A Taylor-like Expansion of a Commutator with a Function of Self-Adjoint, Pairwise Commuting Operators

0.16 MB·English

by  Morten Grud Rasmussen

A Taylor-like Expansion of a Commutator with a
Function of Self-adjoint, Pairwise Commuting
Operators
Morten Grud Rasmussen
2 Department of Mathematical Sciences
1
Aarhus University
0
2
DK-8000 Aarhus

n
a Denmark
J
email:
[email protected]
6
1

]
A Abstract
F
. Let A be a ν-vector of self-adjoint, pairwise commuting operators and B
h
t a bounded operator of class Cn0(A). We prove a Taylor-like expansion of the
a
m commutator [B,f(A)] for a large class of functions f: Rν → R, generalising
theone-dimensional resultwhere Aisjustaself-adjoint operator. Thisisdone
[
using almost analytic extensions and the higher-dimensional Helffer-Sjöstrand

1
formula.
v
8
Keywords: Commutator expansions, functional calculus, almost analytic ex-
0
0 tensions, Helffer-Sjöstrand formula.
2
Mathematical Subject Classification (2010): 47B47
.
5
0
2
1
:
v
i
X
r
a
1
1 Introduction
It is well-known that if A is a self-adjoint operator, B is a bounded operator of class
Cn0(A) in the sense of [1] and f satisfies |f(n)(x)| ≤ C hxis−n for all n, then for
n
0 ≤ t ≤ n , 0 ≤ t ≤ 1 with s+t +t < n ,
1 0 2 1 2 0
n0−1 1
[B,f(A)] = f(k)(A)adk(B)+R (A,B)
k! A n0
X
k=1
where adk(B) is the k’th iterated commutator, R (A,B) ∈ B(H−t2;Ht1) and Ht
A n0 A A A
is defined as D(hAit) equipped with the graph-norm kvk = khAitvk for t ≥ 0 and
t
H−t is the dual space of Ht . This follows relatively easily from using the (one-
A A
dimensional) Helffer-Sjöstrand formula
1
f(A) = ∂¯f˜(z)(A−z)−1dz, (1)
π Z
C
where ∂¯= 1(∂ +i∂ ) and f˜is an almost analytic extension of f, and the identity
2 x y
n0−1 1 k!
[B,f(A)] = ∂¯f˜(z)(−1)k(A−z)−k−1dz
k! π Z
X C
k=1
(−1)n0
+ ∂¯f˜(z)(A−z)−n0 adn0(B)(A−z)−1dz
π Z A
C
when k! ∂¯f˜(z)(−1)k(A− z)−k−1dz is recognised as f(k)(A) using (1). Such com-
π C
mutatorRexpansions where first proved in [7]. See e.g. [4] for details. Due to the
higher complexity of the general Helffer-Sjöstrand formula, these calculations do not
lead directly to the generalised result where A is a vector of self-adjoint, pairwise
commuting operators. However, we will follow the same idea.
The theorem may be viewed as an abstract analogue of pseudo-differential cal-
culus. The one-dimensional version is an often used result, see e.g. [2] and [4].
Apart from the obvious interest in generalising the result to higher dimensions, our
improvement has proven useful in the treatment of models in quantum field theory,
see [6]. In particular, a lemma in [6] whose proof depends on our result, extends the
results of [5] to a larger class of models.
2 The setting and result
In the following, A = (A ,...,A ) is a vector of self-adjoint, pairwise commuting
1 ν
operators acting on a Hilbert space H, and B ∈ B(H) is a bounded operator on H.
We shall use the notion of B being of class Cn0(A) introduced in [1]. For notational
convenience, we adobt the following convention: If 0 ≤ j ≤ ν, then δ denotes the
j
multi-index (0,...,0,1,0,...,0), where the 1 is in the j’th entry.
