Minimal time for the approximate bilinear control of Schrödinger equations

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Minimal time for the approximate bilinear control of Schrödinger equations

Karine Beauchard, Jean-Michel Coron, Holger Teismann

To cite this version:

Karine Beauchard, Jean-Michel Coron, Holger Teismann. Minimal time for the approximate bilinear

control of Schrödinger equations. Mathematical Methods in the Applied Sciences, Wiley, 2018, 41 (5),

pp.1831-1844. �10.1002/mma.4710�. �hal-01333537�

Minimal time for the approximate bilinear control of Schrödinger equations

Karine Beauchard

, Jean-Michel Coron

, Holger Teismann

a

IRMAR and ENS Rennes, Campus de Ker Lann, 35170 Bruz, France, email:

Karine.Beauchard@ens-rennes.fr

b

Sorbonne Universités, UPMC Univ Paris 06, UMR 7598, Laboratoire Jacques-Louis Lions, F-75005, Paris, France, email: coron@ann.jussieu.fr

c

Department of Mathematics and Statistics, Acadia University, Wolfville, NS, Canada, email: hteisman@acadiau.ca

Abstract

We consider a quantum particle in a potential V (x) ( x ∈ R

) in a time- dependent electric eld E(t) (the control). Boscain, Caponigro, Chambrion and Sigalotti proved in [2] that, under generic assumptions on V , this system is approximately controllable on the L

( R

, C ) -sphere, in suciently large time T . In the present article we show that approximate controllability does not hold in arbitrarily small time, no matter what the initial state is. This generalizes our previous result for Gaussian initial conditions. Moreover, we prove that the minimal time can in fact be arbitrarily large.

Keywords: Schrödinger equation, quantum control, minimal time.

1. Introduction and main result

In this article, we consider quantum systems described by the linear Schrödinger equation

i∂

ψ(t, x) = −

∆ + V (x) − hE(t), xi

ψ(t, x), (t, x) ∈ (0, T ) × R

,

ψ(0, x) = ψ

(x), x ∈ R

. (1)

Here, N ∈ N

is the space dimension, h., .i is the euclidian scalar product on

R

, V : x ∈ R

→ R, E : t ∈ (0, T ) → R

and ψ : (t, x) ∈ (0, T ) × R

→ C

are a static potential, a time-dependent electric eld, and the wave function,

respectively. This equation represents a quantum particle trapped by the

potential V and under the inuence of the electric eld E . Planck's constant

and the particle mass have been set to one (the dependence on the physical constants is discussed briey in Section 3.4).

System (1) is a control system in which the control is the electric eld E and the state is the wave function ψ , which belongs to the unit sphere S of L

( R

, C ) . The expression bilinear control refers to the bilinear nature of the term hE(t), xiψ with respect to (E, ψ) .

We are interested in the minimal time required to achieve approximate controllability of the system (1). Since in (1) decoherence is neglected, in re- alistic scenarios the model may only be applicable for small times t (typically on the order of several periods of the ground state). Since, to be practically relevant, controllability results need to be valid for time intervals in which equation (1) remains a reasonable model, quantication of the minimal con- trol time is an important issue.

We consider potentials V that are smooth and subquadratic, i.e.

V ∈ C

( R

) and, ∀α ∈ N

such that |α| > 2, ∂

V ∈ L

( R

) . (2) For this class of potentials there is a classical well-posedness result [8], which we quote from [6].

Proposition 1. Consider V satisfying assumption (2) and E ∈ L

( R , R

) . There exists a strongly continuous map (t, s) ∈ R

7→ U (t, s) , with values in the set of unitary operators on L

( R

, C ) , such that

U(t, t) = Id , U (t, τ )U (τ, s) = U (t, s) , U(t, s)

= U (s, t)

, ∀t, τ, s ∈ R and for every t, s ∈ R, ϕ ∈ L

( R

, C ) , the function ψ(t, x) := U (t, s)ϕ(x) is a weak solution in C

([0, T ], L

( R

, C )) of the rst equation of (1) with initial condition ψ (s, x) = ϕ(x) .

For V satisfying (2), we introduce the operator D(A

) :=

ϕ ∈ L

( R

); −

∆ϕ + V (x)ϕ ∈ L

( R

) , A

ϕ := −

∆ϕ + V (x)ϕ .

For appropriate potentials V , approximate controllability of (1) in S (possi-

bly in large time) is a corollary of a general result by Boscain, Caponigro,

Chambrion, Mason and Sigalotti (the original proof of [7] is generalized in

[2]; inequality (4) below is proved in [7, Proposition 4.6]; an analogous state-

ment for vector valued controls is given in [3, Theorem 2.6]; see also [4] for a

survey of results in this area).

