in the proof of the corresponding part in Theorem 16.


## 6. Unitary Equivalence to a Gauge Theoretic Dirac Operator and a Vanishing Theorem forkerH

In the papers [3, 4], it was shown that, under a suitable condition, the Hamiltonian of the CQS (GCQS) model is unitarily transformed to a Dirac operator which is simpler in a sense. In this section, we show that those structures are unified into a simple general structure.We introduce a class ofΦ(·):
(78)ℱ≔{supx∈ℝd∥Ej(x)∥<∞,j=1,…,dΦ(·)∈ℱs.a.∣e±iΦ(·)/2isstronglydifferentiableandsupx∈ℝd∥Ej(x)∥<∞,j=1,…,d},
where
(79)Ej(x)≔Dje-iΦ(x)/2
denotes the strong partial derivative of e-iΦ(x)/2 in xj. For Φ(·)∈ℱ, one can define a bounded linear operator
(80)Aj≔i∫ℝd⊕‍eiΦ(x)/2Ej(x)dx
on ℋ.Remark 20.
IfΦ(·)∈ℱ sucht that Φ(x) and Φ(x′) commute for a.e. x,x′∈ℝd, then Ej(x)=-ie-iΦ(x)/2DjΦ(x)/2 and hence
(81)Aj=12∫ℝd⊕‍DjΦ(x)dx.Lemma 21.
For eachj=1,…,d, Aj is a bounded self-adjoint operator on ℋ.Proof.
Sinceesssupx∈ℝd∥eiΦ(x)/2Ej(x)∥=esssupx∈ℝd∥Ej(x)∥<∞,Aj is bounded. We have
(82)Aj*=(-i)2∫ℝd⊕‍(DjeiΦ(x)/2)e-iΦ(x)/2dx.
Differentiating the identity eiΦ(x)/2e-iΦ(x)/2=I in xj, we have
(83)eiΦ(x)/2Ej(x)=-(DjeΦ(x)/2)e-iΦ(x)/2.
Hence Aj*=Aj.ForΦ(·)∈ℱ, we define an operator:
(84)H′≔∑j=1d‍αj(-iDj-Aj)+αd+1M=HM-∑j=1d‍αjAj.Lemma 22.
Assume(A.4). Let Φ(·)∈ℱ. Suppose that
(85)αjΦ⊂Φαj,j=1,…,d.
Then H′ is self-adjoint and every core of H0 is a core of H′.Proof.
Under condition(A.4), HM is self-adjoint. By (85), we have αjAj=Ajαj (j=1,…,d). Hence, by Lemma 21, -∑j=1d‍αjAj is a bounded self-adjoint operator. Hence the Kato-Rellich theorem yields the desired result.We note that, if one regardsA≔(A1,…,Ad) as a (noncommutative) gauge potential, then H′ is a gauge theoretic Dirac operator with gauge potential A.Let(86)U≔eiΦ/2,
which is unitary. The following theorem shows that, under a suitable condition, H is unitarily equivalent to a gauge theoretic Dirac operator H′.Theorem 23.
Assume(A.1)–(A.4) and (85). Let Φ(·)∈ℱ. Then
(87)UHU-1=H′.Proof.
We have(88)UHU-1=-i∑j=1d‍(UαjU-1)UDjU-1+Uαd+1eiΦU-1(UMU-1).
By (85) and Lemma 2, UαjU-1=αj. By(A.3)and Lemma 2, UMU-1=M. By(A.1)and Lemma 2(iv), Uαd+1eiΦU-1=αd+1e-iΦ/2eiΦeiΦ/2=αd+1. Moreover,
(89)UDjU-1=Dj-iAj.
Hence (87) holds.The following theorem gives another sufficient condition forkerH to be trivial.Theorem 24.
Assume(A.1)–(A.4) and (85). Let Φ(·)∈ℱ and
(90)∑j=1d‍ess
supx∈ℝd∥Ej(x)∥<μ0.
Then kerH={0}and 0∈ρ(H).Proof.
We writeH′=HM+X with X≔-∑j=1d‍αjAj. Then
(91)∥X∥≤∑j=1d‍∥αjAj∥≤∑j=1d‍esssupx∈ℝd∥Ej(x)∥.
By this estimate and (90), ∥XHM-1∥<1. Hence H′ is bijective and 0∈ρ(H′). In particular, ker H′={0}. On the other hand, (87) implies thatρ(H′)=ρ(H)and ker H=U-1 ker H′. Thus 0∈ρ(H) and ker H={0}.