Definition 1. Let n ∈ N∪{∞}. Assume that the multi-commutator form defined
0
iteratively by ad0(B) = B and adα(B) = [adα−δj(B),A ] as a form on D(A ), where
A A A j j
2
α ≥ δ is a multi-index and 1 ≤ j ≤ ν, can be represented by a bounded operator
j
also denoted by adα(B), for all multi-indices α, |α| < n +1. Then B is said to be
A 0
of class Cn0(A) and we write B ∈ Cn0(A).
Remark 2. The definition of adα(B) does not depend on the order of the iteration
A
since the A are pairwise commuting. We call |α| the degree of adα(B).
j A
In the following, Hs := D(|H|s) for s ≥ 0 will be used to denote the scale of
A
spaces associated to A. For negative s, we define Hs := (H−s)∗.
A A
Theorem 3. Assume that B ∈ Cn0(A) for some n ≥ n+ 1 ≥ 1, 0 ≤ t ≤ n+1,
0 1
0 ≤ t ≤ 1 and that {f } satisfies
2 λ λ∈I
∀α∃C : |∂αf (x)| ≤ C hxis−|α|
α λ α
uniformly in λ for some s ∈ R such that t +t +s < n+1. Then
1 2
n
1
[B,f (A)] = ∂αf (A) adα(B)+R (A,B)
λ α! λ A λ,n
X
|α|=1
as an identity on D(hAis), where R (A,B) ∈ B(H−t2,Ht1) and there exist a con-
λ,n A A
stant C independent of A, B and λ such that
kR (A,B)k ≤ C kadα(B)k.
λ,n B(H−t2,Ht1) A
A A X
|α|=n+1
Remark 4. A similar statement holds with the adα(B) and ∂αf (A) interchanged
A λ
at the cost of a sign correction given by (−1)|α|−1, and the corresponding remainder
term R′ (A,B) ∈ B(H−t1,Ht2). This can be seen either by proving it analogously
λ,n A A
or by taking the adjoint equation and replacing B by −B.
Remark 5. If k ≤ t and n ≥ n + 1 + k, then R (A,B) can be replaced by
1 0 λ,n
Rk (A,B) ∈ B(H−t2+k,Ht1−k). This can be seen by commuting |A − z|−2 and
λ,n A A
adα(B) in the terms of the remainder, see page 8.
A
3 The Proof
Let z ∈ Cν, Imz 6= 0, 1 ≤ ℓ ≤ ν and g,g : Rν → C be given as g(t) = |t−z|−2 and
ℓ
g (t) = t −z¯. Write for 2β ≤ α
ℓ ℓ ℓ
Tβ(t,z) := (−2)|α−β||α−β|!(t−Rez)α−2β|t−z|−2|α−β|.
α 2|β|β!(α−2β)!
Lemma 6. Let g be as above and α be any multi-index. Then
∂αg(t) = α!Tβ(t,z)|t−z|−2.
α
X
2β≤α
3
Proof. For brevity, we will write αi or βi for α + δ or β + δ , respectively. The
i i
formula is obviously true for |α| ≤ 1. Now assume that we have proven the formula
for |α| ≤ k. Let |α| = k and 0 ≤ i ≤ ν be arbitrary. It suffices to prove the formula
for αi. One easily verifies using the chain rule that
(∂δign)(t) = −2n(t −Rez )|t−z|−2n−2. (2)
i i
Now by the induction hypothesis, we see that
∂α+δig(t) = ∂δi (−2)|α−β|α!|α−β|!(t−Rez)α−2β|t−z|−2|α−β|−2
t 2|β|β!(α−2β)!
X
2β≤α
= (−2)|α−β|α!|α−β|!(∂δi(t−Rez)α−2β)|t−z|−2|α−β|−2 (3)
2|β|β!(α−2β)! t
X
2β≤α
+ (−2)|α−β|α!|α−β|!(t−Rez)α−2β(∂δi|t−z|−2|α−β|−2). (4)
2|β|β!(α−2β)! t
X
2β≤α
For the sake of clarity, we will now consider each sum independently.