Theorem 1. Let m ∈ {1, ..., N } and assume that

• there exists a Hilbert basis (φ

)

of L

( R

, C ) composed of eigenvec- tors of A

: A

φ

= λ

φ

and x

φ

∈ L

( R

) , ∀k ∈ N,

• R

R^N

x

φ

(x)φ

(x)dx = 0 for every j, k ∈ N such that λ

= λ

and j 6= k ,

• for every j, k ∈ N, there exists a nite number of integers p

, ..., p

∈ N such that

p

= j, p

= k, R

R^N

x

φ

(x)φ

(x)dx 6= 0, ∀l = 1, ..., r − 1 ,

|λ

− λ

| 6= |λ

− λ

|, ∀1 6 l 6 r − 1, L, M ∈ N with {L, M } 6= {p

, p

}.

Then, for every > 0 and ψ

, ψ

∈ S, there exist a time T > 0 and a piecewise constant function u : [0, T ] → R such that the solution of (1) with E(t) = u(t)e

satises

kψ(T ) − ψ

k

< . (3) Moreover, for every δ > 0 , the existence of a piecewise constant function u : [0, T ] → (−δ, δ) such that the solution of (1) with E(t) = u(t)e

satises (3) implies that

T > 1 δ sup

k∈N

| |hφ

, ψ

i| − |hφ

, ψ

i| | −

kx

φ

k

. (4)

In Theorem 1, the time T is not known a priori and may be large. Note

that the lower bound on the control time in (4) goes to zero when δ → +∞ .

As a result, Theorem 1 gives no information about the control time if the

controls are allowed to be arbitrarily large; in particular, it does not preclude

the possibility of approximate controllability in arbitrarily small time. In

our previous work [1], we proved that, for potentials V satisfying (2), and

for particular (Gaussian) initial conditions, approximate controllability does

not hold in arbitrarily small time even with large controls. Specically, we

proved the following result.

Theorem 2. Assume that V satises assumption (2). Let b > 0 , x

, x ˙

∈ R

and ψ

∈ S be dened by

ψ

(x) := b

C

e

where C

:=

Z

R^N

e

dy

. Moreover, let ψ

∈ S be a state that does not have a Gaussian prole in the sense that

|ψ

(.)| 6= det(S)

C

e

√

S(.−γ)k²

, ∀γ ∈ R

, S ∈ M

( R ) symmetric positive . Then there exist positive numbers T

= T

(kV

k

, kV

k

, b, ψ

) and δ = δ(kV

k

, b, ψ

) such that, for every E ∈ C

([0, T

], R

) (piecewise continuous functions [0, T

] → R

), the solution ψ of (1) satises

kψ(t) − ψ

k

> δ , ∀t ∈ [0, T

] .

The goal of the present article is to generalize this result to arbitrary initial conditions ψ

, and to demonstrate that the minimal control time can in fact become arbitrarily large. Specically, we prove the following

Theorem 3. Assume that V satises assumption (2) and let ψ

∈ H

( R

)∩

L

(kxkdx) ∩ S.

1. There exists ψ

∈ S, T

> 0 and δ > 0 such that, for every E ∈ L

((0, T

), R

) , the solution ψ of (1) satises

kψ(t) − ψ

k

> δ , ∀t ∈ [0, T

] . 2. Moreover, if V is of the form

V (x) = W (x) , ∀x ∈ R

, (5)

then T

>

for every ∈ (0, 1) , for some positive constant C = C(ψ

, ψ

, W ) .

Remark 1. If V satises (2) and the assumptions of Theorem 1 (which hold

generically, this fact may be proved as in [10]), then system (1) is approx-

imately controllable in S in large time but not in small time T < T

. A

characterization of the minimal time required for -approximate controllabil-

ity is an open problem.

Remark 2. In part 2 of the theorem the demonstration that the minimal control time can become innitely large is accomplished by a particular choice (rescaling) of the potential. In a forthcoming paper we will investigate the conditions on V under which this can also be accomplished by a suitable choice of initial and/or target states. (See also Section 3.4)

The remainder of this paper is devoted to the proof of Theorem 3. The next section contains some notation and auxiliary results, whereas Section 3 contains the proof proper. There is also a brief discussion of the dependence on Planck's constant (Section 3.4) and an appendix containing the proof of a functional-analytic lemma needed in the argument. We refer to our previous article [1] for bibliographical comments.

2. Notation and auxiliary results

Denote by M

( R ) the set of N × N matrices with coecients in R, GL

( R ) the group of its invertible matrices and I

its identity element;

Tr (M ) the trace and M

the transposition of a matrix M ∈ M

( R ) ; S

( R ) the set of symmetric matrices in M

( R ) ; k.k the Euclidean norm on R

and the associated operator norm on M

( R ) ; x(t) := ˙

(t) , x(t) := ¨

(t) , for a function x of the scalar variable t and D

χ the Hessian matrix of a function χ : R

→ C, D

χ(x) =

∂²χ

∂yi∂yj

16i,j6N

.