## 7. Essential Spectrum ofH

In this section, we consider the essential spectrum ofH. For a self-adjoint operator S on a Hilbert space, we denote by σess(S) the essential spectrum of S.Lemma 25.
Letdim𝒦<∞ and m>0 be a constant. Let V(·):ℝd→𝔅(ℂNd⊗𝒦) be Borel measurable satisfying the following conditions.(i)
The operatorV≔∫ℝd⊕‍V(x)dx is relatively bounded with respect toH0.(ii)
lim|x|→∞∥V(x)∥=0.(iii)
The operatorHm+V on ℋ is self-adjoint.Then
(92)σess(Hm+V)=(-∞,-m]∪[m,∞).Proof.
For eachR>0, we denote by χR the characteristic function of the set {x∈ℝd∣|x|<R}. As in the case of the 3-dimensional free Dirac operator (cf. [6, Lemma4.6]), one can show that |Hm|-kχR is compact for all k>0 as an operator on L2(ℝd;ℂNd). Sincedim𝒦<∞, it follows that|Hm|-kχRis compact as an operator onℋ. Hence, for all z∈ℂ∖ℝ, (Hm-z)-1χR=((Hm-z)-1|Hm|)|Hm|-1χR is compact. Since Hm+V is self-adjoint with D(Hm+V)=D(Hm)=D(H0) and V is closed, it follows from the closed graph theorem that V(Hm+V-z)-1 is bounded (this can be shown by direct estimates too). Therefore we have
(93)(Hm+V-z)-1-(Hm-z)-1=-(Hm-z)-1V(Hm+V-z)-1=-WR-XR,
whereWR≔(Hm-z)-1χR[V(Hm+V-z)-1], XR=(Hm-z)-1(1-χR)V(Hm+V-z)-1. By the fact mentioned previously, WR is compact. We have
(94)∥XR∥≤1|Imz|2∥(1-χR)V∥.
By condition (ii), for every ε>0, there exists a constant R>0 such that, for all a.e. x∈ℝd with |x|≥R, ∥V(x)∥<ε, that is, esssup|x|≥R∥V(x)∥≤ε, which implies that ∥(1-χR)V∥≤ε. Hence limR→∞∥XR∥=0. Therefore (Hm+V-z)-1-(Hm-z)-1 is compact. Hence, by Weyl’s essential spectrum theorem (e.g., [10, TheoremXIII.14]), σess(Hm+V)=σess(Hm). On the other hand, as in the case of the 3-dimensional free Dirac operator [6, Theorem1.1], one can show that σ(Hm)=σess(Hm)=(-∞,-m]∪[m,∞). Thus (92) holds.Theorem 26.
Letdim𝒦<∞. Assume (A.1)–(A.4). Suppose that there exists a constant m∈ℝ satisfying
(95)lim|x|→∞∥M(x)-m∥=0,(96)lim|x|→∞∥sinΦ(x)2∥=0.
Then
(97)σess(H)=(-∞,-m]∪[m,∞).Proof.
We write(98)H=Hm+V1+V2
with
(99)V1≔αd+1(M-m),V2≔αd+1(eiΦ-1)M.
It is obvious that V1 and V2 are relatively bounded with respect to H0 and
(100)lim|x|→∞∥V1(x)∥=lim|x|→∞∥M(x)-m∥=0.
As for V2, we have
(101)∥V2(x)∥≤∥M(x)∥∥eiΦ(x)-1∥≤2∥M(x)∥∥sinΦ(x)2∥.
Hence, by (95) and (96), we have lim|x|→∞∥V2(X)∥=0. Therefore lim|x|→∞∥V1(x)+V2(x)∥=0. Thus we can apply Lemma 25 to obtain (97).IfΦ(·) is in the class ℱ introduced in Section 6, then we can obtain a sufficient condition for (97) to hold.Theorem 27.
Letdim𝒦<∞. Assume (A.1)–(A.4) (85) and (95). Let Φ(·)∈ℱ. Suppose that
(102)lim|x|→∞∥Ej(x)∥=0.
Then (97) holds.Proof.
By (87), we have σess(H)=σess(H′). Hence we need only to prove
(103)σess(H′)=(-∞,-m]∪[m,∞).
We write
(104)H′=Hm+αd+1(M-m)-∑j=1d‍αjAj.
We have lim|x|→∞∥αd+1(M(x)-m)∥=0. Moreover, ∥αjAj(x)∥≤∥Ej(x)∥. Hence lim|x|→∞∥-∑j=1d‍αjAj(x)∥=0. Thus we can apply Lemma 25 to obtain (103).


## 8. Bounds on the Number of Discrete Eigenvalues

In this section, in view of Theorem26, we consider the number of eigenvalues ofHin the interval (-m,m) and establish upper bounds on it. This aspect has been considered in the CQS model [3] as well as the GCQS model [4]. In this paper, we take another method, which is an extension of the method used in [13] where the number of eigenvalues of the three-dimensional Dirac operator Hm+V with a scalar potential V:ℝ3→ℝ in (-m,m) is considered. This extension is not difficult. But, for the sake of completeness, we present some details of it. One easily notes that the problem under consideration can be studied in a more general frame work as in Lemma 25. Hence we first discuss the general case.


