# Essential Self-Adjointness of Anticommutative Operators

Authors: Toshimitsu Takaesu
Journal: Journal of Mathematics
 (2014)

---

## Abstract

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.


---

## Body

## 1. Introduction and Main Theorem

In this paper, we consider the essential self-adjointness of anticommutative symmetric operators. LetH 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.LetH and X be linear operators on ℋ. Assume the following conditions.(
C
.
1
)
H is symmetric and X is self-adjoint.
(
C
.
2
)
There existsδ
X
>
0 such that, for all Ψ
∈
𝒟
(
X
),
(1)
δ
X
(
Ψ
,
Ψ
)
≤
|
(
Ψ
,
X
Ψ
)
|
.
(
C
.
3
)
X has a core 𝒟
0 satisfying 𝒟
0
⊂
𝒟
(
H
), and there exist constants a
≥
0 and b
≥
0 such that, for all Ψ
∈
𝒟
0,
(2)
∥
H
Ψ
∥
≤
a
∥
X
Ψ
∥
+
b
∥
Ψ
∥
.Theorem A (Glimm-Jaffe-Nelson commutator theorem).
LetH and X be operators satisfying (
C
.
1
)
–
(
C
.
3
). Suppose (i) or (ii) as follows. (i)
There exists a constantc
1
≥
0 such that, for all Ψ
∈
𝒟
0,
(3)
|
(
H
Ψ
,
X
Ψ
)
-
(
X
Ψ
,
H
Ψ
)
|
≤
c
1
|
(
Ψ
,
X
Ψ
)
|
.
(ii)
There exists a constantc
2
≥
0 such that, for all Ψ
∈
𝒟
0,
(4)
c
2
|
(
Ψ
,
X
Ψ
)
|
≤
|
(
H
Ψ
,
X
Ψ
)
-
(
X
Ψ
,
H
Ψ
)
|
.
Then, H is essentially self-adjoint on 𝒟
0.Remark 1.
In the commutator theorem, condition (i) is usually supposed. It is also proven under condition (ii) in a similar way to Theorem2.The idea of the proof of the commutator theorem is as follows. LetX and Y be symmetric operators on a Hilbert space. Then, the real part and the imaginary part of the inner product (
X
Ψ
,
Y
Ψ
) for Ψ
∈
𝒟
(
X
Y
)
∩
𝒟
(
Y
X
) are expressed by
(5)
Re
(
X
Ψ
,
Y
Ψ
)
=
1
2
(
Ψ
,
{
X
,
Y
}
Ψ
)
,
Im
⁡
(
X
Ψ
,
Y
Ψ
)
=
1
2
i
(
Ψ
,
[
X
,
Y
]
Ψ
)
,
respectively, where {
X
,
Y
}
=
X
Y
+
Y
X and [
X
,
Y
]
=
X
Y
-
Y
X. 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. (I)
There exists a constantd
1
≥
0 such that, for all Ψ
∈
𝒟
0,
(6)
|
(
H
Ψ
,
X
Ψ
)
+
(
X
Ψ
,
H
Ψ
)
|
≤
d
1
|
(
Ψ
,
X
Ψ
)
|
.
(II)
There exists a constantd
2
≥
0 such that, for all Ψ
∈
𝒟
0,
(7)
d
2
|
(
Ψ
,
X
Ψ
)
|
≤
|
(
H
Ψ
,
X
Ψ
)
+
(
X
Ψ
,
H
Ψ
)
|
.
Then, H is essentially self-adjoint on 𝒟
0.Proof of Theorem2.
We show that, for somez
∈
C
∖
R, dim
⁡
ker
⁡
(
(
H
↾
𝒟
0
)
*
+
z
♯
)
=
0 where z
♯
=
z
,
z
*. Let Ψ
∈
𝒟
(
(
H
↾
𝒟
0
)
*
) and let Ξ
=
X
-
1
Ψ. Since (
(
H
↾
𝒟
0
)
*
)
*
=
H
↾
𝒟
0
¯, we have
(8)
Re
(
Ξ
,
(
(
H
↾
𝒟
0
)
*
+
z
♯
)
Ψ
)
=
1
2
(
(
H
↾
𝒟
0
¯
Ξ
,
X
Ξ
)
+
(
X
Ξ
,
H
↾
𝒟
0
¯
Ξ
)
)
+
Re
z
(
Ξ
,
X
Ξ
)
.
First, we assume that (I) holds. Let z
∈
C
∖
R satisfy |
Re
z
|
>
d
1
/
2. Since 𝒟
0 is a core of X, it follows from (
C
.
3
) and (I) that 𝒟
(
X
)
⊂
𝒟
(
H
↾
𝒟
0
¯
) and for all Φ
∈
𝒟
(
X
),
(9)
|
(
H
↾
𝒟
0
¯
Φ
,
X
Φ
)
+
(
X
Φ
,
H
↾
𝒟
0
¯
Φ
)
|
≤
d
1
|
(
Φ
,
X
Φ
)
|
.
By (8) and (9), we have
(10)
|
Re
(
Ξ
,
(
(
H
↾
𝒟
0
)
*
+
z
♯
)
Ψ
)
|
≥
(
|
Re
z
|
-
d
1
2
)
|
Ξ
,
X
Ξ
|
≥
δ
X
(
|
Re
z
|
-
d
1
2
)
(
Ξ
,
Ξ
)
.
Since Ψ
∈
ker
⁡
(
(
H
↾
𝒟
0
)
*
+
z
♯
), we have Ξ
=
X
-
1
Ψ
=
0 from (10). Then, we have Ψ
=
0. Next, we suppose that (II) follows. Let z
∈
C
∖
R satisfy |
Re
z
|
<
d
2
/
2. Since 𝒟
0 is a core of X, it also follows from (
C
.
3
) and (II) that 𝒟
(
X
)
⊂
𝒟
(
H
↾
𝒟
0
¯
) and for all Φ
∈
𝒟
(
X
),
(11)
d
2
|
(
Φ
,
X
Φ
)
|
≤
|
(
H
↾
𝒟
0
¯
Φ
,
X
Φ
)
+
(
X
Ψ
,
H
↾
𝒟
0
¯
Φ
)
|
.
Then from (8) and (11), we have
(12)
|
Re
(
Ξ
,
(
(
H
↾
𝒟
0
)
*
+
z
♯
)
Ψ
)
|
≥
(
d
2
2
-
|
Re
z
|
)
|
Ξ
,
X
Ξ
|
≥
δ
X
(
d
2
2
-
|
Re
z
|
)
(
Ξ
,
Ξ
)
.
Since Ψ
∈
ker
⁡
(
(
H
↾
𝒟
0
)
*
+
z
♯
) and Ξ
=
X
-
1
Ψ, we have Ψ
=
0 from (12). Thus, the proof is obtained.