The goal of this section is to prove the following result, which will be used in the proof of Theorem 3.

Proposition 2. Let T, L, R > 0 , B :=

σ ∈ M

( R ); kσ − I

k 6

, (6) K := n

M ∈ C

([0, T ], M

( R )) L-Lipschitz ; kM (t)k 6 R, ∀t ∈ [0, T ] o . (7) For φ

∈ S and M ∈ K , let the function χ

: [0, T ] × R

→ C be dened as the unique solution of

i∂

χ(τ, y) +

Tr

M (τ)

M (τ )D

χ(τ, y)

= 0, (τ, y) ∈ [0, T ] × R

,

χ(0, y) = φ

(y), y ∈ R

. (8)

1. For every φ

∈ S, the set V (φ

) := n

f ∈ S ; ∃(τ, σ, M, α) ∈ [0, T ] × B × K × R

such that

|f (x)| = √

det(σ)

χ

0

(τ, σ

[x − α])

for a.e. x ∈ R

o

is a strict and closed subset of S (w.r.t. the strong L

( R

) -topology).

2. For φ

, φ

∈ S then V(φ

) ⊂ V(φ

) + B

0, kφ

− φ

k

. Remark 3. It is clear that the unique solution χ

∈ C

([0, T ], L

( R

)) of (8) satises kχ

0

(τ, .)k

≡ 1 and is given by χ c

0

(τ, η) = φ b

(η)e

, for a.e. η ∈ R

, ∀τ ∈ [0, T ], (9) where the hat denotes the Fourier transform, dened by

f ˆ (η) = Z

R^N

e

f(y)dy , ∀f ∈ L

( R

).

The proof of Proposition 2 will use the following facts, proved in the appendix.

Lemma 1. Let (f

)

∈ L

( R

)

that converges to a function f in L

( R

) . 1. If (α

)

∈ R

satises kα

k −→

n→∞

+∞ , then τ

f

−→

n→∞

0 in D

( R

) .

2. If (α

)

converges to α in R

then τ

f

−→

n→∞

τ

f in L

( R

) . 3. If (M

)

converges toward M in GL

( R ) then f

◦ M

−→

n→∞

f ◦ M in L

( R

) .

Proof of Proposition 2:

Step 1: Let (f

)

∈ V(φ

)

and (τ

, σ

, M

, α

)

associated pa- rameters in [0, T ] × B × K × R

. We prove that a subsequence of (τ

|f

|)

converges in S (in the strong L

( R

) -topology). By Ascoli's theorem, there exists (τ

, σ

, M

) in [0, T ] ×B ×K such that, up to extract- ing a subsequence, (τ

, σ

) −→

n→∞

(τ

, σ

) in [0, T ]×B and M

(τ) −→

n→∞

M

(τ) uniformly with respect to τ ∈ [0, T ] . Let

k

(y) := 1

p det (σ

) χ

0

(τ

, y) , for a.e. y ∈ R

, ∀n ∈ N ∪ {∞}.

By (9), we have k b

(η) = 1

p det (σ

) φ b

(η)e

for a.e η ∈ R

, ∀n ∈ N ∪ {∞}, and by the dominated convergence theorem, k b

−→

n→∞

k c

in L

( R

) . Thus, Plancherel's theorem shows that k

−→

n→∞

k

in L

( R

) , which gives |k

| −→

n→∞

|k

| in L

( R

) , and nally τ

|f

| = |k

| ◦ σ

−→

n→∞

|k

| ◦ σ

in L

( R

) , by Lemma 1.3.

Step 2: We prove that V(φ

) is a strict subset of S. Working by contradiction, we assume that S = V (φ

) and consider the sequence (f

)

⊂ S, dened by f

(x) := √

nθ(nx) where θ(x) := π

e

2

2

. By

Step 1, there exist a subsequence (n

)

, a sequence (α

)

of R

and h ∈ S such that τ

f

−→

n→∞

h in L

( R

) , and thus in D

( R

) . However, for every ϕ ∈ C

( R

) , we have

R

R^N

τ

f

(x)ϕ(x)dx =

R

R^N

√ n

θ(n

y)ϕ(y + α

)dy 6

kθk

kϕk

, thus τ

f

−→

n→∞

0 in D

( R

) . Therefore h = 0 , which is impossible, since h ∈ S.

Step 3: We prove that V (φ

) is closed in S. Let (f

)

∈ V (φ

)

and f ∈ S be such that f

−→

n→∞

f in L

( R

) . We use the same notation as in Step 1.

Step 3.1: We prove that (α

)

is bounded in R

. Working by contra- diction, we may assume w.l.o.g. that kα

k −→

n→∞

∞ . Since |f

| −→

n→∞

|f| in L

( R

) , Lemma 1.1 implies τ

|f

| −→

n→∞

0 in D

( R

) , which contradicts Step 1.

