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C O L L O Q U I U M M A T H E M A T I C U M

VOL. 79 1999 NO. 1

CHARGE TRANSFER SCATTERING IN A CONSTANT ELECTRIC FIELD

BY

LECH Z I E L I ´ N S K I (PARIS)

We prove the asymptotic completeness of the quantum scattering for a Stark Hamiltonian with a time dependent interaction potential, created by N classical particles moving in a constant electric field.

1. Introduction. We consider a model describing the quantum dy- namics of a light particle (such as an electron) in collisions with some heavy particles (such as some ions) obeying the laws of classical dynamics. Thus only the light particle is considered a quantum particle, while the heavy par- ticles follow some classical trajectories R ∋ t 7→ χ k (t) ∈ R d . If V k denotes the quantum interaction potential between the quantum particle and the kth classical particle, the total quantum time-dependent interaction V (t) is the operator of multiplication by

(1.1) V (t, x) = X

1≤k≤N

V k (x − χ k (t)),

and the total time-dependent Hamiltonian H(t) is a self-adjoint operator in L 2 (R d ),

(1.2) H(t) = H 0 + V (t, x),

where H 0 denotes the free motion Hamiltonian. The subject of scattering theory is to describe the large time behaviour of the evolution propagator {U (t, t 0 )} t≥t 0 of H(t), that is, the family of unitary operators in L 2 (R d ) satisfying

(1.3) i d

dt U (t, t 0 )ϕ = H(t)U (t, t 0 )ϕ, U (t 0 , t 0 )ϕ = ϕ, for ϕ from the domain of H 0 .

The first papers describing such a model considered the case of linear classical trajectories and H 0 the Laplace operator [10, 25, 26]. The papers

1991 Mathematics Subject Classification: 81U10, 47A40, 47N50.

[37]

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[7, 29] deal with classical trajectories which are only asymptotically lin- ear and the papers [30, 31, 32] deal with the dispersive case when H 0 is a more general elliptic operator. We note that all these papers consider the hypothesis that different classical trajectories have different asymp- totic velocities lim t→∞ χ k (t), which implies the separation of trajectories:

k (t) − χ k (t)| ≥ ct with c > 0 if k 6= k .

The aim of this paper is to consider the situation arising in the presence of a constant electric field E ∈ R d \ {0}, when the free motion Hamiltonian for a particle of mass m > 0 and charge q 6= 0 has the form

h 0 (x, p) = p 2

2m − qE · x

and the Hamilton equations ˙p(t) = qE, ˙x(t) = p(t)/m have the solutions of the form

p(t) = qEt + mυ, x(t) = qE

2m t 2 + υt + ω,

where υ = p(0)/m ∈ R d and ω = x(0) ∈ R d . Thus the above solutions of the Hamilton equations describe the motion that is free in the directions orthogonal to the constant field E and uniformly accelerated in the direction parallel to E.

We shall consider only the simplest situation when different classical trajectories have different asymptotic accelerations lim t→∞ χ ′′ k (t). More pre- cisely we begin by assuming the following separation condition: there exist constants T 0 , c > 0, such that for t ≥ T 0 ,

(1.4) |χ k (t) − χ k (t)| ≥ ct 2 if 1 ≤ k < k ≤ N.

Let m k , q k be the mass and the charge of the kth classical particle and assume that χ(t) = (χ 1 (t), . . . , χ N (t)) is a solution of the Newton equations (1.5) m k χ ′′ k (t) = q k E − X

k ∈{1,...,N }\{k}

∇V k,k (χ k (t) − χ k (t)),

where the classical interaction potentials V k,k satisfy the decay condition (1.6) |∇V k,k (x)| ≤ C 0 |x| −1−µ 0 for |x| ≥ C 0

with C 0 , µ 0 > 0.

It is clear that (1.4)–(1.6) imply

(1.7) χ ′′ k (t) = z k + O(t −2(1+µ 0 ) ) with z k = q k

m k

E

as t → ∞, i.e. z k = (q k /m k )E = lim t→∞ χ ′′ k (t) is the asymptotic acceleration of the trajectory χ k . Since (1.7) means that dt d k (t) − z k t) = O(t −2−2µ 0 ), the limit

υ k = lim

t→∞ (χ k (t) − z k t)

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exists and introducing e χ k by the relation

(1.8) χ k (t) = 1 2 z k t 2 + υ k t + e χ k (t), we have

(1.9) χ e ′′ k (t) = O(t −2−2µ 0 ), χ e k (t) = O(t −1−2µ 0 ) as t → ∞.

The Hamiltonian of the free motion for a quantum particle of mass m 0 >

0 and charge q 0 6= 0 has the form

(1.10) H 0 = p 2

2m 0

− q 0 E · x, where p = (p 1 , . . . , p d ) = (−i∂ x 1 , . . . , −i∂ x d ).

For quantum interactions V k we assume that for some constants C, b C, ε 0

> 0,

V k (x)(1 + p 2 ) −1+ε 0 is a compact operator in L 2 (R d ), (1.11a)

|∂ x α V k (x)| ≤ C for |x · E| ≥ b C and |α| ≤ 2, (1.11b)

and V k = V k l + V k s with real valued functions V k l , V k s , such that for some µ > 0 we have

|∂ x α V k l (x)| ≤ C(1 + |x|) −µ−|α| for x ∈ R d and |α| ≤ 1, (1.11c)

|∂ x α V k s (x)| ≤ C(1 + |x|) −µ+(|α|−1)/2 for |x · E| ≥ b C (1.11d)

and |α| ≤ 1.

Theorem 1. Let U (t, t 0 ) be defined by (1.3) with H(t) given by (1.1), (1.2), (1.10). For k = 0, 1, . . . , N , let z k = q k E/m k be such that z k 6= z k if 0 ≤ k < k ≤ N . Assume that the trajectories χ k (t) have the form (1.8) with e

χ k (t) satisfying (1.9) for some µ 0 > 0. If V k = V k l + V k s satisfy (1.11a–d) for some µ > 0, ε 0 > 0, then the limit

(1.12)

Ω(t 0 )ψ = lim

t→∞ U (t, t 0 ) e −itH 0 −iS(t) ψ with S(t) =

t

\

1

dτ X

1≤k≤N

V k l 1 2 z 0 τ 2 − χ k (τ )  ,

exists in the norm of L 2 (R d ) for every ψ ∈ L 2 (R d ). Moreover , the asymp- totic completeness holds, i.e. the wave operator Ω(t 0 ) defined by (1.12) is unitary.

We recall the result of I. M. Sigal [20] (cf. also [3, 4, 5]) which guarantees the absence of eigenvalues for 2-body Stark Hamiltonians H k = H 0 + V k (x).

This allows us to neglect bound states and the asymptotic completeness

formulated in Theorem 1 implies that for every ϕ ∈ L 2 (R d ) there exists ψ ∈

L 2 (R d ) such that ϕ = Ω(t 0 )ψ. Thus U (t, t 0 )ϕ−e −itH 0 −iS(t) ψ → 0 as t → ∞,

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which means that the asymptotic behaviour of U (t, t 0 )ϕ is asymptotically the same as for the free evolution (modulo a phase factor e −iS(t) ).

We note that the approach used in the proof below comes from recent developments of scattering theory of N -body systems ([6, 8, 21]). We also mention the references [9, 12, 15–17, 19, 23, 24, 27, 28, 33] concerning Stark scattering in the 2-body case and [1, 2, 13, 14, 18, 22] in the N -body case.

