Quadratic exponential vectors

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Quadratic exponential vectors

Luigi Accardi, Ameur Dhahri

To cite this version:

Luigi Accardi, Ameur Dhahri. Quadratic exponential vectors. 2008. �hal-00424043�

Quadratic exponential vectors

Luigi Accardi, Ameur Dhahri

Volterra Center, University of Roma Tor Vergata Via Columbia 2, 00133 Roma, Italy e-mail:accardi@volterra.uniroma2.it

ameur@volterra.uniroma2.it

Abstract

We give a necessary and sufficient condition for the existence of a quadratic exponential vector with test function in L²(R^d)∩L^∞(R^d).

We prove the linear independence and totality, in the quadratic Fock space, of these vectors. Using a technique different from the one used in [2], we also extend, to a more general class of test functions, the explicit form of the scalar product between two such vectors.

1 Introduction

Exponential vectors play a fundamental role in first order quantization. It is therefore natural to expect that their quadratic analogues, first introduced in [2b], will play such a role in the only example, up to now, of nonlinear renormalized quantization whose structure is explicitly known: the quadratic Fock functor.

The canonical nature of the objects involved justifies a detailed study of their structure.

Let us emphasize that the nonlinearity introduces substantially new features with respect to the linear case so that, contrarily to what happens in the usual deformations of the commutation relations, quadratic quantization is not a simple variant of the first order one, but interesting new phenomena arise.

For example usual exponential vectors can be defined for any square inte- grable test function, but this is by far not true for quadratic ones. This poses the problem to characterise those test functions for which quadratic

exponential vectors can be defined. This is the first problem solved in the present paper (see Theorem (3)).

Another problem is the linear independence of the quadratic exponential vectors. This very useful property, in the first order case, is a simple conse- quence of the linear independence of the complex exponential functions. In the quadratic case the proof is more subtle. This is the second main result of the present paper (see Theorem (5)).

For notations and terminology we refer to the papers cited in the bibliogra- phy. We simply recall that the algebra of the renormalized square of white noise (RSWN) with test function algebra

A :=L²(R^d)∩L^∞(R^d)

is the ∗-Lie-algebra, with self–adjoint central element denoted 1, generators {B⁺_f, Bh, Ng,1 : f, g, h∈L²(R^d)∩L^∞(R^d)}

involution

(B_f⁺)^∗ =Bf , N_f^∗ =Nf¯

and commutation relations

[Bf, B_g⁺] = 2chf, gi+ 4Nf g¯ , [Na, B_f⁺] = 2B_af⁺, c >0 [B_f⁺, B_g⁺] = [Bf, Bg] = [Na, Na^′] = 0

for alla,a^′,f,g ∈L²(R^d)∩L^∞(R^d). Such algebra admits a unique, up to uni- tary isomorphism,∗–representation (in the sense defined in [7]) characterized by the existence of a cyclic vector Φ satisfying

BhΦ =NgΦ = 0 : g, h∈L²(R^d)∩L^∞(R^d)}

(see [3] for more detailed properties of this representation).

2 Factorisation properties of the quadratic exponential vectors

Recall from [2b] that, if a quadratic exponential vector with test function f ∈L²(R^d)∩L^∞(R^d) exists, it is given by

Ψ(f) = X

n≥0

B⁺ⁿ_f Φ n!

where by definition

Ψ(0) :=B_f⁺⁰Φ := Φ

(notice the absence of the square root in the denominator).

In this section, after proving some simple consequences of the commutation relations, we give a direct proof for the factorization property of the expo- nential vectors.

Lemma 1 For allf, g ∈L²(R^d)∩L^∞(R^d), one has

[Nf, B_g⁺ⁿ] = 2nB_g⁺⁽ⁿ⁻¹⁾B_{f g}⁺, (1) [Bf, B_g⁺ⁿ] = 2nchf, giB_g⁺⁽ⁿ⁻¹⁾+ 4nB⁺⁽ⁿ⁻¹⁾_g Nf g¯ + 4n(n−1)B_g⁺⁽ⁿ⁻²⁾B_{f g}⁺_¯2. (2)

Proof. The mutual commutativity of the creators implies that

[Nf, B_g⁺ⁿ] =

n−1

X

i=0

B_g⁺ⁱ[Nf, B⁺_g]B_g^+(n−i−1)