Step 3.2: We prove that f ∈ V(φ

) . Since (α

)

is bounded, some subsequence converges to some α

∈ R

; w.l.o.g. α

−→

n→∞

α

. From Step 1, we know that, up to potentially extracting a subsequence, τ

|f

| =

|k

| ◦ σ

−→

n→∞

|k

| ◦ σ

in L

( R

) . Therefore, by Lemma 1.2, |f

| =

τ

[τ

|f

|] −→

n→∞

τ

[|k

| ◦ σ

] in L

( R

) . By uniqueness of the limit,

|f | = τ

[|k

| ◦ σ

] , i.e. f ∈ V(φ

) .

Step 4: We prove that V (φ

) ⊂ V(φ

) + B

0, kφ

− φ

k

Let f

∈ V (φ

) . Then, there exists (τ, σ, M, α) ∈ [0, T ] × B × K × R

and a . measurable function θ : R

→ R such that

f

(x) = e

p det(σ) χ

τ, σ

[x − α]

for a.e. x ∈ R

. Let

f

(x) := e

p det(σ) χ

0

τ, σ

[x − α]

for a.e. x ∈ R

. Then, f

∈ V(φ

) and, by (9) and Plancherel's theorem,

R

R^N

|(f

− f

)(x)|

dx = R

R^N

|χ

1

(τ, y) − χ

0

(τ, y)|

dy

= (2π)

R

R^N

| χ c

1

(τ, η) − χ c

0

(τ, η)|

dη

= (2π)

R

R^N

|( φ b

− φ b

)(η)|

dη

= R

R^N

|(φ

− φ

)(x)|

dx ,

which gives the conclusion. 2

3. Proof of Theorem 3

In the whole section, the following quantities are kept xed.

• V satisfying (2),

• ψ

∈ S ∩ H

( R

) ∩ L

(kxkdx) ,

• x

, x ˙

∈ R

dened by x

:=

Z

R^N

x|ψ

(x)|dx , x ˙

:= −i Z

R^N

∇

ψ

(x)ψ

(x)dx

• φ

∈ S dened by

φ

(x) := ψ

(x + x

)e

.

Our strategy to prove Theorem 3 relies on approximate solutions, which are centred at the classical (Newtonian) trajectories. Accordingly, these ap- proximate solutions ψ e

(dened in eq. (20) below) depend on the classical trajectories x

: R → R

and certain functions Q

, σ

: R → M

( C ) , which satisfy the ODEs (10) below. The remainder of this section is orga- nized as follows. In Section 3.1, we prove a preliminary result for the solutions of ODEs (10). In Section 3.2, we introduce the explicit approximate solu- tion ψ e

and prove that the error kψ

− ψ e

k

can be bounded uniformly with respect to E ∈ L

( R , R

) . Finally, Section 3.3 contains the proof of Theorem 3.

3.1. ODEs for x

, Q

and σ

For E ∈ L

( R , R

) , let x

∈ C

( R , R

) , Q

, σ

∈ C

((T

, T

), M

( R )) and τ

∈ C

((T

, T

), R ) be the maximal solutions of



 

 

d²x^E_c

dt²

(t) + ∇V [x

(t)] = E(t), x

(0) = x

,

dx^E_c

dt

(0) = ˙ x

,

dt

(t) + Q

(t)

+ V

[x

(t)] = 0, Q

(0) = 0,

dt

(t) = Q

(t)σ

(t), σ

(0) = I

,

(

dt

(t) =

, τ

(0) = 0,

(10)

where ∇V and V

denote the gradient and Hessian matrix of V , respectively.

Note that

• x

(t) is dened for every t ∈ R because ∇V is globally Lipschitz by assumption (2);

• x

is twice derivable almost everywhere and satises the rst equality of (10) for almost every t ∈ R;

• Q

(t) ∈ S

( R ) and σ

(t) ∈ GL

( R ) for every t ∈ (T

, T

) . A priori, the maximal interval (T

, T

) may depend on E .

Proposition 3. 1. There exists T

= T

(kV

k

) > 0 such that, for every E ∈ L

( R , R

) ,

T

> T

, (11)

kQ

(t)k 6 1

2 , ∀t ∈ [0, T

] , (12) kσ

(t) − I

k 6 1

2 , ∀t ∈ [0, T

] , (13)

|τ

(t) − t| 6 t

2 , ∀t ∈ [0, T

] . (14) 2. Moreover, if V is of the form (5), then T

>

for every ∈ (0, 1) , for

some positive constant C = C(W ) . Proof of Proposition 3: Fix δ ∈ 0,

and choose T

= T

(kV

k

) > 0 such that

T

(δ

+ kV

k

) < δ, 2N T

δ 6 1

4 , e

− 1 6 1

2 (15)

(the third inequality actually follows from the second). Let E ∈ L

( R , R

) . Step 1: We prove (11) and (12). Let

T

:= sup

t ∈ [0, T

); kQ

(s)k 6 δ , ∀s ∈ [0, t] . Working by contradiction, we assume that T

< T

. Then,

δ = kQ

(T

)k =

T_]^E

R

0

Q

(s)

+ V

[x

(s)]

ds 6 T

δ

+ kV

k

< δ by (15) , which is impossible. Therefore, T

> T

and

kQ

(t)k 6 δ for every t ∈ [0, T

], (16) which proves (11) and (12).