(3) = (−2)|α−β|α!|α−β|!(α −2β )(t−Rez)α−2β−δi|t−z|−2|α−β|−2
2|β|β!(α−2β)! i i
X
2β≤α
= 2(β +1)(−2)|αi−βi|α!|αi−βi|!(t−Rez)αi−2βi|t−z|−2|αi−βi|−2
i 2|βi|βi!(αi−2βi)!
X
2β≤α
2βi<αi
= 2β (−2)|αi−β|α!|αi−β|!(t−Rez)αi−2β|t−z|−2|αi−β|−2. (5)
i 2|β|β!(αi−2β)!
X
2β≤α+δi
Using (2), we see that (4) equals
(−2)|α−β|α!|α−β|!(t−Rez)α−2β(−2)(|α−β|+1)(t −Rez )|t−z|−2|α−β|−4
2|β|β!(α−2β)! i i
X
2β≤α
= (α +1−2β )(−2)|αi−β|α!|αi−β|!(t−Rez)αi−2β|t−z|−2|αi−β|−2
i i 2|β|β!(αi−2β)!
X
2β≤α
= (−2)|αi−β|αi!|αi−β|!(t−Rez)αi−2β|t−z|−2|αi−β|−2 (6)
2|β|β!(αi−2β)!
X
2β≤α
− 2β (−2)|αi−β|α!|αi−β|!(t−Rez)αi−2β|t−z|−2|αi−β|−2. (7)
i 2|β|β!(αi−2β)!
X
2β≤α
Now (7) cancels (5) except for possible terms with 2β = α+δ :
i
(5)+(7) = (−2)|αi−β|αi!|αi−β|!(t−Rez)αi−2β|t−z|−2|αi−β|−2. (8)
2|β|β!(αi−2β)!
X
2β=α+δi
Adding (6) and (8) finishes the induction.
Lemma 7. Let B ∈ Cn0(A) for some n ≥ 1 and let n ∈ N and α be a multi-index
0 0 0
satisfying |α |+n+1 ≤ n . Then
0 0
n
1
[adα0(B),g(A)] = ∂αg(A)adα0+α(B)+Rg(A,adα0(B)), (9)
A α! A n A
X
|α|=1
4
where
Rg(A,adα0(B))
n A
ν
= βi+1 Tβ+δi (A,z)adα0+α+2δi(B)|A−z|−2 (10)
X X |α+δi−β| α+2δi A
|α|=n−1 i=1
2β≤α
ν
+ βi+1 Tβ+δi (A,z)(A −z¯)adα0+α+δi(B)|A−z|−2 (11)
X X |α+δi−β| α+2δi i i A
|α|=n i=1
2β≤α
ν
+ βi+1 Tβ+δi (A,z)adα0+α+δi(B)(A −z )|A−z|−2. (12)
X X |α+δi−β| α+2δi A i i
|α|=n i=1
2β≤α
Proof. The proof goes by induction. One may check by inspection of the following
identity that the statement is true for n = 0.
ν
[adα0(B),|A−z|−2] = − |A−z|−2(A −z¯)adα0+δi(B)|A−z|−2
A i i A
X
i=1
(13)
ν
− |A−z|−2adα0+δi(B)(A −z )|A−z|−2.
A i i
X
i=1
Now assume that we have proven the formula for k ≤ n, |α | + n + 2 ≤ n . We
0 0
will now show that this implies that the formula holds for k = n+1. We begin by
noting two useful identities.