In Section 2 we begin by describing in Lemma 2.1 asymptotic concentra- tion of the free evolution trajectories e −itH 0 ϕ on classical Stark trajectories.

Then it is easy to prove the existence of the wave operator Ω(t 0 ) given by (1.12). Clearly Ω(t 0 ) is an isometric injection and in order to prove the asymptotic completeness it suffices to prove the existence of the limit (1.12 ) Ω(t 0 ) ϕ = lim

t→∞ e itH 0 +iS(t) U (t, t 0 )ϕ

for every ϕ ∈ L 2 (R d ). Indeed, if Ω(t 0 ) given by (1.12 ) exists, then applying the chain rule we get Ω(t 0 )Ω(t 0 ) ϕ = ϕ, that is, Ω(t 0 ) is surjective and hence unitary.

To begin the proof of the existence of (1.12 ) we assume for simplicity V k s = 0 and introduce the auxiliary observable η t . This observable is used in Proposition 3.2 to introduce an energy cut-off, similarly to the “boosted Hamiltonian” of Graf [7]. However, instead of Enss approach used in Graf [7], our next step is based on the existence of the wave operators Ω k (t) of Proposition 3.7 (similar to the Deift–Simon operators of the N -body theory developed in Graf [8]). Then Proposition 3.7 allows us to localize and “distinguish” interactions of different classical charges, reducing the problem to the 2-body problem when the number of classical charges is N = 1.

The situation N = 1 is studied in Section 4 using the ideas of the Mourre estimate. More precisely, knowing that z 0 ·p is the conjugate operator for H 0

(i.e. we have the positive commutator [iH 0 , z 0 · p] = z 0 2 I), we find the propa- gation estimate of Proposition 4.3 using a suitable cut-off g 1 (z 0 ·p/t) instead of z 0 · p. Finally, in Section 5 we sketch the idea allowing one to modify the observable η t in order to recover all the previous results in the case of inter- action potentials with singularities, V k s 6= 0.

2. Preliminary estimates. For U ⊂ R d , C 0 (U ) is the set of smooth

functions with compact support in U . We write a t = O(f (t)) if there is a

constant C > 0 such that ka t k ≤ Cf (t), where k · k is the norm of L 2 (R d )

or the norm of bounded operators B(L 2 (R d )). The analogous notation will

be used when a t = (a 1 t , . . . , a d t ) assuming ka t k = (ka 1 t k 2 + . . . + ka d t k 2 ) 1/2 .

Moreover, a t = b t +O(f (t)) means a t −b t = O(f (t)). For Z ⊂ R, 1 Z denotes

the characteristic function of Z on R.

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Assume that V 0 is a real function satisfying

(2.1) |∂ t nx α V 0 (t, x)| ≤ Ct −2µ−2|α|−n for |α| + n ≤ 1, and denote by U 0 (t, t 0 ) the evolution propagator of the Hamiltonian (2.2) H 0 (t) = H 0 + V 0 (t, x),

where H 0 is given by (1.10). By rescaling we may assume further on that m 0 = 1.

Let y t = (y 1 t , . . . , y t d ), w t = (w 1 t , . . . , w d t ) be systems of d commuting self-adjoint operators,

(2.3) y t = 2x

t 2 − z 0 , w t = p t − z 0 .

Lemma 2.1. Let U 0 (t, t 0 ), y t , w t be as above and ϕ ∈ C 0 (R d ). Then (2.4) w t U 0 (t, t 0 )ϕ = O(t −1 ), y t U 0 (t, t 0 )ϕ = O(t −1 )

and for every κ > 0 and j = 1, . . . , d one has 1 [κ;∞[ (|y j t |)U 0 (t, t 0 )ϕ = O(t −1 ).

P r o o f. Define U 0 (t, t 0 ) = U t 0 , f (t) = U t 0∗ pU t 0 ϕ and g(t) = U t 0∗ xU t 0 ϕ.

Then

f (t) = U t 0∗ [iH 0 (t), p]U t 0 ϕ = z 0 ϕ + O(t −2(1+µ) ), g (t) = U t 0∗ [iH 0 (t), x]U t 0 ϕ = f (t)

= f (t 0 ) +

t

\

t 0

f (τ ) dτ = tz 0 ϕ + O(1), hence w t U t 0 ϕ = t −1 U t 0 (f (t) − z 0 tϕ) = O(t −1 ). Moreover,

g(t) = g(t 0 ) +

t

\

t 0

g (τ ) dτ = 1 2 z 0 t 2 ϕ + O(t), and x− 1 2 z 0 t 2 

U t 0 ϕ = U t 0 g(t)− 1 2 z 0 t 2 ϕ 

= O(t) implies the second estimate (2.4). Finally, using κ 2 1 [κ;∞[ (|λ|) ≤ λ 2 and the second estimate (2.4) we obtain

2 1 [κ;∞[ (|y j t |)U t 0 ϕ, U t 0 ϕ) ≤ ((y t j ) 2 U t 0 ϕ, U t 0 ϕ) = ky t j U t 0 ϕk 2 = O(t −2 ).

Note that (1.9) implies the existence of

(2.5) lim

t→∞ χ e k (t) = ω k with χ e k (t) = ω k + O(t −2µ 0 ), hence

(2.5 ) χ k (t) = z k t + υ k + O(t −1−2µ 0 ), χ k (t) = 1 2 z k t 2 + υ k t + O(1).

By rotation of the coordinate system we may assume further on that E =

(E 1 , 0, . . . , 0) with E 1 ∈ R \ {0}, hence z k = (z 1 k , 0, . . . , 0) with z k 1 =

(6)

E 1 q k /m k . Further, we set

(2.6) τ = 16 1 min{|z k 1 − z k 1 | : 0 ≤ k < k ≤ N }.

Fix J 0 ∈ C 0 (] − 4τ ; 4τ [) such that 0 ≤ J 0 ≤ 1, J 0 = 1 on [−2τ ; 2τ ], define J 0 = 1 − J 0 and let

V 0k (t, x) = J 0 (4x 1 /t 2 − 2z 1 k )V k l (x − χ k (t)) (2.7)

= J 0 (2y 1 t − 2e z k )V k l (x − χ k (t)) where we have set e z k = z 1 k − z 0 1 . Then we have

Proposition 2.2. Let V 0 = P

1≤k≤N V 0k , where V 0k is given by (2.7).

Then (2.1) holds and for every ϕ ∈ L 2 (R d ) the following limits exist:

(2.8)

Ω(t e 0 ) ϕ = lim

t→∞ e itH 0 +iS(t) U 0 (t, t 0 )ϕ, Ω(t e 0 )ϕ = lim

t→∞ U 0 (t, t 0 ) e −itH 0 −iS(t) ϕ.

P r o o f. Since χ k (t) = 1 2 z k t 2 + O(t) there is T 0 such that for t ≥ T 0 we have

J 0 (4x 1 /t 2 − 2z k 1 ) 6= 0 ⇒ |4x 1 /t 2 − 2z 1 k | ≥ 2τ

⇒ |x − χ k (t)| ≥

x 1 − 1 2 z k 1 t 2

1 2 z k t 2 − χ k (t)

1 2 τ t 2 − C t ≥ 1 4 τ t 2 and applying (1.11) we find

(2.9) |x − χ k (t)| ≥ 1 4 τ t 2 ⇒ |(∂ α V k l )(x − χ k (t))| ≤ Ct −2(µ+|α|) if |α| ≤ 1.