### 8.1. A General Case

LetV be as in Lemma 25 and
(105)H(V)≔Hm+V.
Then, by (92), an eigenvalue of H(V) in (-m,m) (if it exists) is an isolated eigenvalue of H(V) with finite multiplicity. For each λ∈(0,m2), we denote by N(λ,V) the number of eigenvalues in the interval (-m2-λ,m2-λ).We first note an elementary fact:Theorem 28.
Suppose that the assumption of Lemma25 holds and that ∥V(x)∥≤λ/4m for a.e. x∈ℝd. Then N(λ,V)=0.Proof.
Suppose thatN(λ,V)≥1. Then, it follows from the definition of N≔N(λ,V) that there exists anN-dimensional subspace E of ℋ such that
(106)∥H(V)ψ∥≤m2-λ∥ψ∥,∀ψ∈E.
Hence
(107)∥Hmψ∥≤∥H(V)ψ∥+∥Vψ∥≤(m2-λ+λ4m)∥ψ∥≤m2-λ2∥ψ∥.
Hence ∥Hmψ∥2≤(m2-(λ/2))∥ψ∥2, which is equivalent to ∥(-Δ+(λ/2))1/2ψ∥2≤0. This implies that ψ=0. But this is a contradiction.In view of Theorem28, we define, for each λ>0, Vλ:ℝd→𝔅(ℂNd⊗𝒦) by
(108)Vλ(x)≔{V(x)if∥V(x)∥>λ4m,0otherwise.For eachλ>0, the operator
(109)Rλ≔(-Δ+λ2)-1/2
is a bounded self-adjoint operator. Since V is H0-bounded, where H0 is defined by (10) and (H0+i)Rλ is bounded, it follows that VRλ and VλRλ are bounded operators on ℋ. Also H0Rλ is bounded with ∥H0Rλ∥≤1. Hence the following operators Tλj (j=1,2,3,5) are in 𝔅(ℋ):
(110)Tλ1≔(H0Rλ)*VλRλ,Tλ2≔(VλRλ)*H0Rλ,Tλ3≔mαd+1RλVλRλ,Tλ4≔mRλVλRλαd+1,Tλ5≔(VλRλ)*VλRλ.
We set
(111)v(x)≔∥V(x)∥,vλ(x)≔∥Vλ(x)∥,a.e.x∈ℝd.For a compact operatorA on a Hilbert space, we denote the nonincreasing sequence of the singular values of A (repeated with multiplicity) by μn(A) (n∈ℕ). For f∈Lp(ℝd), we set ∥f∥Lp≔(∫ℝd‍|f(x)|pdx)1/p.Lemma 29.
Letd≥3 and suppose that the assumption of Lemma 25 holds and v∈Ld(ℝd)∩Ld/2(ℝd). Then, for all j=1,2,3,4,5, Tλjis compact. Moreover, there exists a constant C>0 independent of V and λ>0 such that, for all n∈ℕ,
(112)μn(Tλj)≤C∥vλ∥Ldn-1/d(j=1,2),μn(Tλj)≤C∥vλ1/2∥Ld2n-2/d(j=3,4),μn(Tλ5)≤C∥vλ∥Ld2n-2/d.Proof.
By the weak Hausdorff-Young inequality (e.g., [11, page 32]) and the condition d≥3, one can easily see that the Fourier transform gλ of the function: ℝd∋k↦(k2+λ/2)-1/2 is in Lwp′(ℝd) (the weak Lp′ space on ℝd) with 1/p′=1-1/d and ∥gλ∥p′,w≤cd, where ∥·∥p′,w denotes the “pseudo” norm of Lwp′(ℝd) and cd is a constant independent of λ>0. By Cwikel’s theorem [14, Section 3] and the condition v∈Ld(ℝd), which implies thatvλ∈Ld(ℝd),vλRλis compact as an operator on L2(ℝd) and
(113)μn(vλRλ)≤K1∥vλ∥Ldn-1/d,n∈ℕ,
where K1>0 is a constant independent of V, λ>0 and n∈ℕ. Since dim𝒦<∞, it follows that vλRλ is compact also as an operator on ℋ. Let
(114)Bλ(x)≔{V(x)v(x)ifv(x)>λ4m,0otherwise.
Then Bλ is bounded with ∥Bλ(x)∥≤1. We have VλRλ=BλvλRλ. Hence VλRλ is compact. This shows that all Tλj (j=1,2,3,4,5) are compact.
In general, for all compact operatorsA and bounded operators B on a Hilbert space
(115)μn(BA)≤∥B∥μn(A).
(e.g., see [15, Theorem1.6].). Hence
(116)μn(VλRλ)≤∥Bλ∥μn(vλRλ)≤K1∥vλ∥Ldn-1/d.
Therefore
(117)μn(Tλ1)≤∥H0Rλ∥K1∥v∥Ldn-1/d≤K1∥vλ∥Ldn-1/d.
Similarly one can show that Tλ2, is compact and
(118)μn(Tλ2)≤K1∥vλ∥Ldn-1/d,
where we have use the fact that μn(A)=μn(A*) for all compact operators on a Hilbert space [15, (1.3)].
As forTλ3, we write
(119)Tλ3=mαd+1Rλvλ1/2Bλvλ1/2Rλ.