Step 2: We prove (13). We have kσ

(t)k =

I

+ Z

0

Q

(s)σ

(s)ds

(16)

6 1 + Z

0

δkσ

(s)kds , ∀t ∈ [0, T

] ,

thus, by Gronwall Lemma, kσ

(t)k 6 e

for every t ∈ [0, T

] and kσ

(t) − I

k =

Z

Q

(s)σ

(s)ds

(16)

6 Z

0

δe

6 e

− 1 , ∀t ∈ [0, T

] , which, together with (15) implies (13).

Step 3: We prove (14). By Liouville's formula, we have 1

det[σ

(t)]

= exp

−2 Z

0

Tr [Q

(s)]ds

, ∀t ∈ [0, T

] . (17) Moreover,

2

Z

Tr [Q

(s)]ds 6 2

Z

N kQ

(s)kds 6 2N T

δ 6 1

4 , (18) by (16) and (15). Thus, by (17), and (18),

1

det[σ

(t)]

− 1 6 1

4 e

< 1

2 , ∀t ∈ [0, T

] and so

|τ

(t) − t| =

Z

1

det[σ

(s)]

− 1

ds 6 t

2 , ∀t ∈ [0, T

].

Step 4: We prove Statement 2. For ∈ (0, δ) , the argument of Step 1 works with δ replaced by and then T

=

for small enough.

2

Proposition 3 implies that, for every E ∈ L

( R , R

) , the function τ

is C

and increasing, i.e. a bijection from [0, T

] to [0, τ

] , where τ

:=

τ

(T

) ∈

2

,

. Denoting the inverse function by t

: [0, τ

] → [0, T

] , we can now dene the C

map

M

: [0, τ

] → SL

( R )

τ 7→ det[(σ

◦ t

)(τ )] (σ

◦ t

)(τ)

. (19)

Thanks to Proposition 3, M

has the following properties.

Proposition 4. There exists R, L > 0 such that, for every E ∈ L

( R , R

) , kM

(τ )k 6 R and

dM

dτ (τ )

6 L , ∀τ ∈ [0, τ

] .

3.2. Approximate solution

Let S

, Φ

: (t, x) ∈ [0, T

) × R

→ R be dened by S

(t, x) := R

0 1

2

k x ˙

(s)k

− V [x

(s)] + hx

(s), E(s)i

ds + h x ˙

(t), xi Φ

(t, x) := S

(t, x − x

(t)) +

hQ

(t)[x − x

(t)], x − x

(t)i

and let χ

:= χ

0

(see (19) and (8)). Then we dene an approximate solution to (1) by

ψ e

(t, x) := e

p det[σ

(t)] χ

τ

(t), σ

(t)

[x − x

(t)]

, (20)

for every (t, x) ∈ [0, T

) × R

. Note that ψ e

(t, .) ∈ S for every t ∈ [0, T

) because (see Remark 3)

k ψ e

(t)k

= R

R^N

χ

τ

(t), σ

(t)

[x − x

(t)]

2 dx

det[σ^E(t)]

= R

R^N

χ

τ

(t), y

2

dy = 1 .

Remark 4. For background information on the approximate solutions ψ e

, see the literature cited in [1]. Their derivation may roughly be described as proceeding in two steps: one rst applies a well-known transformation (see e.g. [9]) to remove the control term; then the Schrödinger equation (arising by Taylor expansion) with the time-dependent quadratic potential hV

[x

(t)]x, xi is solved explicitly (up to solutions of (8)). The second step is related to the (generalized) Mehler formula for time-dependent quadratic Hamiltonians; see e.g. Section 3 of [6].

Proposition 5. If φ

∈ S( R

) , then there exists a constant C(φ

, T

) > 0 such that, for every E ∈ L

( R , R

) , the solution ψ

of (1) and the function ψ e

dened by (20) satisfy

k(ψ

− ψ e

)(t)k

6 C(φ

, T

)kV

k

t , ∀t ∈ [0, T

] , where T

is dened in Proposition 3.

Proof of Proposition 5: For simplicity, we write Q , σ , M , τ , χ , S , ψ , ψ e for Q

, σ

, M

, τ

, χ

0

, S

, ψ

and ψ e

.

