The self-adjoint extensions of symmetric operators satisfying anticommutation relations are considered. It is proven that an anticommutative
type of the Glimm-Jaffe-Nelson commutator theorem follows. Its application to an abstract Dirac operator is also considered.

1. Introduction and Main Theorem

In this paper, we consider the essential self-adjointness of anticommutative symmetric operators. Let H be a symmetric operator on a Hilbert space ℋ; that is, H satisfies H⊂H*. It is said that H is self-adjoint if H=H* and H is essentially self-adjoint if its closure H¯ is self-adjoint. We are interested in conditions under which a symmetric operator is essentially self-adjoint. The Glimm-Jaffe-Nelson commutator theorem (e.g., [1, Theorem 2.32], [2, Theorem X.36]) is one criterion for the essential self-adjointness of commutative symmetric operators. The commutator theorem shows that if a symmetric operator H and a self-adjoint operator X obey a commutation relation on a dense subspace 𝒟, which is a core of X, then H is essentially self-adjoint on 𝒟. Historically, Glimm and Jaffe [3] and Nelson [4] investigate the commutator theorem for quantum field models. Faris and Lavine [5] apply it to quantum mechanical models and Fröhlich [6] considers a generalization of the commutator theorem and proves that a multiple commutator formula follows. Here, we overview the commutator theorem.

Let H and X be linear operators on ℋ. Assume the following conditions.

H is symmetric and X is self-adjoint.

There exists δX>0 such that, for all Ψ∈𝒟(X),
(1)δX(Ψ,Ψ)≤|(Ψ,XΨ)|.

X has a core 𝒟0 satisfying 𝒟0⊂𝒟(H), and there exist constants a≥0 and b≥0 such that, for all Ψ∈𝒟0,
(2)∥HΨ∥≤a∥XΨ∥+b∥Ψ∥.

Theorem A (Glimm-Jaffe-Nelson commutator theorem).

Let H and X be operators satisfying (C.1)–(C.3). Suppose (i) or (ii) as follows.

There exists a constant c1≥0 such that, for all Ψ∈𝒟0,
(3)|(HΨ,XΨ)-(XΨ,HΨ)|≤c1|(Ψ,XΨ)|.

There exists a constant c2≥0 such that, for all Ψ∈𝒟0,
(4)c2|(Ψ,XΨ)|≤|(HΨ,XΨ)-(XΨ,HΨ)|.
Then, H is essentially self-adjoint on 𝒟0.

Remark 1.

In the commutator theorem, condition (i) is usually supposed. It is also proven under condition (ii) in a similar way to Theorem 2.

The idea of the proof of the commutator theorem is as follows. Let X and Y be symmetric operators on a Hilbert space. Then, the real part and the imaginary part of the inner product (XΨ,YΨ) for Ψ∈𝒟(XY)∩𝒟(YX) are expressed by
(5)Re(XΨ,YΨ)=12(Ψ,{X,Y}Ψ),Im(XΨ,YΨ)=12i(Ψ,[X,Y]Ψ),
respectively, where {X,Y}=XY+YX and [X,Y]=XY-YX. In the proof of the commutator theorem, the imaginary part is estimated. In Theorem 2, we prove that an anticommutative symmetric operator is essentially self-adjoint on a dense subspace by estimating the real part.

Theorem 2.

Assume (C.1)–(C.3). In addition, suppose that (I) or (II) holds.

There exists a constant d1≥0 such that, for all Ψ∈𝒟0,
(6)|(HΨ,XΨ)+(XΨ,HΨ)|≤d1|(Ψ,XΨ)|.

There exists a constant d2≥0 such that, for all Ψ∈𝒟0,
(7)d2|(Ψ,XΨ)|≤|(HΨ,XΨ)+(XΨ,HΨ)|.
Then, H is essentially self-adjoint on 𝒟0.

Proof of Theorem <xref ref-type="statement" rid="thm1">2</xref>.