=

n−1

X

i=0

B_g⁺ⁱ(2B_{f g}⁺)B_g^+(n−i−1) = 2nB_g⁺⁽ⁿ⁻¹⁾B_{f g}⁺ which is (1). To prove (2) consider the identity

[Bf, B_g⁺ⁿ] =

n−1

X

i=0

B_g⁺ⁱ[Bf, B_g⁺]B_g^+(n−i−1)

=

n−1

X

i=0

B_g⁺ⁱ(2chf, gi+ 4Nf g¯ )B_g^+(n−i−1)

= 2nchf, giB_g⁺⁽ⁿ⁻¹⁾+ 4

n−1

X

i=0

B_g⁺ⁱNf g¯ B_g^+(n−i−1). From (1) it follows that

Nf g¯ B_g^+(n−i−1) =B_g^+(n−i−1)Nf g¯ + 2(n−i−1)B_g^+(n−i−2)B_{f g}⁺_¯2.

Therefore, one obtains

[Bf, B_g⁺ⁿ] = 2nchf, giB_g⁺⁽ⁿ⁻¹⁾+ 4nB⁺⁽ⁿ⁻¹⁾_g Nf g¯

+4

n−1

X

i=0

B_g⁺ⁱ(2(n−i−1))B_g^+(n−i−2)B_{f g}⁺_¯2

= 2nchf, giB_g⁺⁽ⁿ⁻¹⁾+ 4nB_g⁺⁽ⁿ⁻¹⁾Nf g¯

+8

n−1

X

i=0

(n−i−1)

!

B_g⁺⁽ⁿ⁻²⁾B_{f g}⁺_¯2

= 2nchf, giB_g⁺⁽ⁿ⁻¹⁾+ 4nB⁺⁽ⁿ⁻¹⁾_g Nf g¯

+8

n−1

X

i=0

(n−i−1)

!

B_g⁺⁽ⁿ⁻²⁾B_{f g}⁺_¯2

= 2nchf, giB_g⁺⁽ⁿ⁻¹⁾+ 4nB⁺⁽ⁿ⁻¹⁾_g Nf g¯

+8

n(n−1)− n(n−1) 2

B_g⁺⁽ⁿ⁻²⁾B_{f g}⁺_¯2

= 2nchf, giB_g⁺⁽ⁿ⁻¹⁾+ 4nB_g⁺⁽ⁿ⁻¹⁾Nf g¯

+4n(n−1)B⁺⁽ⁿ⁻²⁾_g B_{f g}⁺_¯2.

Lemma 2 Let f1, . . . , fk, g1, . . . , gh ∈L²(R^d)∩L^∞(R^d). Then, one has

Bfk. . . Bf1B_g⁺_h. . . B_g⁺₁Φ = 0 for all k > h≥0.

Proof. It is sufficient to prove that

Bfh+1. . . Bf1B⁺_gh. . . B⁺_g1Φ = 0, ∀h ∈N. (3) Suppose by induction that, for h≥0, (3) is satisfied. Then, one has

Bfh+1. . . Bf1B_g⁺_h. . . B_g⁺₁Φ = Bfh+1. . . Bf2[Bf1, B_g⁺_h. . . B_g⁺₁]Φ

=Bfh+1. . . Bf2

1

X

i=h

B_g⁺_h. . . B_g⁺_i+1(2chf1, gii+ 4Nf¯1gi) B_g⁺_i−1. . . B_g⁺₁Φ

= 2c

1

X

i=h

hf₁, gii(B_fh+1. . . Bf2)(B_g⁺_h. . .Bˆ_g⁺_i. . . B_g⁺₁)Φ

+4

1

X

i=h

(Bfh+1. . . Bf2)(B⁺_gh. . . Nf¯1giB⁺_gi−1. . . B_g⁺₁)Φ.

By the induction assumption

1

X

i=h

hf1, gii(Bfh+1. . . Bf2)(B_g⁺_h. . .Bˆ_g⁺_i. . . B_g⁺₁)Φ = 0.