Step 1: Equation satised by ψ e . For a function g(x) := f(Ax) we have ∇

g (x) = A

∇

f (Ax) and ∆

g(x) = Tr [AA

D

f(Ax)] . Thus, for every (t, x) ∈ [0, T

) × R

,

∇

ψ(t, x) = e i

˙

x

(t) + Q(t)[x − x

(t)]

ψ e (t, x)

+σ(t)

∇

χ (τ (t), σ(t)

[x − x

(t)]) √

det[σ(t)]

, and

∆

ψ(t, x) = e

i Tr (Q) − k x ˙

(t) + Q(t)[x − x

(t)]k

ψ(t, x) e +2i

˙

x

(t) + Q(t)[x − x

(t)], σ(t)

∇

χ τ (t), σ(t)

[x − x

(t)] e

p det[σ(t)]

+ Tr h

σ(t)

σ(t)

D

χ τ(t), σ(t)

[x − x

(t)] i e

p det[σ(t)] .

Developing the square in the rst line and using (19) gives, for every (t, x) ∈ [0, T

) × R

,

1

2 ∆

ψ(t, x) = e i

2 Tr [Q(t)] − 1

2 k x ˙

(t)k

− 1

2 kQ(t)[x − x

(t)]k

− h x ˙

(t), Q(t)[x − x

(t)]i ψ(t, x) e +i

σ(t)

( ˙ x

(t) + Q(t)[x − x

(t)]) , ∇

χ τ (t), σ(t)

[x − x

(t)] e

p det[σ(t)]

+ 1 2 Tr h

M [τ (t)]

M[τ (t)]D

χ τ (t), σ(t)

[x − x

(t)] i e

det[σ(t)]

. Moreover, using the relations τ(t) = ˙

and (21)

d dt

h

det[σ(t)] i

= det[σ(t)] Tr [σ(t) ˙ σ(t)] = det[σ(t)] Tr [Q(t)] , d

dt

"

1 p det[σ(t)]

#

= − 1

2 det[σ(t)]

d dt

h

det[σ(t)] i

= − Tr [Q(t)]

2 p

det[σ(t)] ,

d

dt

[σ(t)

[x − x

(t)]] = −σ(t)

σ(t)σ(t) ˙

[x − x

(t)] − σ(t)

x ˙

(t)

= −σ(t)

h

Q(t)[x − x

(t)] + ˙ x

(t) i

,

that hold for every t ∈ [0, T

) , we obtain, for every x ∈ R

and almost every t ∈ (0, T

) ,

i∂

ψ(t, x) = e − 1

2 k x ˙

(t)k

+ V [x

(t)] − hx

(t), E(t)i − h¨ x

(t), x − x

(t)i + k x ˙

(t)k

− 1

2 h Q(t)[x ˙ − x

(t)], x − x

(t)i + hQ(t)[x − x

(t)], x ˙

(t)i

− i

2 Tr [Q(t)]

ψ e (t, x)

+ i

det[σ(t)]

∂

χ τ(t), σ(t)

[x − x

(t)]

e

−i

∇

χ τ(t), σ(t)

[x − x

(t)]

, σ(t)

(Q(t)[x − x

(t)] + ˙ x

(t)) e

p det[σ(t)] . And nally, by (10), for every x ∈ R

and almost every t ∈ (0, T

) ,

i∂

ψ(t, x) = e 1

2 k x ˙

(t)k

+ V [x

(t)] − hx

(t), E(t)i − h∇V [x

(t)], x − x

(t)i

−hE(t), x − x

(t)i + 1

2 kQ(t)[x − x

(t)]k + 1

2 hV

[x

(t)][x − x

(t)], x − x

(t)i +hQ(t)[x − x

(t)], x ˙

(t)i − i

2 Tr [Q(t)]

ψ(t, x) e +i∂

χ τ (t), σ(t)

[x − x

(t)] e

det[σ(t)]

−i

∇

χ τ(t), σ(t)

[x − x

(t)]

, σ(t)

(Q(t)[x − x

(t)] + ˙ x

(t)) e

p det[σ(t)] . Combining (21), (22) and (8) gives for every x ∈ R

and almost every (22) t ∈ (0, T

) ,

i∂

ψ(t, x) + e 1

2 ∆ ψ(t, x) e − V (x) ψ(t, x) + e hE(t), xi ψ(t, x) = e R(t, x) ψ(t, x) e (23) where

R(t, x) := −V (x) + V [x

(t)] + h∇V [x

(t)], x − x

(t)i

+

hV

[x

(t)][x − x

(t)], x − x

(t)i . (24) Step 2: Conclusion. Using Taylor's formula, we get

|R(t, x)| 6 kV

k

3! kx − x

(t)k

, ∀(t, x) ∈ [0, T

) × R

.

Thus, for every t ∈ (0, T

), kR(t) ψ(t)k e

6

∞

3!

R

R^N

kx − x

(t)k

|χ (τ (t), σ(t)

[x − x

(t)])|

=

∞

3!

