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. 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