Therefore, one gets

Bfh+1. . . Bf1B_g⁺_h. . . B_g⁺₁Φ

= 4

1

X

i=h

(Bf_h+1. . . Bf2)(B_g⁺_h. . . B⁺_gi+1Nf¯1giB_g⁺_i−1. . . B_g⁺₁)Φ

= 4

1

X

i=h

(Bfh+1. . . Bf2) B_g⁺_h. . . B_g⁺_i+1

1

X

m=i−1

B_g⁺_i−1. . .[Nf¯1gi, B_g⁺_m]. . . B_g⁺₁

! Φ

= 4

1

X

i=h

Bfh+1. . . Bf2B_g⁺_h. . . B_g⁺_i+1 2

1

X

m=i−1

B_g⁺_i−1. . . B_f⁺_¯

1gigmB_g⁺_m−1. . . B_g⁺₁

! Φ

= 8

1

X

i=h 1

X

m=i−1

Bg_h+1. . . Bf2B_g⁺_h. . . B_g⁺_i+1. . .Bˆ_g⁺_m. . . B_g⁺₁B⁺_f¯1g

igmΦ,

which is equal to 0 by the induction assumption.

Lemma 3 Let I, J ⊂R^d such that I∩J =φ and let f1, . . . , fh, f₁^′, . . . , f_h^′′, g1, . . . , gk, g₁^′, . . . , g_k^′′ ∈L²(R^d)∩L^∞(R^d) such that

supp (fi)⊂I , supp(f_i^′)⊂I , supp (gi)⊂J , supp (g_i^′)⊂J Then, for h6=h^′ or k 6=k^′, one has

hB_f⁺

h. . . B_f⁺₁B_g⁺_k. . . B_g⁺₁Φ, B_f^+′

h′. . . B_f^+′

1B_g^+′

k′. . . B_g^+′

1Φi= 0.

Proof. Lemma (1) and the polarization identity imply that [Bf_h. . . Bf1, B_g^+′

k′ . . . B_g^+′

1] = 0. Therefore, it is sufficient to prove the result for h 6= h^′. Taking eventually the complex conjugate, we can suppose that h > h^′. Under this assumption

hB_f⁺

h. . . B_f⁺₁B_g⁺_k. . . B_g⁺₁Φ, B_f^+′

h′ . . . B_f^+′

1B_g^+′

k′ . . . B_g^+′

1Φi

=hB_g⁺_k. . . B_g⁺₁Φ, B_g^+′

k′ . . . B_g^+′

1(Bf_h. . . Bf1B_f^+′

h′. . . B_f^+′

1)Φi

and the statement follows because, from Lemma 2, one has Bfh. . . Bf1B_f^+′

h′. . . B_f^+′

1Φ = 0.

ForI ⊂R^d, denote HI the closed linear span of the set

{B_f⁺ⁿΦ, n ∈N,f ∈L²(R^d)∩L^∞(R^d) such that supp (f)⊂I}

and H = H_R^d. The space H, denoted Γ₂(L²(R^d)∩L^∞(R^d)), is called the quadratic Fock space with test function algebraL²(R^d)∩L^∞(R^d).

Lemma 4 Let I, J ⊂ R^d such that I∩J =φ and let f1, . . . , fh, f₁^′, . . . , f_h^′, g1, . . . , gk, g₁^′, . . . , g_k^′ ∈L²(R^d)∩L^∞(R^d) such that

supp (fi)⊂I, supp (f_i^′)⊂I, supp (gi)⊂J and supp(g^′_i)⊂J.

Then, one has hB⁺_f

h. . . B⁺_f1B_g⁺_k. . . B_g⁺₁Φ, B_f^+′

h. . . B_f^+′

1B_g^+′

k. . . B⁺_g^′

1Φi

=hB_f⁺_h. . . B_f⁺₁ΦI, B_f^+′

h. . . B_f^+′

1ΦIihB_g⁺_k. . . B_g⁺₁ΦJ, B_g^+′

k. . . B⁺_g^′

1ΦJi. (4) Proof. By induction on h. Forh= 1, one has

hB_f⁺₁B_g⁺_k. . . B_g⁺₁Φ, B_f^+′

1B_g^+′

k. . . B⁺_g^′

1Φi

=hB_g⁺_k. . . B_g⁺₁Φ, B_g^+′

k. . . B_g^+′

1(Bf1B_f^+′

1)Φi

= 2chf1, f₁^′ihB_g⁺_k. . . B⁺_g1Φ, B_g^+′

k. . . B_g^+′

1Φi

=hB_f⁺₁ΦI, B_f^+′

1ΦIihB_g⁺

k. . . B_g⁺₁ΦJ, B_g^+′

k. . . B⁺_g^′

1ΦJi.