Tβ(t,z)|t−z|−2 = − βj+1 Tβ+δj (t,z). (14)
α |α+δj−β| α+2δj
(β +1)Tβ+δi (t,z)2(t −Rez ) = (α +1−2β )Tβ (t,z). (15)
i α+2δi i i i i α+δi
Now using (13) and (14) we see that
ν
(10) = βi+1 Tβ+δi (A,z)|A−z|−2adα0+α+2δi(B) (16)
X X X |α+δi−β| α+2δi A
|α|=n−12β≤α i=1
ν ν
+ βi+1 βj+δij+1 Tβ+δi+δj (A,z)
X X XX |α+δi−β||α+δi+δj−β| α+2δi+2δj (17)
|α|=n−12β≤α i=1 j=1
×(A −z¯ )adα0+α+2δi+δj(B)|A−z|−2
j j A
ν ν
+ βi+1 βj+δij+1 Tβ+δi+δj (A,z)
X X XX |α+δi−β||α+δi+δj−β| α+2δi+2δj (18)
|α|=n−12β≤α i=1 j=1
×adα0+α+2δi+δj(B)(A −z )|A−z|−2,
A j j
and by reordering and reindexing the sum in (16), (17) and (18), we get
ν
(16) = βi Tβ(A,z)|A−z|−2adα0+α(B), (19)
|α−β| α A
X X X
i=1 |α|=n+12β≤α
αi≥2 βi≥1
5
and (17) equals
ν ν
βi βj+1 Tβ+δj (A,z)(A −z¯ )adα0+α+δj(B)|A−z|−2 (20)
X X X X |α−β||α+δj−β| α+2δj j j A
i=1 |α|=n+12β≤α j=1
αi≥2 βi≥1
and similarly for (18) with the factor (A −z¯ )adα0+α+δj(B) replaced by the factor
j j A
adα0+α+δj(B)(A −z ). Note that we may relax the extra conditions on α and β in
A j j
the above statements, as a term with β = 0 contributes nothing.
i
Instead of continuing in the same fashion with (11) and (12), we note using (15)
that
ν
(11)+(12) = βi+1 Tβ+δi (A,z)adα0+α+2δi(B)|A−z|−2 (21)
X X X |α+δi−β| α+2δi A
|α|=n2β≤α i=1
ν
+ αi+1−2βiTβ (A,z)adα0+α+δi(B)|A−z|−2, (22)
X X X |α+δi−β| α+δi A
|α|=n2β≤α i=1
so we may focus our attention on (22):
ν
(22) = αi−2βiTβ(A,z)|A−z|−2adα0+α(B) (23)
|α−β| α A
X X X
i=1 |α|=n+1 2β≤α
αi≥1 2βi<αi
ν ν
+ αi−2βi βj+1 Tβ+δj (A,z)
X X X X |α−β| |α+δj−β| α+2δj
i=1 |α|=n+1 2β≤α j=1 (24)
αi≥1 2βi<αi
×(A −z¯ )adα0+α+δj(B)|A−z|−2.
j j A
ν ν
+ αi−2βi βj+1 Tβ+δj (A,z)
X X X X |α−β| |α+δj−β| α+2δj
i=1 |α|=n+1 2β≤α j=1 (25)
αi≥1 2βi<αi
×adα0+α+δj(B)(A −z )|A−z|−2
A j j
We note again that the additional conditions on α and β are superfluous.
We may now recollect the terms. First we see using Lemma 6:
n n+1
1 1
∂αg(A)adα0+α(B)+(19)+(23) = ∂αg(A)adα0+α(B), (26)
α! A α! A
X X
|α|=1 |α|=1
then
ν
(20)+(24) = βj+1 Tβ+δj (A,z)(A −z¯ )adα0+α+δj(B)|A−z|−2, (27)
X X |α+δj−β| α+2δj j j A
|α|=n+1 j=1
2β≤α
and
ν
(18)+(25) = βj+1 Tβ+δj (A,z)adα0+α+δj(B)(A −z )|A−z|−2, (28)
X X |α+δj−β| α+2δj A j j
|α|=n+1 j=1
2β≤α
6
so adding up, we have proved that (9) equals the sum of (26), (21), (27) and (28)
as stated.
The following lemma plays the same role for g as Lemma 7 plays for g, but
ℓ
contrary to Lemma 7, the proof is trivial.
Lemma 8. Let B ∈ Cn0(A) for some n ≥ 1 and let n ∈ N and α be a multi-index
0 0 0
satisfying |α |+n+1 ≤ n . Then
0 0
n
1
[adα0(B),g (A)] = ∂αg (A)adα0+α(B)+Rgℓ(A,adα0(B)),
A ℓ α! ℓ A n A
X
|α|=1
where Rgℓ(A,adα0(B)) = 0 for n ≥ 1, Rgℓ(A,adα0(B)) = adα0+δℓ(B).
n A 0 A A
The following lemma also follows by induction.