We conclude that V 0 satisfies (2.1) noting that

∂x 1

(J 0 (4x 1 /t 2 − 2z k 1 )) = O(t −2 ), ∂

∂t (J 0 (4x 1 /t 2 − 2z k 1 )) = O(t −1 ).

Since C 0 (R d ) is dense in L 2 (R d ), to obtain the existence of e Ω(t 0 ) ϕ it suffices to consider ϕ ∈ C 0 (R d ) and to check that

(2.10) d

dt (e itH 0 +iS(t) U 0 (t, t 0 )ϕ)

= e itH 0 +iS(t) i(S (t) − V 0 (t, x))U 0 (t, t 0 )ϕ = O(t −1−2µ ).

However, for 1 ≤ k ≤ N we have |z k 1 − z 0 1 | ≥ 16τ , hence J 0 (2z 0 1 − 2z 1 k ) = 1 and

(2.11) V 0 t, 1 2 z 0 t 2 

= X

1≤k≤N

J 0 (2z 0 1 − 2z k 1 )V k l 1 2 z 0 t 2 − χ k (t) 

= S (t).

Thus we may write

V 0 (t, x) − S (t) = V 0 (t, x) − V 0 t, 1 2 z 0 t 2 

= γ t · x − 1 2 z 0 t 2 

= 1 2 γ t · t 2 y t

(7)

with

γ t =

1

\

0

dθ ∇ x V 0 t, (1 − θ)x + 1 2 θz 0 t 2  and (2.1) implies t 2 γ t = O(t −2µ ). Therefore

k(S (t) − V 0 (t, x))U 0 (t, t 0 )ϕk = 1 2 t 2 γ t · y t U 0 (t, t 0 )ϕ (2.12)

≤ Ct −2µ ky t U 0 (t, t 0 )ϕk

and by (2.4) the right hand side of (2.12) is O(t −1−2µ ), i.e. (2.10) follows.

We may use V 0 (t, x) = 0 in Lemma 2.1, hence it is clear that e −itH 0 satisfies the same estimates as U 0 (t, t 0 ), and we obtain the existence of the second limit (2.8) as above with e −itH 0 and U 0 (t, t 0 ) interchanged.

Proof of the existence of Ω(t 0 ). Using the chain rule and the existence of (2.8), we note that it suffices to prove the existence of lim t→∞ U (t, t 0 )

× U 0 (t, t 0 )ϕ, where as before we may assume ϕ ∈ C 0 (R d ). Let J ∈ C 0 (R d ) be such that J(x) = 1 for |x| ≤ τ , J(x) = 0 for |x| ≥ 2τ , 0 ≤ J ≤ 1. Then Lemma 2.1 implies

k(1 − J)(y t )U 0 (t, t 0 )ϕk ≤ k1 [τ ;∞[ (|y t |)U 0 (t, t 0 )ϕk = O(t −1 ), i.e.

t→∞ lim U (t, t 0 ) J(y t )U 0 (t, t 0 )ϕ = lim

t→∞ U (t, t 0 ) U 0 (t, t 0 )ϕ and it suffices to show that

(2.13) d

dt (U (t, t 0 ) J(y t )U 0 (t, t 0 )ϕ)

= U (t, t 0 ) (D H 0 J(y t ) + iJ(y t )(V (t, x) − V 0 (t, x)))U 0 (t, t 0

= O(t −1−2µ ) + O(t −2 ),

where D a t b t = [ia t , b t ] + dt d b t denotes the Heisenberg derivative.

However, a simple calculation gives

(2.14) D H

0 J(y t ) = 2 t

X

1≤j≤d

∂ j J(y t )(w j t − y t j ) + O(t −3 ) and using (2.4) we obtain (D H 0 J(y t ))U 0 (t, t 0 )ϕ = O(t −2 ).

Next for 1 ≤ k ≤ N we have J 0 (2y 1 t − 2e z k ) 6= 0 ⇒ |y t 1 − e z k | < 2τ

⇒ |y t 1 | ≥ |e z k | − 2τ = |z k 1 − z 0 1 | − 2τ ≥ 14τ ⇒ J(y t ) = 0, hence J(y t )J 0 (2y t 1 − 2e z k ) = J(y t ) and

J(y t )(V − V 0 )(t, x) = X

1≤k≤N

J(y t )J 0 (2y t 1 − 2e z k )V k s (x − χ k (t)).

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If T 0 is as at the beginning of the proof of Proposition 2.2, then for t ≥ T 0

we have

J 0 (2y t 1 − 2e z k ) 6= 0 ⇒ |x − χ k (t)| ≥ 1 4 τ t 2 ⇒ |V k s (x − χ k (t))| ≤ Ct −1−2µ . Until the end of Section 4 we assume that V k s = 0, that is, V k = V k l . We now introduce

η 0 t = 1 2

 p 1 t − 2x 1

t 2

 2

+ 1 4

 2x 1 t 2 − z 0 1

 2

+ 1 2

X

2≤j≤d

p 2 j t 2 + I, (2.15)

η t = η 0 t + V (t, x) t 2 . (2.16)

Lemma 2.3. If η t 0 , η t are given by (2.15)–(2.16) and D is defined as below (2.13), then D H(t) η t = D H 0 η 0 t + r t with

(2.17) r t = d dt

 V (t, x) t 2





iV (t, x), x 1 p 1 + p 1 x 1

t 3

 . P r o o f. A simple transformation of the expression (2.15) gives

η t 0 = 1 2

 p 2 1

t 2 − 2 x 1 p 1 + p 1 x 1

t 3 + 4x 2 1 t 4



+ 1 4

 4 x 2 1

t 4 − 4 z 1 0 x 1

t 2 + (z 0 1 ) 2

 + 1

2 X

2≤j≤d

p 2 j t 2 + I

= 1 2

p 2 1

t 2 − x 1 p 1 + p 1 x 1

t 3 +

 1

2 · 4 + 1 4 · 4

 x 2 1 t 4

− z 0 1 x 1

t 2 + (z 0 1 ) 2 4 + 1

2 X

2≤j≤d

p 2 j t 2 + I

= 1 t 2

 1

2 p 2 − z 0 1 x 1



− x 1 p 1 + p 1 x 1

t 3 + 3x 2 1

t 4 + (z 0 1 ) 2 4 + I.

Therefore we may express η 0 t in the following way:

(2.15 ) η 0 t = H 0

t 2 − x 1 p 1 + p 1 x 1 t 3 + 3x 2 1

t 4 + (z 0 1 ) 2 4 + I and compute

D H(t) η t = D H(t)



η t 0 + V (t, x) t 2



= D H(t)

 H(t)

t 2 − x 1 p 1 + p 1 x 1

t 3 + 3x 2 1 t 4



= D H(t)

 H(t) t 2





iV (t, x), x 1 p 1 + p 1 x 1 t 3



+ D H 0



− x 1 p 1 + p 1 x 1

t 3 + 3x 2 1 t 4



(9)

= d dt

 H 0

t 2 + V (t, x) t 2





iV (t, x), x 1 p 1 + p 1 x 1 t 3



+ D H 0



− x 1 p 1 + p 1 x 1

t 3 + 3x 2 1 t 4



= D H 0

 H 0

t 2

 + D H 0



− x 1 p 1 + p 1 x 1

t 3 + 3x 2 1 t 4



+ r t = D H 0 η t 0 + r t . Lemma 2.4. If r t is given by (2.17) then r t = O(t −2 ).

P r o o f. First note that d

dt (t −2 V (t, x)) = t −2 ∂ t V (t, x) − 2t −3 V (t, x) = t −2 ∂ t V (t, x) + O(t −3 ).