Let h≥1 and suppose that (4) holds true. Then, one has hB⁺_f

h+1B_f⁺

h. . . B_f⁺₁B_g⁺_k. . . B_g⁺₁Φ, B_f^+′

h+1B_f^+′

h. . . B_f^+′

1B⁺_g^′

k. . . B_g^+′

1Φi

=hB_f⁺

h. . . B_f⁺₁B⁺_gk. . . B_g⁺₁Φ, Bfh+1B_f^+′

h+1B_f^+′

h. . . B_f^+′

1B_f^+′

1B⁺_g^′

k. . . B_g^+′

1Φi

=

1

X

m=h+1

hB_f⁺

h. . . B_f⁺₁B_g⁺_k. . . B_g⁺₁Φ,(B⁺_g^′

k. . . B_g^+′

1)(B_f^+′

h+1. . .[Bfh+1, B_f^+′

m]. . . B_f^+′

1)Φi

= 2c

1

X

m=h+1

hf_h+1, f_m^′ i hB⁺_f

h. . . B⁺_f1B_g⁺_k. . . B_g⁺₁Φ,(B_g^+′

k. . . B_g^+′

1)(B_f^+′

h+1. . .Bˆ_f^+′

m. . . B_f^+′

1)Φi

+4

1

X

m=h+1

hB⁺_fh. . . B⁺_f1B_g⁺_k. . . B_g⁺₁Φ, (B_g^+′

k. . . B_g^+′

1)(B_f^+′

h+1. . . B_f^+′

m+1Nf¯h+1f_m^′ B_f^+′

m−1. . . B_f^+′

1)Φi.

By the induction assumption

1

X

m=h+1

hfh+1, f_m^′ ihB_f⁺_h. . . B_f⁺₁B_g⁺_k. . . B⁺_g1Φ, B⁺_g^′

k. . . B_g^+′

1B_f^+′

h+1. . .Bˆ_f^+′

m. . . B_f^+′

1Φi

=

1

X

m=h+1

hfh+1, f_m^′ ihB_f⁺_h. . . B_f⁺₁ΦI, B_f^+′

h+1. . .Bˆ_f^+′

m. . . B_f^+′

1ΦIi hB_g⁺

k. . . B_g⁺₁ΦJ, B_g^+′

k. . . B_g^+′

1ΦJi.

Moreover hB⁺_f

h. . . B⁺_f1B_g⁺_k. . . B_g⁺₁Φ, B_g^+′

k. . . B_g^+′

1B_f^+′

h+1. . . B_f^+′

m+1Nf¯h+1f_m^′ B⁺_f^′

m−1. . . B_f^+′

1Φi

=

1

X

i=m−1

hB_f⁺

h. . . B_f⁺₁B_g⁺_k. . . B_g⁺₁Φ, B_g^+′

k. . . B_g^+′

1B_f^+′

h+1. . .Bˆ_f^+′

m. . .[Nf¯h+1f^′

m′, B_f^+′

i]. . . B_f^+′

1Φi

= 2

1

X

i=m−1

hB⁺_f

h. . . B⁺_f1B_g⁺_k. . . B_g⁺₁Φ, B_g^+′

k. . . B_g^+′

1B_g^+′

1B⁺_f^′

h+1. . .Bˆ_f^+′

m. . .Bˆ_f^+′

i. . . B_f⁺₁B_f⁺_¯

h+1f_m^′ f_i^′Φi.

Again by the induction assumption, this term is equal to 2

1

X

i=m−1

hB_f⁺_h. . . B_f⁺₁ΦI, B_f^+′

h+1. . .Bˆ_f^+′

m. . .Bˆ_f^+′

i. . . B_f^+′

1B_f⁺_¯

h+1f_m^′ f_i^′ΦIi hB⁺_gk. . . B_g⁺₁ΦJ, B_g^+′

k. . . B_g^+′

1ΦJi.

Hence, one obtains hB⁺_f

h+1B_f⁺_h. . . B_f⁺₁B_g⁺_k. . . B_g⁺₁Φ, B_f^+′

h+1B_f^+′

h. . . B_f^+′

1B⁺_g^′

k. . . B_g^+′

1Φi

=n X¹

m=h+1

h2chfh+1, f_m^′ ihB_f⁺_h. . . B_f⁺₁ΦI, B_f^+′

h+1. . .Bˆ_f^+′

m. . . B_f^+′

1ΦIi +8

1

X

i=m−1

hB_f⁺_h. . . B_f⁺₁ΦI, B_f^+′

h+1. . .Bˆ_f^+′

m. . .Bˆ_f^+′

i. . . B_f^+′

1B_f⁺_¯

h+1f_m^′ f_i^′ΦIiio hB⁺_gk. . . B_g⁺₁ΦJ, B_g^+′

k. . . B_g^+′

1ΦJi.