## 2. Application of Theorem2

We apply Theorem2 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, τ
2
=
I and, τ
𝒟
(
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
,
Y
Ψ
)
=
(
Y
*
Φ
,
X
Ψ
)
.
Let {
P
j
}
j
=
1
N, N
∈
N, be self-adjoint operators on a Hilbert space ℋ. Set 𝒟
0
=
∩
j
=
1
N
𝒟
(
P
j
). Assume that {
P
j
}
j
=
1
N satisfy the following condition. (
S
.
1
)
𝒟
0 is dense in ℋ. For all Φ
,
Ψ
∈
𝒟
0, [
P
j
,
P
l
]
0
(
Φ
,
Ψ
)
=
0, j
,
l
,
=
1
,
…
,
N.LetM be a bounded self-adjoint operator ℋ satisfying the following condition.(
S
.
2
)
For allΦ
,
Ψ
∈
𝒟
0, [
M
,
P
j
]
0
(
Φ
,
Ψ
)
=
0, j
=
1
,
…
,
N.Let𝒦 be a Hilbert space. Let {
Γ
j
}
j
=
1
N and B be bounded self-adjoint operators on 𝒦 satisfying the following anticommutation relations:(
S
.
3
)
(
i
)
{
Γ
j
,
Γ
k
}
=
2
δ
j
,
k
,
j
,
k
=
1
,
…
,
N
,
(
ii
)
{
Γ
j
,
B
}
=
1
,
j
=
1
,
…
,
N
,
(
iii
)
B
2
=
I.Then, the next assertion holds.Theorem 3.
Letℋ
D
=
𝒦
⊗
ℋ. Assume (
S
.
1
)
–
(
S
.
3
). Then,
(14)
H
D
=
∑
j
=
1
N
‍
Γ
j
⊗
P
j
+
B
⊗
M
is self-adjoint on 𝒟
(
H
D
)
=
𝒦
⊗
𝒟
0.Remark 4.
In the case where{
P
j
}
j
=
1
N 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
(
H
D
)
⊂
𝒟
(
H
D
). Then, from Theorem 3, H
D is an abstract Dirac operator on ℋ
D with the unitary involution I
⊗
B.To prove Theorem3, we show some lemmas.Lemma 5.
Let{
C
j
}
j
=
1
N, N
∈
N, be closed operators on a Hilbert space on 𝒳. Suppose that ∩
j
=
1
N
𝒟
(
C
j
) is dense in 𝒳 and for j
≠
l, (
C
j
Ψ
,
C
l
Ψ
)
+
(
C
l
Ψ
,
C
j
Ψ
)
=
0, Ψ
∈
∩
j
=
1
N
𝒟
(
C
j
). Then C
=
∑
j
=
1
N
C
j is closed.Proof.
We see that(
C
Ψ, C
Ψ
)
=
∑
j
=
1
N
∥
C
j
Ψ
∥
2
≥
(
1
/
N
)
(
∑
j
=
1
N
∥
C
j
Ψ
∥
)
2. Then, ∑
j
=
1
N
∥
C
j
Ψ
∥
≤
N
∥
C
Ψ
∥. 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 operatorL on ℋ such that L
≥
1, 𝒟
(
L
)
=
𝒟
0 and for all Φ
,
Ψ
∈
𝒟
(
L
),
(15)
(
L
Φ
,
L
Ψ
)
=
∑
j
=
1
N
‍
(
P
j
Φ
,
P
j
Ψ
)
+
(
Φ
,
Ψ
)
.Lemma 6.
Assume(
S
.
1
). Then, for all Φ
,
Ψ
∈
𝒟
(
L
),
(16)
[
L
,
P
j
]
0
(
Φ
,
Ψ
)
=
0
,