Thus setting χ k (t) = ( ˙ χ 1 k (t), ˙ χ k (t)) ∈ R × R d−1 and using ˙ χ k (t) = O(1), we have

t 2 r t = ∂ t V (t, x) −



iV (t, x), x 1 p 1 + p 1 x 1 t



+ O(t −1 )

= X

1≤k≤N

∂ x 1 V k (x − χ k (t))

 2x 1

t − ˙χ 1 k (t)



+ O(1).

But 2x 1 /t − ˙ χ 1 k (t) = (2/t)(x 1 − χ 1 k (t)) + O(1) by (2.5 ) and we complete the proof noting that ∂ x 1 V k (x − χ k (t))(x 1 − χ 1 k (t)) = O(1).

Proposition 2.5. If η t is given by (2.16) and D as below (2.13), then (2.18) D H(t) η t = − 3

t

 p 1 t − 2x 1

t 2

 2

− X

2≤j≤d

p 2 j

t 3 + O(t −2 ).

P r o o f. By Lemmas 2.3 and 2.4 it suffices to check that

(2.19) D H

0 η t 0 = − 3 t

 p 1

t − 2x 1

t 2

 2

− X

2≤j≤d

p 2 j t 3 . Now we note that formally

(2.20) D a

t (b t eb t ) = (D a t b t )eb t + (b t D a

t eb t ).

If a t and b t are self-adjoint, then (2.20 ) D a

t (b t ) 2 = b t (D a t b t ) + (D a t b t )b t = 2b t (D a t b t ) + hc,

where m t + hc = 1 2 (m t + m t ) denotes the Hermitian symmetrization of the operator m t . In particular, using

(2.21) D H

0 w t = − w t

t , D H

0 y 1 t = 2

t (w 1 t − y t 1 )

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[where w t , y t are given by (2.3)], we obtain 1

4 D H

0 (y t 1 ) 2 = 1 2 y 1 t D H

0 y t 1 + hc = 1

t y 1 t (w 1 t − y 1 t ) + hc, 1

2 D H

0 (w t 1 − y t 1 ) 2 = (w 1 t − y 1 t )D H 0 (w t 1 − y t 1 ) + hc

= 1

t (w 1 t − y 1 t )(2y t 1 − 3w 1 t ) + hc.

Introducing w t = (w t 2 , . . . , w d t ) = (p 2 /t, . . . , p d /t) we may express (2.15) in the form

η t 0 = 1 2 (w t 1 − y t 1 ) 2 + 1 4 (y 1 t ) 2 + 1 2 |w t | 2 + I and it is clear that 1 2 D H

0 |w t | 2 = − 1 t |w t | 2 . To complete the proof we com- pute

1 2 D H

0 (w 1 t − y 1 t ) 2 + 1 4 D H

0 (y t 1 ) 2

= 1

t (w 1 t − y t 1 )(2y 1 t − 3w 1 t ) + 1

t (w 1 t − y 1 t )y t 1 + hc

= 1

t (w 1 t − y t 1 )(3y 1 t − 3w 1 t ) + hc = − 3

t (w 1 t − y t 1 ) 2 . 3. Propagation estimates. We denote by G(H) the set of operator- valued functions t 7→ M (t) ∈ B(L 2 (R d )) satisfying

(3.1)

T

\

1

dt Re(M (t)U (t, t 0 )ϕ, U (t, t 0 )ϕ) ≤ Ckϕk 2 for all ϕ ∈ L 2 (R d ), all T ≥ 1 and for some constant C > 0.

Sometimes we write M (t) ∈ G(H(t)) instead of M ∈ G(H). We note that if M (t) = O(t −1−ε ) with ε > 0, then M ∈ G(H),

(3.2)

if ( f M ∈ G(H) and M (t) ≤ f M (t) for all t ≥ 1), then M ∈ G(H).

(3.3)

If D H(t) M (t) is well defined, then writing U (t, t 0 )ϕ = ϕ t we have

(3.4)

T

\

1

dt ((D H(t) M (t))ϕ t , ϕ t ) =

T

\

1

dt d

dt (M (t)ϕ t , ϕ t ) = [(M (t)ϕ t , ϕ t )] T 1 and if M (t) = O(1), then D H(t) M (t) ∈ G(H(t)).

Note that η t 0 ≥ I and η t = η 0 t + O(t −2 ), hence for n ≥ 1, t ≥ T 0 , e η n,t = (1 + η t /n) −1 is well defined and satisfies 0 ≤ e η n,t ≤ I. Introducing

(3.5) M 0 (t) = 1

t η e n,t (3(w 1 t − y 1 t ) 2 + |w t | 2 )e η n,t ,

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we find that Proposition 2.5 gives

(3.6) nD H(t) e η n,t = −e η n,t (D H(t) η t )e η n,t = M 0 (t) + O(t −2 ).

It is clear that (3.4), (3.2) and (3.6) give

Corollary 3.1. If M 0 is given by (3.5), then M 0 ∈ G(H).

Proposition 3.2. For every ϕ ∈ L 2 (R d ) we have

n→∞ lim sup

t≥T 0

k(I − e η 2 n,t )U (t, t 0 )ϕk = 0.

P r o o f. First we set U (t, t 0 )ϕ = ϕ t and note that 0 ≤ λ ≤ 1 ⇒ (1 − λ 2 ) 2 ≤ 4(1 − λ), hence

k(I − e η n,t 2 )ϕ t k 2 = ((I − e η 2 n,t ) 2 ϕ t , ϕ t ) ≤ 4((I − e η n,t )ϕ t , ϕ t ).

It remains to note that e η n,T 0 ϕ T 0 → ϕ T 0 as n → ∞, and −nD H(t) η e n,t

−M 0 (t) + Ct −2 ≤ Ct −2 allows us to estimate [((I − e η n,t )ϕ t , ϕ t )] T T 0 = −

T

\

T 0

dt ((D H(t) η e n,t )ϕ t , ϕ t ) ≤

T

\

T 0

dt Ct −2 /n ≤ C/n.

Further on in this section we assume n ≥ 1 fixed and write simply e

η t = e η n,t . As below (2.20 ), M (t)+hc denotes the symmetrization 1 2 (M (t)+

M (t) ).

Lemma 3.3. Let J 0 ∈ C 0 (R). Then M 1 ∈ G(H) if

(3.7) M 1 (t) = 1

t η e t (y 1 t − w t 1 )J 0 (y t 1 )e η t + hc.

P r o o f. Let J ∈ C (R) be such that the derivative J = −J 0 , and set M 1,0 (t) = e η t J(y 1 t )e η t .

Then D H(t) M 1,0 = M 1,1 + M 1,2 with

M 1,1 (t) = e η t (D H(t) J(y 1 t ))e η t = 2M 1 (t) + O(t −3 ), M 1,2 (t) = 2e η t J(y t 1 )D H(t) η e t + hc.

From (3.4) we have D H(t) M 1,0 ∈ G(H) and it is clear that in order to show M 1 ∈ G(H) it suffices to check that −M 1,2 ∈ G(H).

Noting that

w t η e t = O(1), y t 1 η e t = O(1), (w 1 t − y 1 t )e η t = O(1), it is easy to estimate the commutators

n[e η t , w t ] = −e η t0 t + O(t −2 ), w t ]e η t = O(t −2 ), n[e η t , w t 1 − y t 1 ] = e η t [η t 0 + O(t −2 ), y 1 t − w t 1 ]e η t

= e η t  1

4 (y t 1 ) 2 , y t 1 − w 1 t  e

η t + O(t −2 ) = O(t −2 ),

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n[e η t , J(y 1 t )] = −e η t [η 0 t , J(y 1 t )]e η t = O(t −2 ).