Since the term between square brackets is equal to hB_f⁺

h+1. . . B_f⁺₁ΦI, B⁺_f^′

h+1. . . B_f^+′

1ΦIi.

This completes the proof of the lemma.

Lemma 5 Let I, J ⊂R^d such that I∩J =φ. Then, the operator UI,J :HI∪J → HI⊗ HJ

defined by

UI,J(B⁺_fn. . . B_f⁺₁B_g⁺_m. . . B_g⁺₁ΦI∪J) = (B_f⁺_n. . . B_f⁺₁ΦI)⊗(B_g⁺_m. . . B_g⁺₁ΦJ) where supp (fi)⊂I, supp (gi)⊂J, is unitary.

Proof. It is sufficient to prove that for all fn, . . . , f1, f_k^′, . . . , f₁^′ ∈ HI; gm, . . . , g₁, g_h^′, . . . , g^′₁ ∈ HJ

hB⁺_fn. . . B_f⁺₁B_g⁺_m. . . B_g⁺₁Φ_I∪J, B_f^+′

k. . . B_f^+′

1B_g^+′

h. . . B_g^+′

1Φ_I∪Ji

=hB_f⁺_n. . . B_f⁺₁ΦI, B_f^+′

k. . . B_f^+′

1ΦIihB_g⁺_m. . . B_g⁺₁ΦJ, B_g^+′

h. . . B_g^+′

1ΦJi. (5) Note that Lemma 3 implies that (5) holds if n 6= h or m 6= h. Moreover, if n =k and m=h, then from Lemma 4, the identity (5) is also satisfied. This

ends the proof.

Theorem 1 Let I1, . . . , In⊂R^d such that Ii∩Ij =φ for all i6=j. Let UI1∪...∪Ik−1,I_k :H^Sk

i=1Ii → H^Sk−1

i=1Ii ⊗ HI_k

be the operator defined by Lemma 5. Then, the operator UI1,...,In :H^Sn_i=1Ii → HI1 ⊗. . .⊗ HIn

given by

UI1,...,In = (UI1,I2 ⊗1I3 ⊗. . .⊗1In)◦(UI1∪I2,I3⊗1I4 ⊗. . .⊗1In)◦. . . . . .◦(UI1∪...∪In−2,In−1 ⊗1In)◦UI1∪...∪In−1,In

is unitary.

Proof. For allk∈ {2, . . . , n}(I₁∪. . .∪I_k−1)∩Ik=∅. Therefore the operator UI1∪...∪Ik−1,Ik :H^Sk

i=1Ii → H^Sk−1

i=1Ii ⊗ HIk

is unitary because of Lemma 5. In particular UI1,...,In is a unitary operator.

Theorem 2 (factorization property of the quadratic exponential vectors) Let I1, . . . , In ⊂R^d such that Ii∩Ij =φ, for all i6=j and let

UI1,...,In :H^Sn_i=1Ii →Nn

i=1HIi⊗. . .⊗ HIn be the unitary operator defined by Theorem 1.

Then, for all f ∈L²(Sn

i=1Ii)∩L^∞(Sn

i=1Ii) such that Ψ(f) exists, one has UI1,...,InΨ(f) = Ψ(fI1)⊗. . .⊗Ψ(fIn)

where fIi :=f χIi.

Proof. Denote Jk = Sk

i=1Ii. Then, UI1∪...∪Ik−1,Ik = UJk−1,Ik is a unitary operator from HJ_k to HJ_k−1 ⊗ HI_k. Let f ∈ L²(Sn

i=1Ii)∩L^∞(Sn

i=1Ii) be such that Ψ(f) exists. Since

f =

n

X

i=1

fIi =fJn−1 +fIn

it follows that

B_f^+m = (B_f⁺

Jn−1 +B_f⁺

In)^m =

m

X

k=0

C_m^kB_f^+k

Jn−1B_f^+(m−k)

In

where

C_m^k = m!

(m−k)!k!.