j
=
1
,
…
,
N
.Proof.
SinceL is positive and self-adjoint, it follows that L
Ξ
=
∫
0
∞
‍
(
1
/
λ
)
(
L
+
λ
)
-
1
L
Ξ, Ξ
∈
𝒟
(
L
). Then, for all Φ
,
Ψ
∈
𝒟
(
L
),
(17)
[
L
,
P
j
]
0
(
Φ
,
Ψ
)
=
∫
0
∞
‍
1
λ
[
(
L
+
λ
)
-
1
L
,
P
j
]
0
(
Φ
,
Ψ
)
d
λ
=
∫
0
∞
‍
λ
[
P
j
,
L
]
0
(
(
L
+
λ
)
-
1
Φ
,
(
L
+
λ
)
-
1
Ψ
)
d
λ
=
∑
l
=
1
N
‍
∫
0
∞
‍
λ
n
n
n
n
n
n
×
{
[
P
j
,
P
l
]
0
(
(
L
+
λ
)
-
1
Φ
,
P
l
(
L
+
λ
)
-
1
Ψ
)
n
n
n
n
n
n
n
n
n
+
[
P
j
,
P
l
]
0
(
P
l
(
L
+
λ
)
-
1
Φ
,
(
L
+
λ
)
-
1
Ψ
)
}
d
λ
.
By (
S
.
1
) and (17), we have [
L
,
P
j
]
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
), ∥
P
j
Ψ
∥
≤
∥
L
Ψ
∥, j
=
1
,
…
,
N. Hence, it follows that [
L
,
P
j
]
0
(
Φ
,
Ψ
)
=
0 for all Φ
,
Ψ
∈
𝒟
(
L
). Thus, the proof is obtained.Proof of Theorem3.
SinceB
⊗
M is bounded, it is enough to show that H
=
∑
j
=
1
N
Γ
j
⊗
P
j 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
,
X
Ψ
)
|
=
(
Ψ
,
(
I
⊗
L
)
Ψ
)
≥
(
Ψ
,
Ψ
)
.
Then, (
C
.
2
) is satisfied. Since 𝒟
0
=
𝒟
(
L
), it follows that 𝒟
(
H
)
=
𝒟
(
X
). Then, by (
S
.
3
), we see that, for all Ψ
∈
𝒟
(
H
),
(19)
∥
H
Ψ
∥
2
=
∑
j
=
1
N
‍
(
(
I
⊗
P
j
)
Ψ
,
(
I
⊗
P
j
)
Ψ
)
≤
∥
(
I
⊗
L
)
Ψ
∥
2
=
∥
X
Ψ
∥
2
.
Then, ∥
H
Ψ
∥
≤
∥
X
Ψ
∥ for all Ψ
∈
𝒟
(
H
), and hence (
C
.
3
) is satisfied. By Lemma 6, it is seen that, for all Ψ
∈
𝒟
(
H
),
(20)
(
H
Ψ
,
X
Ψ
)
+
(
X
Ψ
,
H
Ψ
)
=
∑
j
=
1
N
‍
(
(
(
Γ
j
⊗
P
j
)
Ψ
,
(
B
⊗
L
)
Ψ
)
n
n
n
n
n
n
+
(
(
B
⊗
L
)
Ψ
,
(
Γ
j
⊗
P
j
)
Ψ
)
)
=
∑
j
=
1
N
‍
(
(
I
⊗
L
)
Ψ
,
(
{
Γ
j
,
B
}
⊗
P
j
)
Ψ
)
.
Then by (
S
.
3
) and (20), we have (
H
Ψ
,
X
Ψ
)
+
(
X
Ψ
,
H
Ψ
)
=
0. Then, from (18), it follows that (
H
Ψ
,
X
Ψ
)
+
(
X
Ψ
,
H
Ψ
)
≤
|
(
Ψ
,
X
Ψ
)
| 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)
(
(
Γ
j
⊗
P
j
)
Ψ
,
(
Γ
l
⊗
P
l
)
Ψ
)
+
(
(
Γ
l
⊗
P
l
)
Ψ
,
(
Γ
j
⊗
P
j
)
Ψ
)
=
(
(
I
⊗
P
l
)
Ψ
,
(
{
Γ
j
,
Γ
l
}
⊗
P
j
)
Ψ
)
=
0
.
Then, from Lemma 5, H
¯
=
H, and hence the proof is obtained.



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*Source: 265349-2014-03-27.xml*