Using (2.18) to express D H(t) η e t in M 1,2 (t) it is easy to see that the above commutator estimates allow us to write

−M 1,2 (t) = 2

t η e t (3(w 1 t − y 1 t )a t (w t 1 − y t 1 ) + w t a t w t )e η t + O(t −2 ) with a t = −n −1 J(y t 1 )e η t + hc, and it is clear that the inequality a t ≤ CI implies

(3.8) −M 1,2 (t) ≤ 2CM 0 (t) + Ct −2

where M 0 is given by (3.5). By Lemma 3.3 the right hand side of (3.8) belongs to G(H) and consequently −M 1,2 ∈ G(H).

Proposition 3.4. Let J 0 ∈ C 0 (R \ {e z 1 , . . . , e z N }) where e z k = z k 1 − z 0 1 . Then M 2 ∈ G(H) if

(3.9) M 2 (t) = 1

t η e t J 0 (y 1 t )y 1 t η e t .

P r o o f. If M 1 is given by (3.7), then M 1 ∈ G(H) and M 2 = 3M 1 + M 3

with

M 3 (t) = 1

t η e t (3w t 1 − 2y t 1 )J 0 (y 1 t )e η t + hc.

Thus it remains to show that M 3 ∈ G(H). But for 1 ≤ k ≤ N , e z k 6∈ supp J 0

and

J 0 (y t 1 ) 6= 0 ⇒ |y 1 t − e z k | = |2x 1 /t 2 − z k 1 | ≥ c > 0

⇒ |x − χ k (t)| ≥

x 1 − 1 2 z k 1 t 2

− C t ≥ 1 2 ct 2 − C t implies

[iV (t, x), w 1 t ]J 0 (y t ) = −∂ x V (t, x)J 0 (y t )t −1 = O(t −3 ).

Therefore introducing

M 3,0 (t) = e η t (y t 1 − w 1 t )J 0 (y t )e η t + hc, we find that D H(t) M 3,0 = M 3,1 + M 3,2 + M 3,3 with

M 3,1 (t) = e η t (D H(t) (y t 1 − w t 1 ))J 0 (y t 1 )e η t = M 3 (t) + O(t −3 ), M 3,2 (t) = e η t (y 1 t − w t 1 )(D H(t) J 0 (y t 1 ))e η t + hc,

M 3,3 (t) = 2e η t (y 1 t − w 1 t )J 0 (y 1 t )D H(t) η e t + hc.

As before, (3.4) gives D H(t) M 3,0 ∈ G(H) and M 3 ∈ G(H) follows if we know that −M 3,2 , −M 3,3 ∈ G(H). To show −M 3,3 ∈ G(H) we note that we may replace M 1,2 by M 3,3 in (3.8) using a t = n −1 J 0 (y 1 t )(w 1 t − y 1 t )e η t + hc ≤ CI to express −M 3,3 similarly to −M 1,2 . Also

−M 3,2 (t) = − 2

t η e t (y t 1 − w 1 t )J 0 (y 1 t )(y t 1 − w 1 t )e η t + O(t −3 )

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≤ CM 0 (t) + Ct −3 ∈ G(H(t)).

We keep the notations J 0 , e z k , V 0k , V 0 , H 0 (t), U 0 (t, t 0 ) introduced in Sec- tion 2. Moreover, for 1 ≤ k ≤ N we denote by U k (t, t 0 ) the evolution propagator of the Hamiltonian

(3.10)

H k (t) = H 0 + V k (t, x) with V k (t, x) = V k (x − χ k (t)) + X

k ∈{1,...,N }\{k}

V 0k (t, x).

Corollary 3.5. If M 0 , M 2 , H k are as above, then M 0 , M 2 ∈ G(H k ).

P r o o f. Define η t k by using V k (t, x) instead of V (t, x) in (2.16). As before we obtain

M 0 k (t) = 1

t η e t k (3(w t 1 − y t 1 ) 2 + |w t | 2 )e η k t ∈ G(H k (t))

with e η t k = (1 + η k t /n) −1 . We recall that |∂ t nx α V 0k (t, x)| ≤ Ct −2µ−2|α|−n for

|α| + n ≤ 1, and reasoning as in the proof of Proposition 3.4 we find M 2 k (t) = 1

t e η k t J 0 (y t 1 )y 1 t η e k t ∈ G(H k (t))

for J 0 ∈ C 0 (R \ {e z 1 , . . . , e z N }). However, η t = η t k + O(t −2 ) implies

((w t 1 −y t 1 ) 2 +|w t | 2 )(e η t − e η k t ) = ((w 1 t −y 1 t ) 2 +|w t | 2 )e η t k (η t −η k t )e η t /n = O(t −2 ), hence

M j (t) = M j k (t) + O(t −2 ) ∈ G(H k (t)), j = 0, 2.

The following well known lemma is the basic tool allowing us to obtain the existence of wave operators (we give its proof in the Appendix):

Lemma 3.6. Let U (t, t 0 ) and e U (t, t 0 ) be the evolution propagators of H(t) = H 0 + V (t) and e H(t) = H 0 + e V (t) respectively. Assume that for M (t) ∈ B(L 2 (R d )) we may define D H 0 M (t) as bounded operators with (3.11) ( e V (t) − V (t))M (t) = O(t −1−ε ) and

D H

0 M (t) = f M (t) + O(t −1−ε )

where ε > 0, and that there exists f M 0 ∈ G(H)∩G( e H) satisfying the estimates (3.11 ) − f M 0 (t) ≤ f M (t) ≤ f M 0 (t) and M f 0 (t) ≥ 0 for all t ≥ 1.

If ϕ ∈ L 2 (R d ) and Ω t = e U (t, t 0 ) M (t)U (t, t 0 ), then the limit lim t→∞t ϕ exists.

Proposition 3.7. Set J(y t 1 ) = 1 − P

1≤k≤N J 0 (y 1 t − e z k ) 2 and define (3.12) Ω 0 (t, t 0 ) = U 0 (t, t 0 ) J(y t 1 )U (t, t 0 ),

Ω k (t, t 0 ) = U k (t, t 0 ) J 0 (y t 1 − e z k )U (t, t 0 ) for k = 1, . . . , N.

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Then for every ϕ ∈ L 2 (R d ), k = 0, 1, . . . , N , the following limits exist:

(3.12 ) Ω k (t 0 )ϕ = lim

t→∞ Ω k (t, t 0 )ϕ.

P r o o f. Consider first the case k = 0. By Proposition 3.2 it suffices to show that

t→∞ lim U 0 (t, t 0 ) J(y t 1 )e η 2 n,t U (t, t 0 )ϕ

exists for every n ≥ 1. Further on n is fixed, we write e η t = e η n,t and we apply Lemma 3.6 with e H(t) = H 0 (t) and M (t) = J(y 1 t )e η 2 t .

We begin by noting that the first condition of (3.11) follows from (3.13) (H(t) − H 0 (t))J(y 1 t ) = X

1≤k≤N

J 0 (2y t 1 − 2e z k )V k l (x − χ k (t))J(y t 1 ) = 0.