Therefore, one obtains UJn−1,In(B⁺_f

Jn−1 +B_f⁺

In)^mΦ =

m

X

k=0

C_m^k(B_f^+k

Jn−1ΦJn−1)⊗B_f^+(n−k)

In ΦIn. This gives

UJn−1,InΨ(f) = Ψ(fJn−1)⊗Ψ(fIn).

Now, one has

(UJn−2,In−1⊗1In)◦(UJn−1,In)Ψ(f) = (UJn−2,In−1Ψ(fJn−1))⊗Ψ(fIn).

In the same way, we prove that

UJn−2,In−1Ψ(fJn−1) = Ψ(fJn−2)⊗Ψ(fIn−1).

Hence, the following identity holds

(UJn−2,In−1⊗1In)◦(UJn−1,In)Ψ(f) = Ψ(fJn−2)⊗Ψ(fIn−1)⊗Ψ(fIn).

Iterating this procedure one finds

(UJ_k,I_k+1 ⊗1I_k+2⊗. . .⊗1In)◦(UJ_k+1,I_k+2⊗1I_k+3⊗. . .⊗1In)

◦. . .◦UJn−1,InΨ(f) = Ψ(fJk)⊗Ψ(fIk+1)⊗. . .⊗Ψ(fIn) or equivalently

UI1,...,InΨ(f) = (UI1,I2 ⊗1I3 ⊗. . .⊗1In)◦(UJ2,I3 ⊗1I4 ⊗. . .⊗1In)

◦. . .◦Ujn−1,InΨ(f) = Ψ(fI1)⊗. . .⊗Ψ(fIn).

3 Condition for the existence of the quadratic exponential vectors

When both Ψ(f) and Ψ(g) exist, the explicit form of their scalar product was determined in [2b], for step functions in R with bounded supports. We further prove that the sufficient condition for the existence of a quadratic exponential vector with a pre-assigned test function, given in [2b], is also a

necessary conditon. Finally, using the factorization property of the quadratic exponential vectors and an approximation argument, we extend the formula for the scalar product to exponential vectors with arbitrary step functions.

Due to the nonlinearity involved in this form of the scalar product, the ap- proximation argument is not as straightforward as in the first order case. A different proof of this result was obtained in [2].

Lemma 6 For alln ≥1 and all f, g, h ∈L²(R^d)∩L^∞(R^d), one has hB_f⁺⁽ⁿ⁻¹⁾Φ, BhB_g⁺ⁿΦi = c

n−1

X

k=0

2^2k+1 n!(n−1)!

((n−k−1)!)²hhf^k, g^k+1i hB^+(n−k−1)_f Φ, B^+(n−k−1)_g Φi. (6) Proof. Let us prove the above lemma by induction. For n = 1, it is clear that identity (6) is satisfied.

Now, let n ≥ 1 and suppose that (6) holds true. Note that, from (2) it follows that

BhB_g⁺⁽ⁿ⁺¹⁾Φ = 2(n+ 1)chh, giB_g⁺ⁿΦ + 4n(n+ 1)B_¯hg⁺2B_g⁺⁽ⁿ⁻¹⁾Φ.

Then, one gets

hB_f⁺ⁿΦ, BhB_g⁺⁽ⁿ⁺¹⁾Φi = 2(n+ 1)chh, gihB_f⁺ⁿΦ, B_g⁺ⁿΦi +4n(n+ 1)hB_f⁺ⁿ⁾Φ, B_¯hg⁺2B_g⁺⁽ⁿ⁻¹⁾Φi

= 2(n+ 1)chf, gihB_f⁺ⁿΦ, B_g⁺ⁿΦi

+4n(n+ 1)hBg⁺⁽ⁿ⁻¹⁾Φ, B¯hg²B_f⁺ⁿΦi. (7) Therefore, by induction assumption, one has

hB_g⁺⁽ⁿ⁻¹⁾Φ, B¯hg²B_f⁺ⁿΦi = c

n−1

X

k=0

2^2k+1 n!(n−1)!

((n−k−1)!)²h(¯hg²)g^k, f^k+1i hB^+(n−k−1)_g Φ, B_f^+(n−k−1)Φi

= c

n

X

k=1

2^2k−1 n!(n−1)!

((n−k)!)²hg^k+1, hf^ki hB^+(n−k)_g Φ, B_f^+(n−k)Φi.