R

R^N

kσ(t)yk

|χ (τ (t), y)|

dy 6

3!

R

R^N

kyk

|χ (τ (t), y)|

dy by (13) 6 CkV

k

R

R^N

D

χ ˆ (τ (t), η)

2

dη 6 CkV

k

R

R^N

D

h

φ ˆ

(η)e

i

2

dη by (9)

for some positive constant C that does not depend on E , V and φ

. We deduce from Leibniz formula, (14) and Proposition 4 that

kR(t) ψ(t)k e

6 C(φ

, T

)

kV

k

, ∀t ∈ [0, T

]

for some positive constant C(φ

, T

) > 0 that is nite because φ

∈ S ( R

) . Note that C(φ

, T

)

is a polynomial function of degree 6 of T

, which will become relevant in Section 3.4. Let U (t, s) be the evolution operator for equation (1) (see Proposition 1). Then,

(ψ − ψ)(t) = e Z

0

U(t, s)[R(s) ψ(s)]ds e in L

( R

) , ∀t ∈ (0, T

) , and U (t, s) is an isometry of L

( R

) for every t > s > 0 , thus k(ψ− ψ)(t)k e

6

Z

kR(s) ψ(s)k e

ds 6 C(φ

, T

)kV

k

t , ∀t ∈ [0, T

] . 2 3.3. Conclusion

Let T

be as in Proposition 3, R, L > 0 be as in Proposition 4 and T :=

3T^∗

2

. By Proposition 2, there exists ψ

∈ S \V(φ

) and δ

:= dist

(ψ

, V(φ

)) is positive.

Step 1: We prove the existence of ψ

∈ S ∩ S ( R

) such that kψ

−ψ

k

< δ

4 , Z

R^N

x|ψ

(x)|dx = x

, −i Z

R^N

∇

ψ

(x)ψ

(x)dx = ˙ x

.

There exists a sequence (ξ

)

in S ∩ S( R

) such that kψ

− ξ

k

+ kψ

− ξ

k

−→

→0

0 . Then

x

:=

Z

R^N

x|ξ

(x)|dx , x ˙

:= −i Z

R^N

∇

ξ

(x)ξ

(x)dx converge, respectively, to x

and x ˙

. Thus, the sequence of functions

x 7→ ξ

(x − x

+ x

)e

converges to ψ

in L

( R

) and gives the conclusion.

Step 2: Distance between the approximate solutions associated to ψ

and ψ

. Step 1 implies that

• for every E ∈ L

( R , R

) , the quantities T

, x

, Q

, σ

, τ

, t

, S

associated with ψ

and ψ

are the same;

• the map φ

dened by the formula

φ

(x) := ψ

(x + x

)e

for a.e. x ∈ R

, satises

kφ

− φ

k

= kψ

− ψ

k

< δ

4 ;

• for every M ∈ C

([0, T ], M

( R )) , the functions χ

and χ

(see (8)) satisfy

kχ

(τ ) − χ

(τ )k

= kφ

− φ

k

< δ

4 , ∀τ ∈ [0, T ];

• for every E ∈ L

( R , R

) , the approximate solution ψ e

(resp. ψ e

) dened by (20) (resp. dened by (20) with φ

replaced by φ

) satisfy k ψ e

(t)− ψ e

(t)k

= kχ

(τ (t))−χ

(τ(t))k

< δ

4 , ∀t ∈ [0, T

].

Step 3: Conclusion of Statement 1. By Proposition 2 (part 2) we have that

kψ

− ψ e

(t)k

> 3δ

4 , ∀t ∈ [0, T

] . (25) Let C(φ

, T

) be as in Proposition 5 and

T

:= min

T

; δ

4C(φ

, T

)kV

k

. (26)

Proposition 5 and (25) imply that, for every t ∈ [0, T

] , kψ

− ψ

(t)k

>

kψ

− ψ e

(t)k

− k ψ e

(t) − ψ

(t)k

− kψ

(t) − ψ

(t)k

>

− C(φ

, T

)kV

k

t − kψ

− ψ

k

>

> 0 .

Step 4: Proof of Statement 2. If V (x) = W (x) , we obtain T

:= min

C(W )

; δ

4C(φ

, T

)

kW

k

,

which behaves like

as → 0 . 2

3.4. Dependence on Planck's constant

The Schrödinger equation for a quantum particle in a (static) potential V

, which is subjected to a time-dependent (and spatially homogeneous) electric eld E

, may, after appropriate rescaling, be written in dimensionless form:







iε∂

Ψ(τ, y) =

− ε

2 ∆

+ V

(y) − hE

(τ), yi

Ψ(τ, y) , (τ, y) ∈ (0, Θ) × R

,

Ψ(0, y) = Ψ

(y) y ∈ R

.