To check (3.13) we note that J 0 (y t 1 − e z k ) 6= 0 ⇒ |y t 1 − e z k | < 4τ and for k 6= k we have |e z k −e z k | = |z k 1 −z 1 k | ≥ 16τ , hence J 0 (y t 1 −e z k ) 6= 0 ⇒ J 0 (y 1 t −e z k ) = 0 for k 6= k. Thus it is clear that J 0 (2y 1 t − 2e z k ) 6= 0 ⇒ |y 1 t − e z k | < 2τ ⇒ J 0 (y t 1 − e z k ) = 1 ⇒ J(y 1 t ) = 1 − J 0 (y 1 t − e z k ) 2 = 0.

Next we find that D H 0 M = f M 1 + f M 2 with M f 1 (t) = (D H 0 J(y 1 t ))e η t 2 = 2

t η e t (w 1 t − y t 1 )J (y t 1 )e η t + hc + O(t −2 ), (3.14)

M f 2 (t) = 2e η t J(y 1 t )D H(t) η e t + hc + O(t −2 ).

(3.15)

Next for k = 1, . . . , N , we have |y t 1 | ≤ 2τ ⇒ |y 1 t − e z k | ≥ 14τ ⇒ J 0 (2y t 1 − 2e z k ) = 0. Therefore J = 1 on [−2τ ; 2τ ] and 0 6∈ supp J allows us to define J 0 ∈ C 0 (R \ {e z 1 , . . . , e z N }) satisfying J 0 (λ)λ = J (λ) 2 and to estimate (3.16) ± (w t 1 − y t 1 )J (y t 1 ) + hc ≤ 2(w 1 t − y 1 t ) 2 + 2J 0 (y t 1 )y 1 t

⇒ ± f M 1 ≤ 4M 0 + 4M 2

with M 0 , M 2 given by (3.5), (3.9). Then similarly to the proof of Lemma 3.3 we find ± f M 2 (t) ≤ CM 0 (t) + Ct −2 , hence it is clear that the hypotheses of Lemma 3.6 hold with f M 0 = C 0 M 0 + 4M 2 ∈ G(H) ∩ G(H k ) by Corollary 3.1, 3.5 and Proposition 3.4.

In the case k = 1, . . . , N , we apply Lemma 3.6 with e H(t) = H k (t) and M (t) = e J(y t 1 )e η t 2 , where e J(λ) = J 0 (λ − e z k ). As before we have

(3.17) (H(t) − H k (t)) e J (y t 1 )

= X

k ∈{1,...,N }\{k}

J 0 (2y t 1 − 2e z k )V k l (x − χ k (t)) e J (y 1 t ) = 0.

Indeed, e J(y t 1 ) 6= 0 ⇒ |y t 1 − e z k | < 4τ ⇒ |y 1 t − e z k | ≥ 2τ for k 6= k ⇒

J 0 (2y t 1 − 2e z k ) = 0 for k 6= k. We complete the proof noting that e J = 0 on

[−2τ ; 2τ ] and (3.14)–(3.16) still hold if J is replaced by e J.

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4. Asymptotic completeness. In order to obtain the asymptotic completeness it remains to prove

Proposition 4.1. If k = 1, . . . , N and ϕ ∈ L 2 (R d ), then

t→∞ lim J 0 (y t 1 − e z k )U k (t, t 0 )ϕ = 0.

Indeed, using Propositions 2.2, 3.7 and 4.1, we can see that via the chain rule,

e itH 0 +iS(t) U (t, t 0 )ϕ = e itH 0 +iS(t) 

J(y 1 t ) + X

1≤k≤N

J 0 (y 1 t − e z k ) 2 

U (t, t 0 )ϕ

= e itH 0 +iS(t) U 0 (t, t 0 )Ω 0 (t, t 0 )ϕ

+ X

1≤k≤N

e itH 0 +iS(t) J 0 (y 1 t − e z k )U k (t, t 0 )Ω k (t, t 0 )ϕ converges to e Ω 0 (t 0 ) Ω 0 (t 0 )ϕ, i.e. the limit (1.12 ) exists.

Before starting the proof of Proposition 4.1 we introduce more notation.

We set

(4.1) H 0k = 1 2 p 2 + e z k x 1 , H e k = H 0k + V k (x − ω k ), where k = 1, . . . , N and ω k is as in (2.5). We define

χ 0 k (t) = 1 2 z k t 2 + υ k t, χ ˙ 0 k (t) = z k t + υ k , (4.2)

H e k (t) = H 0k + e V k (t, x) (4.3)

with

V e k (t, x) = V k (t, x + χ 0 k (t))

= V k (x − e χ k (t)) + X

k ∈{1,...,N }\{k}

V 0k t, x + 1 2 z k t 2 + υ k t  . It is easy to see that V 0k t, x+ 1 2 z k t 2k t 

satisfies estimates (2.1) similarly to V 0k . The following lemma allows us to compare e H k and e H k (t).

Lemma 4.2. (a) We have V k (x − e χ k (t)) = V k (x − ω k ) + O(t −2µ 0 ) and d

dt V k (x − e χ k (t)) = −e χ k (t) · ∇V k (x − e χ k (t)) = O(t −1−2µ 0 ).

(b) If h ∈ C 0 (R) then h( e H k (t)) = h( e H k ) + O(t −2µ 0 ) + O(t −2µ ) and D e H k (t) h( e H k (t)) = d

dt h( e H k (t)) = O(t −1−2µ 0 ) + O(t −1−2µ ).

(c) If g, e h ∈ C 0 (R) then [e h( e H k ), g( e w t )] = O(t −1 ).

We note that our assumptions ∇V k = ∇V k l = O(1) and (1.9) give imme-

diately the indicated estimate of dt d V k (x − e χ k (t)), while the first estimate of

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Lemma 4.2(a) follows by integration. The proof of estimates in (b) and (c) is given in the Appendix.

Proposition 4.3. Let g ∈ C 0 

3 4 |e z k |; 3 4 |e z k | 

and h ∈ C 0 (R). Then (4.4) M f h (t) = 1

t h( e H k (t))g( e w t ) 2 h( e H k (t)) ∈ G( e H k (t)), where we have set e w t = p 1 /t.

P r o o f. Let n ∈ N be such that h ∈ C 0 (]−n; n[). Since ( f M h (t)ϕ, ϕ) = t −1 kg( e w t )h( e H k (t))ϕk 2 , it is clear that f M h 1 +h 2 (t) ≤ 2 f M h 1 (t) + 2 f M h 2 (t).

Thus it suffices to show that for every λ ∈ [−n; n] there is δ > 0 such that M f h (t) ∈ G( e H k (t)) with h ∈ C 0 (]λ − δ; λ + δ[), |h| ≤ 1.

Let g 1 ∈ C (R) satisfy g 1 = −g 2 and set

M 0 (t) = e z k h( e H k (t))g 1 ( e w t )h( e H k (t)).

Let ε = min{1, 2µ 0 , 2µ}. Then Lemma 4.2 allows us to write

D e H k (t) M 0 (t) = e z k h( e H k (t))(D e H k (t) g 1 ( e w t ))h( e H k (t)) + O(t −1−ε )

= e z k h( e H k )(D e H k g 1 ( e w t ))h( e H k ) + O(t −1−ε ).