By the condition v∈Ld/2(ℝd), vλ1/2∈Ld(ℝd). Hence, Cwikel’s theorem again, vλ1/2Rλ is compact and
(120)μn(vλ1/2Rλ)≤K1′∥vλ1/2∥Ldn-1/d,
where K1′>0 is a constant independent of V and λ>0. We have
(121)μn(Tλ3)≤mμn(Rλvλ1/2Bλvλ1/2Rλ).
In general, for all compact operators A and bounded operators D on a Hilbert space,
(122)μ2n+1(A*DA)≤∥D∥μn+1(A)2,μ2n(A*DA)≤∥D∥μn(A)2,
where we have used the fact that, for all compact operators A and B on a Hilbert space,
(123)μn+k+1(AB)≤μn+1(A)μk+1(B),n,k≥0.
Hence
(124)μ2n+1(Tλ3)≤mμn+1(vλ1/2Rλ)2≤m(K1′)2∥vλ1/2∥Ld2(n+1)-2/d,μ2n(Tλ3)≤mμn(vλ1/2Rλ)2≤m(K1′)2∥vλ1/2∥Ld2n-2/d,
which imply that
(125)μn(Tλ3)≤K2′∥vλ1/2∥Ld2n-2/d,n∈ℕ,
where K2′>0 is a constant independent of V,λ, and n. Similarly we have
(126)μn(Tλ4)≤K2′∥vλ1/2∥Ld2n-2/d,μn(Tλ5)≤K3′∥vλ∥Ld2n-2/d,n∈ℕ,
where K3′>0 is a constant independent of V,λ and n. Thus the desired results follow.Theorem 30.
Letd≥3, and suppose that the assumption of Lemma 25 holds and ∥V(·)∥∈Ld(ℝd)∩Ld/2(ℝd). Let λ∈(0,m2). Then, there exists a constant C0>0 independent of V and λ such that
(127)N(λ,V)≤C0∫∥V(x)∥>λ/4m‍(∥V(x)∥d/2+∥V(x)∥d)dx.Proof.
We need only to consider the case whereN≔N(λ,V)≥1. Then there exists anN-dimensional subspaceEofℋsuch that (106) holds for allψ∈E. It is easy to see that ∥(Vλ-V)ϕ∥≤(λ/4m)∥ϕ∥, for all ϕ∈ℋ. Let ψ∈E. Then, as in the proof of Theorem 28, we have ∥(Hm+Vλ)ψ∥2≤(m2-(λ/2))∥ψ∥2, which is equivalent to the following inequality:
(128)∥(-Δ+λ2)1/2ψ∥2+〈H0ψ,Vλψ〉+〈Vλψ,H0ψ〉+m〈αd+1ψ,Vλψ〉+m〈Vλψ,αd+1ψ〉+∥Vλψ∥2≤0.
The subspace F≔(-Δ+λ/2)1/2  E is also N-dimensional. Inequality (128) implies that, for all ϕ∈F,
(129)∥ϕ∥2≤〈ϕ,Tλϕ〉,
where
(130)Tλ≔-∑j=15‍Tλj.
By Lemma 29, Tλ is a compact self-adjoint operator on ℋ. Hence, by the Hilbert-Schmidt theorem, there exists a complete orthonormal system {ϕn}n=1∞ of ℋ and a real sequence {tn}n=1∞ such that Tλϕn=tnϕn and limn→∞   tn=0. Using this fact, one sees that the number of eigenvalues tn of Tλ with tn≥1 is more than or equal to dimF=N. Hence μN(Tλ)≥1. Letkbe the largest natural number not exceeding (N+4)/5. Then 5k-4≤N. Hence 1≤μN(Tλ)≤μ5k-4(Tλ). On the other hand, by a general fact on singular values of the sum of two compact operators (e.g., [15, Theorem1.7]), we have
(131)μ5k-4(Tλ)≤∑j=15‍μk(Tλj).
Using this fact and Lemma 29, we obtain
(132)1≤2C∥vλ∥Ldk-1/d+2C∥vλ1/2∥Ld2k-2/d+C∥vλ∥Ld2k-2/d.
We have k≥N/5. Hence
(133)1≤C′(∥vλ∥LdN-1/d+∥vλ1/2∥Ld2N-2/d+∥vλ∥Ld2N-2/d),
where C′>0 is a constant independent of V,λ, and N. This implies that N≤C0(∥vλ1/2∥Ldd+∥vλ∥Ldd) with a constant C0 independent of V and λ. Thus (127) holds.As in Corollaries 1.2 and  1.3 in [13], we have from Theorem 30 the following results.Corollary 31.
Under the same assumption as in Theorem30, the number N(V) of eigenvalues of H(V) in (-m,m) is finite and
(134)N(V)≤C0∫ℝd‍(∥V(x)∥d/2+∥V(x)∥d)dx.Corollary 32.
Suppose that the assumption of Theorem30 holds. Let λj(j=1,…,N(V)) be the eigenvalues of H(V) in (-m,m), counted with multiplicity, and let γ>0 be such that
(135)fγ(V)≔∫ℝd‍∥V(x)∥γ(∥V(x)∥d/2+∥V(x)∥d)dx<∞.
Then, there exists a constant Cγ>0 such that
(136)∑j=1N(V)‍(m2-λj2)γ≤Cγfγ(V).