Lemma 9. Let B ∈ Cn0(A) for some n ≥ 1. Assume that h ∈ C∞(Rν), 1 ≤ i ≤ k,
0 i
satisfies
n
1
[adα0(B),h (A)] = ∂αh (A)adα0+α(B)+Rhi(A,adα0(B)),
A i α! i A n A
X
|α|=1
where Rhi(A,adα0(B)) is bounded for all n ∈ N and multi-indices α satisfying
n A 0 0
|α |+n+1 ≤ n and ∂αh (A) is bounded for all 1 ≤ |α| ≤ n −1. Then
0 0 i 0
k n k
1
B, h (A) = ∂α h (A)adα(B)
h Y i i X α! (cid:16)Y i(cid:17) A
i=1 |α|=1 i=1
k n j−1 k
1
+ ∂α h (A)Rhj (A,adα(B)) h (A).
X X α! (cid:16)Y i(cid:17) n−|α| A Y i
j=1 |α|=0 i=1 i=j+1
Let n + 1 ≤ n . If we put k = ν + 1, h = g for i 6= ν, h = g and apply
0 i ν ℓ
Lemma 7, 8 and 9 we see that
[B,|A−z|−2ν(A −z¯)]
ℓ ℓ
n
1 (29)
= ∂α |·−z|−2ν(· −z¯) (A)adα(B)+R (A,B),
α! ℓ ℓ A ℓ,n
X (cid:0) (cid:1)
|α|=1
where
R (A,B)
ℓ,n
ν−1 n
1
= ∂α(gj−1)(A)Rg (A,adα(B))|A−z|−2(ν−j)(A −z¯) (30)
α! n−|α| A ℓ ℓ
X X
j=1 |α|=0
1
+ ∂α(gν−1)(A)adα+δℓ(B)|A−z|−2 (31)
α! A
X
|α|=n
n
1
+ ∂α(gν−1g )(A)Rg (A,adα(B)) (32)
α! ℓ n−|α| A
X
|α|=0
7
In the following, we will refer to the terms of R (A,B) as the remainder terms. Let
ℓ,n
0 ≤ t ≤ n+1 and 0 ≤ t ≤ 1. By Hadamard’s three-line lemma and using (10–12),
1 2
(30–32), Lemma 6 and the identity
j j
α!
∂α f = ∂αif ,
(cid:16)Yi=1 i(cid:17) PXαi=α ji=1αi! Yi=1 i
Q
wemayinspectthateachremainderterm(withR (A,B)replacedbytheremainder
ℓ,n
term) and hence R (A,B) satisfies the inequality
ℓ,n
khAit1R (A,B)hAit2k ≤ Chzit1+t2|Imz|−n−2ν. (33)
ℓ,n
We will now use the functional calculus of almost analytic extensions. See e.g. [3]
for details. In the following, we write ∂¯= (∂¯ ,...,∂¯ ) where ∂¯ = 1(∂ +i∂ ) and
1 ν j 2 uj vj
u +v = z ∈ C, z = (z ,...,z ) ∈ Cν. The following proposition is inspired by [4]
j j j 1 n
and [8, Chap. X.2].
Proposition 10. Let s ∈ R and {f } ⊂ C∞(Rν) satisfy
λ λ∈I
∀α ∃C : |∂αf (x)| ≤ C hxis−|α|.
α λ α
There exists a family of almost analytic extensions {f˜} ⊂ C∞(Cν) satisfying
λ λ∈I
(i) supp(f˜) ⊂ {u+iv | u ∈ supp(f ),|v| ≤ Chui}.
λ λ
(ii) ∀ℓ ≥ 0 ∃C : |∂¯f˜(z)| ≤ C hzis−ℓ−1|Imz|ℓ.