We now show that choosing δ > 0 small enough we have (4.5) e z k h( e H k )[iV k (x−ω k ), g 1 ( e w t )]h( e H k ) ≥ − z e 2 k

8t h( e H k )g( e w t ) 2 h( e H k )−Ct −2 . Using (1.11b) and the standard pseudo-differential expansion [(A.1) of Ap- pendix with n = 2 and then with n = 1] we find the following expression of the commutator:

[iV k (x − ω k ), g 1 ( e w t )] = − 1

t ∂ x 1 V k (x − ω k )g 1 ( e w t ) + O(t −2 ) (4.6)

= 1

t g( e w t )∂ x 1 V k (x − ω k )g( e w t ) + O(t −2 ),

and since e H k has no eigenvalues (cf. [20]), 1 [λ−2δ;λ+2δ] ( e H k ) → 0 strongly as δ → 0. As ∂ x 1 V k (x − ω k )1 [−n;n] ( e H k ) is compact, for δ > 0 small enough we have

(4.7) z e k eh( e H k )∂ x 1 V k (x − ω k )e h( e H k ) ≥ − 1 8 e z k 2

if e h ∈ C 0 (]λ−2δ; λ+2δ[), 0 ≤ e h ≤ 1. Using e h such that h = he h and Lemma 4.2(c) we obtain (4.5) from (4.6)–(4.7). Next we note that D H 0k g 1 ( e w t ) =

−t −1 (e z k + e w t )g 1 ( e w t ) = t −1 (e z k + e w t )g( e w t ) 2 and since λ ∈ supp g ⇒ |λ| ≤

3

4 |e z k | ⇒ e z k (e z k + λ) ≥ 1 4 z e k 2 , it is clear that (4.8) z e k h( e H k )(D H 0k g 1 ( e w t ))h( e H k ) ≥ 1

4t z e k 2 h( e H k )g( e w t ) 2 h( e H k ).

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Let M 1 , M 2 denote the left hand sides of (4.5) and (4.8). Then (4.4) follows from

1

8 e z k 2 M f h (t) ≤ (M 1 + M 2 )(t) + Ct −1−ε

= D e H k (t) M 0 (t) + O(t −1−ε ) ∈ G( e H k (t)).

Proof of Proposition 4.1 . Step 1. Introduce

(4.9)

G k (t) = e −iΦ k (t) e −ix· ˙ χ 0 k (t) e ip·χ 0 k (t) where Φ k (t) =

t

\

1

dτ z 0 · χ 0 k (τ ) + 1 2 χ ˙ 0 k (τ ) 2  .

Since e −ix· ˙ χ 0 k (t) p = (p + ˙ χ 0 k (t))e −ix·χ 0 k (t) and e ip·χ 0 k (t) x = (x + χ 0 k (t))e ip·χ 0 k (t) , we compute

G k (t) = − z 0 · χ 0 k (t) − 1 2 χ ˙ 0 k (t) 2 − x · z k + (p + ˙ χ 0 k (t)) · ˙ χ 0 k (t)  iG k (t), iG k (t)H k (t) = 1 2 (p + ˙ χ 0 k (t)) 2 − z 0 · (x + χ 0 k (t)) + V k (t, x + χ 0 k (t)) 

iG k (t)

= e H k (t) + p · ˙ χ 0 k (t) + 1 2 χ ˙ 0 k (t) 2 − z k · x − z 0 · χ 0 k (t)  iG k (t)

= i e H k (t)G k (t) + G k (t).

Thus we have d

dt ( e U k (t, t 0 ) G k (t)U k (t, t 0 )ϕ)

= e U k (t, t 0 ) (G k (t) + i e H k (t)G k (t) − iG k (t)H k (t))U k (t, t 0 )ϕ = 0, which implies

(4.10) U e k (t, t 0 ) = G k (t)U k (t, t 0 )G k (t 0 ) −1 . We write e y t = 2x 1 /t 2 . Then

G k (t)J 0 (y 1 t − e z k ) = J 0 (e y t + 2υ k 1 /t)G k (t) = J 0 (e y t )G k (t) + O(t −1 ) and using (4.10) we obtain

(4.11) lim

t→∞ kJ 0 (y 1 t − e z k )U k (t, t 0 )ϕk = lim

t→∞ kJ 0 (e y t ) e U k (t, t 0 )G k (t 0 )ϕk.

Step 2 . It suffices to show that for every h ∈ C 0 (R) we have

(4.12) lim inf

t→∞ kJ 0 (e y t )h( e H k (t)) e ϕ t k = 0 where we have set e ϕ t = e U k (t, t 0 )G k (t 0 )ϕ.

Indeed, note first that (4.11) is the limit of the norms of ϕ(t) = U k (t, t 0 )

× J 0 (y t 1 − e z k )U k (t, t 0 )ϕ and that ϕ(t) converges in L 2 (R d ), by a reasoning

analogous to the proof of Proposition 3.7. Thus the limits (4.11) exist and

we may replace them by lim inf.

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However, taking h 0 ∈ C 0 (R) such that h 0 = 1 in a neighbourhood of 0, 0 ≤ h 0 ≤ 1, we have h 0 ( e H k (T 0 )/n)ψ → ψ as n → ∞ and by Lemma 4.2(b),

[((I − h 0 ( e H k (t)/n)) 2 ϕ e t , e ϕ t )] T T 0 =

T

\

T 0

dt (D e H k (t) (I − h 0 ( e H k (t)/n)) 2 ϕ e t , e ϕ t )

T

\

T 0

dt C ϕ t −1−2 min{µ,µ 0 } /n ≤ e C ϕ /n, which implies

n→∞ lim sup

t≥T 0

k(I − h 0 ( e H k (t)/n)) e ϕ t k = 0.

Step 3 . Instead of (4.12) it suffices to show that M (t) ∈ G( e H k (t)) with

(4.13) M (t) = 1

t h( e H k (t))J 0 (e y t ) 2 h( e H k (t)).

Indeed, (M (t) e ϕ t , e ϕ t ) ∈ L 1 ([t 0 ; ∞[, dt) implies 0 = lim inf

t→∞ t(M (t) e ϕ t , e ϕ t ) = lim inf

t→∞ kJ 0 (e y t )h( e H k (t)) e ϕ t k 2 .

Step 4 . To complete the proof of Proposition 4.1 it suffices to prove Lemma 4.4. Let g ∈ C 0 

3 4 |e z k |; 3 4 |e z k | 

be such that g = 1 on

 − 3 2 |e z k |; 2 3 |e z k |  . Then

(4.14) (1 − g)( e w t )J 0 (e y t )h( e H k (t)) = O(t −1 ).

Indeed, if g, M , f M h are as before, then Lemma 4.4 and Proposition 4.3 give

M (t) = 1

t h( e H k (t))g( e w t )J 0 (e y t ) 2 g( e w t )h( e H k (t)) + O(t −2 )

≤ f M h (t) + Ct −2 ∈ G( e H k (t)).

P r o o f (of Lemma 4.4). We set J = J 0 and g = 1 − g. Then (4.14) follows if we show

(4.14 ) (−i + e H k ) −1 J(e y t )g( w e t ) 2 J(e y t )(i + e H k ) −1 ≤ Ct −2 . Writing g(λ) 2 = eg(λ) λ 21 4 e z k 2 

eg(λ) we have eg ∈ S 1 −1 (R) ⇒ [J(e y t ), e g( e w t )]

× (1 + | e w t |) = O(t −3 ) (cf. Appendix), and J(e y t ) 6= 0 ⇒ |e y t | ≤ 4τ ≤ 1 4 |e z k | ⇒

−e z k y e t1 4 e z k 2 allows us to estimate J(e y t )eg( e w t ) e w 2 t1 4 z e 2 k 

eg( e w t )J(e y t )

= eg( e w t )J(e y t ) e w 2 t1 4 z e 2 k 

J(e y t )eg( e w t ) + O(t −3 )

≤ eg( e w t )J(e y t )( e w 2 t + e z k y e t )J(e y t )eg( e w t ) + Ct −3

≤ eg( e w t )J(e y t )2t −2 H 0k J(e y t )eg( e w t ) + Ct −3 .