We show that, for some z∈C∖R, dimker((H↾𝒟0)*+z♯)=0 where z♯=z,z*. Let Ψ∈𝒟((H↾𝒟0)*) and let Ξ=X-1Ψ. Since ((H↾𝒟0)*)*=H↾𝒟0¯, we have
(8)Re(Ξ,((H↾𝒟0)*+z♯)Ψ)=12((H↾𝒟0¯Ξ,XΞ)+(XΞ,H↾𝒟0¯Ξ))+Rez(Ξ,XΞ).
First, we assume that (I) holds. Let z∈C∖R satisfy |Rez|>d1/2. Since 𝒟0 is a core of X, it follows from (C.3) and (I) that 𝒟(X)⊂𝒟(H↾𝒟0¯) and for all Φ∈𝒟(X),
(9)|(H↾𝒟0¯Φ,XΦ)+(XΦ,H↾𝒟0¯Φ)|≤d1|(Φ,XΦ)|.
By (8) and (9), we have
(10)|Re(Ξ,((H↾𝒟0)*+z♯)Ψ)|≥(|Rez|-d12)|Ξ,XΞ|≥δX(|Rez|-d12)(Ξ,Ξ).
Since Ψ∈ker((H↾𝒟0)*+z♯), we have Ξ=X-1Ψ=0 from (10). Then, we have Ψ=0. Next, we suppose that (II) follows. Let z∈C∖R satisfy |Rez|<d2/2. Since 𝒟0 is a core of X, it also follows from (C.3) and (II) that 𝒟(X)⊂𝒟(H↾𝒟0¯) and for all Φ∈𝒟(X),
(11)d2|(Φ,XΦ)|≤|(H↾𝒟0¯Φ,XΦ)+(XΨ,H↾𝒟0¯Φ)|.
Then from (8) and (11), we have
(12)|Re(Ξ,((H↾𝒟0)*+z♯)Ψ)|≥(d22-|Rez|)|Ξ,XΞ|≥δX(d22-|Rez|)(Ξ,Ξ).
Since Ψ∈ker((H↾𝒟0)*+z♯) and Ξ=X-1Ψ, we have Ψ=0 from (12). Thus, the proof is obtained.

2. Application of Theorem <xref ref-type="statement" rid="thm1">2</xref>

We apply Theorem 2 to abstract Dirac operator theory [1, 7]. Let ℋ be a Hilbert space. Let H and τ be self-adjoint operators on ℋ. Assume that τ is bounded, τ2=I and, τ𝒟(H)⊂𝒟(H). Then H is called an abstract Dirac operator on ℋ with unitary involution τ. We construct an abstract Dirac operator by weakly commuting operators. Let X and Y be densely defined linear operators on a Hilbert space. The weak commutator of X and Y is defined for Φ∈∩𝒟(X*)∩𝒟(Y*) and for Ψ∈𝒟(X)∩𝒟(Y) by
(13)[X,Y]0(Φ,Ψ)=(X*Φ,YΨ)=(Y*Φ,XΨ).
Let {Pj}j=1N, N∈N, be self-adjoint operators on a Hilbert space ℋ. Set 𝒟0=∩j=1N𝒟(Pj). Assume that {Pj}j=1N satisfy the following condition.

𝒟0 is dense in ℋ. For all Φ,Ψ∈𝒟0, [Pj,Pl]0(Φ,Ψ)=0, j,l,=1,…,N.

Let M be a bounded self-adjoint operator ℋ satisfying the following condition.

For all Φ,Ψ∈𝒟0, [M,Pj]0(Φ,Ψ)=0, j=1,…,N.

Let 𝒦 be a Hilbert space. Let {Γj}j=1N and B be bounded self-adjoint operators on 𝒦 satisfying the following anticommutation relations:

Let ℋD=𝒦⊗ℋ. Assume (S.1)–(S.3). Then,
(14)HD=∑j=1NΓj⊗Pj+B⊗M
is self-adjoint on 𝒟(HD)=𝒦⊗𝒟0.

Remark 4.

In the case where {Pj}j=1N strongly commute, Theorem 3 has been proven ([8, Theorem 4.3], [9, Lemma 6.7]) by strongly anticommuting methods [10, 11].

It is seen that (I⊗B)2=I and (I⊗B)𝒟(HD)⊂𝒟(HD). Then, from Theorem 3, HD is an abstract Dirac operator on ℋD with the unitary involution I⊗B.

To prove Theorem 3, we show some lemmas.

Lemma 5.

Let {Cj}j=1N, N∈N, be closed operators on a Hilbert space on 𝒳. Suppose that ∩j=1N𝒟(Cj) is dense in 𝒳 and for j≠l, (CjΨ,ClΨ)+(ClΨ,CjΨ)=0, Ψ∈∩j=1N𝒟(Cj). Then C=∑j=1NCj is closed.

Proof.