ℓ λ ℓ
Proof. We define a mapping C∞(Rν) ∋ f 7→ f˜ ∈ C∞(Cν) in the following way.
Choose a function κ ∈ C∞(R) which equals 1 in a neighbourhood of 0 and put
0
λ = C , λ = max{max C ,λ +1} for k ≥ 1. Writing z = u+iv ∈ Rν⊕iRν,
0 0 k |α|=k α k−1
we now define
∂αf(u) ν λ v
f˜(z) = (iu)α κ |α| j .
X α! Y (cid:16) hui (cid:17)
α j=1
One can now check that the properties hold.
Remark 11. Note that if we for a χ ∈ C∞(Rν;[0,1]) with χ(0) = 1 define a
0
sequence of functions by f (x) = χ(x)f (x), then
k,λ k λ
[B,f (A)] = lim[B,f (A)]
λ k,λ
k→∞
as a form identity on D(hAis) and we have the dominated pointwise convergence
∂¯f˜ (x) → ∂¯f˜(x) for k → ∞.
k,λ λ
Let {f } satisfy the assumption of Proposition 10 with s < 0. Then the
λ λ∈I
almost analytic extensions provide a functional calculus via the formula
ν
f (A) = C ∂¯f˜(z)(A −z¯)|A−z|−2νdz, (34)
λ ν Z ℓ λ ℓ ℓ
X Cν
ℓ=1
8
where C is a positive constant (again we refer to [3] for details). Note that the
ν
integrals are absolutely convergent by Proposition 10(ii).
Multiplying hAit1R (A,B)hAit2 with ∂¯f˜(z), we get from (33) and Proposi-
ℓ,n λ
tion 10 (ii) that
khAit1∂¯f˜(z)R (A,B)hAit2k ≤ Chzit1+t2+s−n−1−2ν. (35)
λ ℓ,n
Hence, if t +t +s < n+1, hAit1∂¯f˜(z)R (A,B)hAit2 is integrable over Cν. Using
1 2 λ ℓ,n
(29), (34) and (35), we see that
ν
[B,f (A)] = C ∂¯f˜(z)[B,(A −z¯)|A−z|−2ν]dz
λ ν Z ℓ λ ℓ ℓ
X Cν
ℓ=1
ν n
1
= C ∂¯f˜(z) ∂α |·−z|−2ν(· −z¯) (A)dz adα(B)
ν Z ℓ λ α! ℓ ℓ A
Xℓ=1 Cν |Xα|=1 (cid:0) (cid:1)
ν
+C ∂¯f˜(z)R (A,B)dz. (36)
ν Z ℓ λ ℓ,n
X Cν
ℓ=1
We denote (36) by R (A,B). Note that
λ,n
ν
1
∂¯f˜(z) ∂α |t−z|−2ν(t −z¯) dz
Z ℓ λ α! t ℓ ℓ
Xℓ=1 Cν (cid:0) (cid:1)
ν
1 1
= ∂α ∂¯f˜(z)|t−z|−2ν(t −z¯)dz = ∂αf (t),
α! t Z ℓ λ ℓ ℓ α! λ
X Cν
ℓ=1
which implies
n
1
[B,f (A)] = ∂αf (A) adα(B)+R (A,B).
λ α! λ A λ,n
X
|α|=1
We have now proved Theorem 3 in the case s < 0. For the general case, we use
Remark 11 to see that [B,f (A)] = lim [B,f (A)] and clearly, f satisfies the
λ k→∞ k,λ k,λ
assumption of Proposition 10 with the same s, so the estimate corresponding to (35)
is now uniform in k and λ. The pointwise convergence and Lebesgue’s theorem on
dominated convergence now finishes the argument.
Acknowledgements
The author would like to thank J. S. Møller for suggestions, fruitful discussions and
ultimatelyforproposingthisproblem. Partofthisworkwasdonewhileparticipating
in the Summer School on Current Topics in Mathematical Phyics at the Erwin
Schrödinger International Institute for Mathematical Physics (ESI) in Vienna.
9
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