It follows that

4n(n+ 1)hBg⁺⁽ⁿ⁻¹⁾Φ, B¯hg²B_f⁺ⁿΦi = c

n

X

k=1

2^2k+1 n!(n+ 1)!

((n−k)!)² hhf^k, g^k+1i hB_f^+(n−k)Φ, B_g^+(n−k)Φi. (8) Finally, identities (7) and (8) imply that

hB_f⁺ⁿΦ, BhB_g⁺⁽ⁿ⁺¹⁾Φi = c

n

X

k=0

2^2k+1 n!(n+ 1)!

((n−k)!)² hhf^k, g^k+1i hB_f^+(n−k)Φ, B^+(n−k)_g Φi.

This ends the proof.

From the above lemma, we prove the following.

Proposition 1 For all n≥1 and all f, g∈L²(R^d)∩L^∞(R^d), one has hB_f⁺ⁿΦ, B_g⁺ⁿΦi = c

n−1

X

k=0

2^2k+1 n!(n−1)!

((n−k−1)!)² hf^k+1, g^k+1i hB_f^+(n−k−1)Φ, B_g^+(n−k−1)Φi.

Proof. In order to prove the above proposition, it is sufficient to take f =h

in Lemma 6.

As a consequence of the Proposition 1, we give a necessary and sufficient condition for the existence of an exponential vector with a given test function.

Theorem 3 Let f ∈L²(R^d)∩L²(R^d). Then, the quadratic exponential vec- tor Ψ(f) exists if and only if ||f||_∞< ¹₂.

Proof. Sufficiency. Let f ∈L²(R^d)∩L²(R^d) be such that||f||_∞< ¹₂. From the above proposition, one has

||B_f⁺ⁿΦ||² = c

n−1

X

k=0

2^2k+1 n!(n−1)!

((n−k−1)!)²|kf^k+1k²₂kB_f^+(n−k−1)Φk²

= c

n−1

X

k=1

2^2k+1 n!(n−1)!

((n−k−1)!)²|kf^k+1k²₂kB_f^+(n−k−1)Φk²

+2nckfk²₂kB_f⁺⁽ⁿ⁻¹⁾Φk²+ 2nckfk²₂kB_f⁺⁽ⁿ⁻¹⁾Φk²

= c

n−2

X

k=0

2^2k+3 n!(n−1)!

(((n−1)−k−1)!)²|kf^k+2k²₂kB+((n−1)−k−1)

f Φk²

≤(4n(n−1)kfk²_∞)h c

n−2

X

k=0

2^2k+1 (n−1)!(n−2)!

(((n−1)−k−1)!)² kf^k+1k²₂B+((n−1)−k−1)

f Φk²i

. (9)

Note that

||B_f⁺⁽ⁿ⁻¹⁾Φ||² =c

n−2

X

k=0

2^2k+1 (n−1)!(n−2)!

((n−1)−k−1)!)²|kf^k+1k²₂kB+((n−1)−k−1)

f Φk².

This proves that

||B_f⁺ⁿΦ||² ≤h

4n(n−1)kfk²_∞+ 2nkfk²₂i

kB_f⁺⁽ⁿ⁻¹⁾Φk². (10) Finally, one gets

||B_f⁺ⁿΦ||²

(n!)² ≤h4n(n−1)kfk²_∞+ 2nkfk²₂ n²

ikB_f⁺⁽ⁿ⁻¹⁾Φk² ((n−1)!)² . Hence, it is clear that if 4kfk²_∞<1, then the seriesP

n≥0

||B_f⁺ⁿΦ||²

(n!)² converges.

Necessity. Let f ∈L²(R^d)∩L²(R^d) be such that ||f||_∞ ≥ ¹₂. Put J ={x∈R^d, |f(x)| ≥ 1

2}.