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Since [e y t , eg( e w t )] and [ e w 2 t , J(e y t )]eg( e w t ) are O(t −3 ), we obtain (4.14 ) noting that H 0k J(e y t )eg( e w t )(i + e H k ) −1 = J(e y t )eg( e w t )H 0k (i + e H k ) −1 + O(t −1 ) = O(1).

5. Interaction potentials with singularities. Let b C be as in (1.11b) and θ ∈ C 0 (R) be such that θ(x 1 ) = 1 for |x 1 | ≤ b C/|E|. Then

(5.1) kV k (x)θ(x 1 )ϕk ≤ 1 5 kp 2 θ(x 1 )ϕk + Ckϕk ≤ 1 2 kH 0 ϕk + C kϕk and V k (x)(1 − θ)(x 1 ) is bounded. Therefore H 0 + V k (x) is well defined as a self-adjoint operator on the domain of H 0 and the operators H 0 (H 0 + V k (x) +i) −1 , (H 0 +V k (x))(H 0 +i) −1 are bounded. The analogous assertion clearly holds if V k (x) is replaced by V k (x − χ k (t)) or by V (t, x) (using constants locally bounded with respect to t).

Further on β > 0 is fixed small enough. Following [7] or [9] we may state Lemma 5.1. There exist functions u j t ∈ C 0 (R), j = 1, . . . , d, such that for t ≥ 1 one has

u 1 t (λ) = ˙ χ 1 k (t)/t − z 1 0 for λ ∈ [e z k − t −β ; e z k + t −β ], (5.2a)

u 1 t (λ) = λ for λ 6∈ [

1≤k≤N

[e z k − 2t −β ; e z k + 2t −β ], (5.2b)

u j t (λ) = ˙ χ j k (t) for λ ∈ [ ˙ χ j k (t) − t −β ; ˙ χ j k (t) + t −β ], j ≥ 2, (5.2c)

u j t (λ) = λ for λ ∈ [−C + t −β ; C − t −β ] (5.2d)

\ [

1≤k≤N

[ ˙ χ j k (t) − 2t −β ; ˙ χ j k (t) + 2t −β ], j ≥ 2,

(u j t ) (λ) = d

dλ u j t (λ) = 0 for |λ| ≥ C, j ≥ 2, (5.2e)

d dt u j t (λ)

≤ Ct −1−β , (u j t ) (λ) ≥ 0,

|(u j t ) (n) (λ)| =

d n d n λ u j t (λ)

≤ C n t (n−1)β for λ ∈ R, n ≥ 1, (5.2f)

where e χ k (t) = ( ˙ χ 1 k (t), . . . , ˙ χ d k (t)) and C is fixed large enough.

We write a t = O(b t ) if b t ≥ I for t ≥ T 0 and b −1/2 t a t b −1/2 t = O(1). Note that a t = O(b t ) holds if we have a t b −1 t = O(1) and b −1 t a t = O(1). Further, we denote x ⊥ = (x 2 , . . . , x d ), e y t = x ⊥ /t, u t (e y t ) = (u 2 t (x 2 /t), . . . , u d t (x d /t)), u t (e y t ) = ((u 2 t ) (x 2 /t), . . . , (u d t ) (x d /t)),

(5.3) η t = 1 2 |w t | 2 − u t (e y t ) · w t /t + hc + C ⊥ I with C ⊥ > 0 large enough and

(5.3 ) η 0 t = 1 2 (w 1 t − u 1 t (y 1 t )) 2 + 1 4 (y 1 t ) 2 .

(20)

Proposition 5.2. Let η t = η 0 t + η t + V (t)/t 2 . If ε > 0 is small enough, then

(w t ) ≤ C(η t 0 + η t ) θ for 0 ≤ θ ≤ 1, (5.4a)

η t − (η t 0 + η t ) = t −2 V (t) = O(t −2ε0 t + η t )), (5.4b)

D H(t) η t = D H 0 (η 0 t + η t ) + O(t −1−ε η 1−ε t ), (5.4c)

D H

0 η t = − 1 t

X

2≤j≤d

w t j (1 + (u j t ) (x j /t))w j t + O(t −1−ε η 1/2 t ), (5.4d)

D H

0 η t 0 = − 1

t (w 1 t − y 1 t )(1 + 2(u 1 t ) (y t 1 ))(w 1 t − y 1 t ) + O(t −1−ε η t 1/2 ).

(5.4e)

P r o o f. By interpolation it suffices to prove (5.4a) for θ = 1. As u t (e y t ) is bounded, we have |w t | 2 ≤ Cη t . Then using u t (y 1 t ) 2 = (y 1 t ) 2 + O(t −β ) we may estimate

(w 1 t ) 2 = (w t 1 − u 1 t (y t 1 )) 2 + u 1 t (y 1 t ) 2 + 2(w t 1 − u 1 t (y t 1 ))u t (y t 1 ) + hc (5.5)

≤ 2(w 1 t − u 1 t (y 1 t )) 2 + 2u 1 t (y 1 t ) 2 + 1 ≤ 12η 0 t + Ct −β . Thus (5.4b) follows from (5.4a) by the estimate

t −2 e −iχ k (t)·p V k (x)e k (t)·p ≤ Ct −2 (1 + p 2 ) 1−ε ≤ C t −2ε (1 + |w t | 2 ) 1−ε . Next we note that

u 1 t (y 1 t ) + z 1 0 − ˙ χ 1 k (t)/t 6= 0 ⇒ |x 1 − χ 1 k (t)| ≥ 1 2 t 2−β

⇒ ∇V k (x − χ k (t)) = O(t −µ(2−β) ), u t (e y t ) − ˙ χ k (t) 6= 0 ⇒ |x ⊥ − χ k (t)| ≥ 1 2 t 1−β

⇒ ∇V k (x − χ k (t)) = O(t −µ(1−β) ),

hence using the fact that ˙ χ 1 k (t)/t, ˙ χ k (t), u 1 t (y 1 t−1/2 t , u t (e y t ) are O(1), we obtain

x 1 V k (x − χ k (t))(u 1 t (y 1 t ) + z 1 0 − ˙ χ 1 t (t)/t) = O(t −ε η 1/2 t ), (5.6)

∂ x ⊥ V k (x − χ k (t))(u t (e y t ) − ˙ χ k (t)) = O(t −ε ).

(5.6 )

Then reasoning as in the proof of Lemma 2.4 we can see that (5.6)–(5.6 ) imply (5.4c).

Finally, we obtain (5.4d, e) calculating D H

0 u t (e y t ) = 1

t u ⊥′ t (e y t )(p − e y t ) + O(t −2+β ),

−tD H 0 η t 0 + O(t −ε ) = (w 1 t ) 2 − 2(y 1 t − w t 1 )u 1′ t (y t 1 )w t 1 − u 1 t (y t 1 )w t 1 + 2(y t 1 − w 1 t )(u 1′ t u 1 t )(y t 1 ) + (y t 1 − w 1 t )y 1 t

= (w 1 t − y 1 t ) 2

+ 2(w 1 t − y 1 t )(1 + 2u 1′ t )(y t 1 )(w t 1 − y t 1 ) + O(t −ε ).

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