### 8.2. Applications

Now we apply the results in the preceeding section to the Dirac operatorH. For λ∈(0,m2), we denote by N(λ) the number of eigenvalues of H in (-m2-λ,m2-λ).Theorem 33.
Letd≥3 and λ∈(0,m2). Suppose that the assumption of Theorem 26 holds. Let
(137)FM,Φ(x)≔∥M(x)-m∥+2m∥sinΦ(x)2∥,a.e.x∈ℝd.(i)
IfFM,Φ(x)≤λ/4m,a.e.x∈ℝd, then N(λ)=0.(ii)
Suppose thatFM,Φ∈Ld/2(ℝd)∩Ld(ℝd). Then there exists a positive constant C>0 independent of M,Φ, and λ such that
(138)N(λ)≤C∫FM,Φ(x)>λ/4m‍(FM,Φ(x)d/2+FM,Φ(x)d)dx<∞.
Moreover, the numberN0 of eigenvalues of H in (-m,m) obeys
(139)N0≤C∫ℝd‍(FM,Φ(x)d/2+FM,Φ(x)d)dx<∞.Proof.
(i) We can writeH=H(V) with V=αd+1(MeiΦ-m). Hence
(140)∥V(x)∥=∥M(x)eiΦ(x)-m∥≤∥M(x)-m∥+m∥eiΦ(x)-1∥=FM,Φ(x).
Hence, the present assumption implies that ∥V(x)∥≤λ/4m a.e. x∈ℝd. Hence, by Theorem 28, N(λ)=0.
(ii) By (140) and the present assumption, ∥V(·)∥∈Ld(ℝd)∩Ld/2(ℝd). Thus we can apply Theorem 30 to obtain (138). Inequality (139) follows from (138) or Corollary 31.We have from Corollary32 the following fact.Corollary 34.
Letd≥3. Suppose that the assumption of Theorem 26 and FM,Φ∈Ld/2(ℝd)∩Ld(ℝd). Let λj(j=1,…,N0) be the eigenvalues ofHin (-m,m), counted with multiplicity, and let γ>0 be such that
(141)fγ(M,Φ)≔∫ℝd‍FM,Φ(x)γ(FM,Φ(x)d/2+FM,Φ(x)d)dx<∞.
Then, there exists a constant Cγ>0 such that
(142)∑j=1N0‍(m2-λj2)γ≤Cγfγ(M,Φ).We can also use Theorems23 and 27 to obtain another upper bound for N(λ). Let
(143)GM,Φ(x)≔∥M(x)-m-∑j=1d‍αd+1αjeiΦ(x)/2Ej(x)∥,a.e.x∈ℝd.Theorem 35.
Letd≥3, and let λ∈(0,m2). Suppose that the assumption of Theorem 27 holds. Then one has the following. (i)
IfGM,ϕ(x)≤λ/4m for a.e. x∈ℝd, then N(λ)=0.(ii)
Suppose thatGM,Φ∈Ld/2(ℝd)∩Ld(ℝd). Then (138) and (139) with FM,Φ replaced by GM,Φ hold.Proof.
By Theorem23, N(λ) is equal to the number of eigenvalues of H′ in (-m2-λ, m2-λ). One can write H′=Hm+V with V≔αd+1(M-m-∑j=1d‍αd+1αjAj). We have ∥V(x)∥=GM,Φ(x). Thus, in the same way as in the proof of Theorem 30, we obtain the desired results.Theorem35 implies the following result as in Corollary 34.Corollary 36.
Letd≥3. Suppose that the assumption of Theorem 27 holds and GM,Φ∈Ld/2(ℝd)∩Ld(ℝd). Then (142) with FM,Φ replaced by GM,Φ holds for all λ>0 such that ∫ℝd‍GM,Φ(x)γ(GM,Φ(x)d/2+GM,Φ(x)d)dx<∞.