It is clear that |J| > 0. In fact, if |J| = 0, this implies that a.e x ∈ R^d,

|f(x)|< ¹₂ and kfk∞< ¹₂ , against our hypothesis. It follows that

|f(x)| ≥χJ(x)|f(x)| ≥ 1

2χJ(x), (11)

for almost all x ∈ R^d. Note that from Proposition 1, it is easy to show by induction that if

|h(x)| ≥ |g(x)|, a.e

then

kB_h⁺ⁿΦk² ≥ kB_g⁺ⁿΦk². (12) Hence, identities (11) and (12) imply that

X

n≥0

1

(n!)²kB_f⁺ⁿΦk² ≥X

n≥0

1

(n!)²kB⁺ⁿ1 2χJΦk². But, from Proposition 1, one has

kB⁺ⁿ1 2χJΦk²

(n!)² =c|J|

2n +n−1 n

kB⁺⁽ⁿ⁻¹⁾1 2χJ Φk²

((n−1)!)² ≥ n−1 n

kB⁺⁽ⁿ⁻¹⁾1 2χJ Φk² ((n−1)!)² ≥. . . . . .≥ n−1

n

n−2 n−1. . .1

2kB⁺1

2χJΦk² = 1 nkB⁺1

2χJΦk² = 1 4n|J|², which proves that the series P

n≥0 1

(n!)²kB⁺ⁿ1

2χjΦk² diverges. It follows that also the series P

n≥0 1

(n!)²kB⁺ⁿ_f Φk² diverges.

Theorem 3 implies that, whenever the quadratic exponential vectors are well defined, their scalar product

hΨ(f),Ψ(g)i exists. More precisely the following results hold.

Theorem 4 For all f, g∈L²(R^d)∩L^∞(R^d) such that kfk_∞< ¹₂ and kgk_∞ < ¹₂, then the integral

Z

R^d

ln(1−4 ¯f(s)g(s))ds exists. Moreover, one has

hΨ(f),Ψ(g)i=e^{−c2RRdln(1−4 ¯}f(s)g(s))ds. (13) Proof. Proposition 1 implies that, for any pair of step functions f = ρχI, g =σχI, where|ρ|< ¹₂, |σ|< ¹₂ and I ⊂R^d such that its Lebesgue measure

|I|<∞, one has hB_f⁺ⁿΦ, B_g⁺ⁿΦi = c

n−2

X

k=0

2^2k+3 n!(n−1)!

(((n−1)−k−1)!)²hf^k+2, g^k+2i

hB+((n−1)−k−1)

f Φ, B+((n−1)−k−1)

g Φi

+2nchf, gihB_f⁺⁽ⁿ⁻¹⁾Φ, B⁺⁽ⁿ⁻¹⁾_g Φi

= 4n(n−1)¯ρσh c

n−2

X

k=0

2^2k+1 (n−1)!(n−2)!

(((n−1)−k−1)!)²hf^k+1, g^k+1i hB+((n−1)−k−1)

f Φ, B+((n−1)−k−1)

g Φii

+2nc¯ρσ|I|hB_f⁺⁽ⁿ⁻¹⁾Φ, B_g⁺⁽ⁿ⁻¹⁾Φi

= h

4n(n−1)¯ρσ+ 2nc¯ρσ|I|i

hB⁺⁽ⁿ⁻¹⁾_f Φ, B⁺⁽ⁿ⁻¹⁾_g Φi.

This gives

hB_f⁺ⁿΦ, B_g⁺ⁿΦi

(n!)² = 4¯ρσhc|I|

2n +n−1 n

ihB_f⁺⁽ⁿ⁻¹⁾Φ, Bg⁺⁽ⁿ⁻¹⁾Φi

((n−1)!)² . (14) On the other hand, if I, J ⊂R^d such that |I|<∞, |J|<∞ and I∩J =∅, then by factorization property (see Theorem 1), one has

hΨρχI∪J,ΨσχI∪Ji=hΨρχI,ΨσχIihΨρχJ,ΨσχJi. (15) Therefore, if we put

F(ρ, σ,|I|) :=hΨρχI,ΨσχIi.

then from (14), it follows that

F(ρ, σ,|I|+|J|) =hΨρχI∪J,ΨσχI∪Ji.

for all I, J ⊂ R^d such that |I| < ∞, |J| < ∞ and I ∩J = ∅. Moreover, identity (15) implies that

F(ρ, σ,|I|+|J|) =F(ρ, σ,|I|)F(ρ, σ,|J|) Thus, there must exists ξ ∈R such that

F(ρ, σ,|I|) = e^|I|ξ

Put t = |I|. The number ξ is obtained by differentiating at t = 0. Using relation (14) one finds

d dt

t=0

hB_f⁺ⁿΦ, B_g⁺ⁿΦi (n!)² = d

dt t=0

4¯ρσ(ct

2n+n−1

n ). . .4¯ρσ(ct 2n+0)

= c 2

(4¯ρσ)ⁿ

n .