We see that (CΨ, CΨ)=∑j=1N∥CjΨ∥2≥(1/N)(∑j=1N∥CjΨ∥)2. Then, ∑j=1N∥CjΨ∥≤N∥CΨ∥. Then from a closedness criterion (e.g., [1, Theorem B1], [12, Proposition 1]), C is closed.

From an argument of quadratic forms, there exists a self-adjoint operator L on ℋ such that L≥1, 𝒟(L)=𝒟0 and for all Φ,Ψ∈𝒟(L),
(15)(LΦ,LΨ)=∑j=1N(PjΦ,PjΨ)+(Φ,Ψ).

Lemma 6.

Assume (S.1). Then, for all Φ,Ψ∈𝒟(L),
(16)[L,Pj]0(Φ,Ψ)=0,j=1,…,N.

Proof.

Since L is positive and self-adjoint, it follows that LΞ=∫0∞(1/λ)(L+λ)-1LΞ, Ξ∈𝒟(L). Then, for all Φ,Ψ∈𝒟(L),
(17)[L,Pj]0(Φ,Ψ)=∫0∞1λ[(L+λ)-1L,Pj]0(Φ,Ψ)dλ=∫0∞λ[Pj,L]0((L+λ)-1Φ,(L+λ)-1Ψ)dλ=∑l=1N∫0∞λnnnnnn×{[Pj,Pl]0((L+λ)-1Φ,Pl(L+λ)-1Ψ)nnnnnnnnn+[Pj,Pl]0(Pl(L+λ)-1Φ,(L+λ)-1Ψ)}dλ.
By (S.1) and (17), we have [L,Pj]0(Φ,Ψ)=0 for all Φ,Ψ∈𝒟(L). Note that 𝒟(L) is a core of L, since L is self-adjoint. In addition, for all Ψ∈𝒟(L), ∥PjΨ∥≤∥LΨ∥, j=1,…,N. Hence, it follows that [L,Pj]0(Φ,Ψ)=0 for all Φ,Ψ∈𝒟(L). Thus, the proof is obtained.

Proof of Theorem <xref ref-type="statement" rid="thm2">3</xref>.

Since B⊗M is bounded, it is enough to show that H=∑j=1NΓj⊗Pj is self-adjoint. Let X=B⊗L. We show that H and X satisfy (C.1)–(C.3) and (I) in Theorem 2. Since H is symmetric and X is self-adjoint, (C.1) is satisfied. Since σ(B)={±1} and L≥1, we see that, for all Ψ∈𝒟(H),
(18)|(Ψ,XΨ)|=(Ψ,(I⊗L)Ψ)≥(Ψ,Ψ).
Then, (C.2) is satisfied. Since 𝒟0=𝒟(L), it follows that 𝒟(H)=𝒟(X). Then, by (S.3), we see that, for all Ψ∈𝒟(H),
(19)∥HΨ∥2=∑j=1N((I⊗Pj)Ψ,(I⊗Pj)Ψ)≤∥(I⊗L)Ψ∥2=∥XΨ∥2.
Then, ∥HΨ∥≤∥XΨ∥ for all Ψ∈𝒟(H), and hence (C.3) is satisfied. By Lemma 6, it is seen that, for all Ψ∈𝒟(H),
(20)(HΨ,XΨ)+(XΨ,HΨ)=∑j=1N(((Γj⊗Pj)Ψ,(B⊗L)Ψ)nnnnnn+((B⊗L)Ψ,(Γj⊗Pj)Ψ))=∑j=1N((I⊗L)Ψ,({Γj,B}⊗Pj)Ψ).
Then by (S.3) and (20), we have (HΨ,XΨ)+(XΨ,HΨ)=0. Then, from (18), it follows that (HΨ,XΨ)+(XΨ,HΨ)≤|(Ψ,XΨ)| for all Ψ∈𝒟(H). Then (I) is satisfied, and hence H¯ is self-adjoint from Theorem 2. In addition, by (S.1) and (S.3), we see that, for j≠l,
(21)((Γj⊗Pj)Ψ,(Γl⊗Pl)Ψ)+((Γl⊗Pl)Ψ,(Γj⊗Pj)Ψ)=((I⊗Pl)Ψ,({Γj,Γl}⊗Pj)Ψ)=0.
Then, from Lemma 5, H¯=H, and hence the proof is obtained.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

It is a pleasure to thank Professor Akito Suzuki and Professor Fumio Hiroshima for their comments. This work is supported by JSPS Grant 24·1671.

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