## 8.1. A General Case

LetV be as in Lemma 25 and
(105)H(V)≔Hm+V.
Then, by (92), an eigenvalue of H(V) in (-m,m) (if it exists) is an isolated eigenvalue of H(V) with finite multiplicity. For each λ∈(0,m2), we denote by N(λ,V) the number of eigenvalues in the interval (-m2-λ,m2-λ).We first note an elementary fact:Theorem 28.
Suppose that the assumption of Lemma25 holds and that ∥V(x)∥≤λ/4m for a.e. x∈ℝd. Then N(λ,V)=0.Proof.
Suppose thatN(λ,V)≥1. Then, it follows from the definition of N≔N(λ,V) that there exists anN-dimensional subspace E of ℋ such that
(106)∥H(V)ψ∥≤m2-λ∥ψ∥,∀ψ∈E.
Hence
(107)∥Hmψ∥≤∥H(V)ψ∥+∥Vψ∥≤(m2-λ+λ4m)∥ψ∥≤m2-λ2∥ψ∥.
Hence ∥Hmψ∥2≤(m2-(λ/2))∥ψ∥2, which is equivalent to ∥(-Δ+(λ/2))1/2ψ∥2≤0. This implies that ψ=0. But this is a contradiction.In view of Theorem28, we define, for each λ>0, Vλ:ℝd→𝔅(ℂNd⊗𝒦) by
(108)Vλ(x)≔{V(x)if∥V(x)∥>λ4m,0otherwise.For eachλ>0, the operator
(109)Rλ≔(-Δ+λ2)-1/2
is a bounded self-adjoint operator. Since V is H0-bounded, where H0 is defined by (10) and (H0+i)Rλ is bounded, it follows that VRλ and VλRλ are bounded operators on ℋ. Also H0Rλ is bounded with ∥H0Rλ∥≤1. Hence the following operators Tλj (j=1,2,3,5) are in 𝔅(ℋ):
(110)Tλ1≔(H0Rλ)*VλRλ,Tλ2≔(VλRλ)*H0Rλ,Tλ3≔mαd+1RλVλRλ,Tλ4≔mRλVλRλαd+1,Tλ5≔(VλRλ)*VλRλ.
We set
(111)v(x)≔∥V(x)∥,vλ(x)≔∥Vλ(x)∥,a.e.x∈ℝd.For a compact operatorA on a Hilbert space, we denote the nonincreasing sequence of the singular values of A (repeated with multiplicity) by μn(A) (n∈ℕ). For f∈Lp(ℝd), we set ∥f∥Lp≔(∫ℝd‍|f(x)|pdx)1/p.Lemma 29.
Letd≥3 and suppose that the assumption of Lemma 25 holds and v∈Ld(ℝd)∩Ld/2(ℝd). Then, for all j=1,2,3,4,5, Tλjis compact. Moreover, there exists a constant C>0 independent of V and λ>0 such that, for all n∈ℕ,
(112)μn(Tλj)≤C∥vλ∥Ldn-1/d(j=1,2),μn(Tλj)≤C∥vλ1/2∥Ld2n-2/d(j=3,4),μn(Tλ5)≤C∥vλ∥Ld2n-2/d.Proof.
By the weak Hausdorff-Young inequality (e.g., [11, page 32]) and the condition d≥3, one can easily see that the Fourier transform gλ of the function: ℝd∋k↦(k2+λ/2)-1/2 is in Lwp′(ℝd) (the weak Lp′ space on ℝd) with 1/p′=1-1/d and ∥gλ∥p′,w≤cd, where ∥·∥p′,w denotes the “pseudo” norm of Lwp′(ℝd) and cd is a constant independent of λ>0. By Cwikel’s theorem [14, Section 3] and the condition v∈Ld(ℝd), which implies thatvλ∈Ld(ℝd),vλRλis compact as an operator on L2(ℝd) and
(113)μn(vλRλ)≤K1∥vλ∥Ldn-1/d,n∈ℕ,
where K1>0 is a constant independent of V, λ>0 and n∈ℕ. Since dim𝒦<∞, it follows that vλRλ is compact also as an operator on ℋ. Let
(114)Bλ(x)≔{V(x)v(x)ifv(x)>λ4m,0otherwise.
Then Bλ is bounded with ∥Bλ(x)∥≤1. We have VλRλ=BλvλRλ. Hence VλRλ is compact. This shows that all Tλj (j=1,2,3,4,5) are compact.
In general, for all compact operatorsA and bounded operators B on a Hilbert space
(115)μn(BA)≤∥B∥μn(A).
(e.g., see [15, Theorem1.6].). Hence
(116)μn(VλRλ)≤∥Bλ∥μn(vλRλ)≤K1∥vλ∥Ldn-1/d.
Therefore
(117)μn(Tλ1)≤∥H0Rλ∥K1∥v∥Ldn-1/d≤K1∥vλ∥Ldn-1/d.
Similarly one can show that Tλ2, is compact and
(118)μn(Tλ2)≤K1∥vλ∥Ldn-1/d,
where we have use the fact that μn(A)=μn(A*) for all compact operators on a Hilbert space [15, (1.3)].
As forTλ3, we write
(119)Tλ3=mαd+1Rλvλ1/2Bλvλ1/2Rλ.