Here the parameter ε is proportional to the Planck constant ~; so it is natural (27) to assume that ε 1 . Obviously, equation (1) arises from equation (27) by the change of variables

t =

, x =

, t ∈ [0, T ] ⇔ τ ∈ [0, εT ], x, y ∈ R

ψ(t, x) = ε

Ψ(εt, εx), V (x) = V

(εx) , E(t) = εE

(εt)

(the factor ε

could be omitted; it ensures that kΨk

= kψk

= 1 ).

Therefore, the lower bound T

of the minimal time for approximate control- lability of system (1) provides a lower bound Θ

(ε) of the minimal time for system (27). By the change of variables, we have Θ

(ε) = εT

, where T

, in general, depends on ε (when xing V

rst).

Letting δ = ε in (15) shows that T

may be chosen as T

= C(kV

k

)ε

(note that kV

k

= ε

kV

k

). So, by (26),

T

= min

( C(kV

k

)

ε ; δ

4C(φ

, T

)ε

kV

k

)

(28) This seems to imply T

∼ ε

(and hence Θ

() ∼ const.), but this is not correct, since C(φ

, T

) also depends on ε . Indeed, C(φ

, T

)

is a degreesix polynomial in T

with certain φ

dependent coecients C

(φ

) ; i.e.,

ε

C

(φ

, T

) = ε

6

X

j=0

C

(φ

)(T

)

!

∼

6

X

j=0

C

(φ

)ε

!

∼ C

(φ

)

as ε → 0 . So for small ε > 0 , T

is independent of ε , which implies that Θ

(ε) ∼ ε 1 .

One may wonder whether the reasoning of the present paper could be re- ned to obtain stronger estimates on the control time, including, potentially, bounds satisfying Θ

(ε) → ∞ as ε → 0 . Two observations suggest that this may be possible in certain cases:

(a) the length of time interval [0, T

] on which the construction of the approximate solutions ψ e

is valid behaves like ε

and is independent of ε for eq. (27). It may therefore be possible to iterate the construction to enlarge the relevant time intervals; (b) the appearance of the quantity C(φ

, T

)

≈ C(φ

, T

) in the denominator of the second term of (28) suggests that for certain initial conditions ψ

the second term, and hence Θ

, may become large.

This circle of ideas will be the subject of a forthcoming paper.

4. Appendix: proof of Lemma 1

1. We have τ

f

= τ

f +τ

(f −f

) where τ

(f −f

) converges strongly to 0 in L

( R

) and thus in D

( R

) . Therefore, it suces to prove that τ

f −→

n→∞

0 in D

( R

) .

Let ϕ ∈ C

( R

) \ {0} and > 0 . There exists A > 0 such that Z

kyk>A

|f (y)|

dy <

kϕk

.

Let R > 0 be such that Supp (ϕ) ⊂ B(0, R) . We deduce from kα

k −→

n→∞

+∞ the existence of n

∈ N such that B(−α

, R) ∩ B(0, A) = ∅ , ∀n >

n

. Then, for every n > n

,

R

R^N

τ

f (x)ϕ(x)dx

=

R

B(−αn,R)

f (y)ϕ(y + α

)dy 6 kf k

kϕk

< . 2 2. See e.g. [5], Lemma 4.3.

3. We may assume that M = I

. We have f

◦M

= (f

−f)◦M

+f ◦M

where (f

− f) ◦ M

converges to 0 in L

( R

) . Thus, it suces to prove that f ◦ M

−→

n→∞

f in L

( R

) . Let > 0 .

Case 1: f ∈ C

( R

) . There exists R > 0 such that Supp (f) ⊂ B(0, R) and Supp (f ◦M

) = M

Supp (f ) ⊂ B (0, R) for every n ∈ N. By Heine theorem, there exists η > 0 such that

|f (y)− f(z)| <

p R

vol [B(0, 1)] , ∀y, z ∈ R

such that ky −zk < η . We chose n

large enough so that kM

− I

k <

for every n > n

. Then,

kM

x − xk 6 kM

− I

kkxk < η , ∀x ∈ B(0, R) , n > n

. Thus, for n > n

,

kf ◦ M

− f k

= Z

B(0,R)

|f [M

(x)] − f(x)|

dx

< . Case 2: f ∈ L

( R

) . There exists f e ∈ C

( R

) such that kf − f e k

<

. By Case 1, there exists n

∈ N such that k f e ◦ M

− f e k

<

for every n > n

. One may assume that p

det(M

) >

for every n > n

. Then, for n > n

,

kf ◦ M

− f k

6 kf ◦ M

− f e ◦ M

k

+k f e ◦ M

− f e k

+ k f e − f k

6

√

det(Mn)

+ 1

k f e − f k

+

< .

Acknowledgments: The authors were partially supported by the Agence Nationale de la Recherche (ANR) Projet Blanc EMAQS number ANR-2011- BS01-017-01 and (ANR) Projet Blanc Finite4SoS number ANR-15-CE23- 0007.

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