By the condition v∈Ld/2(ℝd), vλ1/2∈Ld(ℝd). Hence, Cwikel’s theorem again, vλ1/2Rλ is compact and
(120)μn(vλ1/2Rλ)≤K1′∥vλ1/2∥Ldn-1/d,
where K1′>0 is a constant independent of V and λ>0. We have
(121)μn(Tλ3)≤mμn(Rλvλ1/2Bλvλ1/2Rλ).
In general, for all compact operators A and bounded operators D on a Hilbert space,
(122)μ2n+1(A*DA)≤∥D∥μn+1(A)2,μ2n(A*DA)≤∥D∥μn(A)2,
where we have used the fact that, for all compact operators A and B on a Hilbert space,
(123)μn+k+1(AB)≤μn+1(A)μk+1(B),n,k≥0.
Hence
(124)μ2n+1(Tλ3)≤mμn+1(vλ1/2Rλ)2≤m(K1′)2∥vλ1/2∥Ld2(n+1)-2/d,μ2n(Tλ3)≤mμn(vλ1/2Rλ)2≤m(K1′)2∥vλ1/2∥Ld2n-2/d,
which imply that
(125)μn(Tλ3)≤K2′∥vλ1/2∥Ld2n-2/d,n∈ℕ,
where K2′>0 is a constant independent of V,λ, and n. Similarly we have
(126)μn(Tλ4)≤K2′∥vλ1/2∥Ld2n-2/d,μn(Tλ5)≤K3′∥vλ∥Ld2n-2/d,n∈ℕ,
where K3′>0 is a constant independent of V,λ and n. Thus the desired results follow.Theorem 30.
Letd≥3, and suppose that the assumption of Lemma 25 holds and ∥V(·)∥∈Ld(ℝd)∩Ld/2(ℝd). Let λ∈(0,m2). Then, there exists a constant C0>0 independent of V and λ such that
(127)N(λ,V)≤C0∫∥V(x)∥>λ/4m‍(∥V(x)∥d/2+∥V(x)∥d)dx.Proof.
We need only to consider the case whereN≔N(λ,V)≥1. Then there exists anN-dimensional subspaceEofℋsuch that (106) holds for allψ∈E. It is easy to see that ∥(Vλ-V)ϕ∥≤(λ/4m)∥ϕ∥, for all ϕ∈ℋ. Let ψ∈E. Then, as in the proof of Theorem 28, we have ∥(Hm+Vλ)ψ∥2≤(m2-(λ/2))∥ψ∥2, which is equivalent to the following inequality:
(128)∥(-Δ+λ2)1/2ψ∥2+〈H0ψ,Vλψ〉+〈Vλψ,H0ψ〉+m〈αd+1ψ,Vλψ〉+m〈Vλψ,αd+1ψ〉+∥Vλψ∥2≤0.
The subspace F≔(-Δ+λ/2)1/2  E is also N-dimensional. Inequality (128) implies that, for all ϕ∈F,
(129)∥ϕ∥2≤〈ϕ,Tλϕ〉,
where
(130)Tλ≔-∑j=15‍Tλj.
By Lemma 29, Tλ is a compact self-adjoint operator on ℋ. Hence, by the Hilbert-Schmidt theorem, there exists a complete orthonormal system {ϕn}n=1∞ of ℋ and a real sequence {tn}n=1∞ such that Tλϕn=tnϕn and limn→∞   tn=0. Using this fact, one sees that the number of eigenvalues tn of Tλ with tn≥1 is more than or equal to dimF=N. Hence μN(Tλ)≥1. Letkbe the largest natural number not exceeding (N+4)/5. Then 5k-4≤N. Hence 1≤μN(Tλ)≤μ5k-4(Tλ). On the other hand, by a general fact on singular values of the sum of two compact operators (e.g., [15, Theorem1.7]), we have
(131)μ5k-4(Tλ)≤∑j=15‍μk(Tλj).
Using this fact and Lemma 29, we obtain
(132)1≤2C∥vλ∥Ldk-1/d+2C∥vλ1/2∥Ld2k-2/d+C∥vλ∥Ld2k-2/d.
We have k≥N/5. Hence
(133)1≤C′(∥vλ∥LdN-1/d+∥vλ1/2∥Ld2N-2/d+∥vλ∥Ld2N-2/d),
where C′>0 is a constant independent of V,λ, and N. This implies that N≤C0(∥vλ1/2∥Ldd+∥vλ∥Ldd) with a constant C0 independent of V and λ. Thus (127) holds.As in Corollaries 1.2 and  1.3 in [13], we have from Theorem 30 the following results.Corollary 31.
Under the same assumption as in Theorem30, the number N(V) of eigenvalues of H(V) in (-m,m) is finite and
(134)N(V)≤C0∫ℝd‍(∥V(x)∥d/2+∥V(x)∥d)dx.Corollary 32.
Suppose that the assumption of Theorem30 holds. Let λj(j=1,…,N(V)) be the eigenvalues of H(V) in (-m,m), counted with multiplicity, and let γ>0 be such that
(135)fγ(V)≔∫ℝd‍∥V(x)∥γ(∥V(x)∥d/2+∥V(x)∥d)dx<∞.
Then, there exists a constant Cγ>0 such that
(136)∑j=1N(V)‍(m2-λj2)γ≤Cγfγ(V).


## 8.2. Applications

Now we apply the results in the preceeding section to the Dirac operatorH. For λ∈(0,m2), we denote by N(λ) the number of eigenvalues of H in (-m2-λ,m2-λ).Theorem 33.
Letd≥3 and λ∈(0,m2). Suppose that the assumption of Theorem 26 holds. Let
(137)FM,Φ(x)≔∥M(x)-m∥+2m∥sinΦ(x)2∥,a.e.x∈ℝd.(i)
IfFM,Φ(x)≤λ/4m,a.e.x∈ℝd, then N(λ)=0.(ii)
Suppose thatFM,Φ∈Ld/2(ℝd)∩Ld(ℝd). Then there exists a positive constant C>0 independent of M,Φ, and λ such that
(138)N(λ)≤C∫FM,Φ(x)>λ/4m‍(FM,Φ(x)d/2+FM,Φ(x)d)dx<∞.
Moreover, the numberN0 of eigenvalues of H in (-m,m) obeys
(139)N0≤C∫ℝd‍(FM,Φ(x)d/2+FM,Φ(x)d)dx<∞.Proof.
(i) We can writeH=H(V) with V=αd+1(MeiΦ-m). Hence
(140)∥V(x)∥=∥M(x)eiΦ(x)-m∥≤∥M(x)-m∥+m∥eiΦ(x)-1∥=FM,Φ(x).
Hence, the present assumption implies that ∥V(x)∥≤λ/4m a.e. x∈ℝd. Hence, by Theorem 28, N(λ)=0.
(ii) By (140) and the present assumption, ∥V(·)∥∈Ld(ℝd)∩Ld/2(ℝd). Thus we can apply Theorem 30 to obtain (138). Inequality (139) follows from (138) or Corollary 31.We have from Corollary32 the following fact.Corollary 34.
Letd≥3. Suppose that the assumption of Theorem 26 and FM,Φ∈Ld/2(ℝd)∩Ld(ℝd). Let λj(j=1,…,N0) be the eigenvalues ofHin (-m,m), counted with multiplicity, and let γ>0 be such that
(141)fγ(M,Φ)≔∫ℝd‍FM,Φ(x)γ(FM,Φ(x)d/2+FM,Φ(x)d)dx<∞.
Then, there exists a constant Cγ>0 such that
(142)∑j=1N0‍(m2-λj2)γ≤Cγfγ(M,Φ).We can also use Theorems23 and 27 to obtain another upper bound for N(λ). Let
(143)GM,Φ(x)≔∥M(x)-m-∑j=1d‍αd+1αjeiΦ(x)/2Ej(x)∥,a.e.x∈ℝd.Theorem 35.
Letd≥3, and let λ∈(0,m2). Suppose that the assumption of Theorem 27 holds. Then one has the following. (i)
IfGM,ϕ(x)≤λ/4m for a.e. x∈ℝd, then N(λ)=0.(ii)
Suppose thatGM,Φ∈Ld/2(ℝd)∩Ld(ℝd). Then (138) and (139) with FM,Φ replaced by GM,Φ hold.Proof.
By Theorem23, N(λ) is equal to the number of eigenvalues of H′ in (-m2-λ, m2-λ). One can write H′=Hm+V with V≔αd+1(M-m-∑j=1d‍αd+1αjAj). We have ∥V(x)∥=GM,Φ(x). Thus, in the same way as in the proof of Theorem 30, we obtain the desired results.Theorem35 implies the following result as in Corollary 34.Corollary 36.
Letd≥3. Suppose that the assumption of Theorem 27 holds and GM,Φ∈Ld/2(ℝd)∩Ld(ℝd). Then (142) with FM,Φ replaced by GM,Φ holds for all λ>0 such that ∫ℝd‍GM,Φ(x)γ(GM,Φ(x)d/2+GM,Φ(x)d)dx<∞.



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